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engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/199_17.4 Circuit Applications.md

@@ -1,232 +0,0 @@
-From Example 17.1,
-
-$$
-v_s(t) = \frac{1}{2} + \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin n\pi t
-$$
-,  $n = 2k - 1$ 
-
-where *ωn* = *nω*0 = *nπ*rad/*s*. Using phasors, we obtain the response **V***o* in the circuit of Fig. 17.20 by voltage division:
-
-$$
-\mathbf{V}_o = \frac{j\omega_n L}{R + j\omega_n L} \mathbf{V}_s = \frac{j2n\pi}{5 + j2n\pi} \mathbf{V}_s
-$$
-
-For the dc component (*ωn* = 0 or *n* = 0)
-
-$$
-\mathbf{V}_s = \frac{1}{2} \qquad \Rightarrow \qquad \mathbf{V}_o = 0
-$$
-
-This is expected, given that the inductor is a short circuit to dc. For the *n*th harmonic,
-
-$$
-V_s = \frac{2}{n\pi} \sqrt{-90^\circ}
-$$
- (17.6.1)
-
-and the corresponding response is
-
-$$
-\mathbf{v}_s = \frac{2n\pi}{n\pi} \frac{7 - 90^{\circ}}{(17.6.1)}
-$$
-\nThe corresponding response is
-
-\n
-$$
-\mathbf{V}_o = \frac{2n\pi/90^{\circ}}{\sqrt{25 + 4n^2\pi^2} \left(\frac{\tan^{-1}2n\pi/5}{\tan^{-1}2n\pi/5}\right)} \left(\frac{2}{n\pi} \left(\frac{7 - 90^{\circ}}{\tan^{-1}2n\pi/5}\right)\right)
-$$
-\n(17.6.2)
-
-\n
-$$
-= \frac{4(-\tan^{-1}2n\pi/5)}{\sqrt{25 + 4n^2\pi^2}}
-$$
-
-In the time domain,
-
-$$
-v_o(t) = \sum_{k=1}^{\infty} \frac{4}{\sqrt{25 + 4n^2 \pi^2}} \cos\left(n\pi t - \tan^{-1}\frac{2n\pi}{5}\right), \qquad n = 2k - 1
-$$
-
-The first three terms (*k* = 1, 2, 3 or *n* = 1, 3, 5) of the odd harmonics in the summation give us
-
-$$
-v_o(t) = 0.4981 \cos(\pi t - 51.49^\circ) + 0.2051 \cos(3\pi t - 75.14^\circ)
-$$
-  
-+ 0.1257 \cos(5\pi t - 80.96^\circ) + ... V
-
-Figure 17.21 shows the amplitude spectrum for output v oltage *vo*(*t*), while that of the input voltage *vs*(*t*) is in Fig. 17.4(a). Notice that the two spectra are close. Why? We observe that the circuit in Fig. 17.20 is a high-pass filter with the corner frequency *ωc* = *R*∕*L* = 2.5 rad/s, which is less than the fundamental frequenc y *ω*0 = *π*rad/s. The dc component is not passed and the first harmonic is slightly attenuated, but higher harmonics are passed. In fa ct, from Eqs. (17.6.1) and (17.6.2), **V***o* is identical to **V***s* for large *n*, which is characteristic of a high-pass filter.
-
-**Figure 17.22** For Practice Prob. 17.6.
-
-the output voltage.
-
-Example 17.7
-
-If the sawtooth waveform in Fig. 17.9 (see Practice Prob. 17.2) is the voltage source *vs*(*t*) in the circuit of Fig. 17.22, find the response *vo*(*t*).
-
-**6CUCE PTODIEIII** 17.0 If the sawtooth waveform in Fig. 17.9 (see Practice  
-voltage source 
-$$
-v_s(t)
-$$
- in the circuit of Fig. 17.22, find  
- $v_s(t)$   
- $v_s(t)$   
-$$
-1F = \frac{1}{v_s(t)}
-$$
-  
-**Answer:**  $v_o(t) = \frac{3}{2} - \frac{3}{\pi} \sum_{n=1}^{\infty} \frac{\sin(2\pi nt - \tan^{-1} 4n\pi)}{n\sqrt{1 + 16n^2 \pi^2}} V.$ 
-
-Find the response *io*(*t*) of the circuit of Fig. 17.23 if the input voltage *v*(*t*) has the Fourier series expansion
-
-$$
-v(t) = 1 + \sum_{n=1}^{\infty} \frac{2(-1)^n}{1 + n^2} (\cos nt - n \sin nt)
-$$
-
-For Example 17.6: Amplitude spectrum of
-
-# **Solution:**
-
-Using Eq. (17.13), we can express the input voltage as
-
-$$
-v(t) = 1 + \sum_{n=1}^{\infty} \frac{2(-1)^n}{\sqrt{1 + n^2}} \cos(nt + \tan^{-1} n)
-$$
-  
-= 1 - 1.414 \cos(t + 45^\circ) + 0.8944 \cos(2t + 63.45^\circ)  
--0.6345 \cos(3t + 71.56^\circ) - 0.4851 \cos(4t + 78.7^\circ) + ...
-
-We notice that *ω*0 = 1, *ωn* = *n* rad/s. The impedance at the source is
-
-$$
-\mathbf{Z} = 4 + j\omega_n 2 \mid 4 = 4 + \frac{j\omega_n 8}{4 + j\omega_n 2} = \frac{8 + j\omega_n 8}{2 + j\omega_n}
-$$
-
-The input current is
-
-$$
-\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{2 + j\omega_n}{8 + j\omega_n 8} \mathbf{V}
-$$
-
-where **V** is the phasor form of the source voltage *v*(*t*). By current division,
-
-> **<sup>I</sup>***o* = \_\_\_\_\_\_\_\_ <sup>4</sup> 4 + *jωn*2 **<sup>I</sup>** = \_\_\_\_\_\_\_\_ **<sup>V</sup>** 4 + *jωn*4
-
-Because *ωn* = *n*, **I***o* can be expressed as
-
-Because 
-$$
-\omega_n = n
-$$
-,  $\mathbf{I}_o$  can be expressed as  
-\n
-$$
-\mathbf{I}_o = \frac{\mathbf{V}}{4\sqrt{1 + n^2 / \tan^{-1} n}}
-$$
-\nFor the dc component ( $\omega_n = 0$  or  $n = 0$ )
-
-$$
-\mathbf{V} = 1 \qquad \Rightarrow \qquad \mathbf{I}_o = \frac{\mathbf{V}}{4} = \frac{1}{4}
-$$
-
-For the *n*th harmonic,
-
-$$
-\mathbf{V} = \frac{2(-1)^n}{\sqrt{1 + n^2}} \frac{1}{\tan^{-1} n}
-$$
-
-so that
-
-$$
-\mathbf{I}_o = \frac{1}{4\sqrt{1 + n^2}/\tan^{-1}n} \frac{2(-1)^n}{\sqrt{1 + n^2}} / \tan^{-1}n = \frac{(-1)^n}{2(1 + n^2)}
-$$
-
-In the time domain,
-
-$$
-i_o(t) = \frac{1}{4} + \sum_{n=1}^{\infty} \frac{(-1)^n}{2(1+n^2)} \cos nt \, \text{A}
-$$
-
-If the input voltage in the circuit of Fig. 17.24 is
-
-$$
-v(t) = \frac{7}{3} + \frac{1}{\pi^2} \sum_{n=1}^{\infty} \left( \frac{1}{n^2} \cos nt - \frac{\pi}{n} \sin nt \right) \text{ V}
-$$
-
-determine the response *io*(*t*).
-
-Practice Problem 17.7
-
-<span id="page-802-0"></span>**Figure 17.24** For Practice Prob. 17.7.
-
-# **17.5** Average Power and RMS Values
-
-Recall the concepts of average power and rms value of a periodic signal that we discussed in Chapter 11. To find the average power absorbed by a circuit due to a periodic excitation, we write the voltage and current in amplitude-phase form [see Eq. (17.10)] as
-
-$$
-v(t) = V_{\text{dc}} + \sum_{n=1}^{\infty} V_n \cos(n\omega_0 t - \theta_n)
-$$
- (17.42)
-
-$$
-i(t) = I_{\text{dc}} + \sum_{m=1}^{\infty} I_m \cos(m\omega_0 t - \phi_m)
-$$
- (17.43)
-
-Following the passi ve sign con vention (Fig. 17.25), the a verage power is
-
-$$
-P = \frac{1}{T} \int_0^T v i \, dt \tag{17.44}
-$$
-
-Substituting Eqs. (17.42) and (17.43) into Eq. (17.44) gives
-
-$$
-P = \frac{1}{T} \int_0^T V_{dc} I_{dc} dt + \sum_{m=1}^{\infty} \frac{I_m V_{dc}}{T} \int_0^T \cos(m\omega_0 t - \phi_m) dt
-$$
-  
-+ 
-$$
-\sum_{n=1}^{\infty} \frac{V_n I_{dc}}{T} \int_0^T \cos(n\omega_0 t - \theta_n) dt
-$$
-(17.45)  
-+ 
-$$
-\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{V_n I_m}{T} \int_0^T \cos(n\omega_0 t - \theta_n) \cos(m\omega_0 t - \phi_m) dt
-$$
-
-The second and third integrals vanish, since we are integrating the cosine over its period. According to Eq. (17.4e), all terms in the fourth inte gral are zero when *m* ≠ *n*. By evaluating the first integral and applying Eq. (17.4g) to the fourth integral for the case *m* = *n*, we obtain
-
-$$
-P = V_{\rm dc} I_{\rm dc} + \frac{1}{2} \sum_{n=1}^{\infty} V_n I_n \cos(\theta_n - \phi_n)
-$$
- (17.46)
-
-This shows that in average-power calculation involving periodic voltage and current, the total average power is the sum of the average powers in each harmonically related voltage and current.
-
-Given a periodic function *f*(*t*), its rms value (or the effective value) is given by
-
-$$
-F_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T f^2(t) \, dt} \tag{17.47}
-$$
-
-# **Figure 17.25**
-
-The voltage polarity reference and current reference direction.
-
-Substituting *f*(*t*) in Eq. (17.10) into Eq. (17.47) and noting that (*a* + *b*) 2 = *a*<sup>2</sup> + 2*ab* + *b*<sup>2</sup> , we obtain
-
-$$
-F_{\text{rms}}^2 = \frac{1}{T} \int_0^T \left[ a_0^2 + 2 \sum_{n=1}^\infty a_0 A_n \cos (n\omega_0 t + \phi_n) \right. \\
-\left. + \sum_{n=1}^\infty \sum_{m=1}^\infty A_n A_m \cos(n\omega_0 t + \phi_n) \cos(m\omega_0 t + \phi_m) \right] dt
-$$
-\n
-$$
-= \frac{1}{T} \int_0^T a_0^2 dt + 2 \sum_{n=1}^\infty a_0 A_n \frac{1}{T} \int_0^T \cos(n\omega_0 t + \phi_n) dt
-$$
-\n
-$$
-+ \sum_{n=1}^\infty \sum_{m=1}^\infty A_n A_m \frac{1}{T} \int_0^T \cos(n\omega_0 t + \phi_n) \cos(m\omega_0 t + \phi_m) dt

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/200_17.5 Average Power and RMS Values.md

@@ -1,187 +0,0 @@
-$$
-\n(17.48)
-
-Distinct integers *n* and *m* have been introduced to handle the product of the tw o series summations. Using the same reasoning as abo ve, we get
-
-> *F*rms <sup>2</sup> = *a*<sup>0</sup> 2 <sup>+</sup>\_\_1 2 ∑ *n*=1 ∞ *A n* 2
-
-or
-
-$$
-F_{\rm rms} = \sqrt{a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} A_n^2}
-$$
- (17.49)
-
-In terms of Fourier coefficients *an* and *bn*, Eq. (17.49) may be written as
-
-coefficients 
-$$
-a_n
-$$
- and  $b_n$ , Eq. (17.49) may be written as  
-\n
-$$
-F_{\text{rms}} = \sqrt{a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} (a_n^2 + b_n^2)}
-$$
-\n(17.50)
-
-If *f*(*t*) is the current through a resistor *R*, then the power dissipated in the resistor is
-
-$$
-P = RF_{\rm rms}^2 \tag{17.51}
-$$
-
-Or if *f*(*t*) is the voltage across a resistor *R*, the power dissipated in the resistor is
-
-$$
-P = \frac{F_{\text{rms}}^2}{R}
-$$
- (17.52)
-
-One can a void specifying the nature of the signal by choosing a 1- Ω resistance. The power dissipated by the 1-Ω resistance is
-
-$$
-P_{1\Omega} = F_{\text{rms}}^2 = a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} (a_n^2 + b_n^2)
-$$
- (17.53)
-
-This result is known as *Parseval's theorem*. Notice that *a* <sup>0</sup> 2 is the power in the dc component, while \_\_1 <sup>2</sup> ( *a n* 2 + *b n* 2 ) is the ac power in the *n*th harmonic. Thus, Parseval's theorem states that the average power in a periodic signal is the sum of the average power in its dc component and the average powers in its harmonics.
-
-Historical note: Named after the French mathematician Marc-Antoine Parseval Deschemes (1755–1836).
-
-# Example 17.8
-
-Determine the a verage po wer supplied to the circuit in Fig. 17.26 if *i*(*t*) = 2 + 10 cos(*t* + 10°) + 6 cos(3*t* + 35°) A.
-
-# **Solution:**
-
-The input impedance of the network is
-
-$$
-\mathbf{Z} = 10 \left\| \frac{1}{j2\omega} = \frac{10(1/j2\omega)}{10 + 1/j2\omega} = \frac{10}{1 + j20\omega}
-$$
-
-Hence,
-
-$$
-\mathbf{V} = \mathbf{IZ} = \frac{10}{10 + 1/j2\omega} = \frac{1}{1 + j20\omega}
-$$
-\n
-$$
-\mathbf{V} = \mathbf{IZ} = \frac{10\mathbf{I}}{\sqrt{1 + 400\omega^2}/\tan^{-1}20\omega}
-$$
-
-For the dc component, *ω* = 0,
-
-$$
-\mathbf{I} = 2 \, A \qquad \Rightarrow \qquad \mathbf{V} = 10(2) = 20 \, \text{V}
-$$
-
-This is expected, because the capacitor is an open circuit to dc and the entire 2-A current flows through the resistor. For *ω* = 1 rad/s,
-
-UATE: The number of two degrees of the number of numbers, we have:
-
-\n
-$$
-\mathbf{I} = 10 \times 10^{\circ} \quad \Rightarrow \quad \mathbf{V} = \frac{10(10 \times 10^{\circ})}{\sqrt{1 + 400} \times 10^{\circ}}
-$$
-\n
-$$
-= 5 \times 10^{\circ}
-$$
-
-For *ω* = 3 rad/s,
-
-$$
-= 5/111
-$$
-  
-and/s,  
-$$
-I = 6/35^{\circ} \Rightarrow V = \frac{10(6/35^{\circ})}{\sqrt{1 + 3600}/\tan^{-1}60}
-$$
-  
-= 1/−54.04°
-
-Thus, in the time domain,
-
-$$
-v(t) = 20 + 5\cos(t - 77.14^{\circ}) + 1\cos(3t - 54.04^{\circ})
-$$
- V
-
-We obtain the average power supplied to the circuit by applying Eq. (17.46), as
-
-$$
-P = V_{\text{dc}}I_{\text{dc}} + \frac{1}{2} \sum_{n=1}^{\infty} V_n I_n \cos(\theta_n - \phi_n)
-$$
-
-To get the proper signs of *θn* and *n*, we have to compare *v* and *i* in this example with Eqs. (17.42) and (17.43). Thus,
-
-$$
-P = 20(2) + \frac{1}{2}(5)(10) \cos[77.14^{\circ} - (-10^{\circ})]
-$$
-$$
-+ \frac{1}{2}(1)(6) \cos[54.04^{\circ} - (-35^{\circ})]
-$$
-$$
-= 40 + 1.247 + 0.05 = 41.5 \text{ W}
-$$
-
-Alternatively, we can find the average power absorbed by the resistor as
-
-$$
-P = \frac{V_{\text{dc}}^2}{R} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{|V_n|^2}{R} = \frac{20^2}{10} + \frac{1}{2} \cdot \frac{5^2}{10} + \frac{1}{2} \cdot \frac{1^2}{10}
-$$
-$$
-= 40 + 1.25 + 0.05 = 41.5 \text{ W}
-$$
-
-which is the same as the power supplied, since the capacitor absorbs no average power.
-
-<span id="page-805-0"></span>The voltage and current at the terminals of a circuit are
-
-*v*(*t*) = 128 + 192 cos 120*πt* + 96 cos(360*πt* − 30°) *i*(*t*) = 4 cos(120*πt* − 10°) + 1.6 cos(360*πt* − 60°)
-
-Find the average power absorbed by the circuit.
-
-**Answer:** 444.7 W.
-
-Find an estimate for the rms value of the voltage in Example 17.7.
-
-# **Solution:**
-
-From Example 17.7, *v*(*t*) is expressed as
-
-$$
-v(t) = 1 - 1.414 \cos(t + 45^\circ) + 0.8944 \cos(2t + 63.45^\circ)
-$$
-$$
-- 0.6345 \cos(3t + 71.56^\circ)
-$$
-$$
-- 0.4851 \cos(4t + 78.7^\circ) + \dots V
-$$
-
-Using Eq. (17.49), we find
-
-$$
-V_{\text{rms}} = \sqrt{a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} A_n^2}
-$$
-  
-=  $\sqrt{1^2 + \frac{1}{2} [(-1.414)^2 + (0.8944)^2 + (-0.6345)^2 + (-0.4851)^2 + \cdots]}$   
-=  $\sqrt{2.7186} = 1.649 \text{ V}$ 
-
-This is only an estimate, as we have not taken enough terms of the series. The actual function represented by the Fourier series is
-
-$$
-v(t) = \frac{\pi e^t}{\sinh \pi}, \qquad -\pi < t < \pi
-$$
-
-with *v*(*t*) = *v*(*t* + *T*). The exact rms value of this is 1.776 V.
-
-Find the rms value of the periodic current
-
-*i*(*t*) = 8 + 30 cos 2*t* − 20 sin 2*t* + 15 cos 4*t* − 10 sin 4*t* A
-
-**Answer:** 29.61 A.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/201_17.6 Exponential Fourier Series.md

@@ -1,311 +0,0 @@
-# **17.6** Exponential Fourier Series
-
-A compact way of expressing the Fourier series in Eq. (17.3) is to put it in exponential form. This requires that we represent the sine and cosine functions in the exponential form using Euler's identity:
-
-$$
-\cos n\omega_0 t = \frac{1}{2} \left[ e^{jn\omega_0 t} + e^{-jn\omega_0 t} \right]
-$$
- (17.54a)
-
-$$
-\sin n\omega_0 t = \frac{1}{2j} \left[ e^{jn\omega_0 t} - e^{-jn\omega_0 t} \right]
-$$
- (17.54b)
-
-Practice Problem 17.9
-
-Practice Problem 17.8
-
-Example 17.9
-
-Substituting Eq. (17.54) into Eq. (17.3) and collecting terms, we obtain
-
-$$
-f(t) = a_0 + \frac{1}{2} \sum_{n=1}^{\infty} \left[ (a_n - jb_n)e^{jn\omega_0 t} + (a_n + jb_n)e^{-jn\omega_0 t} \right]
-$$
- (17.55)
-
-If we define a new coefficient *cn* so that
-
-$$
-c_0 = a_0
-$$
-,  $c_n = \frac{(a_n - jb_n)}{2}$ ,  $c_{-n} = c_n^* = \frac{(a_n + jb_n)}{2}$  (17.56)
-
-then *f*(*t*) becomes
-
-$$
-f(t) = c_0 + \sum_{n=1}^{\infty} (c_n e^{jn\omega_0 t} + c_{-n} e^{-jn\omega_0 t})
-$$
- (17.57)
-
-or
-
-or
-
-$$
-f(t) = \sum_{n = -\infty}^{\infty} c_n^{ejn\omega_0 t}
-$$
- (17.58)
-
-This is the *complex* or *exponential Fourier series* representation of *f*(*t*). Note that this e xponential form is more compact than the sine-cosine form in Eq. (17.3). Although the exponential Fourier series coefficients *cn* can also be obtained from *an* and *bn* using Eq. (17.56), they can also be obtained directly from *f*(*t*) as
-
-$$
-c_n = \frac{1}{T} \int_0^T f(t) e^{-e j n \omega_0 t} dt
-$$
- (17.59)
-
-where *ω*0 = 2*π*∕*T*, as usual. The plots of the magnitude and phase of *cn* versus *nω*0 are called the *complex amplitude spectrum* and *complex phase spectrum* of *f* (*t*), respectively. The two spectra form the comple x frequency spectrum of *f* (*t*).
-
-The exponential Fourier series of a periodic function f(t) describes the spectrum of f(t) in terms of the amplitude and phase angle of ac components at positive and negative harmonic frequencies.
-
-The coefficients of the three forms of Fourier series (sine-cosine form, amplitude-phase form, and exponential form) are related by
-
-$$
-A_n / \underline{\phi_n} = a_n - jb_n = 2c_n \tag{17.60}
-$$
-
-*cn* = ∣*cn*∣⧸*θn* = <sup>√</sup> \_\_\_\_\_\_ *a n* 2 + *b n* 2 \_\_\_\_\_\_\_\_ 2 <sup>⧸</sup> −tan−1 *bn*∕*an* **(17.61)**
-
-if only *an* > 0. Note that the phase *θ<sup>n</sup>* of *cn* is equal to *n*.
-
-In terms of the F ourier complex coefficients *cn*, the rms v alue of a periodic signal *f*(*t*) can be found as
-
-$$
-F_{\text{rms}}^2 = \frac{1}{T} \int_0^T f^2(t) \, dt = \frac{1}{T} \int_0^T f(t) \left[ \sum_{n=-\infty}^{\infty} c_n e^{jn\omega_0 t} \right] \, dt
-$$
-\n
-$$
-= \sum_{n=-\infty}^{\infty} c_n \left[ \frac{1}{T} \int_0^T f(t) e^{jn\omega_0 t} \, dt \right]
-$$
-\n
-$$
-= \sum_{n=-\infty}^{\infty} c_n c_n^* = \sum_{n=-\infty}^{\infty} |c_n|^2
-$$
-\nor
-
-or
-
-$$
-F_{\rm rms} = \sqrt{\sum_{n=-\infty}^{\infty} |c_n|^2}
-$$
- (17.63)
-
-Equation (17.62) can be written as
-
-$$
-F_{\rm rms}^2 = |c_0|^2 + 2 \sum_{n=1}^{\infty} |c_n|^2
-$$
- (17.64)
-
-Again, the power dissipated by a 1-Ω resistance is
-
-$$
-P_{1\Omega} = F_{\text{rms}}^2 = \sum_{n=-\infty}^{\infty} |c_n|^2
-$$
- (17.65)
-
-which is a restatement of P arseval's theorem. The *power spectrum* of the signal *f*(*t*) is the plot of ∣*cn*∣ 2 versus *nω*0. If *f*(*t*) is the voltage across a resistor *R*, the average power absorbed by the resistor is *F*rms <sup>2</sup> ∕*R*; if *f*(*t*) is the current through *R*, the power is *F*rms <sup>2</sup> *R*.
-
-As an illustration, consider the periodic pulse train of Fig. 17.27. Our goal is to obtain its amplitude and phase spectra. The period of the pulse train is *T* = 10, so that *ω*0 = 2*π*∕*T* = *π*∕5. Using Eq. (17.59),
-
-$$
-c_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t)e^{-jn\omega_0 t} dt = \frac{1}{10} \int_{-1}^{1} 10e^{-jn\omega_0 t} dt
-$$
-  
-$$
-= \frac{1}{-jn\omega_0} e^{-jn\omega_0 t} \Big|_{-1}^{1} = \frac{1}{-jn\omega_0} (e^{-jn\omega_0} - e^{jn\omega_0})
-$$
-  
-$$
-= \frac{2}{n\omega_0} \frac{e^{jn\omega_0} - e^{-jn\omega_0}}{2j} = 2 \frac{\sin n\omega_0}{n\omega_0}, \qquad \omega_0 = \frac{\pi}{5}
-$$
-  
-$$
-= 2 \frac{\sin n\pi/5}{n\pi/5}
-$$
- (17.66)
-
-‒11 ‒9 ‒1 1 0 9 11 t 10 f(t) **Figure 17.27** The periodic pulse train.
-
-and
-
-$$
-f(t) = 2 \sum_{n = -\infty}^{\infty} \frac{\sin n\pi/5}{n\pi/5} e^{jn\pi t/5}
-$$
- (17.67)
-
-Notice from Eq. (17.66) that *cn* is the product of 2 and a function of the form sin *x*∕*x*. This function is known as the *sinc function*; we write it as
-
-$$
-\text{sinc}(x) = \frac{\sin x}{x} \tag{17.68}
-$$
-
-Some properties of the sinc function are important here. F or zero argument, the value of the sinc function is unity,
-
-$$
-\text{sinc}(0) = 1\tag{17.69}
-$$
-
-The sinc function is called the sampling function in communication theory, where it is very useful.
-
-This is obtained by applying L'Hopital's rule to Eq. (17.68). For an integral multiple of *π*, the value of the sinc function is zero,
-
-$$
-sinc(n\pi) = 0, \qquad n = 1, 2, 3, ... \tag{17.70}
-$$
-
-Also, the sinc function shows even symmetry. With all this in mind, w e can obtain the amplitude and phase spectra of *f*(*t*). From Eq. (17.66), the magnitude is
-
-$$
-|c_n| = 2 \left| \frac{\sin n\pi/5}{n\pi/5} \right| \tag{17.71}
-$$
-
-while the phase is
-
-$$
-\theta_n = \begin{cases}\n0^\circ, & \sin \frac{n\pi}{5} > 0 \\
-180^\circ, & \sin \frac{n\pi}{5} < 0\n\end{cases}
-$$
-\n(17.72)
-
-Figure 17.28 shows the plot of ∣*cn*∣ versus *n* for *n* varying from −10 to 10, where *n* = *ω*∕*ω*0 is the normalized frequenc y. Figure 17.29 shows the plot of *θn* versus *n*. Both the amplitude spectrum and phase spec trum are called *line spectra,* because the v alues of ∣*cn*∣ and *θn* occur only at discrete v alues of frequencies. The spacing between the lines is *ω*0. The power spectrum, which is the plot of ∣*cn*∣ 2 versus *nω*0, can also be plotted. Notice that the sinc function forms the envelope of the amplitude spectrum.
-
-effect of a circuit on a periodic signal. 2 1.87 │cn│
-
-Examining the input and output spectra allows visualization of the
-
-The amplitude of a periodic pulse train.
-
-# Example 17.10
-
-**Figure 17.28**
-
-Find the exponential Fourier series expansion of the periodic function *f*(*t*) = *et* , 0 < *t* < 2*π* with *f*(*t* + 2*π*) = *f*(*t*).
-
-# **Solution:**
-
-Because *T* = 2*π*, *ω*0 = 2*π*∕*T* = 1. Hence,
-
-$$
-c_n = \frac{1}{T} \int_0^T f(t)e^{-jn\omega_0 t} dt = \frac{1}{2\pi} \int_0^{2\pi} e^t e^{-jnt} dt
-$$
-$$
-= \frac{1}{2\pi} \frac{1}{1 - jn} e^{(1 - jn)t} \Big|_0^{2\pi} = \frac{1}{2\pi(1 - jn)} \left[ e^{2\pi} e^{-j2\pi n} - 1 \right]
-$$
-
-But by Euler's identity,
-
-$$
-e^{-j2\pi n} = \cos 2\pi n - j \sin 2 \pi n = 1 - j0 = 1
-$$
-
-Thus,
-
-$$
-c_n = \frac{1}{2\pi(1 - jn)} \left[ e^{2\pi} - 1 \right] = \frac{85}{1 - jn}
-$$
-
-The complex Fourier series is
-
-$$
-f(t) = \sum_{n = -\infty}^{\infty} \frac{85}{1 - jn} e^{jnt}
-$$
-
-We may want to plot the complex frequency spectrum of *f*(*t*). If we let *cn* = ∣*cn*∣ <sup>⧸</sup>*θn*, then
-
-$$
-|c_n| = \frac{85}{\sqrt{1 + n^2}}, \qquad \theta_n = \tan^{-1} n
-$$
-
-By inserting in negative and positive values of *n*, we obtain the amplitude and the phase plots of *cn* versus *nω*0 = *n*, as in Fig. 17.30.
-
-# **Figure 17.30**
-
-The complex frequency spectrum of the function in Example 17.10: (a) amplitude spectrum, (b) phase spectrum.
-
-Obtain the complex Fourier series of the function in Fig. 17.1.
-
-Practice Problem 17.10
-
-**Answer:** 
-$$
-f(t) = \frac{1}{2} - \sum_{\substack{n=-\infty\\n \neq 0\\n = \text{odd}}} \frac{j}{n\pi} e^{jn\pi t}.
-$$
-
-Find the complex Fourier series of the sawtooth wave in Fig. 17.9. Plot the amplitude and the phase spectra.
-
-Example 17.11
-
-# **Solution:**
-
-From Fig. 17.9, *f*(*t*) = *t*, 0 < *t* < 1, *T* = 1 so that *ω*0 = 2*π*∕*T* = 2*π*. Hence,
-
-$$
-c_n = \frac{1}{T} \int_0^T f(t)e^{-jn\omega_0 t} dt = \frac{1}{T} \int_0^1 t e^{-j2n\pi t} dt
-$$
- (17.11.1)
-
-But
-
-$$
-\int t e^{at} dt = \frac{e^{at}}{a^2} (ax - 1) + C
-$$
-
-Applying this to Eq. (17.11.1) gives
-
-$$
-c_n = \frac{e^{-j2n\pi t}}{(-j2n\pi)^2} (-j2n\pi t - 1) \Big|_0^1
-$$
-  
-= 
-$$
-\frac{e^{-j2n\pi} (-j2n\pi - 1) + 1}{-4n^2 \pi^2}
-$$
- (17.11.2)
-
-Again,
-
-$$
-e^{-j2\pi n} = \cos 2\pi n - j \sin 2\pi n = 1 - j0 = 1
-$$
-
-so that Eq. (17.11.2) becomes
-
-$$
-c_n = \frac{-j2n\pi}{-4n^2\pi^2} = \frac{j}{2n\pi}
-$$
- (17.11.3)
-
-This does not include the case when *n* = 0. When *n* = 0,
-
-$$
-c_0 = \frac{1}{T} \int_0^T f(t)dt = \frac{1}{1} \int_0^1 t \, dt = \frac{t^2}{2} \Big|_1^0 = 0.5 \tag{17.11.4}
-$$
-
-Hence,
-
-$$
-f(t) = 0.5 + \sum_{\substack{n=-\infty\\n\neq 0}}^{\infty} \frac{j}{2n\pi} e^{j2n\pi t}
-$$
- (17.11.5)
-
-and
-
-$$
-|c_n| = \begin{cases} \frac{1}{2|n|\pi}, & n \neq 0\\ 0.5, & n = 0 \end{cases}, \qquad \theta_n = 90^\circ, \qquad n \neq 0 \qquad (17.11.6)
-$$
-
-By plotting ∣*cn*∣ and *θn* for different *n*, we obtain the amplitude spectrum and the phase spectrum shown in Fig. 17.31.
-
-<span id="page-811-0"></span>Obtain the complex Fourier series expansion of *f*(*t*) in Fig. 17.17. Show the amplitude and phase spectra. Practice Problem 17.11
-
-**Answer:** 
-$$
-f(t) = \sum_{\substack{n=-\infty\\n\neq 0}}^{\infty} \frac{j(-1)^n}{n\pi} e^{jn\pi t}
-$$
-. See Fig. 17.32 for the spectra.
-
-# **Figure 17.32**
-
-For Practice Prob. 17.11: (a) amplitude spectrum, (b) phase spectrum.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/202_17.7 Fourier Analysis with PSpice.md

@@ -1,194 +0,0 @@
-# **17.7** Fourier Analysis with PSpice
-
-Fourier analysis is usually performed with *PSpice* in conjunction with transient analysis. Therefore, we must do a transient analysis to perform a Fourier analysis.
-
-To perform the F ourier analysis of a w aveform, we need a circuit whose input is the waveform and whose output is the Fourier decomposition. A suitable circuit is a current (or v oltage) source in series with a 1-Ω resistor as sho wn in Fig. 17.33. The waveform is inputted as *vs*(*t*) using VPULSE for a pulse or VSIN for a sinusoid, and the attributes of the waveform are set over its period *T.* The output V(1) from node 1 is the dc level (*a*0) and the first nine harmonics (*An*) with their corresponding phases *ψn*; that is,
-
-$$
-v_o(t) = a_0 + \sum_{n=1}^{9} A_n \sin(n\omega_0 t + \psi_n)
-$$
- (17.73)
-
-where
-
-$$
-A_n = \sqrt{a_n^2 + b_n^2}, \qquad \psi_n = \phi_n - \frac{\pi}{2}, \qquad \phi_n = \tan^{-1} \frac{b_n}{a_n} \quad (17.74)
-$$
-
-Notice in Eq. (17.74) that the *PSpice* output is in the sine and angle form rather than the cosine and angle form in Eq. (17.10). The *PSpice* output also includes the normalized F ourier coefficients. Each coefficient *an* is normalized by di viding it by the magnitude of the fundamental *a*1, so that the normalized component is *an*∕*a*1. The corresponding phase *ψn* is normalized by subtracting from it the phase *ψ*1 of the fundamental, so that the normalized phase is *ψn* − *ψ*1.
-
-There are tw o types of F ourier analyses of fered by *PSpice for Windows: Discrete Fourier Transform* (DFT) performed by the *PSpice*
-
-**Figure 17.33** Fourier analysis with *PSpice* using: (a) a current source, (b) a voltage source.
-
-program and *Fast Fourier Transform* (FFT) performed by the *PSpice A/D* program. While DFT is an approximation of the exponential Fourier series, FTT is an algorithm for rapid ef ficient numerical computation of DFT. A full discussion of DFT and FTT is beyond the scope of this book.
-
-# **17.7.1** Discrete Fourier Transform
-
-A discrete F ourier transform (DFT) is performed by the *PSpice* pro gram, which tab ulates the harmonics in an output file. To enable a Fourier analysis, we select **Analysis/Setup/Transient** and bring up the Transient dialog box, sho wn in Fig. 17.34. The *Print Step* should be a small fraction of the period *T*, while the *Final Time* could be 6T. The *Center Frequency* is the fundamental frequenc y *f*0 = 1∕*T*. The particular variable whose DFT is desired, V(1) in Fig. 17.34, is entered in the **Output Vars** command box. In addition to filling in the Transient dialog box, **DCLICK** *Enable Fourier*. With the F ourier analysis enabled and the schematic sa ved, run *PSpice* by selecting **Analysis/Simulate** as usual. The program executes a harmonic decomposition into Fourier components of the result of the transient analysis. The results are sent to an output file which can be retrieved by selecting **Analysis/Examine Output**. The output file includes the dc value and the first nine harmonics by default, although you can specify more in the *Number of harmonics* box (see Fig. 17.34).
-
-# **17.7.2** Fast Fourier Transform
-
-A fast Fourier transform (FFT) is performed by the *PSpice A/D* program and displays as a *PSpice A /D* plot the complete spectrum of a t ransient expression. As explained above, we first construct the schematic in Fig. 17.33(b) and enter the attrib utes of the w aveform. We also need to enter the *Print Step* and the *Final Time* in the Transient dialog box. Once this is done, we can obtain the FFT of the waveform in two ways.
-
-One way is to insert a v oltage marker at node 1 in the schematic of the circuit in Fig. 17.33(b). After saving the schematic and selecting **Analysis/Simulate**, the w aveform V(1) will be displayed in the *PSpice A /D* window. Double clicking the FFT icon in the *PSpice A /D* menu will automatically replace the w aveform with its FFT . From the FFT- generated graph, we can obtain the harmonics. In case the FFT generated graph is crowded, we can use the *User Defined* data range (see Fig. 17.35) to specify a smaller range.
-
-| Data Range            | Use Data              |
-|-----------------------|-----------------------|
-| C Auto Range          | $F$ Full              |
-| <b>C</b> User Defined | C Restricted [analog] |
-| OHz<br>to $10Hz$      | 10Hz<br>to TKHz       |
-| Scale                 | Processing Options    |
-| C Linear              | $\nabla$ Fourier      |
-| $C$ Log               | Performance Analysis  |
-
-**Figure 17.35** *X* axis settings dialog box.
-
-| <b>Transient</b>                |        |
-|---------------------------------|--------|
-| <b>Transient Analysis</b>       |        |
-| Print Step:                     | 0.01   |
-| Final Time:                     | 12s    |
-| No-Print Delay:                 |        |
-| Step Ceiling:                   | 10ms   |
-| Detailed Bias Pt.               |        |
-| Skip initial transient solution |        |
-| Fourier Analysis                |        |
-| <b>Ⅳ</b> Enable Fourier         |        |
-| Center Frequency:               | 0.5    |
-| Number of harmonics:            |        |
-| Output Vars.: V(1)              |        |
-|                                 |        |
-| <b>OK</b>                       | Cancel |
-
-**Figure 17.34** Transient dialog box.
-
-Another way of obtaining the FFT of V(1) is to not insert a voltage marker at node 1 in the schematic. After selecting **Analysis/ Simulate**, the *PSpice A /D* window will come up with no graph on it. We select **Trace/Add** and type V(1) in the **Trace Command** box and **DCLICKL OK**. We now select **Plot/X-Axis Settings** to bring up the *X-Axis Setting* dialog box shown in Fig. 17.35 and then select **Fourier/ OK**. This will cause the FFT of the selected trace (or traces) to be dis played. This second approach is useful for obtaining the FFT of any trace associated with the circuit.
-
-A major advantage of the FFT method is that it pro vides graphical output. But its major disadvantage is that some of the harmonics may be too small to see.
-
-In both DFT and FFT , we should let the simulation run for a lar ge number of cycles and use a small value of *Step Ceiling* (in the Transient dialog box) to ensure accurate results. The *Final Time* in the Transient dialog box should be at least five times the period of the signal to allo w the simulation to reach steady state.
-
-Use *PSpice* to determine the Fourier coefficients of the signal in Fig. 17.1.
-
-# **Solution:**
-
-Figure 17.36 shows the schematic for obtaining the Fourier coefficients. With the signal in Fig. 17.1 in mind, we enter the attributes of the volt age source VPULSE as shown in Fig. 17.36. We will solve this example using both the DFT and FFT approaches.
-
-■ **METHOD 1 DFT Approach:** (The voltage marker in Fig. 17.36 is not needed for this method.) From Fig. 17.1, it is evident that *T* = 2 s,
-
-$$
-f_0 = \frac{1}{T} = \frac{1}{2} = 0.5 \text{ Hz}
-$$
-
-So, in the transient dialog box, we select the *Final Time* as 6*T* = 12 s, the *Print Step* as 0.01 s, the *Step Ceiling* as 10 ms, the *Center Frequency*  as 0.5 Hz, and the output variable as V(1). (In fact, Fig. 17.34 is for this particular example.) When *PSpice* is run, the output file contains the following result:
-
-# FOURIER COEFFICIENTS OF TRANSIENT RESPONSE V(1)
-
-# DC COMPONENT = 4.989950E-01
-
-Example 17.12
-
-Schematic for Example 17.12.
-
-| HARMONIC<br>NO | FREQUENCY<br>(HZ) | FOURIER<br>COMPONENT | NORMALIZED<br>COMPONENT | PHASE<br>(DEG) | NORMALIZED<br>PHASE (DEG) |
-|----------------|-------------------|----------------------|-------------------------|----------------|---------------------------|
-| 1              | 5.000E-01         | 6.366E-01            | 1.000E + 00             | -1.809E-01     | 0.000E+00                 |
-| 2              | 1.000E+00         | 2.012E-03            | 3.160E-03               | -9.226E+01     | -9.208E+01                |
-| 3              | 1.500E+00         | 2.122E-01            | 3.333E-01               | -5.427E-01     | -3.619E-01                |
-| 4              | 2.000E+00         | 2.016E-03            | 3.167E-03               | -9.451E+01     | -9.433E+01                |
-| 5              | 2.500E+00         | 1.273E-01            | 1.999E-01               | -9.048E-01     | -7.239E-01                |
-| 6              | 3.000E+00         | 2.024E-03            | 3.180E-03               | -9.676E+01     | -9.658E+01                |
-| 7              | 3.500E+00         | 9.088E-02            | 1.427E-01               | -1.267E+00     | -1.086E+00                |
-| 8              | 4.000E+00         | 2.035E-03            | 3.197E-03               | -9.898E+01     | -9.880E+01                |
-| 9              | 4.500E+00         | 7.065E-02            | 1.110E-01               | -1.630E+00     | -1.449E+00                |
-
-Comparing the result with that in Eq. (17.1.7) (see Example 17.1) or with the spectra in Fig. 17.4 shows a close agreement. From Eq. (17.1.7), the dc component is 0.5 while *PSpice* gives 0.498995. Also, the signal has only odd harmonics with phase *ψn* = −90°, whereas *PSpice* seems to indicate that the signal has even harmonics although the magnitudes of the even harmonics are small.
-
-■ **METHOD 2 FFT Approach:** With voltage marker in Fig. 17.36 in place, we run *PSpice* and obtain the waveform V(1) shown in Fig. 17.37(a) on the *PSpice A/D* window. By double clicking the FFT icon in the *PSpice A /D* menu and changing the X-axis setting to 0 to 10 Hz, we obtain the FFT of V(1) as shown in Fig. 17.37(b). The FFTgenerated graph contains the dc and harmonic components within the selected frequency range. Notice that the magnitudes and frequencies of the harmonics agree with the DFT-generated tabulated values.
-
-**Figure 17.37** (a) Original waveform of Fig. 17.1, (b) FFT of the waveform.
-
-Obtain the Fourier coefficients of the function in Fig. 17.7 using *PSpice*. Practice Problem 17.12
-
-# **Answer:**
-
-FOURIER COEFFICIENTS OF TRANSIENT RESPONSE V(1)
-
-DC COMPONENT = 4.950000E-01
-
-| HARMONIC<br>NO | FREQUENCY<br>(HZ) | FOURIER<br>COMPONENT | NORMALIZED<br>COMPONENT | PHASE<br>(DEG) | NORMALIZED<br>PHASE (DEG) |
-|----------------|-------------------|----------------------|-------------------------|----------------|---------------------------|
-| 1              | 1.000E+00         | 3.184E-01            | 1.000E+00               | -1.782E+02     | 0.000E+00                 |
-| 2              | 2.000E+00         | 1.593E-01            | 5.002E-01               | -1.764E+02     | 1.800E+00                 |
-| 3              | 3.000E+00         | 1.063E-01            | 3.338E-01               | -1.746E+02     | 3.600E+00                 |
-|                |                   |                      |                         |                | (continued)               |
-
-| (continued) |           |           |           |            |           |
-|-------------|-----------|-----------|-----------|------------|-----------|
-| 4           | 4.000E+00 | 7.979E-02 | 2.506E-03 | -1.728E+02 | 5.400E+00 |
-| 5           | 5.000E+00 | 6.392E-01 | 2.008E-01 | -1.710E+02 | 7.200E+00 |
-| 6           | 6.000E+00 | 5.337E-02 | 1.676E-03 | -1.692E+02 | 9.000E+00 |
-| 7           | 7.000E+00 | 4.584E-02 | 1.440E-01 | -1.674E+02 | 1.080E+01 |
-| 8           | 8.000E+00 | 4.021E-02 | 1.263E-01 | -1.656E+02 | 1.260E+01 |
-| 9           | 9.000E+00 | 3.584E-02 | 1.126E-01 | -1.638E+02 | 1.440E+01 |
-|             |           |           |           |            |           |
-
-If *vs* = 12 sin(200 *πt*)*u*(*t*) V in the circuit of Fig. 17.38, find *i*(*t*).
-
-# **Solution:**
-
-- 1. **Define.** Although the problem appears to be clearly stated, it might be advisable to check with the individual who assigned the problem to make sure he or she wants the transient response rather than the steady-state response; in the latter case the problem becomes trivial.
-- 2. **Present.** We are to determine the response *i*(*t*) given the input *vs*(*t*), using *PSpice* and Fourier analysis.
-- 3. **Alternative.** We will use DFT to perform the initial analysis. We will then check using the FFT approach.
-- 4. **Attempt.** The schematic is shown in Fig. 17.39. We may use the DFT approach to obtain the Fourier coefficents of *i*(*t*). Because the period of the input waveform is *T* = 1∕100 = 10 ms, in the Transient dialog box we select *Print Step:* 0.1 ms, *Final Time:* 100 ms, *Center Frequency:* 100 Hz, *Number of harmonics:* 4, and *Output Vars:* I(L1). When the circuit is simulated, the output file includes the following:
-
-FOURIER COEFFICIENTS OF TRANSIENT RESPONSE I(VD)
-
-| HARMONIC<br>NO | FREQUENCY<br>(HZ) | FOURIER<br>COMPONENT | NORMALIZED<br>COMPONENT | PHASE<br>(DEG) | NORMALIZED<br>PHASE (DEG) |
-|----------------|-------------------|----------------------|-------------------------|----------------|---------------------------|
-| 1              | 1.000E+02         | 8.730E-03            | 1.000E+00               | -8.984E+01     | 0.000E+00                 |
-| 2              | 2.000E+02         | 1.017E-04            | 1.165E-02               | -8.306E+01     | 6.783E+00                 |
-| 3              | 3.000E+02         | 6.811E-05            | 7.802E-03               | -8.235E+01     | 7.490E+00                 |
-| 4              | 4.000E+02         | 4.403E-05            | 5.044E-03               | -8.943E+01     | 4.054E+00                 |
-
-With the Fourier coefficients, the Fourier series describing the current *i*(*t*) can be obtained using Eq. (17.73); that is,
-
-$$
-i(t) = 8.5833 + 8.73 \sin(2\pi \cdot 100t - 89.84^{\circ})
-$$
-  
-+ 0.1017 sin(2\pi \cdot 200t - 83.06^{\circ})  
-+ 0.068 sin(2\pi \cdot 300t - 82.35^{\circ}) + \cdots mA
-
-5. **Evaluate.** We can also use the FFT approach to cross-check our result. The current marker is inserted at pin 1 of the inductor as shown in Fig. 17.39. Running *PSpice* will automatically produce the plot of I(L1) in the *PSpice A/D* window, as shown
-
-**Figure 17.39** Schematic of the circuit in Fig. 17.38.
-
-**Figure 17.40** For Example 17.13: (a) plot of *i*(*t*), (b) the FFT of *i*(*t*).
-
-in Fig. 17.40(a). By double clicking the FFT icon and setting the range of the X-axis from 0 to 200 Hz, we generate the FFT of I(L1) shown in Fig. 17.40(b). It is clear from the FFT-generated plot that only the dc component and the first harmonic are visible. Higher harmonics are negligibly small.
-
-One final observation, does the answer make sense? Let us look at the actual transient response, *i*(*t*) = (9.549*e*<sup>−</sup>0.5*<sup>t</sup>* − 9.549) cos(200*πt*)*u*(*t*) mA. The period of the cosine wave is 10 ms while the time constant of the exponential is 2000 ms (2 seconds). So, the answer we obtained by Fourier techniques does agree.
-
- 6. **Satisfactory?** Clearly, we have solved the problem satisfactorily using the specified approach. We can now present our results as a solution to the problem.
-
-A sinusoidal current source of amplitude 4 A and frequency 2 kHz is applied to the circuit in Fig. 17.41. Use *PSpice* to find *v*(*t*). Practice Problem 17.13
-
-**Answer:** *v*(*t*) = −150.72 + 145.5 sin(4*π* ⋅ 103 *t* + 90°) + ⋯ *μ*V. The Fourier components are shown below:
-
-**Figure 17.41** For Practice Prob. 17.13.
-
-FOURIER COEFFICIENTS OF TRANSIENT RESPONSE V(R1:1)
-
-```
-DC COMPONENT = -1.507169E-04
-```
-
-| HARMONIC<br>NO | FREQUENCY<br>(HZ) | FOURIER<br>COMPONENT | NORMALIZED<br>COMPONENT | PHASE<br>(DEG) | NORMALIZED<br>PHASE (DEG) |
-|----------------|-------------------|----------------------|-------------------------|----------------|---------------------------|
-| 1              | 2.000E+03         | 1.455E-04            | 1.000E+00               | 9.006E+01      | 0.000E+00                 |
-| 2              | 4.000E+03         | 1.851E-06            | 1.273E-02               | 9.597E+01      | 5.910E+00                 |
-| 3              | 6.000E+03         | 1.406E-06            | 9.662E-03               | 9.323E+01      | 3.167E+00                 |
-| 4              | 8.000E+03         | 1.010E-06            | 6.946E-02               | 8.077E+01      | -9.292E+00                |
-|                |                   |                      |                         |                |                           |

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/203_17.8 Applications.md

@@ -1,102 +0,0 @@
-# <span id="page-817-0"></span>**17.8** Applications
-
-We demonstrated in Section 17.4 that the F ourier series expansion permits the application of the phasor techniques used in ac analysis to circuits containing nonsinusoidal periodic excitations. The Fourier series has many other practical applications, particularly in communications and signal processing. Typical applications include spectrum analysis, filtering, rectification, and harmonic distortion. We will consider two of these: spectrum analyzers and filters.
-
-# **17.8.1** Spectrum Analyzers
-
-The Fourier series provides the spectrum of a signal. As we have seen, the spectrum consists of the amplitudes and phases of the harmonics versus frequency. By providing the spectrum of a signal *f*(*t*), the Fourier series helps us identify the pertinent features of the signal. It demon strates which frequencies are playing an important role in the shape of the output and which ones are not. F or example, audible sounds ha ve significant components in the frequency range of 20 Hz to 15 kHz, while visible light signals range from 105 to 106 GHz. Table 17.4 presents some other signals and the frequency ranges of their components. A periodic function is said to be *band-limited* if its amplitude spectrum contains only a finite number of coefficients *An* or *cn*. In this case, the F ourier series becomes
-
-$$
-f(t) = \sum_{n=-N}^{N} c_n e^{jn\omega_0 t} = a_0 + \sum_{n=1}^{N} A_n \cos(n\omega_0 t + \phi_n)
-$$
- (17.75)
-
-This shows that we need only 2*N* + 1 terms (namely, *a*0, *A*1, *A*2, …, *AN*, 1, 2, …, *N*) to completely specify *f*(*t*) if *ω*0 is kno wn. This leads to the *sampling theorem:* a band-limited periodic function whose Fourier series contains *N* harmonics is uniquely specified by its values at 2*N* + 1 instants in one period.
-
-A *spectrum analyzer* is an instrument that displays the amplitude of the components of a signal v ersus frequenc y. It sho ws the v arious frequency components (spectral lines) that indicate the amount of energy at each frequency.
-
-It is unlik e an oscilloscope, which displays the entire signal (all components) versus time. An oscilloscope shows the signal in the time domain, while the spectrum analyzer sho ws the signal in the frequenc y domain. There is perhaps no instrument more useful to a circuit analyst than the spectrum analyzer . An analyzer can conduct noise and spuri ous signal analysis, phase checks, electromagnetic interference and filter examinations, vibration measurements, radar measurements, and more. Spectrum analyzers are commercially a vailable in v arious sizes and shapes. Figure 17.42 displays a typical one.
-
-# **17.8.2** Filters
-
-Filters are an important component of electronics and communications systems. Chapter 14 presented a full discussion on passive and active filters. Here, we investigate how to design filters to select the fundamental component (or any desired harmonic) of the input signal and reject other harmonics. This filtering process cannot be accomplished without the
-
-# **TABLE 17.4**
-
-# Frequency ranges of typical signals.
-
-| Signal             | Frequency Range    |
-|--------------------|--------------------|
-| Audible sounds     | 20 Hz to 15 kHz    |
-| AM radio           | 540–1600 kHz       |
-|                    |                    |
-| Video signals      | dc to 4.2 MHz      |
-| (U.S. standards)   |                    |
-| VHF television,    | 54–216 MHz         |
-| FM radio           |                    |
-| UHF television     | 470–806 MHz        |
-| Cellular telephone | 824–891.5 MHz      |
-| Microwaves         | 2.4–300 GHz        |
-| Visible light      | 105<br>–106<br>GHz |
-| X-rays             | 108<br>–109<br>GHz |
-| Short-wave radio   | 3–36 MHz           |
-
-Fourier series expansion of the input signal. For the purpose of illustration, we will consider tw o cases, a lo w-pass filter and a band-pass filter. In Example 17.6, we already looked at a high-pass *RL* filter.
-
-The output of a lo w-pass filter depends on the input signal, the transfer function *H*(*ω*) of the filter, and the corner or half-po wer fre quency *ωc*. We recall that *ωc* = 1∕*RC* for an *RC* passive filter. As shown in Fig. 17.43(a), the low-pass filter passes the dc and low-frequency components, while blocking the high-frequency components. By making *ω<sup>c</sup>* sufficiently large (*ωc* ≫ *ω*0, e.g., making *C* small), a large number of the harmonics can be passed. On the other hand, by making *ωc* sufficiently small (*ωc* ≪ *ω*0), we can block out all the ac components and pass only dc, as shown typically in Fig. 17.43(b). (See Fig. 17.2(a) for the Fourier series expansion of the square wave.)
-
-# **Figure 17.43**
-
-(a) Input and output spectra of a low-pass filter, (b) the low-pass filter passes only the dc component when *ω<sup>c</sup>* ≪ *ω*0.
-
-Similarly, the output of a bandpass filter depends on the input signal, the transfer function of the filter *H*(*ω*), its bandwidth *B*, and its cen ter frequency *ωc*. As illustrated in Fig. 17.44(a), the filter passes all the harmonics of the input signal within a band of frequencies ( *ω*1 < *ω*< *ω*2) centered around *ωc*. We have assumed that *ω*0, 2*ω*0, and 3*ω*0 are within that band. If the filter is made highly selective (*B* ≪ *ω*0) and *ωc* = *ω*0, where *ω*0 is the fundamental frequency of the input signal, the filter passes only the fundamental component (*n* = 1) of the input and blocks out all higher harmonics. As shown in Fig. 17.44(b), with a square wave as input, we obtain a sine wave of the same frequency as the output. (Again, refer to Fig. 17.2(a).)
-
-# **Figure 17.44**
-
-x(t)
-
-1
-
-**Figure 17.45** For example 17.14.
-
-‒1 0 2 3
-
-1 (a)
-
-(a) Input and output spectra of a bandpass filter, (b) the bandpass filter passes only the fundamental component when *B* ≪ *ω*0.
-
-If the sawtooth waveform in Fig. 17.45(a) is applied to an ideal lo w-pass filter with the transfer function shown in Fig. 17.45(b), determine the output.
-
-t
-
-0 *ω*
-
-10 (b)
-
-1
-
-│H│
-
-Example 17.14
-
-The input signal in Fig. 17.45(a) is the same as the signal in Fig. 17.9. From Practice Prob. 17.2, we know that the Fourier series expansion is
-
-$$
-x(t) = \frac{1}{2} - \frac{1}{\pi} \sin \omega_0 t - \frac{1}{2\pi} \sin 2\omega_0 t - \frac{1}{3\pi} \sin 3\omega_0 t - \dots
-$$
-
-In this section, we have used *ω*c for the center frequency of the bandpass filter instead of *ω*0 as in Chapter 14, to avoid confusing *ω*0 with the fundamental frequency of the input signal.
-
-where the period is *T* = 1 s and the fundamental frequency is *ω*0 = 2*π*rad/s. Inasmuch as the corner frequency of the filter is *ωc* = 10 rad/s, only the dc component and harmonics with *nω*0 < 10 will be passed. For *n* = 2, *nω*0 = 4*π* = 12.566 rad/s, which is higher than 10 rad/s, meaning that second and higher harmonics will be rejected. Thus, only the dc and fundamental components will be passed. Hence, the output of the filter is
-
-$$
-y(t) = \frac{1}{2} - \frac{1}{\pi} \sin 2\pi t
-$$
-
-<span id="page-820-0"></span>**Figure 17.46** For Practice Prob. 17.14. Rework Example 17.14 if the low-pass filter is replaced by the ideal bandpass filter shown in Fig. 17.46.
-
-**Answer:** 
-$$
-y(t) = -\frac{1}{3\pi} \sin 3\omega_0 t - \frac{1}{4\pi} \sin 4\omega_0 t - \frac{1}{5\pi} \sin 5\omega_0 t
-$$
-.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/204_17.9 Summary.md

@@ -1,124 +0,0 @@
-# **17.9** Summary
-
-- 1. A periodic function is one that repeats itself every *T* seconds; that is, *f*(*t* ± *nT*) = *f*(*t*), *n* = 1, 2, 3, ….
-- 2. Any nonsinusoidal periodic function *f*(*t*) that we encounter in electrical engineering can be e xpressed in terms of sinusoids using Fourier series:
-
-$$
-f(t) = \underbrace{a_0}_{\text{dc}} + \underbrace{\sum_{n=1}^{\infty} (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t)}_{\text{ac}}
-$$
-
- where *ω*0 = 2*π*∕*T* is the fundamental frequency. The Fourier series resolves the function into the dc component *a*0 and an ac component containing infinitely many harmonically related sinusoids. The Fourier coefficients are determined as
-
-$$
-a_0 = \frac{1}{T} \int_0^T f(t) dt, \qquad a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t dt
-$$
-$$
-b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t dt
-$$
-
- If *f*(*t*) is an e ven function, *bn* = 0, and when *f*(*t*) is odd, *a*0 = 0 and *an* = 0. If *f*(*t*) is half-wave symmetric, *a*0 = *an* = *bn* = 0 for even values of *n*.
-
-3. An alternative to the trigonometric (or sine-cosine) Fourier series is the amplitude-phase form
-
-$$
-f(t) = a_0 + \sum_{n=1}^{\infty} A_n \cos(n\omega_0 t + \phi_n)
-$$
-
-where
-
-$$
-A_n = \sqrt{a_n^2 + b_n^2}
-$$
-,  $\phi_n = -\tan^{-1} \frac{b_n}{a_n}$ 
-
-- 4. Fourier series representation allo ws us to apply the phasor method in analyzing circuits when the source function is a nonsinusoidal periodic function. We use phasor technique to determine the response of each harmonic in the series, transform the responses to the time domain, and add them up.
-- 5. The average-power of periodic voltage and current is
-
-$$
-P = V_{\rm dc} I_{\rm dc} + \frac{1}{2} \sum_{n=1}^{\infty} V_n I_n \cos(\theta_n - \phi_n)
-$$
-
- In other w ords, the total average power is the sum of the a verage powers in each harmonically related voltage and current.
-
-6. A periodic function can also be represented in terms of an exponential (or complex) Fourier series as
-
-$$
-f(t) = \sum_{n = -\infty}^{\infty} c_n e^{jn\omega_0 t}
-$$
-
-where
-
-$$
-c_n = \frac{1}{T} \int_0^T f(t) e^{-jn\omega_0 t} dt
-$$
-
- and *ω*0 = 2*π*∕*T*. The e xponential form describes the spectrum of *f*(*t*) in terms of the amplitude and phase of ac components at posi tive and negative harmonic frequencies. Thus, there are three basic forms of F ourier series representation: the trigonometric form, the amplitude-phase form, and the exponential form.
-
-- 7. The frequency (or line) spectrum is the plot of *An* and *n* or ∣*cn*∣ and *θ<sup>n</sup>* versus frequency.
-- 8. The rms value of a periodic function is given by
-
-$$
-F_{\rm rms} = \sqrt{a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} A_n^2}
-$$
-
-The power dissipated by a 1-Ω resistance is
-
-$$
-P_{1\Omega} = F_{\text{rms}}^2 = a_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} (a_n^2 + b_n^2) = \sum_{n=-\infty}^{\infty} |c_n|^2
-$$
-
-This relationship is known as *Parseval's theorem*.
-
-- 9. Using *PSpice*, a F ourier analysis of a circuit can be performed in conjunction with the transient analysis.
-- 10. Fourier series find application in spectrum analyzers and filters. The spectrum analyzer is an instrument that displays the discrete Fourier spectra of an input signal, so that an analyst can determine the fre quencies and relative energies of the signal's components. Because the Fourier spectra are discrete spectra, filters can be designed for great effectiveness in blocking frequenc y components of a signal that are outside a desired range.
-
-# <span id="page-822-0"></span>Review Questions
-
-- **17.1** Which of the following cannot be a Fourier series?
-  - (a) *t* − *t* 2 \_\_ 2 <sup>+</sup>*<sup>t</sup>* 3 \_\_ 3 − *t* 4 \_\_ 4 <sup>+</sup>*<sup>t</sup>* 5 \_\_ 5 (b) 5 sin *t* + 3 sin 2*t* − 2 sin 3*t* + sin 4*t*
-  - (c) sin *t* − 2 cos 3*t* + 4 sin 4*t* + cos 4*t*
-  - (d) sin *t* + 3 sin 2.7*t* − cos *πt* + 2 tan *πt*
-
-(e) 
-$$
-1 + e^{-j\pi t} + \frac{e^{-j2\pi t}}{2} + \frac{e^{-j3\pi t}}{3}
-$$
-
-**17.2** If *f*(*t*) = *t*, 0 < *t* < *π*, *f*(*t* + *nπ*) = *f*(*t*), the value of *ω*0 is
-
-(a) 1 (b) 2 (c) 
-$$
-\pi
-$$
- (d)  $2\pi$ 
-
-**17.3** Which of the following are even functions?
-
-| 2<br>(a) t + t         | 2<br>(b) t<br>cos t | (c) et2 |
-|------------------------|---------------------|---------|
-| 2<br>4<br>(d) t<br>+ t | (e) sinh t          |         |
-
-**17.4** Which of the following are odd functions?
-
-| (a) sin t + cos t | (b) t sin t         |
-|-------------------|---------------------|
-| (c) t ln t        | 3<br>(d) t<br>cos t |
-| (e) sinh t        |                     |
-
-**17.5** If *f*(*t*) = 10 + 8 cos *t* + 4 cos 3*t* + 2 cos 5*t* + …, the magnitude of the dc component is:
-
-| (a) 10 | (b) 8 | (c) 4 |
-|--------|-------|-------|
-| (d) 2  | (e) 0 |       |
-
-**17.6** If *f*(*t*) = 10 + 8 cos *t* + 4 cos 3*t* + 2 cos 5*t* + …, the angular frequency of the 6th harmonic is
-
-| (a) 12 | (b) 11 | (c) 9 |
-|--------|--------|-------|
-| (d) 6  | (e) 1  |       |
-
-- **17.7** The function in Fig. 17.14 is half-wave symmetric.
-  - (a) True (b) False
-- **17.8** The plot of ∣*cn*∣ versus *nω*0 is called:
-
-(a) complex frequency spectrum (b) complex amplitude spectrum (c) complex phase spectrum

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@@ -1,10 +0,0 @@
-
-- **17.9** When the periodic voltage 2 + 6 sin *ω*0*t* is applied to a 1-Ω resistor, the integer closest to the power (in watts) dissipated in the resistor is:
-  - (a) 5 (b) 8 (c) 20 (d) 22 (e) 40
-- **17.10** The instrument for displaying the spectrum of a signal is known as:
-
-| (a) oscilloscope      | (b) spectrogram          |
-|-----------------------|--------------------------|
-| (c) spectrum analyzer | (d) Fourier spectrometer |
-
-*Answers: 17.1a,d, 17.2b, 17.3b,c,d, 17.4d,e, 17.5a, 17.6d, 17.7a, 17.8b, 17.9d, 17.10c.*

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/206_Problems.md

@@ -1,417 +0,0 @@
-# Problems
-
-# Section 17.2 Trigonometric Fourier Series
-
-**17.1** Evaluate each of the following functions and see if it is periodic. If periodic, find its period.
-
-(a) 
-$$
-f(t) = \cos \pi t + 2 \cos 3\pi t + 3 \cos 5\pi t
-$$
-  
-\n(b)  $y(t) = \sin t + 4 \cos 2 \pi t$   
-\n(c)  $g(t) = \sin 3t \cos 4t$   
-\n(d)  $h(t) = \cos^2 t$   
-\n(e)  $z(t) = 4.2 \sin(0.4\pi t + 10^{\circ}) + 0.8 \sin(0.6\pi t + 50^{\circ})$   
-\n(f)  $p(t) = 10$ 
-
-$$
-(g) q(t) = e^{-\pi t}
-$$
-
-**17.2** Using MATLAB, synthesize the periodic waveform for which the Fourier trigonometric Fourier series is
-
-$$
-f(t) = \frac{1}{2} - \frac{4}{\pi^2} \left( \cos t + \frac{1}{9} \cos 3t + \frac{1}{25} \cos 5t + \cdots \right)
-$$
-
-**17.3** Give the Fourier coefficients *a*0, *an*, and *bn* of the waveform in Fig. 17.47. Plot the amplitude and phase spectra.
-
-For Prob. 17.3.
-
-**17.4** Find the Fourier series expansion of the backward sawtooth waveform of Fig. 17.48. Obtain the amplitude and phase spectra.
-
-For Probs. 17.4 and 17.66.
-
-**17.5** Obtain the Fourier series expansion for the waveform shown in Fig. 17.49.
-
-# **Figure 17.49**
-
-For Prob. 17.5.
-
-**17.6** Find the trigonometric Fourier series for
-
-$$
-f(t) = \begin{cases} 7.5 & 0 < t < \pi \\ 15 & \pi < t < 2\pi \end{cases} \quad \text{and} \quad f(t + 2\pi) = f(t).
-$$
-
-**17.7** Determine the Fourier series of the periodic function in Fig. 17.50.
-
-# **Figure 17.50**
-
-For Prob. 17.7.
-
-**17.8** Using Fig. 17.51, design a problem to help other students better understand how to determine the exponential Fourier series from a periodic wave shape.
-
-\* An asterisk indicates a challenging problem.
-
-**17.9** Determine the Fourier coefficients *an* and *bn* of the first three harmonic terms of the rectified cosine wave in Fig. 17.52.
-
-**17.10** Find the exponential Fourier series for the waveform in Fig. 17.53.
-
-# **Figure 17.53**
-
-For Prob. 17.10.
-
-**17.11** Obtain the exponential Fourier series for the signal in Fig. 17.54.
-
-**\*17.12** A voltage source has a periodic waveform defined over its period as
-
-$$
-v(t) = 120t(2\pi - t) \text{ V}, \qquad 0 < t < 2\pi
-$$
-
-Find the Fourier series for this voltage.
-
-**17.13** Design a problem to help other students better understand obtaining the Fourier series from a periodic function.
-
-**17.14** Find the quadrature (cosine and sine) form of the Fourier series
-
-$$
-f(t) = 7.5 + \sum_{n=1}^{\infty} \frac{37.5}{n^3 + 1} \cos\left(2nt + \frac{n\pi}{4}\right)
-$$
-
-**17.15** Express the Fourier series
-
-$$
-f(t) = 10 + \sum_{n=1}^{\infty} \frac{4}{n^2 + 1} \cos 10nt + \frac{1}{n^3} \sin 10nt
-$$
-
-(a) in a cosine and angle form,
-
-(b) in a sine and angle form.
-
-**17.16** The waveform in Fig. 17.55(a) has the following Fourier series:
-
-$$
-v_1(t) = \frac{1}{2} - \frac{4}{\pi^2} \left( \cos \pi t + \frac{1}{9} \cos 3\pi t + \frac{1}{25} \cos 5\pi t + \cdots \right) \text{V}
-$$
-
-Obtain the Fourier series of *v*2(*t*) in Fig. 17.55(b).
-
-**Figure 17.55**
-
-For Probs. 17.16 and 17.69.
-
-# Section 17.3 Symmetry Considerations
-
-**17.17** Determine if these functions are even, odd, or neither.
-
-(a) 1 + *t* (b) *t* <sup>2</sup> − 1 (c) cos *nπt* sin *nπt* (d) sin2 *πt* (e) *e*<sup>−</sup>*<sup>t</sup>*
-
-**17.18** Determine the fundamental frequency and specify the type of symmetry present in the functions in Fig. 17.56.
-
-(c)
-
-**Figure 17.56** For Probs. 17.18 and 17.63.
-
-**17.19** Obtain the Fourier series for the periodic waveform in Fig. 17.57.
-
-**17.20** Find the Fourier series for the signal in Fig. 17.58. Evaluate *f*(*t*) at *t* = 2 using the first three nonzero harmonics.
-
-**17.21** Determine the trigonometric Fourier series of the signal in Fig. 17.59.
-
-**Figure 17.59** For Prob. 17.21.
-
-**17.22** Calculate the Fourier coefficients for the function in Fig. 17.60.
-
-**Figure 17.60** For Prob. 17.22.
-
-**17.23** Using Fig. 17.61, design a problem to help other students better understand finding the Fourier series of a periodic wave shape.
-
-**Figure 17.61** For Prob. 17.23.
-
-- (a) find the trigonometric Fourier series coefficients *a*2 and *b*2,
-- (b) calculate the magnitude and phase of the component of *f*(*t*) that has *ωn* = 10 rad/s,
-- (c) use the first four nonzero terms to estimate *f*(*π*∕2),
-- (d) show that
-
-$$
-\frac{\pi}{4} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \cdots
-$$
-
-**Figure 17.62** For Probs. 17.24 and 17.60.
-
-**17.25** Determine the Fourier series representation of the function in Fig. 17.63.
-
-**17.26** Find the Fourier series representation of the signal shown in Fig. 17.64.
-
-For Prob. 17.26.
-
-**17.27** For the waveform shown in Fig. 17.65 below,
-
-- (a) specify the type of symmetry it has,
-- (b) calculate *a*3 and *b*3,
-- (c) find the rms value using the first five nonzero harmonics.
-
-**Figure 17.65** For Prob. 17.27.
-
-For Prob. 17.28.
-
-**17.29** Determine the Fourier series expansion of the sawtooth function in Fig. 17.67.
-
-**Figure 17.67**
-
-For Prob. 17.29.
-
-**17.30** (a) If *f*(*t*) is an even function, show that
-
-$$
-c_n = \frac{2}{T} \int_0^{T/2} f(t) \cos n\omega_o t \, dt
-$$
-
-(b) If *f*(*t*) is an odd function, show that
-
-$$
-c_n = -\frac{j2}{T} \int_0^{T/2} f(t) \sin n\omega_o t \, dt
-$$
-
-**17.31** Let *an* and *bn* be the Fourier series coefficients of *f*(*t*) and let *ωo* be its fundamental frequency. Suppose *f*(*t*) is time-scaled to give *h*(*t*) = *f*(*t*). Express the *a*ʹ *<sup>n</sup>* and *b*ʹ *<sup>n</sup>*, and *ω*ʹ *<sup>o</sup>*, of *h*(*t*) in terms of *an*, *bn*, and *ωo* of *f*(*t*).
-
-# Section 17.4 Circuit Applications
-
-**17.32** Find *i*(*t*) in the circuit of Fig. 17.68 given that
-
-$$
-i_s(t) = 3.5 + \sum_{n=1}^{\infty} \frac{4}{n^2} \cos 3nt
-$$
- A
-
-**Figure 17.68** For Prob. 17.32.
-
-**17.33** In the circuit shown in Fig. 17.69, the Fourier series expansion of *vs*(*t*) is
-
-$$
-v_s(t) = 10 + \frac{5}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \sin(n\pi t)
-$$
-
-Find *vo*(*t*).
-
-# **Figure 17.69**
-
-For Prob. 17.33.
-
-**17.34** Using Fig. 17.70, design a problem to help other students better understand circuit responses to a Fourier series.
-
-**Figure 17.70** For Prob. 17.34.
-
-**17.35** If *vs* in the circuit of Fig. 17.71 is the same as function *f*2(*t*) in Fig. 17.56(b), determine the dc component and the first three nonzero harmonics of *vo*(*t*).
-
-**Figure 17.71** For Prob. 17.35.
-
-**17.36** Find the response *io* for the circuit in Fig. 17.72(a), where *vs*(*t*) is shown in Fig. 17.72(b).
-
-Problems **805**
-
-**17.37** If the periodic current waveform in Fig. 17.73(a) is applied to the circuit in Fig. 17.73(b), find *vo*.
-
-**17.38** If the square wave shown in Fig. 17.74(a) is applied to the circuit in Fig. 17.74(b), find the Fourier series for *vo*(*t*).
-
-**Figure 17.74** For Prob. 17.38.
-
-**17.39** If the periodic voltage in Fig. 17.75(a) is applied to the circuit in Fig. 17.75(b), find *io*(*t*).
-
-**Figure 17.75** For Prob. 17.39.
-
-**\*17.40** The signal in Fig. 17.76(a) is applied to the circuit in Fig. 17.76(b). Find *vo*(*t*).
-
-For Prob. 17.40.
-
-**17.41** The full-wave rectified sinusoidal voltage in Fig. 17.77(a) is applied to the low-pass filter in Fig. 17.77(b). Obtain the output voltage *vo*(*t*) of the filter.
-
-# **Figure 17.77**
-
-For Prob. 17.41.
-
-**17.42** The square wave in Fig. 17.78(a) is applied to the circuit in Fig. 17.78(b). Find the Fourier series of *vo*(*t*).
-
-# Section 17.5 Average Power and RMS Values
-
-**17.43** The voltage across the terminals of a circuit is
-
-$$
-v(t) = [30 + 20 \cos(60\pi t + 45^{\circ}) + 10 \cos(120\pi t - 45^{\circ})] \text{ V}
-$$
-
- If the current entering the terminal at higher potential is
-
-$$
-i(t) = 6 + 4\cos(60\pi t + 10^{\circ})
-$$
-$$
-- 2\cos(120\pi t - 60^{\circ})
-$$
- A
-
-find:
-
-(a) the rms value of the voltage,
-
-(b) the rms value of the current,
-
-(c) the average power absorbed by the circuit.
-
-**\*17.44** Design a problem to help other students better
-
-understand how to find the rms voltage across and the rms current through an electrical element given a Fourier series for both the current and the voltage. In addition, have them calculate the average power delivered to the element and the power spectrum.
-
-**17.45** A series *RLC* circuit has *R* = 10 Ω, *L* = 2 mH, and *C* = 40 *μ*F. Determine the effective current and average power absorbed when the applied voltage is
-
-> *v*(*t*) = 100 cos 1000*t* + 50 cos 2000*t* + 25 cos 3000*t* V
-
-**17.46** Use *MATLAB* to plot the following sinusoids for 0 < *t* < 5:
-
-(a) 5 cos 3*t* − 2 cos(3*t* − *π*∕3) (b) 8 sin(*πt* + *π*∕4) + 10 cos(*πt* − *π*∕8)
-
-**17.47** The periodic current waveform in Fig. 17.79 is applied across a 2-kΩ resistor. Find the percentage of the total average power dissipation caused by the dc component.
-
-**Figure 17.79** For Prob. 17.47.
-
-**17.48** For the circuit in Fig. 17.80,
-
-$$
-i(t) = 20 + 16 \cos(10t + 45^{\circ})
-$$
-  
-+ 12 \cos(20t - 60^{\circ}) mA
-
-(a) find *v*(*t*), and
-
-(b) calculate the average power dissipated in the resistor.
-
-Problems **807**
-
-**Figure 17.80**
-
-- For Prob. 17.48.
-- **17.49** (a) For the periodic waveform in Prob. 17.5, find the rms value.
-  - (b) Use the first five harmonic terms of the Fourier series in Prob. 17.5 to determine the effective value of the signal.
-  - (c) Calculate the percentage error in the estimated rms value of *z*(*t*) if
-
-s value of 
-$$
-z(t)
-$$
- if  
-\n% error =  $\left(\frac{\text{estimated value}}{\text{exact value}} - 1\right) \times 100$ 
-
-# Section 17.6 Exponential Fourier Series
-
-- **17.50** Obtain the exponential Fourier series for *f*(*t*) = *t*, −1 < *t* < 1, with *f*(*t* + 2*n*) = *f*(*t*) for all integer values of *n*.
-- **17.51** Design a problem to help other students better understand how to find the exponential Fourier series of a given periodic function.
-- **17.52** Calculate the complex Fourier series for *f*(*t*) = *et* , −*π*< *t* < *π*, with *f*(*t* + 2*πn*) = *f*(*t*) for all integer values of *n*.
-- **17.53** Find the complex Fourier series for *f*(*t*) = *e*<sup>−</sup>*<sup>t</sup>* , 0 < *t* < 1, with *f*(*t* + *n*) = *f*(*t*) for all integer values of *n*.
-- **17.54** Find the exponential Fourier series for the function in Fig. 17.81.
-
-**Figure 17.81** For Prob. 17.54.
-
-**17.55** Obtain the exponential Fourier series expansion of the half-wave rectified sinusoidal current of Fig. 17.82.
-
-# **Figure 17.82**
-
-- For Prob. 17.55.
-- **17.56** The Fourier series trigonometric representation of a periodic function is
-
-$$
-f(t) = 10 + \sum_{n=1}^{\infty} \left( \frac{1}{n^2 + 1} \cos n\pi t + \frac{n}{n^2 + 1} \sin n\pi t \right)
-$$
-
- Find the exponential Fourier series representation of *f*(*t*).
-
-**17.57** The coefficients of the trigonometric Fourier series representation of a function are:
-
-$$
-b_n = 0,
-$$
-  $a_n = \frac{6}{n^3 - 2},$   $n = 0, 1, 2, ...$ 
-
- If *ωn* = 50*n*, find the exponential Fourier series for the function.
-
-**17.58** Find the exponential Fourier series of a function that has the following trigonometric Fourier series coefficients:
-
-$$
-a_0 = \frac{\pi}{4}
-$$
-,  $b_n = \frac{(-1)^n}{n}$ ,  $a_n = \frac{(-1)^n - 1}{\pi n^2}$ 
-
-Take *T* = 2*π*.
-
-**17.59** The complex Fourier series of the function in Fig. 17.83(a) is
-
-$$
-f(t) = \frac{1}{2} - \sum_{n=-\infty}^{\infty} \frac{je^{-j(2n+1)t}}{(2n+1)\pi}
-$$
-
- Find the complex Fourier series of the function *h*(*t*) in Fig. 17.83(b).
-
-**Figure 17.83** For Prob.17.59.
-
-- **17.60** Obtain the complex Fourier coefficients of the signal in Fig. 17.62.
-- **17.61** The spectra of the Fourier series of a function are shown in Fig. 17.84. (a) Obtain the trigonometric Fourier series. (b) Calculate the rms value of the function.
-
-For Prob. 17.61.
-
-- **17.62** The amplitude and phase spectra of a truncated Fourier series are shown in Fig. 17.85.
-  - (a) Find an expression for the periodic voltage using the amplitude-phase form. See Eq. (17.10).
-  - (b) Is the voltage an odd or even function of *t*?
-
-**17.63** Plot the amplitude spectrum for the signal *f*2(*t*) in Fig. 17.56(b). Consider the first five terms.
-
-**17.64** Design a problem to help other students better understand the amplitude and phase spectra of a given Fourier series.
-
-**17.65** Given that
-
-$$
-f(t) = \sum_{\substack{n=1 \ n \equiv odd}}^{\infty} \left( \frac{20}{n^2 \pi^2} \cos 2nt - \frac{3}{n\pi} \sin 2nt \right)
-$$
-
- plot the first five terms of the amplitude and phase spectra for the function.
-
-# Section 17.7 Fourier Analysis with PSpice
-
-- **17.66** Determine the Fourier coefficients for the waveform in Fig. 17.48 using *PSpice* or *MultiSim*.
-- **17.67** Calculate the Fourier coefficients of the signal in Fig. 17.58 using *PSpice* or *MultiSim*.
-- **17.68** Use *PSpice* or *MultiSim* to find the Fourier components of the signal in Prob. 17.7.
-- **17.69** Use *PSpice* or *MultiSim* to obtain the Fourier coefficients of the waveform in Fig. 17.55(a).
-- **17.70** Design a problem to help other students better
-- understand how to use *PSpice* or *MultiSim* to solve circuit problems with periodic inputs.
-- **17.71** Use *PSpice* or *MultiSim* to solve Prob. 17.40.
-
-# Section 17.8 Applications
-
-**17.72** The signal displayed by a medical device can be approximated by the waveform shown in Fig. 17.86. Find the Fourier series representation of the signal.
-
-(a)
-
-<span id="page-831-0"></span>
-
-# **Figure 17.86**
-
-For Prob. 17.72.
-
-- **17.73** A spectrum analyzer indicates that a signal is made up of three components only: 640 kHz at 2 V, 644 kHz at 1 V, 636 kHz at 1 V. If the signal is applied across a 10-Ω resistor, what is the average power absorbed by the resistor?
-- **17.74** A certain band-limited periodic current has only three frequencies in its Fourier series representation: dc, 50 Hz, and 100 Hz. The current may be represented as
-
-$$
-i(t) = 4 + 6 \sin 100\pi t + 8 \cos 100\pi t
-$$
-  
-- 3 sin 200  $\pi t$  – 4 cos 200 $\pi t$  A
-
-# (a) Express i(*t*) in amplitude-phase form.
-
-- (b) If i(*t*) flows through a 2-Ω resistor, how many watts of average power will be dissipated?
-- **17.75** Design a low-pass *RC* filter with a resistance *R* = 2 kΩ. The input to the filter is a periodic rectangular pulse train (see Table 17.3) with *A* = 1 *V*, *T* = 10 ms, and *τ* = 1 ms. Select *C* such that the dc component of the output is 50 times greater than the fundamental component of the output.
-- **17.76** A periodic signal given by *vs*(*t*) = 10 V for 0 < *t* < 1 and 0 V for 1 < *t* < 2 is applied to the high-pass filter in Fig. 17.87. Determine the value of *R* such that the output signal *vo*(*t*) has an average power of at least 70 percent of the average power of the input signal.
-
-**Figure 17.87** For Prob. 17.76.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/207_Comprehensive Problems.md

@@ -1,45 +0,0 @@
-# Comprehensive Problems
-
-**17.77** The voltage across a device is given by
-
-$$
-v(t) = -2 + 10 \cos 4t + 8 \cos 6t + 6 \cos 8t
-$$
-  
-- 5 sin 4t - 3 sin 6t - sin 8t V
-
-Find:
-
-- (a) the period of *v*(*t*),
-- (b) the average value of *v*(*t*),
-- (c) the effective value of *v*(*t*).
-- **17.78** A certain band-limited periodic voltage has only three harmonics in its Fourier series representation. The harmonics have the following rms values: fundamental 40 V, third harmonic 20 V, fifth harmonic 10 V.
-  - (a) If the voltage is applied across a 5-Ω resistor, find the average power dissipated by the resistor.
-  - (b) If a dc component is added to the periodic voltage and the measured power dissipated increases by 5 percent, determine the value of the dc component added.
-- **17.79** Write a program to compute the Fourier coefficients (up to the 10th harmonic) of the square wave in Table 17.3 with *A* = 10 and *T* = 2.
-- **17.80** Write a computer program to calculate the exponential Fourier series of the half-wave rectified sinusoidal
-
-current of Fig. 17.82. Consider terms up to the 10th harmonic.
-
-- **17.81** Consider the full-wave rectified sinusoidal current in Table 17.3. Assume that the current is passed through a 1-Ω resistor.
-  - (a) Find the average power absorbed by the resistor.
-  - (b) Obtain *cn* for *n* = 1, 2, 3, and 4.
-  - (c) What fraction of the total power is carried by the dc component?
-  - (d) What fraction of the total power is carried by the second harmonic (*n* = 2)?
-- **17.82** A band-limited voltage signal is found to have the complex Fourier coefficients presented in the table below. Calculate the average power that the signal would supply a 4-Ω resistor.
-
-| nω0 | ∣cn∣ | θn  |
-|-----|------|-----|
-| 0   | 10.0 | 0°  |
-| ω   | 8.5  | 15° |
-| 2ω  | 4.2  | 30° |
-| 3ω  | 2.1  | 45° |
-| 4ω  | 0.5  | 60° |
-| 5ω  | 0.2  | 75° |
-|     |      |     |
-
-*This page intentionally left blank* 
-
-# **chapter**
-
-18

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/208_Chapter 18 - Fourier Transform.md

@@ -1,29 +0,0 @@
-# <span id="page-833-0"></span>Fourier Transform
-
-*Planning is doing today to mak e us better tomorrow because the future belongs to those who make the hard decisions today.*
-
-—*BusinessWeek*
-
-# Enhancing Your Skills and Your Career
-
-# **Career in Communications Systems**
-
-Communications systems apply the principles of circuit analysis. A communication system is designed to convey information from a source (the transmitter) to a destination (the receiver) via a channel (the propagation medium). Communications engineers design systems for transmitting and receiving information. The information can be in the form of voice, data, or video.
-
-We live in the information age—ne ws, weather, sports, shopping, financial, business inventory, and other sources make information available to us almost instantly via communications systems. Some ob vious examples of communications systems are the telephone network, mobile cellular telephones, radio, cable TV, satellite TV, fax, and radar. Mobile radio, used by police and fire departments, aircraft, and various businesses is another example.
-
-The field of communications is perhaps the fastest growing area in electrical engineering. The merging of the communications field with computer technology in recent years has led to digital data communications networks such as local area netw orks, metropolitan area netw orks, and broadband integrated services digital networks. For example, the Internet (the "information superhighway") allows educators, business people, and others to send electronic mail from their computers w orldwide, log onto remote databases, and transfer files. The Internet has hit t he world like a tidal wave and is drastically changing the w ay people do b usiness, communicate, and get information. This trend will continue.
-
-A communications engineer designs systems that provide high-quality information services. The systems include hardw are for generating, transmitting, and receiving information signals. Communications engineers are employed in numerous communications industries and places where com munications systems are routinely used. More and more government agencies, academic departments, and businesses are demanding faster and more accurate transmission of information. To meet these needs, communications engineers are in high demand. Therefore, the future is in communications and every electrical engineer must prepare accordingly.
-
-Photo by Charles Alexander
-
-# <span id="page-834-0"></span>Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Define the Fourier transform and explain how to use it.
-- 2. Understand the properties of the Fourier transform.
-- 3. Know how to use the Fourier transform in the analysis of circuits.
-- 4. Understand Parseval's theorem.
-- 5. Understand the relationship between the Laplace transform and the Fourier transform.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/209_18.1 Introduction.md

@@ -1,11 +0,0 @@
-# **18.1** Introduction
-
-Fourier series enable us to represent a periodic function as a sum of sinusoids and to obtain the frequenc y spectrum from the series. The Fourier transform allows us to e xtend the concept of a frequenc y spectrum to nonperiodic functions. The transform assumes that a nonperiodic function is a periodic function with an infinite period. Thus, the Fourier transform is an inte gral representation of a nonperiodic function that is analogous to a Fourier series representation of a periodic function.
-
-The F ourier transform is an *integral tr ansform* lik e the Laplace transform. It transforms a function in the time domain into the frequency domain. The Fourier transform is very useful in communications systems and digital signal processing, in situations where the Laplace transform does not apply . While the Laplace transform can only handle circuits with inputs for *t* > 0 with initial conditions, the F ourier transform can handle circuits with inputs for *t* < 0 as well as those for *t* > 0.
-
-We begin by using a Fourier series as a stepping stone in defining the Fourier transform. Then we de velop some of the properties of the Fourier transform. Next, we apply the Fourier transform in analyzing circuits. We discuss Parseval's theorem, compare the Laplace and F ourier transforms, and see ho w the F ourier transform is applied in amplitude modulation and sampling.
-
-**Figure 18.1**
-
-(a) A nonperiodic function, (b) increasing *T* to infinity makes *f*(*t*) become the nonperiodic function in (a).

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/210_18.2 Definition of the Fourier Transform.md

@@ -1,745 +0,0 @@
-# **18.2** Definition of the Fourier Transform
-
-We saw in the previous chapter that a nonsinusoidal periodic function can be represented by a Fourier series, provided that it satisfies the Dirichlet conditions. What happens if a function is not periodic? Unfortunately , there are many important nonperiodic functions—such as a unit step or an exponential function—that we cannot represent by a Fourier series. As we shall see, the Fourier transform allows a transformation from the time to the frequency domain, even if the function is not periodic.
-
-Suppose we w ant to find the Fourier transform of a nonperiodic function *p*(*t*), shown in Fig. 18.1(a). We consider a periodic function *f*(*t*) whose shape over one period is the same as *p*(*t*), as shown in Fig. 18.1(b). If we let the period *T* → ∞, only a single pulse of width *τ* [the desired nonperiodic function in Fig. 18.1(a)] remains, because the adjacent
-
-Effect of increasing *T* on the spectrum of the periodic pulse trains in Fig. 18.1(b) using the appropriately modified Eq. (17.66).
-
-pulses have been moved to infinity. Thus, the function *f*(*t*) is no longer periodic. In other words, *f*(*t*) = *p*(*t*) as *T* → ∞. It is interesting to consider the spectrum of *f*(*t*) for *A* = 10 and *τ* = 0.2 (see Section 17.6). Figure 18.2 shows the effect of increasing *T* on the spectrum. First, we notice that the general shape of the spectrum remains the same, and the frequenc y at which the envelope first becomes zero remains the same. However, the amplitude of the spectrum and the spacing between adjacent components both decrease, while the number of harmonics increases. Thus, over a range of frequencies, the sum of the amplitudes of the harmonics remains almost constant. As the total "strength" or ener gy of the components within a band must remain unchanged, the amplitudes of the harmonics must decrease as *T* increases. Because *f* = 1∕*T*, as *T* increases, *f* or *ω* decreases, so that the discrete spectrum ultimately becomes continuous.
-
-To further understand this connection between a nonperiodic function and its periodic counterpart, consider the e xponential form of a Fourier series in Eq. (17.58), namely,
-
-$$
-f(t) = \sum_{n = -\infty}^{\infty} c_n e^{jn\omega_0 t}
-$$
- (18.1)
-
-where
-
-$$
-c_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t} dt
-$$
- (18.2)
-
-The fundamental frequency is
-
-$$
-\omega_0 = \frac{2\pi}{T} \tag{18.3}
-$$
-
-and the spacing between adjacent harmonics is
-
-$$
-\Delta \omega = (n+1)\omega_0 - n\omega_0 = \omega_0 = \frac{2\pi}{T}
-$$
- (18.4)
-
-Substituting Eq. (18.2) into Eq. (18.1) gives
-
-$$
-f(t) = \sum_{n=-\infty}^{\infty} \left[ \frac{1}{T} \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t} dt \right] e^{jn\omega_0 t}
-$$
-  
-\n
-$$
-= \sum_{n=-\infty}^{\infty} \left[ \frac{\Delta \omega}{2\pi} \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t} dt \right] e^{jn\omega_0 t}
-$$
-  
-\n
-$$
-= \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} \left[ \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t} dt \right] \Delta \omega e^{jn\omega_0 t}
-$$
- (18.5)
-
-If we let *T* → ∞, the summation becomes inte gration, the incremental spacing ∆*ω* becomes the dif ferential separation *dω*, and the discrete harmonic frequency *nω*0 becomes a continuous frequenc y *ω*. Thus, as *T* → ∞,
-
-$$
-\sum_{n=-\infty}^{\infty} \Rightarrow \int_{-\infty}^{\infty}
-$$
-  
-\n
-$$
-\Delta \omega \Rightarrow d\omega \qquad (18.6)
-$$
-  
-\n
-$$
-n\omega_0 \Rightarrow \omega
-$$
-
-so that Eq. (18.5) becomes
-
-$$
-f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \left[ \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt \right] e^{j\omega t} d\omega \tag{18.7}
-$$
-
-Some authors use F(j*ω*) instead of F(*ω*) to represent the Fourier transform.
-
-The term in the brackets is known as the *Fourier transform* of *f*(*t*) and is represented by *F*(*ω*). Thus,
-
-$$
-F(\omega) = \mathcal{F}[f(t)] = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt
-$$
- (18.8)
-
-where is the Fourier transform operator. It is evident from Eq. (18.8) that:
-
-The Fourier transform is an integral transformation of f (t) from the time domain to the frequency domain.
-
-In general, *F*(*ω*) is a complex function; its magnitude is called the *amplitude spectrum*, while its phase is called the *phase spectrum*. Thus, *F*(*ω*) is the *spectrum*.
-
-Equation (18.7) can be written in terms of *F*(*ω*), and we obtain the *inverse Fourier transform* as
-
-$$
-f(t) = \mathcal{F}^{-1}[F(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega
-$$
- (18.9)
-
-The function *f*(*t*) and its transform *F*(*ω*) form the Fourier transform pairs:
-
-$$
-f(t) \qquad \Leftrightarrow \qquad F(\omega) \tag{18.10}
-$$
-
-given that one can be derived from the other.
-
-The F ourier transform *F*(*ω*) e xists when the F ourier inte gral in Eq. (18.8) converges. A sufficient but not necessary condition that *f* (*t*) has a Fourier transform is that it be completely integrable in the sense that
-
-$$
-\int_{-\infty}^{\infty} |f(t)| \, dt < \infty \tag{18.11}
-$$
-
-For example, the Fourier transform of the unit ramp function *tu*(*t*) does not exist, because the function does not satisfy the condition above.
-
-To avoid the complex algebra that explicitly appears in the F ourier transform, it is sometimes expedient to temporarily replace *jω* with *s* and then replace *s* with *jω* at the end.
-
-Find the F ourier transform of the follo wing functions: (a) *δ*(*t* − *t*0), Example 18.1 (b) *e j<sup>ω</sup>*0*<sup>t</sup>* , (c) cos *ω*<sup>0</sup> *t*.
-
-# **Solution:**
-
-(a) For the impulse function,
-
-$$
-F(\omega) = \mathcal{F}[\delta(t - t_0)] = \int_{-\infty}^{\infty} \delta(t - t_0) e^{-j\omega t} dt = e^{-j\omega t_0} \qquad (18.1.1)
-$$
-
-where the sifting property of the impulse function in Eq. (7.32) has been applied. For the special case *t*0 = 0, we obtain
-
-$$
-\mathcal{F}[\delta(t)] = 1 \tag{18.1.2}
-$$
-
-This shows that the magnitude of the spectrum of the impulse function is constant; that is, all frequencies are equally represented in the impulse function.
-
-(b) We can find the Fourier transform of *e j<sup>ω</sup>*0*<sup>t</sup>* in two ways. If we let
-
-$$
-F(\omega) = \delta(\omega - \omega_0)
-$$
-
-then we can find *f*(*t*) using Eq. (18.9), writing
-
-$$
-f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega
-$$
-
-Using the sifting property of the impulse function gives
-
-$$
-f(t) = \frac{1}{2\pi} e^{j\omega_0 t}
-$$
-
-Inasmuch as *F*(*ω*) and *f* (*t*) constitute a F ourier transform pair , so too must 2*πδ*(*ω* − *ω*0) and *e j<sup>ω</sup>*0*<sup>t</sup>* ,
-
-$$
-\mathcal{F}[e^{j\omega_0 t}] = 2\pi\delta(\omega - \omega_0)
-$$
- (18.1.3)
-
-Alternatively, from Eq. (18.1.2),
-
-$$
-\delta(t) = \mathcal{F}^{-1}[1]
-$$
-
-Using the inverse Fourier transform formula in Eq. (18.9),
-
-$$
-\delta(t) = \mathcal{F}^{-1}[1] = \frac{1}{2\pi} \int_{-\infty}^{\infty} 1 e^{j\omega t} d\omega
-$$
-
-or
-
-$$
-\int_{-\infty}^{\infty} e^{j\omega t} d\omega = 2 \pi \delta(t)
-$$
- (18.1.4)
-
-Interchanging variables *t* and *ω* results in
-
-$$
-\int_{-\infty}^{\infty} e^{j\omega t} dt = 2 \pi \delta(\omega)
-$$
- (18.1.5)
-
-Using this result, the Fourier transform of the given function is
-
-$$
-\mathcal{F}[e^{j\omega_0 t}] = \int_{-\infty}^{\infty} e^{j\omega_0 t} e^{-j\omega t} dt = \int_{-\infty}^{\infty} e^{j(\omega_0 - \omega)} dt = 2 \pi \delta(\omega_0 - \omega)
-$$
-
-Because the impulse function is an e ven function, with *δ*(*ω*<sup>0</sup> − *ω*) = *δ*(*ω* − *ω*0),
-
-$$
-\mathcal{F}[e^{j\omega_0 t}] = 2\pi \delta(\omega - \omega_0)
-$$
- (18.1.6)
-
-By simply changing the sign of *ω*0, we readily obtain
-
-$$
-\mathcal{F}[e^{-j\omega_0 t}] = 2\pi \delta(\omega + \omega_0)
-$$
- (18.1.7)
-
-Also, by setting *ω*0 = 0,
-
-$$
-\mathcal{F}[1] = 2\pi \delta(\omega) \tag{18.1.8}
-$$
-
-(c) By using the result in Eqs. (18.1.6) and (18.1.7), we get
-
-$$
-\mathcal{F}[\cos \omega_0 t] = \mathcal{F} \left[ \frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2} \right]
-$$
-  
-=  $\frac{1}{2} \mathcal{F} [e^{j\omega_0 t}] + \frac{1}{2} \mathcal{F} [e^{-j\omega_0 t}]$  (18.1.9)  
-=  $\pi \delta(\omega - \omega_0) + \pi \delta(\omega + \omega_0)$ 
-
-The Fourier transform of the cosine signal is shown in Fig. 18.3.
-
-Fourier transform of *f*(*t*) = cos *ω*0*t*.
-
-Practice Problem 18.1 Determine the Fourier transforms of the follo wing functions: (a) g ate function *g*(*t*) = 10*u*(*t* − 1) − 10*u*(*t* − 2), (b) 12*δ*(*t* − 2), (c) 15 sin *ω*0*t*.
-
-> **Answer:** (a) 10(*e*<sup>−</sup>*j<sup>ω</sup>* − *e*<sup>−</sup>*j*2*<sup>ω</sup>*)∕*jω*, (b) 12*ej*<sup>2</sup>*<sup>ω</sup>*, (c) *j*15*π*[*δ*(*ω* + *ω*0) − *δ*(*ω* − *ω*0)].
-
-Derive the Fourier transform of a single rectangular pulse of width *τ* and Example 18.2 height *A*, shown in Fig. 18.4.
-
-# **Solution:**
-
-$$
-F(\omega) = \int_{-\tau/2}^{\tau/2} Ae^{-j\omega t} dt = -\frac{A}{j\omega} e^{-j\omega t} \Big|_{-\tau/2}^{\tau/2}
-$$
-$$
-= \frac{2A}{\omega} \left( \frac{e^{j\omega \tau/2} - e^{-j\omega \tau/2}}{2j} \right)
-$$
-$$
-= A\tau \frac{\sin \omega \tau/2}{\omega \tau/2} = A\tau \text{ sinc } \frac{\omega \tau}{2}
-$$
-
-If we make *A* = 10 and *τ* = 2 as in Fig. 17.27 (like in Section 17.6), then
-
-*F*(*ω*) = 20 sinc *ω*
-
-whose amplitude spectrum is shown in Fig. 18.5. Comparing Fig. 18.4 with the frequency spectrum of the rectangular pulses in Fig. 17.28, we notice that the spectrum in Fig. 17.28 is discrete and its envelope has the same shape as the Fourier transform of a single rectangular pulse.
-
-# **Figure 18.4**
-
-# **Figure 18.5**
-
-Amplitude spectrum of the rectangular pulse in Fig. 18.4: for Example 18.2.
-
-# 0 t 25 ‒1 f(t) 1 Obtain the Fourier transform of the function in Fig. 18.6. Practice Problem 18.2 **Answer:** 50(cos *ω*− 1) \_\_\_\_\_\_\_\_\_\_\_\_ *<sup>j</sup>ω*.
-
-**Figure 18.6** For Practice Prob. 18.2.
-
-‒25
-
-Obtain the Fourier transform of the "switched-on" e xponential function Example 18.3 shown in Fig. 18.7.
-
-0,
-
-*t* > 0 *t* < 0
-
-# **Solution:**
-
-From Fig. 18.7,
-
-Hence,
-
-$$
-F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt = \int_{0}^{\infty} e^{-at} e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(a+j\omega)t} dt
-$$
-$$
-= \frac{-1}{a+j\omega} e^{-(a+j\omega)t} \Big|_{0}^{\infty} = \frac{1}{a+j\omega}
-$$
-
-*<sup>f</sup>*(*t*) = *e*<sup>−</sup>*atu*(*t*) = {*e*<sup>−</sup>*at*,
-
-For Example 18.3.
-
-<span id="page-840-0"></span>Practice Problem 18.3 Determine the Fourier transform of the "switched-off" exponential function in Fig. 18.8.
-
-Answer: 
-$$
-\frac{37.5}{a - j\omega}
-$$
-.
-
-# **18.3** Properties of the Fourier Transform
-
-We now develop some properties of the Fourier transform that are useful in finding the transforms of complicated functions from the transforms of simple functions. F or each property, we will first state and derive it, and then illustrate it with some examples.
-
-# **Linearity**
-
-If *F*1(*ω*) and *F*2(*ω*) are the F ourier transforms of *f*1(*t*) and *f*2(*t*), respectively, then
-
-$$
-\mathcal{F}[a_1 f_1(t) + a_2 f_2(t)] = a_1 F_1(\omega) + a_2 F_2(\omega)
-$$
- (18.12)
-
-where *a*1 and *a*2 are constants. This property simply states that the Fourier transform of a linear combination of functions is the same as the linear combination of the transforms of the indi vidual functions. The proof of the linearity property in Eq. (18.12) is straightforward. By definition,
-
-$$
-\mathcal{F}[a_1 f_1(t) + a_2 f_2(t)] = \int_{-\infty}^{\infty} [a_1 f_1(t) + a_2 f_2(t)] e^{-j\omega t} dt
-$$
-  
-= 
-$$
-\int_{-\infty}^{\infty} a_1 f_1(t) e^{-j\omega t} dt + \int_{-\infty}^{\infty} a_2 f_2(t) e^{-j\omega t} dt
-$$
-  
-= 
-$$
-a_1 F_1(\omega) + a_2 F_2(\omega)
-$$
- (18.13)
-
-For e xample, sin *ω*0*t* = \_\_1 2*j* (*e<sup>j</sup>ω*0*<sup>t</sup>* − *e*<sup>−</sup>*jω*0*<sup>t</sup>* ). Using the linearity property,
-
-$$
-F[\sin \omega_0 t] = \frac{1}{2j} [\mathcal{F}(e^{j\omega_0 t}) - \mathcal{F}(e^{-j\omega_0 t})]
-$$
-  
-$$
-= \frac{\pi}{j} [\delta(\omega - \omega_0) - \delta(\omega + \omega_0)]
-$$
-(18.14)  
-$$
-= j\pi [\delta(\omega + \omega_0) - \delta(\omega - \omega_0)
-$$
-
-# **Time Scaling**
-
-If *F*(*ω*) = [ *f*(*t*)], then
-
-$$
-\mathcal{F}[f(at)] = \frac{1}{|a|} F(\frac{\omega}{a})
-$$
- (18.15)
-
-where *a* is a constant. Equation (18.15) sho ws that time e xpansion (∣*a*∣ > 1) corresponds to frequency compression, or conversely, time compression (∣*a*∣ < 1) implies frequenc y expansion. The proof of the time-scaling property proceeds as follows.
-
-$$
-\mathcal{F}[f(at)] = \int_{-\infty}^{\infty} f(at)e^{-j\omega t} dt
-$$
- (18.16)
-
-If we let *x* = *at*, so that *dx* = *a dt*, then
-
-$$
-\mathcal{F}[f(at)] = \int_{-\infty}^{\infty} f(x)e^{-j\omega x/a} \frac{dx}{a} = \frac{1}{a}F\left(\frac{\omega}{a}\right)
-$$
- (18.17)
-
-For example, for the rectangular pulse *p*(*t*) in Example 18.2,
-
-$$
-\mathcal{F}[p(t)] = A\tau \operatorname{sinc} \frac{\omega \tau}{2}
-$$
- (18.18a)
-
-Using Eq. (18.15),
-
-$$
-\mathcal{F}[p(2t)] = \frac{A\tau}{2}\operatorname{sinc}\frac{\omega\tau}{4}
-$$
- (18.18b)
-
-It may be helpful to plot *p*(*t*) and *p*(2*t*) and their F ourier transforms. Because
-
-$$
-p(t) = \begin{cases} A, & \frac{\tau}{2} < t < \frac{\tau}{2} \\ 0, & \text{otherwise} \end{cases} \tag{18.19a}
-$$
-
-then replacing every *t* with 2*t* gives
-
-$$
-p(2t) = \begin{cases} A, & -\frac{\tau}{2} < 2t < \frac{\tau}{2} \\ 0, & \text{otherwise} \end{cases} = \begin{cases} A, & -\frac{\tau}{4} < t < \frac{\tau}{4} \\ 0, & \text{otherwise} \end{cases}
-$$
- (18.19b)
-
-showing that *p*(2*t*) is time compressed, as shown in Fig. 18.9(b). To plot both Fourier transforms in Eq. (18.18), we recall that the sinc function has zeros when its argument is *nπ*, where *n* is an integer. Hence, for the transform of *p*(*t*) in Eq. (18.18a), *ωτ*∕2 = 2*πfτ*∕2 = *nπ* → *f* = *n*∕*τ*, and for the transform of *p*(2*t*) in Eq. (18.18b), *ωτ*∕4 = 2*π f τ*∕4 = *nπ* → *f* = 2*n*∕*τ*. The plots of the Fourier transforms are shown in Fig. 18.9, which shows that time compression corresponds with frequency expansion. We should expect this intuitively, because when the signal is squashed in time, we expect it to change more rapidly, thereby causing higher-frequency components to exist.
-
-# **Time Shifting**
-
-If *F*(*ω*) = [ *f*(*t*)], then
-
-$$
-\mathcal{F}[f(t-t_0)] = e^{-j\omega t_0} F(\omega)
-$$
- (18.20)
-
-that is, a delay in the time domain corresponds to a phase shift in the frequency domain. To derive the time shifting property, we note that
-
-$$
-\mathcal{F}[f(t-t_0)] = \int_{-\infty}^{\infty} f(t-t_0) e^{-j\omega t} dt
-$$
- (18.21)
-
-# **Figure 18.9**
-
-The effect of time scaling: (a) transform of the pulse, (b) time compression of the pulse causes frequency expansion.
-
-If we let *x* = *t* − *t*0 so that *dx* = *dt* and *t* = *x* + *t*0, then
-
-$$
-\mathcal{F}[f(t-t_0)] = \int_{-\infty}^{\infty} f(x)e^{-j\omega(x+t_0)} dx
-$$
-  
-=  $e^{-j\omega t_0} \int_{-\infty}^{\infty} f(x)e^{-j\omega x} dx = e^{-j\omega t_0} F(\omega)$  (18.22)
-
-Similarly, [ *f*(*t* + *t*0)] = *e jωt*<sup>0</sup> *F*(*ω*). For example, from Example 18.3,
-
-$$
-\mathcal{F}[e^{-at}u(t)] = \frac{1}{a + j\omega} \tag{18.23}
-$$
-
-The transform of *f*(*t*) = *e*<sup>−</sup>(*t*−2)*u*(*t* − 2) is
-
-$$
-F(\omega) = \mathcal{F}[e^{-(t-2)} u(t-2)] = \frac{e^{-j2\omega}}{1+j\omega}
-$$
- (18.24)
-
-# **Frequency Shifting (or Amplitude Modulation)**
-
-This property states that if *F*(*ω*) = [ *f*(*t*)], then
-
-$$
-\mathcal{F}[f(t)e^{j\omega_0 t}] = F(\omega - \omega_0)
-$$
- (18.25)
-
-meaning, a frequency shift in the frequency domain adds a phase shift to the time function. By definition,
-
-$$
-\mathcal{F}[f(t)e^{j\omega_0 t}] = \int_{-\infty}^{\infty} f(t)e^{j\omega_0 t} e^{-j\omega t} dt
-$$
-  
-= 
-$$
-\int_{-\infty}^{\infty} f(t)e^{-j(\omega - \omega_0)t} dt = F(\omega - \omega_0)
-$$
- (18.26)
-
-For e xample, cos *ω*0*t* = \_1 2 (*e j<sup>ω</sup>*0*<sup>t</sup>* + *e*<sup>−</sup>*jω*0*<sup>t</sup>* ). Using the property in Eq. (18.25),
-
-$$
-\mathcal{F}[f(t)\cos\omega_0 t] = \frac{1}{2}\mathcal{F}[f(t)e^{j\omega_0 t}] + \frac{1}{2}\mathcal{F}[f(t)e^{-j\omega_0 t}]
-$$
-  
-$$
-= \frac{1}{2}F(\omega - \omega_0) + \frac{1}{2}F(\omega + \omega_0)
-$$
- (18.27)
-
-This is an important result in modulation where frequenc y components of a signal are shifted. If, for e xample, the amplitude spectrum of *f*(*t*) is as shown in Fig. 18.10(a), then the amplitude spectrum of *f*(*t*)cos*ω*0*t* will be as shown in Fig. 18.10(b). We will elaborate on amplitude modulation in Section 18.7.1.
-
-Amplitude spectra of: (a) signal *f*(*t*), (b) modulated signal *f*(*t*)cos *ω*0*t*.
-
-# **Time Differentiation**
-
-Given that *F*(*ω*) = [ *f*(*t*)], then
-
-$$
-\mathcal{F}[f'(t)] = j\omega F(\omega)
-$$
- (18.28)
-
-In other words, the transform of the derivative of *f*(*t*) is obtained by multiplying the transform of *f*(*t*) by *jω*. By definition,
-
-$$
-f(t) = \mathcal{F}^{-1}[F(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t} d\omega
-$$
- (18.29)
-
-Taking the derivative of both sides with respect to *t* gives
-
-$$
-f'(t) = \frac{j\omega}{2\pi} \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega = j\omega \mathcal{F}^{-1}[F(\omega)]
-$$
-
-or
-
-$$
-F[f'(t)] = j\omega F(\omega)
-$$
- (18.30)
-
-Repeated applications of Eq. (18.30) give
-
-$$
-\mathcal{F}[f^{(n)}(t)] = (j\omega)^n F(\omega)
-$$
- (18.31)
-
-For example, if *f*(*t*) = *e*<sup>−</sup>*atu*(*t*), then
-
-$$
-f'(t) = -ae^{-at} u(t) + e^{-at} \delta(t) = -af(t) + e^{-at} \delta(t)
-$$
- (18.32)
-
-Taking the Fourier transforms of the first and last terms, we obtain
-
-$$
-j\omega F(\omega) = -aF(\omega) + 1
-$$
-  $\Rightarrow$   $F(\omega) = \frac{1}{a + j\omega}$  (18.33)
-
-which agrees with the result in Example 18.3.
-
-# **Time Integration**
-
-Given that *F*(*ω*) = [ *f*(*t*)], then
-
-$$
-\mathcal{F}\left[\int_{-\infty}^{t} f(\tau) d\tau\right] = \frac{F(\omega)}{j\omega} + \pi F(0)\delta(\omega)
-$$
- (18.34)
-
-that is, the transform of the inte gral of *f*(*t*) is obtained by di viding the transform of *f*(*t*) by *jω* and adding the result to the impulse term that reflects the dc component *F*(0). Someone might ask, "How do we know that when we take the Fourier transform for time integration, we should integrate over the interval [−∞, *t*] and not [−∞, ∞]?" When we integrate over [−∞, ∞], the result does not depend on time an ymore, and the Fourier transform of a constant is what we will eventually get. But when we integrate over [−∞, *t*], we get the inte gral of the function from the past to time *t*, so that the result depends on *t* and we can take the Fourier transform of that.
-
-If *ω* is replaced by 0 in Eq. (18.8),
-
-$$
-F(0) = \int_{-\infty}^{\infty} f(t) dt
-$$
- (18.35)
-
-indicating that the dc component is zero when the integral of *f*(*t*) over all time vanishes. The proof of the time inte gration in Eq. (18.34) will be given later when we consider the convolution property.
-
-For example, we know that [*δ*(*t*)] = 1 and that integrating the impulse function gives the unit step function [see Eq. (7.39a)]. By applying the property in Eq. (18.34), we obtain the F ourier transform of the unit step function as
-
-$$
-\mathcal{F}[u(t)] = \mathcal{F}\left[\int_{-\infty}^{t} \delta(\tau) d\tau\right] = \frac{1}{j\omega} + \pi \delta(\omega)
-$$
- (18.36)
-
-# **Reversal**
-
-If *F*(*ω*) = [ *f*(*t*)], then
-
-$$
-\mathcal{F}[f(-t)] = F(-\omega) = F^*(\omega)
-$$
- (18.37)
-
-where the asterisk denotes the comple x conjugate. This property states that reversing *f*(*t*) about the time axis reverses *F*(*ω*) about the frequency axis. This may be regarded as a special case of time scaling for which *a* = −1 in Eq. (18.15).
-
-For example, 1 = *u*(*t*) + *u*(−*t*). Hence,
-
-$$
-F[1] = F[u(t)] + F[u(-t)]
-$$
-$$
-= \frac{1}{j\omega} + \pi\delta(\omega)
-$$
-$$
-- \frac{1}{j\omega} + \pi\delta(-\omega)
-$$
-$$
-= 2\pi\delta(\omega)
-$$
-
-# **Duality**
-
-This property states that if *F*(*ω*) is the Fourier transform of *f*(*t*), then the Fourier transform of *F*(*t*) is 2*πf*(−*ω*); we write
-
-$$
-\mathcal{F}[f(t)] = F(\omega) \qquad \Rightarrow \qquad \mathcal{F}[F(t)] = 2\pi f(-\omega) \qquad (18.38)
-$$
-
-This expresses the symmetry property of the Fourier transform. To derive this property, we recall that
-
-$$
-f(t) = \mathcal{F}^{-1}[F(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega
-$$
-
-or
-
-$$
-2\pi f(t) = \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega
-$$
- (18.39)
-
-Replacing *t* by −*t* gives
-
-$$
-2\pi f(-t) = \int_{-\infty}^{\infty} F(\omega)e^{-j\omega t} d\omega
-$$
-
-If we interchange *t* and *ω*, we obtain
-
-$$
-2\pi f(-\omega) = \int_{-\infty}^{\infty} F(t)e^{-j\omega t} dt = \mathcal{F}[F(t)]
-$$
- (18.40)
-
-as expected.
-
-For example, if *f*(*t*) = *e* −∣*t*<sup>∣</sup> , then
-
-$$
-F(\omega) = \frac{2}{\omega^2 + 1}
-$$
- (18.41)
-
-By the duality property, the Fourier transform of *F*(*t*) = 2∕(*t* 2 + 1) is
-
-$$
-2\pi f(\omega) = 2\pi e^{-|\omega|} \tag{18.42}
-$$
-
-Figure 18.11 sho ws another e xample of the duality property . It illus trates the f act that if *f*(*t*) = *δ*(*t*) so that *F*(*ω*) = 1, as in Fig. 18.11(a), then the Fourier transform of *F*(*t*) = 1 is 2*πf*(*ω*) = 2*πδ*(*ω*) as shown in Fig. 18.11(b).
-
-# **Convolution**
-
-Recall from Chapter 15 that if *x*(*t*) is the input excitation to a circuit with an impulse function of *h*(*t*), then the output response *y*(*t*) is given by the convolution integral
-
-$$
-y(t) = h(t) * x(t) = \int_{-\infty}^{\infty} h(\lambda) x(t - \lambda) d\lambda
-$$
- (18.43)
-
-Because f(t) is the sum of the signals in Figs. 18.7 and 18.8, F(*ω*) is the sum of the results in Example 18.3 and Practice Prob. 18.3.
-
-A typical illustration of the duality property of the Fourier transform: (a) transform of impulse, (b) transform of unit dc level.
-
-> If *X*(*ω*), *H*(*ω*), and *Y*(*ω*) are the Fourier transforms of *x*(*t*), *h*(*t*), and *y*(*t*), respectively, then
-
-$$
-Y(\omega) = \mathcal{F}[h(t) * x(t)] = H(\omega)X(\omega)
-$$
- (18.44)
-
-which indicates that con volution in the time domain corresponds with multiplication in the frequency domain.
-
-To derive the convolution property, we take the Fourier transform of both sides of Eq. (18.43) to get
-
-$$
-Y(\omega) = \int_{-\infty}^{\infty} \left[ \int_{-\infty}^{\infty} h(\lambda) x(t - \lambda) \, d\lambda \right] e^{-j\omega t} \, dt \tag{18.45}
-$$
-
-Exchanging the order of inte gration and factoring *h*(λ), which does not depend on *t*, we have
-
-$$
-Y(\omega) = \int_{-\infty}^{\infty} h(\lambda) \left[ \int_{-\infty}^{\infty} x(t - \lambda) e^{-j\omega t} dt \right] d\lambda
-$$
-
-For the integral within the brack ets, let *τ* = *t* − λ so that *t* = *τ* + λ and *dt* = *dτ*. Then,
-
-$$
-Y(\omega) = \int_{-\infty}^{\infty} h(\lambda) \left[ \int_{-\infty}^{\infty} x(\tau) e^{-j\omega(\tau+\lambda)} d\tau \right] d\lambda
-$$
-  
-= 
-$$
-\int_{-\infty}^{\infty} h(\lambda) e^{-j\omega\lambda} d\lambda \int_{-\infty}^{\infty} x(\tau) e^{-j\omega\tau} d\tau = H(\omega)X(\omega)
-$$
-(18.46)
-
-as expected. This result e xpands the phasor method be yond what w as done with the Fourier series in the previous chapter.
-
-To illustrate the con volution property , suppose both *h*(*t*) and *x*(*t*) are identical rectangular pulses, as sho wn in Fig. 18.12(a) and 18.12(b). We recall from Example 18.2 and Fig. 18.5 that the Fourier transforms of the rectangular pulses are sinc functions, as sho wn in Fig. 18.12(c) and 18.12(d). According to the con volution property, the product of the sinc functions should gi ve us the con volution of the rectangular pulses in the time domain. Thus, the con volution of the pulses in Fig. 18.12(e) and the product of the sinc functions in Fig. 18.12(f) form a Fourier pair.
-
-In view of the duality property, we expect that if convolution in the time domain corresponds with multiplication in the frequenc y domain,
-
-The important relationship in Eq. (18.46) is the key reason for using the Fourier transform in the analysis of linear systems.
-
-then multiplication in the time domain should ha ve a correspondence in the frequency domain. This happens to be the case. If *f*(*t*) = *f*1(*t*) *f*2(*t*), then
-
-$$
-F(\omega) = \mathcal{F}[f_1(t)f_2(t)] = \frac{1}{2\pi}F_1(\omega) * F_2(\omega)
-$$
- (18.47)
-
-or
-
-$$
-F(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F_1(\lambda) F_2(\omega - \lambda) d\lambda
-$$
- (18.48)
-
-which is convolution in the frequency domain. The proof of Eq. (18.48) readily follows from the duality property in Eq. (18.38).
-
-Let us no w deri ve the time inte gration property in Eq. (18.34). If we replace *x*(*t*) with the unit step function *u*(*t*) and *h*(*t*) with *f*(*t*) in Eq. (18.43), then
-
-$$
-\int_{-\infty}^{\infty} f(\lambda)u(t-\lambda) \, d\lambda = f(t) * u(t) \tag{18.49}
-$$
-
-But by the definition of the unit step function,
-
-∫
-
-$$
-u(t - \lambda) = \begin{cases} 1, & t - \lambda > 0 \\ 0, & t - \lambda > 0 \end{cases}
-$$
-
-We can write this as
-
-$$
-u(t - \lambda) = \begin{cases} 1, & \lambda < t \\ 0, & \lambda > t \end{cases}
-$$
-
-Substituting this into Eq. (18.49) makes the interval of integration change from [−∞, ∞] to [−∞, *t*], and thus Eq. (18.49) becomes
-
-$$
-\int_{-\infty}^{t} f(\lambda) \, d\lambda = u(t) * f(t)
-$$
-
-Taking the Fourier transform of both sides yields
-
-$$
-\mathcal{F}\left[\int_{-\infty}^{t} f(\lambda) d\lambda\right] = U(\omega)F(\omega)
-$$
-\n(18.50)
-
-But from Eq. (18.36), the Fourier transform of the unit step function is
-
-$$
-U(\omega) = \frac{1}{j\omega} + \pi \delta(\omega)
-$$
-
-Substituting this into Eq. (18.50) gives
-
-$$
-\mathcal{F}\left[\int_{-\infty}^{t} f(\lambda) d\lambda\right] = \left(\frac{1}{j\omega} + \pi \delta(\omega)\right) F(\omega)
-$$
-\n
-$$
-= \frac{F(\omega)}{j\omega} + \pi F(0) \delta(\omega)
-$$
-\n(18.51)
-
-which is the time inte gration property of Eq. (18.34). Note that in Eq. (18.51), *F*(*ω*)*δ*(*ω*) = *F*(0)*δ*(*ω*), since *δ*(*ω*) is only nonzero at *ω* = 0.
-
-Table 18.1 lists these properties of the Fourier transform. Table 18.2 presents the transform pairs of some common functions. Note the simi larities between these tables and Tables 15.1 and 15.2.
-
-## **TABLE 18.1**

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/211_18.3 Properties of the Fourier Transform.md

@@ -1,1074 +0,0 @@
-# Properties of the Fourier transform.
-
-| Property        | f(t)                      | F(ω)                                |
-|-----------------|---------------------------|-------------------------------------|
-| Linearity       | a1<br>f1(t) + a2<br>f2(t) | a1F1(ω) + a2F2(ω)                   |
-| Scaling         | f(at)                     | ___1<br>ω<br>__<br>F(<br>a )<br>∣a∣ |
-| Time shift      | f(t − a)                  | e−jωaF(ω)                           |
-| Frequency shift | e jω0t<br>f(t)            | F(ω − ω0)                           |
-
-| TABLE 18.1                | (continued)             |                                     |
-|---------------------------|-------------------------|-------------------------------------|
-| Property                  | f(t)                    | F(ω)                                |
-| Modulation                | cos(ω0t)f(t)            | __1<br>[F(ω + ω0) + F(ω − ω0)]<br>2 |
-| Time differentiation      | df<br>__<br>dt          | jωF(ω)                              |
-|                           | d n<br>f ___<br>dtn     | n<br>( jω)<br>F(ω)                  |
-| Time integration          | t<br>f(t) dt<br>∫<br>−∞ | F(ω) _____<br>jω    + πF(0)δ(ω)     |
-| Frequency differentiation | n<br>t<br>f(t)          | d n ____<br>n<br>( j)<br>dωn F(ω)   |
-| Reversal                  | f(−t)                   | F(−ω)<br>or<br>F*(ω)                |
-| Duality                   | F(t)                    | 2πf(−ω)                             |
-| Convolution in t          | f1(t) *<br>f2(t)        | F1(ω)F2(ω)                          |
-| Convolution in ω          | f1(t)f2(t)              | ___1<br>F1(ω) *<br>F2(ω)<br>2π      |
-
-## **TABLE 18.2**
-
-# Fourier transform pairs.
-
-| f(t)                 | F(ω)                                                  |
-|----------------------|-------------------------------------------------------|
-| δ(t)                 | 1                                                     |
-| 1                    | 2π δ(ω)                                               |
-| u(t)                 | π δ(ω) + ___1<br>jω                                   |
-| u(t + τ) − u(t − τ)  | 2 ______ sin ωτ<br>ω                                  |
-| ∣t∣                  | −2<br>___<br>ω2                                       |
-| sgn(t)               | ___2<br>jω                                            |
-| e−at u(t)            | ______ 1<br>a + jω                                    |
-| eat u(−t)            | ______ 1<br>a − jω                                    |
-| n<br>e−at u(t)<br>t  | __________ n!<br>n+1<br>(a + jω)                      |
-| e−a∣t∣               | _______ 2a<br>a2<br>+ ω2                              |
-| ejω0t                | 2πδ(ω − ω0)                                           |
-| sin ω0t              | jπ[δ(ω + ω0) − δ(ω − ω0)]                             |
-| cos ω0t              | π[δ(ω + ω0) + δ(ω − ω0)]                              |
-| e−at sin ω0tu(t)     | ____________ ω0<br>2<br>2<br>(a + jω)<br>+ ω 0        |
-| e−at cos ω0<br>tu(t) | a + jω<br>____________<br>2<br>2<br>(a + jω)<br>+ ω 0 |
-
-**Figure 18.13** The signum function of Example 18.4.
-
-Example 18.4 Find the Fourier transforms of the following functions: (a) signum function sgn(*t*), shown in Fig. 18.13, (b) the double-sided e xponential *e*<sup>−</sup>*a*∣*t*<sup>∣</sup> , and (c) the sinc function (sin *t*)∕*t*.
-
-# **Solution:**
-
-(a) We can obtain the F ourier transform of the *signum* function in three ways.
-
-■ **METHOD 1** We can write the signum function in terms of the unit step function as
-
-$$
-sgn(t) = f(t) = u(t) - u(-t)
-$$
-
-But from Eq. (18.36),
-
-$$
-U(\omega) = \mathcal{F}[u(t)] = \pi \delta(\omega) + \frac{1}{j\omega}
-$$
-
-Applying this and the reversal property, we obtain
-
-$$
-F[\text{sgn}(t)] = U(\omega) - U(-\omega)
-$$
-$$
-= \left(\pi\delta(\omega) + \frac{1}{j\omega}\right) - \left(\pi\delta(-\omega) + \frac{1}{-j\omega}\right) = \frac{2}{j\omega}
-$$
-
-■ **METHOD 2** Because *δ*(*ω*) = *δ*(−*ω*), another w ay of writing the signum function in terms of the unit step function is
-
-$$
-f(t) = \text{sgn}(t) = -1 + 2u(t)
-$$
-
-Taking the Fourier transform of each term gives
-
-$$
-F(\omega) = -2\pi\delta(\omega) + 2\left(\pi\delta(\omega) + \frac{1}{j\omega}\right) = \frac{2}{j\omega}
-$$
-
-■ **METHOD 3** We can take the derivative of the signum function in Fig. 18.13 and obtain
-
-$$
-f'(t) = 2\delta(t)
-$$
-
-Taking the transform of this,
-
-$$
-j\omega F(\omega) = 2
-$$
-  $\Rightarrow$   $F(\omega) = \frac{2}{j\omega}$ 
-
-as obtained previously.
-
-(b) The double-sided exponential can be expressed as
-
-$$
-f(t) = e^{-a|t|} = e^{-at}u(t) + e^{at}u(-t) = y(t) + y(-t)
-$$
-
-where *y*(*t*) = *e* <sup>−</sup>*atu*(*t*) so that *Y*(*ω*) = 1∕(*a* + *jω*). Applying the re versal property,
-
-$$
-\mathcal{F}[e^{-a|t|}] = Y(\omega) + Y(-\omega) = \left(\frac{1}{a+j\omega} + \frac{1}{a-j\omega}\right) = \frac{2a}{a^2 + \omega^2}
-$$
-
-(c) From Example 18.2,
-
-$$
-\mathcal{F}\left[u\left(t+\frac{\tau}{2}\right)-u\left(t-\frac{\tau}{2}\right)\right]=\tau\frac{\sin(\omega\tau/2)}{\omega\tau/2}=\tau\,\text{sinc}\,\frac{\omega\tau}{2}
-$$
-
-Setting *τ*∕2 = 1 gives
-
-$$
-\mathcal{F}[u(t+1) - u(t-1)] = 2\frac{\sin \omega}{\omega}
-$$
-
-Applying the duality property yields
-
-$$
-\mathcal{F}\left[2\frac{\sin t}{t}\right] = 2\pi \left[U(\omega+1) - U(\omega-1)\right]
-$$
-
-or
-
-$$
-\mathcal{F}\left[\frac{\sin t}{t}\right] = \pi[U(\omega + 1) - U(\omega - 1)]
-$$
-
-Determine the F ourier transforms of these functions: (a) g ate function Practice Problem 18.4 *g*(*t*) = *u*(*t*) − *u*(*t* − 1), (b) *f*(*t*) = 10*t* −2*t u*(*t*), and (c) sawtooth pulse *p*(*t*) = 75*t*[*u*(*t*) − *u*(*t* − 2)].
-
-Answer: (a) 
-$$
-(1 - e^{-j\omega}) \left[ \pi \delta(\omega) + \frac{1}{j\omega} \right]
-$$
-, (b)  $\frac{10}{(2 + j\omega)^2}$ ,  
-(c)  $\frac{75(e^{-j2\omega} - 1)}{\omega^2} + \frac{150j}{\omega} e^{-j2\omega}$ .
-
-Find the Fourier transform of the function in Fig. 18.14. Example 18.5
-
-# **Solution:**
-
-The Fourier transform can be found directly using Eq. (18.8), b ut it is much easier to find it using the derivative property. We can express the function as
-
-$$
-f(t) = \begin{cases} 1+t, & -1 < t < 0 \\ 1-t, & 0 < t < 1 \end{cases}
-$$
-
-Its first derivative is shown in Fig. 18.15(a) and is given by
-
-**Figure 18.15** First and second derivatives of *f*(*t*) in Fig. 18.14; for Example 18.5.
-
-f(t) ‒1 0 1 t 1 **Figure 18.14**
-
-For Example 18.5.
-
-Its second derivative is in Fig. 18.15(b) and is given by
-
-$$
-f''(t) = \delta(t+1) - 2\delta(t) + \delta(t-1)
-$$
-
-Taking the Fourier transform of both sides,
-
-$$
-(j\omega)^2 F(\omega) = e^{j\omega} - 2 + e^{-j\omega} = -2 + 2 \cos \omega
-$$
-
-or
-
-$$
-F(\omega) = \frac{2(1 - \cos \omega)}{\omega^2}
-$$
-
-Example 18.6 Obtain the inverse Fourier transform of:
-
-Obtain the inverse Fourier transform of:  
-\n(a) 
-$$
-F(\omega) = \frac{10j\omega + 4}{(j\omega)^2 + 6j\omega + 8}
-$$
- (b)  $G(\omega) = \frac{\omega^2 + 21}{\omega^2 + 9}$ 
-
-# **Solution:**
-
-(a) To avoid complex algebra, we can replace *jω* with *s* for the moment. Using partial fraction expansion,
-
-$$
-F(s) = \frac{10s + 4}{s^2 + 6s + 8} = \frac{10s + 4}{(s + 4)(s + 2)} = \frac{A}{s + 4} + \frac{B}{s + 2}
-$$
-
-where
-
-$$
-A = (s + 4)F(s)|_{s=-4} = \frac{10s + 4}{(s + 2)}|_{s=-4} = \frac{-36}{-2} = 18
-$$
-$$
-B = (s + 2)F(s)|_{s=-2} = \frac{10s + 4}{(s + 4)}|_{s=-2} = \frac{-16}{2} = -8
-$$
-
-Substituting *A* = 18 and *B* = −8 in *F*(*s*) and *s* with *jω* gives
-
-$$
-F(j\omega) = \frac{18}{j\omega + 4} + \frac{-8}{j\omega + 2}
-$$
-
-With the aid of Table 18.2, we obtain the inverse transform as
-
-$$
-f(t) = (18e^{-4t} - 8e^{-2t})u(t)
-$$
-
-(b) We simplify *G*(*ω*) as
-
-$$
-G(\omega) = \frac{\omega^2 + 21}{\omega^2 + 9} = 1 + \frac{12}{\omega^2 + 9}
-$$
-
-<span id="page-853-0"></span>With the aid of Table 18.2, the inverse transform is obtained as
-
-$$
-g(t) = \delta(t) + 2e^{-3|t|}
-$$
-
-Find the inverse Fourier transform of:  
-\n(a) 
-$$
-H(\omega) = \frac{6(3 + j2\omega)}{(1 + j\omega)(4 + j\omega)(2 + j\omega)}
-$$
-  
-\n(b)  $Y(\omega) = \pi\delta(\omega) + \frac{1}{j\omega} + \frac{2(1 + j\omega)}{(1 + j\omega)^2 + 16}$   
-\n**Answer:** (a)  $h(t) = (2e^{-t} + 3e^{-2t} - 5e^{-4t}) u(t)$ ,  
-\n(b)  $y(t) = (1 + 2e^{-t} \cos 4t)u(t)$ .
-
-# **18.4** Circuit Applications
-
-The Fourier transform generalizes the phasor technique to nonperiodic functions. Therefore, we apply F ourier transforms to circuits with nonsinusoidal excitations in exactly the same way we apply phasor techniques to circuits with sinusoidal excitations. Thus, Ohm's law is still valid:
-
-$$
-V(\omega) = Z(\omega)I(\omega) \tag{18.52}
-$$
-
-where *V*(*ω*) and *I*(*ω*) are the F ourier transforms of the v oltage and current and *Z*(*ω*) is the impedance. We get the same e xpressions for the impedances of resistors, inductors, and capacitors as in phasor analysis, namely,
-
-$$
-\begin{array}{ccc}\nR & \Rightarrow & R \\
-L & \Rightarrow & j\omega L \\
-C & \Rightarrow & \frac{1}{j\omega C}\n\end{array}
-$$
-\n(18.53)
-
-Once we transform the functions for the circuit elements into the fre quency domain and tak e the F ourier transforms of the e xcitations, we can use circuit techniques such as v oltage division, source transforma tion, mesh analysis, node analysis, or Thevenin's theorem, to find the unknown response (current or v oltage). Finally , we tak e the in verse Fourier transform to obtain the response in the time domain.
-
-Although the F ourier transform method produces a response that exists for −∞ < *t* < ∞, F ourier analysis cannot handle circuits with initial conditions.
-
-The transfer function is ag ain defined as the ratio of the output response *Y*(*ω*) to the input excitation *X*(*ω*); that is,
-
-$$
-H(\omega) = \frac{Y(\omega)}{X(\omega)}\tag{18.54}
-$$
-
-Find the inverse Fourier transform of: Practice Problem 18.6
-
-$$
-Y(\omega) = H(\omega)X(\omega) \tag{18.55}
-$$
-
-X(*ω*) H(*ω*) Y(*ω*)
-
-# **Figure 18.17**
-
-The frequenc y domain input-output relationship is portrayed in Fig. 18.17. Equation (18.55) sho ws that if we kno w the transfer func tion and the input, we can readily find the output. The relationship in Eq. (18.54) is the principal reason for using the F ourier transform in circuit analysis. Notice that *H*(*ω*) is identical to *H*(*s*) with *s* = *jω*. Also, if the input is an impulse function [i.e., *x*(*t*) = *δ*(*t*)], then *X*(*ω*) = 1, so that the response is
-
-$$
-Y(\omega) = H(\omega) = \mathcal{F}[h(t)] \tag{18.56}
-$$
-
-indicating that *H*(*ω*) is the Fourier transform of the impulse response *h*(*t*).
-
-2 Ω *v*i (t) 1 F *v*o(t) + ‒ + ‒
-
-**Figure 18.18** For Example 18.7.
-
-# **Solution:**
-
-The Fourier transform of the input voltage is
-
-$$
-V_i(\omega) = \frac{2}{3 + j\omega}
-$$
-
-and the transfer function obtained by voltage division is
-
-$$
-H(\omega) = \frac{V_o(\omega)}{V_i(\omega)} = \frac{1/j\omega}{2 + 1/j\omega} = \frac{1}{1 + j2\omega}
-$$
-
-Hence,
-
-$$
-V_o(\omega) = V_i(\omega)H(\omega) = \frac{2}{(3 + j\omega)(1 + j2\omega)}
-$$
-
-or
-
-$$
-V_o(\omega) = \frac{1}{(3 + j\omega)(0.5 + j\omega)}
-$$
-
-By partial fractions,
-
-$$
-V_o(\omega) = \frac{-0.4}{3 + j\omega} + \frac{0.4}{0.5 + j\omega}
-$$
-
-Taking the inverse Fourier transform yields
-
-$$
-v_o(t) = 0.4(e^{-0.5t} - e^{-3t})u(t)
-$$
-
-1 H *v*i (t) 4 Ω *v*o(t) + ‒ + ‒ **Figure 18.19** For Practice Prob. 18.7.
-
-**Practice Problem 18.7** Determine 
-$$
-v_o(t)
-$$
- in Fig. 18.19 if  $v_i(t) = 5\text{sgn}(t) = (-5 + 10u(t))$  V.
-
-**Answer:** −5 + 10(1 − *e*<sup>−</sup>4*<sup>t</sup>* )*u*(*t*) V.
-
-Using the F ourier transform method, find *io*(*t*) in Fig. 18.20 when Example 18.8 *is*(*t*) = 10 sin 2*t* A.
-
-# **Solution:**
-
-By current division,
-
-$$
-H(\omega) = \frac{I_o(\omega)}{I_s(\omega)} = \frac{2}{2 + 4 + 2/j\omega} = \frac{j\omega}{1 + j\omega^2}
-$$
-
-If *is*(*t*) = 10 sin 2*t*, then
-
-$$
-I_s(\omega) = j\pi 10[\delta(\omega + 2) - \delta(\omega - 2)]
-$$
-
-Hence,
-
-$$
-I_o(\omega) = H(\omega)I_s(\omega) = \frac{10\pi\omega[\delta(\omega - 2) - \delta(\omega + 2)]}{1 + j\omega^2}
-$$
-
-The inverse Fourier transform of *Io*(*ω*) cannot be found using Table 18.2. We resort to the inverse Fourier transform formula in Eq. (18.9) and write
-
-respect to the inverse Fourier transform formula in Eq. (18.9) and  
-\n
-$$
-i_o(t) = \mathcal{F}^{-1}[I_o(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{10\pi\omega[\delta(\omega - 2) - \delta(\omega + 2)]}{1 + j\omega^2} e^{j\omega t} d\omega
-$$
-
-We apply the sifting property of the impulse function, namely,
-
-$$
-\delta(\omega - \omega_0) f(\omega) = f(\omega_0)
-$$
-
-or
-
-$$
-\int_{-\infty}^{\infty} \delta(\omega - \omega_0) f(\omega) \, d\omega = f(\omega_0)
-$$
-
-and obtain
-
-$$
-i_o(t) = \frac{10\pi}{2\pi} \left[ \frac{2}{1+j6} e^{j2t} - \frac{-2}{1-j6} e^{-j2t} \right]
-$$
-  
-=  $10 \left[ \frac{e^{j2t}}{6.082e^{j80.54^\circ}} + \frac{e^{-j2t}}{6.082e^{-j80.54^\circ}} \right]$   
-=  $1.644[e^{j(2t-80.54^\circ)} + e^{-j(2t-80.54^\circ)}]$   
-=  $3.288 \cos(2t - 80.54^\circ)$  A
-
-Find the current *io*(*t*) in the circuit in Fig. 18.21, gi ven that *is*(*t*) = Practice Problem 18.8 50 cos 4*t* A.
-
-**Answer:** 27.95 cos(4*t* + 26.57°) A.
-
-**Figure 18.21** For Practice Prob. 18.8.
-
-# <span id="page-856-0"></span>**18.5** Parseval's Theorem
-
-Parseval's theorem demonstrates one practical use of the F ourier transform. It relates the ener gy carried by a signal to the F ourier transform of the signal. If *p*(*t*) is the power associated with the signal, the energy carried by the signal is
-
-$$
-W = \int_{-\infty}^{\infty} p(t) dt
-$$
- (18.57)
-
-To be able to compare the energy content of current and voltage signals, it is convenient to use a 1-Ω resistor as the base for energy calculation. For a 1-Ω resistor, *p*(*t*) = *v*<sup>2</sup> (*t*) = *i* 2 (*t*) = *f* <sup>2</sup> (*t*), where *f*(*t*) stands for either voltage or current. The energy delivered to the 1-Ω resistor is
-
-$$
-W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) \, dt \tag{18.58}
-$$
-
-Parseval's theorem states that this same ener gy can be calculated in the frequency domain as
-
-$$
-W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega
-$$
- (18.59)
-
-Parseval's theorem states that the total energy delivered to a 1-Ω resistor equals the total area under the square of <sup>f</sup> (t) or 1∕2 *π* times the total area under the square of the magnitude of the Fourier transform of f (t).
-
-Parseval's theorem relates energy associated with a signal to its Fourier transform. It pro vides the ph ysical significance of *F*(*ω*), namely, that ∣*F*(*ω*)∣ 2 is a measure of the ener gy density (in joules per hertz) corre sponding to *f* (*t*).
-
-To deri ve Eq. (18.59), we be gin with Eq. (18.58) and substitute Eq. (18.9) for one of the *f*(*t*)'s. We obtain
-
-$$
-W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) dt = \int_{-\infty}^{\infty} f(t) \left[ \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t} d\omega \right] dt \qquad (18.60)
-$$
-
-The function *f*(*t*) can be mo ved inside the inte gral within the brack ets, since the integral does not involve time:
-
-$$
-W_{1\Omega} = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) F(\omega) e^{j\omega t} d\omega dt
-$$
- (18.61)
-
-Reversing the order of integration,
-
-$$
-W_{1\Omega} = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \left[ \int_{-\infty}^{\infty} f(t) e^{-j(-\omega)t} dt \right] d\omega
-$$
-  
-$$
-= \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) F(-\omega) d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) F^*(\omega) d\omega
-$$
- (18.62)
-
-In fact, ∣F(*ω*)∣ 2 is sometimes known as the energy spectral density of signal f (t). But if *z* = *x* + *jy*, *zz*\* = (*x* + *jy*)(*x* − *jy*) = *x*<sup>2</sup> + *y*<sup>2</sup> = ∣*z*∣ 2 . Hence,
-
-$$
-W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega
-$$
- (18.63)
-
-as expected. Equation (18.63) indicates that the energy carried by a signal can be found by integrating either the square of *f*(*t*) in the time domain or 1 ∕2*π* times the square of *F*(*ω*) in the frequency domain.
-
-Because ∣*F*(*ω*)∣ 2 is an even function, we may integrate from 0 to ∞ and double the result; that is,
-
-$$
-W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) \, dt = \frac{1}{\pi} \int_{0}^{\infty} |F(\omega)|^2 \, d\omega \tag{18.64}
-$$
-
-We may also calculate the energy in any frequency band *ω*<sup>1</sup> < *ω*< *ω*2 as
-
-$$
-W_{1\Omega} = \frac{1}{\pi} \int_{\omega_1}^{\omega_2} |F(\omega)|^2 \, d\omega \tag{18.65}
-$$
-
-Notice that Pa rseval's theorem as stated here applies to nonpe riodic functions. Pa rseval's theorem for periodic functions w as pre sented in Sections 17.5 and 17.6. As evident in Eq. (18.63), Parseval's theorem shows that the ener gy associated with a nonperiodic signal is spread ove r the entire frequenc y spectrum, whereas the ener gy of a periodic signal is concentrated at the frequencies of its harmonic components.
-
-The voltage across a 10- <sup>Ω</sup> resistor is *v*(*t*) = 5*<sup>e</sup>* Example 18.9 <sup>−</sup>3*<sup>t</sup> u*(*t*) V. Find the total energy dissipated in the resistor.
-
-# **Solution:**
-
-- 1. **Define.** The problem is well defined and clearly stated.
-- 2. **Present.** We are given the voltage across the resistor for all time and are asked to find the energy dissipated by the resistor. We note that the voltage is zero for all time less than zero. Thus, we only need to consider the time from zero to infinity.
-- 3. **Alternative.** There are basically two ways to find this answer. The first would be to find the answer in the time domain. We will use the second approach to find the answer using Fourier analysis.
-- 4. **Attempt.** In the time domain,
-
-$$
-W_{10\Omega} = 0.1 \int_{-\infty}^{\infty} f^2(t) dt = 0.1 \int_{0}^{\infty} 25e^{-6t} dt
-$$
-$$
-= 2.5 \frac{e^{-6t}}{-6} \Big|_{0}^{\infty} = \frac{2.5}{6} = 416.7 \text{ mJ}
-$$
-
-5. **Evaluate.** In the frequency domain,
-
-$$
-F(\omega) = V(\omega) = \frac{5}{3 + j\omega}
-$$
-
-so that
-
-$$
-|F(\omega)|^2 = F(\omega)F(\omega)^* = \frac{25}{9 + \omega^2}
-$$
-
-Hence, the energy dissipated is
-
-$$
-W_{10\Omega} = \frac{0.1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega = \frac{0.1}{\pi} \int_{0}^{\infty} \frac{25}{9 + \omega^2} d\omega
-$$
-  
-=  $\frac{2.5}{\pi} \left( \frac{1}{3} \tan^{-1} \frac{\omega}{3} \right) \Big|_{0}^{\infty} = \frac{2.5}{\pi} \left( \frac{1}{3} \right) \left( \frac{\pi}{2} \right) = \frac{2.5}{6} = 416.7 \text{ mJ}$ 
-
-6. **Satisfactory?** We have satisf actorily solv ed the problem and can present the results as a solution to the problem.
-
-Practice Problem 18.9 (a) Calculate the total ener gy absorbed by a 1- <sup>Ω</sup> resistor with *i*(*t*) = 10*e*<sup>−</sup>2∣*t*<sup>∣</sup> A in the time domain. (b) Repeat (a) in the frequency domain.
-
-**Answer:** (a) 50 J, (b) 50 J.
-
-Example 18.10 Calculate the fraction of the total ener gy dissipated by a 1-Ω resistor in the frequency band −10 < *ω*< 10 rad/s when the v oltage across it is *v*(*t*) = *e*<sup>−</sup>2*<sup>t</sup> u*(*t*).
-
-# **Solution:**
-
-Given that *f*(*t*) = *v*(*t*) = *e*<sup>−</sup>2*<sup>t</sup> u*(*t*), then
-
-$$
-F(\omega) = \frac{1}{2 + j\omega} \qquad \Rightarrow \qquad |F(\omega)|^2 = \frac{1}{4 + \omega^2}
-$$
-
-The total energy dissipated by the resistor is
-
-$$
-W_{1\Omega} = \frac{1}{\pi} \int_0^\infty |F(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^\infty \frac{d\omega}{4 + \omega^2}
-$$
-$$
-= \frac{1}{\pi} \left( \frac{1}{2} \tan^{-1} \frac{\omega}{2} \Big|_0^\infty \right) = \frac{1}{\pi} \left( \frac{1}{2} \right) \frac{\pi}{2} = 0.25 \text{ J}
-$$
-
-The energy in the frequencies −10 < *ω*< 10 rad/s is
-
-$$
-W = \frac{1}{\pi} \int_0^{10} |F(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^{10} \frac{d\omega}{4 + \omega^2} = \frac{1}{\pi} \left( \frac{1}{2} \tan^1 \frac{\omega}{2} \Big|_0^{10} \right)
-$$
-$$
-= \frac{1}{2\pi} \tan^{-1} 5 = \frac{1}{2\pi} \left( \frac{78.69^\circ}{180^\circ} \pi \right) = 0.218 \text{ J}
-$$
-
-<span id="page-859-0"></span>Its percentage of the total energy is
-
-$$
-\frac{W}{W_{1\Omega}} = \frac{0.218}{0.25} = 87.4
-$$
- percent
-
-A 2-Ω resistor has *i*(*t*) = 2*e* Practice Problem 18.10 <sup>−</sup>*<sup>t</sup> u*(*t*) A. What percentage of the total energy is in the frequency band −4 < *ω*< 4 rad/s?
-
-**Answer:** 84.4 percent.
-
-# **18.6** Comparing the Fourier and Laplace Transforms
-
-It is worthwhile to take some moments to compare the Laplace and Fourier transforms. The following similarities and differences should be noted:
-
-- 1. The Laplace transform defined in Chapter 15 is one-sided in that the integral is over 0 < *t* < ∞, making it only useful for positi ve-time functions, *f*(*t*), *t* > 0. The Fourier transform is applicable to func tions defined for all time.
-- 2. For a function *f*(*t*) that is nonzero for positive time only (i.e.,
-
-$$
-f(t) = 0, t < 0) \text{ and } \int_0^{\infty} |f(t)| \, dt < \infty \text{, the two transforms are related by}
-$$
-\n
-$$
-F(\omega) = F(s)|_{s=j\omega} \tag{18.66}
-$$
-
-This equation also shows that the Fourier transform can be regarded as a special case of the Laplace transform with *s* = *jω*. Recall that *s* = *σ* + *jω*. Therefore, Eq. (18.66) sho ws that the Laplace trans form is related to the entire *s* plane, whereas the Fourier transform is restricted to the *jω* axis. See Fig. 15.1.
-
-- 3. The Laplace transform is applicable to a wider range of functions than the F ourier transform. F or e xample, the function *tu*(*t*) has a Laplace transform b ut no F ourier transform. But F ourier trans forms exist for signals that are not physically realizable and have no Laplace transforms.
-- 4. The Laplace transform is better suited for the analysis of transient problems involving initial conditions, because it permits the inclu sion of the initial conditions, whereas the F ourier transform does not. The Fourier transform is especially useful for problems in the steady state.
-- 5. The Fourier transform pro vides greater insight into the frequenc y characteristics of signals than does the Laplace transform.
-
-Some of the similarities and dif ferences can be observed by comparing Tables 15.1 and 15.2 with Tables 18.1 and 18.2.
-
- In other words, if all the poles of F(s) lie in the left-hand side of the s plane, then one can obtain the Fourier transform F(*ω*) from the corresponding Laplace transform F(s) by merely replacing s by j*ω*. Note that this is not the case, for example, for u(t) or cos atu(t).
-
-# <span id="page-860-0"></span>**18.7** Applications
-
-Besides its usefulness for circuit analysis, the F ourier transform is used extensively in a variety of fields such as optics, spectroscopy, acoustics, computer science, and electrical engineering. In electrical engineering, it is applied in communications systems and signal processing, where fre quency response and frequency spectra are vital. Here we consider tw o simple applications: amplitude modulation (AM) and sampling.
-
-# **18.7.1** Amplitude Modulation
-
-Electromagnetic radiation or transmission of information through space has become an indispensable part of a modern technological society . However, transmission through space is only efficient and economical at high frequencies (above 20 kHz). To transmit intelligent signals such as for speech and music—contained in the lo w-frequency range of 50 Hz to 20 kHz is e xpensive; it requires a huge amount of po wer and large antennas. A common method of transm itting low-frequency audio information is to transmit a high-frequency signal, called a *carrier*, which is controlled in some w ay to correspond to the audio informa tion. Three characteristics (amplitude, frequency, or phase) of a carrier can be controlled so as to allow it to carry the intelligent signal, called the *modulating signal*. Here we will only consider the control of the carrier's amplitude. This is known as *amplitude modulation*.
-
-Amplitude modulation (AM) is a process whereby the amplitude of the carrier is controlled by the modulating signal.
-
-AM is used in ordinary commercial radio bands and the video portion of commercial television.
-
-Suppose the audio information, such as voice or music (or the modulating signal in general) to be transmitted is *m*(*t*) = *Vm* cos *ωmt*, while the high-frequency carrier is *c*(*t*) = *Vc* cos *ωct*, where *ωc* ≫ *ωm*. Then an AM signal *f*(*t*) is given by
-
-$$
-f(t) = V_c \left[1 + m(t)\right] \cos \omega_c t \tag{18.67}
-$$
-
-Figure 18.22 illustrates the modulating signal *m*(*t*), the carrier *c*(*t*), and the AM signal *f*(*t*). We can use the result in Eq. (18.27) together with the Fourier transform of the cosine function (see Example 18.1 or Table 18.1) to determine the spectrum of the AM signal:
-
-$$
-F(\omega) = \mathcal{F}[V_c \cos \omega_c t] + \mathcal{F}[V_c m(t) \cos \omega_c t]
-$$
-  
-=  $V_c \pi [\delta(\omega - \omega_c) + \delta(\omega + \omega_c)]$   
-+  $\frac{V_c}{2} [M(\omega - \omega_c) + M(\omega + \omega_c)]$  (18.68)
-
-where *M*(*ω*) is the F ourier transform of the modulating signal *m*(*t*). Shown in Fig. 18.23 is the frequenc y spectrum of the AM signal. Fig ure 18.23 indicates that the AM signal consists of the carrier and tw o other sinusoids. The sinusoid with frequenc y *ωc* − *ωm* is kno wn as the *lower sideband*, while the one with frequenc y *ωc* + *ωm* is known as the *upper sideband*.
-
-Frequency spectrum of AM signal.
-
-Notice that we ha ve assumed that the modulating signal is sinu soidal to mak e the analysis easy . In real life, *m*(*t*) is a nonsinusoidal, band-limited signal—its frequency spectrum is within the range between 0 and *ωu* = 2*π fu* (i.e., the signal has an upper frequency limit). Typically, *fu* = 5kHz for AM radio. If the frequenc y spectrum of the modulating signal is as sho wn in Fig. 18.24(a), then the frequenc y spectrum of the AM signal is sho wn in Fig. 18.24(b). Thus, to avoid any interference, carriers for AM radio stations are spaced 10 kHz apart.
-
-At the recei ving end of the transmission, the audio informa tion is recovered from the modulated carrier by a process kno wn as *demodulation*.
-
-Example 18.11 A music signal has frequency components from 15 Hz to 30 kHz. If this signal could be used to amplitude modulate a 1.2-MHz carrier , find the range of frequencies for the lower and upper sidebands.
-
-# **Solution:**
-
-The lo wer sideband is the dif ference of the carrier and modulating frequencies. It will include the frequencies from
-
-$$
-1,200,000 - 30,000 \text{ Hz} = 1,170,000 \text{ Hz}
-$$
-
-to
-
-$$
-1,200,000 - 15
-$$
- Hz = 1,199,985 Hz
-
-The upper sideband is the sum of the carrier and modulating frequencies. It will include the frequencies from
-
-1,200,000 + 15 Hz = 1,200,015 Hz
-
-to
-
-$$
-1,200,000 + 30,000 \text{ Hz} = 1,230,000 \text{ Hz}
-$$
-
-Practice Problem 18.11 If a 2-MHz carrier is modulated by a 4-kHz intelligent signal, determine the frequencies of the three components of the AM signal that results.
-
-**Answer:** 2,004,000 Hz, 2,000,000 Hz, 1,996,000 Hz.
-
-# **Figure 18.25**
-
-(a) Continuous (analog) signal to be sampled, (b) train of impulses, (c) sampled (digital) signal.
-
-(c)
-
-# **18.7.2** Sampling
-
-In analog systems, signals are processed in their entirety . However, in modern digital systems, only samples of signals are required for pro cessing. This is possible as a result of the sampling theorem gi ven in Section 17.8.1. The sampling can be done by using a train of pulses or impulses. We will use impulse sampling here.
-
-Consider the continuous signal *g*(*t*) shown in Fig. 18.25(a). This can be multiplied by a train of impulses *δ*(*t* − *nTs*) shown in Fig. 18.25(b), where *Ts* is the *sampling interval* and *fs* = 1∕*Ts* is the *sampling frequency*  or the *sampling rate*. The sampled signal *gs*(*t*) is therefore
-
-$$
-g_s(t) = g(t) \sum_{n=-\infty}^{\infty} \delta(t - nT_s) = \sum_{n=-\infty}^{\infty} g(nT_s) \delta(t - nT_s)
-$$
-(18.69)
-
-The Fourier transform of this is
-
-$$
-G_{s}(\omega) = \sum_{n=-\infty}^{\infty} g(nT_{s}) \mathcal{F}[\delta(t - nT_{s})] = \sum_{n=-\infty}^{\infty} g(nT_{s})e^{-jn\omega T_{s}}
-$$
-(18.70)
-
-It can be shown that
-
-$$
-\sum_{n=-\infty}^{\infty} g(nT_s)e^{-jn\omega T_s} = \frac{1}{T_s}\sum_{n=-\infty}^{\infty} G(\omega + n\omega_s)
-$$
-(18.71)
-
-<span id="page-863-0"></span>where *ωs* = 2*π*∕*Ts*. Thus, Eq. (18.70) becomes
-
-$$
-G_s(\omega) = \frac{1}{T_s} \sum_{n=-\infty}^{\infty} G(\omega + n\omega_s)
-$$
- (18.72)
-
-This shows that the F ourier transform *Gs*(*ω*) of the sampled signal is a sum of translates of the Fourier transform of the original signal at a rate of 1∕*Ts*.
-
-To ensure optimum recovery of the original signal, what must be the sampling interval? This fundamental question in sampling is answered by an equivalent part of the sampling theorem:
-
-A band-limited signal, with no frequency component higher than W hertz, may be completely recovered from its samples taken at a frequency at least twice as high as 2W samples per second.
-
-In other words, for a signal with bandwidth *W* hertz, there is no loss of information or overlapping if the sampling frequency is at least twice the highest frequency in the modulating signal. Thus,
-
-$$
-\frac{1}{T_s} = f_s \ge 2W\tag{18.73}
-$$
-
-The sampling frequency *fs* = 2*W* is known as the *Nyquist frequency* or rate, and 1∕*fs* is the *Nyquist interval*.
-
-A telephone signal with a cutoff frequency of 5 kHz is sampled at a rate Example 18.12 60 percent higher than the minimum allowed rate. Find the sampling rate.
-
-# **Solution:**
-
-The minimum sample rate is the Nyquist rate = 2*W* = 2 × 5 = 10 kHz. Hence,
-
-$$
-f_s = 1.60 \times 2W = 16 \text{ kHz}
-$$
-
-An audio signal that is band-limited to 12.5 kHz is digitized into 8-bit Practice Problem 18.12 samples. What is the maximum sampling interv al that must be used to ensure complete recovery?
-
-**Answer:** 40 *μ*s.
-
-# **18.8** Summary
-
-1. The Fourier transform con verts a nonperiodic function *f*(*t*) into a transform *F*(*ω*), where
-
-$$
-F(\omega) = \mathcal{F}[f(t)] = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt
-$$
-
-<span id="page-864-0"></span>2. The inverse Fourier transform of *F*(*ω*) is
-
-$$
-f(t) = \mathcal{F}^{-1}[F(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega
-$$
-
-- 3. Important Fourier transform properties and pairs are summarized in Tables 18.1 and 18.2, respectively.
-- 4. Using the F ourier transform method to analyze a circuit in volves finding the Fourier transform of the e xcitation, transforming the circuit element into the frequency domain, solving for the unkno wn response, and transforming the response to the time domain using the inverse Fourier transform.
-- 5. If *H*(*ω*) is the transfer function of a network, then *H*(*ω*) is the Fourier transform of the network's impulse response; that is,
-
-$$
-H(\omega) = \mathcal{F}[h(t)]
-$$
-
- The output *Vo*(*ω*) of the netw ork can be obtained from the input *Vi*(*ω*) using
-
-$$
-V_o(\omega) = H(\omega)V_i(\omega)
-$$
-
-6. Parseval's theorem gives the energy relationship between a function *f*(*t*) and its Fourier transform *F*(*ω*). The 1-Ω energy is
-
-$$
-W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega
-$$
-
- The theorem is useful in calculating energy carried by a signal either in the time domain or in the frequency domain.
-
-7. Typical applications of the Fourier transform are found in amplitude modulation (AM) and sampling. F or AM application, a w ay of determining the sidebands in an amplitude-modulated wave is derived from the modulation property of the F ourier transform. F or sampling application, we found that no information is lost in sampling (required for digital transmission) if the sampling frequency is equal to at least twice the Nyquist rate.
-
-# Review Questions
-
-**18.1** Which of these functions does not have a Fourier transform?
-
-| (a) et  | (b) te−3t   |
-|---------|-------------|
-| u(−t)   | u(t)        |
-| (c) 1∕t | (d) ∣t∣u(t) |
-
-**18.2** The Fourier transform of *e<sup>j</sup>*2*<sup>t</sup>* is:
-
-(a) 
-$$
-\frac{1}{2 + j\omega}
-$$
-   
-\n(b)  $\frac{1}{-2 + j\omega}$    
-\n(c)  $2\pi\delta(\omega - 2)$    
-\n(d)  $2\pi\delta(\omega + 2)$ 
-
-**18.3** The inverse Fourier transform of *e*<sup>−</sup>*j<sup>ω</sup>* \_\_\_\_\_\_ <sup>2</sup> + *<sup>j</sup>ω* is (a) *e*<sup>−</sup>2*<sup>t</sup>* (b) *e*<sup>−</sup>2*<sup>t</sup> u*(*t* − 1) (c) *e*<sup>−</sup>2(*t*−1) (d) *e*<sup>−</sup>2(*t*−1)*u*(*t* − 1)
-
-**18.4** The inverse Fourier transform of *δ*(*ω*) is:
-
-(a) *δ*(*t*) (b) *u*(*t*) (c) 1 (d) 1∕2*π*
-
-**18.5** The inverse Fourier transform of *jω* is:
-
-(a) *δ*ʹ(*t*) (b) *u*ʹ(*t*) (c) 1∕*t* (d) undefined
-
-**18.6** Evaluating the integral ∫ −∞ ∞ 10*δ*(*ω*) \_\_\_\_\_\_ 4 + *ω*<sup>2</sup> *dω* results in:
-
-(a) 0 (b) 2 (c) 2.5 (d) ∞
-
-**18.7** The integral ∫ −∞ ∞ 10*δ*(*ω*− 1) \_\_\_\_\_\_\_\_\_\_ 4 + *ω*<sup>2</sup> *dω* gives: (a) 0 (b) 2 (c) 2.5 (d) ∞ <span id="page-865-0"></span>**18.8** The current through an initially uncharged 1-F capacitor is *δ*(*t*) A. The voltage across the capacitor is:
-
-| (a) u(t) V        | (b) −1∕2 + u(t) V |
-|-------------------|-------------------|
-| (c) e−t<br>u(t) V | (d) δ(t) V        |
-
-**18.9** A unit step current is applied through a 1-H inductor. The voltage across the inductor is:
-
-| (a) u(t) V        | (b) sgn(t) V |
-|-------------------|--------------|
-| (c) e−t<br>u(t) V | (d) δ(t) V   |
-
-# Problems
-
-## † Sections 18.2 and 18.3 Fourier Transform and its Properties
-
-**18.1** Obtain the Fourier transform of the function in Fig. 18.26.
-
-# **Figure 18.26**
-
-For Prob. 18.1.
-
-**18.2** Using Fig. 18.27, design a problem to help other students better understand the Fourier transform given a wave shape.
-
-# **Figure 18.27**
-
-For Prob. 18.2.
-
-**Figure 18.28** For Prob. 18.3.
-
-**18.10** Parseval's theorem is only for nonperiodic functions.
-
-(a) True (b) False
-
-*Answers: 18.1c, 18.2c, 18.3d, 18.4d, 18.5a, 18.6c, 18.7b, 18.8a, 18.9d, 18.10b*
-
-**18.4** Find the Fourier transform of the waveform shown in Fig. 18.29.
-
-† We have marked (with the *MATLAB* icon) the problems where we are asking the student to find the Fourier transform of a wave shape. We do this because you can use *MATLAB* to plot the results as a check.
-
-**18.7** Find the Fourier transforms of the signals in Fig. 18.32.
-
-**18.8** Obtain the Fourier transforms of the signals shown in Fig. 18.33.
-
-**Figure 18.33** For Prob. 18.8.
-
-**Figure 18.34** For Prob. 18.9.
-
-**Figure 18.35** For Prob. 18.10.
-
-**18.11** Find the Fourier transform of the "sine-wave pulse" shown in Fig. 18.36.
-
-**Figure 18.36** For Prob. 18.11.
-
-**18.12** Find the Fourier transform of the following signals.
-
-(a) *f*1(*t*) = *e*<sup>−</sup>3*<sup>t</sup>* sin(10*t*)*u*(*t*) (b) *f*2(*t*) = *e*<sup>−</sup>4*<sup>t</sup>* cos(10*t*)*u*(*t*)
-
-**18.13** Find the Fourier transform of the following signals:
-
-(a) 
-$$
-f(t) = \cos(at - \pi/3)
-$$
-,  $-\infty < t < \infty$   
-\n(b)  $g(t) = u(t + 1) \sin \pi t$ ,  $-\infty < t < \infty$   
-\n(c)  $h(t) = (1 + A \sin at) \cos bt$ ,  $-\infty < t < \infty$ ,  
-\nwhere A, a, and b are constants  
-\n(d)  $i(t) = 1 - t$ ,  $0 < t < 4$ 
-
-- **18.14** Design a problem to help other students better
-  - understand finding the Fourier transform of a variety of time varying functions (do at least three).
-- **18.15** Find the Fourier transforms of the following functions:
-
-(a) 
-$$
-f(t) = \delta(t + 3) - \delta(t - 3)
-$$
-  
-\n(b)  $f(t) = \int_{-\infty}^{\infty} 2\delta(t - 1) dt$   
-\n(c)  $f(t) = \delta(3t) - \delta'(2t)$ 
-
-**18.16** Determine the Fourier transforms of these functions: \*
-
-)
-
-(a) 
-$$
-f(t) = 8/t^2
-$$
-  
-(b)  $g(t) = 4/(4 + t^2)$ 
-
-**18.17** Find the Fourier transforms of:
-
-(a) 2 cos 2*tu*(*t*)
-
-- (b) 0.5 sin 10*tu*(*t*)
-- **18.18** Given that *F*(*ω*) = [ *f*(*t*)], prove the following results, using the definition of Fourier transform:
-
-(a) 
-$$
-\mathcal{F}[f(t - t_0)] = e^{-j\omega t_0} F(\omega)
-$$
-  
-\n(b)  $\mathcal{F}\left[\frac{df(t)}{dt}\right] = j\omega F(\omega)$   
-\n(c)  $\mathcal{F}[f(-t)] = F(-\omega)$   
-\n(d)  $\mathcal{F}[tf(t)] = j\frac{d}{d\omega} F(\omega)$ 
-
-**18.19** Find the Fourier transform of
-
-$$
-f(t) = 2 \cos 2\pi t [u(t) - u(t-1)]
-$$
-
-**18.20** (a) Show that a periodic signal with exponential Fourier series
-
-$$
-f(t) = \sum_{n = -\infty}^{\infty} c_n e^{jn\omega_0 t}
-$$
-
-has the Fourier transform
-
-$$
-F(\omega) = \sum_{n = -\infty}^{\infty} c_n \delta(\omega - n\omega_0)
-$$
-  
-where  $\omega_0 = 2\pi/T$ .
-
-**Figure 18.37**
-
-For Prob. 18.20(b).
-
-**18.21** Show that
-
-$$
-\int_{-\infty}^{\infty} \left( \frac{\sin a\omega}{a\omega} \right)^2 d\omega = \frac{\pi}{a}
-$$
-
-*Hint:* Use the fact that
-
-$$
-\mathcal{F}[u(t+a) - u(t-a)] = 2a\left(\frac{\sin a\omega}{a\omega}\right).
-$$
-
-**18.22** Prove that if *F*(*ω*) is the Fourier transform of *f*(*t*),
-
-$$
-\mathcal{F}[f(t)\sin\omega_0 t] = \frac{j}{2}[F(\omega + \omega_0) - F(\omega - \omega_0)]
-$$
-
-**18.23** If the Fourier transform of *f*(*t*) is
-
-transform of 
-$$
-f(t)
-$$
- is  
-\n
-$$
-F(\omega) = \frac{10}{(2 + j\omega)(5 + j\omega)}
-$$
-
-determine the transforms of the following:
-
-(a) 
-$$
-f(-3t)
-$$
- (b)  $f(2t - 1)$  (c)  $f(t) \cos 2t$   
-(d)  $\frac{d}{dt}f(t)$  (e)  $\int_{-\infty}^{t} f(t) dt$ 
-
-−∞ **18.24** Given that [ *f*(*t*)*t*] = ( *j*∕*ω*)(*e*<sup>−</sup>*j<sup>ω</sup>* − 1), find the Fourier transforms of:
-
-(a) 
-$$
-x(t) = f(t) + 3
-$$
-   
-\n(b)  $y(t) = f(t - 2)$   
-\n(c)  $h(t) = f'(t)$   
-\n(d)  $g(t) = 4f(\frac{2}{3}t) + 10f(\frac{5}{3}t)$ 
-
-**18.25** Obtain the inverse Fourier transform of the following signals.
-
-(a) 
-$$
-G(\omega) = \frac{5}{j\omega - 2}
-$$
-  
-\n(b)  $H(\omega) = \frac{12}{\omega^2 + 4}$   
-\n(c)  $X(\omega) = \frac{10}{(j\omega - 1)(j\omega - 2)}$ 
-
-**18.26** Determine the inverse Fourier transforms of the following:
-
-(a) 
-$$
-F(\omega) = \frac{e^{-j2\omega}}{1 + j\omega}
-$$
-  
-\n(b)  $H(\omega) = \frac{1}{(j\omega + 4)^2}$   
-\n(c)  $G(\omega) = 2u(\omega + 1) - 2u(\omega - 1)$ 
-
-<sup>\*</sup> An asterisk indicates a challenging problem. (c) *G*(*ω*) = 2*u*(*ω* + 1) − 2*u*(*ω*− 1)
-
-**18.27** Find the inverse Fourier transforms of the following functions:
-
-(a) 
-$$
-F(\omega) = \frac{100}{j\omega(j\omega + 10)}
-$$
-  
-\n(b)  $G(\omega) = \frac{10 j\omega}{(-j\omega + 2)(j\omega + 3)}$   
-\n(c)  $H(\omega) = \frac{60}{-\omega^2 + j40\omega + 1300}$   
-\n(d)  $Y(\omega) = \frac{\delta(\omega)}{(j\omega + 1)(j\omega + 2)}$ 
-
-**18.28** Find the inverse Fourier transforms of:
-
-Find the inverse Fourier t  
-\n(a) 
-$$
-\frac{n\delta(\omega)}{(5+j\omega)(2+j\omega)}
-$$
-  
-\n(b)  $\frac{10\delta(\omega+2)}{j\omega(j\omega+1)}$   
-\n(c)  $\frac{20\delta(\omega-1)}{(2+j\omega)(3+j\omega)}$   
-\n(d)  $\frac{5n\delta(\omega)}{5+j\omega} + \frac{5}{j\omega(5+j\omega)}$ 
-
-- **18.29** Determine the inverse Fourier transforms of: \*
-  - (a) *F*(*ω*) = 4*δ*(*ω* + 3) + *δ*(*ω*) + 4*δ*(*ω*− 3)
-  - (b) *G*(*ω*) = 4*u*(*ω* + 2) − 4*u*(*ω*− 2)
-  - (c) *H*(*ω*) = 6 cos 2*ω*
-- **18.30** For a linear system with input *x*(*t*) and output *y*(*t*), find the impulse response for the following cases:
-
-(a) 
-$$
-x(t) = e^{-at}u(t)
-$$
-,  $y(t) = u(t) - u(-t)$   
-\n(b)  $x(t) = e^{-t}u(t)$ ,  $y(t) = e^{-2t}u(t)$   
-\n(c)  $x(t) = \delta(t)$ ,  $y(t) = e^{-at} \sin btu(t)$ 
-
-**18.31** Given a linear system with output *y*(*t*) and impulse response *h*(*t*), find the corresponding input *x*(*t*) for the following cases:
-
-(a) 
-$$
-y(t) = te^{-at}u(t)
-$$
-,  $h(t) = e^{-at}u(t)$   
-\n(b)  $y(t) = u(t+1) - u(t-1)$ ,  $h(t) = \delta(t)$   
-\n(c)  $y(t) = e^{-at}u(t)$ ,  $h(t) = \text{sgn}(t)$ 
-
-**18.32** Determine the functions corresponding to the following Fourier transforms: \*
-
-(a) 
-$$
-F_1(\omega) = \frac{e^{j\omega}}{-j\omega + 1}
-$$
-   
-\n(b)  $F_2(\omega) = 2e^{|\omega|}$   
-\n(c)  $F_3(\omega) = \frac{1}{(1 + \omega^2)^2}$    
-\n(d)  $F_4(\omega) = \frac{\delta(\omega)}{1 + j2\omega}$ 
-
-**18.33** Find *f*(*t*) if: \*
-
-(a) 
-$$
-F(\omega) = 2 \sin \pi \omega [u(\omega + 1) - u(\omega - 1)]
-$$
-  
-(b)  $F(\omega) = \frac{1}{\omega} (\sin 2\omega - \sin \omega) + \frac{j}{\omega} (\cos 2\omega - \cos \omega)$ 
-
-**18.34** Determine the signal *f*(*t*) whose Fourier transform is shown in Fig. 18.38. (*Hint:* Use the duality property.)
-
-# **Figure 18.38**
-
-For Prob. 18.34.
-
-**18.35** A signal *f*(*t*) has Fourier transform
-
-$$
-F(\omega) = \frac{1}{2 + j\omega}
-$$
-
-Determine the Fourier transform of the following signals:
-
-(a) 
-$$
-x(t) = f(3t - 1)
-$$
-  
-\n(b)  $y(t) = f(t) \cos 5t$   
-\n(c)  $z(t) = \frac{d}{dt}f(t)$   
-\n(d)  $h(t) = f(t) * f(t)$   
-\n(e)  $i(t) = tf(t)$ 

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/212_18.4 Circuit Applications.md

@@ -1,77 +0,0 @@
-# Section 18.4 Circuit Applications
-
-**18.36** The transfer function of a circuit is
-
-$$
-H(\omega) = \frac{10}{j\omega + 2}
-$$
-
- If the input signal to the circuit is *vs*(*t*) = *e*<sup>−</sup>4*<sup>t</sup> u*(*t*) V, find the output signal. Assume all initial conditions are zero.
-
-**18.37** Find the transfer function *Io*(*ω*)∕*Is*(*ω*) for the circuit in Fig. 18.39.
-
-**Figure 18.40** For Prob. 18.38.
-
-Problems **847**
-
-**18.39** Given the circuit in Fig. 18.41, with its excitation, determine the Fourier transform of *i*(*t*).
-
-For Prob. 18.39.
-
-**18.40** Determine the current *i*(*t*) in the circuit of Fig. 18.42(b), given the voltage source shown in Fig. 18.42(a).
-
-**Figure 18.42**
-
-For Prob. 18.40.
-
-**18.41** Determine the Fourier transform of *v*(*t*) in the circuit shown in Fig. 18.43.
-
-**Figure 18.43** For Prob. 18.41.
-
-(a) Let *i*(*t*) = sgn(*t*) A. (b) Let *i*(*t*) = 4[*u*(*t*) − *u*(*t* − 1)] A.
-
-# **Figure 18.44**
-
-For Prob. 18.42.
-
-**18.43** Find *vo*(*t*) in the circuit of Fig. 18.45, where *is* = 5*e*<sup>−</sup>*<sup>t</sup> u*(*t*) A.
-
-# **Figure 18.45**
-
-For Prob. 18.43.
-
-**18.44** If the rectangular pulse in Fig. 18.46(a) is applied to the circuit in Fig. 18.46(b), find *vo* at *t* = 1 s.
-
-For Prob. 18.44.
-
-**18.45** Use the Fourier transform to find *i*(*t*) in the circuit of Fig. 18.47 if *vs*(*t*) = 10*e*<sup>−</sup>2*<sup>t</sup> u*(*t*).
-
-# **Figure 18.47** For Prob. 18.45.
-
-For Prob. 18.46.
-
-**18.46** Determine the Fourier transform of *io*(*t*) in the circuit of Fig. 18.48.
-
-**18.47** Find the voltage *vo*(*t*) in the circuit of Fig. 18.49. Let *is*(*t*) = 8*e*<sup>−</sup>*<sup>t</sup> u*(*t*) A.
-
-# **Figure 18.49**
-
-For Prob. 18.47.
-
-**18.48** Find *io*(*t*) in the op amp circuit of Fig. 18.50.
-
-**Figure 18.50** For Prob. 18.48.
-
-**18.49** Use the Fourier transform method to obtain *vo*(*t*) in the circuit of Fig. 18.51.
-
-For Prob. 18.49.
-
-**18.50** Determine *vo*(*t*) in the transformer circuit of Fig. 18.52.
-
-For Prob. 18.50.
-
-**18.51** Find the energy dissipated by the resistor in the circuit of Fig. 18.53.
-
-# **Figure 18.53**
-
-For Prob. 18.51.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/213_18.5 Parseval's Theorem.md

@@ -1,18 +0,0 @@
-# Section 18.5 Parseval's Theorem
-
-**18.52** For 
-$$
-F(\omega) = \frac{3}{3 + j\omega}
-$$
-, find  $J = \int_{-\infty}^{\infty} f^2(t) dt$ .
-
-**18.53** If 
-$$
-f(t) = e^{-2|t|}
-$$
-, find  $J = \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega$ .
-
-- **18.54** Design a problem to help other students better understand finding the total energy in a given signal.
-- **18.55** Let *f* (*t*) = 5*e*<sup>−</sup>(*t*−2)*u*(*t*). Find *F*(*ω*) and use it to find the total energy in *f*(*t*).
-- **18.56** The voltage across a 1-Ω resistor is *v*(*t*) = *te*<sup>−</sup>2*<sup>t</sup> u*(*t*) V. (a) What is the total energy absorbed by the resistor? (b) What fraction of this energy absorbed is in the frequency band −2 ≤ *ω*≤ 2?
-- **18.57** Let *i*(*t*) = 2*et u*(−*t*) A. Find the total energy carried by i(*t*) and the percentage of the 1-Ω energy in the frequency range of −5 < *ω*< 5 rad/s.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/214_18.7 Applications.md

@@ -1,32 +0,0 @@
-# Section 18.6 Applications
-
-**18.58** An AM signal is specified by
-
-*f* (*t*) = 10(1 + 4 cos 200*πt*) cos *π*× 104 *t*
-
-Determine the following:
-
-- (a) the carrier frequency,
-- (b) the lower sideband frequency,
-- (c) the upper sideband frequency.
-- **18.59** For the linear system in Fig. 18.54, when the input voltage is *vi*(*t*) = 2*δ*(*t*) V, the output is *vo*(*t*) = 10*e*<sup>−</sup>2*<sup>t</sup>* − 6*e*<sup>−</sup>4*<sup>t</sup>* V. Find the output when the input is *vi*(*t*) = 4*e*<sup>−</sup>*<sup>t</sup> u*(*t*) V.
-
-**Figure 18.54** For Prob. 18.9.
-
-# <span id="page-871-0"></span>**18.60** A band-limited signal has the following Fourier series representation:
-
-*is*(*t*) = 10 + 8 cos(2*πt* + 30°) + 5 cos(4*πt* − 150°)mA
-
- If the signal is applied to the circuit in Fig. 18.55, find *v*(*t*).
-
-**Figure 18.55** For Prob. 18.60.
-
-**18.61** In a system, the input signal *x*(*t*) is amplitudemodulated by *m*(*t*) = 2 + cos*ω*0*t*. The response *y*(*t*) = *m*(*t*)*x*(*t*). Find *Y*(*ω*) in terms of *X*(*ω*).
-
-- **18.62** A voice signal occupying the frequency band of 0.4 to 3.5 kHz is used to amplitude-modulate a 10-MHz carrier. Determine the range of frequencies for the lower and upper sidebands.
-- **18.63** For a given locality, calculate the number of
-- stations allowable in the AM broadcasting band (540–1600 kHz) without interference with one another.
-- **18.64** Repeat the previous problem for the FM broadcasting band (88–108 MHz), assuming that the carrier frequencies are spaced 200 kHz apart.
-- **18.65** The highest-frequency component of a voice signal is 3.4 kHz. What is the Nyquist rate of the sampler of the voice signal?
-- **18.66** A TV signal is band-limited to 4.5 MHz. If samples are to be reconstructed at a distant point, what is the maximum sampling interval allowable?
-- **18.67** Given a signal *g*(*t*) = sinc(200*π t*), find the Nyquist rate and the Nyquist interval for the signal. \*

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/215_Problems.md

@@ -1,17 +0,0 @@
-# Comprehensive Problems
-
-**18.68** The voltage signal at the input of a filter is *v*(*t*) = 50*e*<sup>−</sup>2∣*t*<sup>∣</sup> V. What percentage of the total 1-Ω energy content lies in the frequency range of 1 < *ω*< 5 rad/s?
-
-**18.69** A signal with Fourier transform
-
-$$
-F(\omega) = \frac{20}{4 + j\omega}
-$$
-
-is passed through a filter whose cutoff frequency is 2 rad/s (i.e., 0 < *ω*< 2). What fraction of the energy in the input signal is contained in the output signal?
-
-*This page intentionally left blank* 
-
-# **chapter**
-
-19

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/216_Chapter 19 - Two-Port Networks.md

@@ -1,33 +0,0 @@
-# <span id="page-873-0"></span>Two-Port Networks
-
-*Never put off till tomorrow what you can do today. Never trouble another for what you can do yourself. Never spend your money before you have it. Never buy what you do not want because it is cheap. Pride costs us more than hunger, thirst, and cold. We seldom repent having eaten too little. Nothing is troublesome that we do willingly. How much pain the evils have cost us that have never happened! Take things always by the smooth handle. When angry, count ten before you speak; if very angry, a hundred.* —Thomas Jefferson
-
-# Enhancing Your Career
-
-# **Career in Education**
-
-While two thirds of all engineers work in private industry, some work in academia and prepare students for engineering careers. The course on circuit analysis you are studying is an important part of the preparation process. If you enjoy teaching others, you may want to consider becoming an engineering educator.
-
-Engineering professors w ork on state-of-the-art research projects, teach courses at graduate and undergraduate levels, and provide services to their professional societies and the community at lar ge. They are expected to make original contrib utions in their areas of specialty. This requires a broad-based education in the fundamentals of electrical engineering and a mastery of the skills necessary for communicating their efforts to others.
-
-If you lik e to do research, to w ork at the frontiers of engineering, to make contributions to technological adv ancement, to invent, consult, and/ or teach, consider a career in engineering education. The best way to start is by talking with your professors and benefiting from their experience.
-
-A solid understanding of mathematics and physics at the undergraduate level is vital to your success as an engineering professor . If you are having difficulty in solving your engineering textbook problems, start correcting any weaknesses you have in your mathematics and physics fundamentals.
-
-Most universities these days require that engineering professors have a doctor's de gree. In addition, some uni versities require that the y be actively involved in research leading to publications in reputable journals. To prepare yourself for a career in engineering education, get as broad an education as possible, because electrical engineering is changing rapidly
-
-Photo by James Watson
-
-<span id="page-874-0"></span>and becoming interdisciplinary. Without doubt, engineering education is a rewarding career. Professors get a sense of satisf action and fulfillment as they see their students graduate, become leaders in their profession, and contribute significantly to the betterment of humanity.
-
-# Learning Objectives
-
-*By using the information and exercises in this chapter you will be able to:*
-
-- 1. Understand the variety of two-port parameters that make analyzing circuits easier.
-- 2. Understand impedance parameters and ho w to use them ef fectively in analyzing certain classes of circuit analysis problems.
-- 3. Understand admittance parameters and how to use them effectively in analyzing certain classes of circuit analysis problems.
-- 4. Understand hybrid parameters and how to use them effectively in analyzing certain classes of circuit analysis problems.
-- 5. Understand transmission parameters and how to use them effectively in analyzing certain classes of circuit analysis problems.
-- 6. Understand the relationships between all two-port parameters.
-- 7. Understand how to interconnect networks using the characteristics of the variety of parametric relationships.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/217_19.1 Introduction.md

@@ -1,59 +0,0 @@
-# **19.1** Introduction
-
-A pair of terminals through which a current may enter or leave a network is known as a *port*. Two-terminal devices or elements (such as resis tors, capacitors, and inductors) result in one-port networks. Most of the circuits we have dealt with so f ar are two-terminal or one-port circuits, represented in Fig. 19.1(a). We have considered the v oltage across or current through a single pair of terminals—such as the tw o terminals of a resistor, a capacitor, or an inductor. We have also studied four-terminal or two-port circuits involving op amps, transistors, and transformers, as shown in Fig. 19.1(b). In general, a network may have *n* ports. A port is an access to the netw ork and consists of a pair of terminals; the current entering one terminal lea ves through the other terminal so that the net current entering the port equals zero.
-
-In this chapter, we are mainly concerned with *two-port* networks (or, simply, *two-ports*).
-
-A two-port network is an electrical network with two separate ports for input and output.
-
-Thus, a two-port network has two terminal pairs acting as access points. As shown in Fig. 19.1(b), the current entering one terminal of a pair leaves the other terminal in the pair. Three-terminal devices such as transistors can be configured into two-port networks.
-
-Our study of tw o-port networks is for at least tw o reasons. First, such netw orks are useful in communications, control systems, po wer systems, and electronics. F or example, they are used in electronics to model transistors and to facilitate cascaded design. Second, knowing the parameters of a two-port network enables us to treat it as a "black box" when embedded within a larger network.
-
-**Figure 19.1** (a) One-port network, (b) two-port network.
-
-<span id="page-875-0"></span>To characterize a tw o-port network requires that we relate the ter minal quantities **V**1, **V**2, **I**1, and **I**2 in Fig. 19.1(b), out of which tw o are independent. The various terms that relate these v oltages and currents are called *parameters*. Our goal in this chapter is to deri ve six sets of these parameters. We will sho w the relationship between these param eters and how two-port networks can be connected in series, parallel, or cascade. As with op amps, we are only interested in the terminal behavior of the circuits. And we will assume that the two-port circuits contain no independent sources, although the y can contain dependent sources. Finally, we will apply some of the concepts developed in this chapter to the analysis of transistor circuits and synthesis of ladder networks.
-
-# **19.2** Impedance Parameters
-
-Impedance and admittance parameters are commonly used in the synthesis of filters. They are also useful in the design and analysis of impedance-matching networks and power distribution networks. We discuss impedance parameters in this section and admittance parameters in the next section.
-
-A tw o-port netw ork may be v oltage-driven as in Fig. 19.2(a) or current-driven as in Fig. 19.2(b). From either Fig. 19.2(a) or (b), the terminal voltages can be related to the terminal currents as
-
-$$
-V_1 = z_{11}I_1 + z_{12}I_2
-$$
-  
-\n
-$$
-V_2 = z_{21}I_1 + z_{22}I_2
-$$
- (19.1)
-
-Reminder: Only two of the four variables (**V**1, **V**2, **I**1, and **I**2) are independent. The other two can be found using Eq. (19.1).
-
-or in matrix form as
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{z}_{11} & \mathbf{z}_{12} \\ \mathbf{z}_{21} & \mathbf{z}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = [\mathbf{z}] \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}
-$$
-(19.2)
-
-where the **z** terms are called the *impedance par ameters,* or simply *z parameters,* and have units of ohms.
-
-The values of the parameters can be e valuated by setting **I**1 = 0 (input port open-circuited) or **I**2 = 0 (output port open-circuited). Thus,
-
-The linear two-port network: (a) driven by voltage sources, (b) driven by current sources.
-
-# **Figure 19.3**
-
-Determination of the *z* parameters: (a) finding **z**11 and **z**21, (b) finding **z**12 and **z**22.
-
-# **Figure 19.4**
-
-Interchanging a voltage source at one port with an ideal ammeter at the other port produces the same reading in a reciprocal two-port.
-
-Because the *z* parameters are obtained by open-circuiting the input or output port, they are also called the *open-circuit impedance parameters.*  Specifically,
-
-- **z**11 = Open-circuit input impedance **z**12 = Open-circuit transfer impedance from port 1 to port 2 **<sup>z</sup>**21 = Open-circuit transfer impedance from port 2 to port 1 **(19.4)**
-- **z**22 = Open-circuit output impedance

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/218_19.2 Impedance Parameters.md

@@ -1,186 +0,0 @@
-
-According to Eq. (19.3), we obtain **z**11 and **z**21 by connecting a voltage **V**1 (or a current source **I**1) to port 1 with port 2 open-circuited as in Fig. 19.3(a) and finding **I**1 and **V**2; we then get
-
-$$
-z_{11} = \frac{V_1}{I_1}, \qquad z_{21} = \frac{V_2}{I_1}
-$$
- (19.5)
-
-Similarly, we obtain **z**12 and **z**22 by connecting a voltage **V**2 (or a current source **I**2) to port 2 with port 1 open-circuited as in Fig. 19.3(b) and finding **I**2 and **V**1; we then get
-
-$$
-z_{12} = \frac{V_1}{I_2}, \qquad z_{22} = \frac{V_2}{I_2}
-$$
- (19.6)
-
-The above procedure pro vides us with a means of calculating or mea suring the *z* parameters.
-
-Sometimes **z**11 and **z**22 are called *driving-point impedances,* while **z**<sup>21</sup> and **z**12 are called *transfer impedances.* A driving-point impedance is the input impedance of a two-terminal (one-port) device. Thus, **z**11 is the input driving-point impedance with the output port open-circuited, while **z**22 is the output driving-point impedance with the input port open-circuited.
-
-When **z**11 = **z**22, the two-port network is said to be *symmetrical.* This implies that the network has mirrorlike symmetry about some center line; that is, a line can be found that divides the network into two similar halves.
-
-When the two-port network is linear and has no dependent sources, the transfer impedances are equal (**z**12 = **z**21), and the two-port is said to be *reciprocal.* This means that if the points of excitation and response are interchanged, the transfer impedances remain the same. As illustrated in Fig. 19.4, a tw o-port is reciprocal if interchanging an ideal v oltage source at one port with an ideal ammeter at the other port gives the same ammeter reading. The reciprocal netw ork yields **V** = **z**12**I** according to Eq. (19.1) when connected as in Fig. 19.4(a), but yields **V** = **z**21**I** when connected as in Fig. 19.4(b). This is possible only if **z**12 = **z**21. Any twoport that is made entirely of resistors, capacitors, and inductors must be reciprocal. A reciprocal network can be replaced by the T-equivalent circuit in Fig. 19.5(a). If the netw ork is not reciprocal, a more general equivalent network is shown in Fig. 19.5(b); notice that this figure follows directly from Eq. (19.1).
-
-It should be mentioned that for some tw o-port netw orks, the *z* parameters do not exist because they cannot be described by Eq. (19.1). As an example, consider the ideal transformer of Fig. 19.6. The defining equations for the two-port network are:
-
-$$
-V_1 = \frac{1}{n} V_2, \qquad I_1 = -nI_2 \tag{19.7}
-$$
-
-Observe that it is impossible to express the voltages in terms of the currents, and vice versa, as Eq. (19.1) requires. Thus, the ideal transformer has no *z* parameters. Ho wever, it does ha ve hybrid parameters, as we shall see in Section 19.4.
-
-Determine the *z* parameters for the circuit in Fig. 19.7. Example 19.1
-
-# **Solution:**
-
-■ **METHOD 1** To determine **z**11 and **z**21, we apply a v oltage source **V**1 to the input port and lea ve the output port open as in Fig. 19.8(a). Then,
-
-$$
-\mathbf{z}_{11} = \frac{\mathbf{V}_1}{\mathbf{I}_1} = \frac{(20 + 40)\mathbf{I}_1}{\mathbf{I}_1} = 60 \ \Omega
-$$
-
-that is, **z**11 is the input impedance at port 1.
-
-$$
-z_{21} = \frac{V_2}{I_1} = \frac{40I_1}{I_1} = 40 \Omega
-$$
-
-To find **z**12 and **z**22, we apply a voltage source **V**2 to the output port and leave the input port open as in Fig. 19.8(b). Then,
-
-$$
-\mathbf{z}_{12} = \frac{\mathbf{V}_1}{\mathbf{I}_2} = \frac{40\mathbf{I}_2}{\mathbf{I}_2} = 40 \ \Omega, \qquad \mathbf{z}_{22} = \frac{\mathbf{V}_2}{\mathbf{I}_2} = \frac{(30 + 40)\mathbf{I}_2}{\mathbf{I}_2} = 70 \ \Omega
-$$
-
-Thus,
-
-$$
-[\mathbf{z}] = \begin{bmatrix} 60 \,\Omega & 40 \,\Omega \\ 40 \,\Omega & 70 \,\Omega \end{bmatrix}
-$$
-
-■ **METHOD 2** Alternatively, as there is no dependent source in the given circuit, **z**12 = **z**21 and we can use Fig. 19.5(a). Comparing Fig. 19.7 with Fig. 19.5(a), we get
-
-$$
-\mathbf{z}_{12} = 40 \ \Omega = \mathbf{z}_{21}
-$$
-\n
-$$
-\mathbf{z}_{11} - \mathbf{z}_{12} = 20 \qquad \Rightarrow \qquad \mathbf{z}_{11} = 20 + \mathbf{z}_{12} = 60 \ \Omega
-$$
-\n
-$$
-\mathbf{z}_{22} - \mathbf{z}_{12} = 30 \qquad \Rightarrow \qquad \mathbf{z}_{22} = 30 + \mathbf{z}_{12} = 70 \ \Omega
-$$
-
-Find the *z* parameters of the two-port network in Fig. 19.9. Practice Problem 19.1
-
-**Answer: z**11 = 12 Ω, **z**12 = **z**21 = **z**22 = 4 Ω.
-
-# **Figure 19.6**
-
-An ideal transformer has no *z* parameters.
-
-**Figure 19.7** For Example 19.1.
-
-# **Figure 19.8**
-
-For Example 19.1: (a) finding **z**11 and **z**21, (b) finding **z**12 and **z**22.
-
-**Figure 19.9** For Practice Prob. 19.1.
-
-# **Solution:**
-
-This is not a reciprocal netw ork. We may use the equi valent circuit in Fig. 19.5(b) b ut we can also use Eq. (19.1) directly . Substituting the given *z* parameters into Eq. (19.1),
-
-$$
-V_1 = 40I_1 + j20I_2 \tag{19.2.1}
-$$
-
-$$
-V_2 = j30I_1 + 50I_2 \tag{19.2.2}
-$$
-
-Because we are looking for **I**1 and **I**2, we substitute
-
-$$
-V_1 = 100/0^\circ
-$$
-,  $V_2 = -10I_2$ 
-
-into Eqs. (19.2.1) and (19.2.2), which become
-
-$$
-100 = 40I_1 + j20I_2 \tag{19.2.3}
-$$
-
-$$
--10I_2 = j30I_1 + 50I_2 \qquad \Rightarrow \qquad I_1 = j2I_2 \tag{19.2.4}
-$$
-
-Substituting Eq. (19.2.4) into Eq. (19.2.3) gives
-
-$$
-100 = j80I_2 + j20I_2 \qquad \Rightarrow \qquad I_2 = \frac{100}{j100} = -j
-$$
-
-From Eq. (19.2.4), **I**1 = *j*2(−*j*) = 2. Thus,
-
-$$
-I_1 = 2/0^\circ A
-$$
-,  $I_2 = 1/-90^\circ A$ 
-
-# Practice Problem 19.2 Calculate **I**1 and **I**2 in the two-port of Fig. 19.11.
-
-**Figure 19.11** For Practice Prob. 19.2.
-
-**Answer:** 800⧸30° mA, 400⧸120° mA.
-
-# <span id="page-879-0"></span>**19.3** Admittance Parameters
-
-In the previous section we saw that impedance parameters may not exist for a tw o-port network. So there is a need for an alternati ve means of describing such a netw ork. This need may be met by the second set of parameters, which we obtain by expressing the terminal currents in terms of the terminal voltages. In either Fig. 19.12(a) or (b), the terminal cur rents can be expressed in terms of the terminal voltages as
-
-$$
-\mathbf{I}_1 = \mathbf{y}_{11}\mathbf{V}_1 + \mathbf{y}_{12}\mathbf{V}_2 \n\mathbf{I}_2 = \mathbf{y}_{21}\mathbf{V}_1 + \mathbf{y}_{22}\mathbf{V}_2
-$$
-\n(19.8)
-
-or in matrix form as
-
-$$
-\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{y}_{11} & \mathbf{y}_{12} \\ \mathbf{y}_{21} & \mathbf{y}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = [\mathbf{y}] \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix}
-$$
-(19.9)
-
-The **y** terms are kno wn as the *admittance par ameters* (or , simply , *y parameters*) and have units of siemens.
-
-The values of the parameters can be determined by setting **V**1 = 0 (input port short-circuited) or **V**2 = 0 (output port short-circuited). Thus,
-
-$$
-\mathbf{y}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} \Big|_{\mathbf{V}_2=0}, \quad \mathbf{y}_{12} = \frac{\mathbf{I}_1}{\mathbf{V}_2} \Big|_{\mathbf{V}_1=0}
-$$
-\n
-$$
-\mathbf{y}_{21} = \frac{\mathbf{I}_2}{\mathbf{V}_1} \Big|_{\mathbf{V}_2=0}, \quad \mathbf{y}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} \Big|_{\mathbf{V}_1=0}
-$$
-\n(19.10)
-
-Because the *y* parameters are obtained by short-circuiting the input or output port, they are also called the *short-circuit admittance parameters.*  Specifically,
-
-- **y**11 = Short-circuit input admittance
-- **y**12 = Short-circuit transfer admittance from port 2 to port 1
-- **y**21 = Short-circuit transfer admittance from port 1 to port 2 **(19.11)**
-- **y**22 = Short-circuit output admittance
-
-Following Eq. (19.10), we obtain **y**11 and **y**21 by connecting a current **I**1 to port 1 and short-circuiting port 2 as in Fig. 19.12(a), finding **V**1 and **I**2, and then calculating
-
-$$
-\mathbf{y}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1}, \qquad \mathbf{y}_{21} = \frac{\mathbf{I}_2}{\mathbf{V}_1}
-$$
-(19.12)
-
-Similarly, we obtain **y**12 and **y**22 by connecting a current source **I**2 to port 2 and short-circuiting port 1 as in Fig. 19.12(b), finding **I**1 and **V**2, and then getting
-
-$$
-y_{12} = \frac{I_1}{V_2}, \qquad y_{22} = \frac{I_2}{V_2}
-$$
- (19.13)

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/219_19.3 Admittance Parameters.md

@@ -1,154 +0,0 @@
-
-This procedure provides us with a means of calculating or measuring the *y* parameters. The impedance and admittance parameters are collectively referred to as *immittance* parameters.
-
-# **Figure 19.12**
-
-Determination of the *y* parameters: (a) finding *y*11 and *y*21, (b) finding *y*12 and *y*22.
-
-For a two-port network that is linear and has no dependent sources, the transfer admittances are equal ( **y**12 = **y**21). This can be proved in the same way as for the *z* parameters. A reciprocal network (**y**12 = **y**21) can be modeled by the Π-equivalent circuit in Fig. 19.13(a). If the network is not reciprocal, a more general equivalent network is shown in Fig. 19.13(b).
-
-**Figure 19.13**
-
-(a) Π-equivalent circuit (for reciprocal case only), (b) general equivalent circuit.
-
-**Figure 19.14** For Example 19.3.
-
-Example 19.3 Obtain the *y* parameters for the Π network shown in Fig. 19.14.
-
-# **Solution:**
-
-■ **METHOD 1** To find **y**11 and **y**21, short-circuit the output port and connect a current source **I**1 to the input port as in Fig. 19.15(a). Because the 8-Ω resistor is short-circuited, the 2-Ω resistor is in parallel with the 4-Ω resistor. Hence,
-
-$$
-\mathbf{V}_1 = \mathbf{I}_1(4 \parallel 2) = \frac{4}{3}\mathbf{I}_1
-$$
-,  $\mathbf{y}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} = \frac{\mathbf{I}_1}{\frac{4}{3}\mathbf{I}_1} = 0.75 \text{ S}$ 
-
-By current division,
-
-$$
--\mathbf{I}_2 = \frac{4}{4+2}\mathbf{I}_1 = \frac{2}{3}\mathbf{I}_1, \qquad \mathbf{y}_{21} = \frac{\mathbf{I}_2}{\mathbf{V}_1} = \frac{-\frac{2}{3}\mathbf{I}_1}{\frac{4}{3}\mathbf{I}_1} = -0.5 \text{ S}
-$$
-
-(a)
-
-To get **y**12 and **y**22, short-circuit the input port and connect a current source **I**2 to the output port as in Fig. 19.15(b). The 4-Ω resistor is shortcircuited so that the 2- and 8-Ω resistors are in parallel.
-
-$$
-\mathbf{V}_2 = \mathbf{I}_2(8 \parallel 2) = \frac{8}{5}\mathbf{I}_2, \qquad \mathbf{y}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} = \frac{\mathbf{I}_2}{\frac{8}{5}\mathbf{I}_2} = \frac{5}{8} = 0.625 \text{ S}
-$$
-
-By current division,
-
-$$
--\mathbf{I}_1 = \frac{8}{8+2} \mathbf{I}_2 = \frac{4}{5} \mathbf{I}_2, \qquad \mathbf{y}_{12} = \frac{\mathbf{I}_1}{\mathbf{V}_2} = \frac{-\frac{4}{5} \mathbf{I}_2}{\frac{8}{5} \mathbf{I}_2} = -0.5 \text{ S}
-$$
-
-as obtained previously.
-
-**Figure 19.15**
-
-For Example 19.3: (a) finding **y**11 and **y**21, (b) finding **y**12 and **y**22.
-
-**Figure 19.16** For Practice Prob. 19.3.
-
-Determine the *y* parameters for the two-port shown in Fig. 19.17. Example 19.4
-
-# **Solution:**
-
-We follow the same procedure as in the previous example. To get **y**11 and **y**21, we use the circuit in Fig. 19.18(a), in which port 2 is short-circuited and a current source is applied to port 1. At node 1,
-
-$$
-\frac{\mathbf{V}_1 - \mathbf{V}_o}{8} = 2\mathbf{I}_1 + \frac{\mathbf{V}_o}{2} + \frac{\mathbf{V}_o - 0}{4}
-$$
-
-But **I**1 = **V**<sup>1</sup> \_\_\_\_\_\_\_ − **V***<sup>o</sup>* 8 ; therefore,
-
-$$
-0 = \frac{V_1 - V_o}{8} + \frac{3V_o}{4}
-$$
-
-$$
-0 = \mathbf{V}_1 - \mathbf{V}_o + 6\mathbf{V}_o \qquad \Rightarrow \qquad \mathbf{V}_1 = -5\mathbf{V}_o
-$$
-
-2 Ω
-
-Solution of Example 19.4: (a) finding **y**11 and **y**21, (b) finding **y**12 and **y**22.
-
-Hence,
-
-$$
-\mathbf{I}_1 = \frac{-5\mathbf{V}_o - \mathbf{V}_o}{8} = -0.75\mathbf{V}_o
-$$
-
-and
-
-$$
-\mathbf{y}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} = \frac{-0.75 \mathbf{V}_o}{-5 \mathbf{V}_o} = 0.15 \text{ S}
-$$
-
-At node 2,
-
-$$
-\frac{\mathbf{V}_o - 0}{4} + 2\mathbf{I}_1 + \mathbf{I}_2 = 0
-$$
-
-$$
--I_2 = 0.25V_o - 1.5V_o = -1.25V_o
-$$
-
-<span id="page-882-0"></span>Hence,
-
-$$
-\mathbf{y}_{21} = \frac{\mathbf{I}_2}{\mathbf{V}_1} = \frac{1.25\mathbf{V}_o}{-5\mathbf{V}_o} = -0.25 \text{ S}
-$$
-
-Similarly, we get **y**12 and **y**22 using Fig. 19.18(b). At node 1,
-
-$$
-\frac{0 - \mathbf{V}_o}{8} = 2\mathbf{I}_1 + \frac{\mathbf{V}_o}{2} + \frac{\mathbf{V}_o - \mathbf{V}_2}{4}
-$$
-  
-But  $\mathbf{I}_1 = \frac{0 - \mathbf{V}_o}{8}$ ; therefore,  
-$$
-0 = -\frac{\mathbf{V}_o}{8} + \frac{\mathbf{V}_o}{2} + \frac{\mathbf{V}_o - \mathbf{V}_2}{4}
-$$
-
-or
-
-$$
-0 = -\mathbf{V}_o + 4\mathbf{V}_o + 2\mathbf{V}_o - 2\mathbf{V}_2 \qquad \Rightarrow \qquad \mathbf{V}_2 = 2.5\mathbf{V}_o
-$$
-
-Hence,
-
-$$
-\mathbf{y}_{12} = \frac{\mathbf{I}_1}{\mathbf{V}_2} = \frac{-\mathbf{V}_o/8}{2.5\mathbf{V}_o} = -0.05 \text{ S}
-$$
-
-At node 2,
-
-$$
-\frac{\mathbf{V}_o - \mathbf{V}_2}{4} + 2\mathbf{I}_1 + \mathbf{I}_2 = 0
-$$
-
-$$
--\mathbf{I}_2 = 0.25\mathbf{V}_o - \frac{1}{4}(2.5\mathbf{V}_o) - \frac{2\mathbf{V}_o}{8} = -0.625\mathbf{V}_o
-$$
-
-Thus,
-
-or
-
-$$
-\mathbf{y}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} = \frac{0.625 \mathbf{V}_o}{2.5 \mathbf{V}_o} = 0.25 \text{ S}
-$$
-
-Notice that **y**<sup>12</sup> ≠ **y**21 in this case, given that the network is not reciprocal.
-
-# Practice Problem 19.4 Obtain the *y* parameters for the circuit in Fig. 19.19.
-
-**Figure 19.19** For Practice Prob. 19.4.
-
-**Answer: y**11 = 312.5 mS, **y**12 = −62.5 mS, **y**21 = 187.5 mS, **y**22 = 62.5 mS.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/220_19.4 Hybrid Parameters.md

@@ -1,355 +0,0 @@
-# **19.4** Hybrid Parameters
-
-The *z* and *y* parameters of a two-port network do not always exist. So there is a need for developing another set of parameters. This third set of parameters is based on making **V**1 and **I**2 the dependent variables. Thus, we obtain
-
-$$
-V_1 = h_{11}I_1 + h_{12}V_2
-$$
-  
-\n
-$$
-I_2 = h_{21}I_1 + h_{22}V_2
-$$
- (19.14)
-
-or in matrix form,
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{h}_{11} & \mathbf{h}_{12} \\ \mathbf{h}_{21} & \mathbf{h}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix} = [\mathbf{h}] \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix}
-$$
-(19.15)
-
-The **h** terms are known as the *hybrid parameters* (or, simply, *h parameters*) because they are a hybrid combination of ratios. They are very useful for describing electronic devices such as transistors (see Section 19.9); it is much easier to measure experimentally the *h* parameters of such devices than to measure their *z* or *y* parameters. In f act, we ha ve seen that the ideal transformer in Fig. 19.6, described by Eq. (19.7), does not ha ve *z* parameters. The ideal transformer can be described by the hybrid parameters, because Eq. (19.7) conforms with Eq. (19.14).
-
-The values of the parameters are determined as
-
-$$
-\mathbf{h}_{11} = \frac{\mathbf{V}_1}{\mathbf{I}_1} \Big|_{\mathbf{V}_2 = 0}, \qquad \mathbf{h}_{12} = \frac{\mathbf{V}_1}{\mathbf{V}_2} \Big|_{\mathbf{I}_1 = 0}
-$$
-\n
-$$
-\mathbf{h}_{21} = \frac{\mathbf{I}_2}{\mathbf{I}_1} \Big|_{\mathbf{V}_2 = 0}, \qquad \mathbf{h}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} \Big|_{\mathbf{I}_1 = 0}
-$$
-\n(19.16)
-
-It is evident from Eq. (19.16) that the parameters **h**11, **h**12, **h**21, and **h**<sup>22</sup> represent an impedance, a voltage gain, a current gain, and an admittance, respectively. This is wh y they are called the h ybrid parameters. To be specific,
-
-$$
-\mathbf{h}_{11} = \text{Short-circuit input impedance}
-$$
-\n
-$$
-\mathbf{h}_{12} = \text{Open-circuit reverse voltage gain}
-$$
-\n
-$$
-\mathbf{h}_{21} = \text{Short-circuit forward current gain}
-$$
-\n
-$$
-\mathbf{h}_{22} = \text{Open-circuit output admittance}
-$$
-\n(19.17)
-
-The procedure for calculating the *h* parameters is similar to that used for the *z* or *y* parameters. We apply a v oltage or current source to the appropriate port, short-circuit or open-circuit the other port, depending on the parameter of interest, and perform re gular circuit analysis. F or reciprocal networks, **h**12 = −**h**21. This can be proved in the same way as we proved that **z**12 = **z**21. Figure 19.20 shows the hybrid model of a twoport network.
-
-A set of parameters closely related to the *h* parameters are the *g parameters* or *inverse hybrid parameters.* These are used to describe the terminal currents and voltages as
-
-$$
-I_1 = g_{11}V_1 + g_{12}I_2
-$$
-  
-\n
-$$
-V_2 = g_{21}V_1 + g_{22}I_2
-$$
-\n(19.18)
-
-**Figure 19.20** The *h*-parameter equivalent network of a two-port network.
-
-or
-
-[
-
-$$
-\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{g}_{11} & \mathbf{g}_{12} \\ \mathbf{g}_{21} & \mathbf{g}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix} = [g] \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix}
-$$
-(19.19)
-
-The values of the *g* parameters are determined as
-
-$$
-\mathbf{g}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} \Big|_{\mathbf{I}_2=0}, \qquad \mathbf{g}_{12} = \frac{\mathbf{I}_1}{\mathbf{I}_2} \Big|_{\mathbf{V}_1=0}
-$$
-\n
-$$
-\mathbf{g}_{21} = \frac{\mathbf{V}_2}{\mathbf{V}_1} \Big|_{\mathbf{I}_2=0}, \qquad \mathbf{g}_{22} = \frac{\mathbf{V}_2}{\mathbf{I}_2} \Big|_{\mathbf{V}_1=0}
-$$
-\n(19.20)
-
-Thus, the inverse hybrid parameters are specifically called
-
-**g**11 = Open-circuit input admittance **<sup>g</sup>**12 = Short-circuit reverse current gain **(19.21) g**21 = Open-circuit forward voltage gain **g**22 = Short-circuit output impedance
-
-Figure 19.21 shows the inverse hybrid model of a tw o-port network. The *g* parameters are frequently used to model field-effect transistors.
-
-Example 19.5 Find the hybrid parameters for the two-port network of Fig. 19.22.
-
-# **Solution:**
-
-To find **h**11 and **h**21, we short-circuit the output port and connect a current source **I**1 to the input port as shown in Fig. 19.23(a). From Fig. 19.23(a),
-
-$$
-\mathbf{V}_1 = \mathbf{I}_1(2 + 3 \parallel 6) = 4\mathbf{I}_1
-$$
-
-Hence,
-
-For Example 19.5.
-
-**Figure 19.23**
-
-**Figure 19.21**
-
-network.
-
-For Example 19.5: (a) computing **h**11 and
-
-**h**21, (b) computing **h**12 and **h**22.
-
-6 Ω
-
-2 Ω 3 Ω
-
-**h**11 = \_\_\_ **V**1 **I**1 = 4 Ω
-
-Also, from Fig. 19.23(a) we obtain, by current division,
-
-$$
--\mathbf{I}_2 = \frac{6}{6+3} \mathbf{I}_1 = \frac{2}{3} \mathbf{I}_1
-$$
-
-Hence,
-
-$$
-\mathbf{h}_{21} = \frac{\mathbf{I}_2}{\mathbf{I}_1} = -\frac{2}{3}
-$$
-
-To obtain **h**12 and **h**22, we open-circuit the input port and connect a voltage source **V**2 to the output port as in Fig. 19.23(b). By voltage division,
-
-$$
-\mathbf{V}_1 = \frac{6}{6+3} \mathbf{V}_2 = \frac{2}{3} \mathbf{V}_2
-$$
-
-Hence,
-
-$$
-\mathbf{h}_{12} = \frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{2}{3}
-$$
-
-Also,
-
-$$
-\mathbf{V}_2 = (3+6)\mathbf{I}_2 = 9\mathbf{I}_2
-$$
-
-The *g*-parameter model of a two-port
-
-Thus,
-
-$$
-\mathbf{h}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} = \frac{1}{9} S
-$$
-
-**Answer:** 
-$$
-\mathbf{h}_{11} = 2.4 \, \Omega
-$$
-,  $\mathbf{h}_{12} = 0.4$ ,  $\mathbf{h}_{21} = -0.4$ ,  $\mathbf{h}_{22} = 200 \, \text{mS}$ .
-
-Determine the Thevenin equivalent at the output port of the circuit in Example 19.6 Fig. 19.25.
-
-# **Solution:**
-
-To find **Z**Th and **V**Th, we apply the normal procedure, keeping in mind the formulas relating the input and output ports of the *h* model. To obtain **Z**Th, remove the 60-V voltage source at the input port and apply a 1-V voltage source at the output port, as shown in Fig. 19.26(a). From Eq. (19.14),
-
-$$
-V_1 = h_{11}I_1 + h_{12}V_2
-$$
- (19.6.1)  
-$$
-I_2 = h_{21}I_1 + h_{22}V_2
-$$
- (19.6.2)
-
-But 
-$$
-V_2 = 1
-$$
-, and  $V_1 = -40I_1$ . Substituting these into Eqs. (19.6.1) and
-
-(19.6.2), we get
-
-$$
--40I_1 = h_{11}I_1 + h_{12} \Rightarrow I_1 = -\frac{h_{12}}{40 + h_{11}}
-$$
-(19.6.3)  
-$$
-I_2 = h_{21}I_1 + h_{22}
-$$
-(19.6.4)
-
-Substituting Eq. (19.6.3) into Eq. (19.6.4) gives
-
-$$
-\mathbf{I}_2 = \mathbf{h}_{21}\mathbf{I}_1 + \mathbf{h}_{22}
-$$
-(19.6.3) into Eq. (19.6.4) gives  
-$$
-\mathbf{I}_2 = \mathbf{h}_{22} - \frac{\mathbf{h}_{21}\mathbf{h}_{12}}{\mathbf{h}_{11} + 40} = \frac{\mathbf{h}_{11}\mathbf{h}_{22} - \mathbf{h}_{21}\mathbf{h}_{12} + \mathbf{h}_{22}40}{\mathbf{h}_{11} + 40}
-$$
-
-Therefore,
-
-$$
-\mathbf{L}_2 = \mathbf{h}_{22} - \frac{\mathbf{L}_1 \cdot \mathbf{L}_2}{\mathbf{h}_{11} + 40} = \frac{1.72 \cdot \mathbf{L}_1 \cdot \mathbf{L}_2}{\mathbf{h}_{11} + 40}
-$$
-$$
-\mathbf{Z}_{\text{Th}} = \frac{\mathbf{V}_2}{\mathbf{I}_2} = \frac{1}{\mathbf{I}_2} = \frac{\mathbf{h}_{11} + 40}{\mathbf{h}_{11}\mathbf{h}_{22} - \mathbf{h}_{21}\mathbf{h}_{12} + \mathbf{h}_{22}40}
-$$
-
-Substituting the values of the *h* parameters,
-
-Substituting the values of the *h* parameters,  
-\n
-$$
-\mathbf{Z}_{\text{Th}} = \frac{1000 + 40}{10^3 \times 200 \times 10^{-6} + 20 + 40 \times 200 \times 10^{-6}}
-$$
-\n
-$$
-= \frac{1040}{20.21} = 51.46 \ \Omega
-$$
-
-To get **V**Th, we find the open-circuit voltage **V**2 in Fig. 19.26(b). At the input port,
-
-$$
--60 + 40I_1 + V_1 = 0 \qquad \Rightarrow \qquad V_1 = 60 - 40I_1 \qquad (19.6.5)
-$$
-
-**Figure 19.26** For Example 19.6: (a) finding **Z**Th, (b) finding **V**Th.
-
-(b)
-
-# **Figure 19.25**
-
-For Example 19.6.
-
-At the output,
-
-$$
-\mathbf{I}_2 = 0 \tag{19.6.6}
-$$
-
-Substituting Eqs. (19.6.5) and (19.6.6) into Eqs. (19.6.1) and (19.6.2), we obtain
-
-$$
-60 - 40\mathbf{I}_1 = \mathbf{h}_{11}\mathbf{I}_1 + \mathbf{h}_{12}\mathbf{V}_2
-$$
-
-or
-
-$$
-60 = (\mathbf{h}_{11} + 40)\mathbf{I}_1 + \mathbf{h}_{12}\mathbf{V}_2
-$$
- (19.6.7)
-
-and
-
-$$
-0 = \mathbf{h}_{21}\mathbf{I}_1 + \mathbf{h}_{22}\mathbf{V}_2 \qquad \Rightarrow \qquad \mathbf{I}_1 = -\frac{\mathbf{h}_{22}}{\mathbf{h}_{21}}\mathbf{V}_2 \tag{19.6.8}
-$$
-
-Now substituting Eq. (19.6.8) into Eq. (19.6.7) gives
-
-$$
-60 = \left[ -(\mathbf{h}_{11} + 40) \frac{\mathbf{h}_{22}}{\mathbf{h}_{21}} + \mathbf{h}_{12} \right] \mathbf{V}_2
-$$
-
-or
-
-$$
-60 = \left[ -(\mathbf{h}_{11} + 40) \frac{22}{\mathbf{h}_{21}} + \mathbf{h}_{12} \right] \mathbf{V}_2
-$$
-$$
-\mathbf{V}_{\text{Th}} = \mathbf{V}_2 = \frac{60}{-(\mathbf{h}_{11} + 40)\mathbf{h}_{22}/\mathbf{h}_{21} + \mathbf{h}_{12}} = \frac{60\mathbf{h}_{21}}{\mathbf{h}_{12}\mathbf{h}_{21} - \mathbf{h}_{11}\mathbf{h}_{22} - 40\mathbf{h}_{22}}
-$$
-
-Substituting the values of the *h* parameters,
-
-$$
-V_{\text{Th}} = \frac{60 \times 10}{-20.21} = -29.69 \text{ V}
-$$
-
-1 Ω 1 H 1 F
-
-**Figure 19.28** For Example 19.7.
-
-Example 19.7 Find the *g* parameters as functions of *s* for the circuit in Fig. 19.28.
-
-# **Solution:**
-
-In the *s* domain,
-
-$$
-1 \text{ H} \Rightarrow sL = s, \quad 1 \text{ F} \Rightarrow \frac{1}{sC} = \frac{1}{s}
-$$
-
-To get **g**11 and **g**21, we open-circuit the output port and connect a voltage source **V**1 to the input port as in Fig. 19.29(a). From the figure,
-
-$$
-\mathbf{I}_1 = \frac{\mathbf{V}_1}{s+1}
-$$
-
-$$
-\mathbf{g}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} = \frac{1}{s+1}
-$$
-
-<span id="page-887-0"></span>By voltage division,
-
-**<sup>V</sup>**2 = \_\_\_\_\_ <sup>1</sup> *s* + 1 **V**<sup>1</sup>
-
-or
-
-$$
-\mathbf{g}_{21} = \frac{\mathbf{V}_2}{\mathbf{V}_1} = \frac{1}{s+1}
-$$
-
-To obtain **g**12 and **g**22, we short-circuit the input port and connect a current source **I**2 to the output port as in Fig. 19.29(b). By current division,
-
-$$
-\mathbf{I}_1 = -\frac{1}{s+1} \mathbf{I}_2
-$$
-
-**g**12 = \_\_ **I**1 **I**2 = − \_\_\_\_\_ <sup>1</sup> *s* + 1
-
-Also,
-
-or
-
-**V**2 = **I**2( \_\_1 *s* + *s* ‖ 1)
-
-or
-
-$$
-\mathbf{g}_{22} = \frac{\mathbf{V}_2}{\mathbf{I}_2} = \frac{1}{s} + \frac{s}{s+1} = \frac{s^2 + s + 1}{s(s+1)}
-$$
-
-Thus,
-
-$$
-[\mathbf{g}] = \begin{bmatrix} \frac{1}{s+1} & -\frac{1}{s+1} \\ \frac{1}{s+1} & \frac{s^2+s+1}{s(s+1)} \end{bmatrix}
-$$
-
-For the ladder network in Fig. 19.30, determine the *g* parameters in the *s* domain.
-
-## **Answer:** [**g**]<sup>=</sup>[ \_\_\_\_\_\_\_\_\_\_ *s* + 2 *s* 2 + 3*s* + 1 <sup>−</sup>\_\_\_\_\_\_\_\_\_\_ <sup>1</sup> *s* 2 + 3*s* + 1 \_\_\_\_\_\_\_\_\_\_ 1 *s* 2 + 3*s* + 1 *<sup>s</sup>*(*s* + 2) \_\_\_\_\_\_\_\_\_\_ *s* 2 + 3*s* + 1 ].

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/221_19.5 Transmission Parameters.md

@@ -1,319 +0,0 @@
-# **19.5** Transmission Parameters
-
-Because there are no restrictions on which terminal voltages and currents should be considered independent and which should be dependent v ariables, we expect to be able to generate many sets of parameters. Another
-
-**V**<sup>1</sup> **V**<sup>2</sup>
-
-(a)
-
-1/s s
-
-**I**1
-
-+
-
-+ ‒ **I I**<sup>1</sup> <sup>2</sup> = 0
-
-1/s s
-
-1 Ω
-
-+
-
-Determining the *g* parameters in the *s* domain for the circuit in Fig. 19.28.
-
-Practice Problem 19.7
-
-For Practice Prob. 19.7.
-
-+
-
-‒
-
-set of parameters relates the variables at the input port to those at the output port. Thus,
-
-$$
-\mathbf{V}_1 = \mathbf{A}\mathbf{V}_2 - \mathbf{B}\mathbf{I}_2
-$$
-  
-$$
-\mathbf{I}_1 = \mathbf{C}\mathbf{V}_2 - \mathbf{D}\mathbf{I}_2
-$$
- (19.22)
-
-[ **V**1 **<sup>I</sup>**<sup>1</sup> ] <sup>=</sup>[ **A B C D**] [ **V**2 −**I**<sup>2</sup> ] = [**T**] [ **V**2 −**I**<sup>2</sup> ] **(19.23)**
-
-Equations (19.22) and (19.23) relate the input variables (**V**1 and **I**1) to the output variables (**V**2 and −**I**2). Notice that in computing the transmission parameters, −**I**2 is used rather than **I**2, because the current is considered to be leaving the network, as shown in Fig. 19.31, as opposed to enter ing the network as in Fig. 19.1(b). This is done merely for conventional reasons; when you cascade two-ports (output to input), it is most logical to think of **I**2 as leaving the two-port. It is also customary in the power industry to consider **I**2 as leaving the two-port.
-
-The two-port parameters in Eqs. (19.22) and (19.23) provide a measure of how a circuit transmits v oltage and current from a source to a load. They are useful in the analysis of transmission lines (such as cable and fiber) because they e xpress sending-end v ariables ( **V**1 and **I**1) in terms of the receiving-end variables (**V**2 and −**I**2). For this reason, the y are called *transmission parameters.* They are also known as **ABCD** parameters. They are used in the design of telephone systems, micro wave networks, and radars.
-
-The transmission parameters are determined as
-
-$$
-\mathbf{A} = \frac{\mathbf{V}_1}{\mathbf{V}_2} \Big|_{\mathbf{I}_2 = 0}, \qquad \mathbf{B} = -\frac{\mathbf{V}_1}{\mathbf{I}_2} \Big|_{\mathbf{V}_2 = 0}
-$$
-\n
-$$
-\mathbf{C} = \frac{\mathbf{I}_1}{\mathbf{V}_2} \Big|_{\mathbf{I}_2 = 0}, \qquad \mathbf{D} = -\frac{\mathbf{I}_1}{\mathbf{I}_2} \Big|_{\mathbf{V}_2 = 0}
-$$
-\n(19.24)
-
-Thus, the transmission parameters are called, specifically,
-
-**A** = Open-circuit voltage ratio **B** = Negative short-circuit transfer impedance **C** = Open-circuit transfer admittance **(19.25) D** = Negative short-circuit current ratio
-
-**A** and **D** are dimensionless, **B** is in ohms, and **C** is in siemens. Because the transmission parameters provide a direct relationship between input and output variables, they are very useful in cascaded networks.
-
-Our last set of parameters may be defined by expressing the variables at the output port in terms of the variables at the input port. We obtain
-
-$$
-\mathbf{V}_2 = \mathbf{a}\mathbf{V}_1 - \mathbf{b}\mathbf{I}_1
-$$
-  
-$$
-\mathbf{I}_2 = \mathbf{c}\mathbf{V}_1 - \mathbf{d}\mathbf{I}_1
-$$
- (19.26)
-
-**Figure 19.31** Terminal variables used to define the **ADCB** parameters.
-
-$$
-\begin{bmatrix} \mathbf{V}_2 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{a} & \mathbf{b} \\ \mathbf{c} & \mathbf{d} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ -\mathbf{I}_1 \end{bmatrix} = [\mathbf{t}] \begin{bmatrix} \mathbf{V}_1 \\ -\mathbf{I}_1 \end{bmatrix}
-$$
-(19.27)
-
-The parameters **a**, **b**, **c**, and **d** are called the *inverse transmission,* or *t, parameters.* They are determined as follows:
-
-$$
-\mathbf{a} = \frac{\mathbf{V}_2}{\mathbf{V}_1} \bigg|_{\mathbf{I}_1 = 0}, \qquad \mathbf{b} = -\frac{\mathbf{V}_2}{\mathbf{I}_1} \bigg|_{\mathbf{V}_1 = 0}
-$$
-\n
-$$
-\mathbf{c} = \frac{\mathbf{I}_2}{\mathbf{V}_1} \bigg|_{\mathbf{I}_1 = 0}, \qquad \mathbf{d} = -\frac{\mathbf{I}_2}{\mathbf{I}_1} \bigg|_{\mathbf{V}_1 = 0}
-$$
-\n(19.28)
-
-From Eq. (19.28) and from our experience so far, it is evident that these parameters are known individually as
-
-> **a** = Open-circuit voltage gain **<sup>b</sup>** = Negative short-circuit transfer impedance **(19.29) c** = Open-circuit transfer admittance **d** = Negative short-circuit current gain
-
-While **a** and **d** are dimensionless, **b** and **c** are in ohms and siemens, respectively.
-
-In terms of the transmission or in verse transmission parameters, a network is reciprocal if
-
-AD – BC = 1, 
-$$
-ad - bc = 1
-$$
- (19.30)
-
-These relations can be pro ved in the same w ay as the transfer imped ance relations for the *z* parameters. Alternatively, we will be able to use Table 19.1 a little later to deri ve Eq. (19.30) from the f act that **z**12 = **z**<sup>21</sup> for reciprocal networks.
-
-Find the transmission parameters for the two-port network in Fig. 19.32. Example 19.8
-
-# **Solution:**
-
-To determine **A** and **C**, we leave the output port open as in Fig. 19.33(a) so that **I**2 = 0 and place a voltage source **V**1 at the input port. We have
-
-$$
-V_1 = (10 + 20)I_1 = 30I_1
-$$
- and  $V_2 = 20I_1 - 3I_1 = 17I_1$ 
-
-Thus,
-
-$$
-A = \frac{V_1}{V_2} = \frac{30I_1}{17I_1} = 1.765, \qquad C = \frac{I_1}{V_2} = \frac{I_1}{17I_1} = 0.0588 \text{ S}
-$$
-
-To obtain **B** and **D**, we short-circuit the output port so that **V**2 = 0 as shown in Fig. 19.33(b) and place a voltage source **V**1 at the input port. At node *a* in the circuit of Fig. 19.33(b), KCL gives
-
-$$
-\frac{\mathbf{V}_1 - \mathbf{V}_a}{10} - \frac{\mathbf{V}_a}{20} + \mathbf{I}_2 = 0
-$$
- (19.8.1)
-
-**Figure 19.33**
-
-For Example 19.8: (a) finding **A** and **C**, (b) finding **B** and **D**.
-
-But **V***a* = 3**I**1 and **I**1 = (**V**1 − **V***a*)∕10. Combining these gives
-
-$$
-V_a = 3I_1 \t V_1 = 13I_1 \t (19.8.2)
-$$
-
-Substituting **V***a* = 3**I**1 into Eq. (19.8.1) and replacing the first term with **I**1,
-
-$$
-\mathbf{I}_1 - \frac{3\mathbf{I}_1}{20} + \mathbf{I}_2 = 0 \qquad \Rightarrow \qquad \frac{17}{20}\mathbf{I}_1 = -\mathbf{I}_2
-$$
-
-Therefore,
-
-$$
-\mathbf{D} = -\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{20}{17} = 1.176, \qquad \mathbf{B} = -\frac{\mathbf{V}_1}{\mathbf{I}_2} = \frac{-13\mathbf{I}_1}{(-17/20)\mathbf{I}_1} = 15.29 \text{ }\Omega
-$$
-
-Practice Problem 19.8 Find the transmission parameters for the circuit in Fig. 19.16 (see Practice Prob. 19.3).
-
-**Answer: A** = 1.5, **B** = 11 Ω, **C** = 250 mS, **D** = 2.5.
-
-**Figure 19.34** For Example 19.9.
-
-Example 19.9 The **ABCD** parameters of the two-port network in Fig. 19.34 are
-
-| 4     | 20 Ω |
-|-------|------|
-| [     | ]    |
-| 0.1 S | 2    |
-
-The output port is connected to a v ariable load for maximum po wer transfer. Find *RL* and the maximum power transferred.
-
-# **Solution:**
-
-What we need is to find the Thevenin equivalent (**Z**Th and **V**Th) at the load or output port. We find **Z**Th using the circuit in Fig. 19.35(a). Our goal is to get **Z**Th = **V**2∕**I**2. Substituting the given **ABCD** parameters into Eq. (19.22), we obtain
-
-$$
-V_1 = 4V_2 - 20I_2 \tag{19.9.1}
-$$
-
-$$
-I_1 = 0.1V_2 - 2I_2 \tag{19.9.2}
-$$
-
-At the input port, **V**1 = −10**I**1. Substituting this into Eq. (19.9.1) gives
-
-$$
--10\mathbf{I}_1 = 4\mathbf{V}_2 - 20\mathbf{I}_2
-$$
-
-$$
-I_1 = -0.4V_2 + 2I_2 \tag{19.9.3}
-$$
-
-# **Figure 19.35**
-
-Solution of Example 19.9: (a) finding **Z**Th, (b) finding **V**Th, (c) finding *RL* for maximum power transfer.
-
-Setting the right-hand sides of Eqs. (19.9.2) and (19.9.3) equal,
-
-$$
-0.1\mathbf{V}_2 - 2\mathbf{I}_2 = -0.4\mathbf{V}_2 + 2\mathbf{I}_2 \implies 0.5\mathbf{V}_2 = 4\mathbf{I}_2
-$$
-
-Hence,
-
-$$
-Z_{\text{Th}} = \frac{V_2}{I_2} = \frac{4}{0.5} = 8 \ \Omega
-$$
-
-To find **V**Th, we use the circuit in Fig. 19.35(b). At the output port **I**2 = 0 and at the input port **V**1 = 50 − 10**I**1. Substituting these into Eqs. (19.9.1) and (19.9.2),
-
-$$
-50 - 10I_1 = 4V_2 \tag{19.9.4}
-$$
-
-$$
-\mathbf{I}_1 = 0.1 \mathbf{V}_2 \tag{19.9.5}
-$$
-
-Substituting Eq. (19.9.5) into Eq. (19.9.4),
-
-$$
-50 - V_2 = 4V_2 \qquad \Rightarrow \qquad V_2 = 10
-$$
-
-Thus,
-
-$$
-\mathbf{V}_{\mathrm{Th}} = \mathbf{V}_2 = 10 \mathrm{V}
-$$
-
-The equivalent circuit is shown in Fig. 19.35(c). For maximum power transfer,
-
-$$
-R_L = \mathbf{Z}_{\text{Th}} = 8 \ \Omega
-$$
-
-From Eq. (4.24), the maximum power is
-
-$$
-P = I^2 R_L = \left(\frac{\mathbf{V}_{\text{Th}}}{2R_L}\right)^2 R_L = \frac{\mathbf{V}_{\text{Th}}^2}{4R_L} = \frac{100}{4 \times 8} = 3.125 \text{ W}
-$$
-
-Find **I**1 and **I**2 if the transmission parameters for the two-port in Fig. 19.36 Practice Problem 19.9 are
-
-For Practice Prob. 19.9.
-
-**Answer:** 1 A, −0.2 A.
-
-## <span id="page-892-0"></span>**19.6** † Relationships Between Parameters
-
-Because the six sets of parameters relate the same input and output terminal variables of the same two-port network, they should be interrelated. If two sets of parameters exist, we can relate one set to the other set. Let us demonstrate the process with two examples.
-
-Given the *z* parameters, let us obtain the *y* parameters. From Eq. (19.2),
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{z}_{11} & \mathbf{z}_{12} \\ \mathbf{z}_{21} & \mathbf{z}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = [\mathbf{z}] \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}
-$$
-(19.31)
-
-or
-
-$$
-\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = [\mathbf{z}]^{-1} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix}
-$$
-(19.32)
-
-Also, from Eq. (19.9),
-
-$$
-\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{y}_{11} & \mathbf{y}_{12} \\ \mathbf{y}_{21} & \mathbf{y}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = [\mathbf{y}] \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix}
-$$
-(19.33)
-
-Comparing Eqs. (19.32) and (19.33), we see that
-
-[
-
-$$
-[y] = [z]^{-1}
-$$
- (19.34)
-
-The adjoint of the [**z**] matrix is
-
-$$
-\begin{bmatrix} \mathbf{z}_{22} & -\mathbf{z}_{12} \\ -\mathbf{z}_{21} & \mathbf{z}_{11} \end{bmatrix}
-$$
-
-and its determinant is
-
-$$
-\Delta_z = \mathbf{z}_{11}\mathbf{z}_{22} - \mathbf{z}_{12}\mathbf{z}_{21}
-$$
-
-Substituting these into Eq. (19.34), we get
-
-$$
-\begin{bmatrix} \mathbf{y}_{11} & \mathbf{y}_{12} \\ \mathbf{y}_{21} & \mathbf{y}_{22} \end{bmatrix} = \frac{\begin{bmatrix} \mathbf{z}_{22} & -\mathbf{z}_{12} \\ -\mathbf{z}_{21} & \mathbf{z}_{11} \end{bmatrix}}{\Delta_z}
-$$
-(19.35)
-
-Equating terms yields
-
-$$
-y_{11} = \frac{z_{22}}{\Delta_z}
-$$
-,  $y_{12} = -\frac{z_{12}}{\Delta_z}$ ,  $y_{21} = -\frac{z_{21}}{\Delta_z}$ ,  $y_{22} = \frac{z_{11}}{\Delta_z}$  (19.36)
-
-As a second example, let us determine the *h* parameters from the *z* parameters. From Eq. (19.1),
-
-$$
-\mathbf{V}_1 = \mathbf{z}_{11}\mathbf{I}_1 + \mathbf{z}_{12}\mathbf{I}_2 \tag{19.37a}
-$$
-
-$$
-V_2 = z_{21}I_1 + z_{22}I_2 \tag{19.37b}
-$$

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/222_19.6 Relationships Between Parameters.md

@@ -1,185 +0,0 @@
-Making **I**2 the subject of Eq. (19.37b),
-
-$$
-\mathbf{I}_2 = -\frac{\mathbf{z}_{21}}{\mathbf{z}_{22}}\mathbf{I}_1 + \frac{1}{\mathbf{z}_{22}}\mathbf{V}_2
-$$
- (19.38)
-
-Substituting this into Eq. (19.37a),
-
-Eq. (19.3/8),  
-\n
-$$
-\mathbf{V}_1 = \frac{\mathbf{z}_{11}\mathbf{z}_{22} - \mathbf{z}_{12}\mathbf{z}_{21}}{\mathbf{z}_{22}}\mathbf{I}_1 + \frac{\mathbf{z}_{12}}{\mathbf{z}_{22}}\mathbf{V}_2
-$$
-\n(19.39)
-
-Putting Eqs. (19.38) and (19.39) in matrix form,
-
-$$
-\begin{bmatrix}\nV_1 \\
-I_2\n\end{bmatrix} = \begin{bmatrix}\n\frac{\Delta_z}{\mathbf{z}_{22}} & \frac{\mathbf{z}_{12}}{\mathbf{z}_{22}} \\
--\frac{\mathbf{z}_{21}}{\mathbf{z}_{22}} & \frac{1}{\mathbf{z}_{22}}\n\end{bmatrix} \begin{bmatrix}\nI_1 \\
-\overline{V}_2\n\end{bmatrix}
-$$
-\n(19.40)\n  
-\n  
-\n  
-\n  
-\n  
-\n  
-\n  
-\n  
-\n  
-\n  
-\n  
-\n  
-\n  
-\n
-
-From Eq. (19.15),
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{h}_{11} & \mathbf{h}_{12} \\ \mathbf{h}_{21} & \mathbf{h}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix}
-$$
-
-Comparing this with Eq. (19.40), we obtain
-
-$$
-\mathbf{h}_{11} = \frac{\Delta_z}{\mathbf{z}_{22}}, \qquad \mathbf{h}_{12} = \frac{\mathbf{z}_{12}}{\mathbf{z}_{22}}, \qquad \mathbf{h}_{21} = -\frac{\mathbf{z}_{21}}{\mathbf{z}_{22}}, \qquad \mathbf{h}_{22} = \frac{1}{\mathbf{z}_{22}} \quad (19.41)
-$$
-
-Table 19.1 provides the conversion formulas for the six sets of twoport parameters. Given one set of parameters, Table 19.1 can be used to find other parameters. For example, given the *T* parameters, we find the corresponding *h* parameters in the fifth column of the third ro w. Also,
-
-## **TABLE 19.1**
-
-Conversion of two-port parameters.
-
-|   |                     | z                   |                     | y                   | h                   |                     | g                   |                     | T            |                  | t               |                |
-|---|---------------------|---------------------|---------------------|---------------------|---------------------|---------------------|---------------------|---------------------|--------------|------------------|-----------------|----------------|
-| z | z11                 | z12                 | y22<br>___<br>∆y    | y12<br>− ___<br>∆y  | ∆h<br>___<br>h22    | h12<br>___<br>h22   | ___1<br>g11         | g12<br>− ___<br>g11 | A<br>__<br>C | ∆T<br>___<br>C   | d<br>__<br>c    | __1<br>c       |
-|   | z21                 | z22                 | y21<br>− ___<br>∆y  | y11<br>___<br>∆y    | h21<br>− ___<br>h22 | ___1<br>h22         | g21<br>___<br>g11   | ∆g<br>___<br>g11    | __1<br>C     | D<br>__<br>C     | ∆t<br>__<br>c   | __a<br>c       |
-| y | z22<br>___<br>∆z    | z12<br>− ___<br>∆z  | y11                 | y12                 | ___1<br>h11         | h12<br>− ___<br>h11 | ∆g<br>___<br>g22    | g12<br>___<br>g22   | D<br>__<br>B | ∆T<br>− ___<br>B | __a<br>b        | − __1<br>b     |
-|   | z21<br>− ___<br>∆z  | z11<br>___<br>∆z    | y21                 | y22                 | h21<br>___<br>h11   | ∆h<br>___<br>h11    | g21<br>− ___<br>g22 | ___1<br>g22         | − __1<br>B   | A<br>__<br>B     | ∆t<br>− __<br>b | d<br>__<br>b   |
-| h | ∆z<br>___<br>z22    | z12<br>___<br>z22   | ___1<br>y11         | y12<br>− ___<br>y11 | h11                 | h12                 | g22<br>___<br>∆g    | g12<br>− ___<br>∆g  | B<br>__<br>D | ∆T<br>___<br>D   | b<br>__<br>a    | __1<br>a       |
-|   | z21<br>− ___<br>z22 | ___1<br>z22         | y21<br>___<br>y11   | ∆y<br>___<br>y11    | h21                 | h22                 | g21<br>− ___<br>∆g  | g11<br>___<br>∆g    | − __1<br>D   | C<br>__<br>D     | ∆t<br>__<br>a   | __c<br>a       |
-| g | ___1<br>z11         | z12<br>− ___<br>z11 | ∆y<br>___<br>y22    | y12<br>___<br>y22   | h22<br>___<br>∆h    | h12<br>− ___<br>∆h  | g11                 | g12                 | C<br>__<br>A | ∆T<br>− ___<br>A | __c<br>d        | − __1<br>d     |
-|   | z21<br>___<br>z11   | ∆z<br>___<br>z11    | y21<br>− ___<br>y22 | ___1<br>y22         | h21<br>− ___<br>∆h  | h11<br>___<br>∆h    | g21                 | g22                 | __1<br>A     | B<br>__<br>A     | ∆t<br>__<br>d   | b<br>− __<br>d |
-| T | z11<br>___<br>z21   | ∆z<br>___<br>z21    | y22<br>− ___<br>y21 | − ___1<br>y21       | ∆h<br>− ___<br>h21  | h11<br>− ___<br>h21 | ___1<br>g21         | g22<br>___<br>g21   | A            | B                | __d<br>∆t       | __b<br>∆t      |
-|   | ___1<br>z21         | z22<br>___<br>z21   | − ∆y<br>___<br>y21  | y11<br>− ___<br>y21 | h22<br>− ___<br>h21 | − ___1<br>h21       | g11<br>___<br>g21   | ∆g<br>___<br>g21    | C            | D                | __c<br>∆t       | __a<br>∆t      |
-| t | z22<br>___<br>z12   | ∆z<br>___<br>z12    | y11<br>− ___<br>y12 | − ___1<br>y12       | ___1<br>h12         | h11<br>___<br>h12   | − ∆g<br>___<br>g12  | g22<br>− ___<br>g12 | ___ D<br>∆T  | ___B<br>∆T       | a               | b              |
-|   | ___1<br>z12         | z11<br>___<br>z12   | − ∆y<br>___<br>y12  | y22<br>− ___<br>y12 | h22<br>___<br>h12   | ∆h<br>___<br>h12    | g11<br>− ___<br>g12 | − ___1<br>g12       | ___ C<br>∆T  | ___ A<br>∆T      | c               | d              |
-
-**∆***z* = **z**11**z**<sup>22</sup> − **z**12**z**21, **∆***h* = **h**11**h**<sup>22</sup> − **h**12**h**21, **∆***T* = **AD** − **BC**
-
-**∆***y* = **y**11**y**<sup>22</sup> − **y**12**y**21, **∆***g* = **g**11**g**<sup>22</sup> − **g**12**g**21, **∆***t* = **ad** − **bc**
-
-given that **z**21 = **z**12 for a reciprocal netw ork, we can use the table to express this condition in terms of other parameters. It can also be shown that
-
-$$
-[g] = [h]^{-1}
-$$
- (19.42)
-
-but
-
-$$
-[t] \neq [T]^{-1}
-$$
- (19.43)
-
-Example 19.10 Find [**z**] and [**g**] of a two-port network if
-
-$$
-[\mathbf{T}] = \begin{bmatrix} 10 & 1.5 \ \Omega \\ 2 \ \mathrm{S} & 4 \end{bmatrix}
-$$
-
-# **Solution:**
-
-If **A** = 10, **B** = 1.5, **C** = 2, **D** = 4, the determinant of the matrix is
-
-$$
-\Delta_T = \mathbf{AD} - \mathbf{BC} = 40 - 3 = 37
-$$
-
-From Table 19.1,
-
-$$
-\mathbf{z}_{11} = \frac{\mathbf{A}}{\mathbf{C}} = \frac{10}{2} = 5, \qquad \mathbf{z}_{12} = \frac{\Delta_T}{\mathbf{C}} = \frac{37}{2} = 18.5
-$$
-\n
-$$
-\mathbf{z}_{21} = \frac{1}{\mathbf{C}} = \frac{1}{2} = 0.5, \qquad \mathbf{z}_{22} = \frac{\mathbf{D}}{\mathbf{C}} = \frac{4}{2} = 2
-$$
-\n
-$$
-\mathbf{g}_{11} = \frac{\mathbf{C}}{\mathbf{A}} = \frac{2}{10} = 0.2, \qquad \mathbf{g}_{12} = -\frac{\Delta_T}{\mathbf{A}} = -\frac{37}{10} = -3.7
-$$
-\n
-$$
-\mathbf{g}_{21} = \frac{1}{\mathbf{A}} = \frac{1}{10} = 0.1, \qquad \mathbf{g}_{22} = \frac{\mathbf{B}}{\mathbf{A}} = \frac{1.5}{10} = 0.15
-$$
-
-Thus,
-
-[**z**] = [ 5 0.5 18.5 <sup>2</sup> ] <sup>Ω</sup>, [**g**] = [ 0.2 S 0.1 −3.7 0.15 Ω]
-
-Practice Problem 19.10 Determine [**y**] and [**T**] of a two-port network whose *z* parameters are
-
-$$
-\begin{aligned} \n\left[\mathbf{z}\right] &= \begin{bmatrix} 6 & 4 \\ 4 & 6 \end{bmatrix} \Omega\\ \n\text{Answer: } \n\left[\mathbf{y}\right] &= \begin{bmatrix} 0.3 & -0.2 \\ -0.2 & 0.3 \end{bmatrix} \text{S}, \quad \n\left[\mathbf{T}\right] = \begin{bmatrix} 1.5 & 5 \Omega \\ 0.25 \text{ S} & 1.5 \end{bmatrix}. \n\end{aligned}
-$$
-
-Example 19.11 Obtain the *y* parameters of the op amp circuit in Fig. 19.37. Show that the circuit has no *z* parameters.
-
-# **Solution:**
-
-Because no current can enter the input terminals of the op amp, **I**1 = 0, which can be expressed in terms of **V**1 and **V**2 as
-
-$$
-I_1 = 0V_1 + 0V_2 \tag{19.11.1}
-$$
-
-<span id="page-895-0"></span>Comparing this with Eq. (19.8) gives
-
-$$
-\mathbf{y}_{11} = 0 = \mathbf{y}_{12}
-$$
-
-Also,
-
-$$
-\mathbf{V}_2 = R_3 \mathbf{I}_2 + \mathbf{I}_o (R_1 + R_2)
-$$
-
-where **I***o* is the current through *R*1 and *R*2. But **I***o* = **V**1∕*R*1. Hence,
-
-$$
-\mathbf{V}_2 = R_3 \mathbf{I}_2 + \frac{\mathbf{V}_1 (R_1 + R_2)}{R_1}
-$$
-
-which can be written as
-
-$$
-\mathbf{I}_2 = -\frac{(R_1 + R_2)}{R_1 R_3} \mathbf{V}_1 + \frac{\mathbf{V}_2}{R_3}
-$$
-
-Comparing this with Eq. (19.8) shows that
-
-$$
-\mathbf{y}_{21} = -\frac{(R_1 + R_2)}{R_1 R_3}, \quad \mathbf{y}_{22} = \frac{1}{R_3}
-$$
-
-The determinant of the [**y**] matrix is
-
-$$
-\Delta_{y} = \mathbf{y}_{11}\mathbf{y}_{22} - \mathbf{y}_{12}\mathbf{y}_{21} = 0
-$$
-
-Since ∆*y* = 0, the [**y**] matrix has no inverse; therefore, the [**z**] matrix does not exist according to Eq. (19.34). Note that the circuit is not reciprocal because of the active element.
-
-Find the *z* parameters of the op amp circuit in Fig. 19.38. Sho w that the Practice Problem 19.11 circuit has no *y* parameters.
-
-**Answer:** [ **<sup>z</sup>**] = [ *R*1 −*R*<sup>2</sup> 0 0] . Because [ **z**] −1 does not e xist, [ **y**] does not exist.
-
-**Figure 19.38** For Practice Prob. 19.11.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/223_19.7 Interconnection of Networks.md

@@ -1,348 +0,0 @@
-# **19.7** Interconnection of Networks
-
-A large, complex network may be divided into subnetworks for the purposes of analysis and design. The subnetworks are modeled as two-port networks, interconnected to form the original network. The two-port networks may therefore be regarded as building blocks that can be interconnected to form a complex network. The interconnection can be in series, in parallel, or in cascade. Although the interconnected netw ork can be described by an y of the six parameter sets, a certain set of parameters may have a definite advantage. For example, when the netw orks are in series, their indi vidual *z* parameters add up to gi ve the *z* parameters of the larger network. When they are in parallel, their individual *y* parameters add up to gi ve the *y* parameters of the lar ger network. When they are cascaded, their individual transmission parameters can be multiplied together to get the transmission parameters of the larger network.
-
-**Figure 19.37** For Example 19.11.
-
-**Figure 19.39** Series connection of two two-port networks.
-
-Consider the series connection of tw o two-port networks shown in Fig. 19.39. The networks are re garded as being in series because their input currents are the same and their voltages add. In addition, each network has a common reference, and when the circuits are placed in series, the common reference points of each circuit are connected together. For network *Na*,
-
-$$
-\mathbf{V}_{1a} = \mathbf{z}_{11a}\mathbf{I}_{1a} + \mathbf{z}_{12a}\mathbf{I}_{2a}
-$$
-  
-\n
-$$
-\mathbf{V}_{2a} = \mathbf{z}_{21a}\mathbf{I}_{1a} + \mathbf{z}_{22a}\mathbf{I}_{2a}
-$$
- (19.44)
-
-and for network *Nb*,
-
-$$
-V_{1b} = z_{11b}I_{1b} + z_{12b}I_{2b}
-$$
-  
-\n
-$$
-V_{2b} = z_{21b}I_{1b} + z_{22b}I_{2b}
-$$
- (19.45)
-
-We notice from Fig. 19.39 that
-
-**I**1 = **I**<sup>1</sup>*a* = **I**<sup>1</sup>*b*, **I**2 = **I**<sup>2</sup>*a* = **I**<sup>2</sup>*<sup>b</sup>* **(19.46)**
-
-and that
-
-$$
-V_1 = V_{1a} + V_{1b} = (z_{11a} + z_{11b})I_1 + (z_{12a} + z_{12b})I_2
-$$
-  
-\n
-$$
-V_2 = V_{2a} + V_{2b} = (z_{21a} + z_{21b})I_1 + (z_{22a} + z_{22b})I_2
-$$
- (19.47)
-
-Thus, the *z* parameters for the overall network are
-
-$$
-\begin{bmatrix} \mathbf{z}_{11} & \mathbf{z}_{12} \\ \mathbf{z}_{21} & \mathbf{z}_{22} \end{bmatrix} = \begin{bmatrix} \mathbf{z}_{11a} + \mathbf{z}_{11b} & \mathbf{z}_{12a} + \mathbf{z}_{12b} \\ \mathbf{z}_{21a} + \mathbf{z}_{21b} & \mathbf{z}_{22a} + \mathbf{z}_{22b} \end{bmatrix}
-$$
-(19.48)
-
-or
-
-$$
-[\mathbf{z}] = [\mathbf{z}_a] + [\mathbf{z}_b] \tag{19.49}
-$$
-
-showing that the *z* parameters for the o verall network are the sum of the *z* parameters for the individual networks. This can be extended to *n* networks in series. If two two-port networks in the [**h**] model, for example, are connected in series, we use Table 19.1 to convert the **h** to **z** and then apply Eq. (19.49). We finally convert the result back to **h** using Table 19.1.
-
-Two two-port networks are in parallel when their port voltages are equal and the port currents of the larger network are the sums of the individual port currents. In addition, each circuit must ha ve a common reference and when the netw orks are connected together , they must all have their common references tied together. The parallel connection of two two-port networks is shown in Fig. 19.40. For the two networks,
-
-$$
-\mathbf{I}_{1a} = \mathbf{y}_{11a}\mathbf{V}_{1a} + \mathbf{y}_{12a}\mathbf{V}_{2a} \n\mathbf{I}_{2a} = \mathbf{y}_{21a}\mathbf{V}_{1a} + \mathbf{y}_{22a}\mathbf{V}_{2a}
-$$
-\n(19.50)
-
-and
-
-$$
-\mathbf{I}_{1b} = \mathbf{y}_{11b} \mathbf{V}_{1b} + \mathbf{y}_{12b} \mathbf{V}_{2b}
-$$
-  
-\n
-$$
-\mathbf{I}_{2a} = \mathbf{y}_{21b} \mathbf{V}_{1b} + \mathbf{y}_{22b} \mathbf{V}_{2b}
-$$
- (19.51)
-
-But from Fig. 19.40,
-
-$$
-V_1 = V_{1a} = V_{1b}, \qquad V_2 = V_{2a} = V_{2b} \tag{19.52a}
-$$
-
-**I**1 = **I**<sup>1</sup>*a* + **I**<sup>1</sup>*b*, **I**2 = **I**<sup>2</sup>*a* + **I**<sup>2</sup>*<sup>b</sup>* **(19.52b)**
-
-**Figure 19.40** Parallel connection of two two-port networks.
-
-Substituting Eqs. (19.50) and (19.51) into Eq. (19.52b) yields
-
-$$
-\mathbf{I}_1 = (\mathbf{y}_{11a} + \mathbf{y}_{11b})\mathbf{V}_1 + (\mathbf{y}_{12a} + \mathbf{y}_{12b})\mathbf{V}_2 \n\mathbf{I}_2 = (\mathbf{y}_{21a} + \mathbf{y}_{21b})\mathbf{V}_1 + (\mathbf{y}_{22a} + \mathbf{y}_{22b})\mathbf{V}_2
-$$
-\n(19.53)
-
-Thus, the *y* parameters for the overall network are
-
-[
-
-$$
-\begin{bmatrix}\n\mathbf{y}_{11} & \mathbf{y}_{12} \\
-\mathbf{y}_{21} & \mathbf{y}_{22}\n\end{bmatrix} = \begin{bmatrix}\n\mathbf{y}_{11a} + \mathbf{y}_{11b} & \mathbf{y}_{12a} + \mathbf{y}_{12b} \\
-\mathbf{y}_{21a} + \mathbf{y}_{21b} & \mathbf{y}_{22a} + \mathbf{y}_{22b}\n\end{bmatrix}
-$$
-\n(19.54)
-
-or
-
-$$
-[y] = [y_a] + [y_b]
-$$
- (19.55)
-
-showing that the *y* parameters of the overall network are the sum of the *y* parameters of the individual networks. The result can be extended to *n* two-port networks in parallel.
-
-Two networks are said to be *cascaded* when the output of one is the input of the other. The connection of two two-port networks in cascade is shown in Fig. 19.41. For the two networks,
-
-$$
-\begin{bmatrix} \mathbf{V}_{1a} \\ \mathbf{I}_{1a} \end{bmatrix} = \begin{bmatrix} \mathbf{A}_a & \mathbf{B}_a \\ \mathbf{C}_a & \mathbf{D}_a \end{bmatrix} \begin{bmatrix} \mathbf{V}_{2a} \\ -\mathbf{I}_{2a} \end{bmatrix}
-$$
-(19.56)
-
-$$
-\begin{bmatrix} \mathbf{V}_{1b} \\ \mathbf{I}_{1b} \end{bmatrix} = \begin{bmatrix} \mathbf{A}_b & \mathbf{B}_b \\ \mathbf{C}_b & \mathbf{D}_b \end{bmatrix} \begin{bmatrix} \mathbf{V}_{2b} \\ -\mathbf{I}_{2b} \end{bmatrix}
-$$
-(19.57)
-
-From Fig. 19.41,
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_1 \end{bmatrix} = \begin{bmatrix} \mathbf{V}_{1a} \\ \mathbf{I}_{1a} \end{bmatrix}, \quad \begin{bmatrix} \mathbf{V}_{2a} \\ -\mathbf{I}_{2a} \end{bmatrix} = \begin{bmatrix} \mathbf{V}_{1b} \\ \mathbf{I}_{1b} \end{bmatrix}, \quad \begin{bmatrix} \mathbf{V}_{2b} \\ -\mathbf{I}_{2b} \end{bmatrix} = \begin{bmatrix} \mathbf{V}_2 \\ -\mathbf{I}_2 \end{bmatrix}, (19.58)
-$$
-
-Substituting these into Eqs. (19.56) and (19.57),
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_1 \end{bmatrix} = \begin{bmatrix} \mathbf{A}_a & \mathbf{B}_a \\ \mathbf{C}_a & \mathbf{D}_a \end{bmatrix} \begin{bmatrix} \mathbf{A}_b & \mathbf{B}_b \\ \mathbf{C}_b & \mathbf{D}_b \end{bmatrix} \begin{bmatrix} \mathbf{V}_2 \\ -\mathbf{I}_2 \end{bmatrix}
-$$
-(19.59)
-
-Thus, the transmission parameters for the overall network are the product of the transmission parameters for the individual transmission parameters:
-
-$$
-\begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} A_a & B_a \\ C_a & D_a \end{bmatrix} \begin{bmatrix} A_b & B_b \\ C_b & D_b \end{bmatrix}
-$$
- (19.60)
-
-or
-
-$$
-[\mathbf{T}] = [\mathbf{T}_a][\mathbf{T}_b]
-$$
- (19.61)
-
-| I1 | I1a |    | I2a | I1b |    | I2b | I2 |
-|----|-----|----|-----|-----|----|-----|----|
-| +  | +   |    | +   | +   |    | +   | +  |
-| V1 | V1a | Na | V2a | V1b | Nb | V2b | V2 |
-| ‒  | ‒   |    | ‒   | ‒   |    | ‒   | ‒  |
-
-# **Figure 19.41**
-
-Cascade connection of two two-port networks.
-
-It is this property that makes the transmission parameters so useful. Keep in mind that the multiplication of the matrices must be in the order in which the networks *Na* and *Nb* are cascaded.
-
-Example 19.12 Evaluate **V**2∕**V***s* in the circuit in Fig. 19.42.
-
-**Figure 19.42** For Example 19.12.
-
-# **Solution:**
-
-This may be regarded as two two-ports in series. For *Nb*,
-
-$$
-\mathbf{z}_{12b} = \mathbf{z}_{21b} = 10 = \mathbf{z}_{11b} = \mathbf{z}_{22b}
-$$
-
-Thus,
-
-$$
-[\mathbf{z}] = [\mathbf{z}_a] + [\mathbf{z}_b] = \begin{bmatrix} 12 & 8 \\ 8 & 20 \end{bmatrix} + \begin{bmatrix} 10 & 10 \\ 10 & 10 \end{bmatrix} = \begin{bmatrix} 22 & 18 \\ 18 & 30 \end{bmatrix}
-$$
-
-But
-
-$$
-V_1 = z_{11}I_1 + z_{12}I_2 = 22I_1 + 18I_2 \qquad (19.12.1)
-$$
-
-$$
-\mathbf{V}_2 = \mathbf{z}_{21}\mathbf{I}_1 + \mathbf{z}_{22}\mathbf{I}_2 = 18\mathbf{I}_1 + 30\mathbf{I}_2 \tag{19.12.2}
-$$
-
-Also, at the input port
-
-$$
-\mathbf{V}_1 = \mathbf{V}_s - 5\mathbf{I}_1 \tag{19.12.3}
-$$
-
-and at the output port
-
-$$
-V_2 = -20I_2
-$$
-  $\Rightarrow$   $I_2 = -\frac{V_2}{20}$  (19.12.4)
-
-Substituting Eqs. (19.12.3) and (19.12.4) into Eq. (19.12.1) gives
-
-$$
-\mathbf{V}_s - 5\mathbf{I}_1 = 22\mathbf{I}_1 - \frac{18}{20}\mathbf{V}_2 \qquad \Rightarrow \qquad \mathbf{V}_s = 27\mathbf{I}_1 - 0.9\mathbf{V}_2 \tag{19.12.5}
-$$
-
-while substituting Eq. (19.12.4) into Eq. (19.12.2) yields
-
-$$
-\mathbf{V}_2 = 18\mathbf{I}_1 - \frac{30}{20}\mathbf{V}_2 \qquad \Rightarrow \qquad \mathbf{I}_1 = \frac{2.5}{18}\mathbf{V}_2 \tag{19.12.6}
-$$
-
-Substituting Eq. (19.12.6) into Eq. (19.12.5), we get
-
-$$
-\mathbf{V}_s = 27 \times \frac{2.5}{18} \mathbf{V}_2 - 0.9 \mathbf{V}_2 = 2.85 \mathbf{V}_2
-$$
-
-And so,
-
-$$
-\frac{\mathbf{V}_2}{\mathbf{V}_s} = \frac{1}{2.85} = 0.3509
-$$
-
-Find **V**2∕**V***s* in the circuit in Fig. 19.43. Practice Problem 19.12
-
-**Answer:** 0.6799⧸−29.05°.
-
-Find the *y* parameters of the two-port in Fig. 19.44. Example 19.13
-
-# **Solution:**
-
-Let us refer to the upper network as *Na* and the lower one as *Nb*. The two networks are connected in parallel. Comparing *Na* and *Nb* with the circuit in Fig. 19.13(a), we obtain
-
-$$
-y_{12a} = -j4 = y_{21a}
-$$
-,  $y_{11a} = 2 + j4$ ,  $y_{22a} = 3 + j4$ 
-
-or
-
-$$
-[\mathbf{y}_a] = \begin{bmatrix} 2+j4 & -j4 \\ -j4 & 3+j4 \end{bmatrix} \text{S}
-$$
-
-and
-
-$$
-y_{12b} = -4 = y_{21b}
-$$
-,  $y_{11b} = 4 - j2$ ,  $y_{22b} = 4 - j6$ 
-
-or
-
-$$
-[\mathbf{y}_b] = \begin{bmatrix} 4 - j2 & -4 \\ -4 & 4 - j6 \end{bmatrix} \text{S}
-$$
-
-The overall *y* parameters are
-
-$$
-[\mathbf{y}] = [\mathbf{y}_a] + [\mathbf{y}_b] = \begin{bmatrix} 6+j2 & -4-j4 \\ -4-j4 & 7-j2 \end{bmatrix} \text{S}
-$$
-
-**Figure 19.44** For Example 19.13.
-
-Practice Problem 19.13 Obtain the *y* parameters for the network in Fig. 19.45.
-
-**Answer:** 
-$$
-\begin{bmatrix} 27 - j15 & -25 + j10 \ -25 + j10 & 27 - j5 \end{bmatrix}
-$$
- S.
-
-**Figure 19.45** For Practice Prob. 19.13.
-
-**Figure 19.46** For Example 19.14.
-
-Example 19.14 Find the transmission parameters for the circuit in Fig. 19.46.
-
-# **Solution:**
-
-We can regard the given circuit in Fig. 19.46 as a cascade connection of two T networks as shown in Fig. 19.47(a). We can show that a T network, shown in Fig. 19.47(b), has the following transmission parameters [see Prob. 19.52(b)]:
-
-$$
-\mathbf{A} = 1 + \frac{R_1}{R_2}, \qquad \mathbf{B} = R_3 + \frac{R_1(R_2 + R_3)}{R_2}
-$$
-$$
-\mathbf{C} = \frac{1}{R_2}, \qquad \mathbf{D} = 1 + \frac{R_3}{R_2}
-$$
-
-Applying this to the cascaded networks *Na* and *Nb* in Fig. 19.47(a), we get
-
-$$
-\mathbf{A}_a = 1 + 4 = 5, \qquad \mathbf{B}_a = 8 + 4 \times 9 = 44 \text{ }\Omega
-$$
-  
-$$
-\mathbf{C}_a = 1 \text{ S}, \qquad \mathbf{D}_a = 1 + 8 = 9
-$$
-
-or in matrix form,
-
-$$
-[\mathbf{T}_a] = \begin{bmatrix} 5 & 44 \ \Omega \\ 1 \ \mathrm{S} & 9 \end{bmatrix}
-$$
-
-$$
-[1_a] = \begin{bmatrix} 1 & 0 & 9 \end{bmatrix}
-$$
-
-$$
-A_b = 1
-$$
-,  $B_b = 6 \Omega$ ,  $C_b = 0.5 S$ ,  $D_b = 1 + \frac{6}{2} = 4$ 
-
-i.e.,
-
-and
-
-Thus, for the total network in Fig. 19.46,
-
-$$
-\begin{aligned} [\mathbf{T}] &= [\mathbf{T}_a][\mathbf{T}_b] = \begin{bmatrix} 5 & 44 \\ 1 & 9 \end{bmatrix} \begin{bmatrix} 1 & 6 \\ 0.5 & 4 \end{bmatrix} \\ &= \begin{bmatrix} 5 \times 1 + 44 \times 0.5 & 5 \times 6 + 44 \times 4 \\ 1 \times 1 + 9 \times 0.5 & 1 \times 6 + 9 \times 4 \end{bmatrix} \\ &= \begin{bmatrix} 27 & 206 \ \Omega \\ 5.5 \ \text{S} & 42 \end{bmatrix} \end{aligned}
-$$
-
-# **Figure 19.47**
-
-For Example 19.14: (a) Breaking the circuit in Fig. 19.46 into two two-ports, (b) a general T two-port.
-
-<span id="page-901-0"></span>Notice that
-
-$$
-\Delta_{T_a}=\Delta_{T_b}=\Delta_T=1
-$$
-
-showing that the network is reciprocal.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/224_19.8 Computing Two-Port Parameters Using PSpice.md

@@ -1,432 +0,0 @@
-# **19.8** Computing Two-Port Parameters Using PSpice
-
-Hand calculation of the two-port parameters may become difficult when the two-port is complicated. We resort to *PSpice* in such situations. If the circuit is purely resisti ve, *PSpice* dc analysis may be used; otherwise, *PSpice* ac analysis is required at a specific frequency. The key to using *PSpice* in computing a particular two-port parameter is to remember how that parameter is defined and to constrain the appropriate port variable with a 1-A or 1-V source while using an open or short circuit to impose the other necessary constraints. The following two examples illustrate the idea.
-
-Find the *h* parameters of the network in Fig. 19.49. Example 19.15
-
-# **Solution:**
-
-From Eq. (19.16),
-
-$$
-\mathbf{h}_{11} = \frac{\mathbf{V}_1}{\mathbf{I}_1} \Big|_{\mathbf{V}_2 = 0}, \qquad \mathbf{h}_{21} = \frac{\mathbf{I}_2}{\mathbf{I}_1} \Big|_{\mathbf{V}_2 = 0}
-$$
-
-showing that **h**11 and **h**21 can be found by setting **V**2 = 0. Also by setting **I**1 = 1 A, **h**11 becomes **V**1∕1 while **h**21 becomes **I**2∕1. With this in mind, we draw the schematic in Fig. 19.50(a). We insert a 1-A dc current
-
-6 Ω
-
-4i x
-
-+ ‒
-
-5 Ω
-
-**Figure 19.50**
-
-For Example 19.15: (a) computing **h**11 and **h**21, (b) computing **h**12 and **h**22.
-
-source IDC to take care of **I**1 = 1 A, the pseudocomponent VIEWPOINT to display **V**1 and pseudocomponent IPROBE to display **I**2. After saving the schematic, we run *PSpice* by selecting **Analysis/Simulate** and note the values displayed on the pseudocomponents. We obtain
-
-$$
-\mathbf{h}_{11} = \frac{\mathbf{V}_1}{1} = 10 \ \Omega, \qquad \mathbf{h}_{21} = \frac{\mathbf{I}_2}{1} = -0.5
-$$
-
-Similarly, from Eq. (19.16),
-
-Practice Problem 19.15 Obtain the *h* parameters for the network in Fig. 19.51 using *PSpice.*
-
-$$
-\mathbf{h}_{12} = \frac{\mathbf{V}_1}{\mathbf{V}_2} \Big|_{\mathbf{I}_1 = 0}, \qquad \mathbf{h}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} \Big|_{\mathbf{I}_1 = 0}
-$$
-
-indicating that we obtain **h**12 and **h**22 by open-circuiting the input port (**I**1 = 0). By making **V**2 = 1 V, **h**12 becomes **V**1∕1 while **h**22 becomes **I**2∕1. Thus, we use the schematic in Fig. 19.50(b) with a 1-V dc voltage source VDC inserted at the output terminal to take care of **V**2 = 1 V. The pseudocomponents VIEWPOINT and IPR OBE are inserted to display the values of **V**1 and **I**2, respectively. (Notice that in Fig. 19.50(b), the 5- Ω resistor is ignored because the input port is open-circuited and *PSpice*  will not allow such. We may include the 5- Ω resistor if we replace the open circuit with a very large resistor, say, 10 MΩ.) After simulating the schematic, we obtain the values displayed on the pseudocomponents as shown in Fig. 19.50(b). Thus,
-
-$$
-\mathbf{h}_{12} = \frac{\mathbf{V}_1}{1} = 0.8333, \qquad \mathbf{h}_{22} = \frac{\mathbf{I}_2}{1} = 0.1833 \text{ S}
-$$
-
-**Answer:** *h*11 = 4.238 Ω, *h*21 = −0.6190, *h*12 = −0.7143, *h*22 = −0.1429 S.
-
-**Figure 19.52** For Example 19.16.
-
-Example 19.16 Find the *z* parameters for the circuit in Fig. 19.52 at *ω* = 106 rad/s.
-
-# **Solution:**
-
-Notice that we used dc analysis in Example 19.15 because the circuit in Fig. 19.49 is purely resistive. Here, we use ac analysis at *f* = *ω*∕2*π* = 0.15915 MHz, because *L* and *C* are frequency dependent.
-
-In Eq. (19.3), we defined the *z* parameters as
-
-$$
-\mathbf{z}_{11} = \frac{\mathbf{V}_1}{\mathbf{I}_1} \bigg|_{\mathbf{I}_2 = 0}, \qquad \mathbf{z}_{21} = \frac{\mathbf{V}_2}{\mathbf{I}_1} \bigg|_{\mathbf{I}_2 = 0}
-$$
-
-**Figure 19.53** For Example 19.16: (a) circuit for determining **z**11 and **z**21, (b) circuit for determining **z**12 and **z**22.
-
-This suggests that if we let **I**1 = 1 A and open-circuit the output port so that **I**2 = 0, then we obtain
-
-$$
-\mathbf{z}_{11} = \frac{\mathbf{V}_1}{1} \quad \text{and} \quad \mathbf{z}_{21} = \frac{\mathbf{V}_2}{1}
-$$
-
-We realize this with the schematic in Fig. 19.53(a). We insert a 1-A ac current source IAC at the input terminal of the circuit and two VPRINT1 pseudocomponents to obtain **V**1 and **V**2. The attributes of each VPRINT1 are set as *AC* = *yes, MAG* = *yes,* and *PHASE* = *yes* to print the magnitude and phase values of the voltages. We select **Analysis/Setup/AC Sweep**  and enter 1 as *Total Pts,* 0.1519MEG as *Start Freq,* and 0.1519MEG as *Final Freq* in the **AC Sweep and Noise Analysis** dialog box. After saving the schematic, we select **Analysis/Simulate** to simulate it. We obtain **V**1 and **V**2 from the output file. Thus,
-
-$$
-\mathbf{z}_{11} = \frac{\mathbf{V}_1}{1} = 19.70 \underline{/ 175.7^{\circ}} \,\Omega, \qquad \mathbf{z}_{21} = \frac{\mathbf{V}_2}{1} = 19.79 \underline{/ 170.2^{\circ}} \,\Omega
-$$
-
-In a similar manner, from Eq. (19.3),
-
-$$
-\mathbf{z}_{12} = \frac{\mathbf{V}_1}{\mathbf{I}_2} \big|_{\mathbf{I}_1 = 0}, \qquad \mathbf{z}_{22} = \frac{\mathbf{V}_2}{\mathbf{I}_2} \big|_{\mathbf{I}_1 = 0}
-$$
-
-suggesting that if we let **I**2 = 1 A and open-circuit the input port,
-
-$$
-z_{12} = \frac{V_1}{1}
-$$
- and  $z_{22} = \frac{V_2}{1}$ 
-
-This leads to the schematic in Fig. 19.53(b). The only difference between this schematic and the one in Fig. 19.53(a) is that the 1-A ac current source IA C is no w at the output terminal. We run the schematic in Fig. 19.53(b) and obtain **V**1 and **V**2 from the output file. Thus,
-
-$$
-\mathbf{z}_{12} = \frac{\mathbf{V}_1}{1} = 19.70 \underline{\text{/} 175.7^{\circ}} \,\Omega, \qquad \mathbf{z}_{22} = \frac{\mathbf{V}_2}{1} = 19.56 \underline{\text{/} 175.7^{\circ}} \,\Omega
-$$
-
-<span id="page-904-0"></span>Practice Problem 19.16 Obtain the *z* parameters of the circuit in Fig. 19.54 at *f* = 60 Hz.
-
-**Answer:** 
-$$
-z_{11} = 3.987 / 175.5^{\circ} \Omega
-$$
-,  $z_{21} = 0.0175 / -2.65^{\circ} \Omega$ ,  
-\n $z_{12} = 0$ ,  $z_{22} = 0.2651 / 91.9^{\circ} \Omega$ .
-
-# **19.9** Applications
-
-We have seen how the six sets of network parameters can be used to characterize a wide range of two-port networks. Depending on the way two-ports are interconnected to form a larger network, a particular set of parameters may have advantages over others, as we noticed in S ection 19.7. In this section, we will consider tw o important application areas of two-port parameters: transistor circuits and synthesis of ladder networks.
-
-# **19.9.1** Transistor Circuits
-
-The two-port network is often used to isolate a load from the e xcitation of a circuit. For example, the two-port in Fig. 19.55 may represent an amplifier, a filter, or some other netw ork. When the two-port represents an amplifier, expressions for the voltage gain *Av*, the current gain *Ai*, the input impedance *Z*in, and the output impedance *Z*out can be derived with ease. They are defined as follows:
-
-$$
-A_v = \frac{V_2(s)}{V_1(s)}
-$$
-(19.62)
-
-$$
-A_i = \frac{I_2(s)}{I_1(s)}\tag{19.63}
-$$
-
-$$
-Z_{\text{in}} = \frac{V_1(s)}{I_1(s)}
-$$
-(19.64)
-
-$$
-Z_{\text{out}} = \frac{V_2(s)}{I_2(s)} \bigg|_{V_s=0} \tag{19.65}
-$$
-
-Any of the six sets of tw o-port parameters can be used to deri ve the expressions in Eqs. (19.62) to (19.65). Ho wever, the hybrid (*h*) parameters are the most useful for transistors; the y are easily measured and are often provided in the manufacturer's data or spec sheets for transis tors. The *h* parameters pro vide a quick estimate of the performance of transistor circuits. They are used for finding the exact voltage gain, input impedance, and output impedance of a transistor.
-
-**Figure 19.55** Two-port network isolating source and load.
-
-The *h* parameters for transistors have specific meanings expressed by their subscripts. They are listed by the first subscript and related to the general *h* parameters as follows:
-
-$$
-h_i = h_{11}
-$$
-,  $h_r = h_{12}$ ,  $h_f = h_{21}$ ,  $h_o = h_{22}$  (19.66)
-
-The subscripts *i, r, f,* and *o* stand for input, reverse, forward, and output. The second subscript specifies the type of connection used: *e* for common emitter (CE), *c* for common collector (CC), and *b* for common base (CB). Here we are mainly concerned with the common-emitter connec tion. Thus, the four *h* parameters for the common-emitter amplifier are:
-
-$$
-h_{ie} = \text{Base input impedance}
-$$
-  
-\n
-$$
-h_{re} = \text{Reverse voltage feedback ratio}
-$$
-  
-\n
-$$
-h_{fe} = \text{Base-collector current gain}
-$$
-  
-\n
-$$
-h_{oe} = \text{Output admittance}
-$$
-  
-\n(19.67)
-
-These are calculated or measured in the same w ay as the general *h* parameters. Typical values are *hie* = 6 kΩ, *hre* = 1.5 × 10−4, *hfe* = 200, *hoe* = 8 *µ*S. We must keep in mind that these values represent ac characteristics of the transistor, measured under specific circumstances.
-
-Figure 19.56 sho ws the circuit schematic for the common-emitter amplifier and the equivalent hybrid model. From the figure, we see that
-
-$$
-\mathbf{V}_b = h_{ie}\mathbf{I}_b + h_{re}\mathbf{V}_c
-$$
-\n
-$$
-\mathbf{I}_c = h_{fe}\mathbf{I}_b + h_{oe}\mathbf{V}_c
-$$
-\n(19.68a)\n(19.68b)
-
-**Figure 19.56**
-
-Common emitter amplifier: (a) circuit schematic, (b) hybrid model.
-
-Consider the transistor amplifier connected to an ac source and a load as in Fig. 19.57. This is an example of a two-port network embedded within a larger network. We can analyze the hybrid equivalent circuit as usual with Eq. (19.68) in mind. (See Example 19.6.) Recognizing
-
-Transistor amplifier with source and load resistance.
-
-from Fig. 19.57 that **V***c* = −*RL***I***c* and substituting this into Eq. (19.68b) gives
-
-$$
-\mathbf{I}_c = h_{fe}\mathbf{I}_b - h_{oe}R_L\mathbf{I}_c
-$$
-
-$$
-(1 + h_{oe}R_L)\mathbf{I}_c = h_{fe}\mathbf{I}_b \tag{19.69}
-$$
-
-From this, we obtain the current gain as
-
-$$
-A_i = \frac{\mathbf{I}_c}{\mathbf{I}_b} = \frac{h_{fe}}{1 + h_{oe}R_L}
-$$
- (19.70)
-
-From Eqs. (19.68b) and (19.70), we can express **I***b* in terms of **V***c*:
-
-$$
-\mathbf{I}_c = \frac{h_{fe}}{1 + h_{oe}R_L}\mathbf{I}_b = h_{fe}\mathbf{I}_b + h_{oe}\mathbf{V}_c
-$$
-
-or
-
-or
-
-$$
-\mathbf{I}_{b} = \frac{h_{oe} \mathbf{V}_{c}}{h_{fe}} - h_{fe}
-$$
- (19.71)
-
-Substituting Eq. (19.71) into Eq. (19.68a) and dividing by **V***c* gives
-
-1) into Eq. (19.68a) and dividing by 
-$$
-\mathbf{V}_c
-$$
- gives  
-\n
-$$
-\frac{\mathbf{V}_b}{\mathbf{V}_c} = \frac{h_{oe}h_{ie}}{\frac{h_{fe}}{1 + h_{oe}R_L} - h_{fe}} + h_{re}
-$$
-\n
-$$
-= \frac{h_{ie} + h_{ie}h_{oe}R_L - h_{re}h_{fe}R_L}{-h_{fe}R_L} \tag{19.72}
-$$
-
-Thus, the voltage gain is
-
-gain is  
-\n
-$$
-A_{v} = \frac{\mathbf{V}_{c}}{\mathbf{V}_{b}} = \frac{-h_{fe}R_{L}}{h_{ie} + (h_{ie}h_{oe} - h_{re}h_{fe})R_{L}}
-$$
-\n(19.73)
-
-Substituting **V***c* = −*RL***I***c* into Eq. (19.68a) gives
-
-$$
-\mathbf{V}_b = h_{ie}\mathbf{I}_b - h_{re}R_L\mathbf{I}_c
-$$
-
-or
-
-$$
-\frac{\mathbf{V}_b}{\mathbf{I}_b} = h_{ie} - h_{re} R_L \frac{\mathbf{I}_c}{\mathbf{I}_b}
-$$
-(19.74)
-
-Replacing **I***c*∕**I***b* by the current gain in Eq. (19.70) yields the input impedance as
-
-$$
-Z_{\rm in} = \frac{V_b}{I_b} = h_{ie} - \frac{h_{re}h_{fe}R_L}{1 + h_{oe}R_L}
-$$
- (19.75)
-
-The output impedance *Z*out is the same as the Thevenin equivalent at the output terminals. As usual, by removing the voltage source and placing a
-
-**Figure 19.58** Finding the output impedance of the amplifier circuit in Fig. 19.57.
-
-1-V source at the output terminals, we obtain the circuit in Fig. 19.58, from which *Z*out is determined as 1∕**I***c*. Because **V***c* = 1 V, the input loop gives
-
-$$
-h_{re}(1) = -\mathbf{I}_b(R_s + h_{ie}) \qquad \Rightarrow \qquad \mathbf{I}_b = -\frac{h_{re}}{R_s + h_{ie}} \qquad (19.76)
-$$
-
-For the output loop,
-
-$$
-\mathbf{I}_c = \mathbf{h}_{oe}(1) + h_{fe}\mathbf{I}_b \tag{19.77}
-$$
-
-Substituting Eq. (19.76) into Eq. (19.77) gives
-
-) into Eq. (19.77) gives  
-\n
-$$
-\mathbf{I}_c = \frac{(R_s + h_{ie})h_{oe} - h_{re}h_{fe}}{R_s + h_{ie}}
-$$
-\n(19.78)
-
-From this, we obtain the output impedance *Z*out as 1∕**I***c*; that is,
-
-the output impedance 
-$$
-Z_{\text{out}}
-$$
- as  $1/I_c$ ; that is,  
-$$
-Z_{\text{out}} = \frac{R_s + h_{ie}}{(R_s + h_{ie})h_{oe} - h_{re}h_{fe}}
-$$
-(19.79)
-
-Consider the common-emitter amplifier circuit of Fig. 19.59. Determine Example 19.17 the voltage gain, current g ain, input impedance, and output impedance using these *h* parameters:
-
-*hie* = 1 kΩ, *hre* = 2.5 × 10−4, *hfe* = 50, *hoe* = 20 *µ*S
-
-Find the output voltage **V***o*.
-
-For Example 19.17.
-
-# **Solution:**
-
-1. **Define.** In an initial look at this problem, it appears to be clearly stated. However, when we are asked to determine the input impedance and the voltage gain, do they refer to the transistor or the circuit? As far as the current gain and the output impedance are concerned, the y are the same for both cases.
-
-We ask for clarification and are told that we should calculate the input impedance, the output impedance, and the voltage gain
-
-for the circuit and not the transistor. It is interesting to note that the problem can be restated so that it becomes a simple design problem: Given the *h* parameters, design a simple amplifier that has a gain of −60.
-
-- 2. **Present.** Given a simple transistor circuit, an input voltage of 3.2 mV, and the *h* parameters of the transistor, calculate the output voltage.
-- 3. **Alternative.** There are a couple of ways we can approach the prob lem, the most straightforw ard being to use the equi valent circuit shown in Fig. 19.57. Once you ha ve the equivalent circuit you can use circuit analysis to determine the answer. Once you have a solu tion, you can check it by plugging in the answer into the circuit equations to see if they are correct. Another approach is to simplify the right-hand side of the equi valent circuit and w ork backward to see if you obtain approximately the same answer . We will use that approach here.
-- 4. **Attempt.** We note that *R <sup>s</sup>* = 0.8 k Ω and *R <sup>L</sup>* = 1.2 k Ω. We treat the transistor of Fig. 19.59 as a two-port network and apply Eqs. (19.70) to (19.79).
-
-$$
-h_{ie}h_{oe} - h_{re}h_{fe} = 10^3 \times 20 \times 10^{-6} - 2.5 \times 10^{-4} \times 50
-$$
-$$
-= 7.5 \times 10^{-3}
-$$
-$$
-A_v = \frac{-h_{fe}R_L}{h_{ie} + (h_{ie}h_{oe} - h_{re}h_{fe})R_L} = \frac{-50 \times 1200}{1000 + 7.5 \times 10^{-3} \times 1200}
-$$
-$$
-= -59.46
-$$
-
-*Av* is the voltage gain of the amplifier = *Vo* ∕ *Vb*. To calculate the gain of the circuit we need to find *Vo* ∕ *Vs*. We can do this by using the mesh equation for the circuit on the left and Eqs. (19.71) and (19.73).
-
-$$
--\mathbf{V}_s + R_s \mathbf{I}_b + \mathbf{V}_b = 0
-$$
-
-or
-
-$$
--V_s + R_s I_b + V_b = 0
-$$
-  
-$$
-V_s = 800 \frac{20 \times 10^{-6}}{50} - \frac{1}{59.46} V_o
-$$
-  
-$$
-= -0.03047 V_o.
-$$
-
-Thus, the circuit gain is equal to −**32.82.** Now we can calculate the output voltage.
-
-voltage.  
-\n
-$$
-V_o = \text{gain} \times V_s = -105.09 \underline{/0^{\circ}} \text{ mV}.
-$$
-\n
-$$
-A_i = \frac{h_{fe}}{1 + h_{oe}R_L} = \frac{50}{1 + 20 \times 10^{-6} \times 1200} = 48.83
-$$
-\n
-$$
-Z_{in} = h_{ie} - \frac{h_{re}h_{fe}R_L}{1 + h_{oe}R_L}
-$$
-\n
-$$
-= 1000 - \frac{2.5 \times 10^{-4} \times 50 \times 1200}{1 + 20 \times 10^{-6} \times 1200}
-$$
-\n
-$$
-= 985.4 \Omega
-$$
-
-You can modify *Z*in to include the 800-ohm resistor so that
-
-Circuit input impedance = 800 + 985.4 = **1785.4** Ω . (*Rs* + *hie*)*hoe* − *hrehfe*
-
-= (800 + 1000) × 20 × 10−6 − 2.5 × 10−4 × 50 = 23.5 × 10−3
-
-$$
-Z_{\text{out}} = \frac{R_s + h_{ie}}{(R_s + h_{ie})h_{oe} - h_{re}h_{fe}} = \frac{800 + 1000}{23.5 \times 10^{-3}} = 76.6 \text{ k}\Omega
-$$
-
-5. **Evaluate.** In the equi valent circuit, *hoe* represents a resistor of 50,000 Ω. This is in parallel with a load resistor equal to 1.2 k Ω. The size of the load resistor is so small relative to the *hoe* resistor that *hoe* can be neglected. This then leads to
-
-$$
-I_c = h_{fe}I_b = 50I_b
-$$
-,  $V_c = -1200I_c$ ,
-
-and the following loop equation from the left-hand side of the circuit:
-
-−0.0032 + (800 + 1000)I*b* + (0.00025)(−1200)(50)I*b* = 0 I*b* = 0.0032∕(1785) = 1.7927 *µ*A. I*c* = 50 × 1.7927 = 89.64 *µ*A and V*c* = −1200 × 89.64 × 10−6
-
-$$
-= -107.57
-$$
- mV
-
-This is a good approximation to −105.09 mV.
-
-$$
-Voltage gain = -107.57/3.2 = -33.62
-$$
-
-Again, this is a good approximation to 32.82.
-
-Circuit input impedance = 0.032∕1.7927 × 10−6 = **1785** Ω
-
-which clearly compares well with the 1785.4 Ω we obtained before.
-
- For these calculations, we assumed that Zout = ∞ Ω. Our calculations produced 72.6 kΩ. We can test our assumption by calculating the equivalent resistance of this and the load resistance.
-
-72,600 × 1200∕(72,600 + 1200) = 1,180.5 = 1.1805 kΩ
-
-Again, we have a good approximation.
-
-6. **Satisfactory?** We ha ve satisf actorily solv ed the problem and checked the results. We can now present our results as a solution to the problem.
-
-For the transistor amplifier of Fig. 19.60, find the voltage gain, current Practice Problem 19.17 gain, input impedance, and output impedance. Assume that
-
-$$
-h_{ie} = 6 \text{ k}\Omega
-$$
-,  $h_{re} = 1.5 \times 10^{-4}$ ,  $h_{fe} = 200$ ,  $h_{oe} = 8 \mu\text{S}$ 
-
-**Answer:** −123.61 for the transistor and −4.753 for the circuit, 194.17, 6 kΩ for the transistor and 156 kΩ for the circuit, 128.08 kΩ.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/225_19.9 Applications.md

@@ -1,193 +0,0 @@
-# **19.9.2** Ladder Network Synthesis
-
-Another application of tw o-port parameters is the synthesis (or b uilding) of ladder networks, which are found frequently in practice and have
-
-**Figure 19.61** *LC* ladder networks for low-pass filters of: (a) odd order, (b) even order.
-
-particular use in designing passive low-pass filters. Based on our discussion of second-order circuits in Chapter 8, the order of the filter is the order of the characteristic equation describing the filter and is determined by the number of reactive elements that cannot be combined into single elements (e.g., through series or parallel combination). Figure 19.61(a) shows an *LC* ladder network with an odd number of elements (to realize an odd-order filter), while Fig. 19.61(b) shows one with an even number of elements (for realizing an e ven-order filter). When either network is terminated by the load impedance *ZL* and the source impedance *Zs*, we obtain the structure in Fig. 19.62. To make the design less complicated, we will assume that *Zs* = 0. Our goal is to synthesize the transfer function of the *LC* ladder network. We begin by characterizing the ladder network by its admittance parameters, namely,
-
-$$
-I_1 = y_{11}V_1 + y_{12}V_2 \tag{19.80a}
-$$
-
-$$
-I_2 = y_{21}V_1 + y_{22}V_2 \tag{19.80b}
-$$
-
-**Figure 19.62** *LC* ladder network with terminating impedances.
-
-(Of course, the impedance parameters could be used instead of the admittance parameters.) At the input port, **V**1 = **V***s* since **Z***s* = 0. At the output port, **V**2 = **V***o* and **I**2 = −**V**2∕**Z***L* = −**V***o***Y***L*. Thus, Eq. (19.80b) becomes
-
-$$
--\mathbf{V}_o\mathbf{Y}_L=\mathbf{y}_{21}\mathbf{V}_s+\mathbf{y}_{22}\mathbf{V}_o
-$$
-
-or
-
-$$
-H(s) = \frac{V_o}{V_s} = \frac{-y_{21}}{Y_L + y_{22}}
-$$
-(19.81)
-
-We can write this as
-
-$$
-\mathbf{H}(s) = -\frac{\mathbf{y}_{21}/\mathbf{Y}_L}{1 + \mathbf{y}_{22}/\mathbf{Y}_L}
-$$
- (19.82)
-
-We may ignore the ne gative sign in Eq. (19.82) because filter requirements are often stated in terms of the magnitude of the transfer function. The main objective in filter design is to select capacitors and inductors so that the parameters **y**21 and **y**22 are synthesized, thereby realizing the desired transfer function. To achieve this, we tak e advantage of an important property of the *LC* ladder network: All *z* and *y* parameters are ratios of polynomials that contain only e ven powers of *s* or odd powers of *s*—that is, they are ratios of either Od(*s*)∕Ev(*s*) or Ev(*s*)∕Od(*s*), where Od and Ev are odd and even functions, respectively. Let
-
-$$
-\mathbf{H}(s) = \frac{\mathbf{N}(s)}{\mathbf{D}(s)} = \frac{\mathbf{N}_o + \mathbf{N}_e}{\mathbf{D}_o + \mathbf{D}_e}
-$$
-(19.83)
-
-where **N**(*s*) and **D**(*s*) are the numerator and denominator of the transfer function **H**(*s*); **N***o* and **N***e* are the odd and even parts of **N**; **D***o* and **D***e* are the odd and even parts of **D**. Given that **N**(*s*) must be either odd or even, we can write Eq. (19.83) as
-
-$$
-\mathbf{H}(s) = \begin{cases} \frac{\mathbf{N}_o}{\mathbf{D}_o + \mathbf{D}_e}, & (\mathbf{N}_e = 0) \\ \frac{\mathbf{N}_e}{\mathbf{D}_o + \mathbf{D}_e}, & (\mathbf{N}_o = 0) \end{cases}
-$$
-(19.84)
-
-and can rewrite this as
-
-$$
-\mathbf{H}(s) = \begin{cases} \frac{\mathbf{N}_o / \mathbf{D}_e}{1 + \mathbf{D}_o / \mathbf{D}_e}, & (\mathbf{N}_e = 0) \\ \frac{\mathbf{N}_e / \mathbf{D}_o}{1 + \mathbf{D}_e / \mathbf{D}_o}, & (\mathbf{N}_o = 0) \end{cases}
-$$
-(19.85)
-
-Comparing this with Eq. (19.82), we obtain the *y* parameters of the network as
-
-$$
-\frac{\mathbf{y}_{21}}{\mathbf{Y}_L} = \begin{cases} \frac{\mathbf{N}_o}{\mathbf{D}_e}, & (\mathbf{N}_e = 0) \\ \frac{\mathbf{N}_e}{\mathbf{D}_o}, & (\mathbf{N}_o = 0) \end{cases}
-$$
-(19.86)
-
-and
-
-$$
-\frac{\mathbf{y}_{22}}{\mathbf{Y}_L} = \begin{cases} \frac{\mathbf{D}_o}{\mathbf{D}_e}, & (\mathbf{N}_e = 0) \\ \frac{\mathbf{D}_e}{\mathbf{D}_o}, & (\mathbf{N}_o = 0) \end{cases}
-$$
-(19.87)
-
-The following example illustrates the procedure.
-
-Design the *LC* ladder network terminated with a 1-Ω resistor that has the Example 19.18 normalized transfer function
-
-$$
-H(s) = \frac{1}{s^3 + 2s^2 + 2s + 1}
-$$
-
-(This transfer function is for a Butterworth low-pass filter.)
-
-# **Solution:**
-
-The denominator sho ws that this is a third-order netw ork, so that the *LC* ladder netw ork is sho wn in Fig. 19.63(a), with tw o inductors and one capacitor. Our goal is to determine the v alues of the inductors and
-
-capacitor. To achieve this, we group the terms in the denominator into odd or even parts:
-
-$$
-\mathbf{D}(s) = (s^3 + 2s) + (2s^2 + 1)
-$$
-
-so that
-
-$$
-H(s) = (s + 2s) + (2s + 1)
-$$
-$$
-H(s) = \frac{1}{(s^3 + 2s) + (2s^2 + 1)}
-$$
-
-Divide the numerator and denominator by the odd part of the denominator to get
-
-$$
-\mathbf{H}(s) = \frac{\frac{1}{s^3 + 2s}}{1 + \frac{2s^2 + 1}{s^3 + 2s}}
-$$
-(19.18.1)
-
-From Eq. (19.82), when **Y***L* = 1,
-
-$$
-H(s) = \frac{-y_{21}}{1 + y_{22}} \tag{19.18.2}
-$$
-
-Comparing Eqs. (19.19.1) and (19.19.2), we obtain
-
-$$
-y_{21} = -\frac{1}{s^3 + 2s}
-$$
-,  $y_{22} = \frac{2s^2 + 1}{s^3 + 2s}$ 
-
-Any realization of *y*22 will automatically realize *y*21, since *y*22 is the output driving-point admittance, that is, the output admittance of the net work with the input port short-circuited. We determine the values of *L* and *C* in Fig. 19.63(a) that will give us *y*22. Recall that *y*22 is the shortcircuit output admittance. So we short-circuit the input port as shown in Fig. 19.63(b). First we get *L*3 by letting
-
-$$
-Z_A = \frac{1}{y_{22}} = \frac{s^3 + 2s}{2s^2 + 1} = sL_3 + Z_B
-$$
- (19.18.3)
-
-By long division,
-
-$$
-Z_A = 0.5s + \frac{1.5s}{2s^2 + 1}
-$$
- (19.18.4)
-
-Comparing Eqs. (19.18.3) and (19.18.4) shows that
-
-$$
-L_3 = 0.5H,
-$$
-  $Z_B = \frac{1.5s}{2s^2 + 1}$ 
-
-Next, we seek to get *C*2 as in Fig. 19.63(c) and let
-
-$$
-Y_B = \frac{1}{Z_B} = \frac{2s^2 + 1}{1.5s} = 1.333s + \frac{1}{1.5s} = sC_2 + Y_C
-$$
-
-from which *C*2 = 1.33 F and
-
-$$
-Y_C = \frac{1}{1.5s} = \frac{1}{sL_1} \qquad \Rightarrow \qquad L_1 = 1.5 \text{ H}
-$$
-
-Thus, the *LC* ladder netw ork in Fig. 19.63(a) with *L*1 = 1.5 H, *C*2 = 1.333 F, and *L*3 = 0.5 H has been synthesized to provide the given transfer function **H**(*s*). This result can be confirmed by finding **H**(*s*) = **V**2∕**V**<sup>1</sup> in Fig. 19.63(a) or by confirming the required *y*21.
-
-C2
-
-ZB
-
-L<sup>1</sup> L<sup>3</sup>
-
-(a)
-
-L<sup>1</sup> L<sup>3</sup>
-
-C<sup>2</sup> **V**<sup>2</sup> 1 Ω
-
-+
-
-‒
-
-(b)
-
-y<sup>22</sup> =
-
-1 ZA
-
-**Figure 19.63** For Example 19.18.
-
-**V**1 +
-
-‒
-
-<span id="page-913-0"></span>Realize the following transfer function using an *LC* ladder network ter- Practice Problem 19.18 minated in a 1-Ω resistor:
-
-r:  
-$$
-H(s) = \frac{2}{s^3 + s^2 + 4s + 2}
-$$
-
-**Answer:** Ladder network in Fig. 19.63(a) with *L*1 = *L*3 = 1.0 H and *C*2 = 500 mF.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/226_19.10 Summary.md

@@ -1,85 +0,0 @@
-# **19.10** Summary
-
-- 1. A two-port network is one with tw o ports (or tw o pairs of access terminals), known as input and output ports.
-- 2. The six parameters used to model a two-port network are the impedance [**z**], admittance [**y**], hybrid [**h**], inverse hybrid [**g**], transmission [**T**], and inverse transmission [**t**] parameters.
-- 3. The parameters relate the input and output port variables as
-
-$$
-\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = [\mathbf{z}] \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}, \qquad \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = [\mathbf{y}] \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix}, \qquad \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix} = [\mathbf{h}] \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix}
-$$
-$$
-\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix} = [\mathbf{g}] \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix}, \qquad \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_1 \end{bmatrix} = [\mathbf{T}] \begin{bmatrix} \mathbf{V}_2 \\ -\mathbf{I}_2 \end{bmatrix}, \qquad \begin{bmatrix} \mathbf{V}_2 \\ \mathbf{I}_2 \end{bmatrix} = [\mathbf{t}] \begin{bmatrix} \mathbf{V}_1 \\ -\mathbf{I}_1 \end{bmatrix}
-$$
-
-- 4. The parameters can be calculated or measured by short-circuiting or open-circuiting the appropriate input or output port.
-- 5. A two-port network is reciprocal if **z**12 = **z**21, **y**12 = **y**21, **h**12 = −**h**21, **g**12 = −**g**21, ∆*T* = 1 or ∆*t* = 1. Networks that have dependent sources are not reciprocal.
-- 6. Table 19.1 provides the relationships between the six sets of parameters. Three important relationships are
-
-$$
-[y] = [z]^{-1}, \qquad [g] = [h]^{-1}, \qquad [t] \neq [T]^{-1}
-$$
-
-- 7. Two-port netw orks may be connected in series, in parallel, or in cascade. In the series connection the *z* parameters are added, in the parallel connection the *y* parameters are added, and in the cascade connection the transmission parameters are multiplied in the correct order.
-- 8. One can use *PSpice* to compute the tw o-port parameters by con straining the appropriate port v ariables with a 1-A or 1-V source while using an open or short circuit to impose the other necessary constraints.
-- 9. The network parameters are specifically applied in the analysis of transistor circuits and the synthesis of ladder *LC* networks. Network parameters are especially useful in the analysis of transistor circuits because these circuits are easily modeled as tw o-port networks. *LC* ladder networks, important in the design of passive low-pass filters, resemble cascaded T networks and are therefore best analyzed as two-ports.
-
-# <span id="page-914-0"></span>Review Questions
-
-**19.1** For the single-element two-port network in Fig. 19.64(a), **z**11 is:
-
-(a) 0 (b) 5 (c) 10 (d) 20 (e) undefined
-
-# **Figure 19.64**
-
-For Review Questions.
-
-- **19.2** For the single-element two-port network in Fig. 19.64(b), **z**11 is:
-  - (a) 0 (b) 5 (c) 10
-  - (d) 20 (e) undefined
-- **19.3** For the single-element two-port network in Fig. 19.64(a), **y**11 is:
-  - (a) 0 (b) 5 (c) 10
-  - (d) 20 (e) undefined
-- **19.4** For the single-element two-port network in Fig. 19.64(b), **h**21 is:
-  - (a) −0.1 (b) −1 (c) 0 (d) 10 (e) undefined
-- **19.5** For the single-element two-port network in Fig. 19.64(a), **B** is:
-
-| (a) 0  | (b) 5         | (c) 10 |
-|--------|---------------|--------|
-| (d) 20 | (e) undefined |        |
-
-**19.6** For the single-element two-port network in Fig. 19.64(b), **B** is:
-
-| (a) 0  | (b) 5         | (c) 10 |
-|--------|---------------|--------|
-| (d) 20 | (e) undefined |        |
-
-**19.7** When port 1 of a two-port circuit is short-circuited, **I**1 = 4**I**2 and **V**2 = 0.25**I**2. Which of the following is true?
-
-| (a) y11 = 4  | (b) y12 = 16   |
-|--------------|----------------|
-| (c) y21 = 16 | (d) y22 = 0.25 |
-
-**19.8** A two-port is described by the following equations:
-
-**V**1 = 50**I**1 + 10**I**<sup>2</sup> **V**2 = 30**I**1 + 20**I**<sup>2</sup>
-
-Which of the following is *not* true?
-
-(a) **z**12 = 10 (b) **y**12 = −0.0143 (c) **h**12 = 0.5 (d) **A** = 50
-
-**19.9** If a two-port is reciprocal, which of the following is *not* true?
-
-(a) 
-$$
-\mathbf{z}_{21} = \mathbf{z}_{12}
-$$
-  
-\n(b)  $\mathbf{y}_{21} = \mathbf{y}_{12}$   
-\n(c)  $\mathbf{h}_{21} = \mathbf{h}_{12}$   
-\n(d)  $AD = BC + 1$ 
-
-**19.10** If the two single-element two-port networks in Fig. 19.64 are cascaded, then **D** is:
-
-> (a) 0 (b) 0.1 (c) 2 (d) 10 (e) undefined
-
-*Answers: 19.1c, 19.2e , 19.3e , 19.4b, 19.5a, 19.6c, 19.7b, 19.8d, 19.9c, 19.10c.*

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/227_Problems.md

@@ -1,617 +0,0 @@
-# Problems
-
-# Section 19.2 Impedance Parameters
-
-**19.1** Obtain the *z* parameters for the network in Fig. 19.65.
-
-**19.2** Find the impedance parameter equivalent of the network in Fig. 19.66. \*
-
-**Figure 19.66**
-
-For Prob. 19.2.
-
-\* An asterisk indicates a challenging problem.
-
-**Figure 19.67**
-
-For Prob. 19.3.
-
-**Figure 19.68** For Prob. 19.4.
-
-**19.5** Obtain the *z* parameters for the network in Fig. 19.69 as functions of *s*.
-
-# **Figure 19.69**
-
-For Prob. 19.5.
-
-**19.6** Compute the *z* parameters of the circuit in Fig. 19.70.
-
-**19.7** Calculate the *z* parameters of the circuit in Fig. 19.71 as functions of *s*.
-
-For Prob. 19.7 and 19.80.
-
-**19.3** Find the *z* parameters of the circuit in Fig. 19.67. **19.8** Find the *z* parameters of the two-port in Fig. 19.72.
-
-For Prob. 19.8.
-
-**19.9** The *y* parameters of a network are:
-
-$$
-\mathbf{Y} = [\mathbf{y}] = \begin{bmatrix} 0.5 & -0.2 \\ -0.2 & 0.4 \end{bmatrix} S
-$$
-
-Determine the *z* parameters for the network.
-
-**19.10** Construct a two-port that realizes each of the following *z* parameters.
-
-(a) 
-$$
-\left[\mathbf{z}\right] = \begin{bmatrix} 25 & 20 \\ 5 & 10 \end{bmatrix} \Omega
-$$
-  
-\n(b)  $\left[\mathbf{z}\right] = \begin{bmatrix} 1 + \frac{3}{s} & \frac{1}{s} \\ \frac{1}{s} & 2s + \frac{1}{s} \end{bmatrix} \Omega$ 
-
-**19.11** Determine a two-port network that is represented by the following *z* parameters:
-
-$$
-[\mathbf{z}] = \begin{bmatrix} 6+j3 & 5-j2 \\ 5-j2 & 8-j \end{bmatrix} \Omega
-$$
-
-**19.12** For the circuit shown in Fig. 19.73, let
-
-$$
-\begin{bmatrix} \mathbf{z} \end{bmatrix} = \begin{bmatrix} 10 & -6 \\ -4 & 12 \end{bmatrix} \Omega
-$$
-
-Find 
-$$
-I_1
-$$
-,  $I_2$ ,  $V_1$ , and  $V_2$ .
-
-**Figure 19.73** For Prob. 19.12.
-
-**19.13** Determine the average power delivered to *ZL* = 5 + *j*4 in the network of Fig. 19.74. *Note:* The voltage is rms.
-
-# **Figure 19.74**
-
-For Prob. 19.13.
-
-**19.14** For the two-port network shown in Fig. 19.75, show that at the output terminals,
-
-$$
-\mathbf{Z}_{Th} = \mathbf{z}_{22} - \frac{\mathbf{z}_{12}\mathbf{z}_{21}}{\mathbf{z}_{11} + \mathbf{Z}_s}
-$$
-
-and
-
-$$
-\mathbf{V}_{\mathrm{Th}} = \frac{\mathbf{z}_{21}}{\mathbf{z}_{11} + \mathbf{Z}_s} \mathbf{V}_s
-$$
-
-# **Figure 19.75**
-
-For Probs. 19.14 and 19.41.
-
-**19.15** For the two-port circuit in Fig. 19.76,
-
-$$
-\begin{bmatrix} \mathbf{z} \end{bmatrix} = \begin{bmatrix} 40 & 60 \\ 80 & 120 \end{bmatrix} \Omega
-$$
-
-- (a) Find **Z***L* for maximum power transfer to the load.
-- (b) Calculate the maximum power delivered to the load.
-
-# **Figure 19.76**
-
-For Prob. 19.15.
-
-**19.16** For the circuit in Fig. 19.77, at *ω* = 2 rad/s, **z**11 = 10 Ω, **z**12 = **z**21 = *j*6 Ω, **z**22 = 4 Ω. Obtain the Thevenin equivalent circuit at terminals *a*-*b* and calculate *vo*.
-
-**Figure 19.77** For Prob. 19.16.
-
-# Section 19.3 Admittance Parameters
-
-**19.17** Determine the *z* and *y* parameters for the circuit in Fig. 19.78. \*
-
-# **Figure 19.78**
-
-For Prob. 19.17.
-
-**19.18** Calculate the *y* parameters for the two-port in Fig. 19.79.
-
-For Probs. 19.18 and 19.37.
-
-**19.19** Using Fig. 19.80, design a problem to help other students better understand how to find *y* parameters in the *s*-domain.
-
-**Figure 19.80** For Prob. 19.19.
-
-**19.20** Find the *y* parameters for the circuit in Fig. 19.81.
-
-For Prob. 19.20.
-
-Problems **895**
-
-**19.21** Obtain the admittance parameter equivalent circuit of the two-port in Fig. 19.82.
-
-**Figure 19.82**
-
-- For Prob. 19.21.
-- **19.22** Obtain the *y* parameters of the two-port network in Fig. 19.83.
-
-**Figure 19.83** For Prob. 19.22.
-
-**19.23** (a) Find the *y* parameters of the two-port in Fig. 19.84.
-
-(b) Determine **V**2(*s*) for *vs* = 2*u*(*t*) V.
-
-**Figure 19.84** For Prob. 19.23.
-
-**19.24** Find the resistive circuit that represents these *y* parameters:
-
-$$
-[\mathbf{y}] = \begin{bmatrix} \frac{1}{2} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{3}{8} \end{bmatrix} S
-$$
-
-**19.25** Draw the two-port network that has the following *y* parameters:
-
-$$
-[\mathbf{y}] = \begin{bmatrix} 1 & -0.5 \\ -0.5 & 1.5 \end{bmatrix} \mathbf{S}
-$$
-
-**19.26** Calculate [**y**] for the two-port in Fig. 19.85.
-
-**Figure 19.85**
-
-For Prob. 19.26.
-
-**19.27** Find the *y* parameters for the circuit in Fig. 19.86.
-
-# **Figure 19.86**
-
-For Prob. 19.27.
-
-- **19.28** In the circuit of Fig. 19.65, the input port is connected to a 1-A current source and the right hand side of the circuit is left open (*I*<sup>2</sup> = 0). Calculate the power absorbed by the circuit by using the *y* parameters. Confirm your result by direct circuit analysis.
-- **19.29** In the bridge circuit of Fig. 19.87, *I*1 = 20 A and *I*2 = −8 A.
-  - (a) Find *V*1 and *V*2 using *y* parameters.
-  - (b) Confirm the results in part (a) by direct circuit analysis.
-
-# **Figure 19.87**
-
-For Prob. 19.29.
-
-# Section 19.4 Hybrid Parameters
-
-**19.30** Find the *h* parameters for the networks in Fig. 19.88.
-
-**19.31** Determine the hybrid parameters for the network in Fig. 19.89.
-
-# **Figure 19.89**
-
-For Prob. 19.31.
-
-**Figure 19.90** For Prob. 19.32.
-
-**19.33** Obtain the *h* parameters for the two-port of Fig. 19.91.
-
-For Prob. 19.33.
-
-**19.34** Obtain the *h* and *g* parameters of the two-port in Fig. 19.92.
-
-**19.35** Determine the *h* parameters for the network in Fig. 19.93.
-
-For Prob. 19.35.
-
-**19.36** For the two-port in Fig. 19.94,
-
-$$
-[\mathbf{h}] = \begin{bmatrix} 16 \,\Omega & 3 \\ -2 & 0.01 \,\mathrm{S} \end{bmatrix}
-$$
-
-Find:
-
-(a) 
-$$
-V_2/V_1
-$$
-   
-\n(b)  $I_2/I_1$   
-\n(c)  $I_1/V_1$    
-\n(d)  $V_2/I_1$ 
-
-# **Figure 19.94**
-
-For Prob. 19.36.
-
-- **19.37** The input port of the circuit in Fig. 19.79 is connected to a 10-V dc voltage source while the output port is terminated by a 5-Ω resistor. Find the voltage across the 5-Ω resistor by using *h* parameters of the circuit. Confirm your result by using direct circuit analysis.
-- **19.38** The *h* parameters of the two-port of Fig. 19.95 are:
-
-$$
-[\mathbf{h}] = \begin{bmatrix} 600 \,\Omega & 0.04 \\ 30 & 2 \,\text{mS} \end{bmatrix}
-$$
-
- Given the *Zs* = 2 kΩ and *ZL* = 400 Ω, find *Z*in and *Z*out.
-
-**Figure 19.95** For Prob. 19.38.
-
-**19.39** Obtain the *g* parameters for the wye circuit of Fig. 19.96.
-
-Problems **897**
-
-For Prob. 19.39.
-
-**19.40** Using Fig. 19.97, design a problem to help other students better understand how to find *g* parameters in an ac circuit.
-
-# **Figure 19.97**
-
-For Prob. 19.40.
-
-**19.41** For the two-port in Fig. 19.75, show that
-
-$$
-\frac{I_2}{I_1} = \frac{-g_{21}}{g_{11}Z_L + \Delta_g}
-$$
-$$
-\frac{V_2}{V_s} = \frac{g_{21}Z_L}{(1 + g_{11}Z_s)(g_{22} + Z_L) - g_{21}g_{12}Z_s}
-$$
-
-where ∆*g* is the determinant of [**g**] matrix.
-
-**19.42** The *h* parameters of a two-port device are given by
-
-$$
-\mathbf{h}_{11} = 600 \ \Omega, \qquad \mathbf{h}_{12} = 10^{-3}, \qquad \mathbf{h}_{21} = 120,
-$$
-\n
-$$
-\mathbf{h}_{22} = 2 \times 10^{-6} \ \mathrm{S}
-$$
-
-Draw a circuit model of the device including the value of each element.
-
-# Section 19.5 Transmission Parameters
-
-**19.43** Find the transmission parameters for the singleelement two-port networks in Fig. 19.98.
-
-**19.44** Using Fig. 19.99, design a problem to help other students better understand how to find the transmission parameters of an ac circuit.
-
-# **Figure 19.99**
-
-- For Prob. 19.44.
-- **19.45** Find the **ABCD** parameters for the circuit in Fig. 19.100.
-
-# **Figure 19.100**
-
-For Prob. 19.45.
-
-**19.46** Find the transmission parameters for the circuit in Fig. 19.101.
-
-# **Figure 19.101**
-
-For Prob. 19.46.
-
-**19.47** Obtain the **ABCD** parameters for the network in Fig. 19.102.
-
-# **Figure 19.102**
-
-For Prob. 19.47
-
-- **19.48** For a two-port, let **A** = 4, **B** = 30 Ω, **C** = 0.1 S, and **D** = 1.5. Calculate the input impedance **Z**in = **V**1∕**I**1, when:
-  - (a) the output terminals are short-circuited,
-  - (b) the output port is open-circuited,
-  - (c) the output port is terminated by a 10-Ω load.
-
-**19.49** Using impedances in the *s*-domain, obtain the transmission parameters for the circuit in Fig. 19.103.
-
-**19.50** Derive the *s*-domain expression for the *t* parameters of the circuit in Fig. 19.104.
-
-For Prob. 19.50.
-
-**19.51** Obtain the *t* parameters for the network in Fig. 19.105.
-
-**Figure 19.105** For Prob. 19.51.
-
-# Section 19.6 Relationships Between Parameters
-
-**19.52** (a) For the *T* network in Fig. 19.106, show that the *h* parameters are:
-
-$$
-\mathbf{h}_{11} = R_1 + \frac{R_2 R_3}{R_1 + R_3}, \qquad \mathbf{h}_{12} = \frac{R_2}{R_2 + R_3}
-$$
-$$
-\mathbf{h}_{21} = -\frac{R_2}{R_2 + R_3}, \qquad h_{22} = \frac{1}{R_2 + R_3}
-$$
-
-**Figure 19.106** For Prob. 19.52.
-
-(b) For the same network, show that the transmission parameters are:
-
-$$
-\mathbf{A} = 1 + \frac{R_1}{R_2}, \qquad \mathbf{B} = R_3 + \frac{R_1}{R_2}(R_2 + R_3)
-$$
-$$
-\mathbf{C} = \frac{1}{R_2}, \qquad \mathbf{D} = 1 + \frac{R_3}{R_2}
-$$
-
-- **19.53** Through derivation, express the *z* parameters in terms of the **ABCD** parameters.
-- **19.54** Show that the transmission parameters of a two-port may be obtained from the *y* parameters as:
-
-$$
-\mathbf{A} = -\frac{\mathbf{y}_{22}}{\mathbf{y}_{21}}, \qquad \mathbf{B} = -\frac{1}{\mathbf{y}_{21}}
-$$
-$$
-\mathbf{C} = -\frac{\Delta_y}{\mathbf{y}_{21}}, \qquad \mathbf{D} = -\frac{\mathbf{y}_{11}}{\mathbf{y}_{21}}
-$$
-
-**19.55** Prove that the *g* parameters can be obtained from the *z* parameters as
-
-$$
-\mathbf{g}_{11} = \frac{1}{\mathbf{z}_{11}}, \qquad \mathbf{g}_{12} = -\frac{\mathbf{z}_{12}}{\mathbf{z}_{11}},
-$$
-$$
-\mathbf{g}_{21} = \frac{\mathbf{z}_{21}}{\mathbf{z}_{11}}, \qquad \mathbf{g}_{22} = \frac{\Delta_z}{\mathbf{z}_{11}}
-$$
-
-**19.56** For the network of Fig. 19.107, obtain **Vo**∕**Vs**.
-
-# **Figure 19.107**
-
-For Prob. 19.56.
-
-**19.57** Given the transmission parameters
-
-$$
-[\mathbf{T}] = \begin{bmatrix} 3 & 20 \\ 1 & 7 \end{bmatrix}
-$$
-
-obtain the other five two-port parameters.
-
-**19.58** Design a problem to help other students better understand how to develop the *y* parameters and transmission parameters, given equations in terms of the hybrid parameters.
-
-**19.59** Given that
-
-$$
-[\mathbf{g}] = \begin{bmatrix} 0.06 \text{ S} & -0.4 \\ 0.2 & 2 \Omega \end{bmatrix}
-$$
-
-determine:
-
-(a) [**z**] (b) [**y**] (c) [**h**] (d) [**T**]
-
-Problems **899**
-
-**19.60** Design a T network necessary to realize the following *z* parameters at *ω* = 10<sup>6</sup> rad/s.
-
-$$
-\begin{bmatrix} \mathbf{z} \end{bmatrix} = \begin{bmatrix} 4+j3 & 3 \\ 2 & 5-j \end{bmatrix} \mathbf{k} \Omega
-$$
-
-- **19.61** For the bridge circuit in Fig. 19.108, obtain:
-  - (a) the *z* parameters
-  - (b) the *h* parameters
-  - (c) the transmission parameters
-
-# **Figure 19.108**
-
-For Prob. 19.61.
-
-**19.62** Find the *z* parameters of the op amp circuit in Fig. 19.109. Obtain the transmission parameters.
-
-- For Prob. 19.62.
-- **19.63** Determine the *z* parameters of the two-port in Fig. 19.110.
-
-**Figure 19.110** For Prob. 19.63.
-
-**19.64** Determine the *y* parameters at *ω* = 1,000 rad/s for the op amp circuit in Fig. 19.111. Find the corresponding *h* parameters.
-
-# **Figure 19.111**
-
-For Prob. 19.64.
-
-# Section 19.7 Interconnection of Networks
-
-**19.65** What is the *y* parameter presentation of the circuit in Fig. 19.112?
-
-# **Figure 19.112** For Prob. 19.65.
-
-**19.66** In the two-port of Fig. 19.113, let **y**12 = **y**21 = 0, **y**11 = 2 mS, and **y**22 = 10 mS. Find **V***o*∕**V***s*.
-
-# **Figure 19.113** For Prob. 19.66.
-
-**19.67** If three copies of the circuit in Fig. 19.114 are connected in parallel, find the overall transmission
-
-**19.68** Obtain the *h* parameters for the network in Fig. 19.115.
-
-For Prob. 19.68.
-
-**19.69** The circuit in Fig. 19.116 may be regarded as two two-ports connected in parallel. Obtain the *y* parameters as functions of *s*. \*
-
-**19.70** For the parallel-series connection of the two two-ports in Fig. 19.117, find the *g* parameters. \*
-
-**19.71** Determine the *z* parameters for the network in Fig. 19.118. \*
-
-# **Figure 19.118**
-
-For Prob. 19.71.
-
-**19.72** A series-parallel connection of two two-ports is shown in Fig. 19.119. Determine the *z* parameter representation of the network. \*
-
-# **Figure 19.119**
-
-For Prob. 19.72.
-
-- **19.73** Three copies of the circuit shown in Fig. 19.70 are connected in cascade. Determine the *z* parameters.
-- **19.74** Determine the **ABCD** parameters of the circuit in Fig. 19.120 as functions of *s*. (*Hint:* Partition the circuit into subcircuits and cascade them using the results of Prob. 19.43.) \*
-
-# **Figure 19.120**
-
-For Prob. 19.74.
-
-**19.75** For the individual two-ports shown in Fig. 19.121 where, \*
-
-$$
-\begin{bmatrix} \mathbf{z}_a \end{bmatrix} = \begin{bmatrix} 8 & 6 \\ 4 & 5 \end{bmatrix} \Omega \quad \begin{bmatrix} \mathbf{y}_b \end{bmatrix} = \begin{bmatrix} 8 & -4 \\ 2 & 10 \end{bmatrix} S
-$$
-
-- (a) Determine the *y* parameters of the overall two-port.
-- (b) Find the voltage ratio **V***o*∕**V***i* when **Z***L* = 2 Ω.
-
-**Figure 19.121** For Prob. 19.75.
-
-# Section 19.8 Computing Two-Port Parameters Using PSpice
-
-**19.76** Use *PSpice* or *MultiSim* to obtain the *z* parameters of
-
-**Figure 19.122** For Prob. 19.76.
-
-**19.77** Using *PSpice* or *MultiSim,* find the *h* parameters of the network in Fig. 19.123. Take *ω* = 1 rad/s.
-
-**19.78** Obtain the *h* parameters at *ω* = 4 rad/s for the circuit in Fig. 19.124 using *PSpice* or *MultiSim*.
-
-For Prob. 19.78.
-
-**19.79** Use *PSpice* or *MultiSim* to determine the *z* parameters of the circuit in Fig. 19.125. Take *ω* = 2 rad/s.
-
-# **Figure 19.125**
-
-For Prob. 19.79.
-
-- the network in Fig. 19.122. **19.80** Use *PSpice* or *MultiSim* to find the *z* parameters of the circuit in Fig. 19.71.
-  - **19.81** Repeat Prob. 19.26 using *PSpice* or *MultiSim*.
-  - **19.82** Use *PSpice* or *MultiSim* to rework Prob. 19.31.
-  - **19.83** Rework Prob. 19.47 using *PSpice* or *MultiSim*.
-  - **19.84** Using *PSpice* or *MultiSim,* find the transmission parameters for the network in Fig. 19.126.
-
-**Figure 19.126** For Prob. 19.84.
-
-**19.85** At *ω* = 1 rad/s, find the transmission parameters of the network in Fig. 19.127 using *PSpice* or *MultiSim*.
-
-**19.86** Obtain the *g* parameters for the network in Fig. 19.128 using *PSpice* or *MultiSim.*
-
-**Figure 19.128** For Prob. 19.86.
-
-**19.87** For the circuit shown in Fig. 19.129, use *PSpice* or *MultiSim* to obtain the *t* parameters. Assume *ω* = 1 rad/s.
-
-# **Figure 19.129**
-
-For Prob. 19.87.
-
-Section 19.9 Applications
-
-- **19.88** Using the *y* parameters, derive formulas for *Z*in, *Z*out, *Ai*, and *Av* for the common-emitter transistor circuit.
-- **19.89** A transistor has the following parameters in a common-emitter circuit:
-
-$$
-h_{ie} = 2,640 \Omega
-$$
-,  $h_{re} = 2.6 \times 10^{-4}$ 
-
-$$
-h_{fe} = 72
-$$
-,  $h_{oe} = 16 \,\mu\text{S}$ ,  $R_L = 100 \,\text{k}\Omega$ 
-
-What is the voltage amplification of the transistor? How many decibels gain is this?
-
-**19.90** A transistor with
-
-*hfe* = 120, h*ie* = 2 kΩ
-
-$$
-h_{re} = 10^{-4}
-$$
-,  $h_{oe} = 20 \,\mu\text{S}$ 
-
-is used for a CE amplifier to provide an input resistance of 1.5 kΩ.
-
-- (a) Determine the necessary load resistance *RL*.
-- (b) Calculate *Av*, *Ai*, and *Zout* if the amplifier is driven by a 4-mV source having an internal resistance of 600 Ω.
-- (c) Find the voltage across the load.
-- **19.91** For the transistor network of Fig. 19.130,
-
-$$
-h_{fe} = 80
-$$
-,  $h_{ie} = 1.2 \text{ k}\Omega$   
- $h_{re} = 1.5 \times 10^{-4}$ ,  $h_{oe} = 20 \mu\text{S}$ 
-
-Determine the following:
-
-- (a) voltage gain *Av* = *Vo*∕*Vs*,
-- (b) current gain *Ai* = *Io*∕*Ii*,
-- (c) input impedance *Z*in,
-- (d) output impedance *Z*out.
-
-**19.92** Determine *Av*, *Ai*, *Z*in, and *Z*out for the amplifier shown in Fig. 19.131. Assume that \*
-
-$$
-h_{ie} = 4 \text{ k}\Omega, \qquad h_{re} = 10^{-4}
-$$
-  
-$$
-h_{fe} = 100, \qquad h_{oe} = 30 \text{ }\mu\text{S}
-$$
-
-# **Figure 19.131**
-
-For Prob. 19.92.
-
-**19.93** Calculate *Av*, *Ai*, *Z*in, and *Z*out for the transistor network in Fig. 19.132. Assume that \*
-
-$$
-h_{ie} = 2 \text{ k}\Omega, \qquad h_{re} = 2.5 \times 10^{-4}
-$$
-$$
-h_{fe} = 150, \qquad h_{oe} = 10 \text{ }\mu\text{S}
-$$
-
-# **Figure 19.132**
-
-For Prob. 19.93.
-
-**19.94** A transistor in its common-emitter mode is specified by
-
-$$
-[\mathbf{h}] = \begin{bmatrix} 200 \,\Omega & 0 \\ 100 & 10^{-6} \,\mathrm{S} \end{bmatrix}
-$$
-
- Two such identical transistors are connected in cascade to form a two-stage amplifier used at audio frequencies. If the amplifier is terminated by a 4-kΩ resistor, calculate the overall *Av* and *Z*in.
-
-**19.95** Realize an *LC* ladder network such that
-
-C ladder network such  
-$$
-y_{22} = \frac{s^3 + 5s}{s^4 + 10s^2 + 8}
-$$
-
-**19.96** Design an *LC* ladder network to realize a low-pass filter with transfer function
-
-Design an LC ladder network to realize a low filter with transfer function  
-\n
-$$
-H(s) = \frac{1}{s^4 + 2.613s^2 + 3.414s^2 + 2.613s + 1}
-$$
-
-**19.97** Synthesize the transfer function
-
-$$
-H(s) = \frac{V_o}{V_s} = \frac{s^3}{s^3 + 6s + 12s + 24}
-$$
-
-using the *LC* ladder network in Fig. 19.133.
-
-<span id="page-925-0"></span>
-
-# **Figure 19.133**
-
-For Prob. 19.97.
-
-For Prob. 19.98. **19.98** A two-stage amplifier in Fig. 19.134 contains tw<sup>o</sup> identical stages with
-
-$$
-[\mathbf{h}] = \begin{bmatrix} 2 \text{k}\Omega & 0.004 \\ 200 & 500 \,\mu\text{S} \end{bmatrix}
-$$
-
- If **Z***L* = 20 kΩ, find the required value of **V***s* to produce **V***o* = 16 V.
-
-**Figure 19.134**

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/228_Comprehensive Problem.md

@@ -1,60 +0,0 @@
-# Comprehensive Problem
-
-**19.99** Assume that the two circuits in Fig. 19.135 are equivalent. The parameters of the two circuits must be equal. Using this factor and the *z* parameters, derive Eqs. (9.67) and (9.68).
-
-**Figure 19.135** For Prob. 19.99.
-
-# <span id="page-926-0"></span>Appendix A
-
-Simultaneous Equations and Matrix Inversion
-
-In circuit analysis, we often encounter a set of simultaneous equations having the form
-
-$$
-a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1
-$$
-  
-\n
-$$
-a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2
-$$
-  
-\n
-$$
-\vdots \qquad \vdots
-$$
-  
-\n
-$$
-a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = b_n
-$$
-  
-\n(A.1)
-
-where there are *n* unknown *x*1, *x*2, . . . , *xn* to be determined. Equation (A.1) can be written in matrix form as
-
-$$
-\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} b_2 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}
-$$
- (A.2)
-
-This matrix equation can be put in a compact form as
-
-$$
-AX = B \tag{A.3}
-$$
-
-where
-
-$$
-\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}, \quad \mathbf{X} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix}
-$$
- (A.3)
-
-**A** is a square (*n* × *n*) matrix while **X** and **B** are column matrices.
-
-There are several methods for solving Eq. (A.1) or (A.3). These in clude substitution, Gaussian elimination, Cramer's rule, matrix inver sion, and numerical analysis.
-
-# **A.1** Cramer's Rule
-
-In many cases, Cramer's rule can be used to solve the simultaneous equa tions we encounter in circuit analysis. Cramer's rule states that the solution to Eq. (A.1) or (A.3) is

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/229_Appendix A - Simultaneous Equations and Matrix Inversion.md

@@ -1,518 +0,0 @@
-
-$$
-\begin{aligned}\nx_1 &= \frac{\Delta_1}{\Delta} \\
-x_2 &= \frac{\Delta_2}{\Delta} \\
-&\vdots \\
-x_n &= \frac{\Delta_n}{\Delta}\n\end{aligned}
-$$
-\n(A.5)
-
-where the ∆'s are the determinants given by
-
-$$
-\Delta = \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix}, \qquad \Delta_1 = \begin{vmatrix} b_1 & a_{12} & \cdots & a_{1n} \\ b_2 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ b_n & a_{n2} & \cdots & a_{nn} \end{vmatrix}
-$$
-$$
-\Delta_2 = \begin{vmatrix} a_{11} & b_1 & \cdots & a_{1n} \\ a_{21} & b_2 & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{n1} & b_n & \cdots & a_{nn} \end{vmatrix}, \dots, \Delta_n = \begin{vmatrix} a_{11} & a_{12} & \cdots & b_1 \\ a_{21} & a_{22} & \cdots & b_2 \\ \vdots & \vdots & \cdots & \vdots \\ a_{n1} & a_{n2} & \cdots & b_n \end{vmatrix}
-$$
-$$
-(A.6)
-$$
-
-Notice that ∆ is the determinant of matrix A and ∆ *<sup>k</sup>* is the determinant of the matrix formed by replacing the *k*th column of **A** by **B**. It is evi dent from Eq. (A.5) that Cramer's rule applies only when ∆ ≠ 0. When ∆ = 0, the set of equations has no unique solution, because the equations are linearly dependent.
-
-The value of the determinant ∆, for example, can be obtained by expanding along the first row:
-
-$$
-\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn} \end{vmatrix}
-$$
-  
-=  $a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} + \cdots + (-1)^{1+n}a_{1n}M_{1n}$  (A.7)
-
-where the minor *Mij* is an ( *n* − 1) × ( *n* − 1) determinant of the matrix formed by striking out the *i*th row and *j*th column. The value of ∆ may also be obtained by expanding along the first column:
-
-$$
-\Delta = a_{11}M_{11} - a_{21}M_{21} + a_{31}M_{31} + \dots + (-1)^{n+1}a_{n1}M_{n1}
-$$
- (A.8)
-
-We now specifically develop the formulas for calculating the deter minants of 2 × 2 and 3 × 3 matrices, because of their frequent occurrence in this text. For a 2 × 2 matrix,
-
-$$
-\Delta = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}
-$$
- (A.9)
-
-1 *n* 1 *n*
-
-For a 3 × 3 matrix,
-
-$$
-\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11}(-1)^2 \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + a_{21}(-1)^3 \begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix}
-$$
-  
-+  $a_{31}(-1)^4 \begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix}$   
-=  $a_{11}(a_{22}a_{33} - a_{32}a_{23}) - a_{21}(a_{12}a_{33} - a_{32}a_{13})$   
-+  $a_{31}(a_{12}a_{23} - a_{22}a_{13})$  (A.10)
-
-An alternative method of obtaining the determinant of a 3 × 3 matrix is by repeating the first two rows and multiplying the terms diagonally as follows.
-
-$$
-= a_{11}a_{22}a_{33} + a_{21}a_{32}a_{13} + a_{31}a_{12}a_{23} - a_{13}a_{22}a_{31} - a_{23}a_{32}a_{11}
-$$
-  
--a<sub>33</sub>a<sub>12</sub>a<sub>21</sub> (A.11)
-
-# In summary:
-
-The solution of linear simultaneous equations by Cramer's rule boils down to finding
-
-$$
-x_k = \frac{\Delta_k}{\Delta}, \qquad k = 1, 2, \dots, n \tag{A.12}
-$$
-
-where ∆ is the determinant of matrix A and ∆k is the determinant of the matrix formed by replacing the kth column of A by B.
-
-You may not find much need to use Cramer's method described in this appendix, in view of the availability of calculators, computers, and software packages such as *MATLAB*, which can be used easily to solve a set of linear equations. But in case you need to solve the equations by hand, the material covered in this appendix becomes useful. At any rate, it is important to know the mathematical basis of those calculators and software packages.
-
-Example A.1 Solve the simultaneous equations
-
-$$
-4x_1 - 3x_2 = 17, \qquad -3x_1 + 5x_2 = -21
-$$
-
-# **Solution:**
-
-The given set of equations is cast in matrix form as
-
-$$
-\begin{bmatrix} 4 & -3 \ -3 & 5 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 17 \ -21 \end{bmatrix}
-$$
-
-The determinants are evaluated as
-
-$$
-\Delta = \begin{vmatrix} 4 & -3 \\ -3 & 5 \end{vmatrix} = 4 \times 5 - (-3)(-3) = 11
-$$
-  
-\n
-$$
-\Delta_1 = \begin{vmatrix} 17 & -3 \\ -21 & 5 \end{vmatrix} = 17 \times 5 - (-3)(-21) = 22
-$$
-  
-\n
-$$
-\Delta_2 = \begin{vmatrix} 4 & 17 \\ -3 & -21 \end{vmatrix} = 4 \times (-21) - 17 \times (-3) = -33
-$$
-
-One may use other methods, such as matrix inversion and elimination. Only Cramer's method is covered here, because of its simplicity and also because of the availability of powerful calculators.
-
-Hence,
-
-$$
-x_1 = \frac{\Delta_1}{\Delta} = \frac{22}{11} = 2
-$$
-,  $x_2 = \frac{\Delta_2}{\Delta} = \frac{-33}{11} = -3$ 
-
-Find the solution to the following simultaneous equations:
-
-3*x*<sup>1</sup> − *x*2 = 4, −6*x*1 + 18*x*2 = 16
-
-**Answer:** *x*1 = 1.833, *x*2 = 1.5.
-
-Determine *x*1, *x*2, and *x*3 for this set of simultaneous equations:
-
-$$
-25x1 - 5x2 - 20x3 = 50
-$$
-  
-$$
--5x1 + 10x2 - 4x3 = 0
-$$
-  
-$$
--5x1 - 4x2 + 9x3 = 0
-$$
-
-# **Solution:**
-
-In matrix form, the given set of equations becomes
-
-| 25      | −5 | −20    | x1           |   | 50          |  |
-|---------|----|--------|--------------|---|-------------|--|
-| −5      | 10 | −4     | x2           | = | 0           |  |
-| [<br>−5 | −4 | ]<br>9 | [<br>]<br>x3 |   | [<br>]<br>0 |  |
-
-We apply Eq. (A.11) to find the determinants. This requires that we repeat the first two rows of the matrix. Thus,
-
-$$
-\Delta = \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & -4 & 9 \end{vmatrix} = \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & -5 & 10 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\ -5 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 25 & -5 & -20 \\ -5 & 10 & -4 \\
-$$
-
-Similarly,
-
-$$
-\Delta_1 = \begin{vmatrix} 50 & -5 & -20 \\ 0 & 10 & -4 \\ 0 & -4 & 9 \end{vmatrix} = \begin{vmatrix} 50 & -5 & -20 \\ 0 & 10 & -4 \\ 50 & 50 & 5 \\ 0 & 10 & -4 \end{vmatrix} + \begin{vmatrix} 50 & -5 & -20 \\ 0 & 4 & 4 \\ + & 0 & 10 \end{vmatrix}
-$$
-
-Practice Problem A.1
-
-Example A.2
-
-$$
-= 0 + 0 + 1000 - 0 - 0 + 2250 = 3250
-$$
-
-= 0 + 1000 + 0 + 2500 − 0 − 0 = 3500
-
-Hence, we now find
-
-$$
-x_1 = \frac{\Delta_1}{\Delta} = \frac{3700}{125} = 29.6
-$$
-$$
-x_2 = \frac{\Delta_2}{\Delta} = \frac{3250}{125} = 26
-$$
-$$
-x_3 = \frac{\Delta_2}{\Delta} = \frac{3500}{125} = 28
-$$
-
-Obtain the solution of this set of simultaneous equations: Practice Problem A.2
-
-> 3*x*<sup>1</sup> − *x*<sup>2</sup> − 2*x*3 = 1 −*x*1 + 6*x*<sup>2</sup> − 3*x*3 = 0 −2*x*<sup>1</sup> − 3*x*2 + 6*x*3 = 6
-
-**Answer:** *x*1 = 3 = *x*3, *x*2 = 2.
-
-# **A.2** Matrix Inversion
-
-The linear system of equations in Eq. (A.3) can be solved by matrix inversion. In the matrix equation **AX** = **B**, we may invert **A** to get **X**, i.e.,
-
-$$
-\mathbf{X} = \mathbf{A}^{-1} \mathbf{B} \tag{A.13}
-$$
-
-where **A**<sup>−</sup><sup>1</sup> is the inverse of **A**. Matrix inversion is needed in other applications apart from using it to solve a set of equations.
-
-By definition, the inverse of matrix **A** satisfies
-
-$$
-\mathbf{A}^{-1}\mathbf{A} = \mathbf{A}\mathbf{A}^{-1} = \mathbf{I}
-$$
- (A.14)
-
-where **I** is an identity matrix. **A**<sup>−</sup><sup>1</sup> is given by
-
-$$
-A^{-1} = \frac{\text{adj } A}{\text{det } A}
-$$
- (A.15)
-
-where adj **A** is the adjoint of **A** and det **A** = |**A**| is the determinant of **A**. The adjoint of **A** is the transpose of the cofactors of **A**. Suppose we are given an *n* × *n* matrix **A** as
-
-$$
-\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}
-$$
- (A.16)
-
-The cofactors of **A** are defined as
-
-$$
-\mathbf{C} = \text{cof}(\mathbf{A}) = \begin{bmatrix} c_{11} & c_{12} & \cdots & c_{1n} \\ c_{21} & c_{22} & \cdots & c_{2n} \\ \vdots & & & \\ c_{n1} & c_{n2} & \cdots & c_{nn} \end{bmatrix}
-$$
-(A.17)
-
-where the cofactor *cij* is the product of ( −1)*<sup>i</sup>+<sup>j</sup>* and the determinant of the (*n* − 1) × (*n* − 1) submatrix is obtained by deleting the *i*th row and *j*th column from **A**. For example, by deleting the first row and the first column of **A** in Eq. (A.16), we obtain the cofactor *c*11 as
-
-$$
-c_{11} = (-1)^2 \begin{vmatrix} a_{22} & a_{23} & \cdots & a_{2n} \\ a_{32} & a_{33} & \cdots & a_{3n} \\ \vdots & & & \\ a_{n2} & a_{n3} & \cdots & a_{nn} \end{vmatrix}
-$$
- (A.18)
-
-Once the cofactors are found, the adjoint of **A** is obtained as
-
-$$
-adj (A) = \begin{bmatrix} c_{11} & c_{12} & \cdots & c_{1n} \\ c_{21} & c_{22} & \cdots & c_{2n} \\ \vdots & & & \\ c_{n1} & c_{n2} & \cdots & c_{nn} \end{bmatrix}^T = C^T
-$$
- (A.19)
-
-where *T* denotes transpose.
-
- In addition to using the cofactors to find the adjoint of **A**, they are also used in finding the determinant of **A** which is given by
-
-$$
-|\mathbf{A}| = \sum_{j=1}^{n} a_{ij} c_{ij}
-$$
- (A.20)
-
-where *i* is any value from 1 to *n*. By substituting Eqs. (A.19) and (A.20) into Eq. (A.15), we obtain the inverse of **A** as
-
-$$
-\mathbf{A}^{-1} = \frac{\mathbf{C}^T}{|\mathbf{A}|} \tag{A.21}
-$$
-
-For a 2 × 2 matrix, if
-
-$$
-\mathbf{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \tag{A.22}
-$$
-
-its inverse is
-
-$$
-\mathbf{A}^{-1} = \frac{1}{|\mathbf{A}|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
-$$
-(A.23)
-
-For a 3 × 3 matrix, if
-
-$$
-\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}
-$$
- (A.24)
-
-we first obtain the cofactors as
-
-$$
-\mathbf{C} = \begin{bmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{bmatrix}
-$$
- (A.25)
-
-where
-
-$$
-c_{11} = \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix}, \t c_{12} = -\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix}, \t c_{13} = \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix},
-$$
-  
-\n
-$$
-c_{21} = -\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix}, \t c_{22} = \begin{vmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{vmatrix}, \t c_{23} = -\begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{vmatrix},
-$$
-  
-\n
-$$
-c_{31} = \begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix}, \t c_{32} = -\begin{vmatrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{vmatrix}, \t c_{33} = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}
-$$
-  
-\n(A.26)
-
-The determinant of the 3 × 3 matrix can be found using Eq. (A.11). Here, we want to use Eq. (A.20), i.e.,
-
-$$
-|\mathbf{A}| = a_{11}c_{11} + a_{12}c_{12} + a_{13}c_{13}
-$$
- (A.27)
-
-The idea can be extended *n* > 3, but we deal mainly with 2 × 2 and 3 × 3 matrices in this book.
-
-Example A.3 Use matrix inversion to solve the simultaneous equations
-
-2*x*1 + 10*x*2 = 2, −*x*1 + 3*x*2 = 7
-
-# **Solution:**
-
-We first express the two equations in matrix form as
-
-$$
-\begin{bmatrix} 2 & 10 \ -1 & 3 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 2 \ 7 \end{bmatrix}
-$$
-
-or
-
-$$
-AX = B \longrightarrow X = A^{-1}B
-$$
-
-where
-
-$$
-\mathbf{A} = \begin{bmatrix} 2 & 10 \\ -1 & 3 \end{bmatrix}, \qquad \mathbf{X} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, \qquad \mathbf{B} = \begin{bmatrix} 2 \\ 7 \end{bmatrix}
-$$
-
-The determinant of **A** is |**A**| = 2 × 3 − 10(−1) = 16, so the inverse of **A** is
-
-$$
-\mathbf{A}^{-1} = \frac{1}{16} \begin{bmatrix} 3 & -10 \\ 1 & 2 \end{bmatrix}
-$$
-
-Hence,
-
-$$
-\mathbf{X} = \mathbf{A}^{-1} \mathbf{B} = \frac{1}{16} \begin{bmatrix} 3 & -10 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 7 \end{bmatrix} = \frac{1}{16} \begin{bmatrix} -64 \\ 16 \end{bmatrix} = \begin{bmatrix} -4 \\ 1 \end{bmatrix}
-$$
-
-i.e., *x*1 = −4 and *x*2 = 1.
-
-Solve the following two equations by matrix inversion. Practice Problem A.3
-
-$$
-2y_1 - y_2 = 4, \quad y_1 + 3y_2 = 9
-$$
-
-**Answer:** *y*1 *=* 3, *y*2 *=* 2.
-
-Determine *x*1, *x*2, and *x*3 for the following simultaneous equations using Example A.4 matrix inversion.
-
-$$
-x_1 + x_2 + x_3 = 5
-$$
-  
--x<sub>1</sub> + 2x<sub>2</sub> = 9  
-$$
-4x_1 + x_2 - x_3 = -2
-$$
-
-# **Solution:**
-
-In matrix form, the equations become
-
-$$
-\begin{bmatrix} 1 & 1 & 1 \ -1 & 2 & 0 \ 4 & 1 & -1 \ \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 5 \ 9 \ -2 \end{bmatrix}
-$$
-
-or
-
-$$
-AX = B \longrightarrow X = A^{-1}B
-$$
-
-where
-
-$$
-\mathbf{A} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 2 & 0 \\ 4 & 1 & -1 \end{bmatrix}, \quad \mathbf{X} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 5 \\ 9 \\ -2 \end{bmatrix}
-$$
-
-We now find the cofactors
-
-$$
-c_{11} = \begin{vmatrix} 2 & 0 \\ 1 & -1 \end{vmatrix} = -2, \quad c_{12} = -\begin{vmatrix} -1 & 0 \\ 4 & -1 \end{vmatrix} = -1, \quad c_{13} = \begin{vmatrix} -1 & 2 \\ 4 & 1 \end{vmatrix} = -9
-$$
-  
-$$
-c_{21} = -\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = 2, \quad c_{22} = \begin{vmatrix} 1 & 1 \\ 4 & -1 \end{vmatrix} = -5, \quad c_{23} = -\begin{vmatrix} 1 & 1 \\ 4 & 1 \end{vmatrix} = 3
-$$
-  
-$$
-c_{31} = \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} = -2, \quad c_{32} = -\begin{vmatrix} 1 & 1 \\ -1 & 0 \end{vmatrix} = -1, \quad c_{33} = \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} = 3
-$$
-
-The adjoint of matrix **A** is
-
-$$
-adj \mathbf{A} = \begin{bmatrix} -2 & -1 & -9 \\ 2 & -5 & 3 \\ -2 & -1 & 3 \end{bmatrix}^T = \begin{bmatrix} -2 & 2 & -2 \\ -1 & -5 & -1 \\ -9 & 3 & 3 \end{bmatrix}
-$$
-
-We can find the determinant of **A** using any row or column of **A**. Because one element of the second row is 0, we can take advantage of this to find the determinant as
-
-$$
-|\mathbf{A}| = -1c_{21} + 2c_{22} + (0)c_{23} = -1(2) + 2(-5) = -12
-$$
-
-Hence, the inverse of **A** is
-
-$$
-\mathbf{A}^{-1} = \frac{1}{-12} \begin{bmatrix} -2 & 2 & -2 \\ -1 & -5 & -1 \\ -9 & 3 & 3 \end{bmatrix}
-$$
-$$
-\mathbf{X} = \mathbf{A}^{-1} \mathbf{B} = \frac{1}{-12} \begin{bmatrix} -2 & 2 & -2 \\ -1 & -5 & -1 \\ -9 & 3 & 3 \end{bmatrix} \begin{bmatrix} 5 \\ 9 \\ -2 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \\ 2 \end{bmatrix}
-$$
-
-i.e., 
-$$
-x_1 = -1
-$$
-,  $x_2 = 4$ ,  $x_3 = 2$ .
-
-Practice Problem A.4 Solve the following equations using matrix inversion.
-
-$$
-y_1 - y_3 = 1
-$$
-  
-2y<sub>1</sub> + 3y<sub>2</sub> - y<sub>3</sub> = 1  
-$$
-y_1 - y_2 - y_3 = 3
-$$
-
-**Answer:** *y*1 *=* 6*, y*2 *= −*2, *y*3 *=* 5.
-
-# <span id="page-935-0"></span>Appendix B
-
-# Complex Numbers
-
-The ability to manipulate complex numbers is very handy in circuit analysis and in electrical engineering in general. Complex numbers are particularly useful in the analysis of ac circuits. Again, although calculators and computer software packages are now available to manipulate complex numbers, it is still advisable for a student to be familiar with how to handle them by hand.
-
-# **B.1** Representations of Complex Numbers
-
-A complex number *z* may be written in *rectangular form* as
-
-$$
-z = x + jy \tag{B.1}
-$$
-
-where *j* = √ \_\_\_ −1 ; *x* is the *real part* of *z* while *y* is the *imaginary part* of *z*; that is,
-
-$$
-x = \text{Re}(z), \qquad y = \text{Im}(z) \tag{B.2}
-$$
-
-The complex number *z* is shown plotted in the complex plane in Fig. B.1. Because *j* = √ \_\_\_ −1 ,
-
-$$
-\frac{1}{j} = -j
-$$
-\n
-$$
-j^{2} = -1
-$$
-\n
-$$
-j^{3} = j \cdot j^{2} = -j
-$$
-\n
-$$
-j^{4} = j^{2} \cdot j^{2} = 1
-$$
-\n
-$$
-j^{5} = j \cdot j^{4} = j
-$$
-\n
-$$
-\vdots
-$$
-\n
-$$
-n^{n+4} = j^{n}
-$$
-\n(B.3)
-
- The complex plane looks like the two-dimensional curvilinear coordinate space, but it is not.
-
-A second way of representing the complex number *z* is by specifying its magnitude *r* and the angle it makes with the real axis, as Fig. B.1 shows. This is known as the *polar form*. It is given by
-
-*j*
-
-$$
-z = |z| \underline{\theta} = r \underline{\theta}
-$$
- (B.4)
-
-where
-
-$$
-r = \sqrt{x^2 + y^2}
-$$
-,  $\theta = \tan^{-1} \frac{y}{x}$  (B.5a)
-
-or
-
-$$
-x = r \cos \theta, \qquad y = r \sin \theta \tag{B.5b}
-$$
-
-that is,
-
-$$
-z = x + jy = r/\theta = r\cos\theta + jr\sin\theta
-$$
- (B.6)

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/230_Appendix B - Complex Numbers.md

@@ -1,546 +0,0 @@
-In converting from rectangular to polar form using Eq. (B.5), we must exercise care in determining the correct value of . These are the four possibilities:
-
-$$
-z = x + jy, \qquad \theta = \tan^{-1} \frac{y}{x}
-$$
- (1st Quadrant)  
-\n
-$$
-z = -x + jy, \qquad \theta = 180^{\circ} - \tan^{-1} \frac{y}{x}
-$$
- (2nd Quadrant)  
-\n
-$$
-z = -x - jy, \qquad \theta = 180^{\circ} + \tan^{-1} \frac{y}{x}
-$$
- (3rd Quadrant)  
-\n
-$$
-z = x - jy, \qquad \theta = 360^{\circ} - \tan^{-1} \frac{y}{x}
-$$
- (4th Quadrant)
-
-assuming that *x* and *y* are positive.
-
-The third way of representing the complex *z* is the *exponential form*:
-
-$$
-z = re^{j\theta} \tag{B.8}
-$$
-
-This is almost the same as the polar form, because we use the same magnitude *r* and the angle .
-
-The three forms of representing a complex number are summarized as follows.
-
-$$
-z = x + jy, \qquad (x = r \cos \theta, y = r \sin \theta)
-$$
-Rectangular form  
-\n
-$$
-z = r/\theta, \qquad \left(r = \sqrt{x^2 + y^2}, \theta = \tan^{-1}\frac{y}{x}\right)
-$$
-Polar form  
-\n
-$$
-z = re^{j\theta}, \qquad \left(r = \sqrt{x^2 + y^2}, \theta = \tan^{-1}\frac{y}{x}\right)
-$$
-Exponential form  
-\n(B.9)
-
-The first two forms are related by Eqs. (B.5) and (B.6). In Section B.3 we will derive Euler's formula, which proves that the third form is also equivalent to the first two.
-
-Example B.1 Express the following complex numbers in polar and exponential form: (a) *z*1 = 6 + *j*8, (b) *z*2 = 6 − *j*8, (c) *z*3 = −6 + *j*8, (d) *z*4 = −6 − *j*8.
-
-# **Solution:**
-
-Notice that we have deliberately chosen these complex numbers to fall in the four quadrants, as shown in Fig. B.2. (a) For *z*1 = 6 + *j*8 (1st quadrant),
-
-$$
-r_1 = \sqrt{6^2 + 8^2} = 10
-$$
-,  $\theta_1 = \tan^{-1}\frac{8}{6} = 53.13^\circ$ 
-
-Hence, the polar form is 10⧸ 53.13° and the exponential form is 10*e<sup>j</sup>*53.13° . (b) For *z*2 = 6 − *j*8 (4th quadrant),
-
-$$
-r_2 = \sqrt{6^2 + (-8)^2} = 10
-$$
-,  $\theta_2 = 360^\circ - \tan^{-1}\frac{8}{6} = 306.87^\circ$ 
-
-In the exponential form, z = rej so that dz∕d = jre <sup>j</sup> = jz.
-
-so that the polar form is 10 ⧸ 306.87° and the exponential form is 10*e j*306.87*°* . The angle 2 may also be taken as −53.13°, as shown in Fig. B.2, so that the polar form becomes 10 ⧸−53.13° and the exponential form becomes 10*e*<sup>−</sup>*j*53.13*°* .
-
-(c) For *z*3 = −6 + *j*8 (2nd quadrant),
-
-$$
-r_3 = \sqrt{(-6)^2 + 8^2} = 10
-$$
-,  $\theta_3 = 180^\circ - \tan^{-1}\frac{8}{6} = 126.87^\circ$ 
-
-Hence, the polar form is 10⧸ 126.87° and the exponential form is 10*e j*126.87° . (d) For *z*4 = −6 − *j*8 (3rd quadrant),
-
-$$
-r_4 = \sqrt{(-6)^2 + (-8)^2} = 10
-$$
-,  $\theta_4 = 180^\circ + \tan^{-1}\frac{8}{6} = 233.13^\circ$ 
-
-so that the polar form is 10⧸ 233.13° and the exponential form is 10*e <sup>j</sup>*233.13° .
-
-For Example B.1.
-
-Convert the following complex numbers to polar and exponential Practice Problem B.1 forms: (a) *z*1 = 3 − *j*4, (b) *z*2 = 5 + *j*12, (c) *z*3 = −3 − *j*9, (d) *z*4 = −7 + *j*. **Answer:** (a) 5⧸ 306.9°, 5*e <sup>j</sup>*306.9° , (b) 13⧸ 67.38°, 13*ej*67.38° , (c) 9.487⧸ 251.6°, 9.487*e<sup>j</sup>*251.6° , (d) 7.071⧸ 171.9°, 7.071*e <sup>j</sup>*171.9° .
-
-Convert the following complex numbers into rectangular form: Example B.2 (a) 12⧸ −60°, (b) −50⧸ 285°, (c) 8*e <sup>j</sup>*10° , (d) 20*e*<sup>−</sup>*jπ*∕<sup>3</sup> .
-
-# **Solution:**
-
-(a) Using Eq. (B.6),
-
-$$
-12\angle -60^{\circ} = 12\cos(-60^{\circ}) + j12\sin(-60^{\circ}) = 6 - j10.39
-$$
-
-Note that = −60° is the same as = 360° − 60° = 300°. (b) We can write
-
-$$
--50/285^{\circ} = -50 \cos 285^{\circ} - j50 \sin 285^{\circ} = -12.94 + j48.3
-$$
-
-(c) Similarly,
-
-$$
-8e^{j10^{\circ}} = 8 \cos 10^{\circ} + j8 \sin 10^{\circ} = 7.878 + j1.389
-$$
-
-(d) Finally,
-
-$$
-20e^{-j\pi/3} = 20\cos(-\pi/3) + j20\sin(-\pi/3) = 10 - j17.32
-$$
-
-Find the rectangular form of the following complex numbers: Practice Problem B.2 (a) −8⧸ 210°, (b) 40⧸ 305°, (c) 10*e* <sup>−</sup>*j*30° , (d) 50*e jπ*∕<sup>2</sup> .
-
-**Answer:** (a) 6.928 + *j*4, (b) 22.94 −*j*32.77, (c) 8.66 − *j*5, (d) *j*50.
-
-We have used lightface notation for complex numbers—since they are not time- or frequency-dependent whereas we use boldface notation for phasors.
-
-# **B.2** Mathematical Operations
-
-Two complex numbers *z*1 = *x*1 + *jy*1 and *z*2 = *x*2 + *jy*2 are equal if and only if their real parts are equal and their imaginary parts are equal,
-
-$$
-x_1 = x_2, \t y_1 = y_2 \t (B.10)
-$$
-
-The *complex conjugate* of the complex number *z* = *x* + *jy* is
-
-$$
-z^* = x - jy = r \angle -\theta = re^{-j\theta}
-$$
- (B.11)
-
-Thus, the complex conjugate of a complex number is found by replacing every *j* by −*j*.
-
-Given two complex numbers *z*1 = *x*1 + *jy*1 = *r*1<sup>⧸</sup>*θ*<sup>1</sup> and *z*2 = *x*2 + *jy*2 = *r*<sup>2</sup> <sup>⧸</sup>*θ*2, their sum is
-
-$$
-z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2)
-$$
- (B.12)
-
-and their difference is
-
-$$
-z_1 - z_2 = (x_1 - x_2) + j(y_1 - y_2)
-$$
- (B.13)
-
-While it is more convenient to perform addition and subtraction of complex numbers in rectangular form, the product and quotient of the two complex numbers are best done in polar or exponential form. For their product,
-
-$$
-z_1 z_2 = r_1 r_2 / \theta_1 + \theta_2 \tag{B.14}
-$$
-
-Alternatively, using the rectangular form,
-
-$$
-z_1 z_2 = (x_1 + jy_1)(x_2 + jy_2)
-$$
-  
-=  $(x_1 x_2 - y_1 y_2) + j(x_1 y_2 + x_2 y_1)$  (B.15)
-
-For their quotient,
-
-$$
-\frac{z_1}{z_2} = \frac{r_1}{r_2} \underline{\beta_1 - \theta_2}
-$$
- (B.16)
-
-Alternatively, using the rectangular form,
-
-$$
-\frac{z_1}{z_2} = \frac{x_1 + jy_1}{x_2 + jy_2}
-$$
- (B.17)
-
-We rationalize the denominator by multiplying both the numerator and denominator by *z*2\*.
-
-denominator by 
-$$
-z_2^*
-$$
-.  
-\n
-$$
-\frac{z_1}{z_2} = \frac{(x_1 + jy_1)(x_2 - jy_2)}{(x_2 + jy_2)(x_2 - jy_2)} = \frac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2} + \frac{jx_2y_1 - x_1y_2}{x_2^2 + y_2^2}
-$$
-(B.18)
-
-Example B.3 If *A* = 2 + *j*5, *B* = 4 <sup>−</sup> *<sup>j</sup>*6, find: (a) *A*\*(*A* + *B*), (b) (*A* + *B*)∕(*A* − *B*).
-
-# **Solution:**
-
-(a) If *A* = 2 + *j*5, then *A*\* = 2 − *j*5 and
-
-$$
-A + B = (2 + 4) + j(5 - 6) = 6 - j
-$$
-
-so that
-
-$$
-A^*(A + B) = (2 - j5)(6 - j) = 12 - j2 - j30 - 5 = 7 - j32
-$$
-
-(b) Similarly,
-
-$$
-A - B = (2 - 4) + j(5 - -6) = -2 + j11
-$$
-
-Hence,
-
-Hence,  
-\n
-$$
-\frac{A+B}{A-B} = \frac{6-j}{-2+j11} = \frac{(6-j)(-2-j11)}{(-2+j11)(-2-j11)}
-$$
-\n
-$$
-= \frac{-12 - j66 + j2 - 11}{(-2)^2 + 11^2} = \frac{-23 - j64}{125} = -0.184 - j0.512
-$$
-
-Given that *C* = −3 + *j* 7 and *D* = 8 + *j*, calculate: Practice Problem B.3 (a) (*C* − *D*\*)(*C* + *D*\*), (b) *D*<sup>2</sup> ∕*C*\*, (c) 2*CD*∕(*C* + *D*).
-
-**Answer:** (a) −103 − *j*26, (b) −5.19 + *j* 6.776, (c) 6.045 + *j*11.53.
-
-Evaluate: Example B.4
-
-Evaluate:  
-\n(a) 
-$$
-\frac{(2+j5)(8e^{j10^{\circ}})}{2+j4+2(-40^{\circ})}
-$$
- (b)  $\frac{j(3-j4)^{*}}{(-1+j6)(2+j)^{2}}$ 
-
-# **Solution:**
-
-(a) Because there are terms in polar and exponential forms, it may be best to express all terms in polar form:
-
-$$
-2 + j5 = \sqrt{2^2 + 5^2} / \tan^{-1} 5/2 = 5.385 / 68.2^\circ
-$$
-  
-$$
-(2 + j5)(8e^{j10^\circ}) = (5.385 / 68.2^\circ)(8 / 10^\circ) = 43.08 / 78.2^\circ
-$$
-  
-$$
-2 + j4 + 2 / \frac{-40^\circ}{2} = 2 + j4 + 2 \cos(-40^\circ) + j2 \sin(-40^\circ)
-$$
-  
-$$
-= 3.532 + j2.714 = 4.454 / 37.54^\circ
-$$
-
-Thus,
-
-$$
-\frac{(2+j5)(8e^{j10^{\circ}})}{2+j4+2 \angle -40^{\circ}} = \frac{43.08 \angle 78.2^{\circ}}{4.454 \angle 37.54^{\circ}} = 9.672 \angle 40.66^{\circ}
-$$
-
-(b) We can evaluate this in rectangular form, because all terms are in that form. But
-
-$$
-j(3 - j4)* = j(3 + j4) = -4 + j3
-$$
-  
-\n
-$$
-(2 + j)^2 = 4 + j4 - 1 = 3 + j4
-$$
-  
-\n
-$$
-(-1 + j6)(2 + j)^2 = (-1 + j6)(3 + j4) = -3 - 4j + j18 - 24
-$$
-  
-\n
-$$
-= -27 + j14
-$$
-
-Hence,
-
-$$
-= -27 + j14
-$$
-  
-Hence,  
-$$
-\frac{j(3 - j4)^{*}}{(-1 + j6)(2 + j)^{2}} = \frac{-4 + j3}{-27 + j14} = \frac{(-4 + j3)(-27 - j14)}{27^{2} + 14^{2}}
-$$
-$$
-= \frac{108 + j56 - j81 + 42}{925} = 0.1622 - j0.027
-$$
-
-# Practice Problem B.4 Evaluate these complex fractions:
-
-Evaluate these complex fractions:  
-\n(a) 
-$$
-\frac{6/30^{\circ} + j5 - 3}{-1 + j + 2e^{j45^{\circ}}}
-$$
-\n(b) 
-$$
-\left[ \frac{(15 - j7)(3 + j2)^{*}}{(4 + j6)^{*}(3/70^{\circ})} \right]^{*}
-$$
-
-**Answer:** (a) 3.387 ⧸ −5.615°, (b) 2.759 ⧸ −287.6°.
-
-# **B.3** Euler's Formula
-
-Euler's formula is an important result in complex variables. We derive it from the series expansion of *ex* , cos , and sin . We know that
-
-$$
-e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \cdots
-$$
- (B.19)
-
-Replacing *x* by *j* gives
-
-$$
-e^{j\theta} = 1 + j\theta - \frac{\theta^2}{2!} - j\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots
-$$
- (B.20)
-
-Also,
-
-$$
-\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots
-$$
-\n
-$$
-\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots
-$$
-\n(B.21)
-
-so that
-
-$$
-\cos \theta + j \sin \theta = 1 + j\theta - \frac{\theta^2}{2!} - j\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + j\frac{\theta^5}{5!} - \cdots
-$$
- (B.22)
-
-Comparing Eqs. (B.20) and (B.22), we conclude that
-
-$$
-e^{j\theta} = \cos\theta + j\sin\theta
-$$
- (B.23)
-
-This is known as *Euler's formula*. The exponential form of representing a complex number as in Eq. (B.8) is based on Euler's formula. From Eq. (B.23), notice thatθ
-
-$$
-\cos \theta = \text{Re}(e^{j\theta}), \qquad \sin \theta = \text{Im}(e^{j\theta})
-$$
- (B.24)
-
-and that
-
-$$
-|e^{j\theta}| = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1
-$$
-
-Replacing by − in Eq. (B.23) gives
-
-$$
-e^{-j\theta} = \cos\theta - j\sin\theta \tag{B.25}
-$$
-
-Adding Eqs. (B.23) and (B.25) yields
-
-$$
-\cos \theta = \frac{1}{2} (e^{j\theta} + e^{-j\theta})
-$$
- (B.26)
-
-Subtracting Eq. (B.25) from Eq. (B.23) yields
-
-$$
-\sin \theta = \frac{1}{2j} (e^{j\theta} - e^{-j\theta})
-$$
- (B.27)
-
-# **Useful Identities**
-
-The following identities are useful in dealing with complex numbers. If *z* = *x* + *jy* = *r*⧸, then
-
-$$
-zz^* = x^2 + y^2 = r^2
-$$
- (B.28)
-
-$$
-\sqrt{z} = \sqrt{x + jy} = \sqrt{r}e^{j\theta/2} = \sqrt{r} \angle{\theta/2}
-$$
- (B.29)
-
-$$
-zn = (x + jy)n = rn / n\theta = rn ejn\theta = rn (cos n\theta + j sin n\theta)
-$$
- (B.30)
-
-$$
-z^{1/n} = (x + jy)^{1/n} = r^{1/n} \sqrt{\theta/n + 2\pi k/n}
-$$
- (B.31)  
-$$
-k = 0, 1, 2, ..., n - 1
-$$
-
-$$
-\ln(re^{j\theta}) = \ln r + \ln e^{j\theta} = \ln r + j\theta + j2k\pi
-$$
- (B.32)
-
-$$
-(k = \text{integer})
-$$
-
-$$
-\frac{1}{j} = -j
-$$
-\n
-$$
-e^{\pm j\pi} = -1
-$$
-\n(B.33)\n
-$$
-e^{\pm j2\pi} = 1
-$$
-\n
-$$
-e^{j\pi/2} = j
-$$
-\n
-$$
-e^{-j\pi/2} = -j
-$$
-\n
-$$
-Re(e^{(\alpha + j\omega)t}) = Re(e^{at}e^{j\omega t}) = e^{at} \cos \omega t
-$$
-\n
-$$
-Im(e^{(\alpha + j\omega)t}) = Im(e^{at}e^{j\omega t}) = e^{at} \sin \omega t
-$$
-\n(B.34)
-
-If *A* = 6 + *j*8, find: (a) <sup>√</sup> Example B.5 \_\_ *A* , (b) *A*<sup>4</sup> .
-
-# **Solution:**
-
-(a) First, convert A to polar form:
-
-$$
-r = \sqrt{6^2 + 8^2} = 10
-$$
-,  $\theta = \tan^{-1} \frac{8}{6} = 53.13^\circ$ ,  $A = 10/53.13^\circ$ 
-
-Then
-
-$$
-\sqrt{A} = \sqrt{10}/53.13^{\circ}/2 = 3.162/26.56^{\circ}
-$$
-
-(b) Because *A* = 10 ⧸ 53.13°,
-
-$$
-A^4 = r^4 / 4\theta = 10^4 / 4 \times 53.13^\circ = 10,000 / 212.52^\circ
-$$
-
-If *A* = 3 − *j*4, find: (a) *A*<sup>1</sup>∕<sup>3</sup>
-
-**Answer:** (a) 1.71 ⧸ 102.3°, 1.71 ⧸ 222.3°, 1.71 ⧸ 342.3°,
-
-(b) 1.609 + *j*5.356 + *j*2*nπ* (*n* = 0, 1, 2, . . . ).
-
-(3 roots), and (b) ln A. Practice Problem B.5
-
-# <span id="page-942-0"></span>Appendix C
-
-# Mathematical Formulas
-
-This appendix—by no means exhaustive—serves as a handy reference. It does contain all the formulas needed to solve circuit problems in this book.
-
-# **C.1** Quadratic Formula
-
-The roots of the quadratic equation *ax*<sup>2</sup> + *bx* + *c* = 0 are
-
-$$
-x_1, x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
-$$
-
-# **C.2** Trigonometric Identities
-
-| 1<br>csc x = ____<br>sin x<br>cot x = _____ 1<br>tan x |
-|--------------------------------------------------------|
-|                                                        |
-|                                                        |
-|                                                        |
-|                                                        |
-|                                                        |
-| (law of sines)                                         |
-| (law of cosines)                                       |
-| (law of tangents)                                      |
-| ± cos x sin y                                          |
-| ± y) = cos x cos y ∓ sin x sin y                       |
-|                                                        |
-| − y) − cos(x + y)                                      |
-| 2 sin x cos y = sin(x + y) + sin(x<br>− y)             |
-| 2 cos x cos y = cos(x + y) + cos(x<br>− y)             |
-|                                                        |
-| 1 ∓ tan x tan y                                        |
-
-$$
-\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x
-$$
-  
-\n
-$$
-\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}
-$$
-  
-\n
-$$
-\sin^2 x = \frac{1}{2} (1 - \cos 2x)
-$$
-  
-\n
-$$
-\cos^2 x = \frac{1}{2} (1 + \cos 2x)
-$$
-  

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/231_Appendix C - Mathematical Formulas.md

@@ -1,275 +0,0 @@
-\n
-$$
-K_1 \cos x + K_2 \sin x = \sqrt{K_1^2 + K_2^2} \cos \left(x + \tan^{-1} \frac{-K_2}{K_1}\right)
-$$
-  
-\n
-$$
-e^{jx} = \cos x + j \sin x
-$$
- (Euler's formula)
-
-$$
-\cos x = \frac{e^{jx} + e^{-jx}}{2}
-$$
-$$
-\sin x = \frac{e^{jx} - e^{-jx}}{2j}
-$$
-$$
-1 \text{ rad} = 57.296^{\circ}
-$$
-
-# **C.3** Hyperbolic Functions
-
-$$
-\sinh x = \frac{1}{2} (e^x - e^{-x})
-$$
-$$
-\cosh x = \frac{1}{2} (e^x + e^{-x})
-$$
-$$
-\tanh x = \frac{\sinh x}{\cosh x}
-$$
-$$
-\coth x = \frac{1}{\tanh x}
-$$
-$$
-\operatorname{csch} x = \frac{1}{\sinh x}
-$$
-$$
-\operatorname{sech} x = \frac{1}{\cosh x}
-$$
-
- sinh(*x* ± *y*) = sinh *x* cosh *y* ± cosh *x* sinh *y* cosh(*x* ± *y*) = cosh *x* cosh *y* ± sinh *x* sinh *y*
-
-# **C.4** Derivatives
-
-If *U* = *U*(*x*), *V* = *V*(*x*), and *a* = constant,
-
-$$
-\frac{d}{dx}(aU) = a\frac{dU}{dx}
-$$
-$$
-\frac{d}{dx}(UV) = U\frac{dV}{dx} + V\frac{dU}{dx}
-$$
-
-$$
-\frac{d}{dx}\left(\frac{U}{V}\right) = \frac{V\frac{dU}{dx} - U\frac{dV}{dx}}{V^2}
-$$
-$$
-\frac{d}{dx}(aU^n) = naU^{n-1}
-$$
-$$
-\frac{d}{dx}(a^U) = a^U \ln a \frac{dU}{dx}
-$$
-$$
-\frac{d}{dx}(e^U) = e^U \frac{dU}{dx}
-$$
-$$
-\frac{d}{dx}(\sin U) = \cos U \frac{dU}{dx}
-$$
-$$
-\frac{d}{dx}(\cos U) = -\sin U \frac{dU}{dx}
-$$
-
-# **C.5** Indefinite Integrals
-
-If 
-$$
-U = U(x)
-$$
-,  $V = V(x)$ , and  $a = \text{constant}$ ,  
-\n
-$$
-\int a \, dx = ax + C
-$$
-\n
-$$
-\int U \, dV = UV - \int V \, dU \qquad \text{(integration by parts)}
-$$
-\n
-$$
-\int U^n \, dU = \frac{U^{n+1}}{n+1} + C, \qquad n \neq 1
-$$
-\n
-$$
-\int \frac{dU}{U} = \ln U + C
-$$
-\n
-$$
-\int a^U \, dU = \frac{a^U}{\ln a} + C, \qquad a > 0, a \neq 1
-$$
-\n
-$$
-\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C
-$$
-\n
-$$
-\int xe^{ax} \, dx = \frac{e^{ax}}{a^2} (ax - 1) + C
-$$
-\n
-$$
-\int x^2 e^{ax} \, dx = \frac{e^{ax}}{a^3} (a^2 x^2 - 2ax + 2) + C
-$$
-\n
-$$
-\int \ln x \, dx = x \ln x - x + C
-$$
-\n
-$$
-\int \sin ax \, dx = -\frac{1}{a} \cos ax + C
-$$
-\n
-$$
-\int \cos ax \, dx = \frac{1}{a} \sin ax + C
-$$
-\n
-$$
-\int \cos^2 ax \, dx = \frac{x}{2} - \frac{\sin 2ax}{4a} + C
-$$
-\n
-$$
-\int \cos^2 ax \, dx = \frac{x}{2} + \frac{\sin 2ax}{4a} + C
-$$
-
-$$
-\int x \sin ax \, dx = \frac{1}{a^2} (\sin ax - ax \cos ax) + C
-$$
-  
-$$
-\int x \cos ax \, dx = \frac{1}{a^2} (\cos ax + ax \sin ax) + C
-$$
-  
-$$
-\int x^2 \sin ax \, dx = \frac{1}{a^3} (2ax \sin ax + 2 \cos ax - a^2x^2 \cos ax) + C
-$$
-  
-$$
-\int x^2 \cos ax \, dx = \frac{1}{a^3} (2ax \cos ax - 2 \sin ax + a^2x^2 \sin ax) + C
-$$
-  
-$$
-\int e^{ax} \sin bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C
-$$
-  
-$$
-\int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + C
-$$
-  
-$$
-\int \sin ax \sin bx \, dx = \frac{\sin(a - b)x}{2(a - b)} - \frac{\sin(a + b)x}{2(a + b)} + C, \quad a^2 \neq b^2
-$$
-  
-$$
-\int \sin ax \cos bx \, dx = -\frac{\cos(a - b)x}{2(a - b)} - \frac{\cos(a + b)x}{2(a + b)} + C, \quad a^2 \neq b^2
-$$
-  
-$$
-\int \cos ax \cos bx \, dx = \frac{\sin(a - b)x}{2(a - b)} + \frac{\sin(a + b)x}{2(a + b)} + C, \quad a^2 \neq b^2
-$$
-  
-$$
-\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C
-$$
-  
-$$
-\int \frac{x^2 dx}{a^2 + x^2} = \frac{1}{2a^2} \left( \frac{x}{x^2 + a^2} + \frac{1}{a} \tan^{-1} \frac{x}{a} \right) + C
-$$
-
-# **C.6** Definite Integrals
-
-If *m* and *n* are integers,
-
-$$
-\int_{0}^{2\pi} \sin ax \, dx = 0
-$$
-  
-$$
-\int_{0}^{2\pi} \cos ax \, dx = 0
-$$
-  
-$$
-\int_{0}^{\pi} \sin^{2} ax \, dx = \int_{0}^{\pi} \cos^{2} ax \, dx = \frac{\pi}{2}
-$$
-  
-$$
-\int_{0}^{\pi} \sin mx \sin nx \, dx = \int_{0}^{\pi} \cos mx \cos nx \, dx = 0, \quad m \neq n
-$$
-  
-$$
-\int_{0}^{\pi} \sin mx \cos nx \, dx = \begin{cases} 0, & m + n = \text{even} \\ \frac{2m}{m^{2} - n^{2}}, & m + n = \text{odd} \end{cases}
-$$
-  
-$$
-\int_{0}^{2\pi} \sin mx \sin nx \, dx = \int_{-\pi}^{\pi} \sin mx \sin nx \, dx = \begin{cases} 0, & m \neq n \\ \pi, & m = n \end{cases}
-$$
-
-$$
-\int_0^\infty \frac{\sin ax}{x} dx = \begin{cases} \frac{\pi}{2}, & a > 0 \\ 0, & a = 0 \\ -\frac{\pi}{2}, & a < 0 \end{cases}
-$$
-
-# **C.7** L'Hopital's Rule
-
-If *f*(0) = 0 = *h*(0), then
-
-$$
-\lim_{x \to 0} \frac{f(x)}{h(x)} = \lim_{x \to 0} \frac{f'(x)}{h'(x)}
-$$
-
-where the prime indicates differentiation.
-
-# <span id="page-947-0"></span>Appendix D
-
-# Answers to Odd-Numbered Problems
-
-# Chapter 1
-
-- **1.1** (a) −103.84 mC, (b) −198.65 mC, (c) −3.941 C, (d) −26.08 C
-- **1.3** (a) 3*t* + 1 C, (b) *t* 2 + 5*t* mC, (c) 2 sin(10*t* + *π*∕6) + 1 *μ*C, (d) −*e*<sup>−</sup>30*<sup>t</sup>* [0.16 cos 40*t* + 0.12 sin 40*t*] C
-- **1.5** 25 C
-
-1.7 
-$$
-i = \frac{dq}{dt} = \begin{cases} 10 \text{ A}, & 0 < t < 1 \\ -20 \text{ A}, & 1 < t < 2 \\ 0 \text{ A}, & 2 < t < 3 \\ 10 \text{ A}, & 3 < t < 4 \end{cases}
-$$
-
-**1.27** (a) 43.2 kC, (b) 475.2 kJ, (c) 1.188 cents
-
-- **1.29** 39.6 cents
-- **1.31** \$6.451
-- **1.33** 6 C
-- **1.35** 2.333 MWh
-- **1.37** 46.3 A-hour
-- **1.39** 24 cents
-
-See the sketch in Fig. D.1.
-
-**Figure D.1**
-
-For Prob. 1.7.
-
-- **1.9** (a) 10 C, (b) 22.5 C, (c) 30 C
-- **1.11** 3.888 kC, 5.832 kJ
-- **1.13** 123.37 mW, 58.76 mJ
-- **1.15** (a) 2.945 mC, (b) −720*e*<sup>−</sup>4*<sup>t</sup> μ*W, (c) −180 *μ*J
-- **1.17** 10 W absorbed
-- **1.19** −6 A, −150 W, 60 W, 54 W, 36 W
-- **1.21** 2.696 × 1023 electrons, 43,200 C
-- **1.23** \$1.35
-- **1.25** 10.08 cents
-
-# Chapter 2
-
-- **2.1** This is a design problem with several answers.
-- **2.3** 184.3 mm
-- **2.5** *n* = 9, *b* = 15, *l* = 7
-- **2.7** 6 branches and 4 nodes
-- **2.9** 5 A, −8 A, 4 A
-- **2.11** 6 V, 3 V
-- **2.13** 12 A, −10 A, 5 A, −2 A
-- **2.15** 6 V, −4 A
-- **2.17** 2 V, −22 V, 10 V
-- **2.19** −2 A, 12 W, −24 W, 20 W, 16 W
-- **2.21** 4.167 V
-- **2.23** −100 V, 960 W
-- **2.25** 0.1 A, 2 kV, 0.2 kW

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/232_Appendix D - Answers to Odd-Numbered Problems.md

@@ -1,1724 +0,0 @@
-- **2.27** 1 A
-- **2.29** 3.5 Ω
-
-**2.75** 8 Ω
-
-**2.77** (a) Four 20-Ω resistors in parallel
-
-(b) One 300-Ω resistor in series with a 1.8-Ω resistor
-
- (c) Two 24-kΩ resistors in parallel connected in series with two 56-kΩ resistors in parallel
-
-combination of two 56-kΩ resistors
-
-(d) A series combination of a 20-Ω resistor,
-
-and a parallel combination of two 20-Ω resistors
-
-300-Ω resistor, 24-kΩ resistor, and a parallel
-
-|      | 2.31 56 A, 8 A, 48 A, 32 A, 16 A                                                                                           |           | 2.79 75 Ω                                                                    |
-|------|----------------------------------------------------------------------------------------------------------------------------|-----------|------------------------------------------------------------------------------|
-|      | 2.33 3 V, 6 A                                                                                                              |           | 2.81 6.667 kΩ, 5 kΩ                                                          |
-|      | 2.35 32 V, 800 mA                                                                                                          |           | 2.83 3.84 kΩ, ∞ Ω (best answer)                                              |
-|      | 2.37 60 Ω                                                                                                                  |           |                                                                              |
-|      | 2.39 (a) 2.182 Ω, (b) 1.5 kΩ                                                                                               |           |                                                                              |
-|      | 2.41 16 Ω                                                                                                                  | Chapter 3 |                                                                              |
-|      | 2.43 (a) 12 Ω, (b) 16 Ω                                                                                                    | 3.1       | This is a design problem with several answers.                               |
-|      | 2.45 (a) 59.8 Ω, (b) 32.5 Ω                                                                                                | 3.3       | −6 A, −3 A, −2 A, 1 A, −60 V                                                 |
-|      | 2.47 24 Ω                                                                                                                  | 3.5       | −60 V                                                                        |
-| 2.49 | (a) 20 Ω, (b) Ran = 45 Ω, Rbn = 7.5 Ω, Rcn = 15 Ω                                                                          | 3.7       | 20 V                                                                         |
-|      | 2.51 (a) 9.231 Ω, (b) 36.25 Ω                                                                                              | 3.9       | 79.34 mA                                                                     |
-|      | 2.53 (a) 142.32 Ω, (b) 33.33 Ω                                                                                             |           | 3.11 3 V, 293.9 W, 750 mW, 121.5 W                                           |
-|      | 2.55 119.75 mA                                                                                                             |           | 3.13 583.3 V, 100 V                                                          |
-|      | 2.57 32.44 Ω, 1.5413 A                                                                                                     |           | 3.15 29.45 A, 144.6 W, 129.6 W, 12 W                                         |
-| 2.59 | P40W = 102.4 W (means that this immediately burns<br>out), P60W = 9.6 W, P100W = 16 W. The best way to                     |           |                                                                              |
-|      | wire the bulbs is to connect the 100-W bulb in series<br>with a parallel combination of the 60-W bulb and the              |           | 3.17 1.73 A                                                                  |
-|      | 40-W bulb.                                                                                                                 |           | 3.19 10 V, 4.933 V, 12.267 V                                                 |
-|      | 2.61 Use R1 and R3 bulbs                                                                                                   | 3.21      | −15 V, 0 V                                                                   |
-|      | 2.63 0.4 Ω, ≅ 1 W                                                                                                          |           | 3.23 90 V                                                                    |
-|      | 2.65 So, our circuit consists of the meter in series with an<br>18-kΩ resistor.                                            |           | 3.25 25.52 V, 22.05 V, 14.842 V, 15.055 V                                    |
-|      |                                                                                                                            |           | 3.27 625 mV, 375 mV, 1.625 V                                                 |
-|      | 2.67 (a) 4 V, (b) 2.857 V, (c) 28.57%, (d) 6.25%                                                                           | 3.29      | −0.7708 V, 1.209 V, 2.309 V, 0.7076 V                                        |
-|      | 2.69 (a) 6.662 V (with), 6.786 V (without)<br>(b) 24.61 V (with), 26.39 V (without)<br>(c) 62.5 V (with), 75.4 V (without) |           | 3.31 4.97 V, 4.85 V, −0.12 V                                                 |
-|      | 2.71 22.5 Ω                                                                                                                |           | 3.33 (a) and (b) are both planar and can be redrawn as<br>shown in Fig. D.2. |
-|      | 2.73 45 Ω                                                                                                                  |           |                                                                              |
-|      |                                                                                                                            |           |                                                                              |
-
-3 Ω 6 Ω 1 Ω 5 Ω 2 A 2 Ω 4 Ω (a)
-
-# **Figure D.2**
-
-For Prob. 3.33.
-
-- **3.35** 20 V
-- **3.37** 12 V
-- **3.39** This is a design problem with several different answers.
-- **3.41** 1.188 A
-- **3.43** 1.7778 A, 53.33 V
-- **3.45** 8.561 A
-- **3.47** 10 V, 4.933 V, 12.267 V
-- **3.49** 114 V, 36 A
-- **3.51** 233.3 V
-- **3.53** 1.6196 mA, −1.0202 mA, −2.461 mA, 3 mA, −2.423 mA
-- **3.55** −1 A, 0 A, 2 A
-- **3.57** 12 kΩ, 120 V, 80 V
-- **3.59** −4.48 A, −1.0752 kV
-- **3.61** −0.2813
-- **3.63** −4 V, 2.105 A
-- **3.65** 2.17 A, 1.9912 A, 1.8119 A, 2.094 A, 2.249 A
-- **3.67** −30 V
-
-**3.69** ⎡ ⎢ ⎣ 0.35 −0.1 −0.05 −0.1 0.4 −0.2 −0.05 −0.2 0.25 ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ *v*1 *v*2 *v*3 ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ 100 50 −10 ⎤ ⎥ ⎦
-
-**3.71** 6.255 A, 1.9599 A, 3.694 A
-
-3.73 
-$$
-\begin{bmatrix} 30 & -10 & -10 & 0 \ -10 & 40 & -10 & 0 \ -10 & -10 & 50 & -10 \ 0 & 0 & -10 & 20 \ \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ i_3 \\ i_4 \end{bmatrix} = \begin{bmatrix} 15 \\ 0 \\ 25 \\ -10 \end{bmatrix}
-$$
-3.75 -3 A, 0 A, 3 A
-
-**3.77** 3.111 V, 1.4444 V
-
-**3.79** −10.556 V, 20.56 V, 1.3889 V, −43.75 V
-
-**3.81** 26.67 V, 6.667 V, 173.33 V, −46.67 V
-
-**3.83** See Fig. D.3; −12.5 V
-
-# **Figure D.3**
-
-For Prob. 3.83.
-
-- **3.85** 9 Ω
-- **3.87** −5
-- **3.89** 22.5 *μ*A, 12.75 V
-- **3.91** 0.8078 *μ*A, 8.345 V, 48.79 mV
-- **3.93** 1.333 A, 1.333 A, 2.6667 A
-
-# Chapter 4
-
-- **4.1** 600 mA, 250 V
-- **4.3** (a) 0.5 V, 0.5 A, (b) 5 V, 5 A, (c) 5 V, 500 mA
-- **4.5** 4.5 V
-- **4.7** 888.9 mV
-- **4.9** 2 A
-- **4.11** 17.99 V, 1.799 A
-- **4.13** 8.696 V
-- **4.15** 1.875 A, 10.55 W
-- **4.17** −8.571 V
-- **4.19** −16 V
-
-| 4.21                                            | 4.81                                            |
-|-------------------------------------------------|-------------------------------------------------|
-| This is a design problem with multiple answers. | 3.3 Ω, 10 V (Note, values obtained graphically) |
-| 4.23                                            | 4.83                                            |
-| 1 A, 8 W                                        | 8 Ω, 72 V                                       |
-| 4.25                                            | 4.85                                            |
-| −6.6 V                                          | (a) 80 V, 30 kΩ, (b) 32 V                       |
-| 4.27                                            | 4.87                                            |
-| −48 V                                           | (a) 10 mA, 8 kΩ, (b) 9.926 mA                   |
-| 4.29                                            | 4.89                                            |
-| 3 V                                             | (a) 99.99 μA, (b) 99.99 μA                      |
-| 4.31                                            | 4.91                                            |
-| 9.13 V                                          | (a) 150 Ω, 25 Ω, (b) 150 Ω, 250 Ω               |
-| 4.33                                            | 4.93 ____________ Vs                            |
-| 80 V, 33 Ω, 2 A                                 | Rs + (1 + β)Ro                                  |
-| 4.35                                            | 4.95                                            |
-| −125 mV                                         | 10.667 V, 33.33 kΩ                              |
-| 4.37                                            | 4.97                                            |
-| 5 kΩ, 1 mA                                      | 2 kΩ, 5 V                                       |
-| 4.39<br>20 Ω, −84 V                             |                                                 |
-| 4.41<br>4 Ω, −8 V, −2 A                         | Chapter 5                                       |
-| 4.43                                            | 5.1                                             |
-| 10 Ω, 0 V                                       | 60 μV                                           |
-| 4.45                                            | 5.3                                             |
-| 3 Ω, 15 V                                       | 10 V                                            |
-| 4.47                                            | 5.5                                             |
-| 20 V, 20 Ω, 1 A                                 | 0.999990                                        |
-| 4.49                                            | 5.7                                             |
-| 28 Ω, 3.286 V                                   | −100 nV, −10 mV                                 |
-| 4.51                                            | 5.9                                             |
-| (a) 2 Ω, 7 A, (b) 1.5 Ω, 12.667 A               | 2 V, 2 V                                        |
-| 4.53                                            | 5.11                                            |
-| 10 Ω, −3 A                                      | This is a design problem with multiple answers. |
-| 4.55                                            | 5.13                                            |
-| 100 kΩ, −20 mA                                  | 2.7 V, 288 μA                                   |
-| 4.57<br>10 Ω, 166.67 V, 16.667 A                | R____<br>1R3<br>5.15                            |
-| 4.59<br>22.5 Ω, 40 V, 1.7778 A                  | (a) −(R1 + R3 +<br>), (b) −92 kΩ<br>R2          |
-| 4.61                                            | 5.17                                            |
-| 1.2 Ω, 9.6 V, 8 A                               | (a) −2.4, (b) −16, (c) −400                     |
-| 4.63                                            | 5.19                                            |
-| −3.333 Ω, 0 A                                   | −562.5 μA                                       |
-| 4.65<br>V0<br>= 24 − 5I0                        | 5.21<br>−3 V                                    |
-| 4.67<br>25 kΩ, 49 mW                            | Rf<br>___<br>5.23<br>−<br>R1                    |
-| 4.69                                            | 5.25                                            |
-| ∞ (theoretically)                               | 9.375 V                                         |
-| 4.71                                            | 5.27                                            |
-| 8 kΩ, 1.152 W                                   | 2.7 V                                           |
-| 4.73<br>20.77 W                                 | R___2<br>5.29<br>R1                             |
-| 4.75                                            | 5.31                                            |
-| 250 Ω, 12 mW                                    | 4.545 mA                                        |
-| 4.77<br>(a) 3.8 Ω, 4 V, (b) 3.2 Ω, 15 V         |                                                 |
-| 4.79                                            | 5.33                                            |
-| 10 Ω, 167 V                                     | 75 mW, −1 mA                                    |
-
-**5.35** If *Ri* = 60 k, *Rf* = 390 k.
-
-**5.37** −13.6 V
-
-**5.39** 7 V
-
-**5.41** See Fig. D.4.
-
-# **Figure D.4**
-
-For Prob. 5.41.
-
-**5.43** 200 k.
-
-**5.45** This is a design problem with many correct answers. One possible design is shown in Fig. D.5.
-
-# **Figure D.5**
-
-- **5.47** 14.09 V
-- **5.49** *R*1 = *R*3 = 20 kΩ, *R*2 = *R*4 = 80 kΩ
-- **5.51** See Fig. D.6.
-
-# **Figure D.6**
-
-For Prob. 5.51.
-
-**5.53** Proof.
-
-- **5.55** 7.956, 7.956, 1.989
-- **5.57** 6*v*s1 − 6*v*s2
-
-**5.59** −12
-
-**5.61** 7.2 V
-
-- **5.63** \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ *R*2*R*4∕*R*1*R*<sup>5</sup> − *R*4∕*R*<sup>6</sup> 1 − *R*2*R*4∕*R*3*R*<sup>5</sup> **5.65** 2 V **5.67** −1.6 V **5.69** −25.71 mV **5.71** 7.5 V **5.73** 32.4 V
-- **5.75** −2, 200 *μ*A
-- **5.77** −6.686 mV
-- **5.79** −4.992 V
-- **5.81** 343.4 mV, 24.51 *μ*A
-- **5.83** The result depends on your design. Hence, let *RG* = 10 k ohms, *R*1 = 10 k ohms, *R*2 = 20 k ohms, *R*3 = 40 k ohms,
-  - *R*4 = 80 k ohms, *R*5 = 160 k ohms,
-  - *R*6 = 320 k ohms, then,
-
-$$
--v_o = (R_f/R_1)v_1 +
-$$
-  
-=  $v_1 + 0.5v_2 + 0.25v_3 + 0.125v_4$   
-+  $0.0625v_5 + 0.03125v_6$ 
-
-- (a) **|***vo***|** = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1∕8) + (1∕16), which implies, [*v*<sup>1</sup> *v*<sup>2</sup> *v*<sup>3</sup> *v*<sup>4</sup> *v*<sup>5</sup> *v*6] = [**100110**]
-- (b) **|***vo***|** = 0 + (1∕2) + (1∕4) + 0 + (1∕16) + (1∕32) = (27∕32) = **843.75 mV**
-
-(c) This corresponds to [111111].  
-\n
-$$
-|v_o| = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32)
-$$
-  
-\n $= 63/32 = 1.96875$  V
-
-**5.85** *R* = 200 kΩ, 2,000
-
-$$
-5.87 \quad \left(1 + \frac{R_4}{R_3}\right) v_2 - \left[\left(\frac{R_4}{R_3}\right) + \left(\frac{R_2 R_4}{R_1 R_3}\right)\right] v_1
-$$
-\n
-$$
-\text{Let } R_4 = R_1 \text{ and } R_3 = R_2;
-$$
-\n
-$$
-\text{then } v_0 = \left(1 + \frac{R_4}{R_3}\right) (v_2 - v_1)
-$$
-\n
-$$
-\text{a subtractor with a gain of } \left(1 + \frac{R_4}{R_3}\right).
-$$
-
-**5.89** A summer with *v*0 = −*v*<sup>1</sup> − (5∕3)*v*2 where *v*2 = 6 V battery and an inverting amplifier with *v*1 = −12 *vs*.
-
-$$
-5.91\ \ 9
-$$
-
-battery and an inverting amplifier with 
-$$
-v_1 =
-$$
-  
-\n**5.91** 9  
-\n**5.93**  $A = \frac{1}{\left(1 + \frac{R_1}{R_3}\right)R_L - R_1\left(\frac{R_2 + R_L}{R_2 R_3}\right)\left(R_4 + \frac{R_2 R_L}{R_2 + R_L}\right)}$ 
-
-For Prob. 5.45.
-
-# Chapter 6
-
-**6.1** 15(1 − 3*t*)*e*<sup>−</sup>3*<sup>t</sup>* A, 30*t*(1 − 3*t*)*e*<sup>−</sup>6*<sup>t</sup>* W
-
-**6.3** This is a design problem with multiple answers.
-
-6.5 
-$$
-v = \begin{cases} 20 \text{ mA}, & 0 < t < 2 \text{ ms} \\ -20 \text{ mA}, & 2 < t < 6 \text{ ms} \\ 20 \text{ mA}, & 6 < t < 8 \text{ ms} \end{cases}
-$$
-
-**6.7** [0.1*t* 2 + 10] V
-
-**6.9** 13.624 V, 70.66 W
-
-$$
-\textbf{6.11} \quad v(t) = \begin{cases} 10 + 3.75t \text{ V}, & 0 < t < 2s \\ 22.5 - 2.5t \text{ V}, & 2 < t < 4s \\ 12.5 \text{ V}, & 4 < t < 6s \\ 2.5t - 2.5 \text{ V}, & 6 < t < 8s \end{cases}
-$$
-
-- **6.13** *v*1 = 42 V, *v*2 = 48 V
-- **6.15** (a) 125 mJ, 375 mJ, (b) 70.31 mJ, 23.44 mJ
-- **6.17** (a) 3 F, (b) 8 F, (c) 1 F
-- **6.19** 10 *μ*F
-- **6.21** 2.5 *μ*F
-- **6.23** This is a design problem with multiple answers.
-- **6.25** (a) For the capacitors in series,
-
-$$
-Q_1 = Q_2 \rightarrow C_1 v_1 = C_2 v_2 \rightarrow \frac{v_1}{v_2} = \frac{C_2}{C_1}
-$$
-  
-$$
-v_s = v_1 + v_2 = \frac{C_2}{C_1} v_2 + v_2 = \frac{C_1 + C_2}{C_1} v_2
-$$
-  
-$$
-\rightarrow v_2 = \frac{C_1}{C_1 + C_2} v_s
-$$
-  
-Similarly,  $v_1 = \frac{C_2}{C_1 + C_2} v_s$ 
-
-(b) For capacitors in parallel,
-
-$$
-v_1 = v_2 = \frac{Q_1}{C_1} = \frac{Q_2}{C_2}
-$$
-  
-$$
-Q_s = Q_1 + Q_2 = \frac{C_1 Q_2 + Q_2}{C_2} = \frac{C_1 + C_2 Q_2}{C_2}
-$$
-
-or
-
-$$
-Q_2 = \frac{C_2}{C_1 + C_2}
-$$
-$$
-Q_1 = \frac{C_1}{C_1 + C_2} Q_s
-$$
-
-$$
-i = \frac{dQ}{dt} \rightarrow i_1 = \frac{C_1}{C_1 + C_2} i_s,
-$$
-  
-\n
-$$
-i_2 = \frac{C_2}{C_1 + C_2} i_s
-$$
-  
-\n6.27 2.5 µF, 40 µF  
-\n6.29 (a) 1.6 C, (b) 1 C  
-\n
-$$
-1.5t^2 \text{ kV}, \qquad 0 < t < 1s
-$$
-  
-\n6.31 v(t) = 
-$$
-\begin{cases} 1.5t^2 \text{ kV}, \qquad 0 < t < 1s\\ [3t - 1.5] \text{ kV}, \qquad 1 < t < 3s\\ [0.75t^2 - 7.5t + 23.25] \text{ kV}, \qquad 3 < t < 5s \end{cases}
-$$
-  
-\n
-$$
-i_1 = \begin{cases} 18t \text{ mA}, \qquad 0 < t < 1s\\ 18 \text{ mA}, \qquad 1 < t < 3s;\\ [9t - 45] \text{ mA}, \qquad 3 < t < 5s \end{cases}
-$$
-  
-\n
-$$
-i_2 = \begin{cases} 12t \text{ mA}, \qquad 0 < t < 1s\\ 12 \text{ mA}, \qquad 1 < t < 3s\\ [6t - 30] \text{ mA}, \qquad 3 < t < 5s \end{cases}
-$$
-  
-\n6.33 15 V, 10 F  
-\n6.35 3.2 mH  
-\n6.37 4.8 cos 100t V, 96 mJ  
-\n6.39 [-50e^{-2t} + 50 + 20t^2 + 80t] A  
-\n6.41 5.977 A, 35.72 J  
-\n6.43 270 µJ  
-\n6.45 i(t) = 
-$$
-\begin{cases} 100t^2 \text{ A}, \qquad 0 < t < 1s\\ [400 - 400t + 100t^2] \text{ A}, \qquad 1 < t < 2s \end{cases}
-$$
-  
-\n6.47 5  $\Omega$   
-\n6.49 15 mH  
-\n6.53 20 mH  
-\n6.55 (a) 1.4 L, (b) 500 mL  
-\n6.57 6.625 H  
-\n6.59 Proof.
-
-;
-
-**6.61** (a) 6.667 mH, *e*<sup>−</sup>*<sup>t</sup>* mA, 2*e*<sup>−</sup>*<sup>t</sup>* mA (b) −20*e*<sup>−</sup>*<sup>t</sup> μ*V (c) 1.3534 nJ
-
-**6.63** See Fig. D.7.
-
-# **Figure D.7**
-
-For Prob. 6.63.
-
-- **6.65** (a) 40 J, 40 J, (b) 80 J, (c) 5 × 10<sup>−</sup><sup>5</sup> (*e*<sup>−</sup>200*<sup>t</sup>* − 1) + 4 A, 1.25 × 10<sup>−</sup><sup>5</sup> (*e*<sup>−</sup>200*<sup>t</sup>* − 1) − 2 A (d) 6.25 × 10<sup>−</sup><sup>5</sup> (*e*<sup>−</sup>200*<sup>t</sup>* − 1) + 2 A
-- **6.67** 100 cos(50*t*) mV
-- **6.69** See Fig. D.8.
-
-# **Figure D.8**
-
-For Prob. 6.69.
-
-**6.71** By combining a summer with an integrator, we get the circuit shown in Fig. D.9 where *C* = 5 *μ*F, *R*1 = 200 kΩ, *R*2 = 50 kΩ, and *R*3 = 20 kΩ.
-
-$$
-v_o = -\frac{1}{R_1C} \int v_1 dt - \frac{1}{R_2C} \int v_2 dt - \frac{1}{R_2C} \int v_2 dt
-$$
-
-For the given problem, *C* = 2 *μ*F : *R*1 = 500 kΩ, *R*2 = 125 kΩ, *R*3 = 50 kΩ.
-
-**6.73** Consider the op amp as shown in Fig. D.10.
-
-# **Figure D.10** For Prob. 6.73.
-
-Let *va* = *vb* = *v*. At node *a*,
-
-$$
-\frac{0 - v}{R} = \frac{v - v_0}{R} \longrightarrow 2v - v_0 = 0
-$$
-(1)  
-At node *b*, 
-$$
-\frac{v_i - v}{R} = \frac{v - v_0}{R} + C\frac{dv}{dt}
-$$
-$$
-v_i = 2v - v_o + RC\frac{dv}{dt}
-$$
-(2)
-
-Combining Eqs. (1) and (2),
-
-$$
-v_i = v_o - v_o + \frac{RC}{2} \frac{dv_o}{dt}
-$$
- or  $v_o = \frac{2}{RC} \int v_i dt$ 
-
-showing that the circuit is a noninverting integrator.
-
-$$
-6.75 -17.5 \text{ mV}
-$$
-
-**6.77** See Fig. D.11.
-
-**Figure D.11** For Prob. 6.77.
-
-**6.79** See Fig. D.12.
-
-**6.81** See Fig. D.13.
-
-For Prob. 6.81.
-
-- **6.83** Eight groups in parallel with each group made up of four capacitors in series.
-- **6.85** 1.25 mH inductor
-
-# Chapter 7
-
-- **7.1** (a) 0.7143 *μ*F, (b) 5 ms, (c) 3.466 ms
-- **7.3** 1.5 *μ*s
-- **7.5** This is a design problem with multiple answers.
-- **7.7** 15*e*<sup>−</sup>*<sup>t</sup>* V for 0 < *t* < 1 sec, 5.518*e*<sup>−</sup>2(*t*−1) V for 1 sec < *t* < ∞
-- **7.9** 10*e*<sup>−</sup>*<sup>t</sup>*/12 V
-
-- **7.11** 1.2*e*<sup>−</sup>3*<sup>t</sup>* A
-- **7.13** (a) 16 kΩ, 16 H, 1 ms, (b) 126.42 *μ*J
-- **7.15** (a) 10 Ω, 500 ms, (b) 40 Ω, 250 *μ*s
-- **7.17** [−15*e*<sup>−</sup>2*<sup>t</sup>* ] V for all *t* > 0.
-- **7.19** 5*e*<sup>−</sup>5*<sup>t</sup> u*(*t*) A
-- **7.21** 1.618 Ω
-- **7.23** 10*e*<sup>−</sup>4*<sup>t</sup>* V, *t* > 0, 2.5*e*<sup>−</sup>4*<sup>t</sup>* V, *t* > 0
-- **7.25** This is a design problem with multiple answers.
-- **7.27** [5*u*(*t* + 1) + 10*u*(*t*) − 25*u*(*t* −1) + 15*u*(*t* − 2)] V
-- **7.29** (c) *z*(*t*) = cos 4*t δ*(*t* − 1) = cos 4*δ*(*t* − 1) = −0.6536*δ*(*t* − 1), which is sketched below.
-
-**Figure D.14** For Prob. 7.29.
-
-- **7.31** (a) 112 × 10 <sup>−</sup><sup>9</sup> , (b) 7
-- **7.33** 4.5*u*(*t* − 2) A
-- **7.35** (a) −*e* <sup>−</sup>2*<sup>t</sup> u*(*t*) V, (b) 2*e*1.5*<sup>t</sup> u*(*t*) A
-- **7.37** (a) 4 s, (b) 10 V, (c) (10 − 8*e*<sup>−</sup>*t*∕<sup>4</sup> ) *u*(*t*) V
-- **7.39** (a) 4 V, *t* < 0, 20 −16*e* <sup>−</sup>*t*∕<sup>8</sup> , *t* > 0, (b) 4 V, *t* < 0, 12 − 8*e*<sup>−</sup>*t*∕<sup>6</sup> V, *t* > 0.
-  - **7.41** This is a design problem with multiple answers.
-
-7.43 0.8 A, 
-$$
-0.8e^{-t/480}u(t)
-$$
- A
-
-**7.45** [20 −15*e*<sup>−</sup>14.286*<sup>t</sup>* ] *u*(*t*) V
-
-7.47 
-$$
-\begin{cases} 24(1 - e^{-t})V, & 0 < t < 1 \\ 30 - 14.83e^{-(t-1)}V, & t > 1 \end{cases}
-$$
-
-7.49 
-$$
-\begin{cases} 8(1 - e^{-t/5}) \text{ V}, & 0 < t < 1 \\ [-16 + 31.17e^{-(t-1)}] \text{ V}, & t > 1 \end{cases}
-$$
-
-$$
-7.51 \quad V_S = Ri + L\frac{di}{dt}
-$$
-\n
-$$
-\text{or } L\frac{di}{dt} = -R\left(i - \frac{V_S}{R}\right)
-$$
-\n
-$$
-\frac{di}{i - V_S/R} = \frac{-R}{L}dt
-$$
-
-Integrating both sides,
-
-$$
-\ln\left(i - \frac{V_S}{R}\right)|_{I_0}^{i(t)} = \frac{-R}{L}t
-$$
-$$
-\ln\left(\frac{i - V_S/R}{I_0 - V_S/R}\right) = \frac{-t}{\tau}
-$$
-
-$$
-\text{or } \frac{i - V_S/R}{I_0 - V_S/R} = e^{-t/\tau}
-$$
-
-$$
-i(t) = \frac{V_S}{R} + \left(I_0 - \frac{V_S}{R}\right)e^{-t/\tau}
-$$
-
-which is the same as Eq. (7.60).
-
-- **7.53** (a) 5 A, 5*e*<sup>−</sup>*t*∕<sup>2</sup> *u*(*t*) A, (b) 6 A, 6*e*<sup>−</sup>2*t*∕<sup>3</sup> *u*(*t*) A
-- **7.55** 96 V, 96*e*<sup>−</sup>4*<sup>t</sup> u*(*t*) V
-- **7.57** 2.4*e*<sup>−</sup>2*<sup>t</sup> u*(*t*) A, 600*e*<sup>−</sup>5*<sup>t</sup> u*(*t*) mA
-- **7.59** 120*e*<sup>−</sup>4*<sup>t</sup> u*(*t*) volts
-- **7.61** 20*e*<sup>−</sup>8*<sup>t</sup> u*(*t*) V, (10 − 5*e*<sup>−</sup>8*<sup>t</sup>* )*u*(*t*) A
-- **7.63** 2*e*<sup>−</sup>8*<sup>t</sup> u*(*t*) A, −8*e*<sup>−</sup>8*<sup>t</sup> u*(*t*) V
-- **7.65** { 2(1 − *e*<sup>−</sup>2*<sup>t</sup>* )A 1.729*e*<sup>−</sup>2(*t*−1)A 0 < *t* < 1 *t* > 1
-- **7.67** 10*e*–*t*/6*u*(*t*) V
-- **7.69** 48(*e*<sup>−</sup>*t*∕<sup>3000</sup>−1) *u*(*t*) V
-- **7.71** [−5 + 5*e*–*<sup>t</sup>* ]*u*(*t*) V
-- **7.73** −9*e*<sup>−</sup>5*<sup>t</sup> u*(*t*) V
-- **7.75** [20 10*e*–*<sup>t</sup>* ]*u*(*t*) V, 100*μ*A
-- **7.77** See Fig. D.15.
-
-**7.79** [1.75 – 0.75*e*–2*<sup>t</sup>* ]*u*(*t*) A
-
-**7.81** See Fig. D.16.
-
-For Prob. 7.81.
-
-- **7.83** 6.278 m/s
-- **7.85** (a) 659.7 *μ*s, (b) 16.636 s
-- **7.87** 441 mA
-- **7.89** *L* < 200 mH
-- **7.91** 1.271 Ω
-
-# Chapter 8
-
-- **8.1** (a) 2 A, 12 V, (b) −4 A∕s, −5 V∕s, (c) 0 A, 0 V
-- **8.3** (a) 0 A, −10 V, 0 V, (b) 0 A∕s, 8 V∕s, 8 V∕s, (c) 400 mA, 6 V, 16 V
-  - **8.5** (a) 0 A, 0 V, (b) 4 A∕s, 0 V∕s, (c) 2.4 A, 9.6 V
-- **8.7** overdamped
-
-- **8.9** [(10 + 50*t*)*e*<sup>−</sup>5*<sup>t</sup>* ] A **8.11** [(10 + 10*t*)*e*<sup>−</sup>*<sup>t</sup>* ] V
-- **8.13** 120 Ω
-- **8.15** 750 Ω, 200 *μ*F, 25 H
-- **8.17** [21.55*e*<sup>−</sup>2.679*<sup>t</sup>* − 1.55*e*<sup>−</sup>37.32*<sup>t</sup>* ] V
-- **8.19** 24 sin(0.5*t*) V
-- **8.21** 18*e*<sup>−</sup>*<sup>t</sup>* �� 2*e*<sup>−</sup>9*<sup>t</sup>* V
-- **8.23** 40 mF
-- **8.25** This is a design problem with multiple answers.
-- **8.27** [3 − 3(cos(2*t*) + sin(2*t*))*e*<sup>−</sup>2*<sup>t</sup>* ] volts
-- **8.29** (a) 3 − 3 cos 2*t* + sin 2*t* V, (b) 2 − 4*e*<sup>−</sup>*<sup>t</sup>* + *e*<sup>−</sup>4*<sup>t</sup>* A,
-
-- (c) 3 + (2 + 3*t*)*e*<sup>−</sup>*<sup>t</sup>* V, (d) 2 + 2 cos 2*te*<sup>−</sup>*<sup>t</sup>* A
-  - **8.31** 80 V, 40 V
-  - **8.33** [30 + 0.3078*e*<sup>−</sup>4.95*<sup>t</sup>* − 15.308*e*<sup>−</sup>0.05*<sup>t</sup>* ] V
-- **8.35** This is a design problem with multiple answers.
-- **8.37** 5*e*<sup>−</sup>4*<sup>t</sup>* A
-- **8.39** (−60 + [−0.2102*e*<sup>−</sup>47.83*<sup>t</sup>* + 60.21*e*<sup>−</sup>0.167*<sup>t</sup>* ]) V
-- **8.41** [8.7 sin(4.583*t*)*e*–2*<sup>t</sup>* ]*u*(*t*) A
-- **8.43** 8 Ω, 2.075 mF
-- **8.45** [6 − [5 cos(1.3229*t*) + 1.8898 sin(1.3229*t*)]*e*<sup>−</sup>*t*∕<sup>2</sup> ] A, [7.559 sin(1.3229*t*)*e*<sup>−</sup>*t*∕<sup>2</sup> ] V
-  - **8.47** (400*te*<sup>−</sup>10*<sup>t</sup>* ) V
-  - **8.49** {9 + [(3 + 6*t*)*e*–2*<sup>t</sup>* ]} *u*(*t*) A
-  - **8.51** [ <sup>−</sup> *<sup>i</sup>* \_\_\_\_0 *oC* sin(*ot*) ] V where *o* = 1∕ √ \_\_\_ LC
-  - **8.53** (*d*<sup>2</sup> *i*∕*dt*<sup>2</sup> ) + 1.25(*di*∕*dt*) + 400*i* = 200
-  - **8.55** 2*e*–*<sup>t</sup>*/2 A for *t* > 0
-
-- **8.57** (a) *s* 2 + 10*s* + 9 = 0, (b) [–1.75*e* –*t* + 3.75*e* –9*t* ]*u*(*t*) A, [–21*e* –*t* + 45*e* –9*t* ]*u*(*t*) V
-  - **8.59** 48*te*–2*<sup>t</sup>* V
-- **8.61** 2.4 2.667*e*<sup>−</sup>2*<sup>t</sup>* + 0.2667*e*<sup>−</sup>5*<sup>t</sup>* A, 9.6 – 16*e*<sup>−</sup>2*<sup>t</sup>* + 6.4*e*<sup>−</sup>5*<sup>t</sup>* V
-
-$$
-8.63 \frac{d^2 i(t)}{dt^2} = -\frac{v_s}{RCL}
-$$
-
-**8.65** 
-$$
-\frac{d^2v_o}{dt^2} - \frac{v_o}{R^2C^2} = 0, e^{10t} - e^{-10t} \text{ V}
-$$
-
-Note, circuit is unstable.
-
-- **8.67** −*te*<sup>−</sup>*<sup>t</sup> u*(*t*) V
-- **8.69** See Fig. D.17.
-
-**Figure D.17** For Prob. 8.69.
-
-- **8.73** This is a design problem with multiple answers.
-- **8.75** See Fig. D.19.
-
-For Prob. 8.75.
-
-**8.77** See Fig. D.20.
-
-**Figure D.20**  For Prob. 8.77.
-
-- **8.79** 173.61 *μ*F
-- **8.81** 2.533 *μ*H, 625 *μ*F
-
-**8.83** *<sup>d</sup>*<sup>2</sup> \_\_\_*v dt*2 + \_\_ *R L*  \_\_\_ *dv dt* + \_\_\_*<sup>R</sup> LC iD* + \_\_1 *C*  \_\_\_ *diD dt* = \_\_\_ *<sup>v</sup><sup>s</sup> LC* 
-
-# Chapter 9
-
-- **9.1** (a) 50 V, (b) 209.4 ms, (c) 4.775 Hz, (d) 44.48 V, 0.3 rad
-- **9.3** (a) 10 cos(*ωt* − 60°), (b) 9 cos(8*t* + 90°), (c) 20 cos(*ωt* + 135°)
-- **9.5** 30°, *v*1 lags *v*<sup>2</sup>
-- **9.7** Proof
-- **9.9** (a) 50.88 ⧸−15.52°, (b) 60.02 ⧸−110.96°
-- **9.11** (a) 21 ⧸−15° V, (b) 8 ⧸ 160° mA, (c) 120 ⧸−140° V, (d) 60 ⧸−170° mA
-
-**9.13** (a) −1.2749 + *j*0.1520, (b) −2.083, (c) 35 + *j*14
-
-- **9.15** (a) −6 − *j*11, (b) 120.99 + *j*4.415, (c) −1
-- **9.17** 15.62 cos(50*t* − 9.8°) V
-- **9.19** (a) 3.32 cos(20*t* + 114.49°), (b) 64.78 cos(50*t* − 70.89°), (c) 9.44 cos(400*t* − 44.7°)
-- **9.21** (a) *f*(*t*) = 8.324 cos(30*t* + 34.86°), (b) *g*(*t*) = 5.565 cos(*t* − 62.49°), (c) *h*(*t*) = 1.2748 cos(40*t* − 168.69°)
-- **9.23** (a) 320.1 cos(20*t* − 80.11°) A, (b) 36.05 cos(5*t* + 93.69°) A
-- **9.25** (a) 0.8 cos(2*t* − 98.13°) A, (b) 0.745 cos(5*t* − 4.56°) A
-- **9.27** 0.289 cos(377*t* − 92.45°) V
-- **9.29** 2 sin(10<sup>6</sup> *t* − 65°)
-- **9.31** 900.6 cos(2*t* + 51.21°) mA
-- **9.33** 139.64 V
-- **9.35** 11.015 cos(200*t* − 16.7°) A
-- **9.37** (25 − *j*25) mS
-- **9.39** 9.135 + *j*27.47 Ω, 3.972 cos(10*t* − 71.6°) A
-- **9.41** 72.74 cos(*t* − 18.43°) V
-- **9.43** 1.3868 ⧸ 33.69° A
-- **9.45** *j*5 A
-- **9.47** 10.598 cos(2000*t* + 52.63°) mA
-- **9.49** 22.63 sin(200*t* − 45°) V
-- **9.51** 225 cos(2*t* − 53.13°) A
-- **9.53** 23.66⧸−21.67° A
-- **9.55** (2.798 − *j*16.403) Ω
-
-- **9.57** 0.3171 − *j*0.1463 S
-- **9.59** (10 − *j*10) ohms
-- **9.61** 1 + *j*0.5 Ω
-- **9.63** 34.69 − *j*6.93 Ω
-- **9.65** 17.35⧸ 0.9° A, 6.83 + *j*1.094 Ω
-- **9.67** (a) 14.8⧸−20.22° mS, (b) 19.704⧸ 74.56° mS
-- **9.69** 1.661 + *j*0.6647 S
-- **9.71** 1.058 − *j*2.235 Ω
-- **9.73** 0.3796 + *j*1.46 Ω
-- **9.75** Can be achieved by the RL circuit shown in Fig. D.21.
-
-# **Figure D.21**
-
-For Prob. 9.75.
-
-- **9.77** (a) 26.57° lagging, (b) 1 MHz
-- **9.79** (a) 140.2°, (b) leading, (c) 18.43 V
-- **9.81** 1.8 kΩ, 0.1 *μ*F
-- **9.83** 104.17 mH
-- **9.85** Proof
-- **9.87** 34.96⧸−6.54° Ω
-- **9.89** 25 *μ*F
-- **9.91** 4 *μ*F
-- **9.93** 3.592⧸−38.66° A
-
-Chapter 10
-
-- **10.1** 1.9704 cos(10*t* + 5.65°) A
-- **10.3** 3.835 cos(4*t* − 35.02°) V
-- **10.5** 12.398 cos(4 × 10<sup>3</sup> *t* + 4.06°) mA
-- **10.7** 124.08⧸−154° V
-
-- **10.9** 6.154 cos(10<sup>3</sup> *t* + 70.26°) V  **10.11** 199.5⧸ 86.89° mA  **10.13** 29.36⧸ 62.88° A  **10.15** 7.906⧸ 43.49° A  **10.17** 9.25⧸−162.12° A  **10.19** 7.682⧸ 50.19° V  **10.21** (a) 1, 0, − *<sup>j</sup>* \_\_ *R* √ \_\_ \_\_*L C* , (b) 0, 1, *<sup>j</sup>* \_\_ *R* √ \_\_ \_\_*L C*   **10.23** (1 − *ω*<sup>2</sup> *LC*)*Vs* \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_<sup>1</sup>− *ω*<sup>2</sup> *LC* + *jωRC*(2 − *ω*<sup>2</sup> *LC*)  **10.25** 1.4142 cos(2*t* + 45°) A  **10.27** 7.047⧸ 95.24° A, 1.4892⧸ 37.71° A  **10.29** This is a design problem with several different answers.  **10.31** 1.0897⧸ 61.44° A  **10.33** 7.906⧸ 43.49° A  **10.35** 1.971⧸−2.1° A  **10.37** 2.38⧸−96.37° A, 2.38 ⧸143.63° A, 2.38⧸23.63° A  **10.39** 381.4⧸109.6° mA, 344.3⧸124.4° mA, 145.5⧸60.42° mA, 100.5⧸48.5° mA  **10.41** [14.142 sin (2*t* + 45°) + 26.83 cos(4*t* + 26.57°)] V  **10.43** 19.804 cos(2*t* − 129.17°) A  **10.45** 395.6 cos(10*t* + 21.47°) + 149.75 sin(4*t* + 176.57°) mA  **10.47** [4 + 0.504 sin(*t* + 19.1°) + 0.3352 cos(3*t* − 76.43°)] A  **10.49** 883.9 cos(20*t* − 30°) mA  **10.51** 109.3⧸30° mA Appendix D Answers to Odd-Numbered Problems **A-33**
-  - **10.53** 27.44⧸−59.04° V
-  - **10.55** (a) **Z***N* = **Z**Th = 22.63 ⧸−63.43° Ω, **V**Th = 25⧸−150° V, **I***N* = 1.1181⧸−86.6° A, (b) **Z***N* = **Z**Th = 10 ⧸ 26° Ω, **V**Th = 101⧸58° V, **I***N* = 10.176⧸32° A
-  - **10.57** This is a design problem with multiple answers.
-
-| 10.59       | −6 + j38 Ω                                                                          | 11.15 90 W    |  |
-|-------------|-------------------------------------------------------------------------------------|---------------|--|
-|             | 10.61 (−180 + j90) V, (−8 + j6) Ω                                                   |               |  |
-|             | 10.63 11.314⧸15° A, (10 − j10) Ω                                                    |               |  |
-|             | 10.65 This is a design problem with multiple answers.                               | 11.21 19.58 Ω |  |
-|             | 10.67 7.415⧸−84.68° V, 656.5⧸−90.16° mA,<br>11.243 + j1.079 Ω                       |               |  |
-|             | 10.69 j[1/(ωRC)], Vm sin(ωt + 90°) V                                                | 11.25 8.165   |  |
-|             | 10.71 72 cos (2t + 29.52°) V                                                        | 11.27 2.887 A |  |
-|             | 10.73 21.21⧸−45° kΩ                                                                 |               |  |
-|             | 10.75 0.12499⧸180°                                                                  | 11.31 2.944 V |  |
-| 10.77       | R2 + R3 + jωC2R2R3<br>________________________<br>(1 + jωR1C1)(R3 + jωC2R2R3)       | 11.33 5.332 A |  |
-|             | 10.79 35.78⧸−153.44° V                                                              | 11.35 21.6 V  |  |
-|             | 10.81 11.27⧸128.1 V                                                                 |               |  |
-|             | 10.83 6.611 cos (1,000t − 159.2°) V                                                 |               |  |
-|             | 10.85 This is a design problem with multiple answers.                               |               |  |
-| 10.87       | 15.91⧸169.6° V, 5.172⧸−138.6° V, 2.27⧸−152.4° V                                     |               |  |
-| 10.89 Proof |                                                                                     |               |  |
-|             | 10.91 (a) 180 kHz,<br>(b) 40 kΩ                                                     |               |  |
-| 10.93 Proof |                                                                                     |               |  |
-| 10.95 Proof |                                                                                     |               |  |
-|             | Chapter 11                                                                          |               |  |
-|             | (Assume all values of currents and voltages are rms unless<br>otherwise specified.) |               |  |
-| 11.1        | [1.320 + 2.640 cos(100t + 60˚)] kW, 1.320 kW                                        |               |  |
-| 11.3        | 213.4 W                                                                             |               |  |
-| 11.5        | P1Ω = 1.4159 W, P2Ω = 5.097 W,<br>P3H = P0.25F = 0 W                                |               |  |
-| 11.7        | 1 kW                                                                                |               |  |
-| 11.9        | 897 μW                                                                              |               |  |
-|             | 11.11 3.472 W                                                                       |               |  |
-|             | 11.13 28.36 W                                                                       |               |  |
-
-|             | 11.17 20 Ω, 31.25 W                                                                                                                                                                                                                                                                                                                                               |
-|-------------|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
-|             | 11.19 100 Ω, 6.25 W                                                                                                                                                                                                                                                                                                                                               |
-|             | 11.21 19.58 Ω                                                                                                                                                                                                                                                                                                                                                     |
-|             | 11.23 This is a design problem with multiple answers.                                                                                                                                                                                                                                                                                                             |
-| 11.25 8.165 |                                                                                                                                                                                                                                                                                                                                                                   |
-|             | 11.27 2.887 A                                                                                                                                                                                                                                                                                                                                                     |
-|             | 11.29 17.321 A, 3.6 kW                                                                                                                                                                                                                                                                                                                                            |
-|             | 11.31 2.944 V                                                                                                                                                                                                                                                                                                                                                     |
-|             | 11.33 5.332 A                                                                                                                                                                                                                                                                                                                                                     |
-|             | 11.35 21.6 V                                                                                                                                                                                                                                                                                                                                                      |
-|             | 11.37 This is a design problem with multiple answers.                                                                                                                                                                                                                                                                                                             |
-|             | 11.39 (a) 0.8575, 17.794 kW, 10.676 kVAR,<br>(b) 585.1 μF                                                                                                                                                                                                                                                                                                         |
-|             | 11.41 (a) 0.5547 (leading), (b) 0.9304 (lagging)                                                                                                                                                                                                                                                                                                                  |
-|             | 11.43 This is a design problem with multiple answers.                                                                                                                                                                                                                                                                                                             |
-|             | 11.45 (a) 46.9 V, 1.061 A, (b) 20 W                                                                                                                                                                                                                                                                                                                               |
-|             | 11.47 (a) S = (339.4 + j339.4) VA,<br>average power = 339.4 W,<br>reactive power = 339.4 VAR<br>(b) S = (678.8 – j678.8) VA,<br>average power = 678.8 W,<br>reactive power = −678.8 VAR<br>(c) S = (7.637 + j7.637) kVA, average power =<br>7.637 W, reactive power = 7.637 VAR<br>(d) S = (250 + j433) kVA, average power =<br>250 kW, reactive power = 433 kVAR |
-|             | 11.49 (a) 4 + j2.373 kVA,<br>(b) 1.6 – j1.2 kVA,<br>(c) 0.4624 + j1.2705 kVA,<br>(d) 110.77 + j166.16 VA                                                                                                                                                                                                                                                          |
-|             | 11.51 (a) 0.9956 (lagging),<br>(b) 304 W,<br>(c) 28.64 VAR,                                                                                                                                                                                                                                                                                                       |
-
-- (d) 305.3 VA,
-- (e) [304 + *j*28.64] VA
-- **11.53** (a) 47 ⧸29.8° A, (b) 1.0 (lagging)
-
-- **11.55** This is a design problem with multiple answers.
-- **11.57** (219 − *j*145.99) VA
-- **11.59** *j*2 VAR, −*j*2 VAR
-- **11.61** 66.2⧸92.4° A, 6.62⧸−2.4° kVA
-- **11.63** 129.31⧸18.43° A
-- **11.65** 80 *μ*W
-- **11.67** (a) 12.5⧸−36.87° mVA, (b) 78.13 W
-- **11.69** (a) 0.8 (lagging), (b) 6.195 kW, (c) 63.66 *μ*F
-- **11.71** (a) 50.14 + *j*1.7509 mΩ, (b) 0.9994 lagging, (c) 2.392⧸−2° kA
-- **11.73** (a) 12.21 kVA, (b) 50.86⧸−35° A, (c) 4.083 kVAR, 188.03 *μ*F, (d) 43.4⧸−16.26° A
-- **11.75** (a) (32.14 + *j*7.357) kVAR, (b) 0.9748 (lagging), (c) 100.08 *μ*F
-- **11.77** 157.69 W
-- **11.79** 50 mW
-- **11.81** This is a design problem with multiple answers.
-- **11.83** (a) 688.1 W, (b) 840 VA, (c) 481.8 VAR, (d) 0.8191 (lagging)
-- **11.85** (a) 13 A, 21.71⧸ 166.3° A, 9.588⧸−32.43° A, (b) (4.091 + *j*0.617) kVA, (c) 0.9888 (lagging)
-- **11.87** 0.5333
-- **11.89** (a) 12 kVA, 9.36 + *j*7.51 kVA, (b) 2.866 + *j*2.3 Ω
-- **11.91** 0.8182 (lagging), 1.398 *μ*F
-- **11.93** (a) 7.328 kW, 1.196 kVAR, (b) 0.987
-- **11.95** (a) 2.814 kHz, (b) 431.8 mW
-- **11.97** 1.8396 kW
-
-# Chapter 12
-
-**(Assume all values of currents and voltages are rms unless otherwise specified.)**
-
-- **12.1** (a) 231⧸−30°, 231⧸−150°, 231⧸ 90° V, (b) 231⧸ 30°, 231⧸ 150°, 231⧸−90° V
-- **12.3** acb sequence, 100⧸−75° V
-- **12.5** 207.8 cos(*ω*t + 62°) V, 207.8 cos(*ωt* − 58°) V, 207.8 cos(*ω*t −178°) V
-- **12.7** 44⧸ 53.13° A, 44⧸−66.87° A, 44⧸ 173.13° A
-- **12.9** 4.8⧸−36.87° A, 4.8⧸−156.87° A, 4.8⧸ 83.13° A
-- **12.11** 762.1 V, 366.1 A
-- **12.13** 20.43 A, 3.744 kW
-- **12.15** 13.66 A
-- **12.17** 4.8⧸ 53.13° A, 4.8⧸−66.87° A, 4.8⧸ 173.13° A
-- **12.19** 13.915⧸−18.43° A, 13.915⧸−138.43° A, 13.915⧸ 101.57° A, 24.1⧸−48.43° A, 24.1⧸−168.43° A, 24.1⧸71.57° A
-- **12.21** 44⧸−30° A, 76.21⧸−60° A, 0.866
-- **12.23** 106.61⧸ –0.65° V, 106.55⧸ 119.34° V, 106.6⧸ –120.67° V
-- **12.25** 17.742⧸ 4.78° A, 17.742⧸−115.22°A, 17.742⧸124.78° A
-- **12.27** 91.79 V
-- **12.29** [5.197 + *j*4.586] kVA
-- **12.31** (a) 6.144 + *j*4.608 Ω, (b) 36.08 A, (c) 207.2 *μ*F
-- **12.33** 7.69 A, 360.3 V
-- **12.35** (a) 14.61 − *j*5.953 A, (b) [10.081 + *j*4.108] kVA, (c) 0.9261
-- **12.37** 26.24 A, (5.808 − *j*7.744) Ω
-- **12.39** 432 W
-- **12.41** 9.021 A
-- **12.43** 4.373 − *j*1.145 kVA
-- **12.45** 2.109⧸ 24.83° kV
-
-- **12.47** 39.19 A (rms), 0.9982 (lagging)
-- **12.49** (a) 27.65 kW, (b) 9.216 kW
-- **12.51** 2.078⧸ 120° A, 2.078⧸ 90° A, 2.078⧸ 150° A, 2.939⧸ 165° A, 1.0759⧸ 15° A, 2.078⧸ –150° A
-- **12.53** This is a design problem with multiple answers.
-- **12.55** 8⧸−60° A, 28.84⧸ 133.9° A, 21.17⧸−40.89° A, (8.64 + *j*1.6627) kVA
-- **12.57** *Ia* = 3.917⧸−18.1° A, *Ib* = 2.931⧸−130.55° A, *Ic* = 3.895⧸ 117.82° A
-- **12.59** 220.6⧸−34.56°, 214.1⧸−81.49°, 49.91⧸−50.59° V, assuming that *N* is grounded.
-- **12.61** 11.15⧸ 37° A, 230.8⧸−133.4° V, assuming that *N* is grounded.
-- **12.63** 18.67⧸ 158.9° A, 12.38⧸ 144.1° A
-- **12.65** 11.02⧸ 12° A, 11.02⧸−108° A, 11.02⧸ 132° A
-- **12.67** (a) 97.67 kW, 88.67 kW, 82.67 kW, (b) 108.97 A
-- **12.69** I*a* = 94.32⧸−62.05° A, I*b* = 94.32⧸ 177.95° A, I*c* = 94.32⧸ 57.95° A, 28.8 + *j*18.03 kVA
-- **12.71** (a) 2,590 W, 4,808 W, (b) 8,335 VA
-- **12.73** 2,360 W, −632.8 W
-- **12.75** (a) 20 mA, (b) 200 mA
-- **12.77** 520 W
-- **12.79** 37.29⧸−19.65°, 37.29⧸−139.65°, 37.29⧸100.35° A, 484.7⧸ 2.97°, 484.7⧸−117.03°, 484.7⧸ 122.97° V
-- **12.81** 516 V
-- **12.83** 183.42 A
-
-- **12.85** Z*Y* = 2.133 Ω
-- **12.87** 2.77⧸−176.6° A, (4.581 + *j*2.604) kVA, (3.971 + *j*2.64) kVA
-
-# Chapter 13
-
-# **(Assume all values of currents and voltages are rms unless otherwise specified.)**
-
-- **13.1** 20 H
-- **13.3** 300 mH, 100 mH, 50 mH, 0.2887
-- **13.5** (a) 247.4 mH, (b) 48.62 mH
-- **13.7** 1.081⧸ 144.16° V
-- **13.9** 2.074⧸ 21.12° V
-- **13.11** 461.9 cos(600*t* − 80.26°) mA
-- **13.13** [4.308 + *j*4.538] Ω
-- **13.15** (11.251 + *j*18.754) Ω, 970.1⧸−14.04° mA
-- **13.17** [25.07 + *j*25.86] Ω
-- **13.19** See Fig. D.22.
-
-# **Figure D.22**
-
-For Prob. 13.19.
-
-- **13.21** This is a design problem with multiple answers.
-- **13.23** 100 cos(100*t* − 90°) V, 5 J
-- **13.25** 2.2 sin(2*t* − 4.88°) A, 1.5085⧸ 17.9° Ω
-- **13.27** 191.86 W
-- **13.29** 0.9845, 521.6 mJ
-- **13.31** This is a design problem with multiple answers.
-- **13.33** 12.769 + *j* 7.154 Ω
-
-- **13.35** 1.4754⧸−21.41° A, 77.5⧸−134.85° mA, 77⧸−110.41° mA **13.37** (a) 10, (b) 208.3 A, (c) 20.83 A **13.39** 15.7⧸ 20.31° A, 78.5⧸ 20.31° A **13.41** −6 A **13.43** 16.744 V, 66.98 V **13.45** 36.71 mW **13.47** 109.55 cos(3*t* + 5.48°) V **13.49** 0.937 cos(2*t* + 51.34°) A **13.51** [8 − *j*1.5 Ω, 14.743⧸ 10.62° A **13.53** (a) 5, (b) 112.5 W **13.55** 5 Ω **13.57** (a) 25.9⧸ 69.96°, 12.95⧸ 69.96° A (rms), (b) 21.06⧸ 147.4°, 42.12⧸ 147.4°, 42.12⧸ 147.4° V(rms), (c) 1554⧸ 20.04° VA  **13.59** 420.1 W, 283.6 W, 52.52 W  **13.61** 6 A, 0.36 A, −60 V  **13.63** 7.071⧸−45° A, 3.536⧸−45° A, 14.142⧸−45° A  **13.65** 11.05 W  **13.67** (a) 352 V, (b) 14.205 A, (c) 5.682 A  **13.69** 200 V, (4 − *j*4) kΩ, (4 + *j*4) kΩ  **13.71** 0.913, 7.841 A  **13.73** (a) three-phase ∆-Y transformer, (b) 8.66⧸ 156.87° A, 5⧸−83.13° A, (c) 1.8 kW  **13.75** (a) 0.11547, (b) 76.98 A, 15.395 A  **13.77** (a) a single-phase transformer, 1:*n*, *n* = 1∕110, (b) 7.576 mA **13.79** 1.306⧸−68.01° A, 406.8⧸−77.86° mA, 1.336⧸−54.92° A
-- **13.81** 104.5⧸ 13.96° mA, 29.54⧸−143.8° mA, 208.824.4° mA
-
-- **13.83** 1.08⧸ 33.91° A, 15.14⧸−34.21° V
-- **13.85** 100 turns
-- **13.87** 0.5
-- **13.89** 0.5, 41.67 A, 83.33 A
-- **13.91** (a) 1,875 kVA, (b) 7,812 A
-- **13.93** (a) See Fig. D.23(a). (b) See Fig. D.23(b).
-
-# **Figure D.23**
-
-For Prob. 13.93.
-
-**13.95** (a) 1∕60, (b) 139 mA
-
-Chapter 14
-
-$$
-14.1 \quad \frac{1}{1 + j\omega/\omega_o}, \omega_o = \frac{R}{L}
-$$
-
-**14.3** 20*s*∕(*s* <sup>2</sup> + 4*s* + 1)
-
-14.5 
-$$
-\frac{(Ls + R)}{(LCs^2 + RCs + 1)}.
-$$
-
-**14.7** (a) 1.0116, (b) 0.5623, (c) 5.623 × 10<sup>10</sup>
-
-.
-
-**Figure D.24**
-
-For Prob. 14.9.
-
-**14.11** See Fig. D.25.
-
-**Figure D.26** For Prob. 14.13.
-
-**Figure D.25** For Prob. 14.11.
-
-**Figure D.27** For Prob. 14.15.
-
-# **Figure D.28**
-
-For Prob. 14.17.
-
-**14.19** See Fig. D.29.
-
-**14.21** See Fig. D.30.
-
-**14.21** See Fig. D.30.  
-**14.23** 
-$$
-\frac{1,000j\omega}{(1+j\omega)(10+j\omega)^2}
-$$
-
- (It should be noted that this function could also have a minus sign out in front and still be correct. The magnitude plot does not contain this information. It can only be obtained from the phase plot.)
-
-- **14.25** 2 kΩ, 2 − *j*0.75 kΩ, 2 − *j*0.3 kΩ, 2 + *j*0.3 kΩ, 2 + *j*0.75 kΩ
-- **14.27** *R* = 1 Ω, *L* = 0.1 H, *C* = 25 mF
-- **14.29** 4.082 krad/s, 105.55 rad/s, 38.67
-- **14.31** 0.5, 0.25 nF, 10 kΩ
-- **14.33** 125, 5 Mrad/s
-- **14.35** 250 *μ*F, 40, 400 krad/s
-- **14.37** 2 kΩ, (1.4212 + *j*53.3) Ω, (8.85 + *j*132.74) Ω, (8.85 − *j*132.74) Ω, (1.4212 − *j*53.3) Ω
-- **14.39** 4.841 krad/s
-
-**Figure D.30** For Prob. 14.21.
-
-**14.41** This is a design problem with multiple answers.
-
-14.43 
-$$
-\sqrt{\frac{1}{LC} - \frac{R^2}{L^2}}, \frac{1}{\sqrt{LC}}
-$$
-
-- **14.45** 447.2 rad/s, 1.067 rad/s, 419.1
-- **14.47** 796 kHz
-- **14.49** This is a design problem with multiple answers.
-- **14.51** 1.256 kΩ
-- **14.53** 18.045 kΩ. 2.872 H, 10.5
-- **14.55** 1.56 kHz < *f* < 1.62 kHz, 25
-- **14.57** (a) 1 rad/s, 3 rad/s, (b) 1 rad/s, 3 rad/s
-- **14.59** 2.408 krad/s, 15.811 krad/s
-
-**14.61** (a) 
-$$
-\frac{1}{1 + j\omega RC}
-$$
-  
-(b) 
-$$
-\frac{j\omega RC}{1 + j\omega RC}
-$$
-
-- **14.63** 10 MΩ, 100 kΩ
-- **14.65** Proof
-- **14.67** If *Rf* = 20 kΩ, then *Ri* = 80 kΩ and *C* = 15.915 nF.
-- **14.69** Let *R* = 10 kΩ, then *Rf* = 25 kΩ, *C* = 7.96 nF.
-- **14.71** *Kf* = 2 × 10<sup>−</sup><sup>4</sup> , *Km* = 5 × 10<sup>−</sup><sup>3</sup>
-- **14.73** 9.6 MΩ, 32 *μ*H, 0.375 pF
-
-- **14.75** 200 Ω, 400 *μ*H, 1 *μ*F
-- **14.77** (a) 1,200 H, 0.5208 *μ*F, (b) 2 mH, 312.5 nF, (c) 8 mH, 7.81 pF
-
-**14.79** (a) 
-$$
-8s + 5 + \frac{10}{s}
-$$
-,  
-(b)  $0.8s + 50 + \frac{10^4}{s}$ , 111.8 rad/s
-
-- **14.81** (a) 0.4 Ω, 0.4 H, 1 mF, 1 mS, (b) 0.4 Ω, 0.4 mH, 1 *μ*F, 1 mS
-- **14.83** 0.1 pF, 0.5 pF, 1 MΩ, 2 MΩ
-- **14.85** See Fig. D.31.
-- **14.87** See Fig. D.32; high-pass filter, *f*0 = 1.2 Hz.
-- **14.89** See Fig. D.33.
-- **14.91** See Fig. D.34; *fo* = 800 Hz.
-- **14.93** \_\_\_\_\_\_\_\_\_ −*RCs* + 1 *RCs* + 1
-- **14.95** (a) 0.541 MHz < *fo* < 1.624 MHz, (b) 67.98, 204.1
-- **14.97** *<sup>s</sup>* 3 *LRLC*1*C*<sup>2</sup> \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*sRiC*1 + 1)(*<sup>s</sup>* 2 *LC*2 + *sRLC*<sup>2</sup> + 1) + *s* 2 *LC*1(*sRLC*2 + 1)
-- **14.99** 8.165 MHz, 4.188 × 10<sup>6</sup> rad/s
-- **14.101** 1.061 kΩ
-
-**14.101** 1.061 kΩ  
-**14.103** 
-$$
-\frac{R_2(1+sCR_1)}{R_1+R_2+sCR_1R_2}
-$$
-
-**Figure D.31**
-
-For Prob. 14.85.
-
-For Prob. 14.87.
-
-For Prob. 14.89.
-
-For Prob. 14.91.
-
-**15.13** (a) 
-$$
-\frac{s^2 - 1}{(s^2 + 1)^2}
-$$
-,  
-\n(b)  $\frac{2(s + 1)}{(s^2 + 2s + 2)^2}$ ,  
-\n(c)  $\tan^{-1}\left(\frac{\beta}{s}\right)$   
-\n**15.15**  $5\frac{1 - e^{-s} - se^{-s}}{s^2(1 - e^{-3s})}$ 
-
-**15.17** This is a design problem with multiple answers.
-
-15.19 
-$$
-\frac{1}{1 - e^{-2s}}
-$$
-  
-15.21 
-$$
-\frac{(2\pi s - 1 + e^{-2\pi s})}{2\pi s^2 (1 - e^{-2\pi s})}
-$$
-
-**15.23** (a) 
-$$
-\frac{(1 - e^{-s})^2}{s(1 - e^{-2s})}
-$$
-  
-(b) 
-$$
-\frac{2(1 - e^{-2s}) - 4se^{-2s}(s + s^2)}{s^3(1 - e^{-2s})}
-$$
-
-**15.25** (a) 18 and 0, (b) 18 and 0
-
-- **15.27** (a) *u*(*t*) + 2*e*<sup>−</sup>*<sup>t</sup> u*(*t*), (b) 3*δ*(*t*) − 11*e*<sup>−</sup>4*<sup>t</sup> u*(*t*), (c) (2*e*<sup>−</sup>*<sup>t</sup>* − 2*e*<sup>−</sup>3*<sup>t</sup>* )*u*(*t*), (d) (3*e*<sup>−</sup>4*<sup>t</sup>* − 3*e*<sup>−</sup>2*<sup>t</sup>* + 6*te*<sup>−</sup>*2<sup>t</sup>* )*u*(*t*)
-- **15.29** [1 + 2*e*<sup>−</sup>*<sup>t</sup>* cos (*t* + 90°)] *u*(*t*)
-- **15.31** (a) (−5*e*<sup>−</sup>*<sup>t</sup>* + 20*e*<sup>−</sup>2*<sup>t</sup>* − 15*e*<sup>−</sup>3*<sup>t</sup>* )*u*(*t*) (b) (−*e*<sup>−</sup>*<sup>t</sup>* <sup>+</sup>(1 + 3*<sup>t</sup>* <sup>−</sup> *t* 2 \_\_ 2 )*e*<sup>−</sup>2*<sup>t</sup>* ) *u*(*t*), (c) (−0.2*e*<sup>−</sup>2*<sup>t</sup>* + 0.2*e*<sup>−</sup>*<sup>t</sup>* cos(2*t*) + 0.4*e*<sup>−</sup>*<sup>t</sup>* sin(2*t*))*u*(*t*)
-- **15.33** (a) (3*e*<sup>−</sup>*<sup>t</sup>* + 3 sin(*t*) − 3 cos(*t*))*u*(*t*), (b) cos(*t* −*π*)*u*(*t* − *π*), (c) 8 [1 − *e*<sup>−</sup>*<sup>t</sup>* − *te*<sup>−</sup>*<sup>t</sup>* − 0.5*t* 2 *e*−*t* ]*u*(*t*)
-
-**15.35** (a) 
-$$
-[2e^{-(t-6)} - e^{-2(t-6)}]u(t-6)
-$$
-,  
-\n(b)  $\frac{4}{3}u(t)[e^{-t} - e^{-4t}] - \frac{1}{3}u(t-2)[e^{-(t-2)} - e^{-4(t-2)}]$ ,  
-\n(c)  $\frac{1}{13}u(t-1)[-3e^{-3(t-1)} + 3\cos 2(t-1) + 2\sin 2(t-1)]$ 
-
-**15.37** (a) 
-$$
-(2 - e^{-2t})u(t)
-$$
-,  
-\n(b)  $[0.4e^{-3t} + 0.6e^{-t} \cos t + 0.8e^{-t} \sin t]u(t)$ ,  
-\n(c)  $e^{-2(t-4)} u(t-4)$ ,  
-\n(d)  $\left(\frac{10}{3} \cos t - \frac{10}{3} \cos 2t\right)u(t)$ 
-
-**15.39** (a) 
-$$
-(-1.6e^{-t} \cos 4t - 4.05e^{-t} \sin 4t + 3.6e^{-2t} \cos 4t + (3.45e^{-2t} \sin 4t) u(t),
-$$
-  
-\n(b)  $[0.08333 \cos 3t + 0.02778 \sin 3t + 0.0944e^{-0.551t} - 0.1778e^{-5.449t}]u(t)$ 
-
-$$
-\mathbf{15.41} \quad z(t) = \begin{cases} 8t, & 0 < t < 2 \\ 16 - 8t, & 2 < t < 6 \\ -16, & 6 < t < 8 \\ 8t - 80, & 8 < t < 12 \\ 112 - 8t, & 12 < t < 14 \\ 0, & \text{otherwise} \end{cases}
-$$
-
-**15.43** (a) 
-$$
-y(t) = \begin{cases} \frac{1}{2}t^2, & 0 < t < 1 \\ -\frac{1}{2}t^2 + 2t - 1, & 1 < t < 2 \\ 1, & t > 2 \\ 0, & \text{otherwise} \end{cases}
-$$
-
-(b) 
-$$
-y(t) = 2(1 - e^{-t}), t > 0,
-$$
-
-(c) 
-$$
-y(t) = \begin{cases} \frac{1}{2}t^2 + t + \frac{1}{2}, & -1 < t < 0 \\ \frac{1}{2}t^2 + t + \frac{1}{2}, & 0 < t < 2 \\ \frac{1}{2}t^2 - 3t + \frac{9}{2}, & 2 < t < 3 \\ 0, & \text{otherwise} \end{cases}
-$$
-
-**15.45** 
-$$
-(4e^{-2t} - 8te^{-2t})u(t)
-$$
-  
-\n**15.47** (a)  $[-6e^{-t} + 12e^{-2t}]u(t)$ , (b)  $[6e^{-t} - 6e^{-2t}]$   
-\n**15.49** (a)  $\left(\frac{t}{a}(e^{at} - 1) - \frac{1}{a^2} - \frac{e^{at}}{a^2}(at - 1)\right)u(t)$ ,  
-\n(b)  $[0.5 \cos(t)(t + 0.5 \sin(2t)) - 0.5 \sin(t)(\cos(t) - 1)]u(t)$ 
-
-**15.51** [12.5*e*<sup>−</sup>*<sup>t</sup>* − 7.5*e*–3*<sup>t</sup>* ]*u*(*t*)
-
-**15.53** cos(*t*) + sin(*t*) or 1.4142 cos(*t* − 45°)
-
-$$
-15.55\ \left(\frac{1}{40} + \frac{1}{20}e^{-2t} - \frac{3}{104}e^{-4t} - \frac{3}{65}e^{-t}\cos(2t) - \frac{2}{65}e^{-t}\sin(2t)\right)u(t)
-$$
-
-- **15.57** This is a design problem with multiple answers.
-- **15.59** [−7.5*e*<sup>−</sup>*<sup>t</sup>* + 36*e*<sup>−</sup>2*<sup>t</sup>* − 31.5*e*<sup>−</sup>3*<sup>t</sup>* ]*u*(*t*)
-- **15.61** (a) [3 + 3.162 cos (2*t* − 161.12°)]*u*(*t*) volts, (b) [2 − 4*e*<sup>−</sup>*<sup>t</sup>* + *e*<sup>−</sup>4*<sup>t</sup>* ]*u*(*t*) amps, (c) [3 + 2*e*<sup>−</sup>*<sup>t</sup>* + 3*te*<sup>−</sup>*<sup>t</sup>* ]*u*(*t*) volts, (d) [2 + 2*e*<sup>−</sup>*<sup>t</sup>* cos(2*t*)]*u*(*t*) amps
-
-# Chapter 16
-
-- **16.1** [(7 + 35*t*)*e*<sup>−</sup>5*<sup>t</sup>* ] *u*(*t*) A
-- **16.3** [(20 + 20*t*)*e*<sup>−</sup>*<sup>t</sup>* ]*u*(*t*) V
-- **16.5** 750 Ω, 25 H, 200 *μ*F
-- **16.7** [6 + 12*e*<sup>−</sup>*<sup>t</sup>* cos(2*t*) + 2 sin(2*t*))]*u*(*t*) A
-- **16.9** [3 + 5.924*e*<sup>−</sup>1.5505*<sup>t</sup>* − 1.4235*e*<sup>−</sup>6.45*<sup>t</sup>* ]*u*(*t*) mA
-- **16.11** 20.83 Ω, 80 *μ*F
-- **16.13** This is a design problem with multiple answers.
-- **16.15** 120 Ω
-- **16.17** 7.5 (*e*<sup>−</sup>2*<sup>t</sup>* <sup>−</sup>\_\_\_2 √ \_\_ 7 *e*<sup>−</sup>0.5*<sup>t</sup>* sin ( √ \_\_ \_\_\_7 2 *t* )) *u*(*t*) A
-- **16.19** [−2.333*e*<sup>−</sup>*t*∕<sup>2</sup> + 2.333*e*<sup>−</sup>2*<sup>t</sup>* ]*u*(*t*) volts
-- **16.21** [10.776 *e*<sup>−</sup>2.679*<sup>t</sup>* − 0.774*e*<sup>−</sup>37.32*<sup>t</sup>* ]*u*(*t*) volts
-- **16.23** 24 cos(0.5*t* + 90°)*u*(*t*) volts
-- **16.25** [45*e*<sup>−</sup>*<sup>t</sup>* − 5*e*<sup>−</sup>9*<sup>t</sup>* ]*u*(*t*) volts
-- **16.27** [30 − 15.309*e*<sup>−</sup>0.05051*<sup>t</sup>* + 0.3078*e*<sup>−</sup>4.949*<sup>t</sup>* ]*u*(*t*) volts
-- **16.29** 17.5 cos(8*t* + 90°)*u*(*t*) amps
-- **16.31** [−16 + 66.67*e*<sup>−</sup>0.8*<sup>t</sup>* cos(0.6*t* − 53.13°)]*u*(*t*) volts, 13.333*e*<sup>−</sup>0.8*<sup>t</sup>* [cos(0.6*t* + 90°)]*u*(*t*) amps
-- **16.33** This is a design problem with multiple answers.
-- **16.35** [9.091*e*<sup>−</sup>*<sup>t</sup>* + 19.653*e*<sup>−</sup>0.0625*<sup>t</sup>* cos(0.7044*t* − 117.55°] *u*(*t*) V.
-- **16.37** [−60 + 60.21*e*<sup>−</sup>0.1672*<sup>t</sup>* − 0.21*e*<sup>−</sup>47.84*<sup>t</sup>* ]*u*(*t*) volts
-- **16.39** [4.364*e*<sup>−</sup>2*<sup>t</sup>* cos(4.583*t* − 90°)]*u*(*t*) amps
-- **16.41** [100*te*<sup>−</sup>10*<sup>t</sup>* ]*u*(*t*) volts
-- **16.43** [9 + 9*e*<sup>−</sup>2*<sup>t</sup>* + 6*te*<sup>−</sup>2*<sup>t</sup>* ]*u*(*t*) amps
-- **16.45** [*io*∕(*ωC*)] cos(*ωt* + 90°)*u*(*t*) volts
-- **16.47** [60 − 40*e*<sup>−</sup>0.6*<sup>t</sup>* cos(0.2*t*) − sin(0.2*t*))]*u*(*t*) A
-- **16.49** [1.0714*e*<sup>−</sup>2*<sup>t</sup>* − 2.572*e*<sup>−</sup>0.5*<sup>t</sup>* cos(1.25*t*) + 4.791*e*<sup>−</sup>0.5*<sup>t</sup>* sin(1.25*t*)]*u*(*t*) A
-
-- **16.51** [−12 + 41.17*e*<sup>−</sup>15.125*<sup>t</sup>* cos(4.608*t* − 73.06°)]*u*(*t*) amps
-- **16.53** [11.547*e*<sup>−</sup>*<sup>t</sup>* cos(1.7321*t* + 30°)]*u*(*t*) volts
-- **16.55** [5−4*e*<sup>−</sup>*<sup>t</sup>* − 1*e*<sup>−</sup>6*<sup>t</sup>* ]*u*(*t*) amps, [2*e*<sup>−</sup>*<sup>t</sup>* − 2*e*<sup>−</sup>6*<sup>t</sup>* ]*u*(*t*) amps
-- **16.57** (a) (3∕*s*)[1 − *e*<sup>−</sup>*<sup>s</sup>* ], (b) [(2 − 2*e*<sup>−</sup>1.5*<sup>t</sup>* )*u*(*t*) − (2 − 2*e*<sup>−</sup>1.5(*t*−1))*u*(*t* − 1)] V
-- **16.59** [5*e*<sup>−</sup>*<sup>t</sup>* − 10*e*<sup>−</sup>*t*∕<sup>2</sup> cos (*t*∕2)]*u*(*t*) V
-- **16.61** [2.333 − 2.38*e*<sup>−</sup>1.2306*<sup>t</sup>* + 2.033*e*<sup>−</sup>0.6347*<sup>t</sup>* cos(1.4265*t* + 88.68°)]*u*(*t*) V
-- **16.63** [7.5*e*<sup>−</sup>4*<sup>t</sup>* cos (2*t*) + 345*e*<sup>−</sup>4*<sup>t</sup>* sin (2*t*)]*u*(*t*) V, [6 − 9*e*<sup>−</sup>4*<sup>t</sup>* cos (2*t*) − 17.062*e*<sup>−</sup>4*<sup>t</sup>* sin (2*t*)]*u*(*t*) A
-- **16.65** {110.1*e*<sup>−</sup>3*<sup>t</sup>* + 192*te*<sup>−</sup>3*<sup>t</sup>* − 10.1 cos(4*t*) + 34.58 sin(4*t*)}*u*(*t*) V
-- **16.67** [*e*<sup>10</sup>*<sup>t</sup>* − *e*<sup>−</sup>10*<sup>t</sup>* ] *u*(*t*) volts; this is an unstable circuit!
-- **16.69** 240(*s* + 1)∕[*s*(*s* + 3)(3*s* <sup>2</sup> + 8*s* + 1)], −120(*s* −1)∕ [*s*(*s* + 3)(3*s* <sup>2</sup> + 8*s* + 1)]
-
-**16.71** 
-$$
-160[2e^{-1.5t} - e^{-t}]u(t)
-$$
- A
-
-$$
-16.73 \ \ \frac{120s^2}{s^2+4}
-$$
-
-$$
-16.75 \quad 6 + \frac{1.5s}{2(s+3)} - \frac{3s(s+2)}{s^2 + 4s + 20} - \frac{18s}{s^2 + 4s + 20}
-$$
-
-$$
-16.77 \ \ \frac{9s}{3s^2+9s+2}
-$$
-
-**16.79** (a) 
-$$
-\frac{s^2 - 3}{3s^2 + 2s - 9}
-$$
-, (b)  $\frac{-3}{2s}$ 
-
-- **16.81** −1∕(*RLCs*<sup>2</sup> ) **16.83** (a) \_\_ *R L e*<sup>−</sup>*Rt*∕*<sup>L</sup> u*(*t*), (b) (1 − *e*<sup>−</sup>*Rt*∕*<sup>L</sup>* )*u*(*t*) **16.85** [9*e*<sup>−</sup>*<sup>t</sup>* − 9*e*<sup>−</sup>2*<sup>t</sup>* − 6*te*<sup>−</sup>2*<sup>t</sup>* ]*u*(*t*)
-- **16.87** This is a design problem with multiple answers.
-
-$$
-16.89 \begin{bmatrix} v'_{C} \\ i'_{L} \end{bmatrix} = \begin{bmatrix} -0.25 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} v'_{C} \\ i'_{L} \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} v_{s} \\ i_{s} \end{bmatrix};
-$$
-$$
-v_{o}(t) = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} v_{C} \\ i_{L} \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_{s} \\ i_{s} \end{bmatrix}
-$$
-$$
-16.91 \begin{bmatrix} x'_{1} \\ x'_{2} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -3 & -4 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} z(t);
-$$
-$$
-y(t) = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} + \begin{bmatrix} 0 \end{bmatrix} z(t)
-$$
-
-**16.93** 
-$$
-\begin{bmatrix} x'_{1} \\ x'_{2} \\ x'_{3} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -11 & -6 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} z(t);
-$$
-$$
-y(t) = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} + \begin{bmatrix} 0 \end{bmatrix} z(t)
-$$
-
-**16.95** 
-$$
-[-2.4 + 4.4e^{-3t}\cos(t) - 0.8e^{-3t}\sin(t)]u(t),
-$$
-$$
-[-1.2 - 0.8e^{-3t}\cos(t) + 0.6e^{-3t}\sin(t)]u(t)
-$$
-
-- **16.97** (a) 7(*e*<sup>−</sup>*<sup>t</sup>* − *e*<sup>−</sup>4*<sup>t</sup>* )*u*(*t*), (b) The system is stable.
-- **16.99** 500 *μ*F, 333.3 H
-- **16.101** 100 *μ*F
-
- **16.103** −100, 400, 2 × 104
-
-**16.105** If you let *L* = *R*<sup>2</sup> *C* then *Vo*/*Io* = *sL*.
-
-Chapter 17
-
-**17.1** (a) periodic, 2, (b) not periodic, (c) periodic, 2 *π*, (d) periodic, *π*, (e) periodic, 10, (f) not periodic, (g) not periodic
-
-**17.3** See Fig. D.35.
-
-**Figure D.36** For Prob. 17.7.
-
-- **17.9** *a*0 = 0.7958, *a*1 = 1.25, *a*2 = 0.5305, *a*3 = 0, *b*1 = 0 = *b*2 = *b*<sup>3</sup>
-- **17.11** <sup>∑</sup>*<sup>n</sup>*=−∞ ∞ <sup>75</sup> 4*n*<sup>2</sup> *π*2 [2 − 2 cos(*nπ*∕2) − 2*j* sin(*nπ*∕2) <sup>+</sup>*jnπ* cos(*nπ*∕2) + *nπ* (sin(*nπ*∕2))]*ejn π t*∕<sup>2</sup>
-- **17.13** This is a design problem with multiple answers.
-
-17.15 (a) 
-$$
-10 + \sum_{n=1}^{\infty} \sqrt{\frac{16}{(n^2 + 1)^2} + \frac{1}{n^6}}
-$$
-  
-\n $\cos\left(10nt - \tan^{-1}\frac{n^2 + 1}{4n^3}\right)$ ,  
-\n(b)  $10 + \sum_{n=1}^{\infty} \sqrt{\frac{16}{(n^2 + 1)} + \frac{1}{n^6}}$   
-\n $\sin\left(10nt + \tan^{-1}\frac{4n^3}{n^2 + 1}\right)$ 
-
- **17.17** (a) neither odd nor even, (b) even, (c) odd, (d) even, (e) neither odd nor even
-
-$$
-17.19 \frac{5}{n^2 \omega_o^2} \sin n\pi/2 - \frac{10}{n\omega_o} (\cos \pi n - \cos n\pi/2)
-$$
-$$
-- \frac{5}{n^2 \omega_o^2} (\sin \pi n - \sin n\pi/2) - \frac{2}{n\omega_o} \cos n\pi - \frac{\cos \pi n/2}{n\omega_o}
-$$
-$$
-17.21 \frac{5}{2} + \sum_{n=1}^{\infty} \frac{40}{n^2} \Big[ 1 - \cos \Big( \frac{n\pi}{2} \Big) \Big] \cos \Big( \frac{n\pi t}{2} \Big)
-$$
-
-17.21 
-$$
-\frac{5}{2} + \sum_{n=1}^{\infty} \frac{40}{n^2 \pi^2} \left[1 - \cos\left(\frac{n\pi}{2}\right)\right] \cos\left(\frac{n\pi}{2}\right)
-$$
-  
-17.23 This is a design problem with multiple
-
-answers.
-
-17.25  
-\n
-$$
-\sum_{n=1}^{\infty} \left\{ \left[ \frac{6}{\pi^2 n^2} \left( \cos \left( \frac{2\pi n}{3} \right) - 1 \right) + \frac{4}{\pi n} \sin \left( \frac{2\pi n}{3} \right) \right] \cos \left( \frac{2\pi n}{3} \right) \right\}
-$$
-\n
-$$
-+ \left[ \frac{6}{\pi^2 n^2} \sin \left( \frac{2\pi n}{3} \right) - \frac{4}{n \pi} \cos \left( \frac{2\pi n}{3} \right) \right] \sin \left( \frac{2\pi n}{3} \right)
-$$
-
-**17.27** (a) odd, (b) −0.315, (c) 2.681
-
-**17.29** 
-$$
-2\sum_{k=1}^{\infty} \left[\frac{2}{n^2 \pi} \cos(nt) - \frac{1}{n} \sin(nt)\right], n = 2k - 1
-$$
-
-**17.31** 
-$$
-\omega'_{o} = \frac{2\pi}{T'} = \frac{2\pi}{T/\alpha} = \alpha \omega_{o}
-$$
-$$
-a'_{n} = \frac{2}{T'} \int_{0}^{T'} f(\alpha t) \cos n\omega'_{o} t \, dt
-$$
-Let  $\alpha t = \lambda$ ,  $dt = d\lambda/\alpha$ , and  $\alpha T' = T$ . Then
-$$
-a'_{n} = \frac{2\alpha}{T} \int_{0}^{T} f(\lambda) \cos n\omega_{o} \lambda \, d\lambda/\alpha = a_{n}
-$$
-Similarly,  $b'_{n} = b_{n}$ 
-
-**17.33** 
-$$
-v_o(t) = \sum_{n=1}^{\infty} A_n \sin(n \pi t - \theta_n) \text{ V},
-$$
-  
-\n
-$$
-A_n = \frac{10(4 - 2n^2 \pi^2)}{\sqrt{(20 - 10n^2 \pi^2)^2 - 64n^2 \pi^2}},
-$$
-\n
-$$
-\theta_n = 90^\circ - \tan^{-1} \left(\frac{8n \pi}{20 - 10n^2 \pi^2}\right)
-$$
-
-17.35 
-$$
-\frac{3}{8} + \sum_{n=1}^{\infty} A_n \cos\left(\frac{2\pi n}{3} + \theta_n\right)
-$$
-, where  
-$$
-A_n = \frac{\frac{6}{n\pi} \sin\frac{2n\pi}{3}}{\sqrt{9\pi^2 n^2 + (2\pi^2 n^2/3 - 3)^2}},
-$$
-$$
-\theta_n = \frac{\pi}{2} - \tan^{-1}\left(\frac{2n\pi}{9} - \frac{1}{n\pi}\right)
-$$
-
-17.37 
-$$
-\sum_{n=1}^{\infty} \frac{2(1 - \cos \pi n)}{\sqrt{1 + n^2 \pi^2}} \cos (n \pi t - \tan^{-1} n \pi)
-$$
-
-$$
-17.39 \frac{1}{10} + \frac{400}{\pi} \sum_{k=1}^{\infty} I_n \sin(n\pi t - \theta_n), n = 2k - 1,
-$$
-$$
-\theta_n = 90^\circ + \tan^{-1} \frac{2n^2 \pi^2 - 1,200}{802n\pi},
-$$
-$$
-I_n = \frac{1}{n\sqrt{(804n\pi)^2 + (2n^2 \pi^2 - 1,200)}}
-$$
-
-17.41 
-$$
-\frac{200}{\pi} + \sum_{n=1}^{\infty} A_n \cos(2nt + \theta_n)
-$$
- where  
-$$
-A_n = \frac{2,000}{\pi (4n^2 - 1)\sqrt{16n^2 - 40n + 29}}
-$$
- and  
-$$
-\theta_n = 90^\circ - \tan^{-1}(2n - 2.5)
-$$
-
-**17.43** (a) 33.91 V, (b) 6.782 A, (c) 203.1 W
-
-**17.45** 4.263 A, 181.7 W
-
-**17.47** 10%
-
-17.49
-
-\n(a) 
-$$
-3.162
-$$
-,
-
-\n(b)  $3.065$ ,
-
-\n(c)  $3.068\%$ 
-
-**17.51** This is a design problem with multiple answers.
-
-17.53 
-$$
-\sum_{n=-\infty}^{\infty} \frac{0.6321e^{j2n\pi t}}{1+j2n\pi}
-$$
-
-17.55 
-$$
-\sum_{n=-\infty}^{\infty} \frac{1+e^{-jn\pi}}{2\pi(1-n^2)} e^{jnt}
-$$
-
-$$
-17.57 -3 + \sum_{n=\infty, n\neq 0}^{\infty} \frac{3}{n^3 - 2} e^{j50nt}
-$$
-
-17.59 
-$$
--\sum_{\substack{n=-\infty\\n\neq 0}}^{\infty} \frac{j4e^{-j(2n+1)\pi t}}{(2n+1)\pi}
-$$
-
-**17.61** (a) 6 + 2.571 cos *t* − 3.83 sin *t* + 1.638 cos 2*t* − 1.147 sin 2*t* + 0.906 cos 3*t* − 0.423 sin 3*t* + 0.47 cos 4*t* − 0.171 sin 4*t*, (b) 6.828
-
-**17.63** See Fig. D.37.
-
-**Figure D.37**
-
-**17.67** DC COMPONENT = 2.000396E + 00
-
-| HARMONIC<br>NO | FREQUENCY<br>(HZ) | FOURIER<br>COMPONENT | NORMALIZED<br>COMPONENT | PHASE<br>(DEG) | NORMALIZED<br>PHASE (DEG) |
-|----------------|-------------------|----------------------|-------------------------|----------------|---------------------------|
-| 1              | 1.667E-01         | 2.432E+00            | 1.000E+00               | -8.996E+01     | 0.000E+00                 |
-| 2              | 3.334E-01         | 6.576E-04            | 2.705E-04               | -8.932E+01     | 6.467E-01                 |
-| 3              | 5.001E-01         | 5.403E-01            | 2.222E-01               | 9.011E+01      | 1.801E+02                 |
-| 4              | 6.668E+01         | 3.343E-04            | 1.375E-04               | 9.134E+01      | 1.813E+02                 |
-| 5              | 8.335E-01         | 9.716E-02            | 3.996E-02               | -8.982E+01     | 1.433E-01                 |
-| 6              | 1.000E+00         | 7.481E-06            | 3.076E-06               | -9.000E+01     | -3.581E-02                |
-| 7              | 1.167E+00         | 4.968E-02            | 2.043E-01               | -8.975E+01     | 2.173E-01                 |
-| 8              | 1.334E+00         | 1.613E-04            | 6.634E-05               | -8.722E+01     | 2.748E+00                 |
-| 9              | 1.500E+00         | 6.002E-02            | 2.468E-02               | -9.032E+01     | 1.803E+02                 |
-
-| 17.69 HARMONIC<br>NO | FREQUENCY<br>(HZ) | FOURIER<br>COMPONENT | NORMALIZED<br>COMPONENT | PHASE<br>(DEG) | NORMALIZED<br>PHASE (DEG) |
-|----------------------|-------------------|----------------------|-------------------------|----------------|---------------------------|
-| 1                    | 5.000E-01         | 4.056E-01            | 1.000E+00               | -9.090E+01     | 0.000E+00                 |
-| 2                    | 1.000E+00         | 2.977E-04            | 7.341E-04               | -8.707E+01     | 3.833E+00                 |
-| 3                    | 1.500E+00         | 4.531E-02            | 1.117E-01               | -9.266E+01     | -1.761E+00                |
-| 4                    | 2.000E+00         | 2.969E-04            | 7.320E-04               | -8.414E+01     | 6.757E+00                 |
-| 5                    | 2.500E+00         | 1.648E-02            | 4.064E-02               | -9.432E+01     | -3.417E+00                |
-| 6                    | 3.000E+00         | 2.955E-04            | 7.285E-04               | -8.124E+01     | 9.659E+00                 |
-| 7                    | 3.500E+00         | 8.535E-03            | 2.104E-02               | -9.581E+01     | -4.911E+00                |
-| 8                    | 4.000E+00         | 2.935E-04            | 7.238E-04               | -7.836E+01     | 1.254E+01                 |
-| 9                    | 4.500E+00         | 5.258E-03            | 1.296E-02               | -9.710E+01     | -6.197E+00                |
-
-TOTAL HARMONIC DISTORTION = 1.214285+01 PERCENT
-
-**17.71** See Fig. D.39.
-
-**17.73** 300 mW
-
-**17.75** 24.59 mF
-
-**17.77** (a) *π*, (b) −2 V, (c) 11.02 V
-
-**17.79** See below for the program in *MATLAB* and the results. % for problem 17.79 a = 10; c = 4.\**a*∕pi for n = 1:10 b(n) = c/(2\*n-1); end diary n, b
-
-| n  | bn      |  |  |  |
-|----|---------|--|--|--|
-| 1  | 12.7307 |  |  |  |
-| 2  | 4.2430  |  |  |  |
-| 3  | 2.5461  |  |  |  |
-| 4  | 1.8187  |  |  |  |
-| 5  | 1.414   |  |  |  |
-| 6  | 1.1573  |  |  |  |
-| 7  | 0.9793  |  |  |  |
-| 8  | 0.8487  |  |  |  |
-| 9  | 0.7488  |  |  |  |
-| 10 | 0.6700  |  |  |  |
-
-diary off
-
-**17.81** (a) 
-$$
-\frac{A^2}{2}
-$$
-, (b)  $|c_1| = 2A/(3\pi)$ ,  $|c_2| = 2A/(15\pi)$ ,  
- $|c_3| = 2A/(35\pi)$ ,  $|c_4| = 2A/(63\pi)$  (c) 81.1%  
-(d) 0.72%
-
-# Chapter 18
-
-Chapter 18  
-\n**18.1** 
-$$
-\frac{14(\cos 2\omega - \cos \omega)}{j\omega}
-$$
-\n**18.3** 
-$$
-\frac{j8}{\omega^2} (2\omega \cos 2\omega - \sin 2\omega)
-$$
-
-**18.5** <sup>6</sup>*<sup>j</sup> \_\_ ω*<sup>−</sup> 6*j \_\_\_ ω*2 sin *ω*
-
-**18.7** (a) 
-$$
-\frac{2 - e^{-j\omega} - e^{-j2\omega}}{j\omega}
-$$
-, (b)  $\frac{5e^{-j2\omega}}{\omega^2} (1 + j\omega^2) - \frac{5}{\omega^2}$ 
-
-**18.9** (a) 
-$$
-\frac{10}{\omega} \sin 2\omega + \frac{20}{\omega} \sin \omega
-$$
-,  
-\n(b)  $\frac{10}{\omega^2} - \frac{10e^{-j\omega}}{\omega^2} (1 + j\omega)$   
-\n**18.11**  $\frac{11\pi}{\omega^2 - \pi^2} (e^{-j\omega^2} - 1)$ 
-
-**18.13** (a) 
-$$
-\pi e^{-j\pi/3} \delta(\omega - a) + \pi e^{j\pi/3} \delta(\omega + a)
-$$
-,  
-\n(b)  $\frac{e^{j\omega}}{\omega^2 - 1}$ , (c)  $\pi[\delta(\omega + b) + \delta(\omega - b)]$   
-\n $+ \frac{j\pi A}{2} [\delta(\omega + a + b) - \delta(\omega - a + b) + \delta(\omega + a - b) - \delta(\omega - a - b)],$   
-\n(d)  $\frac{1}{\omega^2} - \frac{e^{-j4\omega}}{j\omega} - \frac{e^{-j4\omega}}{\omega^2} (j4\omega + 1)$ 
-
-**18.15** (a) 
-$$
-2j \sin 3\omega
-$$
-, (b)  $\frac{2e^{-j\omega}}{j\omega}$ , (c)  $\frac{1}{3} - \frac{j\omega}{2}$ 
-
-**18.17** (a) 
-$$
-\pi[\delta(\omega + 2) + \delta(\omega - 2)] - \frac{2j\omega}{\omega^2 - 4}
-$$
-,  
-(b)  $\frac{j\pi}{4}[\delta(\omega + 10) - \delta(\omega - 10)] - \frac{5}{\omega^2 - 100}$ 
-
-$$
-18.19 \ \frac{2j\omega}{\omega^2 - 4\pi^2} (e^{-j\omega} - 1)
-$$
-
-**18.21** Proof
-
-**18.21 Proof**  
-\n**18.23** (a) 
-$$
-\frac{30}{(6 - j\omega)(15 - j\omega)}
-$$
-,  
-\n(b)  $\frac{20e^{-j\omega/2}}{(4 + j\omega)(10 + j\omega)}$ ,  
-\n(c)  $\frac{5}{[2 + j(\omega + 2)][5 + j(\omega + 2)]}$  +  
-\n $\frac{5}{[2 + j(\omega - 2)][5 + j(\omega - 2)]}$   
-\n(d)  $\frac{j\omega 10}{(2 + j\omega)(5 + j\omega)}$ ,  
-\n(e)  $\frac{10}{j\omega(2 + j\omega)(5 + j\omega)} + \pi\delta(\omega)$ 
-
-**18.25** (a) 5*e*<sup>2</sup>*<sup>t</sup> u*(*t*), (b) 6*e*<sup>−</sup>2*<sup>t</sup>* , (c) (−10*et u*(*t*) + 10*e*<sup>2</sup>*<sup>t</sup>* )*u*(*t*)
-
-**18.27** (a) 
-$$
-5 \operatorname{sgn}(t) - 10e^{-10t} u(t)
-$$
-,  
-\n(b)  $4e^{2t}u(-t) - 6e^{-3t}u(t)$ ,  
-\n(c)  $2e^{-20t} \sin(30t) u(t)$ , (d)  $\frac{1}{4} \pi$ 
-
-**18.29** (a) 
-$$
-\frac{1}{2\pi}(1 + 8 \cos 3t)
-$$
-, (b)  $\frac{4 \sin 2t}{\pi t}$ ,  
-(c)  $3\delta(t + 2) + 3\delta(t - 2)$ 
-
-**18.31** (a) 
-$$
-x(t) = e^{-at}u(t)
-$$
-,  
-\n(b)  $x(t) = u(t+1) - u(t-1)$ ,  
-\n(c)  $x(t) = \frac{1}{2}\delta(t) - \frac{a}{2}e^{-at}u(t)$ 
-
-**18.33** (a) 
-$$
-\frac{2j \sin t}{t^2 - \pi^2}
-$$
-, (b)  $u(t-1) - u(t-2)$ 
-
-**18.35** (a) 
-$$
-\frac{e^{-j\omega/3}}{6+j\omega}
-$$
-, (b)  $\frac{1}{2} \left[ \frac{1}{2+j(\omega+5)} + \frac{1}{2+j(\omega-5)} \right]$ ,  
-(c)  $\frac{j\omega}{2+j\omega}$ , (d)  $\frac{1}{(2+j\omega)^2}$ , (e)  $\frac{1}{(2+j\omega)^2}$ 
-
-$$
-18.37 \ \frac{j\omega}{4+j3\omega}
-$$
-
-**18.39** 5 × 10<sup>3</sup> \_\_\_\_\_\_\_\_ 106 + *jω* ( \_\_\_1 *jω* + \_\_\_1 *<sup>ω</sup>*<sup>2</sup> <sup>−</sup> \_\_\_1 *<sup>ω</sup>*<sup>2</sup> *<sup>e</sup>*<sup>−</sup>*j<sup>ω</sup>* )
-
-**18.41** <sup>2</sup>*jω*(4.5 + *j*2*ω*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (2 + *jω*)(4 − 2*ω*<sup>2</sup> + *jω*) 
-
-**18.43** 1000(*e*<sup>−</sup>1*<sup>t</sup>* − *e*<sup>−</sup>1.25*<sup>t</sup>* )*u*(*t*) V
-
-**18.45** 5(*e*<sup>−</sup>*<sup>t</sup>* − *e*<sup>−</sup>*2*<sup>t</sup> )*u*(*t*) A
-
-- **18.47** 16(*e*<sup>−</sup>*<sup>t</sup>* − *e*<sup>−</sup>*2*<sup>t</sup> )*u*(*t*) V
-- **18.49** 0.542 cos (*t* + 13.64°) V
-- **18.51** 16.667 J
-- **18.53** *π*
-- **18.55** 682.5 J
-- **18.57** 2 J, 87.43%
-- **18.59** (16*e*<sup>−</sup>*<sup>t</sup>* − 20*e*<sup>−</sup>*2*<sup>t</sup> + 4*e*<sup>−</sup>4<sup>t</sup> )*u*(*t*) V
-- **18.61** 2*X*(*ω*) + 0.5*X*(*ω* + *ω*0) + 0.5*X*(*ω* − *ω*0)
-- **18.63** 106 stations
-- **18.65** 6.8 kHz
-- **18.67** 200 Hz, 5 ms
-
-**18.69** 35.24%
-
-Chapter 19  
-\n19.1 
-$$
-\begin{bmatrix} 30 & 10 \\ 10 & 30 \end{bmatrix}
-$$
- Ω  
-\n19.3  $\begin{bmatrix} 10 & -j10 \\ -j10 & -j10 \end{bmatrix}$  Ω  
-\n19.5  $\begin{bmatrix} 10(s+2) & 10 \\ 10 & 10 \end{bmatrix}$   
-\n19.7  $\begin{bmatrix} 20(s+0.5) & -30 \\ -10 & -20 \end{bmatrix}$  Ω  
-\n19.9  $\begin{bmatrix} 2.5 & 1.25 \\ 1.25 & 3.125 \end{bmatrix}$  Ω
-
-**19.11** See Fig. D.40.
-
-# **Figure D.40**
-
-For Prob. 19.11.
-
-**19.13** 329.9 W
-
-**19.15** 24 Ω, 1.536 kW
-
-- **19.17** [ 9.6 −0.8 −0.8 8.4 ] Ω and [ 0.105 0.01 0.01 0.12] <sup>S</sup>
-- **19.19** This is a design problem with multiple answers.
-- **19.21** See Fig. D.41.
-
-**Figure D.41** For Prob. 19.21.
-
-**19.23** 
-$$
-\begin{bmatrix} s+2 & -(s+1) \ -(s+1) & \frac{s^2+s+1}{s} \end{bmatrix}, \frac{0.8(s+1)}{s^2+1.8s+1.2}
-$$
-
-**19.25** See Fig. D.42.
-
-**Figure D.42** For Prob. 19.25.
-
-- **19.27** [0.25 5 0.025 0.6 ]<sup>S</sup>
-- **19.29** (a) 44 V, 16 V, (b) same
-- **19.31** [3.8 <sup>Ω</sup> −3.6 0.4 0.2 S] **19.33** [(3.077 + *j*1.2821) <sup>Ω</sup> −0.3846 + j0.2564 0.3846 − *j*0.2564 (76.9 + 282.1) mS]
-- **19.35** [ <sup>2</sup><sup>Ω</sup> −0.5 0.5 0 ]
-
-**19.37** 3.571 V
-
-**19.39** 
-$$
-g_{11} = \frac{1}{R_1 + R_2}, g_{12} = -\frac{R_2}{R_1 + R_2}
-$$
-  
- $g_{21} = \frac{R_2}{R_1 + R_2}, g_{22} = R_3 + \frac{R_1 R_2}{R_1 + R_2}$ 
-
-**19.41** Proof
-
-**19.43** (a) 
-$$
-\begin{bmatrix} 1 & \mathbf{Z} \\ 0 & 1 \end{bmatrix}
-$$
-, (b)  $\begin{bmatrix} 1 & 0 \\ \mathbf{Y} & 1 \end{bmatrix}$   
-\n**19.45**  $\begin{bmatrix} 1 & (20+j20) \Omega \\ j100 \mu s & 1 \end{bmatrix}$   
-\n**19.47**  $\begin{bmatrix} 0.3235 & 1.176 \Omega \\ 0.02941 S & 0.4706 \end{bmatrix}$ 
-
-$$
-19.49 \begin{bmatrix} \frac{2s+1}{s} & \frac{1}{s} \Omega \\ \frac{(s+1)(3s+1)}{s} S & 2 + \frac{1}{s} \end{bmatrix}
-$$
-  
-$$
-19.51 \begin{bmatrix} 2 & 2+j5 \\ j & -2+j \end{bmatrix}
-$$
-
-**19.53** 
-$$
-z_{11} = \frac{A}{C}
-$$
-,  $z_{12} = \frac{AD - BC}{C}$ ,  $z_{21} = \frac{1}{C}$ ,  $z_{22} = \frac{D}{C}$ 
-
-**19.55** Proof
-
-**19.57** 
-$$
-\begin{bmatrix} 3 & 1 \\ 1 & 7 \end{bmatrix} \Omega
-$$
-,  $\begin{bmatrix} \frac{7}{20} & \frac{-1}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{bmatrix} S$ ,  $\begin{bmatrix} \frac{20}{7} \Omega & \frac{1}{7} \\ \frac{-1}{7} & \frac{1}{7} S \end{bmatrix}$ ,  $\begin{bmatrix} \frac{1}{3} S & \frac{-1}{3} \\ \frac{1}{3} & \frac{20}{3} \Omega \end{bmatrix}$ ,  $\begin{bmatrix} 7 & 20 \Omega \\ 1 S & 3 \end{bmatrix}$ 
-
-**19.59** 
-$$
-\begin{bmatrix} 16.667 & 6.667 \\ 3.333 & 3.333 \end{bmatrix} \Omega, \begin{bmatrix} 0.1 & -0.2 \\ -0.1 & 0.5 \end{bmatrix} S,
-$$
-$$
-\begin{bmatrix} 10 \Omega & 2 \\ -1 & 0.3 \Omega \end{bmatrix}, \begin{bmatrix} 5 \Omega & 10 \Omega \\ 0.3 \Omega & 1 \end{bmatrix}
-$$
-
-**19.61** (a) 
-$$
-\begin{bmatrix} 5 & 4 \\ 3 & 3 \\ 4 & 5 \\ 3 & 3 \end{bmatrix}
-$$
-  $\Omega$ , (b)  $\begin{bmatrix} 5 & 2 & 4 \\ 3 & 5 \\ -4 & 3 & 5 \\ 5 & 5 & 5 \end{bmatrix}$ , (c)  $\begin{bmatrix} 5 & 3 & 0 \\ 4 & 4 & 0 \\ 3 & 5 & 5 \\ 4 & 4 & 4 \end{bmatrix}$ 
-
-**19.63** 
-$$
-\begin{bmatrix} 0.8 & 2.4 \\ 2.4 & 7.2 \end{bmatrix} \Omega
-$$
-
-$$
-19.65\begin{bmatrix} \frac{0.5}{3} & -\frac{1}{-0.5} \\ -\frac{0.5}{3} & \frac{2}{5/6} \end{bmatrix} S
-$$
-
-**19.67** [ <sup>4</sup> 0.1576 S 63.29 Ω 4.994 ]
-
-19.69 
-$$
-\begin{bmatrix} \frac{s+1}{s+2} & \frac{-(3s+2)}{2(s+2)} \\ \frac{-(3s+2)}{2(s+2)} & \frac{5s^2+4s+4}{2s(s+2)} \end{bmatrix}
-$$
-
-**19.71** 
-$$
-\begin{bmatrix} 2 & -3.334 \ 3.334 & 20.22 \end{bmatrix} \Omega
-$$
-  
-\n**19.73** 
-$$
-\begin{bmatrix} 14.628 & 3.141 \ 5.432 & 19.625 \end{bmatrix} \Omega
-$$
-  
-\n**19.75** (a) 
-$$
-\begin{bmatrix} 0.3015 & -0.1765 \ 0.0588 & 19.625 \end{bmatrix} S
-$$
-, (b) -0.0051  
-\n**19.77** 
-$$
-\begin{bmatrix} 0.9488/-161.6^{\circ} \\ 0.3163/-161.6^{\circ} \end{bmatrix} \begin{bmatrix} 0.3163/18.42^{\circ} \\ 0.9488/-161.6^{\circ} \end{bmatrix}
-$$
-  
-\n**19.79** 
-$$
-\begin{bmatrix} 4.669/-136.7^{\circ} \\ 2.53/-108.4^{\circ} \end{bmatrix} \begin{bmatrix} 2.53/-108.4^{\circ} \\ 1.789/-153.4^{\circ} \end{bmatrix} \Omega
-$$
-  
-\n**19.81** 
-$$
-\begin{bmatrix} 1.5 & -0.5 \\ 3.5 & 1.5 \end{bmatrix} S
-$$
-  
-\n**19.83** 
-$$
-\begin{bmatrix} 0.3235 & 1.1765 \\ 0.02941 S & 0.4706 \end{bmatrix}
-$$
-  
-\n**19.85** 
-$$
-\begin{bmatrix} 1.581/71.59^{\circ} \\ 1.587 \end{bmatrix} \begin{bmatrix} -\frac{1}{3} \\ 5.661 \times 10^{-4} \end{bmatrix}
-$$
-
-5.661 × 10<sup>−</sup><sup>4</sup>]
-
-*j* S
-
-$$
-19.87 \begin{bmatrix} -j1.765 & -j1.765 \Omega \\ j888.2 \text{ S} & j888.2 \end{bmatrix}
-$$
-
-**19.89** −1,613, 64.15 dB
-
-**19.91** (a) −25.64 for the transistor and −9.615 for the circuit. (b) 74.07, (c) 1.2 kΩ, (d) 51.28 kΩ
-
-**19.93** −17.74, 144.5, 31.17 Ω, −6.148 MΩ
-
-**19.95** See Fig. D.43.
-
-**Figure D.43**
-
-For Prob. 19.95.
-
-**19.97** 250 mF, 333.3 mH, 500 mF
-
-**19.99** Proof

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/233_Selected Bibliography.md

@@ -1,65 +0,0 @@
-# <span id="page-979-0"></span>Selected Bibliography
-
-- Aidala, J. B., and L. Katz. *Transients in Electric Circuits.*  Englewood Cliffs, NJ: Prentice Hall, 1980.
-- Angerbaur, G. J. *Principles of DC and AC Circuits.* 3rd ed. Albany, NY: Delman Publishers, 1989.
-- Attia, J. O. *Electronics and Circuit Analysis Using MATLAB.*  Boca Raton, FL: CRC Press, 1999.
-- Balabanian, N. *Electric Circuits.* New York: McGraw-Hill, 1994.
-- Bartkowiak, R. A. *Electric Circuit Analysis.* New York: Harper & Row, 1985.
-- Blackwell, W. A., and L. L. Grigsby. *Introductory Network Theory.* Boston, MA: PWS Engineering, 1985.
-- Bobrow, L. S. *Elementary Linear Circuit Analysis.* 2nd ed. New York: Holt, Rinehart & Winston, 1987.
-- Boctor, S. A. *Electric Circuit Analysis.* 2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1992.
-- Boylestad, R. L. *Introduction to Circuit Analysis.* 10th ed. Columbus, OH: Merrill, 2000.
-- Budak, A. *Circuit Theory Fundamentals and Applications.*  2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1987.
-- Carlson, B. A. *Circuit: Engineering Concepts and Analysis of Linear Electric Circuits.* Boston, MA: PWS Publishing, 1999.
-- Chattergy, R. *Spicey Circuits: Elements of Computer-Aided Circuit Analysis.* Boca Raton, FL: CRC Press, 1992.
-- Chen, W. K. *The Circuit and Filters Handbook.* Boca Raton, FL: CRC Press, 1995.
-- Choudhury, D. R. *Networks and Systems.* New York: John Wiley & Sons, 1988.
-- Ciletti, M. D. *Introduction to Circuit Analysis and Design.*  New York: Oxford University Press, 1995.
-- Cogdeil, J. R. *Foundations of Electric Circuits.* Upper Saddle River, NJ: Prentice Hall, 1998.
-- Cunningham, D. R., and J. A. Stuller. *Circuit Analysis.* 2nd ed. New York: John Wiley & Sons, 1999.
-- Davis, A., (ed.). *Circuit Analysis Exam File.* San Jose, CA: Engineering Press, 1986.
-- Davis, A. M. *Linear Electric Circuit Analysis.* Washington, DC: Thomson Publishing, 1998.
-- DeCarlo, R. A., and P. M. Lin. *Linear Circuit Analysis.* 2nd ed. New York: Oxford University Press, 2001.
-- Del Toro, V. *Engineering Circuits.* Englewood Cliffs, NJ: Prentice Hall, 1987.
-- Dorf, R. C., and J. A. Svoboda. *Introduction to Electric Circuits.* 4th ed. New York: John Wiley & Sons, 1999.
-- Edminister, J. *Schaum's Outline of Electric Circuits.* 3rd ed. New York: McGraw-Hill, 1996.
-
-- Floyd, T. L. *Principles of Electric Circuits.* 7th ed. Upper Saddle River, NJ: Prentice Hall, 2002.
-- Franco, S. *Electric Circuits Fundamentals.* Fort Worth, FL: Saunders College Publishing, 1995.
-- Goody, R. W. *Microsim PSpice for Windows.* Vol. 1. 2nd ed. Upper Saddle River, NJ: Prentice Hall, 1998.
-- Harrison, C. A. *Transform Methods in Circuit Analysis.*  Philadelphia, PA: Saunders, 1990.
-- Harter, J. J., and P. Y Lin. *Essentials of Electric Circuits.*  2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1986.
-- Hayt, W H., and J. E. Kemmerly. *Engineering Circuit Analysis.* 6th ed. New York: McGraw-Hill, 2001.
-- Hazen, M. E. *Fundamentals of DC and AC Circuits.*  Philadel phia, PA: Saunders, 1990.
-- Hostetter, G. H. *Engineering Network Analysis.* New York: Harper & Row, 1984.
-- Huelsman, L. P. *Basic Circuit Theory.* 3rd ed. Englewood Cliffs, NJ: Prentice Hall, 1991.
-- Irwin, J. D. *Basic Engineering Circuit Analysis.* 7th ed. New York: John Wiley & Sons, 2001.
-- Jackson, H. W., and P. A. White. *Introduction to Electric Circuits.* 7th ed. Englewood Cliffs, NJ: Prentice Hall, 1997.
-- Johnson, D. E. et al. *Electric Circuit Analysis.* 3rd ed. Upper Saddle River, NJ: Prentice Hall, 1997.
-- Karni, S. *Applied Circuit Analysis.* New York: John Wiley & Sons, 1988.
-- Kraus, A. D. *Circuit Analysis.* St. Paul, MN: West Publishing, 1991.
-- Madhu, S. *Linear Circuit Analysis.* 2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1988.
-- Mayergoyz, I. D., and W. Lawson. *Basic Electric Circuits Theory.* San Diego, CA: Academic Press, 1997.
-- Mottershead, A. *Introduction to Electricity and Electronics: Conventional and Current Version.* 3rd ed. Englewood Cliffs, NJ: Prentice Hall, 1990.
-- Nasar, S. A. *3000 Solved Problems in Electric Circuits. (Schaum's Outline)* New York: McGraw-Hill, 1988.
-- Neudorfer, P. O., and M. Hassul. *Introduction to Circuit Analysis.* Englewood Cliffs, NJ: Prentice Hall, 1990.
-- Nilsson, J. W., and S. A. Riedel. *Electric Circuits.* 5th ed. Reading, MA: Addison-Wesley, 1996.
-- O'Malley, J. R. *Basic Circuit Analysis. (Schaum's Outline)*  New York: McGraw-Hill, 2nd ed., 1992.
-- Parrett, R. *DC-AC Circuits: Concepts and Applications.*  Englewood Cliffs, NJ: Prentice Hall, 1991.
-- Paul, C. R. *Analysis of Linear Circuits.* New York: McGraw-Hill, 1989.
-
-Poularikas, A. D., (ed.). *The Transforms and Applications Handbook.* Boca Raton, FL: CRC Press, 2nd ed., 1999.
-
-- Ridsdale, R. E. *Electric Circuits.* 2nd ed. New York: McGraw-Hill, 1984.
-- Sander, K. F. *Electric Circuit Analysis: Principles and Applications.* Reading, MA: Addison-Wesley, 1992.
-- Scott, D. *Introduction to Circuit Analysis: A Systems Approach.* New York: McGraw-Hill, 1987.
-- Smith, K. C., and R. E. Alley. *Electrical Circuits: An Introduction.* New York: Cambridge University Press, 1992.
-- Stanley, W. D. *Transform Circuit Analysis for Engineering and Technology.* 3rd ed. Upper Saddle River, NJ: Prentice Hall, 1997.
-- Strum, R. D., and J. R. Ward. *Electric Circuits and Networks.*  2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1985.
-
-- Su, K. L. *Fundamentals of Circuit Analysis.* Prospect Heights, IL: Waveland Press, 1993.
-- Thomas, R. E., and A. J. Rosa. *The Analysis and Design of Linear Circuits.* 3rd ed. New York: John Wiley & Sons, 2000.
-- Tocci, R. J. *Introduction to Electric Circuit Analysis.* 2nd ed. Englewood Cliffs, NJ: Prentice Hall, 1990.
-- Tuinenga, P. W. *SPICE: A Guide to Circuit Simulation.*  Englewood Cliffs, NJ: Prentice Hall, 1992.
-- Whitehouse, J. E. *Principles of Network Analysis.* Chichester, U.K.: Ellis Horwood, 1991.
-- Yorke, R. *Electric Circuit Theory.* 2nd ed. Oxford, U.K.: Pergamon Press, 1986.

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/234_Index.md

@@ -1,149 +0,0 @@
-# <span id="page-981-0"></span>Index
-
-Note: Page numbers followed by f or t represent figures or tables respectively.
-
-# A
-
-ABCD parameters, 866 Abc sequence, 505, 505f ac (alternating current), 7–8, 7f, 368, 369 ac bridge circuit, 396–400 Acb sequence, 505, 505f ac circuits, 369 AC power analysis, 455–486 apparent power, 468–471 average power, 457–462 complex power, 471–475 conservation of ac power, 475–478 effective value, 465–466 electricity consumption cost, 484–486 instantaneous power, 456–457 maximum average power transfer, 462–465 power factor, 469–471 power factor correction, 479–481 power measurement, 481–484 rms value, 466–468 Active element, 14 Active filters, 635, 640–646 AC voltage, 10 Additivity property, 127 Admittance, 385–387 Admittance parameters, 857–860 Air-core transformers, 566 Alexander, Charles K., 125, 311 Alternating current (ac), 7–8, 7f, 368, 369 American Institute of Electrical Engineers (AIEE), 13 Ampere, Andre-Marie, 7 Amplitude modulation (AM), 820–821, 838–840 Amplitude-phase form, 761 Amplitude spectrum, 762, 814 Analog computers, 235–238 Apparent power, 468–471 Audio transformers, 566f Automobile ignition circuit, 296–297, 296f Automobile ignition system, 351–353 Autotransformers, 579–582, 579f Average power, 780–783 Axial lead inductor, 224f
-
-# B
-
-Bacon, Francis, 3 Bailey, P. J., 251 Balanced, 156 Balanced delta-delta connection, 512–514 Balanced delta-wye circuit, 391 Balanced delta-wye connection, 514–517 Balanced load, 506 Balanced networks, 53 Balanced three-phase voltages, 503–506 Balanced wye-delta connection, 510–512 Balanced wye-wye connection, 507–510 Band-pass filters, 636, 636f, 637–638, 641–643, 642f Band-reject filter, 643–644, 643f Band-stop filters, 636, 636f, 638 Bandwidth, 629 Bandwidth of rejection, 638 Bardeen, John, 106 Barkhausen criteria, 437–438 Bell, Alexander Graham, 616 Bell Laboratories, 106, 143 Binary weighted ladder, 194 Bipolar junction transistor (BJT), 105–106 Bode plots, 615, 617–627 Branch, 35, 35f Brattain, Walter, 106 Braun, Karl Ferdinand, 17 Break frequency, 619 Brush Electric Company, 13 Bunsen, Robert, 38 Buxton, W. J. Wilmont, 79 Byron, Lord, 173
-
-# C
-
-Capacitance, 215 Capacitance multiplier, 435–437, 435f Capacitors, 214–223. *See also* Inductors analog computer, 235–238 characteristics, 230 current-voltage relationship, 216–217 defined, 214 differentiator, 233–234 integrator, 232–233
-
-# **I-2** Index
-
-Capacitors (*continued*) properties, 232 series and parallel, 220–223 types, 216 Careers communications systems, 811 in computer engineering, 251 in control systems, 611 in education, 851–852 in electromagnetics, 553 in electronic instrumentation, 173 in electronics, 79 engineering, 311 in power systems, 455 in software engineering, 411 Cascaded networks, 875 Cascaded op amp circuits, 189–192 Cathode-ray tube (CRT), 16, 17f Cesium, 38 Characteristic equation, 318 Charge. *See* Electric charge Chassis ground, 81, 81f Choke, 224 Circuit analysis, 720–723 Circuit applications, 776–780, 831–833 Circuit element models, 715–720 Circuit theorems, 125–158 linearity, 126–127 maximum power transfer, 148–150 Norton's theorem, 143–148 with *PSpice,* 150–153 resistance measurement, 156–158 source modeling, 153–155 source transformation, 133–136 superposition principle, 129–133 Thevenin's theorem, 137–143, 147–148 Closed-loop gain, 176 Coefficient of coupling, 563–564 Coil, 224 Common-base current gain, 107 Common-emitter current gain, 107 Communication skills, 125 Communications systems, careers in, 811 Complete response, 273–274 Completing the square, 690 Complex amplitude spectrum, 784 Complex conjugate, A–12 Complex numbers, 374, A–9 to A–15 Complex phase spectrum, 784 Complex poles, 690–691 Complex power, 471–475 Computer engineering careers, 251 Conductance, 33, 386 Conductance matrix, 99 Conductively coupled, 554 Conservation of ac power, 475–478 Consumption cost, electricity, 484–486 Control systems, career in, 611
-
-Convolution, 823–826 Convolution integral, 695–703 Copper wound dry power transformer, 566f Corner frequency, 619 Cramer's rule, 80, A to A–4 Critically damped case source-free parallel *RLC* circuits, 325 source-free series *RLC* circuit, 319–320 step response of parallel *RLC* circuits, 335 step response series *RLC* circuits, 330 Crossover network, 659–661, 660f Current divider, 46 Current-division principle, 390 Cutoff frequency, 636–637 Cyclic frequency, 370
-
-# D
-
-DAC (digital-to-analog converter), 194–195, 194f, 195t Damped frequency, 321 Damped natural frequency, 321, 352 d'Arsonval meter movement, 60, 60f Darwin, Francis, 213 Datum node, 81 dc (direct current), 7–8, 7f, 368 DC meters, design of, 59–62 DC transistor circuits (application), 105–107 DC voltage, 10 Decibel scale, 615–617 Definite integrals, A–19 to A–20 Delay circuits, 291–293 Delta function, 265 Delta to wye conversion, 52–53, 390 Demodulation, 838 Derivatives, A–17 to A–18 Deschemes, Marc-Antoine Parseval, 781 Difference op amp, 185–188, 185f Differential equations, 674 Differentiator, 233–234 Digital meter, 61–62 Digital-to-analog converter (DAC), 194–195, 194f, 195t Dinger, J. E., 411 Direct current (dc), 7–8, 7f, 368 Dirichlet, P. G. L., 760 Dirichlet conditions, 760 Discrete Fourier Transform (DFT), 789–790 Dot convention, 557–558 Driving-point impedances, 854 Duality, 348–350, 823
-
-# E
-
-Earth ground, 81 Edison, Thomas Alva, 13, 57, 368, 502, 503 Education, careers in, 851–852 Effective value, 465–466 Electrical engineer, 79
-
-Electrical isolation, 579, 590 Electrical lighting systems (application), 57–58 Electric charge, 6 Electric circuits, 4–5, 4–5f elements in, 14–16 Electric current, 6–7, 6f flow of, 8, 8f Electricity bills (application), 18, 18t Electricity consumption cost, 484–486 Electrolytic capacitors, 216, 216f Electromagnetic induction, 215 Electromagnetics, careers in, 553 Electronic instrumentation, career in, 173 Electronics, 79 Elements, 4, 14–16 Energy, 10–12 Energy sink, 59 Energy source, 59 Equivalent capacitance of parallel-connected capacitors, 221 of series-connected capacitors, 222 Equivalent conductance, 45 Equivalent impedance, 388–390 Equivalent inductance of parallel inductors, 229 of series-connected inductors, 229 Equivalent resistance of parallel resistors, 45 of series resistors, 44 Ethics, 501 Euler's formula, A–14 to A–15 Euler's identity, 321, 783 Even symmetry, 766–768 Excitation, 126 Exponential form, A–10 Exponential Fourier series, 783–789 External electromotive force (emf), 9
-
-# F
-
-Faraday, Michael, 215, 455 Faraday's law, 572 Fast Fourier Transform (FFT), 790–791 Field-effect transistors (FETs), 105–106 Film capacitors, 216, 216f Filters active, 635, 640–646 band-pass, 636, 636f, 637–638 band-stop, 538, 636, 636f defined, 635 design of, 795–798 high-pass, 636, 636f, 637 low-pass, 636–637, 636f passive, 635–640 First-order circuits, 251–297 automobile ignition circuit, 296–297, 296f defined, 252
-
-delay circuits, 291–293 natural response, 253–254 op amp, 282–287 photoflash unit, 293–294 relay circuits, 294–296 singularity functions, 263–271 source-free *RC* circuit, 253–257 source-free *RL* circuit, 257–263 step response of an *RC* circuit, 271–277 step response of an *RL* circuit, 278–282 time constant, 254–255 transient analysis with *PSpice,* 287–291 First-order differential equation, 253 First-order high-pass filters, 641 First-order low-pass filter, 641 Fixed capacitor, 216 Fixed resistors, 32, 32f Forced response, 273 Fourier, Jean Baptiste Joseph, 758 Fourier analysis, 369, 760 Fourier coefficients, 759 Fourier cosine series, 767 Fourier series, 757–798 average power, 780–783 circuit applications, 776–780 defined, 759 exponential, 783–789 filters, 795–798 Gibbs phenomenon, 764 Parseval's theorem, 781 *PSpice,* 789–794 RMS value, 780–783 sinc function, 785 sine, 769 spectrum analyzers, 795 symmetry considerations. *See* Symmetry trigonometric series, 759–766 Fourier theorem, 759 Fourier transform, 811–841 amplitude modulation, 820–821, 838–840 circuit applications, 831–833 convolution, 823–826 defined, 812–818 duality, 823 frequency shifting, 820–821 inverse, 814 linearity, 818 pairs, 827t Parseval's theorem, 834–837 reversal, 822–823 time differentiation, 821–822 time integration, 822 time scaling, 818–819 time shifting, 819–820 *vs.* Laplace transform, 837 Franklin, Benjamin, 6, 611 Frequency differentiation, 682 Frequency domain, 378, 387–388
-
-# **I-4** Index
-
-Frequency mixer, 656 Frequency of rejection, 638 Frequency response, 611–661 active filters, 640–646 bode plots, 617–627 crossover network, 659–661, 660f decibel scale, 615–617 defined, 612 *MATLAB,* 653–655 parallel resonance, 632–635 passive filters, 635–640 radio receiver, 655–657 scaling, 646–649 series resonance, 627–632 touch-tone telephone, 658–659 transfer function, 612–615 using *PSpice,* 650–653 Frequency scaling, 648–649 Frequency shift/shifting, 679–680, 820–821 Frequency spectrum, 762 Full-wave rectified sine, 772 Fundamental angular frequency, 759
-
-# G
-
-Ganged tuning, 656 Gate function, 267 Generalized node, 87 General second-order circuits, 337–341 Gibbs, Josiah Willard, 764 Gibbs phenomenon, 764 Ground, 81, 81f Ground-fault circuit interrupter (GFCI), 540 Györgyi, Albert Szent, 757
-
-# H
-
-Half-power frequencies, 629 Half-wave rectified sine, 772 Half-wave symmetry, 770–776 Heaviside's theorem, 689 Henry, Joseph, 224, 225 Herbert, G., 501 Hertz, Heinrich Rudorf, 370 Heterodyne circuit, 656 Higher potential, in resistor, 81 High-pass filters, 636, 636f, 637 High-Q circuit, 630 Homogeneity property, 126 Hybrid parameters, 860–865 Hyperbolic functions, A–17 Hysteresis, 375
-
-# I
-
-Ibn, Al Halif Omar, 455 Ideal autotransformers, 579–582, 579f Ideal dependent/controlled source, 15, 15f Ideal independent source, 14, 15f Ideal op amp, 178–179, 178f Ideal transformers, 571–578 IEEE (Institute of Electrical and Electronics Engineers), 13, 79, 251, 375, 554 Imaginary part, A–9 Immittance parameters, 857 Impedance, 385–387 combinations, 388–394 Impedance matching, 574, 591 Impedance parameters, 853–856 Impedance triangle, 473, 473f Indefinite integrals, A–18 to A–19 Inductance, 224 Inductance, mutual, 555–561 Inductive, 385 Inductors, 224–231, 224f. *See also* Capacitors analog computer, 235–238 characteristics, 230 defined, 224 differentiator, 233–234 integrator, 232–233 linear, 225 nonlinear, 225 parallel, 228–231 properties, 232 series, 228–231 Inspection, of circuit, 98–99 Instantaneous power, 11, 456–457 Institute of Electrical and Electronics Engineers (IEEE), 13, 79, 251, 375, 554 Institute of Radio Engineers (IRE), 13 Instrumentation amplifier (IA), 185, 187–188, 188f, 196–197, 196f Integral transform, 812 Integrator, 232–233 Integrodifferential equations, 703–705 International Electrical Exhibition, 13 International System of Units (SI), 5, 5t Inverse Fourier transform, 814 Inverse hybrid parameters, 861 Inverse Laplace transform, 676, 688–695 Inverse transmission, 867 Inverting op amp, 179–181 Isolation transformer, 573
-
-# J
-
-Jefferson, Thomas, 851
-
-# K
-
-Kirchhoff, Gustav Robert, 38 Kirchhoff's current law (KCL), 37–39, 81–82, 97, 412–415 Kirchhoff's voltage law (KVL), 39–40, 87, 92, 96, 387–388, 415–419
-
-Knowledge capturing integrated design environment" (KCIDE), *see* the web site associated with this book.
-
-# L
-
-Lagging power factor, 469 Lamme, B. G., 368 Laplace, Pierre Simon, 674 Laplace transform, 673–705, 714 circuit analysis, 720–723 circuit element models, 715–720 convolution integral, 695–703 defined, 675 Fourier transform *vs.,* 837 frequency differentiation, 682 frequency shift, 679–680 integrodifferential equations, 703–705 inverse, 676, 688–695 linearity, 678 network stability, 735–738 network synthesis, 738–743 one-sided, 676 properties of, 677–688, 685t sampling, 840–841 scaling, 678 state variables, 728–735 steps in applying, 715 time differentiation, 680 time integration, 681–682 time periodicity, 682–684 time shift, 678–679 transfer functions, 724–728 two-sided, 676 Law of conservation of charge, 6 Law of conservation of energy, 11 Law of cosines, A–16 Law of sines, A–16 Law of tangents, A–16 Leading power factor, 469 Least significant bit (LSB), 194 L'Hopital's rule, 786, A–20 Lighting systems (application), 57–58 Linear capacitor, 216 Linear circuit, 127, 127f Linearity, 126–127, 678, 818 Linear resistor, 33, 33f Linear transformers, 565–571, 566f Line spectra, 786 Line-to-line voltages, 508 Load, 59, 137 Loading effect, 154 Local oscillator, 656 Loop analysis. *See* Mesh analysis Loops, 35f, 36 Loosely coupled, 564 Lower potential, in resistor, 81 Low-pass filters, 636–637, 636f
-
-# M
-
-Magnetically coupled circuits, 553–594 energy in coupled circuit, 562–565 ideal autotransformers, 579–582 linear transformers, 565–571 mutual inductance, 555–561 power distribution, 593–594 *PSpice,* 584–589 three-phase transformers, 582–584 Magnitude scaling, 647 *Maple,* 82, 760 *Mathcad,* 82, 760 *MATLAB,* 80, 653–655 Maximum average power transfer, 462–465 Maximum power theorem, 148 Maximum power transfer, 148–150 Maxwell, James Clerk, 370, 554 Megger tester, 156 Mesh, defined, 91 Mesh analysis, 91–93 with current sources, 96–98 by inspection, 98–99 KVL and, 415–419 nodal analysis *vs.,* 102–103 steps, 92 Mesh-current method. *See* Mesh analysis Method of algebra, 691 Milliohmmeter, 156 Morse, Samuel F. B., 62 Most significant bit (MSB), 194 Multidisciplinary teams, 367 Multimeter, 59 Mutual inductance, 555–561 Mutual voltage, 556
-
-# N
-
-Natural frequencies, 319 Natural response, 253–254 Negative current flow, 8, 8f Negative sequence, 505, 505f Neper frequency, 319 Network function. *See* Transfer function Network stability, 735–738 Network synthesis, 738–743 Nodal analysis by inspection, 98–99 KCL and, 412–415 steps, 80–82 with voltage sources, 86–87 *vs.* mesh analysis, 102–103 Nodes, 35–36, 35f Node-voltage method, 80–82 Noninverting op amp, 181–183 Nonlinear capacitor, 216 Nonlinear resistor, 33, 33f
-
-# **I-6** Index
-
-Nonplanar circuit, 91, 91f Norton, E. L., 143 Norton equivalent circuits, 424–428, 424f Norton's theorem, 143–148 Notch filter, 643–644, 643f *npn* transistors, 106–107, 106f Nyquist frequency, 841 Nyquist interval, 841
-
-# O
-
-Odd symmetry, 768–770 Ohm, Georg Simon, 31 Ohm's law, 30–34 One-sided Laplace transform, 676 Open circuit, 32, 32f Open-circuit impedance parameters, 854 Open delta, 583 Open-loop voltage gain, 176 Operational amplifier (op amp), 173–197 ac circuit, 429–430 cascaded circuits, 189–192 defined, 174 difference, 185–188, 185f feedback, 176 first-order circuits, 282–287 ideal, 178–179, 178f instrumentation amplifier, 185, 187–188, 188f, 196–197, 196f inverting, 179–181 noninverting, 181–183 parameters, range of, 176t *PSpice,* analysis with, 192–193 second-order circuits, 342–344 summing, 183–185, 183f terminals, 175 Oscillators, 437–439 Overdamped case source-free parallel *RLC* circuits, 325 source-free series *RLC* circuit, 319 step response of parallel *RLC* circuits, 335 step response series *RLC* circuits, 330
-
-# P
-
-```
-Parallel, electric circuit, 36
-Parallel capacitors, 220–223
-Parallel inductors, 228–231
-Parallel resistors, 44–47, 44f
-Parallel resonance, 632–635
-Parallel RLC circuits
-  source-free, 324–329
-  step response, 334–337
-Parameters
-  ABCD, 866
-  admittance, 857–860
-```
-
-hybrid, 860–865 immittance, 857 impedance, 853–856 inverse transmission, 867 relationships between, 870–873 short-circuit admittance, 857 transmission, 865–869 Parseval's theorem, 781, 834–837 Partial fraction expansion, 688 Passive elements, 14 Passive filters, 635–640 Passive sign convention, 11 Period, 369, 370 Periodic function, 370, 759 Phase sequence, 505 Phase-shifting circuit, 394–396 Phase spectrum, 762, 814 Phase voltages, 504 Phasor diagram, 377, 378f Phasor relationships for circuit elements, 383–384 Phasors, 374–382. *See also* Sinusoids Photoflash unit, 293–294 Planar circuit, 91, 91f *pnp* transistors, 106, 106f Poisson, Simeon, 674 Polar form, A–9 Poles, 613, 618 complex, 690–691 first-order, 688–689 repeated, 689–690 Polyester capacitors, 216, 216f Polyphase, 502 Port, 852 Positive current flow, 8, 8f Positive sequence, 505, 505f Potential difference. *See* Voltage Potentiometer (pot), 32, 32f, 59, 59f Power, 10–12 Power distribution system, 593–594, 593f Power factor, 469–471 Power factor angle, 469 Power factor correction, 479–481 Power grid, 593 Power measurement, 481–484 Power spectrum, 785 Power systems, careers in, 455 Power triangle, 473, 473f Primary winding, 566 Principle of current division, 46 Principle of voltage division, 44 Problem solving technique, 19–20 *PSpice,* 80 ac analysis using, 431–435 analysis of *RLC* circuits, 344–347 circuit analysis with, 103–105 circuit theorems with, 150–153 Fourier analysis, 789–794
-
-frequency response, 650–653 magnetically coupled circuits, 584–589 operational amplifier analysis with, 192–193 three-phase circuits, 527–532 transient analysis with, 287–291 two-port networks, 279–282
-
-# Q
-
-Quadratic formulas, A–16 Quadrature power, 472 Quality factor, 629–630 *Quattro Pro,* 82
-
-# R
-
-Radio receiver, 655–657 Ragazzini, John, 174 *RC* circuits delay, 291–293 source-free, 253–257 step response, 271–277 *RC* phase-shifting circuits, 394–396 Reactance, 385 Reactive load, 459 Reactive power, 472 Real part of complex numbers, A–9 Reciprocal network, 854 Rectangular form of complex numbers, A–9 Rectangular pulse train, 772 Reference node, 81, 81f Reflected impedance, 567, 574 Relay circuits, 294–296 Relay delay time, 295 Residential wiring, 538–540, 539f Residue method, 689 Residues, 689 Resistance, 30, 32, 385 equivalent, 44 measurement, 156–158 Resistance bridge, 156 Resistance matrix, 99 Resistive load, 459 Resistivity, 30, 31t Resistors characteristics, 230 fixed, 32, 32f linear, 33, 33f nonlinear, 33, 33f Ohm's law, 30–34 parallel, 44–47, 44f series, 43–44 variable, 32, 33f Resonance, 627–628 Resonant frequency, 319, 628 Resonant peak, 627
-
-Response, 126 Reversal, 822–823 Right-hand rule, 554 *RLC* circuits source-free parallel, 324–329 source-free series, 317–324 step response of parallel, 334–337 step response of series, 329–334 *RL* circuits, 257–263 RMS value, 466–468, 780–783 Rolloff frequency, 637 Rotor, 503 Rubidium, 38
-
-# S
-
-Sampling, 266, 840–841 Sampling frequency, 840 Sampling function, 785 Sampling interval, 840 Sampling rate, 840 Sampling theorem, 795 Sawtooth function, 268 Sawtooth wave, 772 Scaling, 646–649, 678 frequency, 648–649 magnitude, 647 Schockley, William, 106 Scott, C. F., 368 Secondary winding, 566 Second-order circuits, 311–354 automobile ignition system, 351–353 characteristic equation, 318 defined, 312 duality, 348–350 general, 337–341 initial/final values, 313–317 op amp circuits, 342–344 *PSpice,* 344–347 second-order differential equation, 318 smoothing circuits, 353–354 source-free parallel *RLC* circuits, 324–329 source-free series *RLC* circuits, 317–324 step response of parallel *RLC* circuit, 334–337 step response of series *RLC* circuit, 329–334 Second-order differential equation, 318 Self-inductance, 555 Series, electric circuit, 36 Series capacitors, 220–223 Series inductors, 228–231 Series resistors, 43–44 Series resonance, 627–632 Series *RLC* circuits source-free, 317–324 step response, 329–334 Short circuit, 32, 32f
-
-# **I-8** Index
-
-Short-circuit admittance parameters, 857 Sifting, 266 Signal, 9 Simultaneous equations, A to A–4 Sinc function, 785 Single-phase three-wire system, 502, 502f Singularity functions, 263–271 Sinusoidal steady-state analysis, 411–439 capacitance multiplier, 435–437, 435f mesh analysis, 415–419 nodal analysis, 412–415 Norton equivalent circuits, 424–428, 424f op amp ac circuits, 429–430 oscillators, 437–439 *PSpice,* 431–435 source transformation, 422–424 superposition theorem, 419–422 Thevenin equivalent circuits, 424–428, 424f Sinusoidal steady-state response, 369 Sinusoids, 368–374 SI units, 5, 5t Smoothing circuits, 353–354 Software engineering, career in, 411 Solenoidal wound inductor, 224f Source-free parallel *RLC* circuits, 324–329 Source-free *RC* circuit, 253–257 Source-free *RL* circuit, 257–263 Source-free series *RLC* circuits, 317–324 Source modeling, 153–155 Source transformation, 133–136, 422–424 Spectrum analyzers, 795 Sprague, Frank, 13 Square wave, 772 Stability, network, 735–738 Standard form, 618 State variables, 728–735 Stator, 503 Steady-state response, 274 Steinmetz, Charles P., 713 Steinmetz, Charles Proteus, 374, 375 Step-down autotransformer, 579, 579f Step-down transformers, 573 Step response of *RC* circuit, 271–277 of *RL* circuit, 278–282 Step response of parallel *RLC* circuits, 334–337 Step response series *RLC* circuits, 329–334 Step-up autotransformer, 579–580, 579f Step-up transformer, 573 Storage elements, 214 Strength, of impulse function, 265 Summing op amp, 183–185, 183f Superheterodyne receiver, 656 Supermesh, 96 Supernode, 87 Superposition, 129–133 Superposition integral, 698
-
-Superposition theorem, 419–422 Susceptance, 386 Switching functions. *See* Singularity functions Symmetrical network, 854 Symmetry even, 766–768 half-wave, 770–776 odd, 768–770 System, 714 System design, 213
-
-# T
-
-Terminals, 175 Tesla, Nikola, 368, 503 Thevenin, M. Leon, 137 Thevenin equivalent circuit, 137, 137f Thevenin equivalent circuits, 424–428, 424f Thevenin's theorem, 137–143, 147–148 Thompson, Elihu, 13 Three-phase circuits, 501–540 balanced delta-delta connection, 512–514 balanced delta-wye connection, 514–517 balanced three-phase voltages, 503–506 balanced wye-delta connection, 510–512 balanced wye-wye connection, 507–510 importance of, 502–503 power in balanced system, 517–523 power measurement, 533–538 *PSpice,* 527–532 residential wiring, 538–540, 539f unbalanced three-phase systems, 523–526 Three-phase transformers, 582–584 Three-stage cascaded connection, 189, 189f Three-wattmeter method, 533, 533f Tightly coupled, 564 Time constant, 254–255 Time differentiation, 680, 821–822 Time integration, 681–682, 822 Time periodicity, 682–684 Time scaling, 818–819 Time shift/shifting, 678–679, 819–820 Toroidal inductor, 224f Total response, 273 Touch-tone telephone, 658–659 Transfer functions, 612–615, 724–728 Transfer impedances, 854 Transformation ratio, 572 Transformer bank, 582 Transformers, 555 air-core, 566 ideal, 571–578 isolation, 573 as isolation device, 590–591 linear, 565–571, 566f as matching device, 591–592 step-down, 573
-
-## Index **I-9**
-
-step-up, 573 three-phase, 582–584 Transient analysis with *PSpice,* 287–291 Transient response, 274 Transistor, 105–107, 106f Transistor circuits, 882–887 Transmission parameters, 865–869 Triangular wave, 772 Trigonometric Fourier series, 759–766 Trigonometric identities, A–16 to A–17 Turns ratio, 572 TV picture tube, 16, 17f Two-phase three-wire system, 502f Two-port networks, 851–891 admittance parameters, 857–860 defined, 852 hybrid parameters, 860–865 impedance parameters, 853–856 interconnection of networks, 873–879 inverse transmission parameters, 867 *PSpice,* 879–882 reciprocal network, 854 relationships between parameters, 870–873 symmetrical network, 854 transistor circuits, 882–887 transmission parameters, 865–869 Two-sided Laplace transform, 676 Two-wattmeter method, 533–534, 533f
-
-# U
-
-Unbalanced three-phase systems, 523–526 Undamped natural frequency, 319 Underdamped case source-free parallel *RLC* circuits, 325 source-free series *RLC* circuit, 321–322 step response of parallel *RLC* circuits, 335 step response series *RLC* circuits, 330
-
-United States Electric Lighting Company, 13 Unit impulse function, 265, 265f Unit ramp function, 266, 266f Unit step function, 264 Unity gain amplifier, 182 Unloaded source, 154
-
-# V
-
-Variable capacitor, 216 Variable resistors, 32, 33f Volta, Alessandro Antonio, 10 Voltage, 9–10, 9f Voltage divider, 44 Voltage-division relationship, 389 Voltage drop, 9, 9f Voltage follower, 182, 182f Voltage rise, 9, 9f Volt-ampere reactive (VAR), 472 Volt-ohm meter (VOM), 59
-
-# W
-
-Watson, James A., 713 Watson, Thomas A., 616 Wattmeter, 481–482, 482f, 533 Westinghouse, George, 368 Weston, Edward, 13 Wheatstone, Charles, 156 Wheatstone bridge, 156 Wien-bridge oscillator, 437, 437f Winding capacitance, 226 Winding resistance, 226 Wye-delta transformations, 51–53, 390
-
-# Z
-
-Zeros, transfer function, 613, 618 Zworykin, Vladimir K., 17

engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/archive/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku.pdf

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-# Book Title: Fundamentals of Electric Circuits
-> Author: Charles K. Alexander and Matthew N.O. Sadiku | Category: Engineering | Pages: 990 | Year: 2015
-
-## Overview & Metadata
-- Document ID: fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku
-- Category: Engineering
-- Tags: 6th, alexander, and, charles, circuits, electric, fundamentals, matthew
-- Primary Markdown HF Raw URL: https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku.md
-- Original PDF HF Download: https://huggingface.co/datasets/learner20011/CloverTexts-Data/resolve/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku.pdf
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-- Book LLM Text URL: https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/llm.txt
-- Online Web Reader: /book?id=fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku
-
-## Table of Contents
-- Cover - Page 1
-- Copyright - Page 3
-- Dedication - Page 4
-- Contents - Page 6
-- Preface - Page 12
-- Acknowledgments - Page 16
-- About the Authors - Page 22
-- PART 1: DC Circuits - Page 25
-  - Chapter 1: Basic Concepts - Page 26
-    - 1.1 Introduction - Page 27
-    - 1.2 Systems of Units - Page 28
-    - 1.3 Charge and Current - Page 29
-    - 1.4 Voltage - Page 32
-    - 1.5 Power and Energy - Page 33
-    - 1.6 Circuit Elements - Page 37
-    - 1.7 Applications - Page 39
-      - 1.7.1 TV Picture Tube - Page 39
-      - 1.7.2 Electricity Bills - Page 41
-    - 1.8 Problem Solving - Page 42
-    - 1.9 Summary - Page 45
-    - Review Questions - Page 46
-    - Problems - Page 47
-    - Comprehensive Problems - Page 49
-  - Chapter 2: Basic Laws - Page 52
-    - 2.1 Introduction - Page 53
-    - 2.2 Ohm's Law - Page 53
-    - 2.3 Nodes, Branches, and Loops - Page 58
-    - 2.4 Kirchhoff's Laws - Page 60
-    - 2.5 Series Resistors and Voltage Division - Page 66
-    - 2.6 Parallel Resistors and Current Division - Page 67
-    - 2.7 Wye-Delta Transformations - Page 74
-      - Delta to Wye Conversion - Page 75
-      - Wye to Delta Conversion - Page 76
-    - 2.8 Applications - Page 80
-      - 2.8.1 Lighting Systems - Page 80
-      - 2.8.2 Design of DC Meters - Page 82
-    - 2.9 Summary - Page 86
-    - Review Questions - Page 87
-    - Problems - Page 88
-    - Comprehensive Problems - Page 100
-  - Chapter 3: Methods of Analysis - Page 102
-    - 3.1 Introduction - Page 103
-    - 3.2 Nodal Analysis - Page 103
-    - 3.3 Nodal Analysis with Voltage Sources - Page 109
-    - 3.4 Mesh Analysis - Page 114
-    - 3.5 Mesh Analysis with Current Sources - Page 119
-    - 3.6 Nodal and Mesh Analyses by Inspection - Page 121
-    - 3.7 Nodal Versus Mesh Analysis - Page 125
-    - 3.8 Circuit Analysis with PSpice - Page 126
-    - 3.9 Applications: DC Transistor Circuits - Page 128
-    - 3.10 Summary - Page 133
-    - Review Questions - Page 134
-    - Problems - Page 135
-    - Comprehensive Problem - Page 147
-  - Chapter 4: Circuit Theorems - Page 148
-    - 4.1 Introduction - Page 149
-    - 4.2 Linearity Property - Page 149
-    - 4.3 Superposition - Page 152
-    - 4.4 Source Transformation - Page 156
-    - 4.5 Thevenin's Theorem - Page 160
-    - 4.6 Norton's Theorem - Page 166
-    - 4.7 Derivations of Thevenin's and Norton's Theorems - Page 170
-    - 4.8 Maximum Power Transfer - Page 171
-    - 4.9 Verifying Circuit Theorems with PSpice - Page 173
-    - 4.10 Applications - Page 176
-      - 4.10.1 Source Modeling - Page 176
-      - 4.10.2 Resistance Measurement - Page 179
-    - 4.11 Summary - Page 181
-    - Review Questions - Page 182
-    - Problems - Page 183
-    - Comprehensive Problems - Page 194
-  - Chapter 5: Operational Amplifiers - Page 196
-    - 5.1 Introduction - Page 197
-    - 5.2 Operational Amplifiers - Page 197
-    - 5.3 Ideal Op Amp - Page 201
-    - 5.4 Inverting Amplifier - Page 202
-    - 5.5 Noninverting Amplifier - Page 204
-    - 5.6 Summing Amplifier - Page 206
-    - 5.7 Difference Amplifier - Page 208
-    - 5.8 Cascaded Op Amp Circuits - Page 212
-    - 5.9 Op Amp Circuit Analysis with PSpice - Page 215
-    - 5.10 Applications - Page 217
-      - 5.10.1 Digital-to-Analog Converter - Page 217
-      - 5.10.2 Instrumentation Amplifiers - Page 219
-    - 5.11 Summary - Page 220
-    - Review Questions - Page 222
-    - Problems - Page 223
-    - Comprehensive Problems - Page 234
-  - Chapter 6: Capacitors and Inductors - Page 236
-    - 6.1 Introduction - Page 237
-    - 6.2 Capacitors - Page 237
-    - 6.3 Series and Parallel Capacitors - Page 243
-    - 6.4 Inductors - Page 247
-    - 6.5 Series and Parallel Inductors - Page 251
-    - 6.6 Applications - Page 254
-      - 6.6.1 Integrator - Page 255
-      - 6.6.2 Differentiator - Page 256
-      - 6.6.3 Analog Computer - Page 258
-    - 6.7 Summary - Page 261
-    - Review Questions - Page 262
-    - Problems - Page 263
-    - Comprehensive Problems - Page 272
-  - Chapter 7: First-Order Circuits - Page 274
-    - 7.1 Introduction - Page 275
-    - 7.2 The Source-Free RC Circuit - Page 276
-    - 7.3 The Source-Free RL Circuit - Page 280
-    - 7.4 Singularity Functions - Page 286
-    - 7.5 Step Response of an RC Circuit - Page 294
-    - 7.6 Step Response of an RL Circuit - Page 301
-    - 7.7 First-Order Op Amp Circuits - Page 305
-    - 7.8 Transient Analysis with PSpice - Page 310
-    - 7.9 Applications - Page 314
-      - 7.9.1 Delay Circuits - Page 314
-      - 7.9.2 Photoflash Unit - Page 316
-      - 7.9.3 Relay Circuits - Page 317
-      - 7.9.4 Automobile Ignition Circuit - Page 319
-    - 7.10 Summary - Page 320
-    - Review Questions - Page 321
-    - Problems - Page 322
-    - Comprehensive Problems - Page 332
-  - Chapter 8: Second-Order Circuits - Page 334
-    - 8.1 Introduction - Page 335
-    - 8.2 Finding Initial and Final Values - Page 336
-    - 8.3 The Source-Free Series RLC Circuit - Page 340
-    - 8.4 The Source-Free Parallel RLC Circuit - Page 347
-    - 8.5 Step Response of a Series RLC Circuit - Page 352
-    - 8.6 Step Response of a Parallel RLC Circuit - Page 357
-    - 8.7 General Second-Order Circuits - Page 360
-    - 8.8 Second-Order Op Amp Circuits - Page 365
-    - 8.9 PSpice Analysis of RLC Circuits - Page 367
-    - 8.10 Duality - Page 371
-    - 8.11 Applications - Page 374
-      - 8.11.1 Automobile Ignition System - Page 374
-      - 8.11.2 Smoothing Circuits - Page 376
-    - 8.12 Summary - Page 377
-    - Review Questions - Page 378
-    - Problems - Page 379
-    - Comprehensive Problems - Page 388
-- PART 2: AC Circuits - Page 389
-  - Chapter 9: Sinusoids and Phasors - Page 390
-    - 9.1 Introduction - Page 391
-    - 9.2 Sinusoids - Page 392
-    - 9.3 Phasors - Page 397
-    - 9.4 Phasor Relationships for Circuit Elements - Page 406
-    - 9.5 Impedance and Admittance - Page 408
-    - 9.6 Kirchhoff's Laws in the Frequency Domain - Page 410
-    - 9.7 Impedance Combinations - Page 411
-    - 9.8 Applications - Page 417
-      - 9.8.1 Phase-Shifters - Page 417
-      - 9.8.2 AC Bridges - Page 419
-
-## Available Raw Markdown Chunks (Direct Hugging Face Raw URLs - 0 Bot Protection)
-- [001_Cover.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/001_Cover.md) — Fundamentals of Electric Circuits
-- [002_Copyright.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/002_Copyright.md) — • Fourier and Laplace Transforms Coverage
-- [003_Preface.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/003_Preface.md) — Zekeriya Aliyazicioglu, California State Polytechnic University— Pomona
-- [004_About the Authors.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/004_About%20the%20Authors.md) — About the Authors
-- [005_PART 1 - DC Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/005_PART%201%20-%20DC%20Circuits.md) — Fundamentals of Electric Circuits
-- [006_Chapter 1 - Basic Concepts.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/006_Chapter%201%20-%20Basic%20Concepts.md) — Basic Concepts
-- [007_1.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/007_1.1%20Introduction.md) — 1.1 Introduction
-- [008_1.2 Systems of Units.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/008_1.2%20Systems%20of%20Units.md) — 1.3 Charge and Current
-- [009_1.4 Voltage.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/009_1.4%20Voltage.md) — 1.4 Voltage
-- [010_1.5 Power and Energy.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/010_1.5%20Power%20and%20Energy.md) — 1.5 Power and Energy
-- [011_1.6 Circuit Elements.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/011_1.6%20Circuit%20Elements.md) — 1.6 Circuit Elements
-- [012_1.7 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/012_1.7%20Applications.md) — 1.7 Applications2
-- [013_1.8 Problem Solving.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/013_1.8%20Problem%20Solving.md) — 1.8 Problem Solving
-- [014_1.9 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/014_1.9%20Summary.md) — 1.9 Summary
-- [015_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/015_Review%20Questions.md) — Problems
-- [016_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/016_Problems.md) — Comprehensive Problems
-- [017_Chapter 2 - Basic Laws.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/017_Chapter%202%20-%20Basic%20Laws.md) — Basic Laws 2
-- [018_2.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/018_2.1%20Introduction.md) — 2.1 Introduction
-- [019_2.2 Ohm's Law.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/019_2.2%20Ohm's%20Law.md) — 2.2 Ohm's Law
-- [020_2.3 Nodes, Branches, and Loops.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/020_2.3%20Nodes%2C%20Branches%2C%20and%20Loops.md) — 2.3 Nodes, Branches, and Loops
-- [021_2.4 Kirchhoff's Laws.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/021_2.4%20Kirchhoff's%20Laws.md) — 2.4 Kirchhoff's Laws
-- [022_2.5 Series Resistors and Voltage Division.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/022_2.5%20Series%20Resistors%20and%20Voltage%20Division.md) — Figure 2.30
-- [023_2.6 Parallel Resistors and Current Division.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/023_2.6%20Parallel%20Resistors%20and%20Current%20Division.md) — Solution:
-- [024_2.7 Wye-Delta Transformations.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/024_2.7%20Wye-Delta%20Transformations.md) — 2.7 Wye-Delta Transformations
-- [025_2.8 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/025_2.8%20Applications.md) — 2.8 Applications
-- [026_2.9 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/026_2.9%20Summary.md) — 2.9 Summary
-- [027_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/027_Review%20Questions.md) — Figure 2.65
-- [028_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/028_Problems.md) — Problems
-- [029_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/029_Comprehensive%20Problems.md) — Comprehensive Problems
-- [030_Chapter 3 - Methods of Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/030_Chapter%203%20-%20Methods%20of%20Analysis.md) — Methods of Analysis
-- [031_3.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/031_3.1%20Introduction.md) — Solution:
-- [032_3.2 Nodal Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/032_3.2%20Nodal%20Analysis.md) — 3.3 Nodal Analysis with Voltage Sources
-- [033_3.4 Mesh Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/033_3.4%20Mesh%20Analysis.md) — 3.5 Mesh Analysis with Current Sources
-- [034_3.6 Nodal and Mesh Analyses by Inspection.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/034_3.6%20Nodal%20and%20Mesh%20Analyses%20by%20Inspection.md) — 3.6 Nodal and Mesh Analyses by Inspection
-- [035_3.7 Nodal Versus Mesh Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/035_3.7%20Nodal%20Versus%20Mesh%20Analysis.md) — 3.7 Nodal Versus Mesh Analysis
-- [036_3.8 Circuit Analysis with PSpice.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/036_3.8%20Circuit%20Analysis%20with%20PSpice.md) — 3.8 Circuit Analysis with PSpice
-- [037_3.9 Applications - DC Transistor Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/037_3.9%20Applications%20-%20DC%20Transistor%20Circuits.md) — Applications: DC Transistor Circuits
-- [038_3.10 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/038_3.10%20Summary.md) — Review Questions
-- [039_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/039_Review%20Questions.md) — Figure 3.49
-- [040_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/040_Problems.md) — Problems
-- [041_Comprehensive Problem.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/041_Comprehensive%20Problem.md) — chapter
-- [042_Chapter 4 - Circuit Theorems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/042_Chapter%204%20-%20Circuit%20Theorems.md) — Circuit Theorems
-- [043_4.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/043_4.1%20Introduction.md) — Figure 4.1
-- [044_4.3 Superposition.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/044_4.3%20Superposition.md) — Solution:
-- [045_4.5 Thevenin's Theorem.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/045_4.5%20Thevenin's%20Theorem.md) — 4.7 Derivations of Thevenin's and Norton's Theorems
-- [046_4.7 Derivations of Thevenin's and Norton's Theorems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/046_4.7%20Derivations%20of%20Thevenin's%20and%20Norton's%20Theorems.md) — 148 Chapter 4 Circuit Theorems
-- [047_4.8 Maximum Power Transfer.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/047_4.8%20Maximum%20Power%20Transfer.md) — 4.8 Maximum Power Transfer
-- [048_4.9 Verifying Circuit Theorems with PSpice.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/048_4.9%20Verifying%20Circuit%20Theorems%20with%20PSpice.md) — 4.9 Verifying Circuit Theorems with PSpice
-- [049_4.10 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/049_4.10%20Applications.md) — 4.10 Applications
-- [050_4.11 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/050_4.11%20Summary.md) — 4.11 Summary
-- [051_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/051_Review%20Questions.md) — Problems
-- [052_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/052_Comprehensive%20Problems.md) — Comprehensive Problems
-- [053_Chapter 5 - Operational Amplifiers.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/053_Chapter%205%20-%20Operational%20Amplifiers.md) — Operational Amplifiers
-- [054_5.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/054_5.1%20Introduction.md) — 5.1 Introduction
-- [055_5.2 Operational Amplifiers.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/055_5.2%20Operational%20Amplifiers.md) — 5.2 Operational Amplifiers
-- [056_5.3 Ideal Op Amp.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/056_5.3%20Ideal%20Op%20Amp.md) — 5.3 Ideal Op Amp
-- [057_5.4 Inverting Amplifier.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/057_5.4%20Inverting%20Amplifier.md) — 5.4 Inverting Amplifier
-- [058_5.5 Noninverting Amplifier.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/058_5.5%20Noninverting%20Amplifier.md) — Solution:
-- [059_5.6 Summing Amplifier.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/059_5.6%20Summing%20Amplifier.md) — 5.6 Summing Amplifier
-- [060_5.7 Difference Amplifier.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/060_5.7%20Difference%20Amplifier.md) — 5.7 Difference Amplifier
-- [061_5.8 Cascaded Op Amp Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/061_5.8%20Cascaded%20Op%20Amp%20Circuits.md) — 5.8 Cascaded Op Amp Circuits
-- [062_5.9 Op Amp Circuit Analysis with PSpice.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/062_5.9%20Op%20Amp%20Circuit%20Analysis%20with%20PSpice.md) — 5.9 Op Amp Circuit Analysis with PSpice
-- [063_5.10 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/063_5.10%20Applications.md) — Section 5.10 Applications
-- [064_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/064_Problems.md) — Comprehensive Problems
-- [065_Chapter 6 - Capacitors and Inductors.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/065_Chapter%206%20-%20Capacitors%20and%20Inductors.md) — Capacitors and Inductors
-- [066_6.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/066_6.1%20Introduction.md) — 6.1 Introduction
-- [067_6.2 Capacitors.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/067_6.2%20Capacitors.md) — 6.2 Capacitors
-- [068_6.3 Series and Parallel Capacitors.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/068_6.3%20Series%20and%20Parallel%20Capacitors.md) — 6.3 Series and Parallel Capacitors
-- [069_6.4 Inductors.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/069_6.4%20Inductors.md) — 6.4 Inductors
-- [070_6.5 Series and Parallel Inductors.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/070_6.5%20Series%20and%20Parallel%20Inductors.md) — 6.5 Series and Parallel Inductors
-- [071_6.6 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/071_6.6%20Applications.md) — 6.6 Applications
-- [072_6.7 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/072_6.7%20Summary.md) — 6.7 Summary
-- [073_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/073_Review%20Questions.md) — Review Questions
-- [074_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/074_Problems.md) — Problems
-- [075_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/075_Comprehensive%20Problems.md) — chapter
-- [076_Chapter 7 - First-Order Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/076_Chapter%207%20-%20First-Order%20Circuits.md) — First-Order Circuits
-- [077_7.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/077_7.1%20Introduction.md) — 7.1 Introduction
-- [078_7.2 The Source-Free RC Circuit.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/078_7.2%20The%20Source-Free%20RC%20Circuit.md) — 7.3 The Source-Free RL Circuit
-- [079_7.4 Singularity Functions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/079_7.4%20Singularity%20Functions.md) — 7.4 Singularity Functions
-- [080_7.5 Step Response of an RC Circuit.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/080_7.5%20Step%20Response%20of%20an%20RC%20Circuit.md) — 7.5 Step Response of an RC Circuit
-- [081_7.6 Step Response of an RL Circuit.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/081_7.6%20Step%20Response%20of%20an%20RL%20Circuit.md) — Solution:
-- [082_7.7 First-Order Op Amp Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/082_7.7%20First-Order%20Op%20Amp%20Circuits.md) — 7.7 † First-Order Op Amp Circuits
-- [083_7.8 Transient Analysis with PSpice.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/083_7.8%20Transient%20Analysis%20with%20PSpice.md) — 7.8 Transient Analysis with PSpice
-- [084_7.9 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/084_7.9%20Applications.md) — 7.9 †Applications
-- [085_7.10 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/085_7.10%20Summary.md) — 7.10 Summary
-- [086_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/086_Review%20Questions.md) — Review Questions
-- [087_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/087_Problems.md) — Problems
-- [088_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/088_Comprehensive%20Problems.md) — Comprehensive Problems
-- [089_Chapter 8 - Second-Order Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/089_Chapter%208%20-%20Second-Order%20Circuits.md) — Figure 8.1
-- [090_8.2 Finding Initial and Final Values.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/090_8.2%20Finding%20Initial%20and%20Final%20Values.md) — Solution:
-- [091_8.3 The Source-Free Series RLC Circuit.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/091_8.3%20The%20Source-Free%20Series%20RLC%20Circuit.md) — 8.3 The Source-Free Series RLC Circuit
-- [092_8.4 The Source-Free Parallel RLC Circuit.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/092_8.4%20The%20Source-Free%20Parallel%20RLC%20Circuit.md) — Critically Damped Case (α = ω0)
-- [093_8.5 Step Response of a Series RLC Circuit.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/093_8.5%20Step%20Response%20of%20a%20Series%20RLC%20Circuit.md) — 8.5 Step Response of a Series RLC Circuit
-- [094_8.6 Step Response of a Parallel RLC Circuit.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/094_8.6%20Step%20Response%20of%20a%20Parallel%20RLC%20Circuit.md) — 8.6 Step Response of a Parallel RLC Circuit
-- [095_8.7 General Second-Order Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/095_8.7%20General%20Second-Order%20Circuits.md) — Solution:
-- [096_8.8 Second-Order Op Amp Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/096_8.8%20Second-Order%20Op%20Amp%20Circuits.md) — Example 8.12
-- [097_8.9 PSpice Analysis of RLC Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/097_8.9%20PSpice%20Analysis%20of%20RLC%20Circuits.md) — Solution:
-- [098_8.10 Duality.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/098_8.10%20Duality.md) — 8.10 Duality
-- [099_8.11 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/099_8.11%20Applications.md) — Section 8.11 Applications
-- [100_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/100_Comprehensive%20Problems.md) — PART TWO
-- [101_PART 2 - AC Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/101_PART%202%20-%20AC%20Circuits.md) — AC Circuits
-- [102_Chapter 9 - Sinusoids and Phasors.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/102_Chapter%209%20-%20Sinusoids%20and%20Phasors.md) — Sinusoids and Phasors
-- [103_9.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/103_9.1%20Introduction.md) — 9.1 Introduction
-- [104_9.2 Sinusoids.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/104_9.2%20Sinusoids.md) — 9.2 Sinusoids
-- [105_9.3 Phasors.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/105_9.3%20Phasors.md) — 9.3 Phasors
-- [106_9.4 Phasor Relationships for Circuit Elements.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/106_9.4%20Phasor%20Relationships%20for%20Circuit%20Elements.md) — Figure 9.9
-- [107_9.5 Impedance and Admittance.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/107_9.5%20Impedance%20and%20Admittance.md) — Solution:
-- [108_9.6 Kirchhoff's Laws in the Frequency Domain.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/108_9.6%20Kirchhoff's%20Laws%20in%20the%20Frequency%20Domain.md) — Figure 9.17
-- [109_9.7 Impedance Combinations.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/109_9.7%20Impedance%20Combinations.md) — 9.7 Impedance Combinations
-- [110_9.8 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/110_9.8%20Applications.md) — 9.8 Applications
-- [111_9.9 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/111_9.9%20Summary.md) — Review Questions
-- [112_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/112_Review%20Questions.md) — Section 9.3 Phasors
-- [113_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/113_Comprehensive%20Problems.md) — Figure 9.90
-- [114_Chapter 10 - Sinusoidal Steady-State Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/114_Chapter%2010%20-%20Sinusoidal%20Steady-State%20Analysis.md) — Sinusoidal Steady-State Analysis
-- [115_10.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/115_10.1%20Introduction.md) — Figure 10.2
-- [116_10.3 Mesh Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/116_10.3%20Mesh%20Analysis.md) — 10.3 Mesh Analysis
-- [117_10.4 Superposition Theorem.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/117_10.4%20Superposition%20Theorem.md) — Example 10.5
-- [118_10.5 Source Transformation.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/118_10.5%20Source%20Transformation.md) — 10.5 Source Transformation
-- [119_10.6 Thevenin and Norton Equivalent Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/119_10.6%20Thevenin%20and%20Norton%20Equivalent%20Circuits.md) — Example 10.9
-- [120_10.7 Op Amp AC Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/120_10.7%20Op%20Amp%20AC%20Circuits.md) — Section 10.7 Op Amp AC Circuits
-- [121_10.9 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/121_10.9%20Applications.md) — Section 10.9 Applications
-- [122_Chapter 11 - AC Power Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/122_Chapter%2011%20-%20AC%20Power%20Analysis.md) — AC Power Analysis
-- [123_11.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/123_11.1%20Introduction.md) — Figure 11.1
-- [124_11.3 Maximum Average Power Transfer.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/124_11.3%20Maximum%20Average%20Power%20Transfer.md) — 11.3 Maximum Average Power Transfer
-- [125_11.4 Effective or RMS Value.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/125_11.4%20Effective%20or%20RMS%20Value.md) — 11.4 Effective or RMS Value
-- [126_11.5 Apparent Power and Power Factor.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/126_11.5%20Apparent%20Power%20and%20Power%20Factor.md) — For Practice Prob. 11.8. 11.5 Apparent Power and Power Factor
-- [127_11.6 Complex Power.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/127_11.6%20Complex%20Power.md) — For Practice Prob. 11.10. 11.6 Complex Power
-- [128_11.7 Conservation of AC Power.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/128_11.7%20Conservation%20of%20AC%20Power.md) — 11.7 Conservation of AC Power
-- [129_11.8 Power Factor Correction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/129_11.8%20Power%20Factor%20Correction.md) — 11.8 Power Factor Correction
-- [130_11.9 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/130_11.9%20Applications.md) — 11.9 Applications
-- [131_11.10 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/131_11.10%20Summary.md) — 11.10 Summary
-- [132_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/132_Review%20Questions.md) — | (a) voltmeter           | (b) ammeter  |
-- [133_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/133_Problems.md) — Problems1
-- [134_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/134_Comprehensive%20Problems.md) — Comprehensive Problems
-- [135_Chapter 12 - Three-Phase Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/135_Chapter%2012%20-%20Three-Phase%20Circuits.md) — Three-Phase Circuits
-- [136_12.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/136_12.1%20Introduction.md) — 12.1 Introduction
-- [137_12.2 Balanced Three-Phase Voltages.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/137_12.2%20Balanced%20Three-Phase%20Voltages.md) — 12.2 Balanced Three-Phase Voltages
-- [138_12.3 Balanced Wye-Wye Connection.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/138_12.3%20Balanced%20Wye-Wye%20Connection.md) — 12.4 Balanced Wye-Delta Connection
-- [139_12.4 Balanced Wye-Delta Connection.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/139_12.4%20Balanced%20Wye-Delta%20Connection.md) — 12.6 Balanced Delta-Wye Connection
-- [140_12.7 Power in a Balanced System.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/140_12.7%20Power%20in%20a%20Balanced%20System.md) — 12.7 Power in a Balanced System
-- [141_12.8 Unbalanced Three-Phase Systems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/141_12.8%20Unbalanced%20Three-Phase%20Systems.md) — 12.8 Unbalanced Three-Phase Systems
-- [142_12.9 PSpice for Three-Phase Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/142_12.9%20PSpice%20for%20Three-Phase%20Circuits.md) — 12.9 PSpice for Three-Phase Circuits
-- [143_12.10 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/143_12.10%20Applications.md) — Section 12.10 Applications
-- [144_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/144_Problems.md) — Comprehensive Problems
-- [145_Chapter 13 - Magnetically Coupled Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/145_Chapter%2013%20-%20Magnetically%20Coupled%20Circuits.md) — Magnetically Coupled Circuits
-- [146_13.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/146_13.1%20Introduction.md) — 13.1 Introduction
-- [147_13.2 Mutual Inductance.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/147_13.2%20Mutual%20Inductance.md) — Alternatively,
-- [148_13.3 Energy in a Coupled Circuit.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/148_13.3%20Energy%20in%20a%20Coupled%20Circuit.md) — Figure 13.15
-- [149_13.4 Linear Transformers.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/149_13.4%20Linear%20Transformers.md) — 13.4 Linear Transformers
-- [150_13.5 Ideal Transformers.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/150_13.5%20Ideal%20Transformers.md) — 13.5 Ideal Transformers
-- [151_13.6 Ideal Autotransformers.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/151_13.6%20Ideal%20Autotransformers.md) — Figure 13.42
-- [152_13.7 Three-Phase Transformers.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/152_13.7%20Three-Phase%20Transformers.md) — 13.7 Three-Phase Transformers
-- [153_13.8 PSpice Analysis of Magnetically Coupled Circuits.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/153_13.8%20PSpice%20Analysis%20of%20Magnetically%20Coupled%20Circuits.md) — 13.8 PSpice Analysis of Magnetically Coupled Circuits
-- [154_13.9 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/154_13.9%20Applications.md) — 13.9 Applications
-- [155_13.10 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/155_13.10%20Summary.md) — Review Questions
-- [156_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/156_Review%20Questions.md) — Problems1
-- [157_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/157_Comprehensive%20Problems.md) — Comprehensive Problems
-- [158_Chapter 14 - Frequency Response.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/158_Chapter%2014%20-%20Frequency%20Response.md) — Frequency 14 Response
-- [159_14.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/159_14.1%20Introduction.md) — 14.1 Introduction
-- [160_14.2 Transfer Function.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/160_14.2%20Transfer%20Function.md) — 14.2 Transfer Function
-- [161_14.3 The Decibel Scale.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/161_14.3%20The%20Decibel%20Scale.md) — 14.3 The Decibel Scale
-- [162_14.4 Bode Plots.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/162_14.4%20Bode%20Plots.md) — 14.4 Bode Plots
-- [163_14.5 Series Resonance.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/163_14.5%20Series%20Resonance.md) — 14.5 Series Resonance
-- [164_14.6 Parallel Resonance.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/164_14.6%20Parallel%20Resonance.md) — Solution:
-- [165_14.7 Passive Filters.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/165_14.7%20Passive%20Filters.md) — Figure 14.29 14.7 Passive Filters For Practice Prob. 14.9
-- [166_14.8 Active Filters.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/166_14.8%20Active%20Filters.md) — 14.8 Active Filters
-- [167_14.9 Scaling.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/167_14.9%20Scaling.md) — 14.9 Scaling
-- [168_14.10 Frequency Response Using PSpice.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/168_14.10%20Frequency%20Response%20Using%20PSpice.md) — 14.10 Frequency Response Using PSpice
-- [169_14.11 Computation Using MATLAB.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/169_14.11%20Computation%20Using%20MATLAB.md) — 14.11 Computation Using MATLAB
-- [170_14.12 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/170_14.12%20Applications.md) — 14.12 Applications
-- [171_14.13 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/171_14.13%20Summary.md) — 14.13 Summary
-- [172_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/172_Review%20Questions.md) — Problems
-- [173_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/173_Problems.md) — Section 14.5 Series Resonance
-- [174_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/174_Comprehensive%20Problems.md) — Comprehensive Problems
-- [175_PART 3 - Advanced Circuit Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/175_PART%203%20-%20Advanced%20Circuit%20Analysis.md) — Advanced Circuit Analysis
-- [176_Chapter 15 - Introduction to the Laplace Transform.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/176_Chapter%2015%20-%20Introduction%20to%20the%20Laplace%20Transform.md) — Introduction to the Laplace Transform
-- [177_15.2 Definition of the Laplace Transform.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/177_15.2%20Definition%20of%20the%20Laplace%20Transform.md) — 15.2 Definition of the Laplace Transform
-- [178_15.3 Properties of the Laplace Transform.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/178_15.3%20Properties%20of%20the%20Laplace%20Transform.md) — 15.3 Properties of the Laplace Transform
-- [179_15.4 The Inverse Laplace Transform.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/179_15.4%20The%20Inverse%20Laplace%20Transform.md) — 15.4.1 Simple Poles
-- [180_15.5 The Convolution Integral.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/180_15.5%20The%20Convolution%20Integral.md) — 15.5 The Convolution Integral
-- [181_15.6 Application to Integrodifferential Equations.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/181_15.6%20Application%20to%20Integrodifferential%20Equations.md) — 15.6 Application to Integrodifferential Equations
-- [182_15.7 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/182_15.7%20Summary.md) — Problems
-- [183_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/183_Problems.md) — Figure 15.26
-- [184_Chapter 16 - Applications of the Laplace Transform.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/184_Chapter%2016%20-%20Applications%20of%20the%20Laplace%20Transform.md) — Applications of the Laplace Transform
-- [185_16.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/185_16.1%20Introduction.md) — 16.1 Introduction
-- [186_16.2 Circuit Element Models.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/186_16.2%20Circuit%20Element%20Models.md) — Figure 16.3
-- [187_16.3 Circuit Analysis.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/187_16.3%20Circuit%20Analysis.md) — Solution:
-- [188_16.4 Transfer Functions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/188_16.4%20Transfer%20Functions.md) — Solution:
-- [189_16.5 State Variables.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/189_16.5%20State%20Variables.md) — For Practice Prob. 16.9. 16.5 State Variables
-- [190_16.6 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/190_16.6%20Applications.md) — 16.6.2 Network Synthesis
-- [191_16.7 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/191_16.7%20Summary.md) — 16.7 Summary
-- [192_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/192_Review%20Questions.md) — Review Questions
-- [193_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/193_Problems.md) — Problems
-- [194_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/194_Comprehensive%20Problems.md) — Comprehensive Problems
-- [195_Chapter 17 - The Fourier Series.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/195_Chapter%2017%20-%20The%20Fourier%20Series.md) — The Fourier Series
-- [196_17.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/196_17.1%20Introduction.md) — 17.1 Introduction
-- [197_17.2 Trigonometric Fourier Series.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/197_17.2%20Trigonometric%20Fourier%20Series.md) — Solution:
-- [198_17.3 Symmetry Considerations.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/198_17.3%20Symmetry%20Considerations.md) — 17.3 Symmetry Considerations
-- [199_17.4 Circuit Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/199_17.4%20Circuit%20Applications.md) — Solution:
-- [200_17.5 Average Power and RMS Values.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/200_17.5%20Average%20Power%20and%20RMS%20Values.md) — Example 17.8
-- [201_17.6 Exponential Fourier Series.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/201_17.6%20Exponential%20Fourier%20Series.md) — 17.6 Exponential Fourier Series
-- [202_17.7 Fourier Analysis with PSpice.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/202_17.7%20Fourier%20Analysis%20with%20PSpice.md) — 17.7 Fourier Analysis with PSpice
-- [203_17.8 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/203_17.8%20Applications.md) — 17.8 Applications
-- [204_17.9 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/204_17.9%20Summary.md) — 17.9 Summary
-- [205_Review Questions.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/205_Review%20Questions.md) — - 17.9 When the periodic voltage 2 + 6 sin ω0t is applied to a 1-Ω resistor, the integer clo
-- [206_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/206_Problems.md) — Problems
-- [207_Comprehensive Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/207_Comprehensive%20Problems.md) — Comprehensive Problems
-- [208_Chapter 18 - Fourier Transform.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/208_Chapter%2018%20-%20Fourier%20Transform.md) — Fourier Transform
-- [209_18.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/209_18.1%20Introduction.md) — 18.1 Introduction
-- [210_18.2 Definition of the Fourier Transform.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/210_18.2%20Definition%20of%20the%20Fourier%20Transform.md) — 18.2 Definition of the Fourier Transform
-- [211_18.3 Properties of the Fourier Transform.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/211_18.3%20Properties%20of%20the%20Fourier%20Transform.md) — Properties of the Fourier transform.
-- [212_18.4 Circuit Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/212_18.4%20Circuit%20Applications.md) — Section 18.4 Circuit Applications
-- [213_18.5 Parseval's Theorem.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/213_18.5%20Parseval's%20Theorem.md) — Section 18.5 Parseval's Theorem
-- [214_18.7 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/214_18.7%20Applications.md) — Section 18.6 Applications
-- [215_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/215_Problems.md) — Comprehensive Problems
-- [216_Chapter 19 - Two-Port Networks.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/216_Chapter%2019%20-%20Two-Port%20Networks.md) — Two-Port Networks
-- [217_19.1 Introduction.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/217_19.1%20Introduction.md) — 19.1 Introduction
-- [218_19.2 Impedance Parameters.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/218_19.2%20Impedance%20Parameters.md) — Solution:
-- [219_19.3 Admittance Parameters.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/219_19.3%20Admittance%20Parameters.md) — Figure 19.12
-- [220_19.4 Hybrid Parameters.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/220_19.4%20Hybrid%20Parameters.md) — 19.4 Hybrid Parameters
-- [221_19.5 Transmission Parameters.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/221_19.5%20Transmission%20Parameters.md) — 19.5 Transmission Parameters
-- [222_19.6 Relationships Between Parameters.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/222_19.6%20Relationships%20Between%20Parameters.md) — Solution:
-- [223_19.7 Interconnection of Networks.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/223_19.7%20Interconnection%20of%20Networks.md) — 19.7 Interconnection of Networks
-- [224_19.8 Computing Two-Port Parameters Using PSpice.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/224_19.8%20Computing%20Two-Port%20Parameters%20Using%20PSpice.md) — 19.8 Computing Two-Port Parameters Using PSpice
-- [225_19.9 Applications.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/225_19.9%20Applications.md) — 19.9.2 Ladder Network Synthesis
-- [226_19.10 Summary.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/226_19.10%20Summary.md) — 19.10 Summary
-- [227_Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/227_Problems.md) — Problems
-- [228_Comprehensive Problem.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/228_Comprehensive%20Problem.md) — Comprehensive Problem
-- [229_Appendix A - Simultaneous Equations and Matrix Inversion.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/229_Appendix%20A%20-%20Simultaneous%20Equations%20and%20Matrix%20Inversion.md) — In summary:
-- [230_Appendix B - Complex Numbers.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/230_Appendix%20B%20-%20Complex%20Numbers.md) — Solution:
-- [231_Appendix C - Mathematical Formulas.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/231_Appendix%20C%20-%20Mathematical%20Formulas.md) — C.3 Hyperbolic Functions
-- [232_Appendix D - Answers to Odd-Numbered Problems.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/232_Appendix%20D%20-%20Answers%20to%20Odd-Numbered%20Problems.md) — Figure D.2
-- [233_Selected Bibliography.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/233_Selected%20Bibliography.md) — Selected Bibliography
-- [234_Index.md](https://huggingface.co/datasets/learner20011/CloverTexts-Data/raw/main/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/234_Index.md) — Index

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/001_Cover.md

@@ -1,828 +0,0 @@
-# LINEAR SYSTEMS AND SIGNALS
-
-### **THE OXFORD SERIES IN ELECTRICAL AND COMPUTER ENGINEERING**
-
-**Adel S. Sedra,** Series Editor
-
-Allen and Holberg, *CMOS Analog Circuit Design, 3rd edition* Boncelet, *Probability, Statistics, and Random Signals* Bobrow, *Elementary Linear Circuit Analysis, 2nd edition* Bobrow, *Fundamentals of Electrical Engineering, 2nd edition* Campbell, *Fabrication Engineering at the Micro- and Nanoscale, 4th edition* Chen, *Digital Signal Processing* Chen, *Linear System Theory and Design, 4th edition* Chen, *Signals and Systems, 3rd edition* Comer, *Digital Logic and State Machine Design, 3rd edition* Comer, *Microprocessor-Based System Design* Cooper and McGillem, *Probabilistic Methods of Signal and System Analysis, 3rd edition* Dimitrijev, *Principles of Semiconductor Device, 2nd edition* Dimitrijev, *Understanding Semiconductor Devices* Fortney, *Principles of Electronics: Analog & Digital* Franco, *Electric Circuits Fundamentals* Ghausi, *Electronic Devices and Circuits: Discrete and Integrated* Guru and Hiziroglu, ˘ *Electric Machinery and Transformers, 3rd edition* Houts, *Signal Analysis in Linear Systems* Jones, *Introduction to Optical Fiber Communication Systems* Krein, *Elements of Power Electronics, 2nd Edition* Kuo, *Digital Control Systems, 3rd edition* Lathi and Green, *Linear Systems and Signals, 3rd edition* Lathi and Ding, *Modern Digital and Analog Communication Systems, 5th edition* Lathi, *Signal Processing and Linear Systems* Martin, *Digital Integrated Circuit Design* Miner, *Lines and Electromagnetic Fields for Engineers* Mitra, *Signals and Systems* Parhami, *Computer Architecture* Parhami, *Computer Arithmetic, 2nd edition* Roberts and Sedra, *SPICE, 2nd edition* Roberts, Taenzler, and Burns, *An Introduction to Mixed-Signal IC Test and Measurement, 2nd edition* Roulston, *An Introduction to the Physics of Semiconductor Devices* Sadiku, *Elements of Electromagnetics, 7th edition* Santina, Stubberud, and Hostetter, *Digital Control System Design, 2nd edition* Sarma, *Introduction to Electrical Engineering* Schaumann, Xiao, and Van Valkenburg, *Design of Analog Filters, 3rd edition* Schwarz and Oldham, *Electrical Engineering: An Introduction, 2nd edition* Sedra and Smith, *Microelectronic Circuits, 7th edition* Stefani, Shahian, Savant, and Hostetter, *Design of Feedback Control Systems, 4th edition* Tsividis, *Operation and Modeling of the MOS Transistor, 3rd edition* Van Valkenburg, *Analog Filter Design* Warner and Grung, *Semiconductor Device Electronics* Wolovich, *Automatic Control Systems* Yariv and Yeh, Photonics: *Optical Electronics in Modern Communications, 6th edition* Zak, ˙ *Systems and Control*
-
-# LINEAR SYSTEMS AND SIGNALS
-
-THIRD EDITION
-
-**B. P. Lathi and R. A. Green**
-
-New York Oxford OXFORD UNIVERSITY PRESS 2018
-
-Oxford University Press is a department of the University of Oxford. It furthers the University's objective of excellence in research, scholarship, and education by publishing worldwide.
-
-Oxford New York Auckland Cape Town Dar es Salaam Hong Kong Karachi Kuala Lumpur Madrid Melbourne Mexico City Nairobi New Delhi Shanghai Taipei Toronto
-
-With offices in Argentina Austria Brazil Chile Czech Republic France Greece Guatemala Hungary Italy Japan Poland Portugal Singapore South Korea Switzerland Thailand Turkey Ukraine Vietnam
-
-Copyright c 2018 by Oxford University Press
-
-For titles covered by Section 112 of the US Higher Education Opportunity Act, please visit [www.oup.com/us/he](http://www.oup.com/us/he) for the latest information about pricing and alternate formats.
-
-Published by Oxford University Press. 198 Madison Avenue, New York, NY 10016 <http://www.oup.com>
-
-Oxford is a registered trademark of Oxford University Press.
-
-All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of Oxford University Press.
-
-Library of Congress Cataloging-in-Publication Data Names: Lathi, B. P. (Bhagwandas Pannalal), author. | Green, R. A. (Roger A.), author. Title: Linear systems and signals / B.P. Lathi and R.A. Green. Description: Third Edition. | New York : Oxford University Press, [2018] | Series: The Oxford Series in Electrical and Computer Engineering Identifiers: LCCN 2017034962 | ISBN 9780190200176 (hardcover : acid-free paper) Subjects: LCSH: Signal processing–Mathematics. | System analysis. | Linear time invariant systems. | Digital filters (Mathematics) Classification: LCC TK5102.5 L298 2017 | DDC 621.382/2–dc23 LC record available at<https://lccn.loc.gov/2017034962>
-
-ISBN 978–0–19–020017–6
-
-Printing number: 9 8 7 6 5 4 3 2 1
-
-Printed by R.R. Donnelly in the United States of America
-
-# <span id="page-6-0"></span>**CONTENTS**
-
-[PREFACE](#page-16-0) xv
-
-### B BACKGROUND
-
-- [B.1 Complex Numbers](#page-20-0) 1
-  - [B.1-1 A Historical Note](#page-20-0) 1
-  - [B.1-2 Algebra of Complex Numbers](#page-24-0) 5
-- [B.2 Sinusoids](#page-35-0) 16
-  - [B.2-1 Addition of Sinusoids](#page-37-0) 18
-  - [B.2-2 Sinusoids in Terms of Exponentials](#page-39-0) 20
-- [B.3 Sketching Signals](#page-39-0) 20
-  - [B.3-1 Monotonic Exponentials](#page-39-0) 20
-  - [B.3-2 The Exponentially Varying Sinusoid](#page-41-0) 22
-- [B.4 Cramer's Rule](#page-42-0) 23
-- [B.5 Partial Fraction Expansion](#page-44-0) 25
-  - [B.5-1 Method of Clearing Fractions](#page-45-0) 26
-  - [B.5-2 The Heaviside "Cover-Up" Method](#page-46-0) 27
-  - [B.5-3 Repeated Factors of](#page-50-0) *Q*(*x*) 31
-  - [B.5-4 A Combination of Heaviside "Cover-Up" and Clearing Fractions](#page-51-0) 32
-  - [B.5-5 Improper](#page-53-0) *F*(*x*) with *m* = *n* 34
-  - [B.5-6 Modified Partial Fractions](#page-54-0) 35
-- [B.6 Vectors and Matrices](#page-55-0) 36
-  - [B.6-1 Some Definitions and Properties](#page-56-0) 37
-  - [B.6-2 Matrix Algebra](#page-57-0) 38
-- [B.7 MATLAB: Elementary Operations](#page-61-0) 42
-  - [B.7-1 MATLAB Overview](#page-61-0) 42
-  - [B.7-2 Calculator Operations](#page-62-0) 43
-  - [B.7-3 Vector Operations](#page-64-0) 45
-  - [B.7-4 Simple Plotting](#page-65-0) 46
-  - [B.7-5 Element-by-Element Operations](#page-67-0) 48
-  - [B.7-6 Matrix Operations](#page-68-0) 49
-  - [B.7-7 Partial Fraction Expansions](#page-72-0) 53
-- [B.8 Appendix: Useful Mathematical Formulas](#page-73-0) 54
-  - [B.8-1 Some Useful Constants](#page-73-0) 54
-
-- <span id="page-7-0"></span>[B.8-2 Complex Numbers](#page-73-0) 54
-- [B.8-3 Sums](#page-73-0) 54
-- [B.8-4 Taylor and Maclaurin Series](#page-74-0) 55
-- [B.8-5 Power Series](#page-74-0) 55
-- [B.8-6 Trigonometric Identities](#page-74-0) 55
-- [B.8-7 Common Derivative Formulas](#page-75-0) 56
-- [B.8-8 Indefinite Integrals](#page-76-0) 57
-- [B.8-9 L'Hôpital's Rule](#page-77-0) 58
-- [B.8-10 Solution of Quadratic and Cubic Equations](#page-77-0) 58
-- *[References](#page-77-0)* 58 *[Problems](#page-78-0)* 59
-
-### 1 SIGNALS AND SYSTEMS
-
-- [1.1 Size of a Signal](#page-83-0) 64
-  - [1.1-1 Signal Energy](#page-84-0) 65
-  - [1.1-2 Signal Power](#page-84-0) 65
-- [1.2 Some Useful Signal Operations](#page-90-0) 71
-  - [1.2-1 Time Shifting](#page-90-0) 71
-  - [1.2-2 Time Scaling](#page-92-0) 73
-  - [1.2-3 Time Reversal](#page-95-0) 76
-  - [1.2-4 Combined Operations](#page-96-0) 77
-- [1.3 Classification of Signals](#page-97-0) 78
-  - [1.3-1 Continuous-Time and Discrete-Time Signals](#page-97-0) 78
-  - [1.3-2 Analog and Digital Signals](#page-97-0) 78
-  - [1.3-3 Periodic and Aperiodic Signals](#page-98-0) 79
-  - [1.3-4 Energy and Power Signals](#page-101-0) 82
-  - [1.3-5 Deterministic and Random Signals](#page-101-0) 82
-- [1.4 Some Useful Signal Models](#page-101-0) 82
-  - [1.4-1 The Unit Step Function](#page-102-0) *u*(*t*) 83
-  - [1.4-2 The Unit Impulse Function](#page-105-0) δ(*t*) 86
-  - [1.4-3 The Exponential Function](#page-108-0) *est* 89
-- [1.5 Even and Odd Functions](#page-111-0) 92
-  - [1.5-1 Some Properties of Even and Odd Functions](#page-111-0) 92
-  - [1.5-2 Even and Odd Components of a Signal](#page-112-0) 93
-- [1.6 Systems](#page-114-0) 95
-- [1.7 Classification of Systems](#page-116-0) 97
-  - [1.7-1 Linear and Nonlinear Systems](#page-116-0) 97
-  - [1.7-2 Time-Invariant and Time-Varying Systems](#page-121-0) 102
-  - [1.7-3 Instantaneous and Dynamic Systems](#page-122-0) 103
-  - [1.7-4 Causal and Noncausal Systems](#page-123-0) 104
-  - [1.7-5 Continuous-Time and Discrete-Time Systems](#page-126-0) 107
-  - [1.7-6 Analog and Digital Systems](#page-128-0) 109
-  - [1.7-7 Invertible and Noninvertible Systems](#page-128-0) 109
-  - [1.7-8 Stable and Unstable Systems](#page-129-0) 110
-
-- <span id="page-8-0"></span>[1.8 System Model: Input–Output Description](#page-130-0) 111
-  - [1.8-1 Electrical Systems](#page-130-0) 111
-  - [1.8-2 Mechanical Systems](#page-133-0) 114
-  - [1.8-3 Electromechanical Systems](#page-137-0) 118
-- [1.9 Internal and External Descriptions of a System](#page-138-0) 119
-- [1.10 Internal Description: The State-Space Description](#page-140-0) 121
-- [1.11 MATLAB: Working with Functions](#page-145-0) 126
-  - [1.11-1 Anonymous Functions](#page-145-0) 126
-  - [1.11-2 Relational Operators and the Unit Step Function](#page-147-0) 128
-  - [1.11-3 Visualizing Operations on the Independent Variable](#page-149-0) 130
-  - [1.11-4 Numerical Integration and Estimating Signal Energy](#page-150-0) 131
-- [1.12 Summary](#page-152-0) 133
-
-*[References](#page-154-0)* 135 *[Problems](#page-155-0)* 136
-
-### 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS
-
-- [2.1 Introduction](#page-169-0) 150
-- [2.2 System Response to Internal Conditions: The Zero-Input Response](#page-170-0) 151 [2.2-1 Some Insights into the Zero-Input Behavior of a System](#page-180-0) 161
-- [2.3 The Unit Impulse Response](#page-182-0) *h*(*t*) 163
-- [2.4 System Response to External Input: The Zero-State Response](#page-187-0) 168
-  - [2.4-1 The Convolution Integral](#page-189-0) 170
-  - [2.4-2 Graphical Understanding of Convolution Operation](#page-197-0) 178
-  - [2.4-3 Interconnected Systems](#page-209-0) 190
-  - [2.4-4 A Very Special Function for LTIC Systems:](#page-212-0)
-    - The Everlasting Exponential *est* 193
-  - [2.4-5 Total Response](#page-214-0) 195
-- [2.5 System Stability](#page-215-0) 196
-  - [2.5-1 External \(BIBO\) Stability](#page-215-0) 196
-  - [2.5-2 Internal \(Asymptotic\) Stability](#page-217-0) 198
-  - [2.5-3 Relationship Between BIBO and Asymptotic Stability](#page-218-0) 199
-- [2.6 Intuitive Insights into System Behavior](#page-222-0) 203
-  - [2.6-1 Dependence of System Behavior on Characteristic Modes](#page-222-0) 203
-  - [2.6-2 Response Time of a System: The System Time Constant](#page-224-0) 205
-  - [2.6-3 Time Constant and Rise Time of a System](#page-225-0) 206
-  - [2.6-4 Time Constant and Filtering](#page-226-0) 207
-  - [2.6-5 Time Constant and Pulse Dispersion \(Spreading\)](#page-228-0) 209
-  - [2.6-6 Time Constant and Rate of Information Transmission](#page-228-0) 209
-  - [2.6-7 The Resonance Phenomenon](#page-229-0) 210
-- [2.7 MATLAB: M-Files](#page-231-0) 212
-  - [2.7-1 Script M-Files](#page-232-0) 213
-  - [2.7-2 Function M-Files](#page-233-0) 214
-
-- [2.7-3 For-Loops](#page-234-0) 215
-- [2.7-4 Graphical Understanding of Convolution](#page-236-0) 217
-- <span id="page-9-0"></span>[2.8 Appendix: Determining the Impulse Response](#page-239-0) 220
-- [2.9 Summary](#page-240-0) 221
-
-*[References](#page-242-0)* 223 *[Problems](#page-242-0)* 223
-
-### 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS
-
-- [3.1 Introduction](#page-256-0) 237
-  - [3.1-1 Size of a Discrete-Time Signal](#page-257-0) 238
-- [3.2 Useful Signal Operations](#page-259-0) 240
-- [3.3 Some Useful Discrete-Time Signal Models](#page-264-0) 245
-  - [3.3-1 Discrete-Time Impulse Function](#page-264-0) δ[*n*] 245
-  - [3.3-2 Discrete-Time Unit Step Function](#page-265-0) *u*[*n*] 246
-  - [3.3-3 Discrete-Time Exponential](#page-266-0) γ *<sup>n</sup>* 247
-  - [3.3-4 Discrete-Time Sinusoid cos](#page-270-0)(*n*+θ ) 251
-  - [3.3-5 Discrete-Time Complex Exponential](#page-271-0) *ej<sup>n</sup>* 252
-- [3.4 Examples of Discrete-Time Systems](#page-272-0) 253
-  - [3.4-1 Classification of Discrete-Time Systems](#page-281-0) 262
-- [3.5 Discrete-Time System Equations](#page-284-0) 265
-
-[3.5-1 Recursive \(Iterative\) Solution of Difference Equation](#page-285-0) 266
-
-- [3.6 System Response to Internal Conditions: The Zero-Input Response](#page-289-0) 270
-- [3.7 The Unit Impulse Response](#page-296-0) *h*[*n*] 277
-  - [3.7-1 The Closed-Form Solution of](#page-297-0) *h*[*n*] 278
-- [3.8 System Response to External Input: The Zero-State Response](#page-299-0) 280
-  - [3.8-1 Graphical Procedure for the Convolution Sum](#page-307-0) 288
-  - [3.8-2 Interconnected Systems](#page-313-0) 294
-  - [3.8-3 Total Response](#page-316-0) 297
-- [3.9 System Stability](#page-317-0) 298
-  - [3.9-1 External \(BIBO\) Stability](#page-317-0) 298
-  - [3.9-2 Internal \(Asymptotic\) Stability](#page-318-0) 299
-  - [3.9-3 Relationship Between BIBO and Asymptotic Stability](#page-320-0) 301
-- [3.10 Intuitive Insights into System Behavior](#page-324-0) 305
-- [3.11 MATLAB: Discrete-Time Signals and Systems](#page-325-0) 306
-  - [3.11-1 Discrete-Time Functions and Stem Plots](#page-325-0) 306
-  - [3.11-2 System Responses Through Filtering](#page-327-0) 308
-  - [3.11-3 A Custom Filter Function](#page-329-0) 310
-  - [3.11-4 Discrete-Time Convolution](#page-330-0) 311
-- [3.12 Appendix: Impulse Response for a Special Case](#page-332-0) 313
-- [3.13 Summary](#page-332-0) 313
-
-*[Problems](#page-333-0)* 314
-
-### <span id="page-10-0"></span>4 CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM
-
-- [4.1 The Laplace Transform](#page-349-0) 330
-  - [4.1-1 Finding the Inverse Transform](#page-357-0) 338
-- [4.2 Some Properties of the Laplace Transform](#page-368-0) 349
-  - [4.2-1 Time Shifting](#page-368-0) 349
-  - [4.2-2 Frequency Shifting](#page-372-0) 353
-  - [4.2-3 The Time-Differentiation Property](#page-373-0) 354
-  - [4.2-4 The Time-Integration Property](#page-375-0) 356
-  - [4.2-5 The Scaling Property](#page-376-0) 357
-  - [4.2-6 Time Convolution and Frequency Convolution](#page-376-0) 357
-- [4.3 Solution of Differential and Integro-Differential Equations](#page-379-0) 360
-  - [4.3-1 Comments on Initial Conditions at 0](#page-382-0)<sup>−</sup> and at 0<sup>+</sup> 363
-  - [4.3-2 Zero-State Response](#page-385-0) 366
-  - [4.3-3 Stability](#page-390-0) 371
-  - [4.3-4 Inverse Systems](#page-392-0) 373
-- [4.4 Analysis of Electrical Networks: The Transformed Network](#page-392-0) 373
-  - [4.4-1 Analysis of Active Circuits](#page-401-0) 382
-- [4.5 Block Diagrams](#page-405-0) 386
-- [4.6 System Realization](#page-407-0) 388
-  - [4.6-1 Direct Form I Realization](#page-408-0) 389
-  - [4.6-2 Direct Form II Realization](#page-409-0) 390
-  - [4.6-3 Cascade and Parallel Realizations](#page-412-0) 393
-  - [4.6-4 Transposed Realization](#page-415-0) 396
-  - [4.6-5 Using Operational Amplifiers for System Realization](#page-418-0) 399
-- [4.7 Application to Feedback and Controls](#page-423-0) 404
-  - [4.7-1 Analysis of a Simple Control System](#page-425-0) 406
-- [4.8 Frequency Response of an LTIC System](#page-431-0) 412
-  - [4.8-1 Steady-State Response to Causal Sinusoidal Inputs](#page-437-0) 418
-- [4.9 Bode Plots](#page-438-0) 419
-  - [4.9-1 Constant](#page-441-0) *Ka*1*a*2/*b*1*b*<sup>3</sup> 422
-  - [4.9-2 Pole \(or Zero\) at the Origin](#page-441-0) 422
-  - [4.9-3 First-Order Pole \(or Zero\)](#page-443-0) 424
-  - [4.9-4 Second-Order Pole \(or Zero\)](#page-445-0) 426
-  - [4.9-5 The Transfer Function from the Frequency Response](#page-454-0) 435
-- [4.10 Filter Design by Placement of Poles and Zeros of](#page-455-0) *H*(*s*) 436
-  - [4.10-1 Dependence of Frequency Response on Poles](#page-455-0) and Zeros of *H*(*s*) 436
-  - [4.10-2 Lowpass Filters](#page-458-0) 439
-  - [4.10-3 Bandpass Filters](#page-460-0) 441
-  - [4.10-4 Notch \(Bandstop\) Filters](#page-460-0) 441
-  - [4.10-5 Practical Filters and Their Specifications](#page-463-0) 444
-- [4.11 The Bilateral Laplace Transform](#page-464-0) 445
-
-- [4.11-1 Properties of the Bilateral Laplace Transform](#page-470-0) 451
-- [4.11-2 Using the Bilateral Transform for Linear System Analysis](#page-471-0) 452
-- <span id="page-11-0"></span>[4.12 MATLAB: Continuous-Time Filters](#page-474-0) 455
-  - [4.12-1 Frequency Response and Polynomial Evaluation](#page-475-0) 456
-  - [4.12-2 Butterworth Filters and the](#page-478-0) Find Command 459
-  - [4.12-3 Using Cascaded Second-Order Sections for Butterworth](#page-480-0) Filter Realization 461
-  - [4.12-4 Chebyshev Filters](#page-482-0) 463
-- [4.13 Summary](#page-485-0) 466
-
-*[References](#page-487-0)* 468 *[Problems](#page-487-0)* 468
-
-### 5 DISCRETE-TIME SYSTEM ANALYSIS USING THE *z*-TRANSFORM
-
-- 5.1 The *z*[-Transform](#page-507-0) 488
-  - [5.1-1 Inverse Transform by Partial Fraction Expansion and Tables](#page-514-0) 495
-  - 5.1-2 Inverse *z*[-Transform by Power Series Expansion](#page-518-0) 499
-- [5.2 Some Properties of the](#page-520-0) *z*-Transform 501
-  - [5.2-1 Time-Shifting Properties](#page-520-0) 501
-  - 5.2-2 *z*[-Domain Scaling Property \(Multiplication by](#page-524-0) γ *<sup>n</sup>*) 505
-  - 5.2-3 *z*[-Domain Differentiation Property \(Multiplication by](#page-525-0) *n*) 506
-  - [5.2-4 Time-Reversal Property](#page-525-0) 506
-  - [5.2-5 Convolution Property](#page-526-0) 507
-- 5.3 *z*[-Transform Solution of Linear Difference Equations](#page-529-0) 510
-  - [5.3-1 Zero-State Response of LTID Systems: The Transfer Function](#page-533-0) 514
-  - [5.3-2 Stability](#page-537-0) 518
-  - [5.3-3 Inverse Systems](#page-538-0) 519
-- [5.4 System Realization](#page-538-0) 519
-- 5.5 Frequency Response of Discrete-Time Systems [526](#page-545-0)
-  - [5.5-1 The Periodic Nature of Frequency Response](#page-551-0) 532
-  - [5.5-2 Aliasing and Sampling Rate](#page-555-0) 536
-- [5.6 Frequency Response from Pole-Zero Locations](#page-557-0) 538
-- [5.7 Digital Processing of Analog Signals](#page-566-0) 547
-- [5.8 The Bilateral](#page-573-0) *z*-Transform 554
-  - [5.8-1 Properties of the Bilateral](#page-578-0) *z*-Transform 559
-  - 5.8-2 Using the Bilateral *z*[-Transform for Analysis of LTID Systems](#page-579-0) 560
-- [5.9 Connecting the Laplace and](#page-582-0) *z*-Transforms 563
-- [5.10 MATLAB: Discrete-Time IIR Filters](#page-584-0) 565
-  - [5.10-1 Frequency Response and Pole-Zero Plots](#page-585-0) 566
-  - [5.10-2 Transformation Basics](#page-586-0) 567
-  - [5.10-3 Transformation by First-Order Backward Difference](#page-587-0) 568
-  - [5.10-4 Bilinear Transformation](#page-588-0) 569
-  - [5.10-5 Bilinear Transformation with Prewarping](#page-589-0) 570
-  - [5.10-6 Example: Butterworth Filter Transformation](#page-590-0) 571
-
-[5.10-7 Problems Finding Polynomial Roots](#page-591-0) 572
-
-[5.10-8 Using Cascaded Second-Order Sections to Improve Design](#page-591-0) 572
-
-<span id="page-12-0"></span>[5.11 Summary](#page-593-0) 574
-
-*[References](#page-594-0)* 575 *[Problems](#page-594-0)* 575
-
-### 6 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER SERIES
-
-- [6.1 Periodic Signal Representation by Trigonometric Fourier Series](#page-612-0) 593
-  - [6.1-1 The Fourier Spectrum](#page-617-0) 598
-  - [6.1-2 The Effect of Symmetry](#page-626-0) 607
-  - [6.1-3 Determining the Fundamental Frequency and Period 609](#page-628-0)
-- [6.2 Existence and Convergence of the Fourier Series](#page-631-0) 612
-  - [6.2-1 Convergence of a Series](#page-632-0) 613
-  - [6.2-2 The Role of Amplitude and Phase Spectra in Waveshaping](#page-634-0) 615
-- [6.3 Exponential Fourier Series](#page-640-0) 621
-  - [6.3-1 Exponential Fourier Spectra](#page-643-0) 624
-  - [6.3-2 Parseval's Theorem](#page-651-0) 632
-  - [6.3-3 Properties of the Fourier Series](#page-654-0) 635
-- [6.4 LTIC System Response to Periodic Inputs](#page-656-0) 637
-- [6.5 Generalized Fourier Series: Signals as Vectors](#page-660-0) 641
-  - [6.5-1 Component of a Vector](#page-661-0) 642
-  - [6.5-2 Signal Comparison and Component of a Signal](#page-662-0) 643
-  - [6.5-3 Extension to Complex Signals](#page-664-0) 645
-  - [6.5-4 Signal Representation by an Orthogonal Signal Set](#page-666-0) 647
-- [6.6 Numerical Computation of](#page-678-0) *Dn* 659
-- [6.7 MATLAB: Fourier Series Applications](#page-680-0) 661
-  - [6.7-1 Periodic Functions and the Gibbs Phenomenon](#page-680-0) 661
-  - [6.7-2 Optimization and Phase Spectra](#page-683-0) 664
-- [6.8 Summary](#page-686-0) 667 *[References](#page-687-0)* 668 *[Problems](#page-688-0)* 669
-
-## 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM
-
-- [7.1 Aperiodic Signal Representation by the Fourier Integral](#page-699-0) 680 [7.1-1 Physical Appreciation of the Fourier Transform](#page-706-0) 687
-- [7.2 Transforms of Some Useful Functions](#page-708-0) 689
-  - [7.2-1 Connection Between the Fourier and Laplace Transforms](#page-719-0) 700
-- [7.3 Some Properties of the Fourier Transform](#page-720-0) 701
-- [7.4 Signal Transmission Through LTIC Systems](#page-740-0) 721
-  - [7.4-1 Signal Distortion During Transmission](#page-742-0) 723
-  - [7.4-2 Bandpass Systems and Group Delay](#page-745-0) 726
-
-### <span id="page-13-0"></span>xii Contents
-
-- [7.5 Ideal and Practical Filters](#page-749-0) 730
-- [7.6 Signal Energy](#page-752-0) 733
-- [7.7 Application to Communications: Amplitude Modulation](#page-755-0) 736
-  - [7.7-1 Double-Sideband, Suppressed-Carrier \(DSB-SC\) Modulation](#page-756-0) 737
-  - [7.7-2 Amplitude Modulation \(AM\)](#page-761-0) 742
-  - [7.7-3 Single-Sideband Modulation \(SSB\)](#page-765-0) 746
-  - [7.7-4 Frequency-Division Multiplexing](#page-768-0) 749
-- [7.8 Data Truncation: Window Functions](#page-768-0) 749
-  - [7.8-1 Using Windows in Filter Design](#page-774-0) 755
-- [7.9 MATLAB: Fourier Transform Topics](#page-774-0) 755
-  - [7.9-1 The Sinc Function and the Scaling Property](#page-776-0) 757
-  - [7.9-2 Parseval's Theorem and Essential Bandwidth](#page-777-0) 758
-  - [7.9-3 Spectral Sampling](#page-778-0) 759
-  - [7.9-4 Kaiser Window Functions](#page-779-0) 760
-- [7.10 Summary](#page-781-0) 762
-
-*[References](#page-782-0)* 763 *[Problems](#page-783-0)* 764
-
-### 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE
-
-- [8.1 The Sampling Theorem](#page-795-0) 776 [8.1-1 Practical Sampling](#page-800-0) 781
-- [8.2 Signal Reconstruction](#page-804-0) 785
-  - [8.2-1 Practical Difficulties in Signal Reconstruction](#page-807-0) 788
-  - [8.2-2 Some Applications of the Sampling Theorem](#page-815-0) 796
-- [8.3 Analog-to-Digital \(A/D\) Conversion](#page-818-0) 799
-- [8.4 Dual of Time Sampling: Spectral Sampling](#page-821-0) 802
-- [8.5 Numerical Computation of the Fourier Transform:](#page-824-0) The Discrete Fourier Transform 805
-  - [8.5-1 Some Properties of the DFT](#page-837-0) 818
-  - [8.5-2 Some Applications of the DFT](#page-839-0) 820
-- [8.6 The Fast Fourier Transform \(FFT\)](#page-843-0) 824
-- [8.7 MATLAB: The Discrete Fourier Transform](#page-846-0) 827
-  - [8.7-1 Computing the Discrete Fourier Transform](#page-846-0) 827
-  - [8.7-2 Improving the Picture with Zero Padding](#page-848-0) 829
-  - [8.7-3 Quantization](#page-850-0) 831
-- [8.8 Summary](#page-853-0) 834
-
-*[References](#page-854-0)* 835 *[Problems](#page-854-0)* 835
-
-### <span id="page-14-0"></span>9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS
-
-- [9.1 Discrete-Time Fourier Series \(DTFS\)](#page-864-0) 845
-  - [9.1-1 Periodic Signal Representation by Discrete-Time Fourier Series](#page-865-0) 846
-  - [9.1-2 Fourier Spectra of a Periodic Signal](#page-867-0) *x*[*n*] 848
-- [9.2 Aperiodic Signal Representation](#page-874-0)
-
-by Fourier Integral 855
-
-- [9.2-1 Nature of Fourier Spectra](#page-877-0) 858
-- [9.2-2 Connection Between the DTFT and the](#page-885-0) *z*-Transform 866
-- [9.3 Properties of the DTFT](#page-886-0) 867
-- [9.4 LTI Discrete-Time System Analysis by DTFT](#page-897-0) 878
-  - [9.4-1 Distortionless Transmission](#page-899-0) 880
-  - [9.4-2 Ideal and Practical Filters](#page-901-0) 882
-- [9.5 DTFT Connection with the CTFT](#page-902-0) 883
-  - [9.5-1 Use of DFT and FFT for Numerical Computation of the DTFT](#page-904-0) 885
-- [9.6 Generalization of the DTFT to the](#page-905-0) *z*-Transform 886
-- [9.7 MATLAB: Working with the DTFS and the DTFT](#page-908-0) 889
-  - [9.7-1 Computing the Discrete-Time Fourier Series](#page-908-0) 889
-  - [9.7-2 Measuring Code Performance](#page-910-0) 891
-  - [9.7-3 FIR Filter Design by Frequency Sampling](#page-911-0) 892
-- [9.8 Summary](#page-917-0) 898
-
-*[Reference](#page-917-0)* 898 *[Problems](#page-918-0)* 899
-
-### 10 STATE-SPACE ANALYSIS
-
-- [10.1 Mathematical Preliminaries](#page-928-0) 909
-  - [10.1-1 Derivatives and Integrals of a Matrix](#page-928-0) 909
-  - [10.1-2 The Characteristic Equation of a Matrix:](#page-929-0)
-    - The Cayley–Hamilton Theorem 910
-  - [10.1-3 Computation of an Exponential and a Power of a Matrix](#page-931-0) 912
-- [10.2 Introduction to State Space](#page-932-0) 913
-- [10.3 A Systematic Procedure to Determine State Equations](#page-935-0) 916
-  - [10.3-1 Electrical Circuits](#page-935-0) 916
-  - [10.3-2 State Equations from a Transfer Function](#page-938-0) 919
-- [10.4 Solution of State Equations](#page-945-0) 926
-  - [10.4-1 Laplace Transform Solution of State Equations](#page-946-0) 927
-  - [10.4-2 Time-Domain Solution of State Equations](#page-952-0) 933
-- [10.5 Linear Transformation of a State Vector](#page-958-0) 939
-  - [10.5-1 Diagonalization of Matrix](#page-962-0) **A** 943
-- [10.6 Controllability and Observability](#page-966-0) 947
-  - [10.6-1 Inadequacy of the Transfer Function Description of a System](#page-972-0) 953
-
-<span id="page-15-0"></span>[10.7 State-Space Analysis of Discrete-Time Systems](#page-972-0) 953
-
-[10.7-1 Solution in State Space](#page-974-0) 955
-
-10.7-2 The *z*[-Transform Solution](#page-978-0) 959
-
-[10.8 MATLAB: Toolboxes and State-Space Analysis](#page-980-0) 961
-
-10.8-1 *z*[-Transform Solutions to Discrete-Time, State-Space Systems](#page-980-0) 961
-
-[10.8-2 Transfer Functions from State-Space Representations](#page-983-0) 964
-
-[10.8-3 Controllability and Observability of Discrete-Time Systems](#page-984-0) 965
-
-[10.8-4 Matrix Exponentiation and the Matrix Exponential](#page-987-0) 968
-
-[10.9 Summary](#page-988-0) 969
-
-*[References](#page-989-0)* 970 *[Problems](#page-989-0)* 970
-
-[INDEX](#page-994-0) 975
-
-# <span id="page-16-0"></span>**[PREFACE](#page-6-0)**
-
-This book, *Linear Systems and Signals,* presents a comprehensive treatment of signals and linear systems at an introductory level. Following our preferred style, it emphasizes a physical appreciation of concepts through heuristic reasoning and the use of metaphors, analogies, and creative explanations. Such an approach is much different from a purely deductive technique that uses mere mathematical manipulation of symbols. There is a temptation to treat engineering subjects as a branch of applied mathematics. Such an approach is a perfect match to the public image of engineering as a dry and dull discipline. It ignores the physical meaning behind various derivations and deprives students of intuitive grasp and the enjoyable experience of logical uncovering of the subject matter. In this book, we use mathematics not so much to prove axiomatic theory as to support and enhance physical and intuitive understanding. Wherever possible, theoretical results are interpreted heuristically and are enhanced by carefully chosen examples and analogies.
-
-This third edition, which closely follows the organization of the second edition, has been refined in many ways. Discussions are streamlined, adding or trimming material as needed. Equation, example, and section labeling is simplified and improved. Computer examples are fully updated to reflect the most current version of MATLAB. Hundreds of added problems provide new opportunities to learn and understand topics. We have taken special care to improve the text without the topic creep and bloat that commonly occurs with each new edition of a text.
-
-## **NOTABLE FEATURES**
-
-The notable features of this book include the following.
-
-- 1. Intuitive and heuristic understanding of the concepts and physical meaning of mathematical results are emphasized throughout. Such an approach not only leads to deeper appreciation and easier comprehension of the concepts, but also makes learning enjoyable for students.
-- 2. Often, students lack an adequate background in basic material such as complex numbers, sinusoids, hand-sketching of functions, Cramer's rule, partial fraction expansion, and matrix algebra. We include a background chapter that addresses these basic and pervasive topics in electrical engineering. Response by students has been unanimously enthusiastic.
-- 3. There are hundreds of worked examples in addition to drills (usually with answers) for students to test their understanding. Additionally, there are over 900 end-of-chapter problems of varying difficulty.
-- 4. Modern electrical engineering practice requires the use of computer calculation and simulation, most often using the software package MATLAB. Thus, we integrate
-
-MATLAB into many of the worked examples throughout the book. Additionally, each chapter concludes with a section devoted to learning and using MATLAB in the context and support of book topics. Problem sets also contain numerous computer problems.
-
-- 5. The discrete-time and continuous-time systems may be treated in sequence, or they may be integrated by using a parallel approach.
-- 6. The summary at the end of each chapter proves helpful to students in summing up essential developments in the chapter.
-- 7. There are several historical notes to enhance students' interest in the subject. This information introduces students to the historical background that influenced the development of electrical engineering.
-
-## **ORGANIZATION**
-
-The book may be conceived as divided into five parts:
-
-- 1. Introduction (Chs. B and 1).
-- 2. Time-domain analysis of linear time-invariant (LTI) systems (Chs. 2 and 3).
-- 3. Frequency-domain (transform) analysis of LTI systems (Chs. 4 and 5).
-- 4. Signal analysis (Chs. 6, 7, 8, and 9).
-- 5. State-space analysis of LTI systems (Ch. 10).
-
-The organization of the book permits much flexibility in teaching the continuous-time and discrete-time concepts. The natural sequence of chapters is meant to integrate continuous-time and discrete-time analysis. It is also possible to use a sequential approach in which all the continuous-time analysis is covered first (Chs. 1, 2, 4, 6, 7, and 8), followed by discrete-time analysis (Chs. 3, 5, and 9).
-
-## **SUGGESTIONS FOR USING THIS BOOK**
-
-The book can be readily tailored for a variety of courses spanning 30 to 45 lecture hours. Most of the material in the first eight chapters can be covered at a brisk pace in about 45 hours. The book can also be used for a 30-lecture-hour course by covering only analog material (Chs. 1, 2, 4, 6, 7, and possibly selected topics in Ch. 8). Alternately, one can also select Chs. 1 to 5 for courses purely devoted to systems analysis or transform techniques. To treat continuous- and discrete-time systems by using an integrated (or parallel) approach, the appropriate sequence of chapters is 1, 2, 3, 4, 5, 6, 7, and 8. For a sequential approach, where the continuous-time analysis is followed by discrete-time analysis, the proper chapter sequence is 1, 2, 4, 6, 7, 8, 3, 5, and possibly 9 (depending on the time available).
-
-## **MATLAB**
-
-MATLAB is a sophisticated language that serves as a powerful tool to better understand engineering topics, including control theory, filter design, and, of course, linear systems and signals. MATLAB's flexible programming structure promotes rapid development and analysis. Outstanding visualization capabilities provide unique insight into system behavior and signal character.
-
-As with any language, learning MATLAB is incremental and requires practice. This book provides two levels of exposure to MATLAB. First, MATLAB is integrated into many examples throughout the text to reinforce concepts and perform various computations. These examples utilize standard MATLAB functions as well as functions from the control system, signal-processing, and symbolic math toolboxes. MATLAB has many more toolboxes available, but these three are commonly available in most engineering departments.
-
-A second and deeper level of exposure to MATLAB is achieved by concluding each chapter with a separate MATLAB section. Taken together, these eleven sections provide a self-contained introduction to the MATLAB environment that allows even novice users to quickly gain MATLAB proficiency and competence. These sessions provide detailed instruction on how to use MATLAB to solve problems in linear systems and signals. Except for the very last chapter, special care has been taken to avoid the use of toolbox functions in the MATLAB sessions. Rather, readers are shown the process of developing their own code. In this way, those readers without toolbox access are not at a disadvantage. All of this book's MATLAB code is available for download at the OUP companion website [www.oup.com/us/lathi.](http://www.oup.com/us/lathi)
-
-## **CREDITS AND ACKNOWLEDGMENTS**
-
-The portraits of Gauss, Laplace, Heaviside, Fourier, and Michelson have been reprinted courtesy of the Smithsonian Institution Libraries. The likenesses of Cardano and Gibbs have been reprinted courtesy of the Library of Congress. The engraving of Napoleon has been reprinted courtesy of Bettmann/Corbis. The many fine cartoons throughout the text are the work of Joseph Coniglio, a former student of Dr. Lathi.
-
-Many individuals have helped us in the preparation of this book, as well as its earlier editions. We are grateful to each and every one for helpful suggestions and comments. Book writing is an obsessively time-consuming activity, which causes much hardship for an author's family. We both are grateful to our families for their enormous but invisible sacrifices.
-
-> *B. P. Lathi R. A. Green*
-
-<span id="page-20-0"></span>
-
-# **[BACKGROUND](#page-6-0)**
-
-The topics discussed in this chapter are not entirely new to students taking this course. You have already studied many of these topics in earlier courses or are expected to know them from your previous training. Even so, this background material deserves a review because it is so pervasive in the area of signals and systems. Investing a little time in such a review will pay big dividends later. Furthermore, this material is useful not only for this course but also for several courses that follow. It will also be helpful later, as reference material in your professional career.
-
-## **[B.1 COMPLEX](#page-6-0) NUMBERS**
-
-*Complex numbers* are an extension of ordinary numbers and are an integral part of the modern number system. Complex numbers, particularly *imaginary numbers,* sometimes seem mysterious and unreal. This feeling of unreality derives from their unfamiliarity and novelty rather than their supposed nonexistence! Mathematicians blundered in calling these numbers "imaginary," for the term immediately prejudices perception. Had these numbers been called by some other name, they would have become demystified long ago, just as irrational numbers or negative numbers were. Many futile attempts have been made to ascribe some physical meaning to imaginary numbers. However, this effort is needless. In mathematics we assign symbols and operations any meaning we wish as long as internal consistency is maintained. The history of mathematics is full of entities that were unfamiliar and held in abhorrence until familiarity made them acceptable. This fact will become clear from the following historical note.
-
-### **[B.1-1 A Historical Note](#page-6-0)**
-
-Among early people the number system consisted only of natural numbers (positive integers) needed to express the number of children, cattle, and quivers of arrows. These people had no need for fractions. Whoever heard of two and one-half children or three and one-fourth cows!
-
-However, with the advent of agriculture, people needed to measure continuously varying quantities, such as the length of a field and the weight of a quantity of butter. The number system, therefore, was extended to include fractions. The ancient Egyptians and Babylonians knew how
-
-#### 2 CHAPTER B BACKGROUND
-
-to handle fractions, but *Pythagoras* discovered that some numbers (like the diagonal of a unit square) could not be expressed as a whole number or a fraction. Pythagoras, a number mystic, who regarded numbers as the essence and principle of all things in the universe, was so appalled at his discovery that he swore his followers to secrecy and imposed a death penalty for divulging this secret [1]. These numbers, however, were included in the number system by the time of Descartes, and they are now known as *irrational numbers*.
-
-Until recently, *negative numbers* were not a part of the number system. The concept of negative numbers must have appeared absurd to early man. However, the medieval Hindus had a clear understanding of the significance of positive and negative numbers [2, 3]. They were also the first to recognize the existence of absolute negative quantities [4]. The works of *Bhaskar* (1114–1185) on arithmetic (*L*¯*ilavat* ¯ ¯*i*) and algebra (*B*¯*ijaganit*) not only use the decimal system but also give rules for dealing with negative quantities. Bhaskar recognized that positive numbers have two square roots [5]. Much later, in Europe, the men who developed the banking system that arose in Florence and Venice during the late Renaissance (fifteenth century) are credited with introducing a crude form of negative numbers. The seemingly absurd subtraction of 7 from 5 seemed reasonable when bankers began to allow their clients to draw seven gold ducats while their deposit stood at five. All that was necessary for this purpose was to write the difference, 2, on the debit side of a ledger [6].
-
-Thus, the number system was once again broadened (generalized) to include negative numbers. The acceptance of negative numbers made it possible to solve equations such as *x*+5=0, which had no solution before. Yet for equations such as *x*<sup>2</sup> + 1 = 0, leading to *x*<sup>2</sup> = −1, the solution could not be found in the real number system. It was therefore necessary to define a completely new kind of number with its square equal to −1. During the time of Descartes and Newton, imaginary (or complex) numbers came to be accepted as part of the number system, but they were still regarded as algebraic fiction. The Swiss mathematician *Leonhard Euler* introduced the notation *<sup>i</sup>* (for *imaginary*) around 1777 to represent √−1. Electrical engineers use the notation *j* instead of *i* to avoid confusion with the notation *i* often used for electrical current. Thus,
-
-$$
-j^2 = -1 \qquad \text{and} \qquad \sqrt{-1} = \pm j
-$$
-
-This notation allows us to determine the square root of any negative number. For example,
-
-$$
-\sqrt{-4} = \sqrt{4} \times \sqrt{-1} = \pm 2j
-$$
-
-When imaginary numbers are included in the number system, the resulting numbers are called *complex numbers*.
-
-### ORIGINS OF COMPLEX NUMBERS
-
-Ironically (and contrary to popular belief), it was not the solution of a quadratic equation, such as *x*<sup>2</sup> + 1 = 0, but a cubic equation with real roots that made imaginary numbers plausible and acceptable to early mathematicians. They could dismiss √−1 as pure nonsense when it appeared as a solution to *x*<sup>2</sup> + 1 = 0 because this equation has no real solution. But in 1545, *Gerolamo Cardano* of Milan published *Ars Magna* (The Great Art), the most important algebraic work of the Renaissance. In this book, he gave a method of solving a general cubic equation in which a root of a negative number appeared in an intermediate step. According to his method, the solution to a third-order equation†
-
-$$
-x^3 + ax + b = 0
-$$
-
-is given by
-
-$$
-x = \sqrt[3]{-\frac{b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}} + \sqrt[3]{-\frac{b}{2} - \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}}
-$$
-
-For example, to find a solution of *x*<sup>3</sup> + 6*x* − 20 = 0, we substitute *a* = 6,*b* = −20 in the foregoing equation to obtain
-
-$$
-x = \sqrt[3]{10 + \sqrt{108}} + \sqrt[3]{10 - \sqrt{108}} = \sqrt[3]{20.392} - \sqrt[3]{0.392} = 2
-$$
-
-We can readily verify that 2 is indeed a solution of *x*<sup>3</sup> + 6*x* − 20 = 0. But when Cardano tried to solve the equation *x*<sup>3</sup> −15*x* −4 = 0 by this formula, his solution was
-
-$$
-x = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}
-$$
-
-What was Cardano to make of this equation in the year 1545? In those days, negative numbers were themselves suspect, and a square root of a negative number was doubly preposterous! Today, we know that
-
-$$
-(2 \pm j)^3 = 2 \pm j11 = 2 \pm \sqrt{-121}
-$$
-
-Therefore, Cardano's formula gives
-
-$$
-x = (2+j) + (2-j) = 4
-$$
-
-We can readily verify that *x* = 4 is indeed a solution of *x*<sup>3</sup> − 15*x* − 4 = 0. Cardano tried to explain halfheartedly the presence of √−121 but ultimately dismissed the whole enterprise as being "as subtle as it is useless." A generation later, however, *Raphael Bombelli* (1526–1573), after examining Cardano's results, proposed acceptance of imaginary numbers as a necessary vehicle that would transport the mathematician from the *real* cubic equation to its *real* solution. In other words, although we begin and end with real numbers, we seem compelled to move into an unfamiliar world of imaginaries to complete our journey. To mathematicians of the day, this proposal seemed incredibly strange [7]. Yet they could not dismiss the idea of imaginary numbers so easily because this concept yielded the real solution of an equation. It took two more centuries for the full importance of complex numbers to become evident in the works of Euler, Gauss, and Cauchy. Still, Bombelli deserves credit for recognizing that such numbers have a role to play in algebra [7].
-
-<sup>†</sup> This equation is known as the *depressed cubic* equation. A general cubic equation
-
-*<sup>y</sup>*<sup>3</sup> <sup>+</sup>*py*<sup>2</sup> <sup>+</sup>*qy*+*<sup>r</sup>* <sup>=</sup> <sup>0</sup>
-
-can always be reduced to a depressed cubic form by substituting *y* = *x* − (*p*/3). Therefore, any general cubic equation can be solved if we know the solution to the depressed cubic. The depressed cubic was independently solved, first by *Scipione del Ferro* (1465–1526) and then by *Niccolo Fontana* (1499–1557). The latter is better known in the history of mathematics as *Tartaglia* ("Stammerer"). Cardano learned the secret of the depressed cubic solution from Tartaglia. He then showed that by using the substitution *y* = *x*−(*p*/3), a general cubic is reduced to a depressed cubic.
-
-#### 4 CHAPTER B BACKGROUND
-
-In 1799 the German mathematician *Karl Friedrich Gauss,* at the ripe age of 22, proved the fundamental theorem of algebra, namely that every algebraic equation in one unknown has a root in the form of a complex number. He showed that every equation of the *n*th order has exactly *n* solutions (roots), no more and no less. Gauss was also one of the first to give a coherent account of complex numbers and to interpret them as points in a complex plane. It is he who introduced the term *complex numbers* and paved the way for their general and systematic use. The number system was once again broadened or generalized to include imaginary numbers. Ordinary (or real) numbers became a special case of generalized (or complex) numbers.
-
-The utility of complex numbers can be understood readily by an analogy with two neighboring countries *X* and *Y*, as illustrated in Fig. B.1. If we want to travel from City *a* to City *b* (both in
-
-Gerolamo Cardano Karl Friedrich Gauss
-
-**Figure B.1** Use of complex numbers can reduce the work.
-
-<span id="page-24-0"></span>Country *X*), the shortest route is through Country *Y*, although the journey begins and ends in Country *X*. We may, if we desire, perform this journey by an alternate route that lies exclusively in *X*, but this alternate route is longer. In mathematics we have a similar situation with real numbers (Country *X*) and complex numbers (Country *Y*). Most real-world problems start with real numbers, and the final results must also be in real numbers. But the derivation of results is considerably simplified by using complex numbers as an intermediary. It is also possible to solve any real-world problem by an alternate method, using real numbers exclusively, but such procedures would increase the work needlessly.
-
-### **[B.1-2 Algebra of Complex Numbers](#page-6-0)**
-
-A complex number (*a*,*b*) or *a* + *jb* can be represented graphically by a point whose Cartesian coordinates are (*a*,*b*) in a complex plane (Fig. B.2). Let us denote this complex number by *z* so that
-
-$$
-z = a + jb \tag{B.1}
-$$
-
-This representation is the Cartesian (or rectangular) form of complex number *z*. The numbers *a* and *b* (the abscissa and the ordinate) of *z* are the *real part* and the *imaginary part*, respectively, of *z*. They are also expressed as
-
-$$
-Re z = a \qquad \text{and} \qquad Im z = b
-$$
-
-Note that in this plane all real numbers lie on the horizontal axis, and all imaginary numbers lie on the vertical axis.
-
-Complex numbers may also be expressed in terms of polar coordinates. If (*r*, θ ) are the polar coordinates of a point *z* = *a*+*jb* (see Fig. B.2), then
-
-$$
-a = r \cos \theta
-$$
- and  $b = r \sin \theta$ 
-
-Consequently,
-
-$$
-z = a + jb = r\cos\theta + jr\sin\theta = r(\cos\theta + j\sin\theta)
-$$
- (B.2)
-
-*Euler's formula* states that
-
-$$
-e^{j\theta} = \cos\theta + j\sin\theta \tag{B.3}
-$$
-
-To prove Euler's formula, we use a Maclaurin series to expand *ej*<sup>θ</sup> , cos θ, and sin θ:
-
-$$
-e^{j\theta} = 1 + j\theta + \frac{(j\theta)^2}{2!} + \frac{(j\theta)^3}{3!} + \frac{(j\theta)^4}{4!} + \frac{(j\theta)^5}{5!} + \frac{(j\theta)^6}{6!} + \cdots
-$$
-  
-\n
-$$
-= 1 + j\theta - \frac{\theta^2}{2!} - j\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + j\frac{\theta^5}{5!} - \frac{\theta^6}{6!} - \cdots
-$$
-  
-\n
-$$
-\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} + \cdots
-$$
-  
-\n
-$$
-\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots
-$$
-
-Clearly, it follows that *ej*<sup>θ</sup> = cos θ +*j*sin θ. Using Eq. (B.3) in Eq. (B.2) yields
-
-$$
-z = re^{i\theta} \tag{B.4}
-$$
-
-This representation is the polar form of complex number *z*.
-
-Summarizing, a complex number can be expressed in rectangular form *a* + *jb* or polar form *rej*<sup>θ</sup> with
-
-$$
-a = r \cos \theta
-$$
-  
-\n
-$$
-b = r \sin \theta
-$$
- and 
-$$
-r = \sqrt{a^2 + b^2}
-$$
-  
-\n
-$$
-\theta = \tan^{-1} \left(\frac{b}{a}\right)
-$$
- (B.5)
-
-Observe that *r* is the distance of the point *z* from the origin. For this reason, *r* is also called the *magnitude* (or *absolute value*) of *z* and is denoted by |*z*|. Similarly, θ is called the angle of *z* and is denoted by *z*. Therefore, we can also write polar form of Eq. (B.4) as
-
-$$
-z = |z|e^{j\angle z}
-$$
- where  $|z| = r$  and  $\angle z = \theta$ 
-
-Using polar form, we see that the reciprocal of a complex number is given by
-
-$$
-\frac{1}{z} = \frac{1}{re^{j\theta}} = \frac{1}{r}e^{-j\theta} = \frac{1}{|z|}e^{-j\sqrt{z}}
-$$
-
-### CONJUGATE OF A COMPLEX NUMBER
-
-We define *z*∗, the *conjugate* of *z* = *a*+*jb*, as
-
-$$
-z^* = a - jb = re^{-j\theta} = |z|e^{-j\angle z}
-$$
- (B.6)
-
-The graphical representations of a number *z* and its conjugate *z*<sup>∗</sup> are depicted in Fig. B.2. Observe that *z*<sup>∗</sup> is a mirror image of *z* about the horizontal axis. *To find the conjugate of any number, we need only replace j with* −*j in that number* (which is the same as changing the sign of its angle).
-
-The sum of a complex number and its conjugate is a real number equal to twice the real part of the number:
-
-$$
-z + z^* = (a + jb) + (a - jb) = 2a = 2 \operatorname{Re} z
-$$
-
-Thus, we see that the real part of complex number *z* can be computed as
-
-$$
-\text{Re}\,z = \frac{z + z^*}{2} \tag{B.7}
-$$
-
-Similarly, the imaginary part of complex number *z* can be computed as
-
-$$
-\operatorname{Im} z = \frac{z - z^*}{2j} \tag{B.8}
-$$
-
-The product of a complex number *z* and its conjugate is a real number |*z*| 2, the square of the magnitude of the number:
-
-$$
-zz^* = |z|e^{j\angle z}|z|e^{-j\angle z} = |z|^2
-$$
- (B.9)
-
-### UNDERSTANDING SOME USEFUL IDENTITIES
-
-In a complex plane, *rej*<sup>θ</sup> represents a point at a distance *r* from the origin and at an angle θ with the horizontal axis, as shown in Fig. B.3a. For example, the number −1 is at a unit distance from the origin and has an angle π or −π (more generally, π plus any integer multiple of 2π), as seen from Fig. B.3b. Therefore,
-
-$$
--1 = e^{j(\pi + 2\pi n)} \qquad n \text{ integer}
-$$
-
-The number 1, on the other hand, is also at a unit distance from the origin, but has an angle 0 (more generally, 0 plus any integer multiple of 2π). Therefore,
-
-$$
-1 = e^{j2\pi n} \qquad n \text{ integer}
-$$
- (B.10)
-
-The number *j* is at a unit distance from the origin and its angle is <sup>π</sup> <sup>2</sup> (more generally, <sup>π</sup> <sup>2</sup> plus any integer multiple of 2π), as seen from Fig. B.3b. Therefore,
-
-$$
-j = e^{j(\frac{\pi}{2} + 2\pi n)} \qquad n \text{ integer}
-$$
-
-Similarly,
-
-$$
--j = e^{j(-\frac{\pi}{2} + 2\pi n)} \qquad n \text{ integer}
-$$
-
-Notice that the angle of any complex number is only known within an integer multiple of 2π.
-
-This discussion shows the usefulness of the graphic picture of *rej*<sup>θ</sup> . This picture is also helpful in several other applications. For example, to determine the limit of *e*(α+*j*ω)*<sup>t</sup>* as *t* → ∞, we note that
-
-*ej*ω*<sup>t</sup>*
-
-$$
-e^{(\alpha+j\omega)t} = e^{\alpha t} e^{j\omega t}
-$$
-\n
-$$
-\lim_{\rho \to 0} \frac{e^{j\theta}}{\log \rho} = \lim_{\rho \to 0} \frac{e^{j\theta}}{\log \rho} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta}
-$$
-\n
-$$
-\lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta}
-$$
-\n
-$$
-\lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{1}{\sqrt{\frac{\pi}{2}}} e^{j\theta} = \lim_{\rho \to 0} \frac{
-$$
-
-**Figure B.3** Understanding some useful identities in terms of *rej*<sup>θ</sup> .
-
-### 8 CHAPTER B BACKGROUND
-
-Now the magnitude of *ej*ω*<sup>t</sup>* is unity regardless of the value of ω or *t* because *ej*ω*<sup>t</sup>* = *rej*<sup>θ</sup> with *r* = 1. Therefore, *e*α*<sup>t</sup>* determines the behavior of *e*(α+*j*ω)*<sup>t</sup>* as *t* → ∞ and
-
-$$
-\lim_{t \to \infty} e^{(\alpha + j\omega)t} = \lim_{t \to \infty} e^{\alpha t} e^{j\omega t} = \begin{cases} 0 & \alpha < 0 \\ \infty & \alpha > 0 \end{cases}
-$$
-
-In future discussions, you will find it very useful to remember *rej*<sup>θ</sup> as a number at a distance *r* from the origin and at an angle θ with the horizontal axis of the complex plane.
-
-### A WARNING ABOUT COMPUTING ANGLES WITH CALCULATORS
-
-From the Cartesian form *a* + *jb*, we can readily compute the polar form *rej*<sup>θ</sup> [see Eq. (B.5)]. Calculators provide ready conversion of rectangular into polar and vice versa. However, if a calculator computes an angle of a complex number by using an inverse tangent function θ = tan−1(*b*/*a*), proper attention must be paid to the quadrant in which the number is located. For instance, θ corresponding to the number −2 − *j*3 is tan−<sup>1</sup>(−3/−2). This result is not the same as tan−<sup>1</sup>(3/2). The former is −123.7◦, whereas the latter is 56.3◦. A calculator cannot make this distinction and can give a correct answer only for angles in the first and fourth quadrants.† A calculator will read tan−<sup>1</sup>(−3/−2) as tan−<sup>1</sup>(3/2), which is clearly wrong. When you are computing inverse trigonometric functions, if the angle appears in the second or third quadrant, the answer of the calculator is off by 180◦. The correct answer is obtained by adding or subtracting 180◦ to the value found with the calculator (either adding or subtracting yields the correct answer). For this reason, it is advisable to draw the point in the complex plane and determine the quadrant in which the point lies. This issue will be clarified by the following examples.
-
-### **EXAMPLE B.1 Cartesian to Polar Form**
-
-Express the following numbers in polar form: **(a)** 2+*j*3, **(b)** −2+*j*1, **(c)** −2−*j*3, and **(d)** 1−*j*3.
-
-**(a)**
-
-$$
-|z| = \sqrt{2^2 + 3^2} = \sqrt{13}
-$$
-  $\angle z = \tan^{-1}(\frac{3}{2}) = 56.3^{\circ}$ 
-
-In this case the number is in the first quadrant, and a calculator will give the correct value of 56.3◦. Therefore (see Fig. B.4a), we can write
-
-$$
-2 + j3 = \sqrt{13} e^{j56.3^{\circ}}
-$$
-
-**(b)**
-
-$$
-|z| = \sqrt{(-2)^2 + 1^2} = \sqrt{5}
-$$
-  $\angle z = \tan^{-1}(\frac{1}{-2}) = 153.4^{\circ}$ 
-
-In this case the angle is in the second quadrant (see Fig. B.4b), and therefore the answer given by the calculator, tan−<sup>1</sup>(1/−2) = −26.6◦, is off by 180◦. The correct answer is
-
-<sup>†</sup> Calculators with two-argument inverse tangent functions will correctly compute angles.
-
-(−26.6±180)◦ = 153.4◦ or −206.6◦. Both values are correct because they represent the same angle. It is a common practice to choose an angle whose numerical value is less than 180◦. Such a value is called the *principal value* of the angle, which in this case is 153.4◦. Therefore,
-
-$$
--2 + j1 = \sqrt{5}e^{j153.4^{\circ}}
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/002_Half title.md

@@ -1,309 +0,0 @@
-**(c)**
-
-$$
-|z| = \sqrt{(-2)^2 + (-3)^2} = \sqrt{13}
-$$
-  $\angle z = \tan^{-1}\left(\frac{-3}{-2}\right) = -123.7^{\circ}$ 
-
-In this case the angle appears in the third quadrant (see Fig. B.4c), and therefore the answer obtained by the calculator (tan−1(−3/−2) = 56.3◦) is off by 180◦. The correct answer is (56.3 ± 180)◦ = 236.3◦ or −123.7◦. We choose the principal value −123.7◦ so that (see Fig. B.4c)
-
-$$
--2 - j3 = \sqrt{13}e^{-j123.7^{\circ}}
-$$
-
-**(d)**
-
-$$
-|z| = \sqrt{1^2 + (-3)^2} = \sqrt{10}
-$$
-  $\angle z = \tan^{-1}\left(\frac{-3}{1}\right) = -71.6^{\circ}$ 
-
-In this case the angle appears in the fourth quadrant (see Fig. B.4d), and therefore the answer given by the calculator, tan−<sup>1</sup>(−3/1) = −71.6◦, is correct (see Fig. B.4d):
-
-$$
-1 - j3 = \sqrt{10}e^{-j71.6^{\circ}}
-$$
-
-### 10 CHAPTER B BACKGROUND
-
-We can easily verify these results using the MATLAB abs and angle commands. To obtain units of degrees, we must multiply the radian result of the angle command by 180 <sup>π</sup> . Furthermore, the angle command correctly computes angles for all four quadrants of √ the complex plane. To provide an example, let us use MATLAB to verify that −2 + *j*1 = 5*e<sup>j</sup>*153.4◦ = 2.2361*e<sup>j</sup>*153.4◦ .
-
->> abs(-2+1j) ans = 2.2361 >> angle(-2+1j)\*180/pi ans = 153.4349
-
-One can also use the cart2pol command to convert Cartesian to polar coordinates. Readers, particularly those who are unfamiliar with MATLAB, will benefit by reading the overview in Sec. B.7.
-
-### **EXAMPLE B.2 Polar to Cartesian Form**
-
-Represent the following numbers in the complex plane and express them in Cartesian form: **(a)** 2*ej*π/3, **(b)** 4*e*−*j*3π/4, **(c)** 2*ej*π/2, **(d)** 3*e*−*j*3<sup>π</sup> , **(e)** 2*ej*4<sup>π</sup> , and **(f)** 2*e*−*j*4<sup>π</sup> .
-
-**(a)** 2*ej*π/<sup>3</sup> = 2(cos π/3+*j*sin π/3) = 1+*j* <sup>√</sup>3 (see Fig. B.5a) **(b)** 4*e*−*j*3π/<sup>4</sup> = 4(cos 3π/4−*j*sin 3π/4) = −2 <sup>√</sup>2−*j*<sup>2</sup> <sup>√</sup>2 (see Fig. B.5b) **(c)** 2*ej*π/<sup>2</sup> = 2(cos π/2+*j*sin π/2) = 2(0+*j*1) = *j*2 (see Fig. B.5c) **(d)** 3*e*−*j*3<sup>π</sup> = 3(cos 3π −*j*sin 3π ) = 3(−1+*j*0) = −3 (see Fig. B.5d) **(e)** 2*ej*4<sup>π</sup> = 2(cos 4π +*j*sin 4π ) = 2(1+*j*0) = 2 (see Fig. B.5e) **(f)** 2*e*−*j*4<sup>π</sup> = 2(cos 4π −*j*sin 4π ) = 2(1−*j*0) = 2 (see Fig. B.5f)
-
-We can readily verify these results using MATLAB. First, we use the exp function to represent a number in polar form. Next, we use the real and imag commands to determine the real and imaginary components of that number. To provide an example, let us use MATLAB to verify the result of part (a): 2*ej*π/<sup>3</sup> = 1+*j* <sup>√</sup><sup>3</sup> <sup>=</sup> <sup>1</sup>+*j*1.7321.
-
-```
->> real(2*exp(1j*pi/3))
-    ans = 1.0000
->> imag(2*exp(1j*pi/3))
-    ans = 1.7321
-```
-
-Since MATLAB defaults to Cartesian form, we could have verified the entire result in one step.
-
-```
->> 2*exp(1j*pi/3)
-    ans = 1.0000 + 1.7321i
-```
-
-One can also use the pol2cart command to convert polar to Cartesian coordinates.
-
-### ARITHMETICAL OPERATIONS, POWERS, AND ROOTS OF COMPLEX NUMBERS
-
-To conveniently perform addition and subtraction, complex numbers should be expressed in Cartesian form. Thus, if
-
-$$
-z_1 = 3 + j4 = 5e^{j53.1^{\circ}}
-$$
-
-and
-
-$$
-z_2 = 2 + j3 = \sqrt{13}e^{j56.3^{\circ}}
-$$
-
-then
-
-$$
-z_1 + z_2 = (3 + j4) + (2 + j3) = 5 + j7
-$$
-
-### 12 CHAPTER B BACKGROUND
-
-If *z*<sup>1</sup> and *z*<sup>2</sup> are given in polar form, we would need to convert them into Cartesian form for the purpose of adding (or subtracting). Multiplication and division, however, can be carried out in either Cartesian or polar form, although the latter proves to be much more convenient. This is because if *z*<sup>1</sup> and *z*<sup>2</sup> are expressed in polar form as
-
-$$
-z_1 = r_1 e^{j\theta_1}
-$$
- and  $z_2 = r_2 e^{j\theta_2}$ 
-
-then
-
-$$
-z_1 z_2 = (r_1 e^{j\theta_1})(r_2 e^{j\theta_2}) = r_1 r_2 e^{j(\theta_1 + \theta_2)}
-$$
-
-and
-
-$$
-\frac{z_1}{z_2} = \frac{r_1 e^{j\theta_1}}{r_2 e^{j\theta_2}} = \frac{r_1}{r_2} e^{j(\theta_1 - \theta_2)}
-$$
-
-Moreover,
-
-$$
-z^n = (re^{j\theta})^n = r^n e^{jn\theta}
-$$
-
-and
-
-$$
-z^{1/n} = (re^{i\theta})^{1/n} = r^{1/n}e^{i\theta/n}
-$$
- (B.11)
-
-This shows that the operations of multiplication, division, powers, and roots can be carried out with remarkable ease when the numbers are in polar form.
-
-Strictly speaking, there are *n* values for *z*1/*<sup>n</sup>* (the *n*th root of *z*). To find all the *n* roots, we reexamine Eq. (B.11):
-
-$$
-z^{1/n} = [re^{j\theta}]^{1/n} = [re^{j(\theta + 2\pi k)}]^{1/n} = r^{1/n}e^{j(\theta + 2\pi k)/n} \qquad k = 0, 1, 2, \dots, n-1
-$$
- (B.12)
-
-The value of *z*1/*<sup>n</sup>* given in Eq. (B.11) is the *principal value* of *z*1/*<sup>n</sup>*, obtained by taking the *n*th root of the principal value of *z*, which corresponds to the case *k* = 0 in Eq. (B.12).
-
-### **EXAMPLE B.3 Multiplication and Division of Complex Numbers**
-
-Using both polar and Cartesian forms, determine *z*1*z*<sup>2</sup> and *z*1/*z*<sup>2</sup> for the numbers
-
-$$
-z_1 = 3 + j4 = 5e^{j53.1^{\circ}}
-$$
- and  $z_2 = 2 + j3 = \sqrt{13}e^{j56.3^{\circ}}$ 
-
-**Multiplication: Cartesian Form**
-
-$$
-z_1 z_2 = (3+j4)(2+j3) = (6-12) + j(8+9) = -6+j17
-$$
-
-**Multiplication: Polar Form**
-
-$$
-z_1 z_2 = (5e^{j53.1^{\circ}})(\sqrt{13}e^{j56.3^{\circ}}) = 5\sqrt{13}e^{j109.4^{\circ}}
-$$
-
-**Division: Cartesian Form**
-
-$$
-\frac{z_1}{z_2} = \frac{3+j4}{2+j3}
-$$
-
-To eliminate the complex number in the denominator, we multiply both the numerator and the denominator of the right-hand side by 2−*j*3, the denominator's conjugate. This yields
-
-$$
-\frac{z_1}{z_2} = \frac{(3+j4)(2-j3)}{(2+j3)(2-j3)} = \frac{18-j1}{2^2+3^2} = \frac{18-j1}{13} = \frac{18}{13} - j\frac{1}{13}
-$$
-
-**Division: Polar Form**
-
-$$
-\frac{z_1}{z_2} = \frac{5e^{j53.1^{\circ}}}{\sqrt{13}e^{j56.3^{\circ}}} = \frac{5}{\sqrt{13}}e^{j(53.1^{\circ} - 56.3^{\circ})} = \frac{5}{\sqrt{13}}e^{-j3.2^{\circ}}
-$$
-
-It is clear from this example that multiplication and division are easier to accomplish in polar form than in Cartesian form.
-
-These results are also easily verified using MATLAB. To provide one example, let us use Cartesian forms in MATLAB to verify that *z*1*z*<sup>2</sup> = −6+*j*17.
-
-```
->> z1 = 3+4j; z2 = 2+3j;
->> z1*z2
-    ans = -6.0000 + 17.0000i
-```
-
-As a second example, let us use polar forms in MATLAB to verify that *z*1/*z*<sup>2</sup> = 1.3868*e*−*j*3.2◦ . Since MATLAB generally expects angles be represented in the natural units of radians, we must use appropriate conversion factors in moving between degrees and radians (and vice versa).
-
-```
->> z1 = 5*exp(1j*53.1*pi/180); z2 = sqrt(13)*exp(1j*56.3*pi/180);
->> abs(z1/z2)
-    ans = 1.3868
->> angle(z1/z2)*180/pi
-    ans = -3.2000
-```
-
-### **EXAMPLE B.4 Working with Complex Numbers**
-
-For *z*<sup>1</sup> = 2*ej*π/<sup>4</sup> and *z*<sup>2</sup> = 8*ej*π/3, find the following: **(a)** 2*z*<sup>1</sup> −*z*2, **(b)** 1/*z*1, **(c)** *z*1/*z*<sup>2</sup> <sup>2</sup>, and **(d)** <sup>√</sup><sup>3</sup> *<sup>z</sup>*2.
-
-**(a)** Since subtraction cannot be performed directly in polar form, we convert *z*<sup>1</sup> and *z*<sup>2</sup> to Cartesian form:
-
-$$
-z_1 = 2e^{j\pi/4} = 2\left(\cos\frac{\pi}{4} + j\sin\frac{\pi}{4}\right) = \sqrt{2} + j\sqrt{2}
-$$
-$$
-z_2 = 8e^{j\pi/3} = 8\left(\cos\frac{\pi}{3} + j\sin\frac{\pi}{3}\right) = 4 + j4\sqrt{3}
-$$
-
-Therefore,
-
-$$
-2z_1 - z_2 = 2(\sqrt{2} + j\sqrt{2}) - (4 + j4\sqrt{3}) = (2\sqrt{2} - 4) + j(2\sqrt{2} - 4\sqrt{3}) = -1.17 - j4.1
-$$
-
-**(b)**
-
-$$
-\frac{1}{z_1} = \frac{1}{2e^{j\pi/4}} = \frac{1}{2}e^{-j\pi/4}
-$$
-
-**(c)**
-
-$$
-\frac{z_1}{z_2^2} = \frac{2e^{j\pi/4}}{(8e^{j\pi/3})^2} = \frac{2e^{j\pi/4}}{64e^{j2\pi/3}} = \frac{1}{32}e^{j(\pi/4 - 2\pi/3)} = \frac{1}{32}e^{-j(5\pi/12)}
-$$
-
-**(d)** There are three cube roots of 8*ej*(π/3) = 8*ej*(π/3+2π*k*) , *k* = 0, 1, 2.
-
-$$
-\sqrt[3]{z_2} = z_2^{1/3} = \left[8e^{i(\pi/3 + 2\pi k)}\right]^{1/3} = 8^{1/3} \left(e^{i[(6\pi k + \pi)/3]}\right)^{1/3} = \begin{cases} 2e^{i\pi/9} & k = 0\\ 2e^{i7\pi/9} & k = 1\\ 2e^{i13\pi/9} & k = 2 \end{cases}
-$$
-
-The value corresponding to *k* = 0 is termed the *principal value*.
-
-### **EXAMPLE B.5 Standard Forms of Complex Numbers**
-
-Consider *X*(ω), a complex function of a real variable ω:
-
-$$
-X(\omega) = \frac{2 + j\omega}{3 + j4\omega}
-$$
-
-**(a)** Express *X*(ω) in Cartesian form, and find its real and imaginary parts.
-
-**(b)** Express *X*(ω) in polar form, and find its magnitude |*X*(ω)| and angle *X*(ω).
-
-$$
-X(\omega) = \frac{(2+j\omega)(3-j4\omega)}{(3+j4\omega)(3-j4\omega)} = \frac{(6+4\omega^2) - j5\omega}{9+16\omega^2} = \frac{6+4\omega^2}{9+16\omega^2} - j\frac{5\omega}{9+16\omega^2}
-$$
-
-This is the Cartesian form of *X*(ω). Clearly, the real and imaginary parts *Xr*(ω) and *Xi*(ω) are given by
-
-$$
-X_r(\omega) = \frac{6 + 4\omega^2}{9 + 16\omega^2}
-$$
- and  $X_i(\omega) = \frac{-5\omega}{9 + 16\omega^2}$ 
-
-**<sup>(</sup>a)** To obtain the real and imaginary parts of *X*(ω), we must eliminate imaginary terms in the denominator of *X*(ω). This is readily done by multiplying both the numerator and the denominator of *X*(ω) by 3−*j*4ω, the conjugate of the denominator 3+*j*4ω so that
-
-$$
-(\mathbf{b})
-$$
-
-$$
-X(\omega) = \frac{2 + j\omega}{3 + j4\omega} = \frac{\sqrt{4 + \omega^2} e^{j\tan^{-1}(\omega/2)}}{\sqrt{9 + 16\omega^2} e^{j\tan^{-1}(4\omega/3)}} = \sqrt{\frac{4 + \omega^2}{9 + 16\omega^2}} e^{j\tan^{-1}(\omega/2) - \tan^{-1}(4\omega/3)}
-$$
-
-This is the polar representation of *X*(ω). Observe that
-
-$$
-|X(\omega)| = \sqrt{\frac{4 + \omega^2}{9 + 16\omega^2}} \quad \text{and} \quad \angle X(\omega) = \tan^{-1}\left(\frac{\omega}{2}\right) - \tan^{-1}\left(\frac{4\omega}{3}\right)
-$$
-
-### LOGARITHMS OF COMPLEX NUMBERS
-
-To take the natural logarithm of a complex number *z*, we first express *z* in general polar form as
-
-$$
-z = re^{j\theta} = re^{j(\theta \pm 2\pi k)}
-$$
-  $k = 0, 1, 2, 3, ...$ 
-
-Taking the natural logarithm, we see that
-
-$$
-\ln z = \ln \left( r e^{j(\theta \pm 2\pi k)} \right) = \ln r \pm j(\theta + 2\pi k) \qquad k = 0, 1, 2, 3, \dots
-$$
-
-The value of ln*z* for *k* = 0 is called the *principal value* of ln*z* and is denoted by Ln*z*. In this way, we see that
-
-$$
-\ln 1 = \ln(1e^{\pm j2\pi k}) = \pm j2\pi k \qquad k = 0, 1, 2, 3, \dots
-$$
-$$
-\ln(-1) = \ln[1e^{\pm j\pi(2k+1)}] = \pm j(2k+1)\pi \qquad k = 0, 1, 2, 3, \dots
-$$
-$$
-\ln j = \ln(e^{j\pi(1\pm 4k)/2}) = j\frac{\pi(1\pm 4k)}{2} \qquad k = 0, 1, 2, 3, \dots
-$$
-$$
-j^j = e^{j\ln j} = e^{-\pi(1\pm 4k)/2} \qquad k = 0, 1, 2, 3, \dots
-$$
-
-In all of these cases, setting *k* = 0 yields the principal value of the expression.
-
-We can further our logarithm skills by noting that the familiar properties of logarithms hold for complex arguments. Therefore, we have
-
-$$
-log(z1z2) = log z1 + log z2
-$$
-$$
-log(z1/z2) = log z1 - log z2
-$$
-$$
-a(z1+z2) = az1 × az2
-$$
-$$
-zc = ecln z
-$$
-$$
-az = ezln a
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/003_B.2 SINUSOIDS.md

@@ -1,186 +0,0 @@
-## <span id="page-35-0"></span>**[B.2 SINUSOIDS](#page-6-0)**
-
-Consider the sinusoid
-
-$$
-x(t) = C\cos(2\pi f_0 t + \theta)
-$$
- (B.13)
-
-We know that
-
-$$
-\cos \varphi = \cos (\varphi + 2n\pi)
-$$
-  $n = 0, \pm 1, \pm 2, \pm 3, ...$ 
-
-Therefore, cos ϕ repeats itself for every change of 2π in the angle ϕ. For the sinusoid in Eq. (B.13), the angle 2π*f*0*t*+θ changes by 2π when *t* changes by 1/*f*0. Clearly, this sinusoid repeats every 1/*f*<sup>0</sup> seconds. As a result, there are *f*<sup>0</sup> repetitions per second. This is the *frequency* of the sinusoid, and the repetition interval *T*<sup>0</sup> given by
-
-$$
-T_0 = \frac{1}{f_0}
-$$
- (B.14)
-
-is the *period*. For the sinusoid in Eq. (B.13), *C* is the *amplitude, f*<sup>0</sup> is the *frequency* (in hertz), and θ is the phase. Let us consider two special cases of this sinusoid when θ = 0 and θ = −π/2 as follows:
-
-$$
-x(t) = C\cos 2\pi f_0 t \qquad (\theta = 0)
-$$
-
-and
-
-$$
-x(t) = C\cos(2\pi f_0 t - \pi/2) = C\sin 2\pi f_0 t \qquad (\theta = -\pi/2)
-$$
-
-The angle or phase can be expressed in units of degrees or radians. Although the radian is the proper unit, in this book we shall often use the degree unit because students generally have a better feel for the relative magnitudes of angles expressed in degrees rather than in radians. For example, we relate better to the angle 24◦ than to 0.419 radian. Remember, however, when in doubt, use the radian unit and, above all, be consistent. In other words, in a given problem or an expression, do not mix the two units.
-
-It is convenient to use the variable ω<sup>0</sup> (*radian frequency*) to express 2π*f*0:
-
-$$
-\omega_0 = 2\pi f_0 \tag{B.15}
-$$
-
-With this notation, the sinusoid in Eq. (B.13) can be expressed as
-
-$$
-x(t) = C\cos{(\omega_0 t + \theta)}
-$$
-
-in which the period *T*<sup>0</sup> and frequency ω<sup>0</sup> are given by [see Eqs. (B.14) and (B.15)]
-
-$$
-T_0 = \frac{1}{\omega_0/2\pi} = \frac{2\pi}{\omega_0} \quad \text{and} \quad \omega_0 = \frac{2\pi}{T_0}
-$$
-
-Although we shall often refer to ω<sup>0</sup> as the frequency of the signal cos(ω0*t*+θ ), it should be clearly understood that ω<sup>0</sup> is the *radian frequency*; the *hertzian frequency* of this sinusoid is *f*<sup>0</sup> = ω0/2π ).
-
-The signals *C*cos ω0*t* and *C*sin ω0*t* are illustrated in Figs. B.6a and B.6b, respectively. A general sinusoid *C*cos(ω0*t*+θ ) can be readily sketched by shifting the signal *C*cos ω0*t* in Fig. B.6a by the appropriate amount. Consider, for example,
-
-$$
-x(t) = C\cos{(\omega_0 t - 60^\circ)}
-$$
-
-**Figure B.6** Sketching a sinusoid.
-
-This signal can be obtained by shifting (delaying) the signal *C*cos ω0*t* (Fig. B.6a) to the right by a phase (angle) of 60◦. We know that a sinusoid undergoes a 360◦ change of phase (or angle) in one cycle. A quarter-cycle segment corresponds to a 90◦ change of angle. We therefore shift (delay) the signal in Fig. B.6a by two-thirds of a quarter-cycle segment to obtain *C*cos(ω0*t* − 60◦), as shown in Fig. B.6c.
-
-Observe that if we delay *C*cos ω0*t* in Fig. B.6a by a quarter-cycle (angle of 90◦ or π/2 radians), we obtain the signal *C*sin ω0*t*, depicted in Fig. B.6b. This verifies the well-known trigonometric identity
-
-$$
-C\cos{(\omega_0 t - \pi/2)} = C\sin{\omega_0 t}
-$$
-
-#### <span id="page-37-0"></span>18 CHAPTER B BACKGROUND
-
-Alternatively, if we advance *C*sin ω0*t* by a quarter-cycle, we obtain *C*cos ω0*t*. Therefore,
-
-$$
-C\sin(\omega_0 t + \pi/2) = C\cos\omega_0 t
-$$
-
-These observations mean that sin ω0*t* lags cos ω0*t* by 90◦(π/2 radians) and that cos ω0*t* leads sin ω0*t* by 90◦.
-
-### **[B.2-1 Addition of Sinusoids](#page-6-0)**
-
-Two sinusoids having the same frequency but different phases add to form a single sinusoid of the same frequency. This fact is readily seen from the well-known trigonometric identity
-
-*C*cos θ cos ω0*t* −*C*sin θ sin ω0*t* = *C*cos(ω0*t* +θ )
-
-Setting *a* = *C*cos θ and *b* = −*C*sin θ, we see that
-
-$$
-a\cos\omega_0 t + b\sin\omega_0 t = C\cos(\omega_0 t + \theta)
-$$
- (B.16)
-
-From trigonometry, we know that
-
-$$
-C = \sqrt{a^2 + b^2} \qquad \text{and} \qquad \theta = \tan^{-1}\left(\frac{-b}{a}\right) \tag{B.17}
-$$
-
-Equation (B.17) shows that *C* and θ are the magnitude and angle, respectively, of a complex number *a* − *jb*. In other words, *a* − *jb* = *Cej*<sup>θ</sup> . Hence, to find *C* and θ, we convert *a* − *jb* to polar form and the magnitude and the angle of the resulting polar number are *C* and θ, respectively.
-
-The process of adding two sinusoids with the same frequency can be clarified by using *phasors* to represent sinusoids. We represent the sinusoid *C*cos(ω0*t*+θ ) by a phasor of length *C* at an angle θ with the horizontal axis. Clearly, the sinusoid *a*cos ω0*t* is represented by a horizontal phasor of length *a*(θ = 0), while *b*sin ω0*t* = *b*cos(ω0*t* −π/2) is represented by a vertical phasor of length *b* at an angle −π/2 with the horizontal (Fig. B.7). Adding these two phasors results in a phasor of length *C* at an angle θ, as depicted in Fig. B.7. From this figure, we verify the values of *C* and θ found in Eq. (B.17). Proper care should be exercised in computing θ, as explained on page 8 ("A Warning About Computing Angles with Calculators").
-
-**Figure B.7** Phasor addition of sinusoids.
-
-### **EXAMPLE B.6 Addition of Sinusoids**
-
-In the following cases, express *x*(*t*) as a single sinusoid:
-
-**(a)** *<sup>x</sup>*(*t*) <sup>=</sup> cos <sup>ω</sup>0*<sup>t</sup>* <sup>−</sup> <sup>√</sup>3 sin <sup>ω</sup>0*<sup>t</sup>*
-
-**(b)** *x*(*t*) = −3 cos ω0*t* +4 sin ω0*t*
-
-**(a)** In this case, *<sup>a</sup>* <sup>=</sup> 1 and *<sup>b</sup>* = −√3. Using Eq. (B.17) yields
-
-$$
-C = \sqrt{1^2 + (\sqrt{3})^2} = 2
-$$
- and  $\theta = \tan^{-1}(\frac{\sqrt{3}}{1}) = 60^\circ$ 
-
-Therefore,
-
-$$
-x(t) = 2\cos{(\omega_0 t + 60^\circ)}
-$$
-
-We can verify this result by drawing phasors corresponding to the two sinusoids. The sinusoid cos ω0*t* is represented by a phasor of unit length at a zero angle with the horizontal. The phasor sin ω0*t* is represented by a unit phasor at an angle of −90◦ with the horizontal. Therefore, − <sup>√</sup>3 sin <sup>ω</sup>0*<sup>t</sup>* is represented by a phasor of length <sup>√</sup>3 at 90◦ with the horizontal, as depicted in Fig. B.8a. The two phasors added yield a phasor of length 2 at 60◦ with the horizontal (also shown in Fig. B.8a).
-
-**Figure B.8** Phasor addition of sinusoids.
-
-Alternately, we note that *a*−*jb* = 1+*j* <sup>√</sup><sup>3</sup> <sup>=</sup> <sup>2</sup>*ej*π/3. Hence, *<sup>C</sup>* <sup>=</sup> 2 and <sup>θ</sup> <sup>=</sup> π/3. Observe that a phase shift of ±π amounts to multiplication by −1. Therefore, *x*(*t*) can also be expressed alternatively as
-
-$$
-x(t) = -2\cos(\omega_0 t + 60^\circ \pm 180^\circ) = -2\cos(\omega_0 t - 120^\circ) = -2\cos(\omega_0 t + 240^\circ)
-$$
-
-In practice, the principal value, that is, −120◦, is preferred.
-
-**(b)** In this case, *a* = −3 and *b* = 4. Using Eq. (B.17) yields
-
-$$
-C = \sqrt{(-3)^2 + 4^2} = 5
-$$
- and  $\theta = \tan^{-1}\left(\frac{-4}{-3}\right) = -126.9^{\circ}$ 
-
-<span id="page-39-0"></span>Observe that
-
-$$
-\tan^{-1}\left(\frac{-4}{-3}\right) \neq \tan^{-1}\left(\frac{4}{3}\right) = 53.1^{\circ}
-$$
-
-Therefore,
-
-$$
-x(t) = 5\cos\left(\omega_0 t - 126.9^\circ\right)
-$$
-
-This result is readily verified in the phasor diagram in Fig. B.8b. Alternately, *a*−*jb* = −3−*j*4 = 5*e*−*j*126.9◦ , a fact readily confirmed using MATLAB.
-
->> C = abs(-3+4j) C=5 >> theta = angle(-3+4j)\*180/pi theta = 126.8699
-
-```
-Hence, C = 5 and θ = −126.8699◦.
-```
-
-We can also perform the reverse operation, expressing *C*cos(ω0*t* +θ ) in terms of cos ω0*t* and sin ω0*t* by again using the trigonometric identity
-
-*C*cos(ω0*t* +θ ) = *C*cos θ cos ω0*t* −*C*sin θ sin ω0*t*
-
-For example,
-
-$$
-10\cos\left(\omega_0 t - 60^\circ\right) = 5\cos\omega_0 t + 5\sqrt{3}\sin\omega_0 t
-$$
-
-### **B.2-2 Sinusoids in Terms of Exponentials**
-
-From Eq. (B.3), we know that *ej*<sup>ϕ</sup> = cos ϕ + *j*sin ϕ and *e*−*j*<sup>ϕ</sup> = cos ϕ − *j*sin ϕ. Adding these two expressions and dividing by 2 provide an expression for cosine in terms of complex exponentials, while subtracting and scaling by 2*j* provide an expression for sine. That is,
-
-$$
-\cos \varphi = \frac{1}{2} (e^{j\varphi} + e^{-j\varphi})
-$$
- and  $\sin \varphi = \frac{1}{2j} (e^{j\varphi} - e^{-j\varphi})$  (B.18)

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/004_B.3 SKETCHING SIGNALS.md

@@ -1,110 +0,0 @@
-## **[B.3 SKETCHING](#page-6-0) SIGNALS**
-
-In this section, we discuss the sketching of a few useful signals, starting with exponentials.
-
-### **[B.3-1 Monotonic Exponentials](#page-6-0)**
-
-The signal *e*−*at* decays monotonically, and the signal *eat* grows monotonically with *t* (assuming *a* > 0), as depicted in Fig. B.9. For the sake of simplicity, we shall consider an exponential *e*−*at* starting at *t* = 0, as shown in Fig. B.10a.
-
-The signal *e*−*at* has a unit value at *t* = 0. At *t* = 1/*a*, the value drops to 1/*e* (about 37% of its initial value), as illustrated in Fig. B.10a. This time interval over which the exponential reduces by
-
-**Figure B.9** Monotonic exponentials.
-
-**Figure B.10** Sketching **(a)** *e*−*at* and **(b)** *e*−2*<sup>t</sup>* .
-
-a factor *e* (i.e., drops to about 37% of its value) is known as the *time constant* of the exponential. Therefore, the time constant of *e*−*at* is 1/*a*. Observe that the exponential is reduced to 37% of its initial value over any time interval of duration 1/*a*. This can be shown by considering any set of instants *t*<sup>1</sup> and *t*<sup>2</sup> separated by one time constant so that
-
-$$
-t_2 - t_1 = \frac{1}{a}
-$$
-
-Now the ratio of *e*−*at*<sup>2</sup> to *e*−*at*<sup>1</sup> is given by
-
-$$
-\frac{e^{-at_2}}{e^{-at_1}} = e^{-a(t_2 - t_1)} = \frac{1}{e} \approx 0.37
-$$
-
-We can use this fact to sketch an exponential quickly. For example, consider
-
-$$
-x(t) = e^{-2t}
-$$
-
-The time constant in this case is 0.5. The value of *x*(*t*) at *t* = 0 is 1. At *t* = 0.5 (one time constant), it is 1/*e* (about 0.37). The value of *x*(*t*) continues to drop further by the factor 1/*e* (37%) over the next half-second interval (one time constant). Thus, *x*(*t*) at *t* = 1 is (1/*e*)2. Continuing in this <span id="page-41-0"></span>manner, we see that *x*(*t*) = (1/*e*)<sup>3</sup> at *t* = 1.5, and so on. A knowledge of the values of *x*(*t*) at *t* = 0, 0.5, 1, and 1.5 allows us to sketch the desired signal, as shown in Fig. B.10b.†
-
-For a monotonically growing exponential *eat*, the waveform increases by a factor *e* over each interval of 1/*a* seconds.
-
-### **[B.3-2 The Exponentially Varying Sinusoid](#page-6-0)**
-
-We now discuss sketching an exponentially varying sinusoid
-
-$$
-x(t) = Ae^{-at}\cos{(\omega_0 t + \theta)}
-$$
-
-Let us consider a specific example:
-
-$$
-x(t) = 4e^{-2t}\cos{(6t - 60^\circ)}
-$$
-
-We shall sketch 4*e*−2*<sup>t</sup>* and cos(6*t* −60◦) separately and then multiply them:
-
-- **(a) Sketching 4***e***−2***<sup>t</sup>* **.** This monotonically decaying exponential has a time constant of 0.5 second and an initial value of 4 at *t* = 0. Therefore, its values at *t* = 0.5, 1, 1.5, and 2 are 4/*e*, 4/*e*2, 4/*e*3, and 4/*e*4, or about 1.47, 0.54, 0.2, and 0.07, respectively. Using these values as a guide, we sketch 4*e*−2*<sup>t</sup>* , as illustrated in Fig. B.11a.
-- **(b) Sketching cos***(***6***<sup>t</sup>* **<sup>−</sup> <sup>60</sup>◦***)***.** The procedure for sketching cos(6*<sup>t</sup>* <sup>−</sup> <sup>60</sup>◦) is discussed in Sec. B.2 (Fig. B.6c). Here, the period of the sinusoid is *T*<sup>0</sup> = 2π/6 ≈ 1, and there is a phase delay of 60◦, or two-thirds of a quarter-cycle, which is equivalent to a delay of about (60/360)(1) ≈ 1/6 seconds (see Fig. B.11b).
-- **(c) Sketching 4***e***−2***<sup>t</sup>* **cos***(***6***<sup>t</sup>* **<sup>−</sup> <sup>60</sup>◦***)***.** We now multiply the waveforms in steps (a) and (b). This multiplication amounts to forcing the sinusoid 4 cos(6*t* −60◦) to decrease exponentially with a time constant of 0.5. The initial amplitude (at *t* = 0) is 4, decreasing to 4/*e* (=1.47) at *t* = 0.5, to 1.47/*e*(=0.54) at *t* = 1, and so on. This is depicted in Fig. B.11c. Note that when cos(6*t* −60◦) has a value of unity (peak amplitude),
-
-$$
-4e^{-2t}\cos{(6t - 60^\circ)} = 4e^{-2t}
-$$
-
-Therefore, 4*e*−2*<sup>t</sup>* cos(6*t*−60◦) touches 4*e*−2*<sup>t</sup>* at the instants at which the sinusoid cos(6*t* −60◦) is at its positive peaks. Clearly, 4*e*−2*<sup>t</sup>* is an envelope for positive amplitudes of 4*e*−2*<sup>t</sup>* cos(6*t* − 60◦). Similar argument shows that 4*e*−2*<sup>t</sup>* cos(6*t* − 60◦) touches −4*e*−2*<sup>t</sup>* at its negative peaks. Therefore, −4*e*−2*<sup>t</sup>* is an envelope for negative amplitudes of 4*e*−2*<sup>t</sup>* cos(6*t* − 60◦). Thus, to sketch 4*e*−2*<sup>t</sup>* cos(6*t* − 60◦), we first draw the envelopes 4*e*−2*<sup>t</sup>* and −4*e*−2*<sup>t</sup>* (the mirror image of 4*e*−2*<sup>t</sup>* about the horizontal axis), and then sketch the sinusoid cos(6*t* − 60◦), with these envelopes acting as constraints on the sinusoid's amplitude (see Fig. B.11c).
-
-In general, *Ke*−*at* cos(ω0*t* + θ ) can be sketched in this manner, with *Ke*−*at* and −*Ke*−*at* constraining the amplitude of cos(ω0*t* +θ ).
-
-<sup>†</sup> If we wish to refine the sketch further, we could consider intervals of half the time constant over which the signal decays by a factor 1/ <sup>√</sup>*e*. Thus, at *<sup>t</sup>* <sup>=</sup> 0.25, *<sup>x</sup>*(*t*) <sup>=</sup> <sup>1</sup>/ <sup>√</sup>*e*, and at *<sup>t</sup>* <sup>=</sup> 0.75, *<sup>x</sup>*(*t*) <sup>=</sup> <sup>1</sup>/*<sup>e</sup>* <sup>√</sup>*e*, and so on.
-
-<span id="page-42-0"></span>**Figure B.11** Sketching an exponentially varying sinusoid.
-
-## **[B.4 CRAMER'S](#page-6-0) RULE**
-
-Cramer's rule offers a very convenient way to solve simultaneous linear equations. Consider a set of *n* linear simultaneous equations in *n* unknowns *x*1, *x*2,..., *xn*:
-
-$$
-a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = y_1
-$$
-  
-\n
-$$
-a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = y_2
-$$
-  
-\n
-$$
-\vdots
-$$
-  
-\n
-$$
-a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n = y_n
-$$
-  
-\n(B.19)
-
-These equations can be expressed in matrix form as
-
-$$
-\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}
-$$
- (B.20)
-
-We denote the matrix on the left-hand side formed by the elements *aij* as **A**. The determinant of **A** is denoted by |**A**|. If the determinant |**A**| is not zero, Eq. (B.19) has a unique solution given by Cramer's formula
-
-$$
-x_k = \frac{|\mathbf{D}_k|}{|\mathbf{A}|} \qquad k = 1, 2, \dots, n
-$$
- (B.21)
-
-where |**D***k*| is obtained by replacing the *k*th column of |**A**| by the column on the right-hand side of Eq. (B.20) (with elements *y*1, *y*2,..., *yn*).
-
-We shall demonstrate the use of this rule with an example.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/005_B.4 CRAMER’S RULE.md

@@ -1,49 +0,0 @@
-### **EXAMPLE B.7 Using Cramer's Rule to Solve a System of Equations**
-
-Use Cramer's rule to solve the following simultaneous linear equations in three unknowns:
-
-$$
-2x_1 + x_2 + x_3 = 3
-$$
-  
-$$
-x_1 + 3x_2 - x_3 = 7
-$$
-  
-$$
-x_1 + x_2 + x_3 = 1
-$$
-
-In matrix form, these equations can be expressed as
-
-| ⎡<br>2 | 1 | ⎤<br>1  | ⎡<br>x1 | ⎤   | ⎡<br>⎤<br>3 |
-|--------|---|---------|---------|-----|-------------|
-| 1<br>⎣ | 3 | −1<br>⎦ | x2<br>⎣ | ⎦ = | 7<br>⎣<br>⎦ |
-| 1      | 1 | 1       | x3      |     | 1           |
-
-Here,
-
-$$
-|\mathbf{A}| = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} = 4
-$$
-
-Since |**A**| = 4 = 0, a unique solution exists for *x*1, *x*2, and *x*3. This solution is provided by Cramer's rule [Eq. (B.21)] as follows:
-
-$$
-x_1 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 3 & 1 & 1 \\ 7 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \frac{8}{4} = 2
-$$
-
-$$
-x_2 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 2 & 3 & 1 \\ 1 & 7 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \frac{4}{4} = 1
-$$
-  
-$$
-x_3 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 2 & 1 & 3 \\ 1 & 3 & 7 \\ 1 & 1 & 1 \end{vmatrix} = \frac{-8}{4} = -2
-$$
-
-<span id="page-44-0"></span>MATLAB is well suited to compute Cramer's formula, so these results are easy to verify. To provide an example, let us verify that *x*<sup>1</sup> = 2 using MATLAB's det command to compute the needed matrix determinants.
-
-```
->> x1 = det([3 1 1;7 3 -1;1 1 1])/det([2 1 1;1 3 -1;1 1 1])
-    x1 = 2.0000
-```

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/006_B.5 PARTIAL FRACTION EXPANSION.md

@@ -1,602 +0,0 @@
-## **[B.5 PARTIAL](#page-6-0) FRACTION EXPANSION**
-
-In the analysis of linear time-invariant systems, we encounter functions that are ratios of two polynomials in a certain variable, say, *x*. Such functions are known as *rational functions*. A rational function *F*(*x*) can be expressed as
-
-$$
-F(x) = \frac{b_m x^m + b_{m-1} x^{m-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0} = \frac{P(x)}{Q(x)}
-$$
-(B.22)
-
-The function *F*(*x*) is *improper* if *m* ≥ *n* and *proper* if *m* < *n*. † An improper function can always be separated into the sum of a polynomial in *x* and a proper function. Consider, for example, the function
-
-$$
-F(x) = \frac{2x^3 + 9x^2 + 11x + 2}{x^2 + 4x + 3}
-$$
-
-Because this is an improper function, we divide the numerator by the denominator until the remainder has a lower degree than the denominator.
-
-$$
-x^{2} + 4x + 3 \quad \begin{array}{c} 2x + 1 \\ 2x^{3} + 9x^{2} + 11x + 2 \\ 2x^{3} + 8x^{2} + 6x \\ x^{2} + 5x + 2 \\ x^{2} + 4x + 3 \\ x - 1 \end{array}
-$$
-
-<sup>†</sup> Some sources classify *<sup>F</sup>*(*x*) as strictly proper if *<sup>m</sup>* <sup>&</sup>lt; *<sup>n</sup>*, proper if *<sup>m</sup>* <sup>≤</sup> *<sup>n</sup>*, and improper if *<sup>m</sup>* <sup>&</sup>gt; *<sup>n</sup>*.
-
-### <span id="page-45-0"></span>26 CHAPTER B BACKGROUND
-
-Therefore, *F*(*x*) can be expressed as
-
-$$
-F(x) = \frac{2x^3 + 9x^2 + 11x + 2}{x^2 + 4x + 3} = \underbrace{2x + 1}_{\text{polynomial in } x} + \underbrace{\frac{x - 1}{x^2 + 4x + 3}}_{\text{proper function}}
-$$
-
-A proper function can be further expanded into partial fractions. The remaining discussion in this section is concerned with various ways of doing this.
-
-### **[B.5-1 Method of Clearing Fractions](#page-6-0)**
-
-A rational function can be written as a sum of appropriate partial fractions with unknown coefficients, which are determined by clearing fractions and equating the coefficients of similar powers on the two sides. This procedure is demonstrated by the following example.
-
-### **EXAMPLE B.8 Method of Clearing Fractions**
-
-Expand the following rational function *F*(*x*) into partial fractions:
-
-$$
-F(x) = \frac{x^3 + 3x^2 + 4x + 6}{(x+1)(x+2)(x+3)^2}
-$$
-
-This function can be expressed as a sum of partial fractions with denominators (*x* + 1), (*x* +2),(*x* +3), and (*x* +3)2, as follows:
-
-$$
-F(x) = \frac{x^3 + 3x^2 + 4x + 6}{(x+1)(x+2)(x+3)^2} = \frac{k_1}{x+1} + \frac{k_2}{x+2} + \frac{k_3}{x+3} + \frac{k_4}{(x+3)^2}
-$$
-
-To determine the unknowns *k*1, *k*2, *k*3, and *k*4, we clear fractions by multiplying both sides by (*x* +1)(*x* +2)(*x* +3)<sup>2</sup> to obtain
-
-$$
-x^{3} + 3x^{2} + 4x + 6 = k_{1}(x^{3} + 8x^{2} + 21x + 18) + k_{2}(x^{3} + 7x^{2} + 15x + 9)
-$$
-  
-+  $k_{3}(x^{3} + 6x^{2} + 11x + 6) + k_{4}(x^{2} + 3x + 2)$   
-=  $x^{3}(k_{1} + k_{2} + k_{3}) + x^{2}(8k_{1} + 7k_{2} + 6k_{3} + k_{4})$   
-+  $x(21k_{1} + 15k_{2} + 11k_{3} + 3k_{4}) + (18k_{1} + 9k_{2} + 6k_{3} + 2k_{4})$ 
-
-Equating coefficients of similar powers on both sides yields
-
-$$
-k_1 + k_2 + k_3 = 1
-$$
-  
-\n
-$$
-8k_1 + 7k_2 + 6k_3 + k_4 = 3
-$$
-  
-\n
-$$
-21k_1 + 15k_2 + 11k_3 + 3k_4 = 4
-$$
-  
-\n
-$$
-18k_1 + 9k_2 + 6k_3 + 2k_4 = 6
-$$
-
-<span id="page-46-0"></span>Solution of these four simultaneous equations yields
-
-$$
-k_1 = 1
-$$
-,  $k_2 = -2$ ,  $k_3 = 2$ ,  $k_4 = -3$ 
-
-Therefore,
-
-$$
-F(x) = \frac{1}{x+1} - \frac{2}{x+2} + \frac{2}{x+3} - \frac{3}{(x+3)^2}
-$$
-
-Although this method is straightforward and applicable to all situations, it is not necessarily the most efficient. We now discuss other methods that can reduce numerical work considerably.
-
-### **[B.5-2 The Heaviside "Cover-Up" Method](#page-6-0)**
-
-### DISTINCT FACTORS OF *Q*(*x*)
-
-We shall first consider the partial fraction expansion of *F*(*x*) = *P*(*x*)/*Q*(*x*), in which all the factors of *Q*(*x*) are distinct (not repeated). Consider the proper function
-
-$$
-F(x) = \frac{b_m x^m + b_{m-1} x^{m-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0} \qquad m < n
-$$
-$$
-= \frac{P(x)}{(x - \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n)}
-$$
-
-As seen in Ex. B.8, *F*(*x*) can be expressed as the sum of partial fractions
-
-$$
-F(x) = \frac{k_1}{x - \lambda_1} + \frac{k_2}{x - \lambda_2} + \dots + \frac{k_n}{x - \lambda_n}
-$$
- (B.23)
-
-To determine the coefficient *k*1, we multiply both sides of Eq. (B.23) by *x*−λ<sup>1</sup> and then let *x* = λ1. This yields
-
-$$
-(x - \lambda_1)F(x)|_{x = \lambda_1} = k_1 + \frac{k_2(x - \lambda_1)}{(x - \lambda_2)} + \frac{k_3(x - \lambda_1)}{(x - \lambda_3)} + \dots + \frac{k_n(x - \lambda_1)}{(x - \lambda_n)}\bigg|_{x = \lambda_1}
-$$
-
-On the right-hand side, all the terms except *k*<sup>1</sup> vanish. Therefore,
-
-$$
-k_1 = (x - \lambda_1)F(x)|_{x = \lambda_1}
-$$
-
-Similarly, we can show that
-
-$$
-k_r = (x - \lambda_r)F(x)|_{x = \lambda_r} \qquad r = 1, 2, \dots, n
-$$
- (B.24)
-
-This procedure also goes under the name *method of residues*.
-
-### **EXAMPLE B.9 Heaviside "Cover-Up" Method**
-
-Expand the following rational function *F*(*x*) into partial fractions:
-
-$$
-F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)} = \frac{k_1}{x+1} + \frac{k_2}{x-2} + \frac{k_3}{x+3}
-$$
-
-To determine *k*1, we let *x* = −1 in (*x* + 1)*F*(*x*). Note that (*x* + 1)*F*(*x*) is obtained from *F*(*x*) by omitting the term (*x* + 1) from its denominator. Therefore, to compute *k*<sup>1</sup> corresponding to the factor (*x* + 1), we cover up the term (*x* + 1) in the denominator of *F*(*x*) and then substitute *x* = −1 in the remaining expression. [Mentally conceal the term (*x* + 1) in *F*(*x*) with a finger and then let *x* = −1 in the remaining expression.] The steps in covering up the function
-
-$$
-F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}
-$$
-
-are as follows.
-
-**Step 1.** Cover up (conceal) the factor (*x* +1) from *F*(*x*):
-
-$$
-\frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}
-$$
-
-**Step 2.** Substitute *x* = −1 in the remaining expression to obtain *k*1:
-
-$$
-k_1 = \frac{2 - 9 - 11}{(-1 - 2)(-1 + 3)} = \frac{-18}{-6} = 3
-$$
-
-Similarly, to compute *k*2, we cover up the factor (*x* − 2) in *F*(*x*) and let *x* = 2 in the remaining function, as follows:
-
-$$
-k_2 = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}\bigg|_{x=2} = \frac{8+18-11}{(2+1)(2+3)} = \frac{15}{15} = 1
-$$
-
-and
-
-$$
-k_3 = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}\bigg|_{x=-3} = \frac{18 - 27 - 11}{(-3+1)(-3-2)} = \frac{-20}{10} = -2
-$$
-
-Therefore,
-
-$$
-F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)} = \frac{3}{x+1} + \frac{1}{x-2} - \frac{2}{x+3}
-$$
-
-### COMPLEX FACTORS OF *Q*(*x*)
-
-The procedure just given works regardless of whether the factors of *Q*(*x*) are real or complex. Consider, for example,
-
-$$
-F(x) = \frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)}
-$$
-  
-=  $\frac{k_1}{x+1} + \frac{k_2}{x+2-j3} + \frac{k_3}{x+2+j3}$  (B.25)
-
-where
-
-$$
-k_1 = \left[ \frac{4x^2 + 2x + 18}{(x+1)\ (x^2 + 4x + 13)} \right]_{x=-1} = 2
-$$
-
-Similarly,
-
-$$
-k_2 = \left[ \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)} \right]_{x=-2+j3} = 1+j2 = \sqrt{5}e^{j63.43^{\circ}}
-$$
-$$
-k_3 = \left[ \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)} \right]_{x=-2-j3} = 1-j2 = \sqrt{5}e^{-j63.43^{\circ}}
-$$
-
-Therefore,
-
-$$
-F(x) = \frac{2}{x+1} + \frac{\sqrt{5}e^{i63.43^{\circ}}}{x+2-j3} + \frac{\sqrt{5}e^{-i63.43^{\circ}}}{x+2+j3}
-$$
-
-The coefficients *k*<sup>2</sup> and *k*<sup>3</sup> corresponding to the complex-conjugate factors are also conjugates of each other. This is generally true when the coefficients of a rational function are real. In such a case, we need to compute only one of the coefficients.
-
-### QUADRATIC FACTORS
-
-Often we are required to combine the two terms arising from complex-conjugate factors into one quadratic factor. For example, *F*(*x*) in Eq. (B.25) can be expressed as
-
-$$
-F(x) = \frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{k_1}{x+1} + \frac{c_1x + c_2}{x^2 + 4x + 13}
-$$
-
-The coefficient *k*<sup>1</sup> is found by the Heaviside method to be 2. Therefore,
-
-$$
-\frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{2}{x+1} + \frac{c_1x + c_2}{x^2 + 4x + 13}
-$$
-(B.26)
-
-The values of *c*<sup>1</sup> and *c*<sup>2</sup> are determined by clearing fractions and equating the coefficients of similar powers of *x* on both sides of the resulting equation. Clearing fractions on both sides of Eq. (B.26) yields
-
-$$
-4x2 + 2x + 18 = 2(x2 + 4x + 13) + (c1x + c2)(x + 1)
-$$
-  
-= (2+c<sub>1</sub>)x<sup>2</sup> + (8+c<sub>1</sub>+c<sub>2</sub>)x + (26+c<sub>2</sub>)
-
-Equating terms of similar powers yields *c*<sup>1</sup> = 2, *c*<sup>2</sup> = −8, and
-
-$$
-\frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{2}{x+1} + \frac{2x - 8}{x^2 + 4x + 13}
-$$
-
-### SHORTCUTS
-
-The values of *c*<sup>1</sup> and *c*<sup>2</sup> in Eq. (B.26) can also be determined by using shortcuts. After computing *k*<sup>1</sup> = 2 by the Heaviside method as before, we let *x* = 0 on both sides of Eq. (B.26) to eliminate *c*1. This gives us
-
-$$
-\frac{18}{13} = 2 + \frac{c_2}{13} \qquad \Rightarrow \qquad c_2 = -8
-$$
-
-To determine *c*1, we multiply both sides of Eq. (B.26) by *x* and then let *x* → ∞. Remember that when *x* → ∞, only the terms of the highest power are significant. Therefore,
-
-$$
-4 = 2 + c_1 \qquad \Rightarrow \qquad c_1 = 2
-$$
-
-In the procedure discussed here, we let *x* = 0 to determine *c*<sup>2</sup> and then multiply both sides by *x* and let *x* → ∞ to determine *c*1. However, nothing is sacred about these values (*x* = 0 or *x* = ∞). We use them because they reduce the number of computations involved. We could just as well use other convenient values for *x*, such as *x* = 1. Consider the case
-
-$$
-F(x) = \frac{2x^2 + 4x + 5}{x(x^2 + 2x + 5)} = \frac{k}{x} + \frac{c_1x + c_2}{x^2 + 2x + 5}
-$$
-
-We find *k* = 1 by the Heaviside method in the usual manner. As a result,
-
-$$
-\frac{2x^2 + 4x + 5}{x(x^2 + 2x + 5)} = \frac{1}{x} + \frac{c_1x + c_2}{x^2 + 2x + 5}
-$$
-(B.27)
-
-If we try letting *x* = 0 to determine *c*<sup>1</sup> and *c*2, we obtain ∞ on both sides. So let us choose *x* = 1. This yields
-
-$$
-\frac{11}{8} = 1 + \frac{c_1 + c_2}{8} \qquad \text{or} \qquad c_1 + c_2 = 3
-$$
-
-We can now choose some other value for *x*, such as *x* = 2, to obtain one more relationship to use in determining *c*<sup>1</sup> and *c*2. In this case, however, a simple method is to multiply both sides of Eq. (B.27) by *x* and then let *x* → ∞. This yields
-
-$$
-2 = 1 + c_1 \qquad \Rightarrow \qquad c_1 = 1
-$$
-
-Since *c*<sup>1</sup> +*c*<sup>2</sup> = 3, we see that *c*<sup>2</sup> = 2 and therefore,
-
-$$
-F(x) = \frac{1}{x} + \frac{x+2}{x^2 + 2x + 5}
-$$
-
-## <span id="page-50-0"></span>**[B.5-3 Repeated Factors of](#page-6-0)** *Q(x)*
-
-If a function *F*(*x*) has a repeated factor in its denominator, it has the form
-
-$$
-F(x) = \frac{P(x)}{(x - \lambda)^r (x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_j)}
-$$
-
-Its partial fraction expansion is given by
-
-$$
-F(x) = \frac{a_0}{(x - \lambda)^r} + \frac{a_1}{(x - \lambda)^{r-1}} + \dots + \frac{a_{r-1}}{(x - \lambda)}
-$$
-  
-+ 
-$$
-\frac{k_1}{x - \alpha_1} + \frac{k_2}{x - \alpha_2} + \dots + \frac{k_j}{x - \alpha_j}
-$$
- (B.28)
-
-The coefficients *k*1, *k*2,..., *kj* corresponding to the unrepeated factors in this equation are determined by the Heaviside method, as before [Eq. (B.24)]. To find the coefficients *a*0,*a*1, *a*2,...,*ar*−1, we multiply both sides of Eq. (B.28) by (*x* −λ)*<sup>r</sup>* . This gives us
-
-$$
-(x - \lambda)^r F(x) = a_0 + a_1(x - \lambda) + a_2(x - \lambda)^2 + \dots + a_{r-1}(x - \lambda)^{r-1} + k_1 \frac{(x - \lambda)^r}{x - \alpha_1} + k_2 \frac{(x - \lambda)^r}{x - \alpha_2} + \dots + k_n \frac{(x - \lambda)^r}{x - \alpha_n}
-$$
-(B.29)
-
-If we let *x* = λ on both sides of Eq. (B.29), we obtain
-
-$$
-(x - \lambda)^r F(x)|_{x = \lambda} = a_0
-$$
-
-Therefore, *a*<sup>0</sup> is obtained by concealing the factor (*x*−λ)*<sup>r</sup>* in *F*(*x*) and letting *x* =λ in the remaining expression (the Heaviside "cover-up" method). If we take the derivative (with respect to *x*) of both sides of Eq. (B.29), the right-hand side is *a*1+ terms containing a factor (*x*−λ) in their numerators. Letting *x* = λ on both sides of this equation, we obtain
-
-$$
-\frac{d}{dx}\left[ (x - \lambda)^r F(x) \right] \Big|_{x = \lambda} = a_1
-$$
-
-Thus, *a*<sup>1</sup> is obtained by concealing the factor (*x*−λ)*<sup>r</sup>* in *F*(*x*), taking the derivative of the remaining expression, and then letting *x* = λ. Continuing in this manner, we find
-
-$$
-a_j = \frac{1}{j!} \left. \frac{d^j}{dx^j} \left[ (x - \lambda)^r F(x) \right] \right|_{x = \lambda}
-$$
- (B.30)
-
-Observe that (*x* − λ)*<sup>r</sup> F*(*x*) is obtained from *F*(*x*) by omitting the factor (*x* − λ)*<sup>r</sup>* from its denominator. Therefore, the coefficient *aj* is obtained by concealing the factor (*x* − λ)*<sup>r</sup>* in *F*(*x*), taking the *j*th derivative of the remaining expression, and then letting *x* = λ (while dividing by *j*!).
-
-### <span id="page-51-0"></span>**EXAMPLE B.10 Partial Fraction Expansion with Repeated Factors**
-
-Expand *F*(*x*) into partial fractions if
-
-$$
-F(x) = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)}
-$$
-
-The partial fractions are
-
-$$
-F(x) = \frac{a_0}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{k}{x+2}
-$$
-
-The coefficient *k* is obtained by concealing the factor (*x* + 2) in *F*(*x*) and then substituting *x* = −2 in the remaining expression:
-
-$$
-k = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)}\bigg|_{x=-2} = 1
-$$
-
-To find *a*0, we conceal the factor (*x* +1)<sup>3</sup> in *F*(*x*) and let *x* = −1 in the remaining expression:
-
-$$
-a_0 = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)}\bigg|_{x=-1} = 2
-$$
-
-To find *a*1, we conceal the factor (*x* + 1)<sup>3</sup> in *F*(*x*), take the derivative of the remaining expression, and then let *x* = −1:
-
-$$
-a_1 = \frac{d}{dx} \left[ \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} \right] \Big|_{x=-1} = 1
-$$
-
-Similarly,
-
-$$
-a_2 = \frac{1}{2!} \frac{d^2}{dx^2} \left[ \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} \right] \Big|_{x=-1} = 3
-$$
-
-Therefore,
-
-$$
-F(x) = \frac{2}{(x+1)^3} + \frac{1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2}
-$$
-
-### **[B.5-4 A Combination of Heaviside "Cover-Up" and Clearing Fractions](#page-6-0)**
-
-For multiple roots, especially of higher order, the Heaviside expansion method, which requires repeated differentiation, can become cumbersome. For a function that contains several repeated and unrepeated roots, a hybrid of the two procedures proves to be the best. The simpler coefficients are determined by the Heaviside method, and the remaining coefficients are found by clearing fractions or shortcuts, thus incorporating the best of the two methods. We demonstrate this procedure by solving Ex. B.10 once again by this method.
-
-In Ex. B.10, coefficients *k* and *a*<sup>0</sup> are relatively simple to determine by the Heaviside expansion method. These values were found to be *k*<sup>1</sup> = 1 and *a*<sup>0</sup> = 2. Therefore,
-
-$$
-\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{1}{x+2}
-$$
-
-We now multiply both sides of this equation by (*x* +1)3(*x* +2) to clear the fractions. This yields
-
-$$
-4x3 + 16x2 + 23x + 13 = 2(x+2) + a1(x+1)(x+2) + a2(x+1)2(x+2) + (x+1)3
-$$
-  
-= (1+a<sub>2</sub>)x<sup>3</sup> + (a<sub>1</sub>+4a<sub>2</sub>+3)x<sup>2</sup> + (5+3a<sub>1</sub>+5a<sub>2</sub>)x + (4+2a<sub>1</sub>+2a<sub>2</sub>+1)
-
-Equating coefficients of the third and second powers of *x* on both sides, we obtain
-
-$$
-\begin{array}{ccc} 1 + a_2 = 4 \\ a_1 + 4a_2 + 3 = 16 \end{array} \implies \begin{array}{c} a_1 = 1 \\ a_2 = 3 \end{array}
-$$
-
-We may stop here if we wish because the two desired coefficients, *a*<sup>1</sup> and *a*2, are now determined. However, equating the coefficients of the two remaining powers of *x* yields a convenient check on the answer. Equating the coefficients of the *x*<sup>1</sup> and *x*<sup>0</sup> terms, we obtain
-
-$$
-23 = 5 + 3a_1 + 5a_2
-$$
-  
-$$
-13 = 4 + 2a_1 + 2a_2 + 1
-$$
-
-These equations are satisfied by the values *a*<sup>1</sup> = 1 and *a*<sup>2</sup> = 3, found earlier, providing an additional check for our answers. Therefore,
-
-$$
-F(x) = \frac{2}{(x+1)^3} + \frac{1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2}
-$$
-
-which agrees with the earlier result.
-
-### A COMBINATION OF HEAVISIDE "COVER-UP" AND SHORTCUTS
-
-In Ex. B.10, after determining the coefficients *a*<sup>0</sup> = 2 and *k* = 1 by the Heaviside method as before, we have
-
-$$
-\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{1}{x+2}
-$$
-
-There are only two unknown coefficients, *a*<sup>1</sup> and *a*2. If we multiply both sides of this equation by *x* and then let *x* → ∞, we can eliminate *a*1. This yields
-
-$$
-4 = a_2 + 1 \quad \Longrightarrow \quad a_2 = 3
-$$
-
-Therefore,
-
-$$
-\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2}
-$$
-
-#### <span id="page-53-0"></span>34 CHAPTER B BACKGROUND
-
-There is now only one unknown *a*1, which can be readily found by setting *x* equal to any convenient value, say, *x* = 0. This yields
-
-$$
-\frac{13}{2} = 2 + a_1 + 3 + \frac{1}{2} \implies a_1 = 1
-$$
-
-which agrees with our earlier answer.
-
-There are other possible shortcuts. For example, we can compute *a*<sup>0</sup> (coefficient of the highest power of the repeated root), subtract this term from both sides, and then repeat the procedure.
-
-## **[B.5-5 Improper](#page-6-0)** *<sup>F</sup>(x)* **with** *<sup>m</sup>* **<sup>=</sup>** *<sup>n</sup>*
-
-A general method of handling an improper function is indicated in the beginning of this section. However, for the special case of when the numerator and denominator polynomials of *F*(*x*) have the same degree (*m* = *n*), the procedure is the same as that for a proper function. We can show that for
-
-$$
-F(x) = \frac{b_n x^n + b_{n-1} x^{n-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0}
-$$
-  
-=  $b_n + \frac{k_1}{x - \lambda_1} + \frac{k_2}{x - \lambda_2} + \dots + \frac{k_n}{x - \lambda_n}$ 
-
-the coefficients *k*1, *k*2,..., *kn* are computed as if *F*(*x*) were proper. Thus,
-
-$$
-k_r = (x - \lambda_r)F(x)|_{x = \lambda_r}
-$$
-
-For quadratic or repeated factors, the appropriate procedures discussed in Secs. B.5-2 or B.5-3 should be used as if *F*(*x*) were proper. In other words, when *m* = *n*, the only difference between the proper and improper case is the appearance of an extra constant *bn* in the latter. Otherwise, the procedure remains the same. The proof is left as an exercise for the reader.
-
-### **EXAMPLE B.11 Partial Fraction Expansion of Improper Rational Function**
-
-Expand *F*(*x*) into partial fractions if
-
-$$
-F(x) = \frac{3x^2 + 9x - 20}{x^2 + x - 6} = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)}
-$$
-
-Here, *m* = *n* = 2 with *bn* = *b*<sup>2</sup> = 3. Therefore,
-
-$$
-F(x) = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} = 3 + \frac{k_1}{x - 2} + \frac{k_2}{x + 3}
-$$
-
-<span id="page-54-0"></span>in which
-
-$$
-k_1 = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} \bigg|_{x=2} = \frac{12 + 18 - 20}{(2 + 3)} = \frac{10}{5} = 2
-$$
-
-and
-
-$$
-k_2 = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)}\bigg|_{x=-3} = \frac{27 - 27 - 20}{(-3 - 2)} = \frac{-20}{-5} = 4
-$$
-
-Therefore,
-
-$$
-F(x) = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} = 3 + \frac{2}{x - 2} + \frac{4}{x + 3}
-$$
-
-### **[B.5-6 Modified Partial Fractions](#page-6-0)**
-
-In finding the inverse *z*-transform (Ch. 5), we require partial fractions of the form *kx*/(*x* −λ*i*)*<sup>r</sup>* rather than *k*/(*x* −λ*i*)*<sup>r</sup>* . This can be achieved by expanding *F*(*x*)/*x* into partial fractions. Consider, for example,
-
-$$
-F(x) = \frac{5x^2 + 20x + 18}{(x+2)(x+3)^2}
-$$
-
-Dividing both sides by *x* yields
-
-$$
-\frac{F(x)}{x} = \frac{5x^2 + 20x + 18}{x(x+2)(x+3)^2}
-$$
-
-Expansion of the right-hand side into partial fractions as usual yields
-
-$$
-\frac{F(x)}{x} = \frac{5x^2 + 20x + 18}{x(x+2)(x+3)^2} = \frac{a_1}{x} + \frac{a_2}{x+2} + \frac{a_3}{(x+3)} + \frac{a_4}{(x+3)^2}
-$$
-
-Using the procedure discussed earlier, we find *a*<sup>1</sup> = 1, *a*<sup>2</sup> = 1, *a*<sup>3</sup> = −2, and *a*<sup>4</sup> = 1. Therefore,
-
-$$
-\frac{F(x)}{x} = \frac{1}{x} + \frac{1}{x+2} - \frac{2}{x+3} + \frac{1}{(x+3)^2}
-$$
-
-Now multiplying both sides by *x* yields
-
-$$
-F(x) = 1 + \frac{x}{x+2} - \frac{2x}{x+3} + \frac{x}{(x+3)^2}
-$$
-
-This expresses *F*(*x*) as the sum of partial fractions having the form *kx*/(*x* −λ*i*)*<sup>r</sup>* .
-
-## <span id="page-55-0"></span>**[B.6 VECTORS AND](#page-6-0) MATRICES**
-
-An entity specified by *n* numbers in a certain order (ordered *n*-tuple) is an *n*-dimensional *vector*. Thus, an ordered *n*-tuple (*x*1, *x*2, ..., *xn*) represents an *n*-dimensional vector **x**. A vector may be represented as a row (*row vector*):
-
-$$
-\mathbf{x} = [x_1 \quad x_2 \quad \cdots \quad x_n]
-$$
-
-or as a column (*column vector*):
-
-$$
-\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}
-$$
-
-Simultaneous linear equations can be viewed as the transformation of one vector into another. Consider, for example, the *m* simultaneous linear equations
-
-$$
-y_1 = a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n
-$$
-  
-\n
-$$
-y_2 = a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n
-$$
-  
-\n
-$$
-\vdots
-$$
-  
-\n
-$$
-y_m = a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n
-$$
-  
-\n(B.31)
-
-If we define two column vectors **x** and **y** as
-
-$$
-\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \quad \text{and} \quad \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix}
-$$
-
-then Eq. (B.31) may be viewed as the relationship or the function that transforms vector **x** into vector **y**. Such a transformation is called a *linear transformation* of vectors. To perform a linear transformation, we need to define the array of coefficients *aij* appearing in Eq. (B.31). This array is called a *matrix* and is denoted by **A** for convenience:
-
-$$
-\mathbf{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{array} \right]
-$$
-
-A matrix with *m* rows and *n* columns is called a matrix of order (*m*,*n*) or an (*m* × *n*) matrix. For the special case of *m* = *n*, the matrix is called a *square matrix* of order *n*.
-
-It should be stressed at this point that a matrix is not a number such as a determinant, but an array of numbers arranged in a particular order. It is convenient to abbreviate the representation of matrix **A** with the form (*aij*)*<sup>m</sup>*×*<sup>n</sup>*, implying a matrix of order *m* × *n* with *aij* as its *ij*th element. In practice, when the order *m* × *n* is understood or need not be specified, the notation can be <span id="page-56-0"></span>abbreviated to (*aij*). Note that the first index *i* of *aij* indicates the row and the second index *j* indicates the column of the element *aij* in matrix **A**.
-
-Equation (B.31) may now be expressed in a matrix form as
-
-$$
-\begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}
-$$
- or  $y = Ax$  (B.32)
-
-At this point, we have not defined the multiplication of a matrix by a vector. The quantity **Ax** is not meaningful until such an operation has been defined.

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/007_B.6 VECTORS AND MATRICES.md

@@ -1,214 +0,0 @@
-### **[B.6-1 Some Definitions and Properties](#page-6-0)**
-
-A square matrix whose elements are zero everywhere except on the main diagonal is a *diagonal matrix*. An example of a diagonal matrix is
-
-| ⎡<br>2 | 0 | 0 | ⎤ |
-|--------|---|---|---|
-| 0<br>⎣ | 1 | 0 | ⎦ |
-| 0      | 0 | 5 |   |
-
-A diagonal matrix with unity for all its diagonal elements is called an *identity matrix* or a *unit matrix*, denoted by **I**. This is a square matrix:
-
-|        | ⎡<br>1      | 0 | 0 | ··· | ⎤<br>0      |
-|--------|-------------|---|---|-----|-------------|
-|        | 0<br>⎢      | 1 | 0 | ··· | 0<br>⎥      |
-| I<br>= | ⎢<br>0<br>⎢ | 0 | 1 | ··· | ⎥<br>0<br>⎥ |
-|        | ⎢<br>⎢<br>⎣ |   |   | ··· | ⎥<br>⎥<br>⎦ |
-|        | 0           | 0 | 0 | ··· | 1           |
-
-The order of the unit matrix is sometimes indicated by a subscript. Thus, **I***<sup>n</sup>* represents the *n*×*n* unit matrix (or identity matrix). However, we shall omit the subscript since order is easily understood by context.
-
-A matrix having all its elements zero is a *zero matrix*.
-
-A square matrix **A** is a *symmetric matrix* if *aij* = *aji* (symmetry about the main diagonal).
-
-Two matrices of the same order are said to be *equal* if they are equal element by element. Thus, if
-
-$$
-\mathbf{A} = (a_{ij})_{m \times n} \quad \text{and} \quad \mathbf{B} = (b_{ij})_{m \times n}
-$$
-
-then **A** = **B** only if *aij* = *bij* for all *i* and *j*.
-
-If the rows and columns of an *m*×*n* matrix **A** are interchanged so that the elements in the *i*th row now become the elements of the *i*th column (for *i* = 1, 2,...,*m*), the resulting matrix is called the *transpose* of **<sup>A</sup>** and is denoted by **<sup>A</sup>***<sup>T</sup>* . It is evident that **<sup>A</sup>***<sup>T</sup>* is an *<sup>n</sup>*×*<sup>m</sup>* matrix. For example, if
-
-$$
-\mathbf{A} = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 3 \end{bmatrix}, \quad \text{then} \quad \mathbf{A}^T = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 2 & 3 \end{bmatrix}
-$$
-
-### <span id="page-57-0"></span>38 CHAPTER B BACKGROUND
-
-Using the abbreviated notation, if **<sup>A</sup>** <sup>=</sup> (*aij*)*<sup>m</sup>*×*<sup>n</sup>*, then **<sup>A</sup>***<sup>T</sup>* <sup>=</sup> (*aji*)*<sup>n</sup>*×*<sup>m</sup>*. Intuitively, further notice that (**A***<sup>T</sup>* )*<sup>T</sup>* <sup>=</sup> **<sup>A</sup>**.
-
-### **[B.6-2 Matrix Algebra](#page-6-0)**
-
-We shall now define matrix operations, such as addition, subtraction, multiplication, and division of matrices. The definitions should be formulated so that they are useful in the manipulation of matrices.
-
-### ADDITION OF MATRICES
-
-For two matrices **A** and **B**, both of the same order (*m*×*n*),
-
-$$
-\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \text{ and } \mathbf{B} = \begin{bmatrix} b_{11} & b_{12} & \cdots & b_{1n} \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ b_{m1} & b_{m2} & \cdots & b_{mn} \end{bmatrix}
-$$
-
-we define the sum **A**+**B** as
-
-$$
-\mathbf{A} + \mathbf{B} = \begin{bmatrix} (a_{11} + b_{11}) & (a_{12} + b_{12}) & \cdots & (a_{1n} + b_{1n}) \\ (a_{21} + b_{21}) & (a_{22} + b_{22}) & \cdots & (a_{2n} + b_{2n}) \\ \vdots & \vdots & \ddots & \vdots \\ (a_{m1} + b_{m1}) & (a_{m2} + b_{m2}) & \cdots & (a_{mn} + b_{mn}) \end{bmatrix}
-$$
-
-or
-
-$$
-\mathbf{A} + \mathbf{B} = (a_{ij} + b_{ij})_{m \times n}
-$$
-
-Note that two matrices can be added only if they are of the same order.
-
-### MULTIPLICATION OF A MATRIX BY A SCALAR
-
-We multiply a matrix **A** by a scalar *c* as follows:
-
-$$
-c\mathbf{A} = c \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} = \begin{bmatrix} ca_{11} & ca_{12} & \cdots & ca_{1n} \\ ca_{21} & ca_{22} & \cdots & ca_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ ca_{m1} & ca_{m2} & \cdots & ca_{mn} \end{bmatrix} = \mathbf{A}c
-$$
-
-Thus, we also observe that the scalar *c* and the matrix **A** commute: *c***A** = **A***c*.
-
-### MATRIX MULTIPLICATION
-
-We define the product
-
-$$
-AB = C
-$$
-
-in which *cij*, the element of **C** in the *i*th row and *j*th column, is found by adding the products of the elements of **A** in the *i*th row multiplied by the corresponding elements of **B** in the *j*th column. Thus,
-
-$$
-c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \dots + a_{in}b_{nj} = \sum_{k=1}^{n} a_{ik}b_{kj}
-$$
- (B.33)
-
-This result is expressed as follows:
-
-Note carefully that if this procedure is to work, the number of columns of **A** must be equal to the number of rows of **B**. In other words, **AB**, the product of matrices **A** and **B**, is defined only if the number of columns of **A** is equal to the number of rows of **B**. If this condition is not satisfied, the product **AB** is not defined and is meaningless. When the number of columns of **A** is equal to the number of rows of **B**, matrix **A** is said to be *conformable* to matrix **B** for the product **AB**. Observe that if **A** is an *m* × *n* matrix and **B** is an *n* × *p* matrix, **A** and **B** are conformable for the product, and **C** is an *m*×*p* matrix.
-
-We demonstrate the use of the rule in Eq. (B.33) with the following examples.
-
-$$
-\begin{bmatrix} 2 & 3 \\ 1 & 1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 1 & 2 \\ 2 & 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 8 & 9 & 5 & 7 \\ 3 & 4 & 2 & 3 \\ 5 & 10 & 4 & 7 \end{bmatrix}
-$$
-$$
-\begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = 8
-$$
-
-In both cases, the two matrices are conformable. However, if we interchange the order of the first matrices as follows:
-
-| 1 | 3 | 1 | !⎡<br>2 | 2      | 3 | ⎤ |
-|---|---|---|---------|--------|---|---|
-|   |   |   |         | 1<br>⎣ | 1 | ⎦ |
-| 2 | 1 | 1 | 1       | 3      | 1 |   |
-
-the matrices are no longer conformable for the product. It is evident that, in general,
-
-### **AB** = **BA**
-
-Indeed, **AB** may exist and **BA** may not exist, or vice versa, as in our examples. We shall see later that for some special matrices, **AB** = **BA**. When this is true, matrices **A** and **B** are said to *commute*. We re-emphasize that in general, matrices do not commute.
-
-### 40 CHAPTER B BACKGROUND
-
-In the matrix product **AB**, matrix **A** is said to be *postmultiplied* by **B** or matrix **B** is said to be *premultiplied* by **A**. We may also verify the following relationships:
-
-$$
-(A + B)C = AC + BC
-$$
-$$
-C(A + B) = CA + CB
-$$
-
-We can verify that any matrix **A** premultiplied or postmultiplied by the identity matrix **I** remains unchanged:
-
-**AI** = **IA** = **A**
-
-Of course, we must make sure that the order of **I** is such that the matrices are conformable for the corresponding product.
-
-We give here, without proof, another important property of matrices:
-
-$$
-|\mathbf{A}\mathbf{B}| = |\mathbf{A}||\mathbf{B}|
-$$
-
-where |**A**| and |**B**| represent determinants of matrices **A** and **B**.
-
-### MULTIPLICATION OF A MATRIX BY A VECTOR
-
-Consider Eq. (B.32), which represents Eq. (B.31). The right-hand side of Eq. (B.32) is a product of the *m*×*n* matrix **A** and a vector **x**. If, for the time being, we treat the vector **x** as if it were an *n*×1 matrix, then the product **Ax**, according to the matrix multiplication rule, yields the right-hand side of Eq. (B.31). Thus, we may multiply a matrix by a vector by treating the vector as if it were an *n* × 1 matrix. Note that the constraint of conformability still applies. Thus, in this case, **xA** is not defined and is meaningless.
-
-### MATRIX INVERSION
-
-To define the inverse of a matrix, let us consider the set of equations represented by Eq. (B.32) when *m* = *n*:
-
-$$
-\begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}
-$$
- (B.34)
-
-We can solve this set of equations for *x*1, *x*2, ... , *xn* in terms of *y*1, *y*2, ... , *yn* by using Cramer's rule [see Eq. (B.21)]. This yields
-
-$$
-\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} \frac{|\mathbf{D}_{11}|}{|\mathbf{A}|} & \frac{|\mathbf{D}_{21}|}{|\mathbf{A}|} & \cdots & \frac{|\mathbf{D}_{n1}|}{|\mathbf{A}|} \\ \frac{|\mathbf{D}_{12}|}{|\mathbf{A}|} & \frac{|\mathbf{D}_{22}|}{|\mathbf{A}|} & \cdots & \frac{|\mathbf{D}_{n2}|}{|\mathbf{A}|} \\ \vdots & \vdots & \cdots & \vdots \\ \frac{|\mathbf{D}_{1n}|}{|\mathbf{A}|} & \frac{|\mathbf{D}_{2n}|}{|\mathbf{A}|} & \cdots & \frac{|\mathbf{D}_{nn}|}{|\mathbf{A}|} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}
-$$
-(B.35)
-
-in which |**A**| is the determinant of the matrix **A** and |**D***ij*| is the *cofactor* of element *aij* in the matrix **A**. The cofactor of element *aij* is given by (−1)*<sup>i</sup>*+*<sup>j</sup>* times the determinant of the (*n* − 1) × (*n* − 1) matrix that is obtained when the *i*th row and the *j*th column in matrix **A** are deleted.
-
-We can express Eq. (B.34) in compact matrix form as
-
-$$
-y = Ax
-$$
- (B.36)
-
-We now define **A**−<sup>1</sup> , the inverse of a square matrix **A**, with the property
-
-> **A**−<sup>1</sup> **A** = **I** (unit matrix)
-
-Then, premultiplying both sides of Eq. (B.36) by **A**−<sup>1</sup> , we obtain
-
-$$
-\mathbf{A}^{-1}\mathbf{y} = \mathbf{A}^{-1}\mathbf{A}\mathbf{x} = \mathbf{I}\mathbf{x} = \mathbf{x}
-$$
-
-or
-
-$$
-\mathbf{x} = \mathbf{A}^{-1} \mathbf{y} \tag{B.37}
-$$
-
-A comparison of Eq. (B.37) with Eq. (B.35) shows that
-
-$$
-A^{-1} = \frac{1}{|A|} \begin{bmatrix} |D_{11}| & |D_{21}| & \cdots & |D_{n1}| \\ |D_{12}| & |D_{22}| & \cdots & |D_{n2}| \\ \vdots & \vdots & \cdots & \vdots \\ |D_{1n}| & |D_{2n}| & \cdots & |D_{nn}| \end{bmatrix}
-$$
-
-One of the conditions necessary for a unique solution of Eq. (B.34) is that the number of equations must equal the number of unknowns. This implies that the matrix **A** must be a square matrix. In addition, we observe from the solution as given in Eq. (B.35) that if the solution is to exist, |**A**| = 0.† Therefore, the inverse exists only for a square matrix and only under the condition that the determinant of the matrix be nonzero. A matrix whose determinant is nonzero is a *nonsingular* matrix. Thus, an inverse exists only for a nonsingular, square matrix. Since **A**−<sup>1</sup> **<sup>A</sup>** <sup>=</sup> **<sup>I</sup>** <sup>=</sup> **AA**−<sup>1</sup> , we further note that the matrices **A** and **A**−<sup>1</sup> commute.‡
-
-The operation of matrix division can be accomplished through matrix inversion.
-
-### **EXAMPLE B.12 Computing the Inverse of a Matrix**
-
-Let us find **A**−<sup>1</sup> if
-
-**A** = ⎡ ⎣ 211 123 321 ⎤ ⎦
-
-<sup>†</sup> These two conditions imply that the number of equations is equal to the number of unknowns and that all the equations are independent.
-
-<sup>‡</sup> To prove **AA**−<sup>1</sup> <sup>=</sup> **<sup>I</sup>**, notice first that we define **<sup>A</sup>**−1**<sup>A</sup>** <sup>=</sup> **<sup>I</sup>**. Thus, **IA** <sup>=</sup> **AI** <sup>=</sup> **<sup>A</sup>**(**A**−1**A**) <sup>=</sup> (**AA**−1)**A**. Subtracting (**AA**−1)**A**, we see that **IA**−(**AA**−1)**<sup>A</sup>** <sup>=</sup> 0 or (**I**−**AA**−1)**<sup>A</sup>** <sup>=</sup> 0. This requires **AA**−<sup>1</sup> <sup>=</sup> **<sup>I</sup>**.
-
-<span id="page-61-0"></span>Here,
-
-$$
-|\mathbf{D}_{11}| = -4, \t |\mathbf{D}_{12}| = 8, \t |\mathbf{D}_{13}| = -4 |\mathbf{D}_{21}| = 1, \t |\mathbf{D}_{22}| = -1, \t |\mathbf{D}_{23}| = -1 |\mathbf{D}_{31}| = 1, \t |\mathbf{D}_{32}| = -5, \t |\mathbf{D}_{33}| = 3 \text{and } |\mathbf{A}| = -4. \text{ Therefore, } \mathbf{A}^{-1} = -\frac{1}{4} \begin{bmatrix} -4 & 1 & 1 \\ 8 & -1 & -5 \\ -4 & -1 & 3 \end{bmatrix}
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/008_B.7 MATLAB - ELEMENTARY OPERATIONS.md

@@ -1,424 +0,0 @@
-## **[B.7 MATLAB: ELEMENTARY](#page-6-0) OPERATIONS**
-
-### **[B.7-1 MATLAB Overview](#page-6-0)**
-
-Although MATLAB´l (a registered trademark of The MathWorks, Inc.) is easy to use, it can be intimidating to new users. Over the years, MATLAB has evolved into a sophisticated computational package with thousands of functions and thousands of pages of documentation. This section provides a brief introduction to the software environment.
-
-When MATLAB is first launched, its command window appears. When MATLAB is ready to accept an instruction or input, a command prompt (>>) is displayed in the command window. Nearly all MATLAB activity is initiated at the command prompt.
-
-Entering instructions at the command prompt generally results in the creation of an object or objects. Many classes of objects are possible, including functions and strings, but usually objects are just data. Objects are placed in what is called the MATLAB workspace. If not visible, the workspace can be viewed in a separate window by typing workspace at the command prompt. The workspace provides important information about each object, including the object's name, size, and class.
-
-Another way to view the workspace is the whos command. When whos is typed at the command prompt, a summary of the workspace is printed in the command window. The who command is a short version of whos that reports only the names of workspace objects.
-
-Several functions exist to remove unnecessary data and help free system resources. To remove specific variables from the workspace, the clear command is typed, followed by the names of the variables to be removed. Just typing clear removes all objects from the workspace. Additionally, the clc command clears the command window, and the clf command clears the current figure window.
-
-Often, important data and objects created in one session need to be saved for future use. The save command, followed by the desired filename, saves the entire workspace to a file, which has the .mat extension. It is also possible to selectively save objects by typing save followed by the filename and then the names of the objects to be saved. The load command followed by the filename is used to load the data and objects contained in a MATLAB data file (.mat file).
-
-Although MATLAB does not automatically save workspace data from one session to the next, lines entered at the command prompt are recorded in the command history. Previous command lines can be viewed, copied, and executed directly from the command history window. From the <span id="page-62-0"></span>command window, pressing the up or down arrow key scrolls through previous commands and redisplays them at the command prompt. Typing the first few characters and then pressing the arrow keys scrolls through the previous commands that start with the same characters. The arrow keys allow command sequences to be repeated without retyping.
-
-Perhaps the most important and useful command for new users is help. To learn more about a function, simply type help followed by the function name. Helpful text is then displayed in the command window. The obvious shortcoming of help is that the function name must first be known. This is especially limiting for MATLAB beginners. Fortunately, help screens often conclude by referencing related or similar functions. These references are an excellent way to learn new MATLAB commands. Typing help help, for example, displays detailed information on the help command itself and also provides reference to relevant functions, such as the lookfor command. The lookfor command helps locate MATLAB functions based on a keyword search. Simply type lookfor followed by a single keyword, and MATLAB searches for functions that contain that keyword.
-
-MATLAB also has comprehensive HTML-based help. The HTML help is accessed by using MATLAB's integrated help browser, which also functions as a standard web browser. The HTML help facility includes a function and topic index as well as full text-searching capabilities. Since HTML documents can contain graphics and special characters, HTML help can provide more information than the command-line help. After a little practice, it is easy to find information in MATLAB.
-
-When MATLAB graphics are created, the print command can save figures in a common file format such as postscript, encapsulated postscript, JPEG, or TIFF. The format of displayed data, such as the number of digits displayed, is selected by using the format command. MATLAB help provides the necessary details for both these functions. When a MATLAB session is complete, the exit command terminates MATLAB.
-
-### **[B.7-2 Calculator Operations](#page-6-0)**
-
-MATLAB can function as a simple calculator, working as easily with complex numbers as with real numbers. Scalar addition, subtraction, multiplication, division, and exponentiation are accomplished using the traditional operator symbols +, -, \*, /, and ^. Since MATLAB predefines <sup>i</sup> <sup>=</sup> <sup>j</sup> <sup>=</sup> √−1, a complex constant is readily created using Cartesian coordinates. For example,
-
->> z = -3-4j z = -3.0000 - 4.0000i
-
-assigns the complex constant −3−*j*4 to the variable *z*.
-
-The real and imaginary components of *z* are extracted by using the real and imag operators. In MATLAB, the input to a function is placed parenthetically following the function name.
-
-$$
-\Rightarrow \quad z\_real = real(z); \ z\_imag = imag(z);
-$$
-
-When a command is terminated with a semicolon, the statement is evaluated but the results are not displayed to the screen. This feature is useful when one is computing intermediate results, and it allows multiple instructions on a single line. Although not displayed, the results z\_real = -3 and z\_imag = -4 are calculated and available for additional operations such as computing |*z*|.
-
-There are many ways to compute the modulus, or magnitude, of a complex quantity. Trigonometry confirms that *z* = −3 − *j*4, which corresponds to a 3-4-5 triangle, has modulus <sup>|</sup>*z*| = |−<sup>3</sup> <sup>−</sup> *<sup>j</sup>*4| = (−3)<sup>2</sup> +(−4)<sup>2</sup> = 5. The MATLAB sqrt command provides one way to compute the required square root.
-
->> z\_mag = sqrt(z\_real^2 + z\_imag^2) z\_mag = 5
-
-In MATLAB, most commands, including sqrt, accept inputs in a variety of forms, including constants, variables, functions, expressions, and combinations thereof.
-
-The same result is also obtained by computing <sup>|</sup>*z*| = <sup>√</sup>*zz*∗. In this case, complex conjugation is performed by using the conj command.
-
->> z\_mag = sqrt(z\*conj(z)) z\_mag = 5
-
-More simply, MATLAB computes absolute values directly by using the abs command.
-
->> z\_mag = abs(z) z\_mag = 5
-
-In addition to magnitude, polar notation requires phase information. The angle command provides the angle of a complex number.
-
->> z\_rad = angle(z) z\_rad = -2.2143
-
-MATLAB expects and returns angles in a radian measure. Angles expressed in degrees require an appropriate conversion factor.
-
->> z\_deg = angle(z)\*180/pi z\_deg = -126.8699
-
-Notice, MATLAB predefines the variable pi = π.
-
-It is also possible to obtain the angle of *z* using a two-argument arc-tangent function, atan2.
-
-```
->> z_rad = atan2(z_imag,z_real)
-    z_rad = -2.2143
-```
-
-Unlike a single-argument arctangent function, the two-argument arctangent function ensures that the angle reflects the proper quadrant. MATLAB supports a full complement of trigonometric functions: standard trigonometric functions cos, sin, tan; reciprocal trigonometric functions sec, csc, cot; inverse trigonometric functions acos, asin, atan, asec, acsc, acot; and hyperbolic variations cosh, sinh, tanh, sech, csch, coth, acosh, asinh, atanh, asech, acsch, and acoth. Of course, MATLAB comfortably supports complex arguments for any trigonometric function. As with the angle command, MATLAB trigonometric functions utilize units of radians.
-
-The concept of trigonometric functions with complex-valued arguments is rather intriguing. The results can contradict what is often taught in introductory mathematics courses. For example, a common claim is that |cos(*x*)| ≤ 1. While this is true for real *x*, it is not necessarily true for complex *x*. This is readily verified by example using MATLAB and the cos function.
-
-```
->> cos(1j)
-   ans = 1.5431
-```
-
-<span id="page-64-0"></span>Problem B.1-19 investigates these ideas further.
-
-Similarly, the claim that it is impossible to take the logarithm of a negative number is false. For example, the principal value of ln(−1) is *j*π, a fact easily verified by means of Euler's equation. In MATLAB, base-10 and base-*e* logarithms are computed by using the log10 and log commands, respectively.
-
->> log(-1) ans = 0 + 3.1416i
-
-### **[B.7-3 Vector Operations](#page-6-0)**
-
-The power of MATLAB becomes apparent when vector arguments replace scalar arguments. Rather than computing one value at a time, a single expression computes many values. Typically, vectors are classified as row vectors or column vectors. For now, we consider the creation of row vectors with evenly spaced, real elements. To create such a vector, the notation a:b:c is used, where a is the initial value, b designates the step size, and c is the termination value. For example, 0:2:11 creates the length-6 vector of even-valued integers ranging from 0 to 10.
-
->> k = 0:2:11 k = 0 2 4 6 8 10
-
-In this case, the termination value does not appear as an element of the vector. Negative and noninteger step sizes are also permissible.
-
->> k = 11:-10/3:0 k = 11.0000 7.6667 4.3333 1.0000
-
-If a step size is not specified, a value of 1 is assumed.
-
->> k = 0:11 k = 0 1 2 3 4 5 6 7 8 9 10 11
-
-Vector notation provides the basis for solving a wide variety of problems.
-
-For example, consider finding the three cube roots of minus one, *w*<sup>3</sup> = −1 = *ej*(π+2π*k*) for integer *k*. Taking the cube root of each side yields *w* = *ej*(π/3+2π*k*/3) . To find the three unique solutions, use any three consecutive integer values of *k* and MATLAB's exp function.
-
->> k = 0:2; w = exp(1j\*(pi/3 + 2\*pi\*k/3)) w = 0.5000 + 0.8660i -1.0000 + 0.0000i 0.5000 - 0.8660i
-
-The solutions, particularly *w* = −1, are easy to verify.
-
-Finding the 100 unique roots of *w*<sup>100</sup> = −1 is just as simple.
-
->> 
-$$
-k = 0.99
-$$
-; w = exp(1j\*(pi/100 + 2\*pi\*k/100));
-
-A semicolon concludes the final instruction to suppress the inconvenient display of all 100 solutions. To view a particular solution, the user must use an index to specify desired elements. <span id="page-65-0"></span>MATLAB indices are integers that increase from a starting value of 1. For example, the fifth element of *w* is extracted using an index of 5.†
-
->> w(5) ans = 0.9603 + 0.2790i
-
-Notice that this solution corresponds to *k* = 4. The independent variable of a function, in this case *k*, rarely serves as the index. Since *k* is also a vector, it can likewise be indexed. In this way, we can verify that the fifth value of *k* is indeed 4.
-
->> k(5) ans = 4
-
-It is also possible to use a vector index to access multiple values. For example, index vector 98:100 identifies the last three solutions corresponding to *k* = [97, 98, 99].
-
->> w(98:100) ans = 0.9877 - 0.1564i 0.9956 - 0.0941i 0.9995 - 0.0314i
-
-Vector representations provide the foundation to rapidly create and explore various signals. Consider the simple 10 Hz sinusoid described by *f*(*t*) = sin(2π10*t* + π/6). Two cycles of this sinusoid are included in the interval 0 ≤ *t* <0.2. A vector *t* is used to uniformly represent 500 points over this interval.
-
->> t = 0:0.2/500:0.2-0.2/500;
-
-Next, the function *f*(*t*) is evaluated at these points.
-
-```
->> f = sin(2*pi*10*t+pi/6)
-```
-
-The value of *f*(*t*) at *t* = 0 is the first element of the vector and is thus obtained by using an index of 1.
-
->> f(1) ans = 0.5000
-
-Unfortunately, MATLAB's indexing syntax conflicts with standard equation notation.‡ That is, the MATLAB indexing command f(1) is not the same as the standard notation *f*(1) = *f*(*t*)|*t*=1. Care must be taken to avoid confusion; remember that the index parameter rarely reflects the independent variable of a function.
-
-## **[B.7-4 Simple Plotting](#page-6-0)**
-
-MATLAB's plot command provides a convenient way to visualize data, such as graphing *f*(*t*) against the independent variable *t*.
-
-```
->> plot(t,f);
-```
-
-<sup>†</sup> Some other programming languages, such as C, begin indexing at 0. Careful attention is warranted.
-
-<sup>‡</sup> MATLAB anonymous functions, considered in Sec. 1.11, are an important and useful exception.
-
-**Figure B.12** *f*(*t*) = sin(2π10*t* +π/6).
-
-Axis labels are added using the xlabel and ylabel commands, where the desired string must be enclosed by single quotation marks. The result is shown in Fig. B.12.
-
->> xlabel('t'); ylabel('f(t)')
-
-The title command is used to add a title above the current axis.
-
-By default, MATLAB connects data points with solid lines. Plotting discrete points, such as the 100 unique roots of *w*<sup>100</sup> = −1, is accommodated by supplying the plot command with an additional string argument. For example, the string 'o' tells MATLAB to mark each data point with a circle rather than connecting points with lines. A full description of the supported plot options is available from MATLAB's help facilities.
-
-```
->> plot(real(w),imag(w),'o');
->> xlabel('Re(w)'); ylabel('Im(w)'); axis equal
-```
-
-The axis equal command ensures that the scale used for the horizontal axis is equal to the scale used for the vertical axis. Without axis equal, the plot would appear elliptical rather than circular. Figure B.13 illustrates that the 100 unique roots of *w*<sup>100</sup> = −1 lie equally spaced on the unit circle, a fact not easily discerned from the raw numerical data.
-
-MATLAB also includes many specialized plotting functions. For example, MATLAB commands semilogx, semilogy, and loglog operate like the plot command but use base-10 logarithmic scales for the horizontal axis, vertical axis, and the horizontal and vertical axes,
-
-**Figure B.13** Unique roots of *w*<sup>100</sup> = −1.
-
-### <span id="page-67-0"></span>48 CHAPTER B BACKGROUND
-
-respectively. Monochrome and color images can be displayed by using the image command, and contour plots are easily created with the contour command. Furthermore, a variety of three-dimensional plotting routines are available, such as plot3, contour3, mesh, and surf. Information about these instructions, including examples and related functions, is available from MATLAB help.
-
-### **[B.7-5 Element-by-Element Operations](#page-6-0)**
-
-Suppose a new function *h*(*t*) is desired that forces an exponential envelope on the sinusoid *f*(*t*), *h*(*t*) = *f*(*t*)*g*(*t*), where *g*(*t*) = *e*−10*<sup>t</sup>* . First, row vector *g*(*t*) is created.
-
->> g = exp(-10\*t);
-
-Given MATLAB's vector representation of *g*(*t*) and *f*(*t*), computing *h*(*t*) requires some form of vector multiplication. There are three standard ways to multiply vectors: inner product, outer product, and element-by-element product. As a matrix-oriented language, MATLAB defines the standard multiplication operator \* according to the rules of matrix algebra: the multiplicand must be conformable to the multiplier. A 1 × *N* row vector times an *N* × 1 column vector results in the scalar-valued inner product. An *N* × 1 column vector times a 1 × *M* row vector results in the outer product, which is an *N* × *M* matrix. Matrix algebra prohibits multiplication of two row vectors or multiplication of two column vectors. Thus, the \* operator is not used to perform element-by-element multiplication.†
-
-Element-by-element operations require vectors to have the same dimensions. An error occurs if element-by-element operations are attempted between row and column vectors. In such cases, one vector must first be transposed to ensure both vector operands have the same dimensions. In MATLAB, most element-by-element operations are preceded by a period. For example, element-by-element multiplication, division, and exponentiation are accomplished using .\*, ./, and .^, respectively. Vector addition and subtraction are intrinsically element-by-element operations and require no period. Intuitively, we know *h*(*t*) should be the same size as both *g*(*t*) and *f*(*t*). Thus, *h*(*t*) is computed using element-by-element multiplication.
-
-The plot command accommodates multiple curves and also allows modification of line properties. This facilitates side-by-side comparison of different functions, such as *h*(*t*) and *f*(*t*). Line characteristics are specified by using options that follow each vector pair and are enclosed in single quotes.
-
->> plot(t,f,'-k',t,h,':k'); >> xlabel('t'); ylabel('Amplitude'); >> legend('f(t)','h(t)');
-
-Here, '-k' instructs MATLAB to plot *f*(*t*) using a solid black line, while ':k' instructs MATLAB to use a dotted black line to plot *h*(*t*). A legend and axis labels complete the plot, as shown in
-
-<sup>&</sup>gt;> h = f.\*g;
-
-<sup>†</sup> While grossly inefficient, element-by-element multiplication can be accomplished by extracting the main diagonal from the outer product of two *N*-length vectors.
-
-<span id="page-68-0"></span>**Figure B.14** Graphical comparison of *f*(*t*) and *h*(*t*).
-
-Fig. B.14. It is also possible, although more cumbersome, to use pull down menus to modify line properties and to add labels and legends directly in the figure window.
-
-### **[B.7-6 Matrix Operations](#page-6-0)**
-
-Many applications require more than row vectors with evenly spaced elements; row vectors, column vectors, and matrices with arbitrary elements are typically needed.
-
-MATLAB provides several functions to generate common, useful matrices. Given integers m, n, and vector x, the function eye(m) creates the *m*×*m* identity matrix; the function ones(m,n) creates the *m* × *n* matrix of all ones; the function zeros(m,n) creates the *m* × *n* matrix of all zeros; and the function diag(x) uses vector x to create a diagonal matrix. The creation of general matrices and vectors, however, requires each individual element to be specified.
-
-Vectors and matrices can be input spreadsheet style by using MATLAB's array editor. This graphical approach is rather cumbersome and is not often used. A more direct method is preferable.
-
-Consider a simple row vector **r**,
-
-$$
-\mathbf{r} = [1 \ 0 \ 0]
-$$
-
-The MATLAB notation a:b:c cannot create this row vector. Rather, square brackets are used to create **r**.
-
->> r = [1 0 0] r=1 0 0
-
-Square brackets enclose elements of the vector, and spaces or commas are used to separate row elements.
-
-Next, consider the 3×2 matrix **A**,
-
-$$
-\mathbf{A} = \left[ \begin{array}{cc} 2 & 3 \\ 4 & 5 \\ 0 & 6 \end{array} \right]
-$$
-
-Matrix **A** can be viewed as a three-high stack of two-element row vectors. With a semicolon to separate rows, square brackets are used to create the matrix.
-
->> A = [2 3;4 5;0 6] A=2 3 4 5 0 6
-
-Each row vector needs to have the same length to create a sensible matrix.
-
-In addition to enclosing string arguments, a single quote performs the complex conjugate transpose operation. In this way, row vectors become column vectors and vice versa. For example, a column vector **c** is easily created by transposing row vector **r**.
-
->> c = r' c=1 0 0
-
-Since vector **r** is real, the complex-conjugate transpose is just the transpose. Had **r** been complex, the simple transpose could have been accomplished by either r.' or (conj(r))'.
-
-More formally, square brackets are referred to as a concatenation operator. A concatenation combines or connects smaller pieces into a larger whole. Concatenations can involve simple numbers, such as the six-element concatenation used to create the 3×2 matrix **A**. It is also possible to concatenate larger objects, such as vectors and matrices. For example, vector **c** and matrix **A** can be concatenated to form a 3×3 matrix **B**.
-
->> B = [c A] B=1 2 3 045 006
-
-Errors will occur if the component dimensions do not sensibly match; a 2×2 matrix would not be concatenated with a 3×3 matrix, for example.
-
-Elements of a matrix are indexed much like vectors, except two indices are typically used to specify row and column.† Element (1, 2) of matrix **B**, for example, is 2.
-
->> B(1,2) ans = 2
-
-Indices can likewise be vectors. For example, vector indices allow us to extract the elements common to the first two rows and last two columns of matrix **B**.
-
-```
->> B(1:2,2:3)
-   ans = 2 3
-        4 5
-```
-
-<sup>†</sup> Matrix elements can also be accessed by means of a single index, which enumerates along columns. Formally, the element from row *m* and column *n* of an *M* × *N* matrix may be obtained with a single index (*n*−1)*M* +*m*. For example, element (1, 2) of matrix **B** is accessed by using the index (2−1)3+1 = 4. That is, B(4) yields 2.
-
-One indexing technique is particularly useful and deserves special attention. A colon can be used to specify all elements along a specified dimension. For example, B(2,:) selects all column elements along the second row of **B**.
-
->> B(2,:) ans = 0 4 5
-
-Now that we understand basic vector and matrix creation, we turn our attention to using these tools on real problems. Consider solving a set of three linear simultaneous equations in three unknowns.
-
-$$
-x_1 - 2x_2 + 3x_3 = 1
-$$
-$$
--\sqrt{3}x_1 + x_2 - \sqrt{5}x_3 = \pi
-$$
-$$
-3x_1 - \sqrt{7}x_2 + x_3 = e
-$$
-
-This system of equations is represented in matrix form according to **Ax** = **y**, where
-
-$$
-\mathbf{A} = \begin{bmatrix} 1 & -2 & 3 \\ -\sqrt{3} & 1 & -\sqrt{5} \\ 3 & -\sqrt{7} & 1 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, \text{ and } \mathbf{y} = \begin{bmatrix} 1 \\ \pi \\ e \end{bmatrix}
-$$
-
-Although Cramer's rule can be used to solve **Ax** = **y**, it is more convenient to solve by multiplying both sides by the matrix inverse of **A**. That is, **x** = **A**−1**Ax** = **A**−1**y**. Solving for **x** by hand or by calculator would be tedious at best, so MATLAB is used. We first create **A** and **y**.
-
->> A = [1 -2 3;-sqrt(3) 1 -sqrt(5);3 -sqrt(7) 1]; y = [1;pi;exp(1)];
-
-The vector solution is found by using MATLAB's inv function.
-
->> 
-$$
-x = inv(A)*y
-$$
-  
- $x = -1.9999$   
- $-3.8998$   
- $-1.5999$ 
-
-It is also possible to use MATLAB's left divide operator x=A\y to find the same solution. The left divide is generally more computationally efficient than matrix inverses. As with matrix multiplication, left division requires that the two arguments be conformable.
-
-Of course, Cramer's rule can be used to compute individual solutions, such as *x*1, by using vector indexing, concatenation, and MATLAB's det command to compute determinants.
-
->> x1 = det([y,A(:,2:3)])/det(A) x1 = -1.9999
-
-Another nice application of matrices is the simultaneous creation of a family of curves. Consider *h*α(*t*) = *e*−α*<sup>t</sup>*sin(2π10*t* + π/6) over 0 ≤ *t* ≤ 0.2. Figure B.14 shows *h*α(*t*) for α = 0 and α = 10. Let's investigate the family of curves *h*α(*t*) for α = [0, 1,..., 10].
-
-An inefficient way to solve this problem is create *h*α(*t*) for each α of interest. This requires 11 individual cases. Instead, a matrix approach allows all 11 curves to be computed simultaneously. First, a vector is created that contains the desired values of α.
-
->> alpha = (0:10);
-
-By using a sampling interval of one millisecond, *t* = 0.001, a time vector is also created.
-
->> t = (0:0.001:0.2)';
-
-The result is a length-201 column vector. By replicating the time vector for each of the 11 curves required, a time matrix T is created. This replication can be accomplished by using an outer product between t and a 1×11 vector of ones.†
-
->> T = t\*ones(1,11);
-
-The result is a 201 × 11 matrix that has identical columns. Right multiplying T by a diagonal matrix created from α, columns of T can be individually scaled and the final result is computed.
-
->> H = exp(-T\*diag(alpha)).\*sin(2\*pi\*10\*T+pi/6);
-
-Here, H is a 201 × 11 matrix, where each column corresponds to a different value of α. That is, **H** = [**h**0,**h**1,...,**h**10], where **h**<sup>α</sup> are column vectors. As shown in Fig. B.15, the 11 desired curves are simultaneously displayed by using MATLAB's plot command, which allows matrix arguments.
-
->> plot(t,H); xlabel('t'); ylabel('h(t)');
-
-This example illustrates an important technique called vectorization, which increases execution efficiency for interpretive languages such as MATLAB. Algorithm vectorization uses matrix and
-
-**Figure B.15** *h*α(*t*) for α = [0, 1,..., 10].
-
-<sup>†</sup> The repmat command provides a more flexible method to replicate or tile objects. Equivalently, T = repmat(t,1,11).
-
-<span id="page-72-0"></span>vector operations to avoid manual repetition and loop structures. It takes practice and effort to become proficient at vectorization, but the worthwhile result is efficient, compact code.†
-
-### **[B.7-7 Partial Fraction Expansions](#page-6-0)**
-
-There are a wide variety of techniques and shortcuts to compute the partial fraction expansion of rational function *F*(*x*) = *B*(*x*)/*A*(*x*), but few are more simple than the MATLAB residue command. The basic form of this command is
-
->> [R,P,K] = residue(B,A)
-
-The two input vectors B and A specify the polynomial coefficients of the numerator and denominator, respectively. These vectors are ordered in descending powers of the independent variable. Three vectors are output. The vector R contains the coefficients of each partial fraction, and vector P contains the corresponding roots of each partial fraction. For a root repeated *r* times, the *r* partial fractions are ordered in ascending powers. When the rational function is not proper, the vector K contains the direct terms, which are ordered in descending powers of the independent variable.
-
-To demonstrate the power of the residue command, consider finding the partial fraction expansion of
-
-$$
-F(x) = \frac{x^5 + \pi}{(x + \sqrt{2})(x - \sqrt{2})^3} = \frac{x^5 + \pi}{x^4 - \sqrt{8x^3 + \sqrt{32x - 4}}}
-$$
-
-By hand, the partial fraction expansion of *F*(*x*) is difficult to compute. MATLAB, however, makes short work of the expansion.
-
->> [R,P,K] = residue([10000 pi],[1 -sqrt(8) 0 sqrt(32) -4]); R.', P.', K R = 7.8888 5.9713 3.1107 0.1112 P = 1.4142 1.4142 1.4142 -1.4142 K = 1.0000 2.8284
-
-Written in standard form, the partial fraction expansion of *F*(*x*) is
-
-$$
-F(x) = x + 2.8284 + \frac{7.8888}{x - \sqrt{2}} + \frac{5.9713}{(x - \sqrt{2})^2} + \frac{3.1107}{(x - \sqrt{2})^3} + \frac{0.1112}{x + \sqrt{2}}
-$$
-
-The signal–processing toolbox function residuez is similar to the residue command and offers more convenient expansion of certain rational functions, such as those commonly encountered in the study of discrete-time systems. Additional information about the residue and residuez commands is available from MATLAB's help facilities.
-
-<sup>†</sup> The benefits of vectorization are less pronounced in recent versions of MATLAB.
-
-## <span id="page-73-0"></span>**[B.8 APPENDIX: USEFUL](#page-6-0) MATHEMATICAL FORMULAS**
-
-We conclude this chapter with a selection of useful mathematical facts.
-
-### **[B.8-1 Some Useful Constants](#page-6-0)**
-
-π ≈ 3.1415926535 *e* ≈ 2.7182818284 1 *<sup>e</sup>* <sup>≈</sup> 0.3678794411 log10 2 ≈ 0.30103 log10 3 ≈ 0.47712
-
-### **[B.8-2 Complex Numbers](#page-7-0)**
-
-$$
-e^{\pm j\pi/2} = \pm j
-$$
-  
-\n
-$$
-e^{\pm j n\pi} = \begin{cases} 1 & n \text{ even} \\ -1 & n \text{ odd} \end{cases}
-$$
-  
-\n
-$$
-e^{\pm j\theta} = \cos \theta \pm j \sin \theta
-$$
-  
-\n
-$$
-a + jb = re^{j\theta} \qquad r = \sqrt{a^2 + b^2}, \theta = \tan^{-1} \left(\frac{b}{a}\right)
-$$
-  
-\n
-$$
-(re^{j\theta})^k = r^k e^{jk\theta}
-$$
-  
-\n
-$$
-(r_1 e^{j\theta_1})(r_2 e^{j\theta_2}) = r_1 r_2 e^{j(\theta_1 + \theta_2)}
-$$
-
-### **[B.8-3 Sums](#page-7-0)**
-
-$$
-\sum_{k=m}^{n} r^{k} = \frac{r^{n+1} - r^{m}}{r - 1} \qquad r \neq 1
-$$
-\n
-$$
-\sum_{k=0}^{n} k = \frac{n(n+1)}{2}
-$$
-\n
-$$
-\sum_{k=0}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}
-$$
-\n
-$$
-\sum_{k=0}^{n} kr^{k} = \frac{r + [n(r-1) - 1]r^{n+1}}{(r-1)^{2}} \qquad r \neq 1
-$$
-\n
-$$
-\sum_{k=0}^{n} k^{2}r^{k} = \frac{r[(1+r)(1-r^{n}) - 2n(1-r)r^{n} - n^{2}(1-r)^{2}r^{n}]}{(1-r)^{3}} \qquad r \neq 1
-$$

engineering/linear-systems-and-signals-3rd-edition-bp-lathi/009_B.8 APPENDIX - USEFUL MATHEMATICAL FORMULAS.md

@@ -1,342 +0,0 @@
-### <span id="page-74-0"></span>**[B.8-4 Taylor and Maclaurin Series](#page-7-0)**
-
-$$
-f(x) = f(a) + \frac{(x-a)}{1!}\dot{f}(a) + \frac{(x-a)^2}{2!}\ddot{f}(a) + \dots = \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!}f^{(k)}(a)
-$$
-$$
-f(x) = f(0) + \frac{x}{1!}\dot{f}(0) + \frac{x^2}{2!}\ddot{f}(0) + \dots = \sum_{k=0}^{\infty} \frac{x^k}{k!}f^{(k)}(0)
-$$
-
-### **[B.8-5 Power Series](#page-7-0)**
-
-$$
-e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots + \frac{x^{n}}{n!} + \dots
-$$
-\n
-$$
-\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots
-$$
-\n
-$$
-\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - \dots
-$$
-\n
-$$
-\tan x = x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + \frac{17x^{7}}{315} + \dots \qquad x^{2} < \pi^{2}/4
-$$
-\n
-$$
-\tanh x = x - \frac{x^{3}}{3} + \frac{2x^{5}}{15} - \frac{17x^{7}}{315} + \dots \qquad x^{2} < \pi^{2}/4
-$$
-\n
-$$
-(1 + x)^{n} = 1 + nx + \frac{n(n-1)}{2!}x^{2} + \frac{n(n-1)(n-2)}{3!}x^{3} + \dots + \binom{n}{k}x^{k} + \dots + x^{n}
-$$
-\n
-$$
-(1 + x)^{n} \approx 1 + nx \qquad |x| \ll 1
-$$
-\n
-$$
-\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \dots \qquad |x| < 1
-$$
-
-### **[B.8-6 Trigonometric Identities](#page-7-0)**
-
-$$
-e^{\pm jx} = \cos x \pm j \sin x
-$$
-  
-\n
-$$
-\cos x = \frac{1}{2} [e^{jx} + e^{-jx}]
-$$
-  
-\n
-$$
-\sin x = \frac{1}{2j} [e^{jx} - e^{-jx}]
-$$
-  
-\n
-$$
-\cos (x \pm \frac{\pi}{2}) = \pm \sin x
-$$
-  
-\n
-$$
-\sin (x \pm \frac{\pi}{2}) = \pm \cos x
-$$
-  
-\n
-$$
-2 \sin x \cos x = \sin 2x
-$$
-  
-\n
-$$
-\sin^2 x + \cos^2 x = 1
-$$
-  
-\n
-$$
-\cos^2 x - \sin^2 x = \cos 2x
-$$
-  
-\n
-$$
-\cos^2 x = \frac{1}{2} (1 + \cos 2x)
-$$
-  
-\n
-$$
-\sin^2 x = \frac{1}{2} (1 - \cos 2x)
-$$
-
-<span id="page-75-0"></span>
-$$
-\cos^3 x = \frac{1}{4} (3 \cos x + \cos 3x)
-$$
-  
-\n
-$$
-\sin^3 x = \frac{1}{4} (3 \sin x - \sin 3x)
-$$
-  
-\n
-$$
-\sin (x \pm y) = \sin x \cos y \pm \cos x \sin y
-$$
-  
-\n
-$$
-\cos (x \pm y) = \cos x \cos y \mp \sin x \sin y
-$$
-  
-\n
-$$
-\tan (x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y}
-$$
-  
-\n
-$$
-\sin x \sin y = \frac{1}{2} [\cos (x - y) - \cos (x + y)]
-$$
-  
-\n
-$$
-\cos x \cos y = \frac{1}{2} [\cos (x - y) + \cos (x + y)]
-$$
-  
-\n
-$$
-\sin x \cos y = \frac{1}{2} [\sin (x - y) + \sin (x + y)]
-$$
-  
-\n
-$$
-\sin x \cos x + b \sin x = C \cos (x + \theta) \qquad C = \sqrt{a^2 + b^2}, \theta = \tan^{-1} (\frac{-b}{a})
-$$
-
-**[B.8-7 Common Derivative Formulas](#page-7-0)**
-
-$$
-\frac{d}{dx}f(u) = \frac{d}{du}f(u)\frac{du}{dx}
-$$
-\n
-$$
-\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}
-$$
-\n
-$$
-\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}
-$$
-\n
-$$
-\frac{dx^n}{dx} = nx^{n-1}
-$$
-\n
-$$
-\frac{d}{dx}\ln(ax) = \frac{1}{x}
-$$
-\n
-$$
-\frac{d}{dx}\log(ax) = \frac{\log e}{x}
-$$
-\n
-$$
-\frac{d}{dx}e^{bx} = be^{bx}
-$$
-\n
-$$
-\frac{d}{dx}a^{bx} = b(\ln a)a^{bx}
-$$
-\n
-$$
-\frac{d}{dx}\sin ax = a\cos ax
-$$
-\n
-$$
-\frac{d}{dx}\cos ax = -a\sin ax
-$$
-\n
-$$
-\frac{d}{dx}\tan ax = \frac{a}{\cos^2 ax}
-$$
-\n
-$$
-\frac{d}{dx}(\sin^{-1}ax) = \frac{a}{\sqrt{1 - a^2x^2}}
-$$
-\n
-$$
-\frac{d}{dx}(\cos^{-1}ax) = \frac{-a}{\sqrt{1 - a^2x^2}}
-$$
-\n
-$$
-\frac{d}{dx}(\tan^{-1}ax) = \frac{a}{1 + a^2x^2}
-$$
-
-### <span id="page-76-0"></span>**[B.8-8 Indefinite Integrals](#page-7-0)**
-
-$$
-\int u dv = uv - \int v du
-$$
-  
-\n
-$$
-\int f(x)\dot{g}(x) dx = f(x)g(x) - \int f(x)g(x) dx
-$$
-  
-\n
-$$
-\int \sin ax dx = -\frac{1}{a} \cos ax \qquad \int \cos ax dx = \frac{1}{a} \sin ax
-$$
-  
-\n
-$$
-\int \sin^2 ax dx = \frac{x}{2} - \frac{\sin 2ax}{4a} \qquad \int \cos^2 ax dx = \frac{x}{2} + \frac{\sin 2ax}{4a}
-$$
-  
-\n
-$$
-\int x \sin ax dx = \frac{1}{a^2} (\sin ax - ax \cos ax)
-$$
-  
-\n
-$$
-\int x \cos ax dx = \frac{1}{a^2} (\cos ax + ax \sin ax)
-$$
-  
-\n
-$$
-\int x^2 \sin ax dx = \frac{1}{a^3} (2ax \sin ax + 2 \cos ax - a^2x^2 \cos ax)
-$$
-  
-\n
-$$
-\int x^2 \cos ax dx = \frac{1}{a^3} (2ax \cos ax - 2 \sin ax + a^2x^2 \sin ax)
-$$
-  
-\n
-$$
-\int \sin ax \sin bx dx = \frac{\sin (a - b)x}{2(a - b)} - \frac{\sin (a + b)x}{2(a + b)} \qquad a^2 \neq b^2
-$$
-  
-\n
-$$
-\int \sin ax \cos bx dx = -\left[ \frac{\cos (a - b)x}{2(a - b)} + \frac{\cos (a + b)x}{2(a + b)} \right] \qquad a^2 \neq b^2
-$$
-  
-\n
-$$
-\int \cos ax \cos bx dx = \frac{\sin (a - b)x}{2(a - b)} + \frac{\sin (a + b)x}{2(a + b)} \qquad a^2 \neq b^2
-$$
-  
-\n
-$$
-\int e^{ax} dx = \frac{e^{ax}}{a^2} (ax - 1)
-$$
-  
-\n
-$$
-\int x^2 e^{ax} dx = \frac{e^{ax}}{a^2} (a^2x^2 - 2ax + 2)
-$$
-  
-\n
-$$
-\int e^{ax} \sin bx dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx)
-$$
-  
-\n
-$$
-\int e^{ax} \cos bx dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx)
-$$
-  
-\n
-$$
-\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \ln(x
-$$
-
-### <span id="page-77-0"></span>**[B.8-9 L'Hôpital's Rule](#page-7-0)**
-
-If lim *f*(*x*)/*g*(*x*) results in the indeterministic form 0/0 or ∞/∞, then
-
-$$
-\lim \frac{f(x)}{g(x)} = \lim \frac{\dot{f}(x)}{\dot{g}(x)}
-$$
-
-### **[B.8-10 Solution of Quadratic and Cubic Equations](#page-7-0)**
-
-Any *quadratic* equation can be reduced to the form
-
-$$
-ax^2 + bx + c = 0
-$$
-
-The solution of this equation is provided by
-
-$$
-x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
-$$
-
-A general *cubic* equation
-
-$$
-y^3 + py^2 + qy + r = 0
-$$
-
-may be reduced to the *depressed cubic* form
-
-$$
-x^3 + ax + b = 0
-$$
-
-by substituting
-
-$$
-y = x - \frac{p}{3}
-$$
-
-This yields
-
-$$
-a = \frac{1}{3}(3q - p^2) \qquad b = \frac{1}{27}(2p^3 - 9pq + 27r)
-$$
-
-Now let
-
-$$
-A = \sqrt[3]{-\frac{b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}} \qquad B = \sqrt[3]{-\frac{b}{2} - \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}}
-$$
-
-The solution of the depressed cubic is
-
-$$
-x = A + B
-$$
-,  $x = -\frac{A+B}{2} + \frac{A-B}{2}\sqrt{-3}$ ,  $x = -\frac{A+B}{2} - \frac{A-B}{2}\sqrt{-3}$ 
-
-and
-
-$$
-y = x - \frac{p}{3}
-$$