diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/199_17.4 Circuit Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/199_17.4 Circuit Applications.md deleted file mode 100644 index f273b1e8ba0af5f1a6a60fc957dfba4000b2ec1f..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/199_17.4 Circuit Applications.md +++ /dev/null @@ -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, - -> **I***o* = \_\_\_\_\_\_\_\_ 4 4 + *jωn*2 **I** = \_\_\_\_\_\_\_\_ **V** 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 - -**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*2 + 2*ab* + *b*2 , 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 -$$ -+ \sum_{n=1}^\infty \sum_{m=1}^\infty A_n A_m \frac{1}{T} \int_0^T \cos(n\omega_0 t + \phi_n) \cos(m\omega_0 t + \phi_m) dt \ No newline at end of file diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/200_17.5 Average Power and RMS Values.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/200_17.5 Average Power and RMS Values.md deleted file mode 100644 index 2e1891e82c9c05b1e8800315c74e159f7e37d08b..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/200_17.5 Average Power and RMS Values.md +++ /dev/null @@ -1,187 +0,0 @@ -$$ -\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 2 = *a*0 2 +\_\_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* 0 2 is the power in the dc component, while \_\_1 2 ( *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. - -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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/201_17.6 Exponential Fourier Series.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/201_17.6 Exponential Fourier Series.md deleted file mode 100644 index 8017257a88d81c7b6618ad3888f1a54a28ea9157..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/201_17.6 Exponential Fourier Series.md +++ /dev/null @@ -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* = √ \_\_\_\_\_\_ *a n* 2 + *b n* 2 \_\_\_\_\_\_\_\_ 2 ⧸ −tan−1 *bn*∕*an* **(17.61)** - -if only *an* > 0. Note that the phase *θn* 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 2 ∕*R*; if *f*(*t*) is the current through *R*, the power is *F*rms 2 *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*∣ ⧸*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. - -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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/202_17.7 Fourier Analysis with PSpice.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/202_17.7 Fourier Analysis with PSpice.md deleted file mode 100644 index 326cc5e128bc843c9e56134f4c74200eb9ddf019..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/202_17.7 Fourier Analysis with PSpice.md +++ /dev/null @@ -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 | -| C User Defined | C Restricted [analog] | -| OHz
to $10Hz$ | 10Hz
to TKHz | -| Scale | Processing Options | -| C Linear | $\nabla$ Fourier | -| $C$ Log | Performance Analysis | - -**Figure 17.35** *X* axis settings dialog box. - -| Transient | | -|---------------------------------|--------| -| Transient Analysis | | -| Print Step: | 0.01 | -| Final Time: | 12s | -| No-Print Delay: | | -| Step Ceiling: | 10ms | -| Detailed Bias Pt. | | -| Skip initial transient solution | | -| Fourier Analysis | | -| Ⅳ Enable Fourier | | -| Center Frequency: | 0.5 | -| Number of harmonics: | | -| Output Vars.: V(1) | | -| | | -| OK | 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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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*−0.5*t* − 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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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 | -| | | | | | | diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/203_17.8 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/203_17.8 Applications.md deleted file mode 100644 index c1d87c9be220cd4b7405983e17dc551b281b7ecb..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/203_17.8 Applications.md +++ /dev/null @@ -1,102 +0,0 @@ -# **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
–106
GHz | -| X-rays | 108
–109
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 *ωc* 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 *ωc* ≪ *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 - -$$ -y(t) = \frac{1}{2} - \frac{1}{\pi} \sin 2\pi t -$$ - -**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:** -$$ -y(t) = -\frac{1}{3\pi} \sin 3\omega_0 t - \frac{1}{4\pi} \sin 4\omega_0 t - \frac{1}{5\pi} \sin 5\omega_0 t -$$ -. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/204_17.9 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/204_17.9 Summary.md deleted file mode 100644 index b899fb8d8115dfb53d2acb7ead89c182cc6ade60..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/204_17.9 Summary.md +++ /dev/null @@ -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 *θn* 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. - -# Review Questions - -- **17.1** Which of the following cannot be a Fourier series? - - (a) *t* − *t* 2 \_\_ 2 +*t* 3 \_\_ 3 − *t* 4 \_\_ 4 +*t* 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
(a) t + t | 2
(b) t
cos t | (c) et2 | -|------------------------|---------------------|---------| -| 2
4
(d) t
+ 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
(d) t
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 \ No newline at end of file diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/205_Review Questions.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/205_Review Questions.md deleted file mode 100644 index e294019cf197f3fa5caa2422f93c8cc36c60dd98..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/205_Review Questions.md +++ /dev/null @@ -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.* diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/206_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/206_Problems.md deleted file mode 100644 index 65a2ad0943f60b0ec39f17d28faaa60913e80430..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/206_Problems.md +++ /dev/null @@ -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* 2 − 1 (c) cos *nπt* sin *nπt* (d) sin2 *πt* (e) *e*−*t* - -**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 *n* and *b*ʹ *n*, and *ω*ʹ *o*, 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*−*t* , 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) - - - -# **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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/207_Comprehensive Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/207_Comprehensive Problems.md deleted file mode 100644 index 4849122d3ab23e5b791b56140df4f4c673567b9b..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/207_Comprehensive Problems.md +++ /dev/null @@ -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 diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/208_Chapter 18 - Fourier Transform.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/208_Chapter 18 - Fourier Transform.md deleted file mode 100644 index 656fb22154a2c65bffb227d99a8319fa84507245..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/208_Chapter 18 - Fourier Transform.md +++ /dev/null @@ -1,29 +0,0 @@ -# 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 - -# 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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/209_18.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/209_18.1 Introduction.md deleted file mode 100644 index 53971aeac621c0fc0d57afe966f977cd6df6d27e..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/209_18.1 Introduction.md +++ /dev/null @@ -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). diff --git 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 b/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 deleted file mode 100644 index 7b352058f4c1584e088ccd299957c58eff55c6cc..0000000000000000000000000000000000000000 --- 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 +++ /dev/null @@ -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 jω*0*t* , (c) cos *ω*0 *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 jω*0*t* 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 jω*0*t* , - -$$ -\mathcal{F}[e^{j\omega_0 t}] = 2\pi\delta(\omega - \omega_0) -$$ - (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*0 − *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*−*jω* − *e*−*j*2*ω*)∕*jω*, (b) 12*ej*2*ω*, (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) \_\_\_\_\_\_\_\_\_\_\_\_ *jω*. - -**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, - -$$ -F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt = \int_{0}^{\infty} e^{-at} e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(a+j\omega)t} dt -$$ -$$ -= \frac{-1}{a+j\omega} e^{-(a+j\omega)t} \Big|_{0}^{\infty} = \frac{1}{a+j\omega} -$$ - -*f*(*t*) = *e*−*atu*(*t*) = {*e*−*at*, - -For Example 18.3. - -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* (*ejω*0*t* − *e*−*jω*0*t* ). 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*0 *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*−(*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 jω*0*t* + *e*−*jω*0*t* ). 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*−*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*∣ , 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** diff --git a/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 b/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 deleted file mode 100644 index fae556cec2b15b74f7dc85c3a05f1ea4ddc8bdae..0000000000000000000000000000000000000000 --- a/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 +++ /dev/null @@ -1,1074 +0,0 @@ -# Properties of the Fourier transform. - -| Property | f(t) | F(ω) | -|-----------------|---------------------------|-------------------------------------| -| Linearity | a1
f1(t) + a2
f2(t) | a1F1(ω) + a2F2(ω) | -| Scaling | f(at) | ___1
ω
__
F(
a )
∣a∣ | -| Time shift | f(t − a) | e−jωaF(ω) | -| Frequency shift | e jω0t
f(t) | F(ω − ω0) | - -| TABLE 18.1 | (continued) | | -|---------------------------|-------------------------|-------------------------------------| -| Property | f(t) | F(ω) | -| Modulation | cos(ω0t)f(t) | __1
[F(ω + ω0) + F(ω − ω0)]
2 | -| Time differentiation | df
__
dt | jωF(ω) | -| | d n
f ___
dtn | n
( jω)
F(ω) | -| Time integration | t
f(t) dt
∫
−∞ | F(ω) _____
jω    + πF(0)δ(ω) | -| Frequency differentiation | n
t
f(t) | d n ____
n
( j)
dωn F(ω) | -| Reversal | f(−t) | F(−ω)
or
F*(ω) | -| Duality | F(t) | 2πf(−ω) | -| Convolution in t | f1(t) *
f2(t) | F1(ω)F2(ω) | -| Convolution in ω | f1(t)f2(t) | ___1
F1(ω) *
F2(ω)
2π | - -## **TABLE 18.2** - -# Fourier transform pairs. - -| f(t) | F(ω) | -|----------------------|-------------------------------------------------------| -| δ(t) | 1 | -| 1 | 2π δ(ω) | -| u(t) | π δ(ω) + ___1
jω | -| u(t + τ) − u(t − τ) | 2 ______ sin ωτ
ω | -| ∣t∣ | −2
___
ω2 | -| sgn(t) | ___2
jω | -| e−at u(t) | ______ 1
a + jω | -| eat u(−t) | ______ 1
a − jω | -| n
e−at u(t)
t | __________ n!
n+1
(a + jω) | -| e−a∣t∣ | _______ 2a
a2
+ 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
2
2
(a + jω)
+ ω 0 | -| e−at cos ω0
tu(t) | a + jω
____________
2
2
(a + jω)
+ 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*−*a*∣*t*∣ , 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* −*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 - -$$ -G(\omega) = \frac{\omega^2 + 21}{\omega^2 + 9} = 1 + \frac{12}{\omega^2 + 9} -$$ - -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*−4*t* )*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. - -# **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*2 (*t*) = *i* 2 (*t*) = *f* 2 (*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 f (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*2 + *y*2 = ∣*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*1 < *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- Ω resistor is *v*(*t*) = 5*e* Example 18.9 −3*t 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- Ω resistor with *i*(*t*) = 10*e*−2∣*t*∣ 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*−2*t u*(*t*). - -# **Solution:** - -Given that *f*(*t*) = *v*(*t*) = *e*−2*t 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 - -$$ -W = \frac{1}{\pi} \int_0^{10} |F(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^{10} \frac{d\omega}{4 + \omega^2} = \frac{1}{\pi} \left( \frac{1}{2} \tan^1 \frac{\omega}{2} \Big|_0^{10} \right) -$$ -$$ -= \frac{1}{2\pi} \tan^{-1} 5 = \frac{1}{2\pi} \left( \frac{78.69^\circ}{180^\circ} \pi \right) = 0.218 \text{ J} -$$ - -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 −*t 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). - -# **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) - -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 - -$$ -F(\omega) = \mathcal{F}[f(t)] = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt -$$ - -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 *ej*2*t* 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*−*jω* \_\_\_\_\_\_ 2 + *jω* is (a) *e*−2*t* (b) *e*−2*t u*(*t* − 1) (c) *e*−2(*t*−1) (d) *e*−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*2 *dω* results in: - -(a) 0 (b) 2 (c) 2.5 (d) ∞ - -**18.7** The integral ∫ −∞ ∞ 10*δ*(*ω*− 1) \_\_\_\_\_\_\_\_\_\_ 4 + *ω*2 *dω* gives: (a) 0 (b) 2 (c) 2.5 (d) ∞ **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
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
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*−3*t* sin(10*t*)*u*(*t*) (b) *f*2(*t*) = *e*−4*t* 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*−*jω* − 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) $G(\omega) = 2u(\omega + 1) - 2u(\omega - 1)$ - -\* 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) $i(t) = tf(t)$ diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/212_18.4 Circuit Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/212_18.4 Circuit Applications.md deleted file mode 100644 index c51bcfd0d1cbf3bf5f98505f394bff18746b5576..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/212_18.4 Circuit Applications.md +++ /dev/null @@ -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*−4*t 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*−*t 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*−2*t 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*−*t 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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/213_18.5 Parseval's Theorem.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/213_18.5 Parseval's Theorem.md deleted file mode 100644 index 40a499ebc244451e711079e094e8168b866c0d67..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/213_18.5 Parseval's Theorem.md +++ /dev/null @@ -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*−(*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*−2*t 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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/214_18.7 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/214_18.7 Applications.md deleted file mode 100644 index 533ba83249c43f3463d514726a20d759f6c65235..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/214_18.7 Applications.md +++ /dev/null @@ -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*−2*t* − 6*e*−4*t* V. Find the output when the input is *vi*(*t*) = 4*e*−*t u*(*t*) V. - -**Figure 18.54** For Prob. 18.9. - -# **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. \* diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/215_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/215_Problems.md deleted file mode 100644 index 54439359961bc69fcd2805d9756f75b31deaf3f4..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/215_Problems.md +++ /dev/null @@ -1,17 +0,0 @@ -# Comprehensive Problems - -**18.68** The voltage signal at the input of a filter is *v*(*t*) = 50*e*−2∣*t*∣ 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 diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/216_Chapter 19 - Two-Port Networks.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/216_Chapter 19 - Two-Port Networks.md deleted file mode 100644 index 31ed2f2fb0fcd8103e834be64ffb3ff20fb4c15f..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/216_Chapter 19 - Two-Port Networks.md +++ /dev/null @@ -1,33 +0,0 @@ -# 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 - -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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/217_19.1 Introduction.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/217_19.1 Introduction.md deleted file mode 100644 index 127a610c9391dfb84e8a184f92614a3e2c043a37..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/217_19.1 Introduction.md +++ /dev/null @@ -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. - -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 **z**21 = Open-circuit transfer impedance from port 2 to port 1 **(19.4)** -- **z**22 = Open-circuit output impedance \ No newline at end of file diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/218_19.2 Impedance Parameters.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/218_19.2 Impedance Parameters.md deleted file mode 100644 index e8fba5c4495f3686b7ddc0df4bb428d256336ba3..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/218_19.2 Impedance Parameters.md +++ /dev/null @@ -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**21 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. - -# **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) \ No newline at end of file diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/219_19.3 Admittance Parameters.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/219_19.3 Admittance Parameters.md deleted file mode 100644 index 1dcb1353349ced4a84749f03ab5f3be810969eea..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/219_19.3 Admittance Parameters.md +++ /dev/null @@ -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**1 \_\_\_\_\_\_\_ − **V***o* 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, - -$$ -\frac{\mathbf{V}_o - 0}{4} + 2\mathbf{I}_1 + \mathbf{I}_2 = 0 -$$ - -$$ --I_2 = 0.25V_o - 1.5V_o = -1.25V_o -$$ - -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**12 ≠ **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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/220_19.4 Hybrid Parameters.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/220_19.4 Hybrid Parameters.md deleted file mode 100644 index cf78c19de4768bcc6425124c8d6077a8e84dab15..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/220_19.4 Hybrid Parameters.md +++ /dev/null @@ -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**22 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 **g**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, - -$$ -\mathbf{I}_1 = \frac{\mathbf{V}_1}{s+1} -$$ - -$$ -\mathbf{g}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} = \frac{1}{s+1} -$$ - -By voltage division, - -**V**2 = \_\_\_\_\_ 1 *s* + 1 **V**1 - -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 = − \_\_\_\_\_ 1 *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**]=[ \_\_\_\_\_\_\_\_\_\_ *s* + 2 *s* 2 + 3*s* + 1 −\_\_\_\_\_\_\_\_\_\_ 1 *s* 2 + 3*s* + 1 \_\_\_\_\_\_\_\_\_\_ 1 *s* 2 + 3*s* + 1 *s*(*s* + 2) \_\_\_\_\_\_\_\_\_\_ *s* 2 + 3*s* + 1 ]. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/221_19.5 Transmission Parameters.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/221_19.5 Transmission Parameters.md deleted file mode 100644 index 59ace8422b851d178a21a2f077127fd42fc1a26f..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/221_19.5 Transmission Parameters.md +++ /dev/null @@ -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**1 **V**2 - -(a) - -1/s s - -**I**1 - -+ - -+ ‒ **I I**1 2 = 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 **I**1 ] =[ **A B C D**] [ **V**2 −**I**2 ] = [**T**] [ **V**2 −**I**2 ] **(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 **b** = 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**21 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. - -## **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), - -$$ -\mathbf{V}_1 = \mathbf{z}_{11}\mathbf{I}_1 + \mathbf{z}_{12}\mathbf{I}_2 \tag{19.37a} -$$ - -$$ -V_2 = z_{21}I_1 + z_{22}I_2 \tag{19.37b} -$$ diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/222_19.6 Relationships Between Parameters.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/222_19.6 Relationships Between Parameters.md deleted file mode 100644 index fbe76a3e1f5bb83cce70a8ac4592cf30c8a533ad..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/222_19.6 Relationships Between Parameters.md +++ /dev/null @@ -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
___
∆y | y12
− ___
∆y | ∆h
___
h22 | h12
___
h22 | ___1
g11 | g12
− ___
g11 | A
__
C | ∆T
___
C | d
__
c | __1
c | -| | z21 | z22 | y21
− ___
∆y | y11
___
∆y | h21
− ___
h22 | ___1
h22 | g21
___
g11 | ∆g
___
g11 | __1
C | D
__
C | ∆t
__
c | __a
c | -| y | z22
___
∆z | z12
− ___
∆z | y11 | y12 | ___1
h11 | h12
− ___
h11 | ∆g
___
g22 | g12
___
g22 | D
__
B | ∆T
− ___
B | __a
b | − __1
b | -| | z21
− ___
∆z | z11
___
∆z | y21 | y22 | h21
___
h11 | ∆h
___
h11 | g21
− ___
g22 | ___1
g22 | − __1
B | A
__
B | ∆t
− __
b | d
__
b | -| h | ∆z
___
z22 | z12
___
z22 | ___1
y11 | y12
− ___
y11 | h11 | h12 | g22
___
∆g | g12
− ___
∆g | B
__
D | ∆T
___
D | b
__
a | __1
a | -| | z21
− ___
z22 | ___1
z22 | y21
___
y11 | ∆y
___
y11 | h21 | h22 | g21
− ___
∆g | g11
___
∆g | − __1
D | C
__
D | ∆t
__
a | __c
a | -| g | ___1
z11 | z12
− ___
z11 | ∆y
___
y22 | y12
___
y22 | h22
___
∆h | h12
− ___
∆h | g11 | g12 | C
__
A | ∆T
− ___
A | __c
d | − __1
d | -| | z21
___
z11 | ∆z
___
z11 | y21
− ___
y22 | ___1
y22 | h21
− ___
∆h | h11
___
∆h | g21 | g22 | __1
A | B
__
A | ∆t
__
d | b
− __
d | -| T | z11
___
z21 | ∆z
___
z21 | y22
− ___
y21 | − ___1
y21 | ∆h
− ___
h21 | h11
− ___
h21 | ___1
g21 | g22
___
g21 | A | B | __d
∆t | __b
∆t | -| | ___1
z21 | z22
___
z21 | − ∆y
___
y21 | y11
− ___
y21 | h22
− ___
h21 | − ___1
h21 | g11
___
g21 | ∆g
___
g21 | C | D | __c
∆t | __a
∆t | -| t | z22
___
z12 | ∆z
___
z12 | y11
− ___
y12 | − ___1
y12 | ___1
h12 | h11
___
h12 | − ∆g
___
g12 | g22
− ___
g12 | ___ D
∆T | ___B
∆T | a | b | -| | ___1
z12 | z11
___
z12 | − ∆y
___
y12 | y22
− ___
y12 | h22
___
h12 | ∆h
___
h12 | g11
− ___
g12 | − ___1
g12 | ___ C
∆T | ___ A
∆T | c | d | - -**∆***z* = **z**11**z**22 − **z**12**z**21, **∆***h* = **h**11**h**22 − **h**12**h**21, **∆***T* = **AD** − **BC** - -**∆***y* = **y**11**y**22 − **y**12**y**21, **∆***g* = **g**11**g**22 − **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 2 ] Ω, [**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 - -$$ -I_1 = 0V_1 + 0V_2 \tag{19.11.1} -$$ - -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:** [ **z**] = [ *R*1 −*R*2 0 0] . Because [ **z**] −1 does not e xist, [ **y**] does not exist. - -**Figure 19.38** For Practice Prob. 19.11. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/223_19.7 Interconnection of Networks.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/223_19.7 Interconnection of Networks.md deleted file mode 100644 index 664de9a7a5809309c53cb318ba189a6d57f03aef..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/223_19.7 Interconnection of Networks.md +++ /dev/null @@ -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**1*a* = **I**1*b*, **I**2 = **I**2*a* = **I**2*b* **(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**1*a* + **I**1*b*, **I**2 = **I**2*a* + **I**2*b* **(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. - -Notice that - -$$ -\Delta_{T_a}=\Delta_{T_b}=\Delta_T=1 -$$ - -showing that the network is reciprocal. diff --git a/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 b/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 deleted file mode 100644 index 07b6b94c331869c3864f279bc5095642a07d6acc..0000000000000000000000000000000000000000 --- a/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 +++ /dev/null @@ -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, - -$$ -\mathbf{z}_{12} = \frac{\mathbf{V}_1}{1} = 19.70 \underline{\text{/} 175.7^{\circ}} \,\Omega, \qquad \mathbf{z}_{22} = \frac{\mathbf{V}_2}{1} = 19.56 \underline{\text{/} 175.7^{\circ}} \,\Omega -$$ - -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 s* = 0.8 k Ω and *R L* = 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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/225_19.9 Applications.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/225_19.9 Applications.md deleted file mode 100644 index 0ff2dac016e9c863d5999f2815bdf7bcd3d0479a..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/225_19.9 Applications.md +++ /dev/null @@ -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**1 in Fig. 19.63(a) or by confirming the required *y*21. - -C2 - -ZB - -L1 L3 - -(a) - -L1 L3 - -C2 **V**2 1 Ω - -+ - -‒ - -(b) - -y22 = - -1 ZA - -**Figure 19.63** For Example 19.18. - -**V**1 + - -‒ - -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. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/226_19.10 Summary.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/226_19.10 Summary.md deleted file mode 100644 index c89252a5ec7353b936815be7f9a603a26dd5cc21..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/226_19.10 Summary.md +++ /dev/null @@ -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. - -# 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**2 **V**2 = 30**I**1 + 20**I**2 - -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.* \ No newline at end of file diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/227_Problems.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/227_Problems.md deleted file mode 100644 index ba5f84543958ba17448217af09359b3f029dd71d..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/227_Problems.md +++ /dev/null @@ -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*2 = 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* = 106 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. - - - -# **Figure 19.133** - -For Prob. 19.97. - -For Prob. 19.98. **19.98** A two-stage amplifier in Fig. 19.134 contains two 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** diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/228_Comprehensive Problem.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/228_Comprehensive Problem.md deleted file mode 100644 index 03553eeef3f15bc1da05c1348ab47b827ba9585c..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/228_Comprehensive Problem.md +++ /dev/null @@ -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. - -# 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 \ No newline at end of file diff --git a/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 b/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 deleted file mode 100644 index 411b11eb0cd19bae01b68388c24d048044fddae9..0000000000000000000000000000000000000000 --- a/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 +++ /dev/null @@ -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 ∆ *k* 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. - -$$ -= a_{11}a_{22}a_{33} + a_{21}a_{32}a_{13} + a_{31}a_{12}a_{23} - a_{13}a_{22}a_{31} - a_{23}a_{32}a_{11} -$$ - --a33a12a21 (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*1 − *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 | | -| [
−5 | −4 | ]
9 | [
]
x3 | | [
]
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*1 − *x*2 − 2*x*3 = 1 −*x*1 + 6*x*2 − 3*x*3 = 0 −2*x*1 − 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**−1 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**−1 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)*i+j* 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. - -$$ -x_1 + x_2 + x_3 = 5 -$$ - --x1 + 2x2 = 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. - -$$ -y_1 - y_3 = 1 -$$ - -2y1 + 3y2 - y3 = 1 -$$ -y_1 - y_2 - y_3 = 3 -$$ - -**Answer:** *y*1 *=* 6*, y*2 *= −*2, *y*3 *=* 5. - -# 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) diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/230_Appendix B - Complex Numbers.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/230_Appendix B - Complex Numbers.md deleted file mode 100644 index 636920024f631faee7f9ce04e7607dd5628f92c9..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/230_Appendix B - Complex Numbers.md +++ /dev/null @@ -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*ej*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 j = 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*−*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 j*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 j*306.9° , (b) 13⧸ 67.38°, 13*ej*67.38° , (c) 9.487⧸ 251.6°, 9.487*ej*251.6° , (d) 7.071⧸ 171.9°, 7.071*e j*171.9° . - -Convert the following complex numbers into rectangular form: Example B.2 (a) 12⧸ −60°, (b) −50⧸ 285°, (c) 8*e j*10° , (d) 20*e*−*jπ*∕3 . - -# **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* −*j*30° , (d) 50*e jπ*∕2 . - -**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⧸*θ*1 and *z*2 = *x*2 + *jy*2 = *r*2 ⧸*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 − *j*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*2 ∕*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) √ Example B.5 \_\_ *A* , (b) *A*4 . - -# **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*1∕3 - -**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 - -# 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*2 + *bx* + *c* = 0 are - -$$ -x_1, x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} -$$ - -# **C.2** Trigonometric Identities - -| 1
csc x = ____
sin x
cot x = _____ 1
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
− y) | -| 2 cos x cos y = cos(x + y) + cos(x
− 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 -$$ -\cos^2 x = \frac{1}{2} (1 + \cos 2x) -$$ - \ No newline at end of file diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/231_Appendix C - Mathematical Formulas.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/231_Appendix C - Mathematical Formulas.md deleted file mode 100644 index 495ff58616968a3ee2355f8d19d28cfa0273a64f..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/231_Appendix C - Mathematical Formulas.md +++ /dev/null @@ -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. - -# 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*−30*t* [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*−4*t 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 \ No newline at end of file diff --git a/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 b/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 deleted file mode 100644 index aaf6f60a95a14c88a79ce8f9ca7376f92e9e34a1..0000000000000000000000000000000000000000 --- a/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 +++ /dev/null @@ -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
out), P60W = 9.6 W, P100W = 16 W. The best way to | | | -| | wire the bulbs is to connect the 100-W bulb in series
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
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)
(b) 24.61 V (with), 26.39 V (without)
(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
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
20 Ω, −84 V | | -| 4.41
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
10 Ω, 166.67 V, 16.667 A | R____
1R3
5.15 | -| 4.59
22.5 Ω, 40 V, 1.7778 A | (a) −(R1 + R3 +
), (b) −92 kΩ
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
V0
= 24 − 5I0 | 5.21
−3 V | -| 4.67
25 kΩ, 49 mW | Rf
___
5.23
−
R1 | -| 4.69 | 5.25 | -| ∞ (theoretically) | 9.375 V | -| 4.71 | 5.27 | -| 8 kΩ, 1.152 W | 2.7 V | -| 4.73
20.77 W | R___2
5.29
R1 | -| 4.75 | 5.31 | -| 250 Ω, 12 mW | 4.545 mA | -| 4.77
(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*5 − *R*4∕*R*6 1 − *R*2*R*4∕*R*3*R*5 **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*1 *v*2 *v*3 *v*4 *v*5 *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*1 − (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*−3*t* A, 30*t*(1 − 3*t*)*e*−6*t* 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*−*t* mA, 2*e*−*t* mA (b) −20*e*−*t 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 × 10−5 (*e*−200*t* − 1) + 4 A, 1.25 × 10−5 (*e*−200*t* − 1) − 2 A (d) 6.25 × 10−5 (*e*−200*t* − 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*−*t* V for 0 < *t* < 1 sec, 5.518*e*−2(*t*−1) V for 1 sec < *t* < ∞ -- **7.9** 10*e*−*t*/12 V - -- **7.11** 1.2*e*−3*t* 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*−2*t* ] V for all *t* > 0. -- **7.19** 5*e*−5*t u*(*t*) A -- **7.21** 1.618 Ω -- **7.23** 10*e*−4*t* V, *t* > 0, 2.5*e*−4*t* 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 −9 , (b) 7 -- **7.33** 4.5*u*(*t* − 2) A -- **7.35** (a) −*e* −2*t u*(*t*) V, (b) 2*e*1.5*t u*(*t*) A -- **7.37** (a) 4 s, (b) 10 V, (c) (10 − 8*e*−*t*∕4 ) *u*(*t*) V -- **7.39** (a) 4 V, *t* < 0, 20 −16*e* −*t*∕8 , *t* > 0, (b) 4 V, *t* < 0, 12 − 8*e*−*t*∕6 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*−14.286*t* ] *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*−*t*∕2 *u*(*t*) A, (b) 6 A, 6*e*−2*t*∕3 *u*(*t*) A -- **7.55** 96 V, 96*e*−4*t u*(*t*) V -- **7.57** 2.4*e*−2*t u*(*t*) A, 600*e*−5*t u*(*t*) mA -- **7.59** 120*e*−4*t u*(*t*) volts -- **7.61** 20*e*−8*t u*(*t*) V, (10 − 5*e*−8*t* )*u*(*t*) A -- **7.63** 2*e*−8*t u*(*t*) A, −8*e*−8*t u*(*t*) V -- **7.65** { 2(1 − *e*−2*t* )A 1.729*e*−2(*t*−1)A 0 < *t* < 1 *t* > 1 -- **7.67** 10*e*–*t*/6*u*(*t*) V -- **7.69** 48(*e*−*t*∕3000−1) *u*(*t*) V -- **7.71** [−5 + 5*e*–*t* ]*u*(*t*) V -- **7.73** −9*e*−5*t u*(*t*) V -- **7.75** [20 10*e*–*t* ]*u*(*t*) V, 100*μ*A -- **7.77** See Fig. D.15. - -**7.79** [1.75 – 0.75*e*–2*t* ]*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*−5*t* ] A **8.11** [(10 + 10*t*)*e*−*t* ] V -- **8.13** 120 Ω -- **8.15** 750 Ω, 200 *μ*F, 25 H -- **8.17** [21.55*e*−2.679*t* − 1.55*e*−37.32*t* ] V -- **8.19** 24 sin(0.5*t*) V -- **8.21** 18*e*−*t* − 2*e*−9*t* 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*−2*t* ] volts -- **8.29** (a) 3 − 3 cos 2*t* + sin 2*t* V, (b) 2 − 4*e*−*t* + *e*−4*t* A, - -- (c) 3 + (2 + 3*t*)*e*−*t* V, (d) 2 + 2 cos 2*te*−*t* A - - **8.31** 80 V, 40 V - - **8.33** [30 + 0.3078*e*−4.95*t* − 15.308*e*−0.05*t* ] V -- **8.35** This is a design problem with multiple answers. -- **8.37** 5*e*−4*t* A -- **8.39** (−60 + [−0.2102*e*−47.83*t* + 60.21*e*−0.167*t* ]) V -- **8.41** [8.7 sin(4.583*t*)*e*–2*t* ]*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*−*t*∕2 ] A, [7.559 sin(1.3229*t*)*e*−*t*∕2 ] V - - **8.47** (400*te*−10*t* ) V - - **8.49** {9 + [(3 + 6*t*)*e*–2*t* ]} *u*(*t*) A - - **8.51** [ − *i* \_\_\_\_0 *oC* sin(*ot*) ] V where *o* = 1∕ √ \_\_\_ LC - - **8.53** (*d*2 *i*∕*dt*2 ) + 1.25(*di*∕*dt*) + 400*i* = 200 - - **8.55** 2*e*–*t*/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*t* V -- **8.61** 2.4 2.667*e*−2*t* + 0.2667*e*−5*t* A, 9.6 – 16*e*−2*t* + 6.4*e*−5*t* 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*−*t 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** *d*2 \_\_\_*v dt*2 + \_\_ *R L* \_\_\_ *dv dt* + \_\_\_*R LC iD* + \_\_1 *C* \_\_\_ *diD dt* = \_\_\_ *vs 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*2 -- **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(106 *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 × 103 *t* + 4.06°) mA -- **10.7** 124.08⧸−154° V - -- **10.9** 6.154 cos(103 *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, − *j* \_\_ *R* √ \_\_ \_\_*L C* , (b) 0, 1, *j* \_\_ *R* √ \_\_ \_\_*L C* **10.23** (1 − *ω*2 *LC*)*Vs* \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_1− *ω*2 *LC* + *jωRC*(2 − *ω*2 *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,
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
________________________
(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,
(b) 40 kΩ | | | -| 10.93 Proof | | | | -| 10.95 Proof | | | | -| | Chapter 11 | | | -| | (Assume all values of currents and voltages are rms unless
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,
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,
(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,
average power = 339.4 W,
reactive power = 339.4 VAR
(b) S = (678.8 – j678.8) VA,
average power = 678.8 W,
reactive power = −678.8 VAR
(c) S = (7.637 + j7.637) kVA, average power =
7.637 W, reactive power = 7.637 VAR
(d) S = (250 + j433) kVA, average power =
250 kW, reactive power = 433 kVAR | -| | 11.49 (a) 4 + j2.373 kVA,
(b) 1.6 – j1.2 kVA,
(c) 0.4624 + j1.2705 kVA,
(d) 110.77 + j166.16 VA | -| | 11.51 (a) 0.9956 (lagging),
(b) 304 W,
(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* 2 + 4*s* + 1) - -14.5 -$$ -\frac{(Ls + R)}{(LCs^2 + RCs + 1)}. -$$ - -**14.7** (a) 1.0116, (b) 0.5623, (c) 5.623 × 1010 - -. - -**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 × 10−4 , *Km* = 5 × 10−3 -- **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** *s* 3 *LRLC*1*C*2 \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*sRiC*1 + 1)(*s* 2 *LC*2 + *sRLC*2 + 1) + *s* 2 *LC*1(*sRLC*2 + 1) -- **14.99** 8.165 MHz, 4.188 × 106 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*−*t u*(*t*), (b) 3*δ*(*t*) − 11*e*−4*t u*(*t*), (c) (2*e*−*t* − 2*e*−3*t* )*u*(*t*), (d) (3*e*−4*t* − 3*e*−2*t* + 6*te*−*2t* )*u*(*t*) -- **15.29** [1 + 2*e*−*t* cos (*t* + 90°)] *u*(*t*) -- **15.31** (a) (−5*e*−*t* + 20*e*−2*t* − 15*e*−3*t* )*u*(*t*) (b) (−*e*−*t* +(1 + 3*t* − *t* 2 \_\_ 2 )*e*−2*t* ) *u*(*t*), (c) (−0.2*e*−2*t* + 0.2*e*−*t* cos(2*t*) + 0.4*e*−*t* sin(2*t*))*u*(*t*) -- **15.33** (a) (3*e*−*t* + 3 sin(*t*) − 3 cos(*t*))*u*(*t*), (b) cos(*t* −*π*)*u*(*t* − *π*), (c) 8 [1 − *e*−*t* − *te*−*t* − 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*−*t* − 7.5*e*–3*t* ]*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*−*t* + 36*e*−2*t* − 31.5*e*−3*t* ]*u*(*t*) -- **15.61** (a) [3 + 3.162 cos (2*t* − 161.12°)]*u*(*t*) volts, (b) [2 − 4*e*−*t* + *e*−4*t* ]*u*(*t*) amps, (c) [3 + 2*e*−*t* + 3*te*−*t* ]*u*(*t*) volts, (d) [2 + 2*e*−*t* cos(2*t*)]*u*(*t*) amps - -# Chapter 16 - -- **16.1** [(7 + 35*t*)*e*−5*t* ] *u*(*t*) A -- **16.3** [(20 + 20*t*)*e*−*t* ]*u*(*t*) V -- **16.5** 750 Ω, 25 H, 200 *μ*F -- **16.7** [6 + 12*e*−*t* cos(2*t*) + 2 sin(2*t*))]*u*(*t*) A -- **16.9** [3 + 5.924*e*−1.5505*t* − 1.4235*e*−6.45*t* ]*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*−2*t* −\_\_\_2 √ \_\_ 7 *e*−0.5*t* sin ( √ \_\_ \_\_\_7 2 *t* )) *u*(*t*) A -- **16.19** [−2.333*e*−*t*∕2 + 2.333*e*−2*t* ]*u*(*t*) volts -- **16.21** [10.776 *e*−2.679*t* − 0.774*e*−37.32*t* ]*u*(*t*) volts -- **16.23** 24 cos(0.5*t* + 90°)*u*(*t*) volts -- **16.25** [45*e*−*t* − 5*e*−9*t* ]*u*(*t*) volts -- **16.27** [30 − 15.309*e*−0.05051*t* + 0.3078*e*−4.949*t* ]*u*(*t*) volts -- **16.29** 17.5 cos(8*t* + 90°)*u*(*t*) amps -- **16.31** [−16 + 66.67*e*−0.8*t* cos(0.6*t* − 53.13°)]*u*(*t*) volts, 13.333*e*−0.8*t* [cos(0.6*t* + 90°)]*u*(*t*) amps -- **16.33** This is a design problem with multiple answers. -- **16.35** [9.091*e*−*t* + 19.653*e*−0.0625*t* cos(0.7044*t* − 117.55°] *u*(*t*) V. -- **16.37** [−60 + 60.21*e*−0.1672*t* − 0.21*e*−47.84*t* ]*u*(*t*) volts -- **16.39** [4.364*e*−2*t* cos(4.583*t* − 90°)]*u*(*t*) amps -- **16.41** [100*te*−10*t* ]*u*(*t*) volts -- **16.43** [9 + 9*e*−2*t* + 6*te*−2*t* ]*u*(*t*) amps -- **16.45** [*io*∕(*ωC*)] cos(*ωt* + 90°)*u*(*t*) volts -- **16.47** [60 − 40*e*−0.6*t* cos(0.2*t*) − sin(0.2*t*))]*u*(*t*) A -- **16.49** [1.0714*e*−2*t* − 2.572*e*−0.5*t* cos(1.25*t*) + 4.791*e*−0.5*t* sin(1.25*t*)]*u*(*t*) A - -- **16.51** [−12 + 41.17*e*−15.125*t* cos(4.608*t* − 73.06°)]*u*(*t*) amps -- **16.53** [11.547*e*−*t* cos(1.7321*t* + 30°)]*u*(*t*) volts -- **16.55** [5−4*e*−*t* − 1*e*−6*t* ]*u*(*t*) amps, [2*e*−*t* − 2*e*−6*t* ]*u*(*t*) amps -- **16.57** (a) (3∕*s*)[1 − *e*−*s* ], (b) [(2 − 2*e*−1.5*t* )*u*(*t*) − (2 − 2*e*−1.5(*t*−1))*u*(*t* − 1)] V -- **16.59** [5*e*−*t* − 10*e*−*t*∕2 cos (*t*∕2)]*u*(*t*) V -- **16.61** [2.333 − 2.38*e*−1.2306*t* + 2.033*e*−0.6347*t* cos(1.4265*t* + 88.68°)]*u*(*t*) V -- **16.63** [7.5*e*−4*t* cos (2*t*) + 345*e*−4*t* sin (2*t*)]*u*(*t*) V, [6 − 9*e*−4*t* cos (2*t*) − 17.062*e*−4*t* sin (2*t*)]*u*(*t*) A -- **16.65** {110.1*e*−3*t* + 192*te*−3*t* − 10.1 cos(4*t*) + 34.58 sin(4*t*)}*u*(*t*) V -- **16.67** [*e*10*t* − *e*−10*t* ] *u*(*t*) volts; this is an unstable circuit! -- **16.69** 240(*s* + 1)∕[*s*(*s* + 3)(3*s* 2 + 8*s* + 1)], −120(*s* −1)∕ [*s*(*s* + 3)(3*s* 2 + 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*2 ) **16.83** (a) \_\_ *R L e*−*Rt*∕*L u*(*t*), (b) (1 − *e*−*Rt*∕*L* )*u*(*t*) **16.85** [9*e*−*t* − 9*e*−2*t* − 6*te*−2*t* ]*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*−*t* − *e*−4*t* )*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*2 *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*3 -- **17.11** ∑*n*=−∞ ∞ 75 4*n*2 *π*2 [2 − 2 cos(*nπ*∕2) − 2*j* sin(*nπ*∕2) +*jnπ* cos(*nπ*∕2) + *nπ* (sin(*nπ*∕2))]*ejn π t*∕2 -- **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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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** 6*j \_\_ ω*− 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*2*t u*(*t*), (b) 6*e*−2*t* , (c) (−10*et u*(*t*) + 10*e*2*t* )*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 × 103 \_\_\_\_\_\_\_\_ 106 + *jω* ( \_\_\_1 *jω* + \_\_\_1 *ω*2 − \_\_\_1 *ω*2 *e*−*jω* ) - -**18.41** 2*jω*(4.5 + *j*2*ω*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (2 + *jω*)(4 − 2*ω*2 + *jω*) - -**18.43** 1000(*e*−1*t* − *e*−1.25*t* )*u*(*t*) V - -**18.45** 5(*e*−*t* − *e*−*2*t )*u*(*t*) A - -- **18.47** 16(*e*−*t* − *e*−*2*t )*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*−*t* − 20*e*−*2*t + 4*e*−4t )*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] S -- **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 ]S -- **19.29** (a) 44 V, 16 V, (b) same -- **19.31** [3.8 Ω −3.6 0.4 0.2 S] **19.33** [(3.077 + *j*1.2821) Ω −0.3846 + j0.2564 0.3846 − *j*0.2564 (76.9 + 282.1) mS] -- **19.35** [ 2Ω −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** [ 4 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 × 10−4] - -*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 diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/233_Selected Bibliography.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/233_Selected Bibliography.md deleted file mode 100644 index 23e0a2774fb0fd37d6137212addd5888ecb0e48d..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/233_Selected Bibliography.md +++ /dev/null @@ -1,65 +0,0 @@ -# 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. 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Oxford, U.K.: Pergamon Press, 1986. diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/234_Index.md b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/234_Index.md deleted file mode 100644 index be0a49dc5caacffb605ea5afef4977fc04036a43..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/234_Index.md +++ /dev/null @@ -1,149 +0,0 @@ -# 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 \ No newline at end of file diff --git a/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.md b/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.md deleted file mode 100644 index db30262bc82f9effeda9883c87325a6a2c541634..0000000000000000000000000000000000000000 --- a/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.md +++ /dev/null @@ -1,46130 +0,0 @@ -# Fundamentals of Electric Circuits - -# Charles K. Alexander - -Department of Electrical and Computer Engineering *Cleveland State University* - -# Matthew N. O. Sadiku - -Department of Electrical and Computer Engineering - -*Prairie View A&M University* - -## FUNDAMENTALS OF ELECTRIC CIRCUITS, SIXTH EDITION - -Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2017 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2013, 2009, and 2007. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. - -Some ancillaries, including electronic and print components, may not be available to customers outside the United States. - -This book is printed on acid-free paper. - -1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5 - -ISBN 978-0-07-802822-9 MHID 0-07-802822-1 - -Senior Vice President, Products & Markets: *Kurt L. Strand* Vice President, General Manager, Products & Markets: *Marty Lange* Vice President, Content Design & Delivery: *Kimberly Meriwether David* Managing Director: *Thomas Timp* Global Brand Manager: *Raghu Srinivasan* Director, Product Development: *Rose Koos* Product Developer: *Vincent Bradshaw* Marketing Manager: *Nick McFadden* Digital Product Analyst: *Patrick Diller* Associate Director of Digital Content: *Amy Bumbaco, Ph.D.* Director, Content Design & Delivery: *Linda Avenarius* Program Manager: *Faye M. Herrig* Content Project Managers: *Melissa M. Leick, Tammy Juran, Sandra Schnee* Buyer: *Sandy Ludovissy* Design: *Studio Montage, Inc.* Content Licensing Specialist: *Lorraine Buczek* Cover Image: *Courtesy NASA/JPL - Caltech* Compositor: *MPS Limited* Printer: *R. R. Donnelley* - -All credits appearing on page or at the end of the book are considered to be an extension of the copyright page. - -## **Library of Congress Cataloging-in-Publication Data** - -Alexander, Charles K., author. - - Fundamentals of electric circuits / Charles K. Alexander, Department of Electrical and Computer Engineering, Cleveland State University, Matthew N. O. Sadiku, Department of Electrical Engineering, Prairie View A&M University. — Sixth edition. - - pages cm Includes index. ISBN 978-0-07-802822-9 (alk. paper) — ISBN 0-07-802822-1 (alk. paper) 1. Electric circuits. I. Sadiku, Matthew N. O., author. II. Title. - - TK454.A452 2017 621.3815—dc23 2015035301 - -The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites. - -Dedicated to our wives, Kikelomo and Hannah, whose understanding and support have truly made this book possible. - -Matthew and Chuck - -*This page intentionally left blank* - -# Contents - -*[Preface xi](#page-11-0) [Acknowledgments xv](#page-15-0) [About the Authors xxi](#page-21-0)* - -# **PART 1** [DC Circuits 2](#page-24-0) **Chapter 1** [Basic Concepts 3](#page-25-0) [1.1 Introduction 4](#page-26-0) [1.2 Systems of Units 5](#page-27-0) [1.3 Charge and Current 6](#page-28-0) [1.4 Voltage 9](#page-31-0) [1.5 Power and Energy 10](#page-32-0) [1.6 Circuit Elements 14](#page-36-0) [1.7 Applications 16](#page-38-0) 1.7.1 TV Picture Tube 1.7.2 Electricity Bills [1.8 Problem Solving 19](#page-41-0) [1.9 Summary 22](#page-44-0) [Review Questions 23](#page-45-0) [Problems 24](#page-46-0) [Comprehensive Problems 26](#page-48-0) - -# **Chapter 2** [Basic Laws 29](#page-51-0) - -- [2.1 Introduction 30](#page-52-0) -- [2.2 Ohm's Law 30](#page-52-0) -- [2.3 Nodes, Branches, and Loops 35](#page-57-0) -- [2.4 Kirchhoff's Laws 37](#page-59-0) -- [2.5 Series Resistors and Voltage Division 43](#page-65-0) -- [2.6 Parallel Resistors and Current Division 44](#page-66-0) -- [2.7 Wye-Delta Transformations 51](#page-73-0) Delta to Wye Conversion Wye to Delta Conversion -- [2.8 Applications 57](#page-79-0) 2.8.1 Lighting Systems [2.8.2 Design of DC Meters](#page-85-0) 2.9 Summary 63 [Review Questions 64](#page-86-0) - -# [Problems 65](#page-87-0) - -[Comprehensive Problems 77](#page-99-0) - -# **Chapter 3** [Methods of Analysis 79](#page-101-0) - -- [3.1 Introduction 80](#page-102-0) -- [3.2 Nodal Analysis 80](#page-102-0) -- [3.3 Nodal Analysis with Voltage Sources 86](#page-108-0) -- [3.4 Mesh Analysis 91](#page-113-0) -- [3.5 Mesh Analysis with Current Sources 96](#page-118-0) -- [3.6 Nodal and Mesh Analyses](#page-120-0) by Inspection 98 -- [3.7 Nodal Versus Mesh Analysis 102](#page-124-0) -- [3.8 Circuit Analysis with](#page-125-0) PSpice 103 -- [3.9 Applications: DC Transistor Circuits 105](#page-127-0) [3.10 Summary 110](#page-132-0) [Review Questions 111](#page-133-0) - - [Problems 112](#page-134-0) [Comprehensive Problem 124](#page-146-0) - -# **Chapter 4** [Circuit Theorems 125](#page-147-0) - -- [4.1 Introduction 126](#page-148-0) -- [4.2 Linearity Property 126](#page-148-0) -- [4.3 Superposition 128](#page-150-0) -- [4.4 Source Transformation 133](#page-155-0) -- [4.5 Thevenin's Theorem 137](#page-159-0) -- [4.6 Norton's Theorem 143](#page-165-0) -- [4.7 Derivations of Thevenin's](#page-169-0) and Norton's Theorems 147 -- [4.8 Maximum Power Transfer 148](#page-170-0) -- [4.9 Verifying Circuit Theorems](#page-172-0) with PSpice 150 -- [4.10 Applications 153](#page-175-0) 4.10.1 Source Modeling [4.10.2 Resistance Measurement](#page-180-0) -- 4.11 Summary 158 [Review Questions 159](#page-181-0) [Problems 160](#page-182-0) [Comprehensive Problems 171](#page-193-0) - -# **Chapter 5** [Operational Amplifiers 173](#page-195-0) - -- [5.1 Introduction 174](#page-196-0) -- [5.2 Operational Amplifiers 174](#page-196-0) - -[5.3 Ideal Op Amp 178](#page-200-0) [5.4 Inverting Amplifier 179](#page-201-0) [5.5 Noninverting Amplifier 181](#page-203-0) [5.6 Summing Amplifier 183](#page-205-0) [5.7 Difference Amplifier 185](#page-207-0) [5.8 Cascaded Op Amp Circuits 189](#page-211-0) [5.9 Op Amp Circuit Analysis with](#page-214-0) PSpice 192 [5.10 Applications 194](#page-216-0) [5.10.1 Digital-to-Analog Converter](#page-219-0) 5.10.2 Instrumentation Amplifiers 5.11 Summary 197 [Review Questions 199](#page-221-0) [Problems 200](#page-222-0) [Comprehensive Problems 211](#page-233-0) - -# **Chapter 6** [Capacitors and Inductors 213](#page-235-0) - -- [6.1 Introduction 214](#page-236-0) -- [6.2 Capacitors 214](#page-236-0) -- [6.3 Series and Parallel Capacitors 220](#page-242-0) -- [6.4 Inductors 224](#page-246-0) -- [6.5 Series and Parallel Inductors 228](#page-250-0) -- [6.6 Applications 231](#page-253-0) 6.6.1 Integrator 6.6.2 Differentiator 6.6.3 Analog Computer -- [6.7 Summary 238](#page-260-0) [Review Questions 239](#page-261-0) [Problems 240](#page-262-0) [Comprehensive Problems 249](#page-271-0) - -# **Chapter 7** [First-Order Circuits 251](#page-273-0) - -- [7.1 Introduction 252](#page-274-0) -- [7.2 The Source-Free](#page-275-0) RC Circuit 253 -- [7.3 The Source-Free](#page-279-0) RL Circuit 257 -- [7.4 Singularity Functions 263](#page-285-0) -- [7.5 Step Response of an](#page-293-0) RC Circuit 271 -- [7.6 Step Response of an](#page-300-0) RL Circuit 278 -- [7.7 First-Order Op Amp Circuits 282](#page-304-0) -- [7.8 Transient Analysis with](#page-309-0) PSpice 287 -- [7.9 Applications 291](#page-313-0) - - 7.9.1 Delay Circuits - - 7.9.2 Photoflash Unit - - 7.9.3 Relay Circuits - - [7.9.4 Automobile Ignition Circuit](#page-319-0) -- 7.10 Summary 297 [Review Questions 298](#page-320-0) [Problems 299](#page-321-0) [Comprehensive Problems 309](#page-331-0) - -# **Chapter 8** [Second-Order Circuits 311](#page-333-0) - -- [8.1 Introduction 312](#page-334-0) -- [8.2 Finding Initial and Final Values 313](#page-335-0) -- [8.3 The Source-Free Series](#page-339-0) RLC Circuit 317 -- [8.4 The Source-Free Parallel](#page-346-0) RLC Circuit 324 -- [8.5 Step Response of a Series](#page-351-0) RLC Circuit 329 -- [8.6 Step Response of a Parallel](#page-356-0) RLC Circuit 334 -- [8.7 General Second-Order Circuits 337](#page-359-0) -- [8.8 Second-Order Op Amp Circuits 342](#page-364-0) -- 8.9 PSpice Analysis of RLC [Circuits 344](#page-366-0) -- [8.10 Duality 348](#page-370-0) -- [8.11 Applications 351](#page-373-0) [8.11.1 Automobile Ignition System](#page-376-0) 8.11.2 Smoothing Circuits -- 8.12 Summary 354 [Review Questions 355](#page-377-0) [Problems 356](#page-378-0) [Comprehensive Problems 365](#page-387-0) - -``` -PART 2 AC Circuits 366 -``` - -# **Chapter 9** [Sinusoids and Phasors 367](#page-389-0) - -- [9.1 Introduction 368](#page-390-0) -- [9.2 Sinusoids 369](#page-391-0) -- [9.3 Phasors 374](#page-396-0) -- [9.4 Phasor Relationships for](#page-405-0) Circuit Elements 383 -- [9.5 Impedance and Admittance 385](#page-407-0) -- [9.6 Kirchhoff's Laws in the Frequency](#page-409-0) Domain 387 -- [9.7 Impedance Combinations 388](#page-410-0) -- [9.8 Applications 394](#page-416-0) 9.8.1 Phase-Shifters 9.8.2 AC Bridges -- [9.9 Summary 400](#page-422-0) [Review Questions 401](#page-423-0) [Problems 401](#page-423-0) [Comprehensive Problems 409](#page-431-0) - -# **Chapter 10** [Sinusoidal Steady-State](#page-433-0) Analysis 411 - -- [10.1 Introduction 412](#page-434-0) -- [10.2 Nodal Analysis 412](#page-434-0) -- [10.3 Mesh Analysis 415](#page-437-0) - -- [10.4 Superposition Theorem 419](#page-441-0) -- [10.5 Source Transformation 422](#page-444-0) [10.6 Thevenin and Norton](#page-446-0) Equivalent Circuits 424 -- [10.7 Op Amp AC Circuits 429](#page-451-0) -- [10.8 AC Analysis Using](#page-453-0) PSpice 431 -- [10.9 Applications 435](#page-457-0) [10.9.1 Capacitance Multiplier](#page-461-0) 10.9.2 Oscillators -- 10.10 Summary 439 [Review Questions 439](#page-461-0) [Problems 441](#page-463-0) - -# **Chapter 11** [AC Power Analysis 455](#page-477-0) - -- [11.1 Introduction 456](#page-478-0) -- [11.2 Instantaneous and Average Power 456](#page-478-0) -- [11.3 Maximum Average Power Transfer 462](#page-484-0) -- [11.4 Effective or RMS Value 465](#page-487-0) [11.5 Apparent Power and](#page-490-0) - - Power Factor 468 -- [11.6 Complex Power 471](#page-493-0) -- [11.7 Conservation of AC Power 475](#page-497-0) -- [11.8 Power Factor Correction 479](#page-501-0) [11.9 Applications 481](#page-503-0) 11.9.1 Power Measurement [11.9.2 Electricity Consumption Cost](#page-508-0) -- 11.10 Summary 486 [Review Questions 488](#page-510-0) [Problems 488](#page-510-0) [Comprehensive Problems 498](#page-520-0) - -# **Chapter 12** [Three-Phase Circuits 501](#page-523-0) - -- [12.1 Introduction 502](#page-524-0) -- [12.2 Balanced Three-Phase Voltages 503](#page-525-0) -- [12.3 Balanced Wye-Wye Connection 507](#page-529-0) -- [12.4 Balanced Wye-Delta Connection 510](#page-532-0) -- [12.5 Balanced Delta-Delta](#page-534-0) Connection 512 -- [12.6 Balanced Delta-Wye Connection 514](#page-536-0) -- [12.7 Power in a Balanced System 517](#page-539-0) -- [12.8 Unbalanced Three-Phase](#page-545-0) Systems 523 -- 12.9 PSpice [for Three-Phase Circuits 527](#page-549-0) -- [12.10 Applications 532](#page-554-0) [12.10.1 Three-Phase Power Measurement](#page-22-0) 12.10.2 Residential Wiring - -[12.11 Summary 541](#page-563-0) [Review Questions 541](#page-563-0) [Problems 542](#page-564-0) [Comprehensive Problems 551](#page-573-0) - -# **Chapter 13** [Magnetically Coupled](#page-575-0) Circuits 553 - -- [13.1 Introduction 554](#page-576-0) -- [13.2 Mutual Inductance 555](#page-577-0) -- [13.3 Energy in a Coupled Circuit 562](#page-584-0) -- [13.4 Linear Transformers 565](#page-587-0) -- [13.5 Ideal Transformers 571](#page-593-0) -- [13.6 Ideal Autotransformers 579](#page-601-0) -- [13.7 Three-Phase Transformers 582](#page-604-0) -- 13.8 PSpice [Analysis of Magnetically](#page-606-0) Coupled Circuits 584 -- [13.9 Applications 589](#page-611-0) - - [13.9.1 Transformer as an Isolation Device](#page-617-0) 13.9.2 Transformer as a Matching Device - - - 13.9.3 Power Distribution -- 13.10 Summary 595 [Review Questions 596](#page-618-0) [Problems 597](#page-619-0) [Comprehensive Problems 609](#page-631-0) - -# **Chapter 14** [Frequency Response 611](#page-633-0) - -- [14.1 Introduction 612](#page-634-0) -- [14.2 Transfer Function 612](#page-634-0) -- [14.3 The Decibel Scale 615](#page-637-0) -- [14.4 Bode Plots 617](#page-639-0) -- [14.5 Series Resonance 627](#page-649-0) -- [14.6 Parallel Resonance 632](#page-654-0) -- [14.7 Passive Filters 635](#page-657-0) - - 14.7.1 Low-Pass Filter - - 14.7.2 High-Pass Filter - - 14.7.3 Band-Pass Filter - - 14.7.4 Band-Stop Filter -- [14.8 Active Filters 640](#page-662-0) 14.8.1 First-Order Low-Pass Filter - - 14.8.2 First-Order High-Pass Filter - - 14.8.3 Band-Pass Filter - - [14.8.4 Band-Reject \(or Notch\) Filter](#page-668-0) -- 14.9 Scaling 646 - - 14.9.1 Magnitude Scaling - - 14.9.2 Frequency Scaling - - [14.9.3 Magnitude and Frequency Scaling](#page-22-0) - -- [14.10 Frequency Response Using](#page-672-0) PSpice 650 -- [14.11 Computation Using](#page-675-0) MATLAB 653 -- [14.12 Applications 655](#page-677-0) - - 14.12.1 Radio Receiver - - [14.12.2 Touch-Tone Telephone](#page-683-0) - - 14.12.3 Crossover Network -- 14.13 Summary 661 [Review Questions 662](#page-684-0) [Problems 663](#page-685-0) [Comprehensive Problems 671](#page-693-0) - -# **PART 3** [Advanced Circuit](#page-694-0) Analysis 672 - -**Chapter 15** [Introduction to the Laplace](#page-695-0) Transform 673 - -- [15.1 Introduction 674](#page-696-0) -- [15.2 Definition of the Laplace](#page-697-0) Transform 675 -- [15.3 Properties of the Laplace](#page-699-0) Transform 677 -- [15.4 The Inverse Laplace Transform 688](#page-710-0) 15.4.1 Simple Poles 15.4.2 Repeated Poles 15.4.3 Complex Poles -- [15.5 The Convolution Integral 695](#page-717-0) -- [15.6 Application to Integrodifferential](#page-725-0) Equations 703 -- [15.7 Summary 706](#page-728-0) [Review Questions 706](#page-728-0) [Problems 707](#page-729-0) - -# **Chapter 16** [Applications of the Laplace](#page-735-0) Transform 713 - -- [16.1 Introduction 714](#page-736-0) -- [16.2 Circuit Element Models 715](#page-737-0) -- [16.3 Circuit Analysis 720](#page-742-0) -- [16.4 Transfer Functions 724](#page-746-0) -- [16.5 State Variables 728](#page-750-0) -- [16.6 Applications 735](#page-757-0) 16.6.1 Network Stability 16.6.2 Network Synthesis -- [16.7 Summary 743](#page-765-0) [Review Questions 744](#page-766-0) [Problems 745](#page-767-0) [Comprehensive Problems 756](#page-778-0) - -# **Chapter 17** [The Fourier Series 757](#page-779-0) - -- [17.1 Introduction 758](#page-780-0) -- [17.2 Trigonometric Fourier Series 759](#page-781-0) -- [17.3 Symmetry Considerations 766](#page-788-0) - - 17.3.1 Even Symmetry - - 17.3.2 Odd Symmetry - - 17.3.3 Half-Wave Symmetry -- [17.4 Circuit Applications 776](#page-798-0) -- [17.5 Average Power and RMS Values 780](#page-802-0) -- [17.6 Exponential Fourier Series 783](#page-805-0) -- [17.7 Fourier Analysis with](#page-811-0) PSpice 789 [17.7.1 Discrete Fourier Transform](#page-817-0) -- 17.7.2 Fast Fourier Transform 17.8 Applications 795 17.8.1 Spectrum Analyzers 17.8.2 Filters -- [17.9 Summary 798](#page-820-0) [Review Questions 800](#page-822-0) [Problems 800](#page-822-0) [Comprehensive Problems 809](#page-831-0) - -# **Chapter 18** [Fourier Transform 811](#page-833-0) - -- [18.1 Introduction 812](#page-834-0) -- [18.2 Definition of the Fourier Transform 812](#page-834-0) -- [18.3 Properties of the Fourier](#page-840-0) Transform 818 -- [18.4 Circuit Applications 831](#page-853-0) -- [18.5 Parseval's Theorem 834](#page-856-0) -- [18.6 Comparing the Fourier and](#page-859-0) Laplace Transforms 837 -- [18.7 Applications 838](#page-860-0) [18.7.1 Amplitude Modulation](#page-863-0) 18.7.2 Sampling -- 18.8 Summary 841 [Review Questions 842](#page-864-0) [Problems 843](#page-865-0) [Comprehensive Problems 849](#page-871-0) - -# **Chapter 19** [Two-Port Networks 851](#page-873-0) - -- [19.1 Introduction 852](#page-874-0) -- [19.2 Impedance Parameters 853](#page-875-0) -- [19.3 Admittance Parameters 857](#page-879-0) -- [19.4 Hybrid Parameters 860](#page-882-0) -- [19.5 Transmission Parameters 865](#page-887-0) -- [19.6 Relationships Between](#page-892-0) Parameters 870 - -## Contents **ix** - -*[Index I-1](#page-981-0)* - -- [19.7 Interconnection of Networks 873](#page-895-0) -- [19.8 Computing Two-Port Parameters](#page-901-0) Using PSpice 879 -- [19.9 Applications 882](#page-904-0) 19.9.1 Transistor Circuits [19.9.2 Ladder Network Synthesis](#page-913-0) -- 19.10 Summary 891 [Review Questions 892](#page-914-0) [Problems 892](#page-914-0) [Comprehensive Problem 903](#page-925-0) -- **Appendix A** [Simultaneous Equations and Matrix](#page-926-0) Inversion A **Appendix B** [Complex Numbers A-9](#page-935-0) **Appendix C** [Mathematical Formulas A-16](#page-942-0) **Appendix D** [Answers to Odd-Numbered](#page-947-0) Problems A-21 *[Selected Bibliography B-1](#page-979-0)* - -*This page intentionally left blank* - -# Preface - -In keeping with our focus on space for covers for our book, we have chosen the NASA Voyager spacecraft for the sixth edition. The reason for this is that like any spacecraft there are many circuits that play criti cal roles in their functionality. The beginning of the Voyager 1 and 2 odyssey began on August 20, 1977, for Voyager 2 and on September 5, 1977, for Voyager 1. Both were launched from NASA's Kennedy Space Center in Florida. The Voyager 1 was launched on a faster orbit so it eventually became the first man-made object to leave our solar system. There is some debate over whether it has actually left the solar system, but it certainly will at some point in time. Voyager 2 and two Pioneer spacecraft will also leave the solar system at some point in time. - -Voyager 1 is still functioning and sending back data, a truly significant achievement for NASA engineers. The design processes that make the Voyager operate so reliably are based on the fundamentals discussed in this textbook. Finally, space is vast so that Voyager 1 will fly past other solar systems; the odds of actually coming into contact with something are so remote that it may virtually fly through the universe forever! For more about Voyager 1, go to NASA's website: www.nasa.gov/. - -# Features - -# New to This Edition - -We have added learning objectives to each chapter to reflect what we believe are the most important items to learn from each chapter. These should help you focus more carefully on what you should be learning. - -There are more than 580 revised end-of-chapter problems, new endof-chapter problems, and revised practice problems. We continue to try and make our problems as practical as possible. - -In addition, we have improved Connect for this edition by increasing the number of problems available substantially. Now, professors may select from more than a thousand problems as they build thier online homework assignments. - -We have also built SmartBook for this edition. With SmartBook, stu dents get the same text as the print version, along with personalized tips on what to study next, thanks to SmartBook's adaptive technology. - -# Retained from Previous Editions - -A course in circuit analysis is perhaps the first exposure students have to electrical engineering. This is also a place where we can enhance some of the skills that they will later need as they learn how to design. An important part of this book is our 121 *design a problem* problems. These problems were developed to enhance skills that are an impor tant part of the design process. We know it is not possible to fully develop a student's design skills in a fundamental course like circuits. To fully develop design skills a student needs a design experience - -normally reserved for their senior year. This does not mean that some of those skills cannot be developed and exercised in a circuits course. The text already included open-ended questions that help students use creativity, which is an important part of learning how to design. We already have some questions that are open-ended but we desired to add much more into our text in this important area and have devel oped an approach to do just that. When we develop problems for the student to solve our goal is that in solving the problem the student learns more about the theory and the problem solving process. Why not have the students design problems like we do? That is exactly what we do in each chapter. Within the normal problem set, we have a set of problems where we ask the student to design a problem to help other students better understand an important concept. This has two very important results. The first will be a better understanding of the basic theory and the second will be the enhancement of some of the student's basic design skills. We are making effective use of the principle of learning by teaching. Essentially we all learn better when we teach a subject. Designing effective problems is a key part of the teaching process. Students should also be encouraged to develop problems, when appropriate, which have nice numbers and do not necessarily overemphasize complicated mathematical manipulations. - -A very important advantage to our textbook, we have a total of 2,481 Examples, Practice Problems, Review Questions, and End-of-Chapter Problems! Answers are provided for all practice problems and the odd numbered end-of-chapter problems. - -The main objective of the sixth edition of this book remains the same as the previous editions—to present circuit analysis in a manner that is clearer, more interesting, and easier to understand than other cir cuit textbooks, and to assist the student in beginning to see the "fun" in engineering. This objective is achieved in the following ways: - -# • **Chapter Openers and Summaries** - -Each chapter opens with a discussion about how to enhance skills which contribute to successful problem solving as well as success ful careers or a career-oriented talk on a subdiscipline of electrical engineering. This is followed by an introduction that links the chap ter with the previous chapters and states the chapter objectives. The chapter ends with a summary of key points and formulas. - -• **Problem-Solving Methodology** - -Chapter 1 introduces a six-step method for solving circuit problems which is used consistently throughout the book and media supple ments to promote best-practice problem-solving procedures. - -• **Student-Friendly Writing Style** - -All principles are presented in a lucid, logical, step-by-step man ner. As much as possible, we avoid wordiness and giving too much detail that could hide concepts and impede overall understanding of the material. - -# • **Boxed Formulas and Key Terms** - -Important formulas are boxed as a means of helping students sort out what is essential from what is not. Also, to ensure that students clearly understand the key elements of the subject matter, key terms are defined and highlighted. - -# • **Margin Notes** - -Marginal notes are used as a pedagogical aid. They serve multiple uses such as hints, cross-references, more exposition, warnings, reminders not to make some particular common mistakes, and prob lem-solving insights. - -# • **Worked Examples** - -Thoroughly worked examples are liberally given at the end of ev ery section. The examples are regarded as a part of the text and are clearly explained without asking the reader to fill in missing steps. Thoroughly worked examples give students a good understanding of the solution process and the confidence to solve problems them selves. Some of the problems are solved in two or three different ways to facilitate a substantial comprehension of the subject mate rial as well as a comparison of different approaches. - -# • **Practice Problems** - -To give students practice opportunity, each illustrative example is immediately followed by a practice problem with the answer. The student can follow the example step-by-step to aid in the solution of the practice problem without flipping pages or looking at the end of the book for answers. The practice problem is also intended to test a student's understanding of the preceding example. It will reinforce their grasp of the material before the student can move on to the next section. Complete solutions to the practice problems are avail able to students on the website. - -# • **Application Sections** - -The last section in each chapter is devoted to practical application aspects of the concepts covered in the chapter. The material covered in the chapter is applied to at least one or two practical problems or devices. This helps students see how the concepts are applied to real-life situations. - -# • **Review Questions** - -Ten review questions in the form of multiple-choice objective items are provided at the end of each chapter with answers. The review questions are intended to cover the little "tricks" that the examples and end-of-chapter problems may not cover. They serve as a self test device and help students determine how well they have mas tered the chapter. - -# • **Computer Tools** - -In recognition of the requirements by ABET ® on integrating computer tools, the use of *PSpice, Multisim, MATLAB, KCIDE for Circuits*, and developing design skills are encouraged in a studentfriendly manner. *PSpice* is covered early on in the text so that stu dents can become familiar and use it throughout the text. Tutorials on all of these are available on Connect. *MATLAB* is also introduced early in the book. - -# • **Design a Problem Problems** - -Finally, *design a problem* problems are meant to help the student de velop skills that will be needed in the design process. - -# • **Historical Tidbits** - -Historical sketches throughout the text provide profiles of important pioneers and events relevant to the study of electrical engineering. - -# • **Early Op Amp Discussion** - - The operational amplifier (op amp) as a basic element is introduced early in the text. - -# • **Fourier and Laplace Transforms Coverage** - - To ease the transition between the circuit course and signals and systems courses, Fourier and Laplace transforms are covered lu cidly and thoroughly. The chapters are developed in a manner that the interested instructor can go from solutions of first-order circuits to Chapter 15. This then allows a very natural progression from Laplace to Fourier to AC. - -# • **Four-Color Art Program** - - An interior design and four-color art program bring circuit drawings to life and enhance key pedagogical elements throughout the text. - -• **Extended Examples** - - Examples worked in detail according to the six-step problem solv ing method provide a road map for students to solve problems in a consistent fashion. At least one example in each chapter is devel oped in this manner. - -• **EC 2000 Chapter Openers** - - Based on ABET's skill-based CRITERION 3, these chapter openers are devoted to discussions as to how students can acquire the skills that will lead to a significantly enhanced career as an engineer. Be cause these skills are so very important to the student while still in college as well after graduation, we use the heading, *"Enhancing your Skills and your Career.* " - -• **Homework Problems** - - There are 580 new or revised end-of-chapter problems and changed practice problems which will provide students with plenty of practice as well as reinforce key concepts. - -• **Homework Problem Icons** - - Icons are used to highlight problems that relate to engineering de sign as well as problems that can be solved using *PSpice, Multisim, KCIDE,* or *MATLAB* . - -# Organization - -This book was written for a two-semester or three-quarter course in linear circuit analysis. The book may also be used for a one-semester course by a proper selection of chapters and sections by the instructor. It is broadly divided into three parts. - -- Part 1, consisting of Chapters 1 to 8, is devoted to dc circuits. It covers the fundamental laws and theorems, circuits techniques, and passive and active elements. -- Part 2, which contains Chapter 9 to 14, deals with ac circuits. It introduces phasors, sinusoidal steady-state analysis, ac power, rms values, three-phase systems, and frequency response. -- Part 3, consisting of Chapters 15 to 19, are devoted to advanced techniques for network analysis. It provides students with a solid introduction to the Laplace transform, Fourier series, Fourier trans form, and two-port network analysis. - -The material in the three parts is more than sufficient for a two-semester course, so the instructor must select which chapters or sections to cover. Sections marked with the dagger sign (†) may be skipped, explained briefly, or assigned as homework. They can be omitted without loss of continuity. Each chapter has plenty of problems grouped according to the sections of the related material and diverse enough that the instructor can choose some as examples and assign some as homework. As stated ear lier, we are using three icons with this edition. We are using to de note problems that either require *PSpice* in the solution process, where the circuit complexity is such that *PSpice* or *Multisim* would make the solution process easier, and where *PSpice* or *Multisim* makes a good check to see if the problem has been solved correctly. We are using to denote problems where *MATLAB* is required in the solution process, where *MATLAB* makes sense because of the problem makeup and its complexity, and where *MATLAB* makes a good check to see if the problem has been solved correctly. Finally, we use to identify problems that help the student develop skills that are needed for engineering design. More difficult problems are marked with an asterisk (\*). - -Comprehensive problems follow the end-of-chapter problems. They are mostly applications problems that require skills learned from that particular chapter. - -# Prerequisites - -As with most introductory circuit courses, the main prerequisites, for a course using this textbook, are physics and calculus. Although familiar ity with complex numbers is helpful in the later part of the book, it is not required. A very important asset of this text is that ALL the mathemati cal equations and fundamentals of physics needed by the student, are included in the text. - -# Acknowledgments - -We would like to express our appreciation for the loving support we have received from our wives (Hannah and Kikelomo), daughters (Christina, Tamara, Jennifer, Motunrayo, Ann, and Joyce), son (Baixi), and our extended family members. We sincerely appreciate the invaluable help given us by Richard Rarick in helping us make the sixth edition a significantly more relevant book. He has checked all the new and revised problems and offered advice on making them more accurate and clear. - -At McGraw-Hill, we would like to thank the following editorial and production staff: Raghu Srinivasan, global brand manager; Vincent Bradshaw, product developer; Nick McFadden, marketing manager; and Melissa Leick, content project manager. - -The sixth edition has benefited greatly from the many outstanding individuals who have offered suggestions for improvements in both the text as well as the various problems. In particular, we thank Nicholas Reeder, Professor of Electronics Engineering Technology, Sinclair Community College, Dayton, Ohio, and Douglas De Boer, Professor of Engineering, Dordt College, Sioux Center, Iowa, for their detailed and careful corrections and suggestions for clarification which have - -## **xvi** Preface - -contributed to making this a better edition. In addition, the follow ing have made important contributions to this edition (in alphabetical order): - -# Zekeriya Aliyazicioglu, *California State Polytechnic University— Pomona* - -Rajan Chandra, *California State Polytechnic University—Pomona* Mohammad Haider, *University of Alabama—Birmingham* John Heathcote, *Reedley College* Peter LoPresti, *University of Tulsa* Robert Norwood, *John Brown University* Aaron Ohta, *University of Hawaii—Manoa* Salomon Oldak, *California State Polytechnic University—Pomona* Hesham Shaalan, *U.S. Merchant Marine Academy* Surendra Singh, *University of Tulsa* - -Finally, we sincerely appreciate the feedback received from instructors and students who used the previous editions. We want this to continue, so please keep sending us e-mails or direct them to the publisher. We can be reached at c.alexander@ieee.org for Charles Alexander and sadiku@ieee .org for Matthew Sadiku. - -C. K. Alexander and M. N. O. Sadiku - -# Supplements - -# Instructor and Student Resources - -Available on Connect are a number of additional instructor and student resources to accompany the text. These include complete solutions for all practice and end-of-chapter problems, solutions in *PSpice* and *Multisim* problems, lecture PowerPoints®, and text image files. In addition, instructors can use COSMOS, a complete online solutions manual organization system to create custom homework, quizzes, and tests using end-of-chapter problems from the text. - -# Knowledge Capturing Integrated Design Environment for Circuits (KCIDE for Circuits) - -This software, developed at Cleveland State University and funded by NASA, is designed to help the student work through a circuits problem in an organized manner using the six-step problem-solving methodology in the text. *KCIDE for Circuits* allows students to work a circuit problem in *PSpice* and *MATLAB*, track the evolution of their solution, and save a record of their process for future reference. In addition, the software automatically generates a Word document and/or a PowerPoint presentation. The software package can be downloaded for free. - -It is hoped that the book and supplemental materials supply the in structor with all the pedagogical tools necessary to effectively present the material. - -# McGraw-Hill Create® - -Craft your teaching resources to match the way you teach! With McGraw-Hill Create, http://create.mheducation.com, you can easily rearrange chapters, combine material from other content sources, and quickly upload content you have written like your course syllabus or teaching notes. Find the content you need in Create by searching through thousands of lead ing McGraw-Hill textbooks. Arrange your book to fit your teaching style. Create even allows you to personalize your book's appearance by select ing the cover and adding your name, school, and course information. Or der a Create book and you'll receive a complimentary print review copy in three to five business days or a complimentary electronic review copy (eComp) via e-mail in minutes. Go to http://create.mheducation.com to day and register to experience how McGraw-Hill Create empowers you to teach *your* students *your* way. - -**Required=Results** - -# **McGraw-Hill Connect® Learn Without Limits** - -Connect is a teaching and learning platform that is proven to deliver better results for students and instructors. - -Connect empowers students by continually adapting to deliver precisely what they need, when they need it and how they need it, so your class time is more engaging and effective. - -88% of instructors who use **Connect** require it; instructor satisfaction **increases** by 38% when **Connect** is required. - -# Analytics - -# **Connect Insight®** - -Connect Insight is Connect's new one-of-a-kind visual analytics dashboard—now available for both instructors and students—that provides at-a-glance information regarding student - -performance, which is immediately actionable. By presenting assignment, assessment, and topical performance results together with a time metric that is easily visible for aggregate or individual results, Connect Insight gives the user the ability to take a just-intime approach to teaching and learning, which was never before available. Connect Insight presents data that empowers students and helps instructors improve class performance in a way that is efficient and effective. - -Mobile - -Connect's new, intuitive mobile interface gives students and instructors flexible and convenient, anytime–anywhere access to all components of the Connect platform. - -Students can view their results for any **Connect** course. - -| | 10.00 | | -|--------------------------------------------------------------------------------------------------------------|------------------------------------------------------------------|----| -| LATE Acounting week 1 quiz | PRACTICE | | -| START EER - dieb, tird - all Countries excellents | | | -| LETT CHEZ-Qui Internations
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Ch Oh. En casa Variabalaria | 18 | - -# Adaptive - -More students earn **A's** and **B's** when they use McGraw-Hill Education **Adaptive** products. - -# **SmartBook®** - -Proven to help students improve grades and study more efficiently, SmartBook contains the same content within the print book, but actively tailors that content to the needs of the individual. SmartBook's adaptive technology provides precise, personalized instruction on what the student should do next, guiding the student to master and remember key concepts, targeting gaps in knowledge and offering customized feedback, driving the student toward comprehension and retention of the subject matter. Available on smartphones and tablets, SmartBook puts learning at the student's fingertips—anywhere, anytime. - -Over **4 billion questions** have been answered making McGraw-Hill Education products more intelligent, reliable & precise. - -THE FIRST AND ONLY **ADAPTIVE READING EXPERIENCE** DESIGNED TO TRANSFORM THE WAY STUDENTS READ - -*This page intentionally left blank* - -# About the Authors - -**Charles K. Alexander** is professor of electrical and computer engineering in the Washkewicz College of Engineering at Cleveland State University, Cleveland, Ohio. He is also the director of the Center for Research in Electronics and Aerospace Technology (CREATE). From 2002 until 2006 he was dean of the Fenn College of Engineering. He has held the position of dean of engineering at Cleveland State University, California State University, Northridge, and Temple University (acting dean for six years). He has held the position of department chair at Temple University and Tennessee Technological University as well as the position of Stocker Visiting Professor (an endowed chair) at Ohio University. He has held faculty status at all of the aforementioned universities. - -Dr. Alexander has secured funding for two centers of research at Ohio University and Cleveland State University. He has been the director of three additional research centers at Temple and Tennessee Tech and has obtained research funding of approximately \$100 million (in today's dollars). He has served as a consultant to 23 private and governmental organizations including the Air Force and the Navy. - -He received the honorary Dr. Eng. from Ohio Northern University (2009), his PhD (1971) and M.S.E.E. (1967) from Ohio University, and the B.S.E.E. (1965) from Ohio Northern University. - -Dr. Alexander has authored many publications, including a work book and a videotape lecture series, and is coauthor of *Fundamentals of Electric Circuits*, currently in its fifth edition, *Engineering Skills for Career Success, Problem Solving Made* ALMOST *Easy*, the fifth edition of the *Standard Handbook of Electronic Engineering,* and *Applied Circuit Analysis,* all with McGraw-Hill. He has delivered more than 500 paper, professional, and technical presentations. - -Dr. Alexander is a Life Fellow of the IEEE and served as its president and CEO in 1997. In addition he has held several volunteer posi tions within the IEEE during his more than 45 years of service. This includes serving from 1991 to 1999 on the IEEE board of directors. - -He has received several local, regional, national, and international awards for teaching and research, including an honorary Doctor of Engineering degree, Fellow of the IEEE, the IEEE-USA Jim Watson Student Professional Awareness Achievement Award, the IEEE Undergraduate Teaching Award, the Distinguished Professor Award, the Distinguished Engineering Education Achievement Award, the Distinguished Engi neering Education Leadership Award, the IEEE Centennial Medal, and the IEEE/RAB Innovation Award. - -Charles K. Alexander - -Matthew N. O. Sadiku - -**Matthew N. O. Sadiku** received his PhD from Tennessee Technological University, Cookeville. From 1984 to 1988, he was an assistant professor at Florida Atlantic University, where he did graduate work in computer science. From 1988 to 2000, he was at Temple University, Philadelphia, Pennsylvania, where he became a full professor. From 2000 to 2002, he was with Lucent/Avaya, Holmdel, New Jersey, as a system engineer and with Boeing Satellite Systems as a senior scientist. He is currently a professor at Prairie View A&M University. - -Dr. Sadiku is the author of more than 240 professional papers and over 60 books, including *Elements of Electromagnetics* (Oxford Uni versity Press, 6th ed., 2015), *Numerical Techniques in Electromagnetics with MATLAB* (CRC, 3rd ed., 2009), and *Metropolitan Area Net works* (CRC Press, 1995). Some of his books have been translated into French, Korean, Chinese (and Chinese Long Form in Taiwan), Italian, Portuguese, and Spanish. He was the recipient of the 2000 McGraw-Hill/ Jacob Millman Award for outstanding contributions in the field of electrical engineering. He was also the recipient of Regents Professor award for 2012 to 2013 by the Texas A&M University System. - -His current research interests are in the areas of numerical modeling of electromagnetic systems and computer communication networks. He is a registered professional engineer and a fellow of the Institute of Electrical and Electronics Engineers (IEEE) "for contributions to computa tional electromagnetics and engineering education." He was the IEEE Region 2 Student Activities Committee Chairman. He was an associ ate editor for *IEEE Transactions on Education* and is a member of the Association for Computing Machinery (ACM). - -# Fundamentals of Electric Circuits - -# **PART ONE** - -# DC Ci rcuits - -# OUTLINE - -- 1 Basic Concepts -- 2 Basic Laws -- 3 Methods of Analysis -- 4 Circuit Theorems -- 5 Operational Amplifiers -- 6 Capacitors and Inductors -- 7 First-Order Circuits -- 8 Second-Order Circuits - -# **chapter** - -1 - -# Basic Concepts - -*Some books are to be tasted, others to be swallowed, and some few to be chewed and digested.* - -—Francis Bacon - -# Enhancing Your Skills and Your Career - -# **ABET EC 2000 criteria (3.a), "an ability to apply knowledge of mathematics, science, and engineering."** - -As students, you are required to study mathematics, science, and engineering with the purpose of being able to apply that kno wledge to the solution of engineering problems. The skill here is the ability to apply the fundamentals of these areas in the solution of a problem. So how do you develop and enhance this skill? - -The best approach is to w ork as man y problems as possible in all of your courses. However, if you are really going to be successful with this, you must spend time analyzing where and when and why you have difficulty in easily arriving at successful solutions. You may be surprised to learn that most of your problem-solving problems are with mathematics rather than your understanding of theory . You may also learn that you start working the problem too soon. Taking time to think about the problem and ho w you should solv e it will al ways save you time and frustration in the end. - -What I have found that works best for me is to apply our sixstep problem-solving technique. Then I carefully identify the areas where I ha ve dif ficulty solving the problem. Many times, my actual deficiencies are in my understanding and ability to use correctly certain mathematical principles. I then return to my fundamental math texts and carefully review the appropriate sections, and in some cases, work some example problems in that text. This brings me to another important thing you should always do: Keep nearby all your basic mathematics, science, and engineering textbooks. - -This process of continually looking up material you thought you had acquired in earlier courses may seem v ery tedious at first; however, as your skills de velop and your kno wledge increases, this process will become easier and easier. On a personal note, it is this very process that led me from being a much less than a verage student to someone who could earn a Ph.D. and become a successful researcher. - -Photo by Charles Alexander - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Understand the different units with which engineers work. -- 2. Understand the relationship between charge and current and how to use both in a variety of applications. -- 3. Understand voltage and how it can be used in a variety of applications. -- 4. Develop an understanding of power and energy and their relationship with current and voltage. -- 5. Begin to understand the volt-amp characteristics of a variety of circuit elements. -- 6. Begin to understand an organized approach to problem solving and how it can be used to assist in your efforts to solve circuit problems. - -# **1.1** Introduction - -Electric circuit theory and electromagnetic theory are the tw o funda mental theories upon which all branches of electrical engineering are built. Many branches of electrical engineering, such as po wer, electric machines, control, electronics, communications, and instrumentation, are based on electric circuit theory . Therefore, the basic electric circuit theory course is the most important course for an electrical engineering student, and al ways an e xcellent starting point for a be ginning student in electrical engineering education. Circuit theory is also v aluable to students specializing in other branches of the ph ysical sciences because circuits are a good model for the study of energy systems in general, and because of the applied mathematics, physics, and topology involved. - -In electrical engineering, we are often interested in communicating or transferring energy from one point to another . To do this requires an interconnection of electrical devices. Such interconnection is referred to as an *electric circuit*, and each component of the circuit is kno wn as an *element.* - -An electric circuit is an interconnection of electrical elements. - -A simple electric circuit is sho wn in Fig. 1.1. It consists of three basic elements: a battery, a lamp, and connecting wires. Such a simple circuit can e xist by itself; it has se veral applications, such as a flashlight, a search light, and so forth. - -A complicated real circuit is displayed in Fig. 1.2, representing the schematic diagram for a radio receiver. Although it seems complicated, this circuit can be analyzed using the techniques we co ver in this book. Our goal in this text is to learn various analytical techniques and computer software applications for describing the behavior of a circuit like this. - -Electric circuits are used in numerous electrical systems to accomplish different tasks. Our objecti ve in this book is not the study of various uses and applications of circuits. Rather, our major concern is the analysis of the circuits. By the analysis of a circuit, we mean a study of the behavior of the - -**Figure 1.1** A simple electric circuit. - - - -# **Figure 1.2** - -Electric circuit of a radio transmitter. - -circuit: How does it respond to a gi ven input? How do the interconnected elements and devices in the circuit interact? - -We commence our study by defining some basic concepts. These concepts include char ge, current, v oltage, circuit elements, po wer, and energy. Before defining these concepts, we must first establish a system of units that we will use throughout the text. - -# **1.2** Systems of Units - -As electrical engineers, we must deal with measurable quantities. Our measurements, ho wever, must be communicated in a standard language that virtually all professionals can understand, irrespecti ve of the country in which the measurement is conducted. Such an international measurement language is the International System of Units (SI), adopted by the General Conference on Weights and Measures in 1960. In this system, there are seven base units from which the units of all other ph ysical quantities can be de rived. Table 1.1 shows six base units and one derived unit (the coulomb) that are related to this text. SI units are commonly used in electrical engineering. - -One great advantage of the SI unit is that it uses prefixes based on the power of 10 to relate larger and smaller units to the basic unit. Table 1.2 shows the SI prefixes and their symbols. For example, the following are expressions of the same distance in meters (m): - -| 600,000,000 mm | 600,000 m | 600 km | -|----------------|-----------|--------| -| | | | - -## **TABLE 1.1** - -Six basic SI units and one derived unit relevant to this text. - -| Quantity | Basic unit | Symbol | -|---------------------------|------------|--------| -| Length | meter | m | -| Mass | kilogram | kg | -| Time | second | s | -| Electric current | ampere | A | -| Thermodynamic temperature | kelvin | K | -| Luminous intensity | candela | cd | -| Charge | coulomb | C | - -## **TABLE 1.2** - -# The SI prefixes. - -| Multiplier | Prefix | Symbol | -|------------|--------|--------| -| 1018 | exa | E | -| 1015 | peta | P | -| 1012 | tera | T | -| 109 | giga | G | -| 106 | mega | M | -| 103 | kilo | k | -| 102 | hecto | h | -| 10 | deka | da | -| 10−1 | deci | d | -| 10−2 | centi | c | -| 10−3 | milli | m | -| 10−6 | micro | μ | -| 10−9 | nano | n | -| 10−12 | pico | p | -| 10−15 | femto | f | -| 10−18 | atto | a | -| | | | - -# **1.3** Charge and Current - -The concept of electric charge is the underlying principle for explaining all electrical phenomena. Also, the most basic quantity in an electric circuit is the *electric charge.* We all experience the effect of electric charge when we try to remove our wool sweater and have it stick to our body or walk across a carpet and receive a shock. - -Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C). - -We know from elementary physics that all matter is made of fundamental building blocks known as atoms and that each atom consists of electrons, protons, and neutrons. We also know that the char ge *e* on an electron is negative and equal in magnitude to 1.602 × 10−19 C, while a proton carries a positi ve charge of the same magnitude as the electron. The presence of equal numbers of protons and electrons leaves an atom neutrally charged. - -The following points should be noted about electric charge: - -- 1. The coulomb is a large unit for charges. In 1 C of charge, there are 1∕(1.602 × 10−19) = 6.24 × 1018 electrons. Thus realistic or laboratory values of charges are on the order of pC, nC, or *μ*C.1 -- 2. According to e xperimental observ ations, the only char ges that occur in nature are inte gral multiples of the electronic char ge *e* = −1.602 × 10−19 C. -- 3. The *law of conservation of charge* states that charge can neither be created nor destroyed, only transferred. Thus, the algebraic sum of the electric charges in a system does not change. - -We now consider the flow of electric char ges. A unique feature of electric charge or electricity is the f act that it is mobile; that is, it can be transferred from one place to another , where it can be con verted to another form of energy. - -When a conducting wire (consisting of se veral atoms) is connected to a battery (a source of electromotive force), the charges are compelled to move; positive charges move in one direction while ne gative charges move in the opposite direction. This motion of char ges creates elec tric current. It is conventional to take the current flow as the movement of positive charges. That is, opposite to the flow of ne gative charges, as Fig. 1.3 illustrates. This con vention w as introduced by Benjamin Franklin (1706–1790), the American scientist and in ventor. Although we now know that current in metallic conductors is due to ne gatively charged electrons, we will follo w the uni versally accepted con vention that current is the net flow of positive charges. Thus, - -Electric current is the time rate of change of charge, measured in amperes (A). - -Mathematically, the relationship between current *i*, charge *q*, and time *t* is - -$$ -i \triangleq \frac{dq}{dt} \tag{1.1} -$$ - -I – + – –– – - -# **Figure 1.3** - -Electric current due to flow of electronic charge in a conductor. - -A convention is a standard way of describing something so that others in the profession can understand what we mean. We will be using IEEE conventions throughout this book. - -1 However, a large power supply capacitor can store up to 0.5 C of charge. - -# Historical - -**Andre-Marie Ampere** (1775–1836), a French mathematician and physicist, laid the foundation of electrodynamics. He defined the electric current and developed a way to measure it in the 1820s. - -Born in Lyons, France, Ampere at age 12 mastered Latin in a few weeks, as he was intensely interested in mathematics and many of the best mathematical works were in Latin. He was a brilliant scientist and a prolific writer. He formulated the laws of electromagnetics. He in vented the electromagnet and the ammeter. The unit of electric current, the ampere, was named after him. - -© Apic/Getty Images - -where current is measured in amperes (A), and - -# 1 ampere = 1 coulomb/second - -The charge transferred between time *t*0 and *t* is obtained by inte grating both sides of Eq. (1.1). We obtain - -$$ -Q \triangleq \int_{t_0}^t i \, dt \tag{1.2} -$$ - -The way we define current as *i* in Eq. (1.1) suggests that current need not be a constant-valued function. As many of the examples and problems in this chapter and subsequent chapters suggest, there can be se veral types of current; that is, charge can vary with time in several ways. - -There are different ways of looking at direct current and alternating current. The best definition is that there are two ways that current can flow: It can always flow in the same direction, where it does not reverse direction, in which case we have *direct current* (dc). These currents can be constant or time varying. If the current flows in both directions, then we have *alternating current* (ac). - -A direct current (dc) flows only in one direction and can be constant or time varying. - -By convention, we will use the symbol *I* to represent a constant current. If the current v aries with respect to time (either dc or ac) we will use the symbol *i*. A common use of this w ould be the output of a rectifier (dc) such as *i*(*t*) = ∣5 sin(377t)∣ amps or a sinusoidal current (ac) such as *i*(*t*) = 160 sin(377t) amps. - -An alternating current (ac) is a current that changes direction with respect to time. - -An example of alternating current (ac) is the current you use in your house to run the air conditioner , refrigerator , w ashing machine, and other electric appliances. Figure 1.4 depicts tw o common e xamples of - -# **Figure 1.4** - -Two common types of current: (a) direct current (dc), (b) alternating current (ac). - -**Figure 1.5** Conventional current flow: (a) positive current flow, (b) negative current flow. - -dc (coming from a battery) and ac (coming from your home outlets). We will consider other types later in the book. - -Once we define current as the movement of charge, we expect current to have an associated direction of flow. As mentioned earlier, the direction of current flow is conventionally taken as the direction of positive charge movement. Based on this convention, a current of 5 A may be represented positively or negatively as shown in Fig. 1.5. In other w ords, a negative current of −5 A flowing in one direction as shown in Fig. 1.5(b) is the same as a current of +5 A flowing in the opposite direction. - -| Example 1.1 | How much charge is represented by 4,600 electrons? | -|----------------------|--------------------------------------------------------------------------------------------------------------------------------------------------------------| -| | Solution:
−19 C. Hence 4,600 electrons will
Each electron has
−1.602 × 10
have −1.602 × 10−19 C/electron × 4,600 electrons =
−7.369 × 10−16 C | -| Practice Problem 1.1 | Calculate the amount of charge represented by 6.667 billion protons. | -| | Answer: 1.0681 × 10−9
C. | -| | | -| Example 1.2 | The total char ge entering a terminal is gi
ven by q
= 5
t sin 4
πt mC.
Calculate the current at t
= 0.5 s. | -| | Solution: | -| | dq___
__d
i
=
dt =
dt (5t sin 4πt) mC/s = (5 sin 4πt
+ 20πt cos 4πt) mA | -| | At t
= 0.5, | -| | i
= 5 sin 2π
+ 10π cos 2π
= 0 + 10π
= 31.42 mA | -| Practice Problem 1.2 | = (10 − 10e−2t
If in Example 1.2, q
) mC, find the current at t
= 1.0 s. | -| | Answer: 2.707 mA. | -| | | -| Example 1.3 | Determine the total charge entering a terminal between t = 1 s and
2 −
t = 2 s if the current passing the terminal is i = (3t
t) A. | - -# **Solution:** - -$$ -Q = \int_{t=1}^{2} i \, dt = \int_{1}^{2} (3t^2 - t) \, dt -$$ -$$ -= \left(t^3 - \frac{t^2}{2}\right)\Big|_{1}^{2} = (8 - 2) - \left(1 - \frac{1}{2}\right) = 5.5 \text{ C} -$$ - -ough an element is - -\n -$$ -i = \begin{cases} 4 \text{ A}, & 0 < t < 1 \\ 4t^2 \text{ A}, & t > 1 \end{cases} -$$ - -Calculate the charge entering the element from *t* = 0 to *t* = 2 s. - -**Answer:** 13.333 C. - -# **1.4** Voltage - -As explained briefly in the previous section, to mo ve the electron in a conductor in a particular direction requires some w ork or energy transfer. This work is performed by an e xternal electromotive force (emf), typically represented by the battery in Fig. 1.3. This emf is also kno wn as *voltage* or *potential difference*. The voltage *vab* between two points *a* and *b* in an electric circuit is the energy (or work) needed to move a unit charge from *b* to *a*; mathematically, - -$$ -v_{ab} \triangleq \frac{dw}{dq} \tag{1.3} -$$ - -where *w* is energy in joules (J) and *q* is charge in coulombs (C). The voltage *vab* or simply *v* is measured in volts (V), named in honor of the Italian physicist Alessandro Antonio Volta (1745–1827), who invented the first voltaic battery. From Eq. (1.3), it is evident that - -1 volt = 1 joule/coulomb = 1 newton-meter/coulomb - -Thus, - -Voltage (or potential difference) is the energy required to move a unit charge from a reference point (−) to another point (+), measured in volts (V). - -Figure 1.6 sho ws the v oltage across an element (represented by a rectangular block) connected to points *a* and *b*. The plus ( +) and minus (−) signs are used to define reference direction or voltage polarity. The *vab* can be interpreted in two ways: (1) Point *a* is at a potential of *vab* volts higher than point *b*, or (2) the potential at point *a* with respect to point *b* is *vab*. It follows logically that in general - -$$ -v_{ab} = -v_{ba} \tag{1.4} -$$ - -For example, in Fig. 1.7, we ha ve two representations of the same v oltage. In Fig. 1.7(a), point *a* is +9 V above point *b*; in Fig. 1.7(b), point *b* is −9 V above point *a*. We may say that in Fig. 1.7(a), there is a 9-V *voltage drop* from *a* to *b* or equivalently a 9-V *voltage rise* from *b* to *a*. In other words, a voltage drop from *a* to *b* is equivalent to a voltage rise from *b* to *a*. - -Current and voltage are the tw o basic variables in electric circuits. The common term *signal* is used for an electric quantity such as a current or a voltage (or even electromagnetic wave) when it is used for conveying - -Polarity of voltage *vab*. - -Two equivalent representations of the same voltage *vab*: (a) Point *a* is 9 V above point *b*; (b) point *b* is −9 V above point *a*. - -# Practice Problem 1.3 - -# Historical - -© UniversalImagesGroup/ Getty Images - -**Alessandro Antonio Volta** (1745–1827), an Italian physicist, in vented the electric battery—which provided the first continuous flow of electricity—and the capacitor. - -Born into a noble family in Como, Italy, Volta was performing electrical experiments at age 18. His invention of the battery in 1796 revolutionized the use of electricity. The publication of his work in 1800 marked the beginning of electric circuit theory. Volta received many honors during his lifetime. The unit of voltage or potential difference, the volt, was named in his honor. - -Keep in mind that electric current is always through an element and that electric voltage is always across the element or between two points. - -information. Engineers prefer to call such v ariables signals rather than mathematical functions of time because of their importance in commu nications and other disciplines. Lik e electric current, a constant v oltage is called a *dc voltage* and is represented by *V*, whereas a sinusoidally time-varying voltage is called an *ac voltage* and is represented by *v*. A dc voltage is commonly produced by a battery; ac v oltage is produced by an electric generator. - -# **1.5** Power and Energy - -Although current and v oltage are the tw o basic v ariables in an electric circuit, the y are not suf ficient by themselves. F or practical purposes, we need to kno w how much *power* an electric de vice can handle. We all know from e xperience that a 100-w att bulb gives more light than a 60-watt bulb. We also know that when we pay our bills to the electric utility companies, we are paying for the electric *energy* consumed over a certain period of time. Thus, power and energy calculations are important in circuit analysis. - -To relate power and ener gy to v oltage and current, we recall from physics that: - -Power is the time rate of expending or absorbing energy, measured in watts (W). - -We write this relationship as - -$$ -p \triangleq \frac{dw}{dt} \tag{1.5} -$$ - -where *p* is power in watts (W), *w* is energy in joules (J), and *t* is time in seconds (s). From Eqs. (1.1), (1.3), and (1.5), it follows that - -$$ -p = \frac{dw}{dt} = \frac{dw}{dq} \cdot \frac{dq}{dt} = vi \tag{1.6} -$$ - -or - -$$ -p = vi \tag{1.7} -$$ - -The power *p* in Eq. (1.7) is a time-varying quantity and is called the *instantaneous power*. Thus, the power absorbed or supplied by an element is the product of the voltage across the element and the current through it. If the power has a + sign, power is being delivered to or absorbed by the element. If, on the other hand, the power has a − sign, power is being supplied by the element. But how do we know when the power has a negative or a positive sign? - -Current direction and voltage polarity play a major role in determining the sign of po wer. It is therefore important that we pay attention to the relationship between current *i* and voltage *v* in Fig. 1.8(a). The voltage polarity and current direction must conform with those sho wn in Fig. 1.8(a) in order for the power to have a positive sign. This is known as the *passive sign convention.* By the passive sign convention, current enters through the positive polarity of the voltage. In this case, *p* = +*vi* or *vi* > 0 implies that the element is absorbing po wer. However, if *p* = −*vi* or *vi* < 0, as in Fig. 1.8(b), the element is releasing or supplying power. - -Passive sign convention is satisfied when the current enters through the positive terminal of an element and p = +vi. If the current enters through the negative terminal, p = −vi. - -Unless otherwise stated, we will follow the passive sign convention throughout this text. For example, the element in both circuits of Fig. 1.9 has an absorbing power of +12 W because a positive current enters the positive terminal in both cases. In Fig. 1.10, ho wever, the element is supplying power of +12 W because a positive current enters the negative terminal. Of course, an absorbing power of −12 W is equivalent to a supplying power of +12 W. In general, - -# +Power absorbed = −Power supplied - -In fact, the *law of conservation of energy* must be obeyed in any electric circuit. For this reason, the algebraic sum of po wer in a circuit, at any instant of time, must be zero: - -$$ -\sum p = 0 \tag{1.8} -$$ - -This again confirms the fact that the total po wer supplied to the circuit must balance the total power absorbed. - -From Eq. (1.6), the energy absorbed or supplied by an element from time *t*0 to time *t* is - -$$ -w = \int_{t_0}^{t} p \, dt = \int_{t_0}^{t} v i \, dt \tag{1.9} -$$ - -# **Figure 1.8** - -Reference polarities for power using the passive sign convention: (a) absorbing power, (b) supplying power. - -When the voltage and current directions conform to Fig. 1.8(b), we have the active sign convention and p = +vi. - -# **Figure 1.9** - -Two cases of an element with an absorbing power of 12 W: (a) *p* = 4 × 3 = 12 W, (b) *p* = 4 × 3 = 12 W. - -# **Figure 1.10** Two cases of an element with a supplying power of 12 W: (a) *p* = −4 × 3 = −12 W, (b) *p* = −4 × 3 = −12 W. - -Energy is the capacity to do work, measured in joules (J). - -The electric po wer utility companies measure ener gy in w att-hours (Wh), where - -$$ -1 \text{ Wh} = 3,600 \text{ J} -$$ - -Example 1.4 An energy source forces a constant current of 2 A for 10 s to flow through a light bulb. If 2.3 kJ is gi ven off in the form of light and heat ener gy, calculate the voltage drop across the bulb. - -# **Solution:** - -The total charge is - -$$ -\Delta q = i \Delta t = 2 \times 10 = 20 \text{ C} -$$ - -The voltage drop is - -$$ -v = \frac{\Delta w}{\Delta q} = \frac{2.3 \times 10^3}{20} = 115 \text{ V} -$$ - -To move charge *q* from point *b* to point *a* requires 25 J. Find the volt age drop *vab* (the voltage at *a* positive with respect to *b*) if: (a) *q* = 5 C, (b) *q* = −10 C. Practice Problem 1.4 - -**Answer:** (a) 5 V, (b) −2.5 V. - -Example 1.5 Find the power delivered to an element at *t* = 3 ms if the current entering its positive terminal is - -*i* = 5 cos 60*π t* A - -and the voltage is: (a) *v* = 3*i*, (b) *v* = 3 *di*∕*dt*. - -# **Solution:** - -(a) The voltage is *v* = 3*i* = 15 cos 60*π t*; hence, the power is - -$$ -p = vi = 75 \cos^2 60 \pi t \,\mathrm{W} -$$ - -$$ -At t = 3 ms, -$$ - -*p* = 75 cos2 (60*π* × 3 × 10−3 ) = 75 cos2 0.18*π* = 53.48 W - -(b) We find the voltage and the power as - -$$ -v = 3\frac{di}{dt} = 3(-60\pi)5 \sin 60\pi t = -900\pi \sin 60\pi t \text{ V} -$$ - -$$ -p = vi = -4500\pi \sin 60\pi t \cos 60\pi t \text{ W} -$$ - -At *t* = 3 ms, - -*p* = −4500*π* sin 0.18*π* cos 0.18*π* W - -= −14137.167 sin 32.4° cos 32.4° = −6.396 kW - -# Historical - -**1884 Exhibition** In the United States, nothing promoted the future of electricity like the 1884 International Electrical Exhibition. Just imagine a world without electricity, a world illuminated by candles and gaslights, a world where the most common transportation was by walking and riding on horseback or by horse-drawn carriage. Into this world an exhibi tion was created that highlighted Thomas Edison and reflected his highly developed ability to promote his inventions and products. His exhibit featured spectacular lighting displays powered by an impressive 100-kW "Jumbo" generator. - -Edward Weston's dynamos and lamps were featured in the United States Electric Lighting Company's display. Weston's well known col lection of scientific instruments was also shown. - -Other prominent exhibitors included Frank Sprague, Elihu Thompson, and the Brush Electric Company of Cleveland. The American Institute of Electrical Engineers (AIEE) held its first technical meeting on October 7–8 at the Franklin Institute during the exhibit. AIEE merged with the Institute of Radio Engineers (IRE) in 1964 to form the Institute of Electrical and Electronics Engineers (IEEE). - -Practice Problem 1.5 - -Source: IEEE History Center - -Find the power delivered to the element in Example 1.5 at *t* = 5 ms if the current remains the same but the voltage is: (a) *v* = 2*i* V, - -(b) -$$ -v = \left(10 + 5 \int_0^t i \, dt \right) V. -$$ - -**Answer:** (a) 17.27 W, (b) 29.7 W. - -Example 1.6 How much energy does a 100-W electric bulb consume in two hours? - -# **Solution:** - -*w* = *pt* = 100 (W) × 2 (h) × 60 (min/h) × 60 (s/min) = 720,000 J = 720 kJ - -This is the same as - -$$ -w = pt = 100 -$$ - W $\times$ 2 h = 200 Wh - -Practice Problem 1.6 - -A home electric heater dra ws 10 A when connected to a 115 V outlet. How much energy is consumed by the heater over a period of 6 hours? - -**Answer:** 6.9 k watt-hours - -# **1.6** Circuit Elements - -As we discussed in Section 1.1, an element is the basic building block of a circuit. An electric circuit is simply an interconnection of the elements. Circuit analysis is the process of determining voltages across (or the currents through) the elements of the circuit. - -There are tw o types of elements found in electric circuits: *passive* elements and *active* elements. An active element is capable of generating energy while a passive element is not. Examples of passive elements are resistors, capacitors, and inductors. Typical active elements include generators, batteries, and operational amplifiers. Our aim in this section is to g ain familiarity with some important acti ve elements. - -The most important acti ve elements are v oltage or current sources that generally deliver power to the circuit connected to them. There are two kinds of sources: independent and dependent sources. - -An ideal independent source is an active element that provides a speci fied voltage or current that is completely independent of other circuit elements. - -**Figure 1.11** - -Symbols for independent voltage sources: (a) used for constant or time-varying voltage, (b) used for constant voltage (dc). - -In other words, an ideal independent voltage source delivers to the circuit whatever current is necessary to maintain its terminal v oltage. Physical sources such as batteries and generators may be regarded as approximations to ideal v oltage sources. Figure 1.11 sho ws the symbols for inde pendent voltage sources. Notice that both symbols in Fig. 1.11(a) and (b) can be used to represent a dc v oltage source, b ut only the symbol in Fig. 1.11(a) can be used for a time-v arying voltage source. Similarly, an ideal independent current source is an active element that provides a specified current completely independent of the voltage across the source. That is, the current source deli vers to the circuit whate ver - -voltage is necessary to maintain the designated current. The symbol for an independent current source is displayed in Fig. 1.12, where the arrow indicates the direction of current *i*. - -An ideal dependent (or controlled) source is an active element in which the source quantity is controlled by another voltage or current. - -Dependent sources are usually designated by diamond-shaped sym bols, as shown in Fig. 1.13. Since the control of the dependent source is achieved by a voltage or current of some other element in the circuit, and the source can be voltage or current, it follows that there are four possible types of dependent sources, namely: - -- 1. A voltage-controlled voltage source (VCVS). -- 2. A current-controlled voltage source (CCVS). -- 3. A voltage-controlled current source (VCCS). -- 4. A current-controlled current source (CCCS). - -Dependent sources are useful in modeling elements such as transistors, operational amplifiers, and integrated circuits. An example of a currentcontrolled voltage source is sho wn on the right-hand side of Fig. 1.14, where the v oltage 10*i* of the v oltage source depends on the current *i* through element *C*. Students might be surprised that the value of the dependent voltage source is 10 *i* V (and not 10 *i* A) because it is a v oltage source. The key idea to keep in mind is that a voltage source comes with polarities (+ −) in its symbol, while a current source comes with an arrow, irrespective of what it depends on. - -It should be noted that an ideal v oltage source (dependent or in dependent) will produce any current required to ensure that the termi nal voltage is as stated, whereas an ideal current source will produce the necessary voltage to ensure the stated current flow. Thus, an ideal source could in theory supply an infinite amount of energy. It should also be noted that not only do sources supply po wer to a circuit, the y can absorb power from a circuit too. For a voltage source, we know the voltage but not the current supplied or drawn by it. By the same token, we know the current supplied by a current source b ut not the v oltage across it. - -i **Figure 1.12** - -Symbols for: (a) dependent voltage source, (b) dependent current source. - -**Figure 1.14** The source on the right-hand side is a current-controlled voltage source. - -Calculate the power supplied or absorbed by each element in Fig. 1.15. Example 1.7 - -# **Solution:** - -We apply the sign convention for power shown in Figs. 1.8 and 1.9. For *p*1, the 5-A current is out of the positive terminal (or into the negative terminal); hence, - -*p*1 = 20(−5) = −100 W Supplied power - -For *p*2 and *p*3, the current flows into the positive terminal of the element in each case. - -> *p*2 = 12(5) = 60 W Absorbed power *p*3 = 8(6) = 48 W Absorbed power - -For Example 1.7. - -For *p*4, we should note that the voltage is 8 V (positive at the top), the same as the voltage for *p*3 since both the passive element and the dependent source are connected to the same terminals. (Remember that v oltage is always measured across an element in a circuit.) Since the current flows out of the positive terminal, - -$$ -p_4 = 8(-0.2I) = 8(-0.2 \times 5) = -8 -$$ - W Supplement power - -We should observ e that the 20-V independent v oltage source and 0.2*I* dependent current source are supplying power to the rest of the network, while the two passive elements are absorbing power. Also, - -*p*1 + *p*2 + *p*3 + *p*4 = −100 + 60 + 48 − 8 = 0 - -In agreement with Eq. (1.8), the total po wer supplied equals the total power absorbed. - -# Practice Problem 1.7 - -**Figure 1.16** For Practice Prob. 1.7. - -Compute the power absorbed or supplied by each component of the circuit in Fig. 1.16. - -**Answer:** *p*1 = −45 W, *p*2 = 18 W, *p*3 = 12 W, *p*4 = 15 W. - -# **1.7** Applications2 - -In this section, we will consider tw o practical applications of the con cepts developed in this chapter. The first one deals with the TV picture tube and the other with how electric utilities determine your electric bill. - -# **1.7.1** TV Picture Tube - -One important application of the motion of electrons is found in both the transmission and reception of TV signals. At the transmission end, a TV camera reduces a scene from an optical image to an electrical signal. Scanning is accomplished with a thin beam of electrons in an iconoscope camera tube. - -At the receiving end, the image is reconstructed by using a cathoderay tube (CR T) located in the TV recei ver.3 The CRT is depicted in Fig. 1.17. Unlike the iconoscope tube, which produces an electron beam of constant intensity, the CRT beam varies in intensity according to the incoming signal. The electron gun, maintained at a high potential, fires the electron beam. The beam passes through two sets of plates for vertical and horizontal deflections so that the spot on the screen where the beam strikes can move right and left and up and down. When the electron beam strikes the fluorescent screen, it gives off light at that spot. Thus, the beam can be made to "paint" a picture on the TV screen. - -2 The dagger sign preceding a section heading indicates the section that may be skipped, explained briefly, or assigned as homework. - -3 Modern TV tubes use a different technology. - -Cathode-ray tube. - -# Historical - -# Karl Ferdinand Braun and Vladimir K. Zworykin - -**Karl Ferdinand Braun** (1850–1918), of the University of Strasbourg, invented the Braun cathode-ray tube in 1879. This then became the basis for the picture tube used for so many years for televisions. It is still the most economical device today, although the price of flat-screen systems is rapidly becoming competitive. Before the Braun tube could be used in television, it took the inventiveness of **Vladimir K. Zworykin** (1889–1982) to develop the iconoscope so that the modern television would become a reality. The iconoscope developed into the orthicon and the image orthicon, which allowed images to be captured and converted into signals that could be sent to the television receiver. Thus, the television camera was born. - -The electron beam in a TV picture tube carries 10 Example 1.8 15 electrons per second. As a design engineer, determine the voltage *Vo* needed to accelerate the electron beam to achieve 4 W. - -# **Solution:** - -The charge on an electron is - -*e* = −1.6 × 10−19 C - -If the number of electrons is *n*, then *q* = *ne* and - -$$ -i = \frac{dq}{dt} = e \frac{dn}{dt} = (-1.6 \times 10^{-19})(10^{15}) = -1.6 \times 10^{-4} \text{ A} -$$ - -The negative sign indicates that the current flows in a direction opposite to electron flow as shown in Fig. 1.18, which is a simplified diagram of the CRT for the case when the vertical deflection plates carry no charge. The beam power is - -$$ -p = V_o i -$$ - or $V_o = \frac{p}{i} = \frac{4}{1.6 \times 10^{-4}} = 25,000 \text{ V}$ - -Thus, the required voltage is 25 kV. - -# Practice Problem 1.8 - -If an electron beam in a TV picture tube carries 10 13 electrons/second and is passing through plates maintained at a potential difference of 30 kV, calculate the power in the beam. - -**Answer:** 48 mW. - -# **1.7.2** Electricity Bills - -The second application deals with ho w an electric utility compan y charges their customers. The cost of electricity depends upon the amount of ener gy consumed in kilo watt-hours (kWh). (Other f actors that affect the cost include demand and po wer factors; we will ignore these for now.) However, even if a consumer uses no energy at all, there is a minimum service char ge the customer must pay because it costs money to stay connected to the po wer line. As ener gy consumption increases, the cost per kWh drops. It is interesting to note the a verage monthly consumption of household appliances for a family of five, shown in Table 1.3. - -## **TABLE 1.3** - -Typical average monthly consumption of household appliances. - -| Appliance | kWh consumed | Appliance | kWh consumed | -|---------------|--------------|-------------------|--------------| -| Water heater | 500 | Washing machine | 120 | -| Freezer | 100 | Stove | 100 | -| Lighting | 100 | Dryer | 80 | -| Dishwasher | 35 | Microwave oven | 25 | -| Electric iron | 15 | Personal computer | 12 | -| TV | 10 | Radio | 8 | -| Toaster | 4 | Clock | 2 | - -A simplified diagram of the cathode-ray tube; for Example 1.8. - -A homeowner consumes 700 kWh in January. Determine the electricity Example 1.9 bill for the month using the following residential rate schedule: - -Base monthly charge of \$12.00. - -First 100 kWh per month at 16 cents/kWh. - -Next 200 kWh per month at 10 cents/kWh. - -Over 300 kWh per month at 6 cents/kWh. - -# **Solution:** - -We calculate the electricity bill as follows. - -Base monthly charge = \$12.00 First 100 kWh @ \$0.16/k Wh = \$16.00 Next 200 kWh @ \$0.10/k Wh = \$20.00 Remaining 400 kWh @ \$0.06/k Wh = \$24.00 Total charge = \$72.00 Average cost = \_\_\_\_\_\_\_\_\_\_\_\_\_\_ \$72 100 + 200 +400 = 10.2 cents/kWh - -Referring to the residential rate schedule in Example 1.9, calculate the average cost per kWh if only 350 kWh are consumed in July when the family is on vacation most of the time. - -Practice Problem 1.9 - -**Answer:** 14.571 cents/kWh. - -# **1.8** Problem Solving - -Although the problems to be solved during one's career will vary in complexity and magnitude, the basic principles to be follo wed remain the same. The process outlined here is the one developed by the authors over many years of problem solving with students, for the solution of engi neering problems in industry, and for problem solving in research. - -We will list the steps simply and then elaborate on them. - -- 1. Carefully **define** the problem. -- 2. **Present** everything you know about the problem. -- 3. Establish a set of **alternative** solutions and determine the one that promises the greatest likelihood of success. -- 4. **Attempt** a problem solution. -- 5. **Evaluate** the solution and check for accuracy. -- 6. Has the problem been solved **satisfactorily**? If so, present the solution; if not, then return to step 3 and continue through the process again. - -1. *Carefully define the problem* . This may be the most important part of the process, because it becomes the foundation for all the rest of the steps. In general, the presentation of engineering problems is - -somewhat incomplete. You must do all you can to make sure you understand the problem as thoroughly as the presenter of the problem understands it. Time spent at this point clearly identifying the problem will save you considerable time and frustration later. As a student, you can clarify a problem statement in a textbook by asking your professor. A problem presented to you in industry may require that you consult several individuals. At this step, it is important to develop questions that need to be addressed before continuing the solution process. If you have such questions, you need to consult with the appropriate individuals or resources to obtain the answers to those questions. With those answers, you can now refine the problem, and use that refinement as the problem statement for the rest of the solution process. - -2. *Present everything you know about the problem*. You are now ready to write down everything you know about the problem and its possible solutions. This important step will save you time and frustration later. - -3. *Establish a set of alternative solutions and determine the one that promises the greatest likelihood of success* . Almost every problem will have a number of possible paths that can lead to a solution. It is highly desirable to identify as many of those paths as possible. At this point, you also need to determine what tools are available to you, such as *PSpice* and *MATLAB* and other software packages that can greatly reduce effort and increase accuracy. Again, we want to stress that time spent carefully defining the problem and investigating alternative ap proaches to its solution will pay big dividends later. Evaluating the al ternatives and determining which promises the greatest likelihood of success may be difficult but will be well worth the effort. Document this process well since you will want to come back to it if the first approach does not work. - -4. *Attempt a problem solution*. Now is the time to actually begin solving the problem. The process you follow must be well documented in order to present a detailed solution if successful, and to evaluate the process if you are not successful. This detailed evaluation may lead to corrections that can then lead to a successful solution. It can also lead to new alternatives to try. Many times, it is wise to fully set up a solution before putting numbers into equations. This will help in checking your results. - -5. *Evaluate the solution and check for accuracy*. You now thoroughly evaluate what you have accomplished. Decide if you have an acceptable solution, one that you want to present to your team, boss, or professor. - -6. *Has the problem been solved satisfactorily? If so, present the solu tion; if not, then return to step 3 and continue through the process again.* Now you need to present your solution or try another alternative. At this point, presenting your solution may bring closure to the process. Often, however, presentation of a solution leads to further refinement of the problem definition, and the process continues. Following this process will eventually lead to a satisfactory conclusion. - -Now let us look at this process for a student taking an electrical and computer engineering foundations course. (The basic process also ap plies to almost e very engineering course.) K eep in mind that although the steps have been simplified to apply to academic types of problems, the process as stated always needs to be followed. We consider a simple example. - -# **Solution:** - -1. *Carefully define the problem*. This is only a simple example, but we can already see that we do not know the polarity on the 3-V source. We have the following options. We can ask the professor what the polarity should be. If we cannot ask, then we need to make a decision on what to do next. If we have time to work the problem both ways, we can solve for the current when the 3-V source is plus on top and then plus on the bottom. If we do not have the time to work it both ways, assume a polarity and then carefully document your decision. Let us assume that the professor tells us that the source is plus on the bottom as shown in Fig. 1.20. - -2. *Present everything you know about the problem*. Presenting all that we know about the problem involves labeling the circuit clearly so that we define what we seek. - -Given the circuit shown in Fig. 1.20, solve for *i*8Ω. - -We now check with the professor , if reasonable, to see if the prob lem is properly defined. - -3. *Establish a set of alternative solutions and determine the one that promises the greatest likelihood of success* . There are essentially three techniques that can be used to solve this problem. Later in the text you will see that you can use circuit analysis (using Kirchhoff's laws and Ohm's law), nodal analysis, and mesh analysis. - -To solve for *i*8Ω using circuit analysis will eventually lead to a solution, but it will likely take more work than either nodal or mesh analysis. To solve for *i*8Ω using mesh analysis will require writing two simultaneous equations to find the two loop currents indicated in Fig. 1.21. Using nodal analysis requires solving for only one unknown. This is the easiest approach. - -**Figure 1.21** Using nodal analysis. - -Therefore, we will solve for *i*8Ω using nodal analysis. - -4. *Attempt a problem solution*. We first write down all of the equations we will need in order to find *i*8Ω. - -$$ -i_{8\Omega} = i_2, -$$ - $i_2 = \frac{v_1}{8},$ $i_{8\Omega} = \frac{v_1}{8}$ - $\frac{v_1 - 5}{2} + \frac{v_1 - 0}{8} + \frac{v_1 + 3}{4} = 0$ - -**Figure 1.19** Illustrative example. - -**Figure 1.20** Problem definition. - -Now we can solve for *v*1. - -$$ -8\left[\frac{v_1 - 5}{2} + \frac{v_1 - 0}{8} + \frac{v_1 + 3}{4}\right] = 0 -$$ - -leads to (4 *v* 1 − 20) + (*v* 1) + (2 *v* 1 + 6) = 0 - -$$ -7v_1 = +14 -$$ -, $v_1 = +2$ V, $i_{8\Omega} = \frac{v_1}{8} = \frac{2}{8} = 0.25$ A - -5. *Evaluate the solution and check for accuracy* . We can now use Kirchhoff's voltage law (KVL) to check the results. - -$$ -i_1 = \frac{v_1 - 5}{2} = \frac{2 - 5}{2} = -\frac{3}{2} = -1.5 \text{ A} -$$ - -$$ -i_2 = i_{8\Omega} = 0.25 \text{ A} -$$ - -$$ -i_3 = \frac{v_1 + 3}{4} = \frac{2 + 3}{4} = \frac{5}{4} = 1.25 \text{ A} -$$ - -*i*1 + *i*2 + *i*3 = −**1.5** + **0.25** + **1.25** = **0** (Checks.) - -Applying KVL to loop 1, - -$$ --5 + v_{2\Omega} + v_{8\Omega} = -5 + (-i_1 \times 2) + (i_2 \times 8) -$$ - -= -5 + [ -(-1.5)2] + (0.25 \times 8) -= -5 + 3 + 2 = 0 (Checks.) - -Applying KVL to loop 2, - -$$ --v_{8\Omega} + v_{4\Omega} - 3 = -(i_2 \times 8) + (i_3 \times 4) - 3 -$$ - -= -(0.25 \times 8) + (1.25 \times 4) - 3 -= -2 + 5 - 3 = 0 (Checks.) - -So we now have a v ery high de gree of confidence in the accuracy of our answer. - -6. *Has the problem been solved satisfactorily? If so, present the solution; if not, then return to step 3 and continue through the process again.* This problem has been solved satisfactorily. - -The current through the 8-Ω resistor is 0.25 A flowing down through the 8-Ω resistor. - -Try applying this process to some of the more difficult problems at the end of the chapter. Practice Problem 1.10 - -# **1.9** Summary - -- 1. An electric circuit consists of electrical elements connected together. -- 2. The International System of Units (SI) is the international mea surement language, which enables engineers to communicate their results. From the se ven principal units, the units of other ph ysical quantities can be derived. - -3. Current is the rate of char ge flow past a gi ven point in a gi ven direction. - -$$ -i = \frac{dq}{dt} -$$ - -4. Voltage is the energy required to move 1 C of char ge from a reference point (−) to another point (+). - -> *vab* = \_\_\_ *dw dq* - -5. Power is the energy supplied or absorbed per unit time. It is also the product of voltage and current. - -$$ -p = \frac{dw}{dt} = vi -$$ - -- 6. According to the passi ve sign con vention, power assumes a posi tive sign when the current enters the positive polarity of the voltage across an element. -- 7. An ideal v oltage source produces a specific potential difference across its terminals re gardless of what is connected to it. An ideal current source produces a specific current through its terminals regardless of what is connected to it. -- 8. Voltage and current sources can be dependent or independent. A dependent source is one whose value depends on some other circuit variable. -- 9. Two areas of application of the concepts covered in this chapter are the TV picture tube and electricity billing procedure. - -# Review Questions - -**1.1** One millivolt is one millionth of a volt. - -``` -(a) True (b) False -``` - -**1.2** The prefix *micro* stands for: - -(a) 106 (b) 103 (c) 10−3 (d) 10−6 - -**1.3** The voltage 2,000,000 V can be expressed in powers of 10 as: - -(a) 2 mV (b) 2 kV (c) 2 MV (d) 2 GV - -**1.4** A charge of 2 C flowing past a given point each second is a current of 2 A. - -(a) True (b) False - -**1.5** The unit of current is: - -(a) coulomb (b) ampere (c) volt (d) joule - -**1.6** Voltage is measured in: - -(a) watts (b) amperes - -(a) True (b) False - -- (c) volts (d) joules per second -- **1.7** A 4-A current charging a dielectric material will accumulate a charge of 24 C after 6 s. - -**1.8** The voltage across a 1.1-kW toaster that produces a current of 10 A is: - -(a) 11 kV (b) 1100 V (c) 110 V (d) 11 V - -**1.9** Which of these is not an electrical quantity? - -| (a) charge | (b) time | (c) voltage | -|-------------|-----------|-------------| -| (d) current | (e) power | | - -- **1.10** The dependent source in Fig. 1.22 is: - - (a) voltage-controlled current source - - (b) voltage-controlled voltage source - - (c) current-controlled voltage source - - (d) current-controlled current source - -**Figure 1.22** For Review Question 1.10. - -*Answers: 1.1b, 1.2d, 1.3c, 1.4a, 1.5b, 1.6c, 1.7a, 1.8c, 1.9b, 1.10d.* - -# Problems - -# Section 1.3 Charge and Current - -- **1.1** How much charge is represented by these number of electrons? - - (a) 6.482 × 1017 - - (b) 1.24 × 1018 - - (c) 2.46 × 1019 - - (d) 1.628 × 1020 -- **1.2** Determine the current flowing through an element if the charge flow is given by - - (a) *q*(*t*) = (3) mC - -(b) -$$ -q(t) = (4t^2 + 20t - 4) -$$ - C - -(c) -$$ -q(t) = (15e^{-3t} - 2e^{-18t}) nC -$$ - -(d) $q(t) = 5t^2(3t^3 + 4) pC$ - -(d) -$$ -q(t) = 5t^2(3t^3 + 4) \text{ pC} -$$ - -(e) $q(t) = 2e^{-3t} \sin(20\pi t) \mu\text{C}$ - -**1.3** Find the charge *q*(*t*) flowing through a device if the current is: - -(a) -$$ -i(t) = 3 -$$ - A, $q(0) = 1$ C -\n(b) $i(t) = (2t + 5)$ mA, $q(0) = 0$ -\n(c) $i(t) = 20 \cos(10t + \pi/6) \mu$ A, $q(0) = 2 \mu$ C -\n(d) $i(t) = 10e^{-30t} \sin 40t$ A, $q(0) = 0$ - -- **1.4** A total charge of 300 C flows past a given cross section of a conductor in 30 seconds. What is the value of the current? -- **1.5** Determine the total charge transferred over the time interval of 0 ≤ *t* ≤ 10 s when *i*(*t*) = \_\_1 2 *t* A. -- **1.6** The charge entering a certain element is shown in Fig. 1.23. Find the current at: - -(a) -$$ -t = 1 -$$ - ms (b) $t = 6$ ms (c) $t = 10$ ms - -**1.7** The charge flowing in a wire is plotted in Fig. 1.24. Sketch the corresponding current. - -# **Figure 1.24** - -For Prob. 1.7. - -**1.8** The current flowing past a point in a device is shown in Fig. 1.25. Calculate the total charge through the point. - -# **Figure 1.25** - -For Prob. 1.8. - -**1.9** The current through an element is shown in Fig. 1.26. Determine the total charge that passed through the element at: - -For Prob. 1.9. - -# Sections 1.4 and 1.5 Voltage, Power, and Energy - -- **1.10** A lightning bolt with 10 kA strikes an object for 15 *μ*s. How much charge is deposited on the object? -- **1.11** A rechargeable flashlight battery is capable of delivering 90 mA for about 12 h. How much charge can it release at that rate? If its terminal voltage is 1.5 V, how much energy can the battery deliver? -- **1.12** If the current flowing through an element is given by - -$$ -i(t) = \begin{cases} 3tA, & 0 \le t < 6 \text{ s} \\ 18A, & 6 \le t < 10 \text{ s} \\ -12A, & 10 \le t < 15 \text{ s} \\ 0, & t \ge 15 \text{ s} \end{cases} -$$ - -Plot the charge stored in the element over 0 < *t* < 20 s. - -**1.13** The charge entering the positive terminal of an element is - -$$ -q = 5\sin 4\pi t \,\mathrm{mC} -$$ - - while the voltage across the element (plus to minus) is - -$$ -v = 3\cos 4\pi t \,\mathrm{V} -$$ - -- (a) Find the power delivered to the element at *t* = 0.3 s. -- (b) Calculate the energy delivered to the element between 0 and 0.6 s. -- **1.14** The voltage *v*(*t*) across a device and the current *i*(*t*) through it are - -$$ -v(t) = 20 \sin(4t) -$$ - V and $i(t) = 10(1 + e^{-2t})$ mA - -Calculate: - -- (a) the total charge in the device at *t* = 1 *s*, *q*(0) = 0. -- (b) the power consumed by the device at *t* = 1 s. -- **1.15** The current entering the positive terminal of a device is *i*(*t*) = 6*e*−2*t* mA and the voltage across the device is *v*(*t*) = 10*di*∕*dt*V. - - (a) Find the charge delivered to the device between *t* = 0 and *t* = 2 s. - - (b) Calculate the power absorbed. - - (c) Determine the energy absorbed in 3 s. - -# Section 1.6 Circuit Elements - -- **1.16** Figure 1.27 shows the current through and the voltage across an element. - - (a) Sketch the power delivered to the element for *t* > 0. - - (b) Fnd the total energy absorbed by the element for the period of 0 < *t* < 4s. - -**Figure 1.27** For Prob. 1.16. - -**1.17** Figure 1.28 shows a circuit with four elements, *p*1 = 60 W absorbed, *p*3 = −145 W absorbed, and *p*4 = 75 W absorbed. How many watts does element 2 absorb? - -# **Figure 1.28** - -For Prob. 1.17. - -**1.18** Find the power absorbed by each of the elements in Fig. 1.29. - -# **Figure 1.29** - -For Prob. 1.18. - -**1.19** Find *I* and the power absorbed by each element in the network of Fig. 1.30. - -# **Figure 1.30** - -For Prob. 1.19. - -**1.20** Find *Vo* and the power absorbed by each element in the circuit of Fig. 1.31. - -For Prob. 1.20. - -# Section 1.7 Applications - -- **1.21** A 60-W incandescent bulb operates at 120 V. How many electrons and coulombs flow through the bulb in one day? -- **1.22** A lightning bolt strikes an airplane with 40 kA for 1.7 ms. How many coulombs of charge are deposited on the plane? -- **1.23** A 1.8-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents/kWh, what is the cost of its operation for 30 days? -- **1.24** A utility company charges 8.2 cents/kWh. If a consumer operates a 60-W light bulb continuously for one day, how much is the consumer charged? -- **1.25** A 1.2-kW toaster takes roughly 4 minutes to heat four slices of bread. Find the cost of operating the toaster twice per day for 2 weeks (14 days). Assume energy costs 9 cents/kWh. -- **1.26** A cell phone battery is rated at 3.85 V and can store 10.78 watt-hours of energy. - - (a) How much average current can it deliver over a period of 3 hours if it is fully discharged at the end of that time? - - (b) How much average power is delivered in part (a)? - - (c) What is the ampere-hour rating of the battery? -- **1.27** A constant current of 3 A for 4 hours is required to charge an automotive battery. If the terminal voltage is 10 + *t*∕2V, where *t* is in hours, - - (a) how much charge is transported as a result of the charging? - -- (b) how much energy is expended? -- (c) how much does the charging cost? Assume electricity costs 9 cents/kWh. -- **1.28** A 150-W incandescent outdoor lamp is connected to a 120-V source and is left burning continuously for an average of 12 hours per day. Determine: - - (a) the current through the lamp when it is lit. - - (b) the cost of operating the light for one non-leap year if electricity costs 9.5 cents per kWh. -- **1.29** An electric stove with four burners and an oven is used in preparing a meal as follows. - -| Burner 1: 20 minutes | Burner 2: 40 minutes | -|----------------------|----------------------| -| Burner 3: 15 minutes | Burner 4: 45 minutes | -| Oven: 30 minutes | | - -If each b urner is rated at 1.2 kW and the o ven at 1.8 kW, and electricity costs 12 cents per kWh, calculate the cost of electricity used in preparing the meal. - -**1.30** Reliant Energy (the electric company in Houston, Texas) charges customers as follows: - -> Monthly charge \$6 First 250 kWh @ \$0.02/kWh All additional kWh @ \$0.07/kWh - -If a customer uses 2,436 kWh in one month, ho w much will Reliant Energy charge? - -**1.31** In a household, a business is run for an average of 6 h/day. The total power consumed by the computer and its printer is 230 W. In addition, a 75-W light runs during the same 6 h. If their utility charges 11.75 cents per kWh, how much do the owners pay every 30 days? - -# Comprehensive Problems - -- **1.32** A telephone wire has a current of 20*μ*A flowing through it. How long does it take for a charge of 15 C to pass through the wire? -- **1.33** A lightning bolt carried a current of 2 kA and lasted for 3 ms. How many coulombs of charge were contained in the lightning bolt? -- **1.34** Figure 1.32 shows the power consumption of a certain household in 1 day. Calculate: - - (a) the total energy consumed in kWh, - - (b) the average power over the total 24 hour period. - -For Prob. 1.34. - -**1.35** The graph in Fig. 1.33 represents the power drawn by an industrial plant between 8:00 and 8:30 a.m. Cal culate the total energy in MWh consumed by the plant. - -For Prob. 1.35. - -**1.36** A battery can be rated in ampere-hours (Ah) or watt hours (Wh). The ampere hours can be obtained from the watt hours by dividing watt hours by a nominal - -voltage of 12 V. If an automobile battery is rated at 20 Ah: - -- (a) What is the maximum current that can be supplied for 15 minutes? -- (b) How many days will it last if it is discharged at a rate of 2 mA? -- **1.37** A total of 2 MJ are delivered to an automobile battery (assume 12 V) giving it an additional charge. How much is that additional charge? Express your answer in ampere-hours. -- **1.38** How much energy does a 10-hp motor deliver in 30 minutes? Assume that 1 horsepower = 746 W. -- **1.39** A 600-W TV receiver is turned on for 4 h with nobody watching it. If electricity costs 10 cents/kWh, how much money is wasted? - -*This page intentionally left blank* - -# **chapter** - -# Basic Laws 2 - -*There are too many people praying for mountains of difficulty to be removed, when what they really need is the courage to climb them!* —Unknown - -# Enhancing Your Skills and Your Career - -# **ABET EC 2000 criteria (3.b), "an ability to design and conduct experiments, as well as to analyze and interpret data."** - -Engineers must be able to design and conduct e xperiments, as well as analyze and interpret data. Most students ha ve spent man y hours per forming experiments in high school and in college. During this time, you have been asked to analyze the data and to interpret the data. Therefore, you should already be skilled in these tw o activities. My recommendation is that, in the process of performing e xperiments in the future, you spend more time in analyzing and interpreting the data in the conte xt of the experiment. What does this mean? - -If you are looking at a plot of voltage versus resistance or current versus resistance or power versus resistance, what do you actually see? Does the curve make sense? Does it agree with what the theory tells you? Does it differ from e xpectation, and, if so, wh y? Clearly, practice with analyzing and interpreting data will enhance this skill. - -Since most, if not all, the e xperiments you are required to do as a student involve little or no practice in designing the experiment, how can you develop and enhance this skill? - -Actually, developing this skill under this constraint is not as difficult as it seems. What you need to do is to take the experiment and analyze it. Just break it down into its simplest parts, reconstruct it trying to under stand why each element is there, and finally, determine what the author of the experiment is trying to teach you. Even though it may not always seem so, every experiment you do w as designed by someone who w as sincerely motivated to teach you something. - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Know and understand the voltage current relationship of resistors (Ohm's law). -- 2. Understand the basic structure of electrical circuits, essentially nodes, loops, and branches. -- 3. Understand Kirchhoff's voltage and current laws and their importance in analyzing electrical circuits. -- 4. Understand series resistances and voltage division, and parallel resistances and current division. -- 5. Know how to convert delta-connected circuits to wye-connected circuits and how to convert wye-connected circuits to deltaconnected circuits. - -# **2.1** Introduction - -Chapter 1 introduced basic concepts such as current, voltage, and power in an electric circuit. To actually determine the v alues of these v ariables in a gi ven circuit requires that we understand some fundamen tal laws that govern electric circuits. These laws, known as Ohm's law and Kirchhoff's laws, form the foundation upon which electric circuit analysis is built. - -In this chapter , in addition to these la ws, we shall discuss some techniques commonly applied in circuit design and analysis. These techniques include combining resistors in series or parallel, voltage division, current division, and delta-to-wye and wye-to-delta transformations. The application of these laws and techniques will be restricted to resistive circuits in this chapter. We will finally apply the laws and techniques to real-life problems of electrical lighting and the design of dc meters. - -# **2.2** Ohm's Law - -Materials in general ha ve a characteristic beha vior of resisting the flow of electric charge. This physical property, or ability to resist current, is known as *resistance* and is represented by the symbol *R*. The resistance of any material with a uniform cross-sectional area *A* depends on *A* and its length ℓ, as sho wn in Fig. 2.1(a). We can represent resistance (as measured in the laboratory), in mathematical form, - -$$ -R = \rho \frac{\ell}{A} \tag{2.1} -$$ - -where *ρ* is known as the *resistivity* of the material in ohm-meters. Good conductors, such as copper and aluminum, ha ve low resistivities, while insulators, such as mica and paper, have high resistivities. Table 2.1 presents the values of *ρ* for some common materials and shows which materials are used for conductors, insulators, and semiconductors. - -The circuit element used to model the current-resisting beha vior of a material is the *resistor*. For the purpose of constructing circuits, resistors are - -**Figure 2.1** (a) Resistor, (b) Circuit symbol for resistance. - -| TABLE 2.1 | | | -|-----------|--|--| -| | | | - -| Material | Resistivity (Ω∙m) | Usage | -|-----------|-------------------|---------------| -| Silver | 1.64 × 10−8 | Conductor | -| Copper | 1.72 × 10−8 | Conductor | -| Aluminum | 2.8 × 10−8 | Conductor | -| Gold | 2.45 × 10−8 | Conductor | -| Carbon | 4 × 10−5 | Semiconductor | -| Germanium | 47 × 10−2 | Semiconductor | -| Silicon | 6.4 × 102 | Semiconductor | -| Paper | 1010 | Insulator | -| Mica | 5 × 1011 | Insulator | -| Glass | 1012 | Insulator | -| Teflon | 3 × 1012 | Insulator | - -Resistivities of common materials. - -usually made from metallic alloys and carbon compounds. The circuit symbol for the resistor is shown in Fig. 2.1(b), where *R* stands for the resistance of the resistor. The resistor is the simplest passive element. - -Georg Simon Ohm (1787–1854), a German ph ysicist, is credited with finding the relationship between current and voltage for a resistor. This relationship is known as *Ohm's law*. - -Ohm's law states that the voltage v across a resistor is directly proportional to the current i flowing through the resistor. - -That is, - -$$ -v \propto i \tag{2.2} -$$ - -Ohm defined the constant of proportionality for a resistor to be the resistance, *R*. (The resistance is a material property which can change if the internal or external conditions of the element are altered, e.g., if there are changes in the temperature.) Thus, Eq. (2.2) becomes - -$$ -v = iR \tag{2.3} -$$ - -# Historical - -**Georg Simon Ohm** (1787–1854), a German physicist, in 1826 experimentally determined the most basic law relating voltage and cur rent for a resistor. Ohm's work was initially denied by critics. - -Born of humble beginnings in Erlangen, Bavaria, Ohm threw himself into electrical research. His efforts resulted in his famous law. He was awarded the Copley Medal in 1841 by the Royal Society of London. In 1849, he was given the Professor of Physics chair by the University of Munich. To honor him, the unit of resistance was named the ohm. - -(b) **Figure 2.2** - -(a) Short circuit (*R* =0), (b) Open circuit (*R* =∞). - -**Figure 2.3** Fixed resistors: (a) wirewound type, (b) carbon film type. © McGraw-Hill Education/Mark Dierker, photographer - -**Figure 2.4** Circuit symbol for: (a) a variable resistor in general, (b) a potentiometer. - -which is the mathematical form of Ohm's law. *R* in Eq. (2.3) is mea sured in the unit of ohms, designated Ω. Thus, - -The resistance R of an element denotes its ability to resist the flow of electric current; it is measured in ohms (Ω). - -We may deduce from Eq. (2.3) that - -$$ -R = \frac{v}{i} \tag{2.4} -$$ - -so that - -$$ -1 \Omega = 1 \text{ V/A} -$$ - -To apply Ohm' s la w as stated in Eq. (2.3), we must pay careful attention to the current direction and v oltage polarity. The direction of current *i* and the polarity of voltage *v* must conform with the passive sign convention, as shown in Fig. 2.1(b). This implies that current flows from a higher potential to a lower potential in order for *v* = *i R*. If current flows from a lower potential to a higher potential, *v* = −*i R*. - -Since the value of *R* can range from zero to infinity, it is important that we consider the tw o extreme possible values of *R*. An element with *R* = 0 is called a *short circuit*, as shown in Fig. 2.2(a). For a short circuit, - -$$ -v = iR = 0 \tag{2.5} -$$ - -showing that the v oltage is zero b ut the current could be an ything. In practice, a short circuit is usually a connecting wire assumed to be a perfect conductor. Thus, - -A short circuit is a circuit element with resistance approaching zero. - -Similarly, an element with *R* =∞ is known as an *open circuit*, as shown in Fig. 2.2(b). For an open circuit, - -ircuit, -\n -$$ -i = \lim_{R \to \infty} \frac{v}{R} = 0 -$$ -\n(2.6) - -indicating that the current is zero though the v oltage could be anything. Thus, - -An open circuit is a circuit element with resistance approaching infinity. - -A resistor is either fixed or v ariable. Most resistors are of the fixed type, meaning their resistance remains constant. The two common types of fixed resistors (wirewound and composition) are shown in Fig. 2.3. The composition resistors are used when large resistance is needed. The circuit symbol in Fig. 2.1(b) is for a fixed resistor. Variable resistors have adjustable resistance. The symbol for a variable resistor is shown in Fig. 2.4(a). A common variable resistor is known as a *potentiometer* or *pot* for short, with the symbol shown in Fig. 2.4(b). The pot is a three-terminal element with a sliding contact or wiper . By sliding the wiper , the resistances be tween the wiper terminal and the fixed terminals v ary. Like fixed resistors, variable resistors can be of either wire wound or composition type, as shown in Fig. 2.5. Although resistors like those in Figs. 2.3 and 2.5 are used in circuit designs, today most circuit components including resistors are either surface mounted or integrated, as typically shown in Fig. 2.6. - -**Figure 2.5** Variable resistors: (a) composition type, (b) slider pot. © McGraw-Hill Education/Mark Dierker, photographer - -It should be pointed out that not all resistors obe y Ohm's law. A resistor that obeys Ohm's law is known as a *linear* resistor. It has a constant resistance and thus its current-voltage characteristic is as illustrated in Fig. 2.7(a): Its *i*-*v* graph is a straight line passing through the ori gin. A *nonlinear* resistor does not obe y Ohm's law. Its resistance varies with current and its *i*-*v* characteristic is typically sho wn in Fig. 2.7(b). Examples of devices with nonlinear resistance are the light bulb and the diode. Although all practical resistors may e xhibit nonlinear beha vior under certain conditions, we will assume in this book that all elements actually designated as resistors are linear. - -A useful quantity in circuit analysis is the reciprocal of resistance *R*, known as *conductance* and denoted by *G*: - -$$ -G = \frac{1}{R} = \frac{i}{v} \tag{2.7} -$$ - -The conductance is a measure of how well an element will conduct electric current. The unit of conductance is the *mho* (ohm spelled backward) or reciprocal ohm, with symbol ℧, the inverted omega. Although engineers often use the mho, in this book we prefer to use the siemens (S), the SI unit of conductance: - -$$ -1 S = 1 \, \mathbf{U} = 1 \, \mathbf{A} / \mathbf{V} \tag{2.8} -$$ - -Thus, - -Conductance is the ability of an element to conduct electric current; it is measured in mhos (℧) or siemens (S). - -The same resistance can be e xpressed in ohms or siemens. F or example, 10 Ω is the same as 0.1 S. From Eq. (2.7), we may write - -$$ -i = Gv \tag{2.9} -$$ - -The power dissipated by a resistor can be e xpressed in terms of *R*. Using Eqs. (1.7) and (2.3), - -$$ -p = vi = i^{2}R = \frac{v^{2}}{R} -$$ - (2.10) - -**Figure 2.6** Resistors in an integrated circuit board. - -**Figure 2.7** The *i*-*v* characteristic of: (a) a linear resistor, (b) a nonlinear resistor. - -The power dissipated by a resistor may also be e xpressed in terms of *G* as - -$$ -p = vi = v^2 G = \frac{i^2}{G} -$$ - (2.11) - -We should note two things from Eqs. (2.10) and (2.11): - -- 1. The power dissipated in a resistor is a nonlinear function of either current or voltage. -- 2. Since *R* and *G* are positive quantities, the power dissipated in a resistor is al ways positive. Thus, a resistor al ways absorbs power from the circuit. This confirms the idea that a resistor is a passive element, incapable of generating energy. - -# **Solution:** - -The voltage across the resistor is the same as the source voltage (30 V) because the resistor and the voltage source are connected to the same pair of terminals. Hence, the current is - -$$ -i = \frac{v}{R} = \frac{30}{5 \times 10^3} = 6 \text{ mA} -$$ - -The conductance is - -$$ -G = \frac{1}{R} = \frac{1}{5 \times 10^3} = 0.2 \text{ mS} -$$ - -We can calculate the power in various ways using either Eqs. (1.7), (2.10), or (2.11). - -*p* = *vi* = 30 (6 × 10 −3) = 180 mW - -or - -$$ -p = i^2 R = (6 \times 10^{-3})^2 5 \times 10^3 = 180 -$$ - mW - -or - -$$ -p = v^2 G = (30)^2 0.2 \times 10^{-3} = 180 -$$ - mW - -**Figure 2.8** For Example 2.2. - -For the circuit shown in Fig. 2.9, calculate the voltage *v*, the conductance *G*, and the power *p*. - -**Answer:** 30 V, 100 *µ*S, 90 mW. - -For Practice Prob. 2.2 - -A voltage source of 20 sin *πt* V is connected across a 5-kΩ resistor. Find Example 2.3 the current through the resistor and the power dissipated. - -# **Solution:** - -$$ -i = \frac{v}{R} = \frac{20 \sin \pi t}{5 \times 10^3} = 4 \sin \pi t \text{ mA} -$$ - -Hence, - -$$ -p = vi = 80 \sin^2 \pi t \text{ mW} -$$ - -A resistor absorbs an instantaneous power of 30 cos Practice Problem 2.3 2 *t* mW when con nected to a voltage source *v* = 15 cos *t* V. Find *i* and *R*. - -**Answer:** 2 cos *t* mA, 7.5 kΩ. - -# **2.3** Nodes, Branches, and Loops - -Since the elements of an electric circuit can be interconnected in se veral ways, we need to understand some basic concepts of network topology. To differentiate between a circuit and a network, we may regard a network as an interconnection of elements or devices, whereas a circuit is a network providing one or more closed paths. The convention, when addressing network topology, is to use the w ord network rather than circuit. We do this even though the word network and circuit mean the same thing when used in this context. In network topology, we study the properties relating to the placement of elements in the network and the geometric configuration of the network. Such elements include branches, nodes, and loops. - -A branch represents a single element such as a voltage source or a resistor. - -In other words, a branch represents an y two-terminal element. The circuit in Fig. 2.10 has five branches, namely, the 10-V voltage source, the 2-A current source, and the three resistors. - -A node is the point of connection between two or more branches. - -A node is usually indicated by a dot in a circuit. If a short circuit (a connecting wire) connects two nodes, the two nodes constitute a single node. The circuit in Fig. 2.10 has three nodes *a*, *b*, and *c*. Notice that - -Practice Problem 2.2 - -**Figure 2.11** The three-node circuit of Fig. 2.10 is redrawn. - -the three points that form node *b* are connected by perfectly conducting wires and therefore constitute a single point. The same is true of the four points forming node *c*. We demonstrate that the circuit in Fig. 2.10 has only three nodes by redrawing the circuit in Fig. 2.11. The two circuits in Figs. 2.10 and 2.11 are identical. However, for the sake of clarity, nodes *b* and *c* are spread out with perfect conductors as in Fig. 2.10. - -A loop is any closed path in a circuit. - -A loop is a closed path formed by starting at a node, passing through a set of nodes, and returning to the starting node without passing through any node more than once. A loop is said to be *independent* if it contains at least one branch which is not a part of an y other independent loop. Independent loops or paths result in independent sets of equations. - -It is possible to form an independent set of loops where one of the loops does not contain such a branch. In Fig. 2.11, *abca* with the 2 Ω resistor is independent. A second loop with the 3 Ω resistor and the current source is independent. The third loop could be the one with the 2Ω resistor in parallel with the 3Ω resistor. This does form an independent set of loops. - -A network with *b* branches, *n* nodes, and *l* independent loops will satisfy the fundamental theorem of network topology: - -$$ -b = l + n - 1 \tag{2.12} -$$ - -As the next two definitions show, circuit topology is of great value to the study of voltages and currents in an electric circuit. - -Two or more elements are in series if they exclusively share a single node and consequently carry the same current. - -Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them. - -Elements are in series when the y are chain-connected or connected se quentially, end to end. F or example, two elements are in series if they share one common node and no other element is connected to that common node. Elements in parallel are connected to the same pair of terminals. Elements may be connected in a w ay that they are neither in series nor in parallel. In the circuit shown in Fig. 2.10, the voltage source and the 5- Ω resistor are in series because the same current will flow through them. The 2-Ω resistor, the 3-Ω resistor, and the current source are in parallel because they are connected to the same two nodes *b* and *c* and consequently have the same voltage across them. The 5-Ω and 2-Ω - -Example 2.4 - -Determine the number of branches and nodes in the circuit shown in Fig. 2.12. Identify which elements are in series and which are in parallel. - -resistors are neither in series nor in parallel with each other. - -# **Solution:** - -Since there are four elements in the circuit, the circuit has four branches: 10 V, 5 Ω, 6 Ω, and 2 A. The circuit has three nodes as identified in Fig. 2.13. The 5-Ω resistor is in series with the 10-V voltage source because the same current would flow in both. The 6-Ω resistor is in parallel with the 2-A current source because both are connected to the same nodes 2 and 3. - -How many branches and nodes does the circuit in Fig. 2.14 have? Identify Practice Problem 2.4 the elements that are in series and in parallel. - -**Answer:** Five branches and three nodes are identified in Fig. 2.15. The 1-Ω and 2-Ω resistors are in parallel. The 4-Ω resistor and 10-V source are also in parallel. - -# **2.4** Kirchhoff's Laws - -Ohm's law by itself is not sufficient to analyze circuits. However, when it is coupled with Kirchhoff's two laws, we have a sufficient, powerful set of tools for analyzing a large variety of electric circuits. Kirchhoff's laws were first introduced in 1847 by the German physicist Gustav Robert Kirchhoff (1824–1887). These laws are formally kno wn as Kirchhoff's current law (KCL) and Kirchhoff's voltage law (KVL). - -Kirchhoff's first law is based on the la w of conservation of charge, which requires that the algebraic sum of charges within a system cannot change. - -Kirchhoff's current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero. - -Mathematically, KCL implies that - -$$ -\sum_{n=1}^{N} i_n = 0 -$$ - (2.13) - -where *N* is the number of branches connected to the node and *in* is the *n*th current entering (or lea ving) the node. By this la w, currents entering a node may be regarded as positive, while currents leaving the node may be taken as negative or vice versa. - -# Historical - -**Gustav Robert Kirchhoff** (1824–1887), a German physicist, stated two basic laws in 1847 concerning the relationship between the cur rents and voltages in an electrical network. Kirchhoff's laws, along with Ohm's law, form the basis of circuit theory. - -Born the son of a lawyer in Konigsberg, East Prussia, Kirchhoff entered the University of Konigsberg at age 18 and later became a lecturer in Berlin. His collaborative work in spectroscopy with German chemist Robert Bunsen led to the discovery of cesium in 1860 and rubidium in 1861. Kirchhoff was also credited with the Kirchhoff law of radiation. Thus, Kirchhoff is famous among engineers, chemists, and physicists. - -To prove KCL, assume a set of currents *i k* (*t*) , *k* = 1, 2,…, flow into a node. The algebraic sum of currents at the node is - -$$ -i_T(t) = i_1(t) + i_2(t) + i_3(t) + \cdots -$$ - (2.14) - -Integrating both sides of Eq. (2.14) gives - -$$ -q_T(t) = q_1(t) + q_2(t) + q_3(t) + \cdots -$$ - (2.15) - -where *qk* (*t*) = ∫ *ik* (*t*) *d t* and *qT* (*t*) = ∫ *iT* (*t*) *d t* .But the law of conservation of electric charge requires that the algebraic sum of electric charges at the node must not change; that is, the node stores no net charge. Thus, *qT* (*t*) = 0 → *iT* (*t*) = 0, confirming the validity of KCL. - -Consider the node in Fig. 2.16. Applying KCL gives - -$$ -i_1 + (-i_2) + i_3 + i_4 + (-i_5) = 0 \tag{2.16} -$$ - -since currents *i*1, *i*3, and *i*4 are entering the node, while currents *i*2 and *i*5 are leaving it. By rearranging the terms, we get - -$$ -i_1 + i_3 + i_4 = i_2 + i_5 \tag{2.17} -$$ - -Equation (2.17) is an alternative form of KCL: - -The sum of the currents entering a node is equal to the sum of the currents leaving the node. - -Note that KCL also applies to a closed boundary . This may be re garded as a generalized case, because a node may be regarded as a closed surface shrunk to a point. In tw o dimensions, a closed boundary is the same as a closed path. As typically illustrated in the circuit of Fig. 2.17, the total current entering the closed surface is equal to the total current leaving the surface. - -A simple application of KCL is combining current sources in parallel. The combined current is the algebraic sum of the current supplied by the indi vidual sources. F or example, the current sources sho wn in - -**Figure 2.16** Currents at a node illustrating KCL. - -**Figure 2.17** Applying KCL to a closed boundary. - -Two sources (or circuits in general) are said to be equivalent if they have the same i-v relationship at a pair of terminals. - -Fig. 2.18(a) can be combined as in Fig. 2.18(b). The combined or equivalent current source can be found by applying KCL to node *a*. - -$$ -I_T + I_2 = I_1 + I_3 -$$ - -or - -$$ -I_T = I_1 - I_2 + I_3 \tag{2.18} -$$ - -A circuit cannot contain two different currents, *I*1 and *I*2, in series, unless *I*1 =*I*2; otherwise KCL will be violated. - -Kirchhoff's second law is based on the principle of conservation of energy: - -Kirchhoff's voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero. - -Expressed mathematically, KVL states that - -$$ -\sum_{m=1}^{M} v_m = 0 -$$ - (2.19) - -where *M* is the number of voltages in the loop (or the number of branches in the loop) and *vm* is the *m*th voltage. - -To illustrate KVL, consider the circuit in Fig. 2.19. The sign on each voltage is the polarity of the terminal encountered first as we travel around the loop. We can start with an y branch and go around the loop either clockwise or counterclockwise. Suppose we start with the v oltage source and go clockwise around the loop as sho wn; then v oltages would be −*v*1, +*v*2, +*v*3, +*v*4, and −*v*5, in that order. For example, as we reach branch 3, the positive terminal is met first; hence, we have +*v*3. For branch 4, we reach the ne gative terminal first; hence, −*v*4. Thus, KVL yields - -$$ --v_1 + v_2 + v_3 - v_4 + v_5 = 0 \tag{2.20} -$$ - -Rearranging terms gives - -$$ -v_2 + v_3 + v_5 = v_1 + v_4 \tag{2.21} -$$ - -which may be interpreted as - -| Sum of voltage drops = Sum of voltage rises | (2.22) | -|---------------------------------------------|--------| -|---------------------------------------------|--------| - -This is an alternative form of KVL. Notice that if we had traveled counterclockwise, the result w ould have been +*v*1, −*v*5, +*v*4, −*v*3, and −*v*2, which is the same as before e xcept that the signs are re versed. Hence, Eqs. (2.20) and (2.21) remain the same. - -When voltage sources are connected in series, KVL can be applied to obtain the total v oltage. The combined v oltage is the algebraic sum of the v oltages of the indi vidual sources. F or example, for the v oltage sources shown in Fig. 2.20(a), the combined or equivalent voltage source in Fig. 2.20(b) is obtained by applying KVL. - -$$ --V_{ab} + V_1 + V_2 - V_3 = 0 -$$ - -# **Figure 2.18** Current sources in parallel: (a) original circuit, (b) equivalent circuit. - -KVL can be applied in two ways: by taking either a clockwise or a counterclockwise trip around the loop. Either way, the algebraic sum of voltages around the loop is zero. - -**Figure 2.19** A single-loop circuit illustrating KVL. - -$$ -V_{ab} = V_1 + V_2 - V_3 \tag{2.23} -$$ - -To avoid violating KVL, a circuit cannot contain tw o different voltages *V*1 and *V*2 in parallel unless *V*1 =*V*2. - -# **Figure 2.20** - -Voltage sources in series: (a) original circuit, (b) equivalent circuit. - -For the circuit in Fig. 2.21(a), find voltages *v*1 and *v*2 Example 2.5 . - -# **Solution:** - -To find *v*1 and *v*2 we apply Ohm's law and Kirchhoff's voltage law. Assume that current *i* flows through the loop as shown in Fig. 2.21(b). From Ohm's law, - -$$ -v_1 = 2i, \qquad v_2 = -3i \tag{2.5.1} -$$ - -Applying KVL around the loop gives - -$$ --20 + v_1 - v_2 = 0 \tag{2.5.2} -$$ - -Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain - -$$ --20 + 2i + 3i = 0 \qquad \text{or} \qquad 5i = 20 \qquad \Rightarrow \qquad i = 4 \text{ A} -$$ - -Substituting *i* in Eq. (2.5.1) finally gives - -$$ -v_1 = 8 \text{ V}, \qquad v_2 = -12 \text{ V} -$$ - -**Answer:** 16 V, −8 V. - -Determine *v* Example 2.6 *o* and *i* in the circuit shown in Fig. 2.23(a). - -# **Figure 2.23** - -For Example 2.6. - -# **Solution:** - -We apply KVL around the loop as shown in Fig. 2.23(b). The result is - -$$ --12 + 4i + 2v_o - 4 + 6i = 0 \tag{2.6.1} -$$ - -Applying Ohm's law to the 6-Ω resistor gives - -$$ -v_o = -6i \tag{2.6.2} -$$ - -Substituting Eq. (2.6.2) into Eq. (2.6.1) yields - -$$ --16 + 10i - 12i = 0 \qquad \Rightarrow \qquad i = -8 \text{ A} -$$ - -and *vo* = 48 V. - -**Answer:** 20 V, −10 V. - -For Example 2.7. - -Find *vo* and *io* in the circuit of Fig. 2.26. - -Practice Problem 2.7 - -Example 2.8 Find currents and voltages in the circuit shown in Fig. 2.27(a). - -For Example 2.8. - -# **Solution:** - -We apply Ohm's law and Kirchhoff's laws. By Ohm's law, - -$$ -v_1 = 8i_1, \qquad v_2 = 3i_2, \qquad v_3 = 6i_3 \tag{2.8.1} -$$ - -Since the voltage and current of each resistor are related by Ohm's law as shown, we are really looking for three things: (*v*1, *v*2, *v*3) or (*i*1, *i*2, *i*3). At node *a*, KCL gives - -$$ -i_1 - i_2 - i_3 = 0 \tag{2.8.2} -$$ - -Applying KVL to loop 1 as in Fig. 2.27(b), - -$$ --30 + v_1 + v_2 = 0 -$$ - -We express this in terms of *i*1 and *i*2 as in Eq. (2.8.1) to obtain - -$$ --30 + 8i_1 + 3i_2 = 0 -$$ - -or - -$$ -i_1 = \frac{(30 - 3i_2)}{8} \tag{2.8.3} -$$ - -Applying KVL to loop 2, - -−*v*2 + *v*3 = 0 ⇒ *v*3 = *v*2 **(2.8.4)** - -as expected since the two resistors are in parallel. We express *v*1 and *v*2 in terms of *i*1 and *i*2 as in Eq. (2.8.1). Equation (2.8.4) becomes - -$$ -6i_3 = 3i_2 -$$ - $\Rightarrow$ $i_3 = \frac{i_2}{2}$ (2.8.5) - -Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives - -$$ -\frac{30 - 3i_2}{8} - i_2 - \frac{i_2}{2} = 0 -$$ - -or *i*2 = 2 A. From the v alue of *i*2, we now use Eqs. (2.8.1) to (2.8.5) to obtain - -$$ -i_1 = 3 -$$ - A, $i_3 = 1$ A, $v_1 = 24$ V, $v_2 = 6$ V, $v_3 = 6$ V - -Find the currents and voltages in the circuit shown in Fig. 2.28. Practice Problem 2.8 - -**Answer:** -$$ -v_1 = 6 -$$ - V, $v_2 = 4$ V, $v_3 = 10$ V, $i_1 = 3$ A, $i_2 = 500$ mA, $i_3 = 2.5$ A. - -# **2.5** Series Resistors and Voltage Division - -The need to combine resistors in series or in parallel occurs so frequently that it warrants special attention. The process of combining the resistors is facilitated by combining two of them at a time. With this in mind, consider the single-loop circuit of Fig. 2.29. The two resistors are in series, since the same current *i* flows in both of them. Applying Ohm's law to each of the resistors, we obtain - -$$ -v_1 = iR_1, \qquad v_2 = iR_2 \tag{2.24} -$$ - -If we apply KVL to the loop (mo ving in the clockwise direction), we have - -$$ --v + v_1 + v_2 = 0 \tag{2.25} -$$ - -Combining Eqs. (2.24) and (2.25), we get - -$$ -v = v_1 + v_2 = i(R_1 + R_2) -$$ - (2.26) - -*v* + – R1 *v*1 R2 *v*2 i + – + – a b - -# **Figure 2.29** A single-loop circuit with two resistors in series. - -or - -$$ -i = \frac{v}{R_1 + R_2} \tag{2.27} -$$ - -**Figure 2.28** For Practice Prob. 2.8. - -Notice that Eq. (2.26) can be written as - -$$ -v = iR_{\text{eq}} \tag{2.28} -$$ - -implying that the two resistors can be replaced by an equivalent resistor *R*eq; that is, - -$$ -R_{\text{eq}} = R_1 + R_2 \tag{2.29} -$$ - -Thus, Fig. 2.29 can be replaced by the equivalent circuit in Fig. 2.30. The two circuits in Figs. 2.29 and 2.30 are equivalent because they exhibit the same voltage-current relationships at the terminals *a*-*b*. An equivalent circuit such as the one in Fig. 2.30 is useful in simplifying the analysis of a circuit. In general, - -The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances. - -For *N* resistors in series then, - -$$ -R_{\text{eq}} = R_1 + R_2 + \dots + R_N = \sum_{n=1}^{N} R_n \tag{2.30} -$$ - -To determine the voltage across each resistor in Fig. 2.29, we substitute Eq. (2.26) into Eq. (2.24) and obtain - -$$ -v_1 = \frac{R_1}{R_1 + R_2} v, \qquad v_2 = \frac{R_2}{R_1 + R_2} v -$$ - (2.31) - -Notice that the source v oltage *v* is divided among the resistors in direct proportion to their resistances; the larger the resistance, the larger the voltage drop. This is called the *principle of voltage division*, and the circuit in Fig. 2.29 is called a *voltage divider*. In general, if a voltage divider has *N* resistors (*R*1, *R*2, … , *RN*) in series with the source voltage *v*, the *n*th resistor (*Rn*) will have a voltage drop of - -... , -$$ -R_N -$$ -) in series with the source voltage *v*, the *n*th -a voltage drop of -$$ -v_n = \frac{R_n}{R_1 + R_2 + \dots + R_N} v -$$ -(2.32) - -# **2.6** Parallel Resistors and Current Division - -Consider the circuit in Fig. 2.31, where tw o resistors are connected in parallel and therefore ha ve the same v oltage across them. From Ohm's law, - -$$ -v = i_1 R_1 = i_2 R_2 -$$ - -$$ -i_1 = \frac{v}{R_1} -$$ -, $i_2 = \frac{v}{R_2}$ (2.33) - -Applying KCL at node *a* gives the total current *i* as - -$$ -i = i_1 + i_2 \tag{2.34} -$$ - -Substituting Eq. (2.33) into Eq. (2.34), we get - -or - -$$ -i = \frac{v}{R_1} + \frac{v}{R_2} = v\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{v}{R_{\text{eq}}} -$$ - (2.35) - -**Figure 2.31** Two resistors in parallel. - -Equivalent circuit of the Fig. 2.29 circuit. - - Resistors in series behave as a single resistor whose resistance is equal to the sum of the resistances of the - -individual resistors. - -**Figure 2.30** - -where *R*eq is the equivalent resistance of the resistors in parallel: - -$$ -\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \tag{2.36} -$$ - -or - -$$ -\frac{1}{R_{\text{eq}}} = \frac{R_1 + R_2}{R_1 R_2} -$$ - -or - -$$ -R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} \tag{2.37} -$$ - -Thus, - -The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum. - -It must be emphasized that this applies only to tw o resistors in parallel. From Eq. (2.37), if *R*1 = *R*2 then *R*eq = *R*1/*R*2. - -We can extend the result in Eq. (2.36) to the general case of a circuit with *N* resistors in parallel. The equivalent resistance is - -$$ -\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_N} -$$ - (2.38) - -Note that *R*eq is always smaller than the resistance of the smallest resistor in the parallel combination. If *R*1 = *R*2 = ⋯=*RN* = *R*, then - -$$ -R_{\text{eq}} = \frac{R}{N} \tag{2.39} -$$ - -For e xample, if four 100- Ω resistors are connected in parallel, their equivalent resistance is 25 Ω. - -It is often more convenient to use conductance rather than resistance when dealing with resistors in parallel. From Eq. (2.38), the equi valent conductance for *N* resistors in parallel is - -$$ -G_{\text{eq}} = G_1 + G_2 + G_3 + \dots + G_N \quad (2.40) -$$ - -where *G*eq = 1/*R*eq, *G*1 = 1/*R*1, *G*2 = 1/*R*2, *G*3 = 1/ *R*3, … , *GN* = 1/ *RN*. Equation (2.40) states: - -The equivalent conductance of resistors connected in parallel is the sum of their individual conductances. - -This means that we may replace the circuit in Fig. 2.31 with that in Fig. 2.32. Notice the similarity between Eqs. (2.30) and (2.40). The equivalent conductance of parallel resistors is obtained the same w ay as the equivalent resistance of series resistors. In the same manner , the equivalent conductance of resistors in series is obtained just the same Conductances in parallel behave as a single conductance whose value is equal to the sum of the individual conductances. - -**Figure 2.32** Equivalent circuit to Fig. 2.31. - -way as the resistance of resistors in parallel. Thus, the equivalent conductance *G*eq of *N* resistors in series (such as shown in Fig. 2.29) is - -$$ -\frac{1}{G_{\text{eq}}} = \frac{1}{G_1} + \frac{1}{G_2} + \frac{1}{G_3} + \dots + \frac{1}{G_N} -$$ - (2.41) - -Given the total current *i* entering node *a* in Fig. 2.31, ho w do we obtain current *i*1 and *i*2? We know that the equi valent resistor has the same voltage, or - -$$ -v = iR_{\text{eq}} = \frac{iR_1R_2}{R_1 + R_2} \tag{2.42} -$$ - -Combining Eqs. (2.33) and (2.42) results in - -$$ -i_1 = \frac{R_2 i}{R_1 + R_2}, \qquad i_2 = \frac{R_1 i}{R_1 + R_2} -$$ - (2.43) - -which shows that the total current *i* is shared by the resistors in in verse proportion to their resistances. This is known as the *principle of current division*, and the circuit in Fig. 2.31 is known as a *current divider*. Notice that the lar ger current flows through the smaller re sistance. - -As an extreme case, suppose one of the resistors in Fig. 2.31 is zero, say *R*2 = 0; that is, *R*2 is a short circuit, as sho wn in Fig. 2.33(a). From Eq. (2.43), *R*2 = 0 implies that *i*1 = 0, *i*2 = *i*. This means that the entire current *i* bypasses *R*1 and flows through the short circuit *R*2 = 0, the path of least resistance. Thus when a circuit is short circuited, as sho wn in Fig. 2.33(a), two things should be kept in mind: - -- 1. The equivalent resistance *R*eq = 0. [See what happens when *R*2 = 0 in Eq. (2.37).] -- 2. The entire current flows through the short circuit. - -As another e xtreme case, suppose *R*2 = ∞, that is, *R*2 is an open circuit, as shown in Fig. 2.33(b). The current still flows through the path of least resistance, *R*1. By taking the limit of Eq. (2.37) as *R*2 → ∞,we obtain *R*eq = *R*1 in this case. - -If we divide both the numerator and denominator by *R*1*R*2, Eq. (2.43) becomes - -$$ -i_1 = \frac{G_1}{G_1 + G_2} i \tag{2.44a} -$$ - -$$ -i_2 = \frac{G_2}{G_1 + G_2} i \tag{2.44b} -$$ - -Thus, in general, if a current divider has *N* conductors (*G*1, *G*2, … , *GN*) in parallel with the source current *i*, the *n*th conductor (*Gn*) will have current - -e current *i*, the *n*th conductor ( -$$ -G_n -$$ -) will have current -\n -$$ -i_n = \frac{G_n}{G_1 + G_2 + \dots + G_N} i -$$ -\n(2.45) - -**Figure 2.33** (a) A shorted circuit, (b) an open circuit. - -In general, it is often convenient and possible to combine resistors in series and parallel and reduce a resisti ve network to a single *equivalent resistance R*eq. Such an equi valent resistance is the resistance between the designated terminals of the netw ork and must e xhibit the same *i*-*v* characteristics as the original network at the terminals. - -Find *R*eq for the circuit shown in Fig. 2.34. Example 2.9 - -# **Solution:** - -To get *R*eq, we combine resistors in series and in parallel. The 6- Ω and 3-Ω resistors are in parallel, so their equivalent resistance is - -$$ -6 \Omega \parallel 3\Omega = \frac{6 \times 3}{6 + 3} = 2 \Omega -$$ - -(The symbol ∥ is used to indicate a parallel combination.) Also, the 1-Ω and 5-Ω resistors are in series; hence their equivalent resistance is - -1 Ω + 5 Ω = 6 Ω - -Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig. 2.35(a), we notice that the tw o 2-Ω resistors are in series, so the equivalent resistance is - -$$ -2\Omega + 2\Omega = 4\Omega -$$ - -This 4-Ω resistor is now in parallel with the 6-Ω resistor in Fig. 2.35(a); their equivalent resistance is - -$$ -4 \Omega || 6 \Omega = \frac{4 \times 6}{4 + 6} = 2.4 \Omega -$$ - -The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig. 2.35(b), the three resistors are in series. Hence, the equivalent resistance for the circuit is - -$$ -R_{\text{eq}} = 4 \ \Omega + 2.4 \ \Omega + 8 \ \Omega = 14.4 \ \Omega -$$ - -By combining the resistors in Fig. 2.36, find *R*eq. Practice Problem 2.9 - -**Answer:** 11 Ω. - -For Practice Prob. 2.9. - -Calculate the equivalent resistance *Rab* Example 2.10 in the circuit in Fig. 2.37. - -# For Example 2.10. - -# **Solution:** - -The 3-Ω and 6-Ω resistors are in parallel because they are connected to the same two nodes *c* and *b*. Their combined resistance is - -$$ -3 \Omega \parallel 6 \Omega = \frac{3 \times 6}{3 + 6} = 2 \Omega -$$ - (2.10.1) - -Similarly, the 12-Ω and 4-Ω resistors are in parallel since the y are connected to the same two nodes *d* and *b*. Hence - -$$ -12 \Omega || 4 \Omega = \frac{12 \times 4}{12 + 4} = 3 \Omega -$$ - (2.10.2) - -Also the 1-Ω and 5-Ω resistors are in series; hence, their equivalent re sistance is - -$$ -1 \Omega + 5 \Omega = 6 \Omega \tag{2.10.3} -$$ - -With these three combinations, we can replace the circuit in Fig. 2.37 with that in Fig. 2.38(a). In Fig. 2.38(a), 3-Ω in parallel with 6-Ω gives 2-Ω, as calculated in Eq. (2.10.1). This 2-Ω equivalent resistance is now in series with the 1-Ω resistance to give a combined resistance of 1 Ω +2Ω=3Ω. Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In Fig. 2.38(b), we combine the 2-Ω and 3-Ω resistors in parallel to get - -$$ -2 \Omega || 3 \Omega = \frac{2 \times 3}{2 + 3} = 1.2 \Omega -$$ - -This 1.2-Ω resistor is in series with the 10-Ω resistor, so that - -$$ -R_{ab} = 10 + 1.2 = 11.2 \ \Omega -$$ - -Find *Rab* for the circuit in Fig. 2.39. - -**Answer:** 19 Ω. - -For Practice Prob. 2.10. - -Find the equivalent conductance *G*eq for the circuit in Fig. 2.40(a). - -# **Solution:** - -The 8-S and 12-S resistors are in parallel, so their conductance is - -$$ -8 S + 12 S = 20 S -$$ - -This 20-S resistor is now in series with 5 S as shown in Fig. 2.40(b) so that the combined conductance is - -$$ -\frac{20 \times 5}{20 + 5} = 4 -$$ - -This is in parallel with the 6-S resistor. Hence, - -$$ -G_{\text{eq}} = 6 + 4 = 10 \text{ S} -$$ - -We should note that the circuit in Fig. 2.40(a) is the same as that in Fig. 2.40(c). While the resistors in Fig. 2.40(a) are expressed in siemens, those in Fig. 2.40(c) are expressed in ohms. To show that the circuits are the same, we find *R*eq for the circuit in Fig. 2.40(c). - -$$ -R_{\text{eq}} = \frac{1}{6} \left\| \left( \frac{1}{5} + \frac{1}{8} \right) \frac{1}{12} \right\| = \frac{1}{6} \left\| \left( \frac{1}{5} + \frac{1}{20} \right) \right\| = \frac{1}{6} \left\| \frac{1}{4} \right\| -$$ -$$ -= \frac{\frac{1}{6} \times \frac{1}{4}}{\frac{1}{6} + \frac{1}{4}} = \frac{1}{10} \Omega -$$ -$$ -G_{\text{eq}} = \frac{1}{R_{\text{eq}}} = 10 \text{ S} -$$ - -This is the same as we obtained previously. - -Calculate *G*eq in the circuit of Fig. 2.41. Practice Problem 2.11 - -# **Answer:** 8 S. - -For Practice Prob. 2.11. - -Find *io* and *vo* in the circuit shown in Fig. 2.42(a). Calculate the power dissipated in the 3-Ω resistor. - -# **Solution:** - -The 6-Ω and 3-Ω resistors are in parallel, so their combined resistance is - -$$ -6 \Omega \parallel 3 \Omega = \frac{6 \times 3}{6 + 3} = 2 \Omega -$$ - -Thus, our circuit reduces to that shown in Fig. 2.42(b). Notice that *vo* is not affected by the combination of the resistors because the resistors are - -# Example 2.11 - -# **Figure 2.40** - -For Example 2.11: (a) original circuit, (b) its equivalent circuit, (c) same circuit as in (a) but resistors are expressed in ohms. - -Example 2.12 - -For Example 2.12: (a) original circuit, (b) its equivalent circuit. - -in parallel and therefore have the same voltage *vo*. From Fig. 2.42(b), we can obtain *vo* in two ways. One way is to apply Ohm's law to get - -$$ -i = \frac{12}{4+2} = 2 \text{ A} -$$ - -and hence, *vo* = 2*i* = 2 × 2 = 4 V. Another way is to apply voltage division, since the 12 V in Fig. 2.42(b) is divided between the 4-Ω and 2-Ω resistors. Hence, - -$$ -v_o = \frac{2}{2+4} (12 \text{ V}) = 4 \text{ V} -$$ - -Similarly, *io* can be obtained in two ways. One approach is to apply Ohm's law to the 3-Ω resistor in Fig. 2.42(a) now that we know *vo*; thus, - -$$ -v_o = 3i_o = 4 \qquad \Rightarrow \qquad i_o = \frac{4}{3} \text{ A} -$$ - -Another approach is to apply current division to the circuit in Fig. 2.42(a) now that we know *i*, by writing - -$$ -i_o = \frac{6}{6+3} i = \frac{2}{3} (2 \text{ A}) = \frac{4}{3} \text{ A} -$$ - -The power dissipated in the 3-Ω resistor is - -$$ -p_o = v_o i_o = 4 \left(\frac{4}{3}\right) = 5.333 \text{ W} -$$ - -Find *v*1 and *v*2 in the circuit sho wn in Fig. 2.43. Also calculate *i*1 and *i*2 and the power dissipated in the 12-Ω and 40-Ω resistors. - -**Answer:** *v*1 = 10 V, *i*1 = 833.3 mA, *p*1 = 8.333 W, *v*2 = 20 V, *i*2 =500 mA, *p*2 = 10 W. - -Practice Problem 2.12 - -# For Practice Prob. 2.12. - -**Figure 2.43** - -# Example 2.13 - -For the circuit sho wn in Fig. 2.44(a), determine: (a) the v oltage *vo*, (b) the power supplied by the current source, (c) the power absorbed by each resistor. - -# **Solution:** - -(a) The 6-k Ω and 12-k Ω resistors are in series so that their combined value is 6 + 12 = 18 kΩ. Thus the circuit in Fig. 2.44(a) reduces to that shown in Fig. 2.44(b). We now apply the current division technique to find *i*1 and *i*2. - -$$ -i_1 = \frac{18,000}{9,000 + 18,000} (30 \text{ mA}) = 20 \text{ mA} -$$ -$$ -i_2 = \frac{9,000}{9,000 + 18,000} (30 \text{ mA}) = 10 \text{ mA} -$$ - -Notice that the voltage across the 9-kΩ and 18-kΩ resistors is the same, and *vo* = 9,000*i*1 = 18,000*i*2 = 180 V, as expected. (b) Power supplied by the source is - -$$ -p_o = v_o i_o = 180(30) \text{ mW} = 5.4 \text{ W} -$$ - -(c) Power absorbed by the 12-kΩ resistor is - -$$ -p = iv = i_2(i_2 R) = i_2^2 R = (10 \times 10^{-3})^2 (12,000) = 1.2 W -$$ - -Power absorbed by the 6-kΩ resistor is - -$$ -p = i_2^2 R = (10 \times 10^{-3})^2 (6,000) = 0.6 W -$$ - -Power absorbed by the 9-kΩ resistor is - -$$ -p = \frac{v_o^2}{R} = \frac{(180)^2}{9,000} = 3.6 \text{ W} -$$ - -or - -$$ -p = v_o i_1 = 180(20) \text{ mW} = 3.6 \text{ W} -$$ - -Notice that the power supplied (5.4 W) equals the power absorbed (1.2 + 0.6 + 3.6 = 5.4 W). This is one way of checking results. - -For the circuit shown in Fig. 2.45, find: (a) *v*1 and *v*2, (b) the power dis- Practice Problem 2.13 sipated in the 3-k Ω and 20-kΩ resistors, and (c) the power supplied by the current source. - -**Answer:** (a) 135 V, 180 V, (b) 2.025 W, 540 mW, (c) 5.4 W. - -# **2.7** Wye-Delta Transformations - -Situations often arise in circuit analysis when the resistors are neither in parallel nor in series. For example, consider the bridge circuit in Fig. 2.46. How do we combine resistors *R*1 through *R*6 when the resistors are neither in series nor in parallel? Man y circuits of the type sho wn in Fig. 2.46 can be simplified by using three-terminal equivalent networks. These are the wye (Y) or tee (T) netw ork shown in Fig. 2.47 and the delta (Δ) or pi (Π) network shown in Fig. 2.48. These networks occur by themselves or as part of a lar ger network. They are used in three-phase networks, electrical filters, and matching networks. Our main interest - -**Figure 2.46** The bridge network. - -Two forms of the same network: (a) Y, (b) T. - -here is in how to identify them when they occur as part of a network and how to apply wye-delta transformation in the analysis of that network. - -# Delta to Wye Conversion - -Suppose it is more convenient to work with a wye network in a place where the circuit contains a delta configuration. We superimpose a wye network on the existing delta network and find the equivalent resistances in the wye network. To obtain the equivalent resistances in the wye network, we compare the tw o networks and mak e sure that the resistance between each pair of nodes in the Δ (or Π) network is the same as the resistance between the same pair of nodes in the Y (or T) network. For terminals 1 and 2 in Figs. 2.47 and 2.48, for example, - -$$ -R_{12}(Y) = R_1 + R_3 \tag{2.46} -$$ - -$$ -R_{12}(\Delta) = R_b || (R_a + R_c) -$$ - -# **Figure 2.48** - -Two forms of the same network: (a) Δ, (b) Π. - -Setting -$$ -R_{12}(Y) = R_{12}(\Delta) -$$ - gives - -$$ -R_{12} = R_1 + R_3 = \frac{R_b (R_a + R_c)}{R_a + R_b + R_c} -$$ - (2.47a) - -Similarly, - -$$ -R_{13} = R_1 + R_2 = \frac{R_c (R_a + R_b)}{R_a + R_b + R_c} -$$ - (2.47b) - -$$ -R_{34} = R_2 + R_3 = \frac{R_a (R_b + R_c)}{R_a + R_b + R_c} -$$ - (2.47c) - -Subtracting Eq. (2.47c) from Eq. (2.47a), we get - -$$ -R_1 - R_2 = \frac{R_c (R_b - R_a)}{R_a + R_b + R_c} -$$ - (2.48) - -Adding Eqs. (2.47b) and (2.48) gives - -$$ -R_1 = \frac{R_b R_c}{R_a + R_b + R_c} \tag{2.49} -$$ - -and subtracting Eq. (2.48) from Eq. (2.47b) yields - -$$ -R_2 = \frac{R_c R_a}{R_a + R_b + R_c} \tag{2.50} -$$ - -Subtracting Eq. (2.49) from Eq. (2.47a), we obtain - -$$ -R_3 = \frac{R_a R_b}{R_a + R_b + R_c} -$$ - (2.51) - -We do not need to memorize Eqs. (2.49) to (2.51). To transform a ∆ network to Y, we create an extra node *n* as shown in Fig. 2.49 and follow this conversion rule: - -Each resistor in the Y network is the product of the resistors in the two adjacent Δ branches, divided by the sum of the three Δ resistors. - -One can follow this rule and obtain Eqs. (2.49) to (2.51) from Fig. 2.49. - -# Wye to Delta Conversion - -To obtain the conversion formulas for transforming a wye network to an equivalent delta network, we note from Eqs. (2.49) to (2.51) that - -the conversion formulas for transforming a wye network to an -it delta network, we note from Eqs. (2.49) to (2.51) that -$$ -R_1 R_2 + R_2 R_3 + R_3 R_1 = \frac{R_a R_b R_c (R_a + R_b + R_c)}{(R_a + R_b + R_c)^2} -$$ -$$ -= \frac{R_a R_b R_c}{R_a + R_b + R_c} -$$ -(2.52) - -Dividing Eq. (2.52) by each of Eqs. (2.49) to (2.51) leads to the follo wing equations: - -$$ -R_a = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1} -$$ - (2.53) - -$$ -R_1 -$$ -\n -$$ -R_b = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2} -$$ -\n(2.54) - -$$ -R_b = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3} -$$ -\n(2.55) - -From Eqs. (2.53) to (2.55) and Fig. 2.49, the con version rule for Y to Δ is as follows: - -Each resistor in the Δ network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor. - -The Y and Δ networks are said to be *balanced* when - -$$ -R_1 = R_2 = R_3 = R_Y, \qquad R_a = R_b = R_c = R_\Delta \tag{2.56} -$$ - -Under these conditions, conversion formulas become - -$$ -R_{\rm Y} = \frac{R_{\rm \Delta}}{3} \qquad \text{or} \qquad R_{\rm \Delta} = 3R_{\rm Y} \tag{2.57} -$$ - -One may w onder wh y *R*Y is less than *R*Δ. Well, we notice that the Y-connection is like a "series" connection while the Δ-connection is like a "parallel" connection. - -Note that in making the transformation, we do not take anything out of the circuit or put in anything new. We are merely substituting different but mathematically equivalent three-terminal network patterns to create a circuit in which resistors are either in series or in parallel, allo wing us to calculate *R*eq if necessary. - -# **Figure 2.49** Superposition of Y and Δ networks as an aid in transforming one to the other. - -Example 2.14 Convert the Δ network in Fig. 2.50(a) to an equivalent Y network. - -**Figure 2.50** For Example 2.14: (a) original Δ network, (b) Y equivalent network. - -# **Solution:** - -Using Eqs. (2.49) to (2.51), we obtain - -$$ -R_1 = \frac{R_b R_c}{R_a + R_b + R_c} = \frac{10 \times 25}{15 + 10 + 25} = \frac{250}{50} = 5 \text{ }\Omega -$$ -\n -$$ -R_2 = \frac{R_c R_a}{R_a + R_b + R_c} = \frac{25 \times 15}{50} = 7.5 \text{ }\Omega -$$ -\n -$$ -R_3 = \frac{R_a R_b}{R_a + R_b + R_c} = \frac{15 \times 10}{50} = 3 \text{ }\Omega -$$ - -The equivalent Y network is shown in Fig. 2.50(b). - -Practice Problem 2.14 Transform the wye network in Fig. 2.51 to a delta network. - -**Answer:** *Ra* = 140 Ω, *Rb* = 70 Ω, *Rc* = 35 Ω. - -**Figure 2.51** For Practice Prob. 2.14. - -Example 2.15 - -Obtain the equivalent resistance *Rab* for the circuit in Fig. 2.52 and use it to find current *i*. - -# **Solution:** - -- 1. **Define.** The problem is clearly defined. Please note, this part normally will deservedly take much more time. -- 2. **Present.** Clearly, when we remove the voltage source, we end up with a purely resistive circuit. Since it is composed of deltas and wyes, we have a more comple x process of combining the elements together . - -We can use wye-delta transformations as one approach to find a solution. It is useful to locate the wyes (there are tw o of them, one at *n* and the other at *c*) and the deltas (there are three: *can, abn, cnb*). - -3. **Alternative.** There are different approaches that can be used to solve this problem. Since the focus of Sec. 2.7 is the wye-delta transfor mation, this should be the technique to use. Another approach would be to solve for the equi valent resistance by injecting one amp into the circuit and finding the voltage between *a* and *b*; we will learn about this approach in Chap. 4. - -The approach we can apply here as a check w ould be to use a wye-delta transformation as the first solution to the problem. Later we can check the solution by starting with a delta-wye transformation. - -4. **Attempt.** In this circuit, there are two Y networks and three Δ networks. Transforming just one of these will simplify the circuit. If we convert the Y network comprising the 5-Ω, 10-Ω, and 20-Ω resistors, we may select - -$$ -R_1 = 10 \Omega, \qquad R_2 = 20 \Omega, \qquad R_3 = 5 \Omega -$$ - -Thus from Eqs. (2.53) to (2.55) we have - -$$ -R_1 = 16 \text{ cm}, \quad R_2 = 26 \text{ cm}, \quad R_3 = 5 \text{ cm} -$$ -\n -$$ -R_a = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1} = \frac{10 \times 20 + 20 \times 5 + 5 \times 10}{10} -$$ -\n -$$ -= \frac{350}{10} = 35 \text{ }\Omega -$$ -\n -$$ -R_b = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2} = \frac{350}{20} = 17.5 \text{ }\Omega -$$ -\n -$$ -R_c = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3} = \frac{350}{5} = 70 \text{ }\Omega -$$ - -With the Y converted to Δ, the equivalent circuit (with the voltage source removed for now) is shown in Fig. 2.53(a). Combining the three pairs of resistors in parallel, we obtain - -**Figure 2.53** Equivalent circuits to Fig. 2.52, with the voltage source removed. - -so that the equivalent circuit is shown in Fig. 2.53(b). Hence, we find - -$$ -R_{ab} = (7.292 + 10.5) \| 21 = \frac{17.792 \times 21}{17.792 + 21} = 9.632 \Omega -$$ - -Then - -$$ -i = \frac{v_s}{R_{ab}} = \frac{120}{9.632} = 12.458 \text{ A} -$$ - -We observe that we have successfully solved the problem. Now we must evaluate the solution. - -5. **Evaluate.** Now we must determine if the answer is correct and then evaluate the final solution. - -It is relatively easy to check the answer; we do this by solving the problem starting with a delta-wye transformation. Let us trans form the delta, *can*, into a wye. - -Let *Rc* = 10 Ω, *Ra* = 5 Ω, and *Rn* = 12.5 Ω. This will lead to (let *d* represent the middle of the wye): - -$$ -R_{ad} = \frac{R_c R_n}{R_a + R_c + R_n} = \frac{10 \times 12.5}{5 + 10 + 12.5} = 4.545 \ \Omega -$$ -$$ -R_{cd} = \frac{R_a R_n}{27.5} = \frac{5 \times 12.5}{27.5} = 2.273 \ \Omega -$$ -$$ -R_{nd} = \frac{R_a R_c}{27.5} = \frac{5 \times 10}{27.5} = 1.8182 \ \Omega -$$ - -This now leads to the circuit sho wn in Figure 2.53(c). Looking at the resistance between *d* and *b*, we have two series combination in parallel, giving us - -parallel, giving us -\n -$$ -R_{db} = \frac{(2.273 + 15)(1.8182 + 20)}{2.273 + 15 + 1.8182 + 20} = \frac{376.9}{39.09} = 9.642 \ \Omega -$$ - -This is in series with the 4.545-Ω resistor, both of which are in parallel with the 30-Ω resistor. This then gives us the equivalent resistance of the circuit. - -tance of the circuit. -\n -$$ -R_{ab} = \frac{(9.642 + 4.545)30}{9.642 + 4.545 + 30} = \frac{425.6}{44.19} = 9.631 \ \Omega -$$ - -This now leads to - -$$ -i = \frac{v_s}{R_{ab}} = \frac{120}{9.631} = 12.46 \text{ A} -$$ - -We note that using tw o variations on the wye-delta transformation leads to the same results. This represents a very good check. - -6. **Satisfactory?** Since we have found the desired answer by deter mining the equi valent resistance of the circuit first and the answer checks, then we clearly ha ve a satisfactory solution. This represents what can be presented to the indi vidual assigning the problem. - -For the bridge network in Fig. 2.54, find *Rab* and *i*. - -**Answer:** 60 Ω, 4 A. - -# **2.8** Applications - -Resistors are often used to model de vices that con vert electrical ener gy into heat or other forms of energy. Such devices include conducting wire, light bulbs, electric heaters, stoves, ovens, and loudspeakers. In this section, we will consider two real-life problems that apply the concepts developed in this chapter: electrical lighting systems and design of dc meters. - -# 2.8.1 Lighting Systems - -Lighting systems, such as in a house or on a Christmas tree, often consist of *N* lamps connected either in parallel or in series, as sho wn in Fig. 2.55. Each lamp is modeled as a resistor . Assuming that all the lamps are identical and *Vo* is the power-line voltage, the voltage across each lamp is *Vo* for the parallel connection and *Vo* /*N* for the series connection. The series connection is easy to manufa cture but is seldom used in practice, for at least two reasons. First, it is less reliable; when a lamp fa ils, all the lamps go out. Second, it is harder to maintain; when a lamp is bad, one must test all the lamps one by one to detect the faulty one. - -**Figure 2.54** For Practice Prob. 2.15. - -So far, we have assumed that connecting wires are perfect conductors (i.e., conductors of zero resistance). In real physical systems, however, the resistance of the connecting wire may be appreciably large, and the modeling of the system must include that resistance. - -# Historical - -**Thomas Alva Edison** (1847–1931) was perhaps the greatest American inventor. He patented 1093 inventions, including such history-making inventions as the incandescent electric bulb, the phonograph, and the first commercial motion pictures. - -Born in Milan, Ohio, the youngest of seven children, Edison received only three months of formal education because he hated school. He was home-schooled by his mother and quickly began to read on his own. In 1868, Edison read one of Faraday's books and found his calling. He moved to Menlo Park, New Jersey, in 1876, where he man aged a well-staffed research laboratory. Most of his inventions came out of this laboratory. His laboratory served as a model for modern research organ iza tions. Because of his diverse interests and the over whelming number of his inventions and patents, Edison began to estab lish manufacturing companies for making the devices he invented. He designed the first electric power station to supply electric light. Formal electrical engineering education began in the mid-1880s with Edison as a role model and leader. - -Library of Congress - -(a) Parallel connection of light bulbs, (b) series connection of light bulbs. - -Example 2.16 - -Three light bulbs are connected to a 9-V battery as shown in Fig. 2.56(a). Calculate: (a) the total current supplied by the battery , (b) the current through each bulb, (c) the resistance of each bulb. - -# **Solution:** - -(a) The total po wer supplied by the battery is equal to the total po wer absorbed by the bulbs; that is, - -$$ -p = 15 + 10 + 20 = 45 -$$ - W - -Since *p* = *V I,* then the total current supplied by the battery is - -$$ -I = \frac{p}{V} = \frac{45}{9} = 5 \text{ A} -$$ - -(b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since *R*1 (20-W bulb) is in parallel with the battery as well as the series com bination of *R*2 and *R*3, - -$$ -V_1 = V_2 + V_3 = 9 \text{ V} -$$ - -The current through *R*1 is - -$$ -I_1 = \frac{p_1}{V_1} = \frac{20}{9} = 2.222 \text{ A} -$$ - -By KCL, the current through the series combination of *R*2 and *R*3 is - -$$ -I_2 = I - I_1 = 5 - 2.222 = 2.778 \text{ A} -$$ - -(c) Since *p* = *I* 2 *R*, - -$$ -R_1 = \frac{p_1}{l_1^2} = \frac{20}{2.222^2} = 4.05 \ \Omega -$$ - -\n -$$ -R_2 = \frac{p_2}{l_2^2} = \frac{15}{2.777^2} = 1.945 \ \Omega -$$ - -\n -$$ -R_3 = \frac{p_3}{l_3^2} = \frac{10}{2.777^2} = 1.297 \ \Omega -$$ - -Refer to Fig. 2.55 and assume there are six light b ulbs that can be con nected in parallel and six dif ferent light bulbs that can be connected in series. In either case, each light bulb is to operate at 40 W. If the voltage at the plug is 115 V for the parallel and series connections, calculate the current through and the voltage across each bulb for both cases. - -**Answer:** 115 V and 347.8 mA (parallel), 19.167 V and 2.087 A (series). - -# 2.8.2 Design of DC Meters - -By their nature, resistors are used to control the flow of current. We take advantage of this property in se veral applications, such as in a poten tiometer (Fig. 2.57). The word *potentiometer*, derived from the w ords *potential* and *meter*, implies that potential can be metered out. The potentiometer (or pot for short) is a three-terminal de vice that operates on the principle of v oltage division. It is essentially an adjustable v oltage divider. As a voltage regulator, it is used as a volume or level control on radios, TVs, and other devices. In Fig. 2.57, - -$$ -V_{\text{out}} = V_{bc} = \frac{R_{bc}}{R_{ac}} V_{\text{in}} -$$ - (2.58) - -where *Rac* = *Rab* + *Rbc*. Thus, *V*out decreases or increases as the sliding contact of the pot moves toward *c* or *a*, respectively. - -Another application where resistors are used to control current flow is in the analog dc meters—the ammeter, voltmeter, and ohmmeter, which measure current, voltage, and resistance, respectively. Each of these meters employs the d'Arsonval meter movement, shown in Fig. 2.58. The movement consists essentially of a movable iron-core coil mounted on a pivot between the poles of a permanent magnet. When current flows through the coil, it creates a torque which causes the pointer to deflect. The amount of current through the coil determines the deflection of the pointer, which is registered on a scale attached to the meter movement. For example, if the meter movement is rated 1 mA, 50 Ω, it w ould take 1 mA to cause a full-scale deflection of the meter movement. By introducing additional circuitry to the d'Arsonval meter mo vement, an am meter, voltmeter, or ohmmeter can be constructed. - -Consider Fig. 2.59, where an analog v oltmeter and ammeter are con nected to an element. The voltmeter measures the voltage across a *load* and - -Practice Problem 2.16 - -**Figure 2.57** The potentiometer controlling potential levels. - - An instrument capable of measuring voltage, current, and resistance is called a multimeter or a volt-ohm meter (VOM). - - A load is a component that is receiving energy (an energy sink), as opposed to a generator supplying energy (an energy source). More about loading will be discussed in Section 4.9.1. - -A d'Arsonval meter movement. - -Connection of a voltmeter and an ammeter to an element. - -is therefore connected in parallel with the element. As shown in Fig. 2.60(a), the voltmeter consists of a d'Arson val movement in series with a resistor whose resistance *Rm* is deliberately made very large (theoretically, infinite), to minimize the current drawn from the circuit. To extend the range of voltage that the meter can measure, series multiplier resistors are often connected with the voltmeters, as shown in Fig. 2.60(b). The multiple-range voltmeter in Fig. 2.60(b) can measure v oltage from 0 to 1 V, 0 to 10 V, or 0 to 100 V, depending on whether the switch is connected to *R*1, *R*2, or *R*3, respectively. - -Let us calculate the multiplier resistor *Rn* for the single-range voltmeter in Fig. 2.60(a), or *Rn* = *R*1, *R*2, or *R*3 for the multiple-range voltmeter in Fig. 2.60(b). We need to determine the value of *Rn* to be connected in series with the internal resistance *Rm* of the voltmeter. In any design, we consider the worst-case condition. In this case, the worst case occurs when the fullscale current *I*fs =*Im* flows through the meter. This should also correspond - -**Figure 2.60** Voltmeters: (a) single-range type, (b) multiple-range type. - -to the maximum v oltage reading or the full-scale v oltage *V*fs. Since the multiplier resistance *Rn* is in series with the internal resistance *Rm*, - -$$ -V_{\text{fs}} = I_{\text{fs}} (R_n + R_m) \tag{2.59} -$$ - -From this, we obtain - -$$ -R_n = \frac{V_{\text{fs}}}{I_{\text{fs}}} - R_m \tag{2.60} -$$ - -Similarly, the ammeter measures the current through the load and is connected in series with it. As shown in Fig. 2.61(a), the ammeter consists of a d'Arsonval movement in parallel with a resistor whose resistance *Rm* is deliberately made very small (theoretically, zero) to minimize the v oltage drop across it. To allow multiple ranges, shunt resistors are often connected in parallel with *Rm* as sho wn in Fig. 2.61(b). The shunt resistors allow the meter to measure in the range 0–10 mA, 0–100 mA, or 0–1 A, depending on whether the switch is connected to *R*1, *R*2, or *R*3, respectively. - -Now our objective is to obtain the multiplier shunt *Rn* for the singlerange ammeter in Fig. 2.61(a), or *Rn* =*R*1, *R*2, or *R*3 for the multiple-range ammeter in Fig. 2.61(b). We notice that *Rm* and *Rn* are in parallel and that at full-scale reading *I* = *I*fs = *Im* + *In*, where *In* is the current through the shunt resistor *Rn*. Applying the current division principle yields - -$$ -I_m = \frac{R_n}{R_n + R_m} I_{\text{fs}} -$$ - -$$ -\sum_{i=1}^{n} x_i -$$ - -$$ -R_n = \frac{I_m}{I_{\text{fs}} - I_m} R_m \tag{2.61} -$$ - -The resistance *Rx* of a linear resistor can be measured in tw o ways. An indirect way is to measure the current *I* that flows through it by connecting an ammeter in series with it and the v oltage *V* across it by con necting a voltmeter in parallel with it, as shown in Fig. 2.62(a). Then - -$$ -R_x = \frac{V}{I} \tag{2.62} -$$ - -The direct method of measuring resistance is to use an ohmmeter . An ohmmeter consists basically of a d'Arson val movement, a v ariable resistor or potentiometer, and a battery, as shown in Fig. 2.62(b). Applying KVL to the circuit in Fig. 2.62(b) gives - -$$ -E = (R + R_m + R_x)I_m -$$ - -or - -$$ -R_x = \frac{E}{I_m} - (R + R_m) -$$ - (2.63) - -The resistor *R* is selected such that the meter gives a full-scale deflection; that is, *Im* = *I*fs when *Rx* = 0. This implies that - -$$ -E = (R + R_m) I_{\text{fs}} \tag{2.64} -$$ - -Substituting Eq. (2.64) into Eq. (2.63) leads to - -$$ -R_x = \left(\frac{I_{\text{fs}}}{I_m} - 1\right)(R + R_m) \tag{2.65} -$$ - -As mentioned, the types of meters we ha ve discussed are kno wn as *analog* meters and are based on the d'Arsonval meter movement. Another type of meter , called a *digital meter*, is based on acti ve circuit elements - -# **Figure 2.61** - -Ammeters: (a) single-range type, (b) multiple-range type. - -# **Figure 2.62** Two ways of measuring resistance: (a) using an ammeter and a voltmeter, (b) using an ohmmeter. - -# Historical - -Library of Congress - -**Samuel F. B. Morse** (1791–1872), an American painter, invented the telegraph, the first practical, commercialized application of electricity. - -Morse was born in Charlestown, Massachusetts, and studied at Yale and the Royal Academy of Arts in London to become an artist. In the 1830s, he became intrigued with developing a telegraph. He had a working model by 1836 and applied for a patent in 1838. The U.S. Senate appro priated funds for Morse to construct a telegraph line between Baltimore and Washington, D.C. On May 24, 1844, he sent the famous first message: "What hath God wrought!" Morse also developed a code of dots and dashes for letters and numbers, for sending messages on the telegraph. The development of the telegraph led to the invention of the telephone. - -such as op amps. For example, a digital multimeter displays measurements of dc or ac voltage, current, and resistance as discrete numbers, instead of using a pointer deflection on a continuous scale as in an analog multimeter. Digital meters are what you would most likely use in a modern lab. However, the design of digital meters is beyond the scope of this book. - -Example 2.17 Following the v oltmeter setup of Fig. 2.60, design a v oltmeter for the following multiple ranges: - -> (a) 0–1 V (b) 0–5 V (c) 0–50 V (d) 0–100 V Assume that the internal resistance *Rm* = 2 kΩ and the full-scale current *I*fs =100*μ*A. - -# **Solution:** - -We apply Eq. (2.60) and assume that *R*1, *R*2, *R*3, and *R*4 correspond with ranges 0–1 V, 0–5 V, 0–50 V, and 0–100 V, respectively. (a) For range 0–1 V, - -$$ -R_1 = \frac{1}{100 \times 10^{-6}} - 2000 = 10,000 - 2000 = 8 \text{ k}\Omega -$$ - -(b) For range 0–5 V, - -$$ -R_2 = \frac{5}{100 \times 10^{-6}} - 2000 = 50,000 - 2000 = 48 \text{ k}\Omega -$$ - -(c) For range 0–50 V, - -$$ -R_3 = \frac{50}{100 \times 10^{-6}} - 2000 = 500,000 - 2000 = 498 \text{ k}\Omega -$$ - -(d) For range 0–100 V, - -$$ -R_4 = \frac{100 \text{ V}}{100 \times 10^{-6}} - 2000 = 1,000,000 - 2000 = 998 \text{ k}\Omega -$$ - -Note that the ratio of the total resistance (*Rn* + *Rm*) to the full-scale voltage *V*fs is constant and equal to 1/*I*fs for the four ranges. This ratio (given in ohms per v olt, or Ω/ V) is known as the *sensitivity* of the v oltmeter. The larger the sensitivity, the better the voltmeter. - -Following the ammeter setup of Fig. 2.61, design an ammeter for the following multiple ranges: (a) 0–1 A (b) 0–100 mA (c) 0–10 mA Take the full-scale meter current as *Im* = 1 mA and the internal resistance of the ammeter as *Rm* = 50 Ω. - -**Answer:** Shunt resistors: 50 mΩ, 505 mΩ, 5.556 Ω. - -# **2.9** Summary - -1. A resistor is a passi ve element in which the v oltage *v* across it is directly proportional to the current *i* through it. That is, a resistor is a device that obeys Ohm's law, - -*v* = *iR* - -where *R* is the resistance of the resistor. - -- 2. A short circuit is a resistor (a perfectly , conducting wire) with zero resistance (*R* = 0). An open circuit is a resistor with infinite resistance (*R* = *∞*). -- 3. The conductance *G* of a resistor is the reciprocal of its resistance: - -$$ -G = \frac{1}{R} -$$ - -4. A branch is a single tw o-terminal element in an electric circuit. A node is the point of connection between tw o or more branches. A loop is a closed path in a circuit. The number of branches *b*, the number of nodes *n*, and the number of independent loops *l* in a network are related as - -$$ -b = l + n - 1 -$$ - -- 5. Kirchhoff's current law (KCL) states that the currents at an y node algebraically sum to zero. In other w ords, the sum of the currents entering a node equals the sum of currents leaving the node. -- 6. Kirchhoff's v oltage la w (KVL) states that the v oltages around a closed path algebraically sum to zero. In other w ords, the sum of voltage rises equals the sum of voltage drops. -- 7. Two elements are in series when the y are connected sequentially , end to end. When elements are in series, the same current flows through them (*i*1 = *i*2). They are in parallel if the y are connected to the same two nodes. Elements in parallel always have the same voltage across them (*v*1 = *v*2). -- 8. When two resistors *R*1 (=1/*G*1) and *R*2 (=1/*G*2) are in series, their equivalent resistance *R*eq and equivalent conductance *G*eq are - -$$ -R_{\text{eq}} = R_1 + R_2, -$$ - $G_{\text{eq}} = \frac{G_1 G_2}{G_1 + G_2}$ - -9. When two resistors *R*1 (=1/*G*1) and *R*2 (=1/*G*2) are in parallel, their equivalent resistance *R*eq and equivalent conductance *G*eq are - -$$ -R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}, \qquad G_{\text{eq}} = G_1 + G_2 -$$ - -Practice Problem 2.17 - -10. The voltage division principle for two resistors in series is - -$$ -v_1 = \frac{R_1}{R_1 + R_2} v -$$ -, $v_2 = \frac{R_2}{R_1 + R_2} v$ - -11. The current division principle for two resistors in parallel is - -$$ -i_1 = \frac{R_2}{R_1 + R_2} i -$$ -, $i_2 = \frac{R_1}{R_1 + R_2} i$ - -12. The formulas for a delta-to-wye transformation are - -$$ -R_1 = \frac{R_b R_c}{R_a + R_b + R_c}, \qquad R_2 = \frac{R_c R_a}{R_a + R_b + R_c} -$$ -$$ -R_3 = \frac{R_a R_b}{R_a + R_b + R_c} -$$ - -13. The formulas for a wye-to-delta transformation are - -The formulas for a wye-to-delta transformation are -\n -$$ -R_a = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_1}, \qquad R_b = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_2} -$$ -\n -$$ -R_c = \frac{R_1 R_2 + R_2 R_3 + R_3 R_1}{R_3} -$$ - -14. The basic laws covered in this chapter can be applied to the problems of electrical lighting and design of dc meters. - -# Review Questions - -| 2.1 | | The reciprocal of resistance is: | | -|-----|--|----------------------------------|--| -| | | | | - -| (a) voltage | (b) current | -|-----------------|--------------| -| (c) conductance | (d) coulombs | - -**2.2** An electric heater draws 10 A from a 120-V line. The resistance of the heater is: - -| (a) 1200 Ω | (b) 120 Ω | -|------------|-----------| -| (c) 12 Ω | (d) 1.2 Ω | - -**2.3** The voltage drop across a 1.5-kW toaster that draws 12 A of current is: - -| (a) 18 kV | (b) 125 V | -|-----------|-------------| -| (c) 120 V | (d) 10.42 V | - -**2.4** The maximum current that a 2W, 80 kΩ resistor can safely conduct is: - -| (a) 160 kA | (b) 40 kA | -|------------|-----------| -| (c) 5 mA | (d) 25 μA | - -**2.5** A network has 12 branches and 8 independent loops. How many nodes are there in the network? - -(a) 19 (b) 17 (c) 5 (d) 4 - -**2.6** The current *I* in the circuit of Fig. 2.63 is: - -| (a) −0.8 A | (b) −0.2 A | -|------------|------------| -| (c) 0.2 A | (d) 0.8 A | - -# **Figure 2.63** - -For Review Question 2.6. - -**2.7** The current *Io* of Fig. 2.64 is: - -(a) -$$ --4 -$$ - A (b) $-2$ A (c) $4$ A (d) $16$ A - -Problems **65** - -**2.8** In the circuit in Fig. 2.65, *V* is: (a) 30 V (b) 14 V (c) 10 V (d) 6 V - -# **Figure 2.65** - -For Review Question 2.8. - -**2.9** Which of the circuits in Fig. 2.66 will give you *Vab* =7V? - -# **Figure 2.66** - -# For Review Question 2.9. - -# Problems - -# Section 2.2 Ohm's Law - -**2.1** Design a problem, complete with a solution, to help students to better understand Ohm's law. Use at least two resistors and one voltage source. Hint, you could use both resistors at once or one at a time, it is up to you. Be creative. - -- **2.10** In the circuit of Fig. 2.67, a decrease in *R*3 leads to a decrease of, select all that apply: - - (a) current through *R*3 - - (b) voltage across *R*3 - - (c) voltage across *R*1 - - (d) power dissipated in *R*2 - - (e) none of the above - -# **Figure 2.67** For Review Question 2.10. - -*Answers: 2.1c, 2.2c, 2.3b, 2.4c, 2.5c, 2.6b, 2.7a, 2.8d, 2.9d, 2.10b, d.* - -- **2.2** Find the hot resistance of a light bulb rated 60 W, 120 V. -- **2.3** A bar of silicon is 4 cm long with a circular cross sec tion. If the resistance of the bar is 240 Ω at room tem perature, what is the cross-sectional radius of the bar? - -- **2.4** (a) Calculate current *i* in Fig. 2.68 when the switch is in position 1. - - (b) Find the current when the switch is in position 2. - -# **Figure 2.68** - -For Prob. 2.4. - -# Section 2.3 Nodes, Branches, and Loops - -**2.5** For the network graph in Fig. 2.69, find the number of nodes, branches, and loops. - -**Figure 2.69** For Prob. 2.5. - -> **2.6** In the network graph shown in Fig. 2.70, determine the number of branches and nodes. - -**2.7** Determine the number of branches and nodes in the circuit of Fig. 2.71. - -**Figure 2.71** For Prob. 2.7. - -# Section 2.4 Kirchhoff's Laws - -**2.8** Design a problem, complete with a solution, to help - -other students better understand Kirchhoff's Current Law. Design the problem by specifying values of *ia*, *ib*, and *ic*, shown in Fig. 2.72, and asking them to solve for values of *i*1, *i*2, and *i*3. Be careful to specify realistic currents. - -# **Figure 2.72** - -For Prob. 2.8. - -**2.9** Find *i*1, *i*2, and *i*3 in Fig. 2.73. - -**Figure 2.73** For Prob. 2.9. - -**2.10** Determine *i*1 and *i*2 in the circuit of Fig. 2.74. - -**Figure 2.74** For Prob. 2.10. - -**2.11** In the circuit of Fig. 2.75, calculate *V*1 and *V*2. - -For Prob. 2.11. - -**2.12** In the circuit in Fig. 2.76, obtain *v*1, *v*2, and *v*3. - -For Prob. 2.12. - -**2.13** For the circuit in Fig. 2.77, use KCL to find the branch currents *I*1 to *I*4. - -**Figure 2.77** For Prob. 2.13. - -**2.14** Given the circuit in Fig. 2.78, use KVL to find the branch voltages *V*1 to *V*4. - -**Figure 2.78** For Prob. 2.14. - -**2.15** Calculate *v* and *ix* in the circuit of Fig. 2.79. - -For Prob. 2.15. - -**2.16** Determine *Vo* in the circuit in Fig. 2.80. - -**2.17** Obtain *v*1 through *v*3 in the circuit of Fig. 2.81. - -**2.18** Find *I* and *V* in the circuit of Fig. 2.82. - -**Figure 2.82** For Prob. 2.18. - -For Prob. 2.19. - -**2.19** From the circuit in Fig. 2.83, find *I*, the power dissipated by the resistor, and the power supplied by each source. - -**2.20** Determine *io* in the circuit of Fig. 2.84. - -- For Prob. 2.20. -- **2.21** Find *Vx* in the circuit of Fig. 2.85. - -**2.22** Find *Vo* in the circuit in Fig. 2.86 and the power absorbed by the dependent source. - -- **2.23** In the circuit shown in Fig. 2.87, determine *Vx* and - - the power absorbed by the 60-Ω resistor. - -**2.24** For the circuit in Fig. 2.88, find *Vo* / *Vs* in terms of *α*, *R*1, *R*2, *R*3, and *R*4. If *R*1 = *R*2 = *R*3 = *R*4, what value of *α* will produce *|Vo* / *Vs|* =10? - -**Figure 2.88** For Prob. 2.24. - -**2.25** For the network in Fig. 2.89, find the current, voltage, and power associated with the 20-kΩ resistor. - -# Sections 2.5 and 2.6 Series and Parallel Resistors - -**2.26** For the circuit in Fig. 2.90, *io* = 3 A. Calculate *ix* and the total power absorbed by the entire circuit. - -**Figure 2.90** For Prob. 2.26. - -**2.27** Calculate *Io* in the circuit of Fig. 2.91. - -**Figure 2.91** For Prob. 2.27. - -## Problems **69** - -**2.28** Design a problem, using Fig. 2.92, to help other students better understand series and parallel circuits. - -**Figure 2.92** For Prob. 2.28. - -**2.29** All resistors (R) in Fig. 2.93 are 10 Ω each. Find *R*eq. - -**Figure 2.93** For Prob. 2.29. - -**2.30** Find *R*eq for the circuit in Fig. 2.94. - -**2.31** For the circuit in Fig. 2.95, determine *i*1 to *i*5. - -For Prob. 2.31. - -**2.32** Find *i*1 through *i*4 in the circuit in Fig. 2.96. - -# **Figure 2.96** - -For Prob. 2.32. - -**2.33** Obtain *v* and *i* in the circuit of Fig. 2.97. - -# **Figure 2.97** - -For Prob. 2.33. - -**2.34** Using series/parallel resistance combination, find the equivalent resistance seen by the source in the circuit of Fig. 2.98. Find the overall absorbed power by the resistor network. - -**Figure 2.98** For Prob. 2.34. - -**2.35** Calculate *Vo* and *Io* in the circuit of Fig. 2.99. - -**2.36** Find *i* and *Vo* in the circuit of Fig. 2.100. - -**2.37** Given the circuit in Fig. 2.101 and that the resistance, *R*eq, looking into the circuit from the left is equal to 100 Ω, determine the value of *R*1. - -**2.39** Evaluate *R*eq looking into each set of terminals for each of the circuits shown in Fig. 2.103. - -**2.40** For the ladder network in Fig. 2.104, find *I* and *R*eq. - -**Figure 2.104** - -For Prob. 2.40. - -# **Figure 2.105** - -For Prob. 2.41. - -**2.42** Reduce each of the circuits in Fig. 2.106 to a single resistor at terminals *a*-*b*. - -For Prob. 2.37. - -**2.38** Find *R*eq and *io* in the circuit of Fig. 2.102. - -# **Figure 2.106** - -For Prob. 2.42. - -**2.43** Calculate the equivalent resistance *Rab* at terminals *a-b* for each of the circuits in Fig. 2.107. - -**2.45** Find the equivalent resistance at terminals *a*-*b* of each circuit in Fig. 2.109. - -**Figure 2.109** For Prob. 2.45. - -**2.44** For the circuits in Fig. 2.108, obtain the equivalent resistance at terminals *a*-*b*. - -**Figure 2.108** For Prob. 2.44. - -**2.46** Find *I* in the circuit of Fig. 2.110. - -For Prob. 2.46. - -**2.47** Find the equivalent resistance *Rab* in the circuit of Fig. 2.111. - -# Section 2.7 Wye-Delta Transformations - -**2.48** Convert the circuits in Fig. 2.112 from Y to Δ. - -**2.50** Design a problem to help other students better understand wye-delta transformations using Fig. 2.114. - -# **Figure 2.114** - -For Prob. 2.50. - -**2.51** Obtain the equivalent resistance at the terminals *a*-*b* for each of the circuits in Fig. 2.115. - -**Figure 2.115** For Prob. 2.51. - -**2.52** For the circuit shown in Fig. 2.116, find the equivalent resistance. All resistors are 3Ω. **\*** - -\* An asterisk indicates a challenging problem. - -## Problems **73** - -**2.53** Obtain the equivalent resistance *R* **2.56** Determine *V* in the circuit of Fig. 2.120. *ab* in each of the circuits of Fig. 2.117. In (b), all resistors have a value of 30 Ω. **\*** - -**2.54** Consider the circuit in Fig. 2.118. Find the equivalent resistance at terminals: (a) *a*-*b*, (b) *c*-*d*. - -**2.55** Calculate *Io* in the circuit of Fig. 2.119. - -For Prob. 2.55. - -For Prob. 2.56. - -**2.57** Find *R*eq **\*** and *I* in the circuit of Fig. 2.121. - -# **Figure 2.121** - -For Prob. 2.57. - -# Section 2.8 Applications - -**2.58** The 150 W light bulb in Fig. 2.122 is rated at 110 volts. Calculate the value of *V*s to make the light bulb operate at its rated conditions. - -**2.59** An enterprising young man travels to Europe carrying three light bulbs he had purchased in North America. The light bulbs he has are a 100-W light bulb, a 60-W light bulb, and a 40-W light bulb. Each light bulb is rated at 110 V. He wishes to connect these to a 220-V system that is found in Europe. For reasons we are not sure of, he connects the 40-W - -light bulb in series with a parallel combination of the 60-W light bulb and the 100-W light bulb as shown in Fig. 2.123. How much power is actually being delivered to each light bulb? What does he see when he first turns on the light bulbs? - -Is there a better way to connect these light bulbs in order to have them work more effectively? - -- **2.60** If the three bulbs of Prob. 2.59 are connected in parallel to the 120-V source, calculate the current through each bulb. -- **2.61** As a design engineer, you are asked to design a -- lighting system consisting of a 70-W power supply and two light bulbs as shown in Fig. 2.124. You must select the two bulbs from the following three available bulbs. - -*R*1 = 80 Ω, cost = \$0.60 (standard size) *R*2 = 90 Ω, cost = \$0.90 (standard size) *R*3 = 100 Ω, cost = \$0.75 (nonstandard size) - -The system should be designed for minimum cost such that *I* lies within the range *I* = 1.2 A ± 5 percent. - -For Prob. 2.61. - -**2.62** A three-wire system supplies two loads *A* and *B* as shown in Fig. 2.125. Load *A* consists of a motor drawing a current of 8 A, while load *B* is a PC drawing 2 A. Assuming 10 h/day of use for 365 days and 6 cents/kWh, calculate the annual energy cost of the system. - -**Figure 2.125** - -- **2.63** If an ammeter with an internal resistance of 100 Ω and a current capacity of 2 mA is to measure 5 A, determine the value of the resistance needed. Calculate the power dissipated in the shunt resistor. -- **2.64** The potentiometer (adjustable resistor) *Rx* in Fig. 2.126 is to be designed to adjust current *ix* from 10 mA to 1 A. Calculate the values of *R* and *Rx* to achieve this. - -**Figure 2.126** For Prob. 2.64. - -- **2.65** Design a circuit that uses a d'Arsonval meter (with an internal resistance of 2 kΩ that requires a current of 5 mA to cause the meter to deflect full scale) to build a voltmeter to read values of voltages up to 100 volts. -- **2.66** A 20-kΩ/V voltmeter reads 10 V full scale. - - (a) What series resistance is required to make the meter read 50 V full scale? - - (b) What power will the series resistor dissipate when the meter reads full scale? -- **2.67** (a) Obtain the voltage *Vo* in the circuit of Fig. 2.127(a). - - (b) Determine the voltage *V*ʹ *o* measured when a voltmeter with 6-kΩ internal resistance is connected as shown in Fig. 2.127(b). - -(c) The finite resistance of the meter introduces an error into the measurement. Calculate the percent error as - -$$ -\left|\frac{V_o - V'_o}{V_o}\right| \times 100\,\% -$$ - -(d) Find the percent error if the internal resistance were 36 kΩ. - -# **Figure 2.127** - -For Prob. 2.67. - -- **2.68** (a) Find the current *I* in the circuit of Fig. 2.128(a). (b) An ammeter with an internal resistance of 1 Ω is inserted in the network to measure *I*ʹas shown in Fig. 2.128(b). What is *I*ʹ? - - (c) Calculate the percent error introduced by the meter as - - *|*×100% - -\_\_\_\_\_ *I*− *I*ʹ *I* - -**2.69** A voltmeter is used to measure *Vo* in the circuit in Fig. 2.129. The voltmeter model consists of an ideal voltmeter in parallel with a 250-kΩ resistor. Let *Vs* = 95 V, *Rs* = 25 kΩ, and *R*1 = 40 kΩ. Calculate *Vo* with and without the voltmeter when - -(a) -$$ -R_2 = 5 \text{ k}\Omega -$$ - (b) $R_2 = 25 \text{ k}\Omega$ - -# **Figure 2.129** For Prob. 2.69. - -- **2.70** (a) Consider the Wheatstone bridge shown in Fig. 2.130. Calculate *va*, *vb*, and *vab*. - - (b) Rework part (a) if the ground is placed at *a* instead of *o*. - -**2.71** Figure 2.131 represents a model of a solar photovoltaic panel. Given that *Vs* = 95 V, *R*1 = 25 Ω, and *iL* =2 A, find *RL*. - -**Figure 2.131** For Prob. 2.71. - -**2.72** Find *Vo* in the two-way power divider circuit in Fig. 2.132. - -For Prob. 2.72. - -**2.73** An ammeter model consists of an ideal ammeter in series with a 20-Ω resistor. It is connected with a current source and an unknown resistor *Rx* as shown in Fig. 2.133. The ammeter reading is noted. When a potentiometer *R* is added and adjusted until the ammeter reading drops to one half its previous reading, then *R* = 65 Ω. What is the value of *Rx* ? - -**2.74** The circuit in Fig. 2.134 is to control the speed of a motor such that the motor draws currents 5 A, 3 A, and 1 A when the switch is at high, medium, and low positions, respectively. The motor can be modeled as a load resistance of 20 mΩ. Determine the series dropping resistances *R*1, *R*2, and *R*3. - -**2.75** Find *Rab* in the four-way power divider circuit in Fig. 2.135. Assume each *R* = 4 Ω. - -**Figure 2.133** For Prob. 2.73. - -**Figure 2.135** For Prob. 2.75. - -# Comprehensive Problems - -**2.76** Repeat Prob. 2.75 for the eight-way divider shown in Fig. 2.136. - -# **Figure 2.136** For Prob. 2.76. - -**2.77** Suppose your circuit laboratory has the following standard commercially available resistors in large quantities: - -1.8 Ω 20 Ω 300 Ω 24 kΩ 56 kΩ - -Using series and parallel combinations and a minimum number of available resistors, how would you obtain the following resistances for an electronic circuit design? - -| (a) 5 Ω | (b) 311.8 Ω | -|-----------|--------------| -| (c) 40 kΩ | (d) 52.32 kΩ | - -**2.78** In the circuit in Fig. 2.137, the wiper divides the potentiometer resistance between *αR* and (1 − *α*)*R*, 0 ≤ α ≤1.Find *vo* / *vs*. - -**2.79** An electric pencil sharpener rated 240 mW, 6 V is connected to a 9-V battery as shown in Fig. 2.138. Calculate the value of the series-dropping resistor *Rx* needed to power the sharpener. - -**2.80** A loudspeaker is connected to an amplifier as shown in Fig. 2.139. If a 10-Ω loudspeaker draws the maximum power of 12 W from the amplifier, determine the maximum power a 4-Ω loudspeaker will draw. - -For Prob. 2.80. - -**2.81** For a specific application, the circuit shown in Fig. 2.140 was designed so that *IL* = 83.33 mA and that *Rin* = 5 kΩ. What are the values of *R*1 and *R*2? - -- **2.82** The pin diagram of a resistance array is shown in Fig. 2.141. Find the equivalent resistance between the following: - - (a) 1 and 2 - - (b) 1 and 3 - - (c) 1 and 4 - -**2.83** Two delicate devices are rated as shown in Fig. 2.142. Find the values of the resistors *R*1 and *R*2 needed to power the devices using a 36-V battery. - -**Figure 2.141** - -For Prob. 2.82. - -# **chapter** - -3 - -# Methods of Analysis - -*No great work is ever done in a hurry. To develop a great scientific discovery, to print a great picture, to write an immortal poem, to become a minister, or a famous general—to do anything great requires time, patience, and perseverance. These things are done by degrees, "little by little."* - -—W. J. Wilmont Buxton - -# Enhancing Your Career - -# **Career in Electronics** - -One area of application for electric circuit analysis is electronics. The term *electronics* was originally used to distinguish circuits of very low current levels. This distinction no longer holds, as power semiconductor devices operate at high le vels of current. Today, electronics is regarded as the science of the motion of char ges in a g as, vacuum, or semicon ductor. Modern electronics in volves transistors and transistor circuits. The earlier electronic circuits were assembled from components. Man y electronic circuits are now produced as integrated circuits, fabricated in a semiconductor substrate or chip. - -Electronic circuits find applications in many areas, such as automation, broadcasting, computers, and instrumentation. The range of devices that use electronic circuits is enormous and is limited only by our imagination. Radio, television, computers, and stereo systems are but a few. - -An electrical engineer usually performs di verse functions and is likely to use, design, or construct systems that incorporate some form of electronic circuits. Therefore, an understanding of the operation and analysis of electronics is essential to the electrical engineer . Electronics has become a specialty distinct from other disciplines within electrical engi neering. Because the field of electronics is ever advancing, an electronics engineer must update his/her knowledge from time to time. The best way to do this is by being a member of a professional organization such as the Institute of Electrical and Electronics Engineers (IEEE). With a membership of over 300,000, the IEEE is the largest professional organization in the world. Members benefit immensely from the numerous magazines, journals, transactions, and conference/symposium proceedings published yearly by IEEE. You should consider becoming an IEEE member. - -Troubleshooting an electronic circuit board. © Brand X Pictures/PunchStock RF - -# Learning Objectives - -By using the information and exercises in this chapter you will be able to: - -- 1. Understand Kirchhoff's current law. -- 2. Understand Kirchhoff's voltage law. -- 3. Develop an understanding of how to use Kirchhoff's current law to write nodal equations and then to solve for unknown node voltages. -- 4. Develop an understanding of how to use Kirchhoff's voltage law to write mesh equations and then to solve for unknown loop currents. -- 5. Explain how to use *PSpice* to solve for unknown node voltages and currents. - -# **3.1** Introduction - -Having understood the fundamental laws of circuit theory (Ohm's law and Kirchhoff's laws), we are now prepared to apply these laws to develop two powerful techniques for circuit analysis: nodal analysis, which is based on a systematic application of Kirchhof f's current law (KCL), and mesh analysis, which is based on a systematic application of Kirchhoff's voltage law (KVL). The two techniques are so important that this chapter should be re garded as the most important in the book. Students are therefore encouraged to pay careful attention. - -With the two techniques to be developed in this chapter, we can analyze any linear circuit by obtaining a set of simultaneous equations that are then solved to obtain the required values of current or voltage. One method of solving simultaneous equations involves Cramer's rule, which allows us to calculate circuit variables as a quotient of determinants. The examples in the chapter will illustrate this method; Appendix A also briefly summarizes the essentials the reader needs to kno w for applying Cramer' s rule. Another method of solving simultaneous equations is to use *MATLAB*, a computer software discussed in Appendix E. - -Also in this chapter, we introduce the use of *PSpice for Windows*, a circuit simulation computer software program that we will use throughout the text. Finally, we apply the techniques learned in this chapter to analyze transistor circuits. - -# **3.2** Nodal Analysis - -Nodal analysis provides a general procedure for analyzing circuits using node voltages as the circuit v ariables. Choosing node v oltages instead of element voltages as circuit variables is convenient and reduces the number of equations one must solve simultaneously. - -To simplify matters, we shall assume in this section that circuits do not contain voltage sources. Circuits that contain voltage sources will be analyzed in the next section. - -In *nodal analysis*, we are interested in finding the node voltages. Given a circuit with n nodes without voltage sources, the nodal analysis of the circuit involves taking the following three steps. - -Nodal analysis is also known as the node-voltage method. - -# Steps to Determine Node Voltages: - -- 1. Select a node as the reference node. Assign voltages *v*1, *v*2, . . . , *vn*− 1 to the remaining *n* − 1 nodes. The voltages are referenced with respect to the reference node. -- 2. Apply KCL to each of the *n* − 1 nonreference nodes. Use Ohm's law to express the branch currents in terms of node voltages. -- 3. Solve the resulting simultaneous equations to obtain the un known node voltages. - -We shall now explain and apply these three steps. - -The first step in nodal analysis is selecting a node as the *reference* or *datum node*. The reference node is commonly called the *ground* since it is assumed to have zero potential. A reference node is indicated b y any of the three symbols in Fig. 3.1. The type of ground in Fig. 3.1(c) is called a *chassis ground* and is used in de vices where the case, enclosure, or chassis acts as a reference point for all circuits. When the potential of the earth is used as reference, we use the *earth ground* in Fig. 3.1(a) or (b). We shall always use the symbol in Fig. 3.1(b). - -Once we ha ve selected a reference node, we assign v oltage designations to nonreference nodes. Consider , for e xample, the circuit in Fig. 3.2(a). Node 0 is the reference node ( *v* = 0), while nodes 1 and 2 are assigned voltages *v*1 and *v*2, respectively. Keep in mind that the node voltages are defined with respect to the reference node. As illustrated in Fig. 3.2(a), each node voltage is the voltage rise from the reference node to the corresponding nonreference node or simply the v oltage of that node with respect to the reference node. - -As the second step, we apply KCL to each nonreference node in the circuit. To avoid putting too much information on the same circuit, the circuit in Fig. 3.2(a) is redra wn in Fig. 3.2(b), where we no w add *i*1, *i*2, and *i*3 as the currents through resistors *R*1, *R*2, and *R*3, respectively. At node 1, applying KCL gives - -$$ -I_1 = I_2 + i_1 + i_2 \tag{3.1} -$$ - -At node 2, - -$$ -I_2 + i_2 = i_3 \tag{3.2} -$$ - -We now apply Ohm's law to express the unknown currents *i*1, *i*2, and *i*3 in terms of node voltages. The key idea to bear in mind is that, since resistance is a passive element, by the passive sign convention, current must always flow from a higher potential to a lower potential. - -Current flows from a higher potential to a lower potential in a resistor. - -We can express this principle as - -$$ -i = \frac{v_{\text{higher}} - v_{\text{lower}}}{R} -$$ - (3.3) - -# **Figure 3.1** - -Common symbols for indicating a reference node, (a) common ground, (b) ground, (c) chassis ground. - -The number of nonreference nodes is equal to the number of independent equations that we will derive. - -**Figure 3.2** Typical circuit for nodal analysis. - -Note that this principle is in agreement with the w ay we defined resistance in Chapter 2 (see Fig. 2.1). With this in mind, we obtain from Fig. 3.2(b), - -$$ -i_1 = \frac{v_1 - 0}{R_1} \quad \text{or} \quad i_1 = G_1 v_1 -$$ - -\n -$$ -i_2 = \frac{v_1 - v_2}{R_2} \quad \text{or} \quad i_2 = G_2 (v_1 - v_2) -$$ - -\n -$$ -i_3 = \frac{v_2 - 0}{R_3} \quad \text{or} \quad i_3 = G_3 v_2 -$$ - -\n(3.4) - -Substituting Eq. (3.4) in Eqs. (3.1) and (3.2) results, respectively, in - -$$ -I_1 = I_2 + \frac{v_1}{R_1} + \frac{v_1 - v_2}{R_2} -$$ - (3.5) - -$$ -I_2 + \frac{v_1 - v_2}{R_2} = \frac{v_2}{R_3} -$$ - (3.6) - -In terms of the conductances, Eqs. (3.5) and (3.6) become - -$$ -I_1 = I_2 + G_1 v_1 + G_2 (v_1 - v_2) -$$ -\n(3.7) - -$$ -I_2 + G_2(v_1 - v_2) = G_3v_2 \tag{3.8} -$$ - -The third step in nodal analysis is to solve for the node voltages. If we apply KCL to *n* − 1 nonreference nodes, we obtain *n* − 1 simulta neous equations such as Eqs. (3.5) and (3.6) or (3.7) and (3.8). F or the circuit of Fig. 3.2, we solve Eqs. (3.5) and (3.6) or (3.7) and (3.8) to obtain the node voltages *v*1 and *v*2 using any standard method, such as the substitution method, the elimination method, Cramer' s rule, or matrix inversion. To use either of the last two methods, one must cast the simultaneous equations in matrix form. For example, Eqs. (3.7) and (3.8) can be cast in matrix form as - -$$ -\begin{bmatrix} G_1 + G_2 & -G_2 \ -G_2 & G_2 + G_3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} I_1 - I_2 \ I_2 \end{bmatrix} -$$ - (3.9) - -which can be solved to get *v*1 and *v*2. Equation 3.9 will be generalized in Section 3.6. The simultaneous equations may also be solv ed using calculators or with softw are packages such as *MATLAB, Mathcad, Maple,* and *Quattro Pro*. - -Example 3.1 Calculate the node voltages in the circuit shown in Fig. 3.3(a). - -# **Solution:** - -Consider Fig. 3.3(b), where the circuit in Fig. 3.3(a) has been prepared for nodal analysis. Notice how the currents are selected for the application of KCL. Except for the branches with current sources, the labeling of the currents is arbitrary but consistent. (By consistent, we mean that if, for example, we assume that *i*2 enters the 4-Ω resistor from the left-hand side, *i*2 must leave the resistor from the right-hand side.) The reference node is selected, and the node voltages *v*1 and *v*2 are now to be determined. - -At node 1, applying KCL and Ohm's law gives - -$$ -i_1 = i_2 + i_3 -$$ - $\Rightarrow$ $5 = \frac{v_1 - v_2}{4} + \frac{v_1 - 0}{2}$ - -Multiplying each term in the last equation by 4, we obtain - -$$ -20 = v_1 - v_2 + 2v_1 -$$ - -Appendix A discusses how to use Cramer's rule. - -$$ -3v_1 - v_2 = 20 \tag{3.1.1} -$$ - -At node 2, we do the same thing and get - -$$ -i_2 + i_4 = i_1 + i_5 -$$ - $\Rightarrow$ $\frac{v_1 - v_2}{4} + 10 = 5 + \frac{v_2 - 0}{6}$ - -Multiplying each term by 12 results in - -$$ -3v_1 - 3v_2 + 120 = 60 + 2v_2 -$$ - -or - -$$ --3v_1 + 5v_2 = 60 \tag{3.1.2} -$$ - -Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve the equations using any method and obtain the values of *v*1 and *v*2. - -■ **METHOD 1** Using the elimination technique, we add Eqs. (3.1.1) and (3.1.2). - -$$ -4v_2 = 80 \qquad \Rightarrow \qquad v_2 = 20 \text{ V} -$$ - -Substituting *v*2 = 20 in Eq. (3.1.1) gives - -$$ -3v_1 - 20 = 20 -$$ - $\Rightarrow$ $v_1 = \frac{40}{3} = 13.333$ V - -■ **METHOD 2** To use Cramer's rule, we need to put Eqs. (3.1.1) and (3.1.2) in matrix form as - -$$ -\begin{bmatrix} 3 & -1 \\ -3 & 5 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 20 \\ 60 \end{bmatrix} -$$ - (3.1.3) - -The determinant of the matrix is - -$$ -\Delta = \begin{bmatrix} 3 & -1 \\ -3 & 5 \end{bmatrix} = 15 - 3 = 12 -$$ - -We now obtain *v*1 and *v*2 as - -$$ -v_1 = \frac{\Delta_1}{\Delta} = \frac{\begin{vmatrix} 20 & -1 \\ 60 & 5 \end{vmatrix}}{\Delta} = \frac{100 + 60}{12} = 13.333 \text{ V} -$$ -$$ -v_2 = \frac{\Delta_2}{\Delta} = \frac{\begin{vmatrix} 3 & 20 \\ -3 & 60 \end{vmatrix}}{\Delta} = \frac{180 + 60}{12} = 20 \text{ V} -$$ - -giving us the same result as did the elimination method. - -If we need the currents, we can easily calculate them from the values of the nodal voltages. - -$$ -i_1 = 5 \text{ A}, -$$ - $i_2 = \frac{v_1 - v_2}{4} = -1.6668 \text{ A},$ $i_3 = \frac{v_1}{2} = 6.666 \text{ A}$ - $i_4 = 10 \text{ A},$ $i_5 = \frac{v_2}{6} = 3.333 \text{ A}$ - -The fact that *i*2 is negative shows that the current flows in the direction opposite to the one assumed. - -Obtain the node voltages in the circuit of Fig. 3.4. Practice Problem 3.1 - -2 - -1= 5 - -2 Ω 6 Ω 10 A - -(a) - -2 Ω 6 Ω 10 A - -2 i 5 - -i i 4= 10 2 - -5 A - -4 Ω - -3 i - -*v*2 *v*1 - -1= 5 i - -5 A - -4 Ω - -1 - -i - -i - -**Answer:** *v*1 = 30 V, *v*2 = −2.5 V. - -14 A 7 A 5 Ω 4 Ω 5 Ω 1 2 - -Example 3.2 Determine the voltages at the nodes in Fig. 3.5(a). - -# **Solution:** - -**Figure 3.4** For Practice Prob. 3.1. - -The circuit in this example has three nonreference nodes, unlike the previous example which has two nonreference nodes. We assign voltages to the three nodes as shown in Fig. 3.5(b) and label the currents. - -At node 1, - -$$ -3 = i_1 + i_x \qquad \Rightarrow \qquad 3 = \frac{v_1 - v_3}{4} + \frac{v_1 - v_2}{2} -$$ - -Multiplying by 4 and rearranging terms, we get - -$$ -3v_1 - 2v_2 - v_3 = 12 \tag{3.2.1} -$$ - -At node 2, - -$$ -i_x = i_2 + i_3 -$$ - $\Rightarrow$ $\frac{v_1 - v_2}{2} = \frac{v_2 - v_3}{8} + \frac{v_2 - 0}{4}$ - -Multiplying by 8 and rearranging terms, we get - -$$ --4v_1 + 7v_2 - v_3 = 0 \tag{3.2.2} -$$ - -At node 3, - -$$ -i_1 + i_2 = 2i_x -$$ - $\Rightarrow$ $\frac{v_1 - v_3}{4} + \frac{v_2 - v_3}{8} = \frac{2(v_1 - v_2)}{2}$ - -Multiplying by 8, rearranging terms, and dividing by 3, we get - -$$ -2v_1 - 3v_2 + v_3 = 0 \tag{3.2.3} -$$ - -We have three simultaneous equations to solve to get the node voltages *v*1, *v*2, and *v*3. We shall solve the equations in three ways. - -■ **METHOD 1** Using the elimination technique, we add Eqs. (3.2.1) and (3.2.3). - -$$ -5v_1 - 5v_2 = 12 -$$ - -or - -$$ -v_1 - v_2 = \frac{12}{5} = 2.4 -$$ - (3.2.4) - -Adding Eqs. (3.2.2) and (3.2.3) gives - -$$ --2v_1 + 4v_2 = 0 \Rightarrow v_1 = 2v_2 \tag{3.2.5} -$$ - -Substituting Eq. (3.2.5) into Eq. (3.2.4) yields - -$$ -2v_2 - v_2 = 2.4 -$$ - $\Rightarrow$ $v_2 = 2.4$ , $v_1 = 2v_2 = 4.8$ V - -From Eq. (3.2.3), we get - -$$ -v_3 = 3v_2 - 2v_1 = 3v_2 - 4v_2 = -v_2 = -2.4 -$$ - V - -Thus, - -$$ -v_1 = 4.8 \text{ V}, \qquad v_2 = 2.4 \text{ V}, \qquad v_3 = -2.4 \text{ V} -$$ - -■ **METHOD 2** To use Cramer's rule, we put Eqs. (3.2.1) to (3.2.3) in matrix form. - -$$ -\begin{bmatrix} 3 & -2 & -1 \ -4 & 7 & -1 \ 2 & -3 & 1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 12 \ 0 \ 0 \end{bmatrix} -$$ - (3.2.6) - -From this, we obtain - -$$ -v_1 = \frac{\Delta_1}{\Delta} -$$ -, $v_2 = \frac{\Delta_2}{\Delta}$ , $v_3 = \frac{\Delta_3}{\Delta}$ - -where Δ, Δ1, Δ2, and Δ3 are the determinants to be calculated as follows. As explained in Appendix A, to calculate the determinant of a 3 by 3 matrix, we repeat the first two rows and cross multiply. - -Similarly, we obtain - -Δ3 = - -ǀ - -− − − - - 2 3 −4 - -−3 −2 7 - - 0 12 0 - -ǀ - -+ + + - -$$ -\Delta_1 = \frac{\begin{vmatrix} 12 & -2 & -1 \\ 0 & 3 & -1 \\ -1 & 12 & 2 \end{vmatrix}}{\begin{vmatrix} 2 & -2 & -1 \\ -1 & 2 & -1 \\ 0 & 7 & -1 \end{vmatrix} + \begin{vmatrix} 1 & -84 + 0 + 0 - 0 - 36 - 0 = 48 \\ + \end{vmatrix}} -$$ -$$ -\Delta_2 = \frac{\begin{vmatrix} 3 & 12 & -1 \\ 3 & 12 & -1 \\ -3 & 3 & -2 \end{vmatrix}}{\begin{vmatrix} 3 & -2 & -12 \\ -3 & 0 & -1 \end{vmatrix} + \begin{vmatrix} 3 & -2 & -12 \\ + \end{vmatrix}} -$$ - -= 0 + 144 + 0 − 168 − 0 − 0 = −24 - -Thus, we find - -$$ -v_1 = \frac{\Delta_1}{\Delta} = \frac{48}{10} = 4.8 \text{ V}, \qquad v_2 = \frac{\Delta_2}{\Delta} = \frac{24}{10} = 2.4 \text{ V} -$$ - - $v_3 = \frac{\Delta_3}{\Delta} = \frac{-24}{10} = -2.4 \text{ V}$ - -as we obtained with Method 1. - -■ **METHOD 3** We now use *MATLAB* to solve the matrix. Equation (3.2.6) can be written as - -$$ -AV = B \qquad \Rightarrow \qquad V = A^{-1}B -$$ - -where **A** is the 3 by 3 square matrix, **B** is the column vector, and **V** is a column vector comprised of *v*1, *v*2, and *v*3 that we want to determine. We use *MATLAB* to determine **V** as follows: - ->>A = -$$ -\begin{bmatrix} 3 & -2 & -1 \\ 4 & 7 & -1 \\ 2 & -3 & 1 \end{bmatrix} -$$ -; ->>B = $\begin{bmatrix} 12 & 0 & 0 \\ 4 & 8 \end{bmatrix}$ ; ->>V = inv(A) \* B -= 4.8000 -= 2.4000 -= 2.4000 - -Thus, *v*1 = 4.8 V, *v*2 = 2.4 V, and *v*3 = −2.4 V, as obtained previously. - -Practice Problem 3.2 Find the voltages at the three nonreference nodes in the circuit of Fig. 3.6. - -4 A 2 Ω 3 Ω 4 Ω 6 Ω i x 4i x 1 3 2 - -**Figure 3.6** For Practice Prob. 3.2. - -**Answer:** *v*1 = 32 V, *v*2 = −25.6 V, *v*3 = 62.4 V. - -# **3.3** Nodal Analysis with Voltage Sources - -We now consider ho w voltage sources af fect nodal analysis. W e use the circuit in Fig. 3.7 for illustration. Consider the following two possibilities. - -■ **CASE 1** If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the non reference node equal to the voltage of the voltage source. In Fig. 3.7, for example, - -$$ -v_1 = 10 \text{ V} \tag{3.10} -$$ - -Thus, our analysis is somewhat simplified by this knowledge of the voltage at this node. - -■ **CASE 2** If the voltage source (dependent or independent) is con nected between two nonreference nodes, the two nonreference nodes - -form a *generalized node* or *supernode*; we apply both KCL and KVL to determine the node voltages. - -A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it. - -In Fig. 3.7, nodes 2 and 3 form a supernode. (W e could have more than two nodes forming a single supernode. F or example, see the circuit in Fig. 3.14.) We analyze a circuit with supernodes using the same three steps mentioned in the pre vious section e xcept that the supernodes are treated dif ferently. Why? Because an essential component of nodal analysis is applying KCL, which requires kno wing the current through each element. There is no way of knowing the current through a voltage source in advance. However, KCL must be satisfied at a supernode like any other node. Hence, at the super node in Fig. 3.7, - -$$ -i_1 + i_4 = i_2 + i_3 \tag{3.11a} -$$ - -or - -$$ -\frac{v_1 - v_2}{2} + \frac{v_1 - v_3}{4} = \frac{v_2 - 0}{8} + \frac{v_3 - 0}{6} -$$ - (3.11b) - -To apply Kirchhoff's voltage law to the supernode in Fig. 3.7, we redraw the circuit as shown in Fig. 3.8. Going around the loop in the clockwise direction gives - -$$ --v_2 + 5 + v_3 = 0 \qquad \Rightarrow \qquad v_2 - v_3 = 5 \tag{3.12} -$$ - -- From Eqs. (3.10), (3.11b), and (3.12), we obtain the node voltages. Note the following properties of a supernode: - - 1. The v oltage source inside the supernode pro vides a constraint equation needed to solve for the node voltages. - - 2. A supernode has no voltage of its own. - - 3. A supernode requires the application of both KCL and KVL. - -**Figure 3.8** Applying KVL to a supernode. - - A supernode may be regarded as a closed surface enclosing the voltage source and its two nodes. - -For Example 3.3. - -For the circuit shown in Fig. 3.9, find the node voltages. - -# **Solution:** - -The supernode contains the 2-V source, nodes 1 and 2, and the 10- Ω resistor. Applying KCL to the supernode as shown in Fig. 3.10(a) gives - -$$ -2 = i_1 + i_2 + 7 -$$ - -Expressing *i*1 and *i*2 in terms of the node voltages - -$$ -2 = \frac{v_1 - 0}{2} + \frac{v_2 - 0}{4} + 7 \qquad \Rightarrow \qquad 8 = 2v_1 + v_2 + 28 -$$ - -or - -$$ -v_2 = -20 - 2v_1 \tag{3.3.1} -$$ - -To get the relationship between *v*1 and *v*2, we apply KVL to the circuit in Fig. 3.10(b). Going around the loop, we obtain - -−*v*1 − 2 + *v*2 = 0 ⇒ *v*2 = *v*1 + 2 **(3.3.2)** - -From Eqs. (3.3.1) and (3.3.2), we write - -$$ -v_2 = v_1 + 2 = -20 - 2v_1 -$$ - -or - -$$ -3v_1 = -22 \qquad \Rightarrow \qquad v_1 = -7.333 \text{ V} -$$ - -and *v*2 = *v*1 + 2 = −5.333 V. Note that the 10-Ω resistor does not make any difference because it is connected across the supernode. - -Applying: (a) KCL to the supernode, (b) KVL to the loop. - -Find the node voltages in the circuit of Fig. 3.12. Example 3.4 - -# **Figure 3.12** For Example 3.4. 20 V 2 Ω 4 Ω 6 Ω 3 Ω 1 Ω *v*x 3*v*x + – + – 10 A 1 4 2 3 + – - -# **Solution:** - -Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL to the two supernodes as in Fig. 3.13(a). At supernode 1-2, - -$$ -i_3 + 10 = i_1 + i_2 -$$ - -Expressing this in terms of the node voltages, - -\_\_\_\_\_\_ *v*3 − *v*2 6 + 10 = \_\_\_\_\_\_ *v*1 − *v*4 3 + \_\_ *v*1 2 - -or - -$$ -5v_1 + v_2 - v_3 - 2v_4 = 60 \tag{3.4.1} -$$ - -At supernode 3-4, - -$$ -i_1 = i_3 + i_4 + i_5 -$$ - $\Rightarrow$ $\frac{v_1 - v_4}{3} = \frac{v_3 - v_2}{6} + \frac{v_4}{1} + \frac{v_3}{4}$ - -or - -$$ -4v_1 + 2v_2 - 5v_3 - 16v_4 = 0 \tag{3.4.2} -$$ - -**Figure 3.13** Applying: (a) KCL to the two supernodes, (b) KVL to the loops. - -We now apply KVL to the branches involving the voltage sources as shown in Fig. 3.13(b). For loop 1, - -$$ --v_1 + 20 + v_2 = 0 \Rightarrow v_1 - v_2 = 20 \tag{3.4.3} -$$ - -For loop 2, - -$$ --v_3 + 3v_x + v_4 = 0 -$$ - -But *vx* = *v*1 − *v*4 so that - -$$ -3v_1 - v_3 - 2v_4 = 0 \tag{3.4.4} -$$ - -For loop 3, - -$$ -v_x - 3v_x + 6i_3 - 20 = 0 -$$ - -But 6*i*3 = *v*3 − *v*2 and *vx* = *v*1 − *v*4. Hence, - -$$ --2v_1 - v_2 + v_3 + 2v_4 = 20 \tag{3.4.5} -$$ - -We need four node v oltages, *v*1, *v*2, *v*3, and *v*4, and it requires only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although the fifth equation is redundant, it can be used to check results. We can solve Eqs. (3.4.1) to (3.4.4) directly using *MATLAB*. We can eliminate one node voltage so that we solv e three simultaneous equations instead of four . From Eq. (3.4.3), *v*2 = *v*1 − 20. Substituting this into Eqs. (3.4.1) and (3.4.2), respectively, gives - -$$ -6v_1 - v_3 - 2v_4 = 80 \tag{3.4.6} -$$ - -and - -$$ -6v_1 - 5v_3 - 16v_4 = 40 \tag{3.4.7} -$$ - -Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as - -3 6 6 −1 −1 −5 −2 −2 −16 ] [ *v*1 *v*3 *v*4 ] = [ 0 80 40] - -Using Cramer's rule gives - -$$ -\Delta = \begin{vmatrix} 3 & -1 & -2 \\ 6 & -1 & -2 \\ 6 & -5 & -16 \end{vmatrix} = -18, \qquad \Delta_1 = \begin{vmatrix} 0 & -1 & -2 \\ 80 & -1 & -2 \\ 40 & -5 & -16 \end{vmatrix} = -480, -$$ - -$$ -\Delta_3 = \begin{vmatrix} 3 & 0 & -2 \\ 6 & 80 & -2 \\ 6 & 40 & -16 \end{vmatrix} = -3120, \qquad \Delta_4 = \begin{vmatrix} 3 & -1 & 0 \\ 6 & -1 & 80 \\ 6 & -5 & 40 \end{vmatrix} = 840 -$$ - -Thus, we arrive at the node voltages as - -[ - -$$ -v_1 = \frac{\Delta_1}{\Delta} = \frac{-480}{-18} = 26.67 \text{ V}, \qquad v_3 = \frac{\Delta_3}{\Delta} = \frac{-3120}{-18} = 173.33 \text{ V}, -$$ - - $v_4 = \frac{\Delta_4}{\Delta} = \frac{840}{-18} = -46.67 \text{ V}$ - -and *v*2 = *v*1 − 20 = 6.667 V. We have not used Eq. (3.4.5); it can be used to cross-check results. - -Find *v*1, *v*2, and *v*3 in the circuit of Fig. 3.14 using nodal analysis. Practice Problem 3.4 - -**Answer:** *v*1 = 7.608 V, *v*2 = −17.39 V, *v*3 = 1.6305 V. - -# **3.4** Mesh Analysis - -Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it. - -Nodal analysis applies KCL to find unknown voltages in a gi ven circuit, while mesh analysis applies KVL to find unknown currents. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is *planar*. A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is *nonplanar*. A circuit may have crossing branches and still be planar if it can be redra wn such that it has no crossing branches. F or example, the circuit in Fig. 3.15(a) has tw o crossing branches, b ut it can be redrawn as in Fig. 3.15(b). Hence, the circuit in Fig. 3.15(a) is planar. However, the circuit in Fig. 3.16 is nonplanar , because there is no w ay to redra w it and a void the branches crossing. Nonplanar circuits can be handled using nodal analysis, but they will not be considered in this text. - -**Figure 3.14** For Practice Prob. 3.4. - - Mesh analysis is also known as loop analysis or the mesh-current method. - -# **Figure 3.15** - -(a) A planar circuit with crossing branches, (b) the same circuit redrawn with no crossing branches. - -To understand mesh analysis, we should first explain more about what we mean by a mesh. - -A mesh is a loop that does not contain any other loops within it. - -**Figure 3.17** A circuit with two meshes. - -In Fig. 3.17, for example, paths *abefa* and *bcdeb* are meshes, but path *abcdefa* is not a mesh. The current through a mesh is known as *mesh current*. In mesh analysis, we are interested in applying KVL to find the mesh currents in a given circuit. - -In this section, we will apply mesh analysis to planar circuits that do not contain current sources. In the next section, we will consider circuits with current sources. In the mesh analysis of a circuit with *n* meshes, we take the following three steps. - -# Steps to Determine Mesh Currents: - -- 1. Assign mesh currents *i*1, *i*2, . . . , *in* to the *n* meshes. -- 2. Apply KVL to each of the *n* meshes. Use Ohm's law to express the voltages in terms of the mesh currents. -- 3. Solve the resulting *n* simultaneous equations to get the mesh currents. - -To illustrate the steps, consider the circuit in Fig. 3.17. The first step requires that mesh currents *i*1 and *i*2 are assigned to meshes 1 and 2. Although a mesh current may be assigned to each mesh in an arbi trary direction, it is conventional to assume that each mesh current flows clockwise. - -As the second step, we apply KVL to each mesh. Applying KVL to mesh 1, we obtain - -$$ --V_1 + R_1 i_1 + R_3 (i_1 - i_2) = 0 -$$ - -or - -$$ -(R_1 + R_3)i_1 - R_3i_2 = V_1 \tag{3.13} -$$ - -−*R*3 *i*1 + (*R*2 + *R*3) *i*2 = −*V*2 **(3.14)** - -For mesh 2, applying KVL gives - -$$ -R_2 i_2 + V_2 + R_3 (i_2 - i_1) = 0 -$$ - -or - -The shortcut way will not apply if one mesh current is assumed clockwise and the other assumed counterclockwise, although this is permissible. Note in Eq. (3.13) that the coefficient of *i*1 is the sum of the resistances in the first mesh, while the coefficient of *i*2 is the negative of the resistance common to meshes 1 and 2. Now observe that the same is true in Eq. (3.14). This can serve as a shortcut way of writing the mesh equa tions. We will exploit this idea in Section 3.6. - -Although path abcdefa is a loop and not a mesh, KVL still holds. This is the reason for loosely using the terms loop analysis and mesh analysis to mean the same thing. - -The direction of the mesh current is arbitrary—(clockwise or counterclockwise)—and does not affect the validity of the solution. - -The third step is to solve for the mesh currents. Putting Eqs. (3.13) and (3.14) in matrix form yields - -$$ -\begin{bmatrix} R_1 + R_3 & -R_3 \ -R_3 & R_2 + R_3 \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \end{bmatrix} = \begin{bmatrix} V_1 \\ -V_2 \end{bmatrix} -$$ - (3.15) - -which can be solved to obtain the mesh currents *i*1 and *i*2. We are at liberty to use any technique for solving the simultaneous equations. According to Eq. (2.12), if a circuit has *n* nodes, *b* branches, and *l* independent loops or meshes, then *l* = *b* − *n* + 1. Hence, *l* independent simultaneous equations are required to solve the circuit using mesh analysis. - -Notice that the branch currents are different from the mesh currents unless the mesh is isolated. To distinguish between the two types of currents, we use *i* for a mesh current and *I* for a branch current. The current elements *I*1, *I*2, and *I*3 are algebraic sums of the mesh currents. It is e vident from Fig. 3.17 that - -$$ -I_1 = i_1, -$$ - $I_2 = i_2,$ $I_3 = i_1 - i_2$ (3.16) - -For the circuit in Fig. 3.18, find the branch currents *I*1, *I*2, and *I*3 using mesh analysis. - -# **Solution:** - -We first obtain the mesh currents using KVL. For mesh 1, - -$$ --15 + 5i_1 + 10(i_1 - i_2) + 10 = 0 -$$ - -or - -$$ -3i_1 - 2i_2 = 1 \t\t(3.5.1) -$$ - -For mesh 2, - -$$ -6i_2 + 4i_2 + 10(i_2 - i_1) - 10 = 0 -$$ - -or - -$$ -i_1 = 2i_2 - 1 \tag{3.5.2} -$$ - -■ **METHOD 1** Using the substitution method, we substitute Eq. (3.5.2) into Eq. (3.5.1), and write - -$$ -6i_2 - 3 - 2i_2 = 1 \qquad \Rightarrow \qquad i_2 = 1 \text{ A} -$$ - -From Eq. (3.5.2), *i*1 = 2*i*2 − 1 = 2 − 1 = 1 A. Thus, - -$$ -I_1 = i_1 = 1 -$$ - A, $I_2 = i_2 = 1$ A, $I_3 = i_1 - i_2 = 0$ - -■ **METHOD 2** To use Cramer's rule, we cast Eqs. (3.5.1) and (3.5.2) in matrix form as - -> [ 3 −1 −2 2 ] [*i*1 *i*2 ] = [ 1 1] - -For Example 3.5. - -# We obtain the determinants - -$$ -\Delta = \begin{vmatrix} 3 & -2 \\ -1 & 2 \end{vmatrix} = 6 - 2 = 4 -$$ - -$$ -\Delta_1 = \begin{vmatrix} 1 & -2 \\ 1 & 2 \end{vmatrix} = 2 + 2 = 4, \qquad \Delta_2 = \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} = 3 + 1 = 4 -$$ - -Thus, - -$$ -i_1 = \frac{\Delta_1}{\Delta} = 1 \text{ A}, \qquad i_2 = \frac{\Delta_2}{\Delta} = 1 \text{ A} -$$ - -as before. - -Practice Problem 3.5 - -Calculate the mesh currents *i*1 and *i*2 of the circuit of Fig. 3.19. - -**Answer:** *i*1 = 4.6 A, *i*2 = 200 mA. - -**Figure 3.19** For Practice Prob. 3.5. - -Example 3.6 Use mesh analysis to find the current *Io* in the circuit of Fig. 3.20. - -# **Solution:** - -We apply KVL to the three meshes in turn. For mesh 1, - -$$ --24 + 10(i_1 - i_2) + 12(i_1 - i_3) = 0 -$$ - -$$ -11i_1 - 5i_2 - 6i_3 = 12 \tag{3.6.1} -$$ - -For mesh 2, - -24*i*2 + 4 (*i*2 − *i*3) + 10 (*i*2 − *i*1) = 0 - -$$ --5i_1 + 19i_2 - 2i_3 = 0 \tag{3.6.2} -$$ - -For mesh 3, - -**Figure 3.20** For Example 3.6. - -$$ -4I_o + 12(i_3 - i_1) + 4(i_3 - i_2) = 0 -$$ - -But at node A, *Io* = *i*1 − *i*2, so that - -$$ -4(i1 - i2) + 12(i3 - i1) + 4(i3 - i2) = 0 -$$ - -or - -$$ --i_1 - i_2 + 2i_3 = 0 \tag{3.6.3} -$$ - -In matrix form, Eqs. (3.6.1) to (3.6.3) become - -$$ -\begin{bmatrix} 11 & -5 & -6 \ -5 & 19 & -2 \ -1 & -1 & 2 \ \end{bmatrix} \begin{bmatrix} i_1 \ i_2 \ i_3 \end{bmatrix} = \begin{bmatrix} 12 \ 0 \ 0 \end{bmatrix} -$$ - -We obtain the determinants as - -−2 - -+ + - -$$ -\Delta_3 = \begin{pmatrix}\n11 & -5 & 12 \\ --5 & 19 & 0 \\ --11 & -5 & 19\n\end{pmatrix} + 60 + 228 = 288 -$$ - -We calculate the mesh currents using Cramer's rule as - -−5 - -− − - -0 - -$$ -i_1 = \frac{\Delta_1}{\Delta} = \frac{432}{192} = 2.25 \text{ A}, \qquad i_2 = \frac{\Delta_2}{\Delta} = \frac{144}{192} = 0.75 \text{ A}, -$$ - - $i_3 = \frac{\Delta_3}{\Delta} = \frac{288}{192} = 1.5 \text{ A}$ - -Thus, *Io* = *i*1 − *i*2 = 1.5 A. - -# Practice Problem 3.6 - -**Figure 3.21** For Practice Prob. 3.6. - -**Figure 3.22** A circuit with a current source. - -Using mesh analysis, find *Io* in the circuit of Fig. 3.21. - -**Answer:** −4 A. - -# **3.5** Mesh Analysis with Current Sources - -Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because the presence of the current sources reduces the number of equations. Consider the following two possible cases. - -■ **CASE 1** When a current source exists only in one mesh: Consider the circuit in Fig. 3.22, for example. We set *i*2 = −5 A and write a mesh equation for the other mesh in the usual way; that is, - -−10 + 4*i*1 + 6(*i*1 − *i*2) = 0 ⇒ *i*1 = −2 A **(3.17)** - -■ **CASE 2** When a current source exists between two meshes: Consider the circuit in Fig. 3.23(a), for example. We create a *supermesh* by excluding the current source and any elements connected in series with it, as shown in Fig. 3.23(b). Thus, - -A supermesh results when two meshes have a (dependent or independent) current source in common. - -(a) Two meshes having a current source in common, (b) a supermesh, created by excluding the current source. - -> As shown in Fig. 3.23(b), we create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh.) Why treat the supermesh differently? Because mesh analysis applies KVL which requires that we know the voltage across each branch—and we do not know the voltage across a current source in advance. However, a supermesh must satisfy KVL like any other mesh. Therefore, applying KVL to the supermesh in Fig. 3.23(b) gives - -−20 + 6*i*1 + 10*i*2 + 4*i*2 = 0 - -$$ -6i_1 + 14i_2 = 20 \tag{3.18} -$$ - -We apply KCL to a node in the branch where the two meshes intersect. Applying KCL to node 0 in Fig. 3.23(a) gives - -$$ -i_2 = i_1 + 6 \tag{3.19} -$$ - -Solving Eqs. (3.18) and (3.19), we get - -$$ -i_1 = -3.2 \text{ A}, \qquad i_2 = 2.8 \text{ A} -$$ - (3.20) - -Note the following properties of a supermesh: - -- 1. The current source in the supermesh pro vides the constraint equation necessary to solve for the mesh currents. -- 2. A supermesh has no current of its own. -- 3. A supermesh requires the application of both KVL and KCL. - -For the circuit in Fig. 3.24, find *i*1 to *i*4 using mesh analysis. Example 3.7 - -**Figure 3.24** - -For Example 3.7. - -# **Solution:** - -Note that meshes 1 and 2 form a supermesh because they have an independent current source in common. Also, meshes 2 and 3 form another supermesh because they have a dependent current source in common. The two supermeshes intersect and form a larger supermesh as shown. Applying KVL to the larger supermesh, - -2*i*1 + 4*i*3 + 8(*i*3 − *i*4) + 6*i*2 = 0 - -or - -$$ -i_1 + 3i_2 + 6i_3 - 4i_4 = 0 \tag{3.7.1} -$$ - -For the independent current source, we apply KCL to node *P*: - -$$ -i_2 = i_1 + 5 \tag{3.7.2} -$$ - -For the dependent current source, we apply KCL to node *Q*: - -$$ -i_2 = i_3 + 3I_o -$$ - -But *Io* = −*i*4, hence, - -or - -$$ -i_2 = i_3 - 3i_4 \tag{3.7.3} -$$ - -Applying KVL in mesh 4, - -$$ -2i_4 + 8(i_4 - i_3) + 10 = 0 -$$ - -$$ -5i_4 - 4i_3 = -5 \tag{3.7.4} -$$ - -From Eqs. (3.7.1) to (3.7.4), - -$$ -i_1 = -7.5 \text{ A}, \qquad i_2 = -2.5 \text{ A}, \qquad i_3 = 3.93 \text{ A}, \qquad i_4 = 2.143 \text{ A} -$$ - -**Figure 3.26** (a) The circuit in Fig. 3.2, (b) the circuit in Fig. 3.17. - -(b) - -Use mesh analysis to determine *i*1, *i*2, and *i*3 in Fig. 3.25. - -**Answer:** -$$ -i_1 = 12.379 \text{ A}, i_2 = 378.9 \text{ mA}, i_3 = 3.284 \text{ A}. -$$ - -# **3.6** Nodal and Mesh Analyses by Inspection - -This section presents a generalized procedure for nodal or mesh analysis. It is a shortcut approach based on mere inspection of a circuit. - -When all sources in a circuit are independent current sources, we do not need to apply KCL to each node to obtain the node-v oltage equations as we did in Section 3.2. We can obtain the equations by mere inspection of the circuit. As an e xample, let us ree xamine the circuit in Fig. 3.2, shown again in Fig. 3.26(a) for convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.2 as - -$$ -\begin{bmatrix} G_1 + G_2 & -G_2 \ -G_2 & G_2 + G_3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} I_1 - I_2 \ I_2 \end{bmatrix} -$$ - (3.21) - -Observe that each of the diagonal terms is the sum of the conductances connected directly to node 1 or 2, while the off-diagonal terms are the negatives of the conductances connected between the nodes. Also, each term on the right-hand side of Eq. (3.21) is the algebraic sum of the currents entering the node. - -In general, if a circuit with independent current sources has *N* nonreference nodes, the node-v oltage equations can be written in terms of the conductances as - -$$ -\begin{bmatrix} G_{11} & G_{12} & \cdots & G_{1N} \\ G_{21} & G_{22} & \cdots & G_{2N} \\ \vdots & \vdots & \vdots & \vdots \\ G_{N1} & G_{N2} & \cdots & G_{NN} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_N \end{bmatrix} = \begin{bmatrix} i_1 \\ i_2 \\ \vdots \\ i_N \end{bmatrix} -$$ - (3.22) - -or simply - -$$ -Gv = i \tag{3.23} -$$ - -where - -*Gkk* = Sum of the conductances connected to node *k* - - *Gk j* = *Gjk* = Negative of the sum of the conductances directly connecting nodes *k* and *j*, *k* ≠ *j* - - *vk*= Unknown voltage at node *k* - - *ik* = Sum of all independent current sources directly connected to node *k*, with currents entering the node treated as positive - -**G** is called the *conductance matrix*; **v** is the output v ector; and **i** is the input vector. Equation (3.22) can be solved to obtain the unknown node voltages. Keep in mind that this is v alid for circuits with only indepen dent current sources and linear resistors. - -Similarly, we can obtain mesh-current equations by inspection when a linear resistive circuit has only independent voltage sources. Consider the circuit in Fig. 3.17, shown again in Fig. 3.26(b) for convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.4 as - -$$ -\begin{bmatrix} R_1 + R_3 & -R_3 \ -R_3 & R_2 + R_3 \end{bmatrix} \begin{bmatrix} i_1 \ i_2 \end{bmatrix} = \begin{bmatrix} v_1 \ -v_2 \end{bmatrix} -$$ - (3.24) - -We notice that each of the diagonal terms is the sum of the resistances in the related mesh, while each of the off-diagonal terms is the negative of the resistance common to meshes 1 and 2. Each term on the right-hand side of Eq. (3.24) is the algebraic sum taken clockwise of all independent voltage sources in the related mesh. - -In general, if the circuit has *N* meshes, the mesh-current equations can be expressed in terms of the resistances as - -$$ -\begin{bmatrix} R_{11} & R_{12} & \cdots & R_{1N} \\ R_{21} & R_{22} & \cdots & R_{2N} \\ \vdots & \vdots & \vdots & \vdots \\ R_{N1} & R_{N2} & \cdots & R_{NN} \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ \vdots \\ i_N \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_N \end{bmatrix} -$$ - (3.25) - -or simply - -$$ -\mathbf{Ri} = \mathbf{v} \tag{3.26} -$$ - -where - -*Rkk* = Sum of the resistances in mesh *k* - -- *Rkj* = *Rjk* = Negative of the sum of the resistances in common with meshes *k* and *j*, *k* ≠ *j* -- *ik* = Unknown mesh current for mesh *k* in the clockwise direction -- *vk* = Sum taken clockwise of all independent voltage sources in mesh *k*, with voltage rise treated as positive - -**R** is called the *resistance matrix* ; **i** is the output v ector; and **v** is the input v ector. We can solv e Eq. (3.25) to obtain the unkno wn mesh currents. - -Example 3.8 Write the node-v oltage matrix equations for the circuit in Fig. 3.27 by inspection. - -**Figure 3.27** For Example 3.8. - -# **Solution:** - -The circuit in Fig. 3.27 has four nonreference nodes, so we need four node equations. This implies that the size of the conductance matrix **G**, is 4 by 4. The diagonal terms of **G**, in siemens, are - -$$ -G_{11} = \frac{1}{5} + \frac{1}{10} = 0.3, \qquad G_{22} = \frac{1}{5} + \frac{1}{8} + \frac{1}{1} = 1.325 -$$ - -$$ -G_{33} = \frac{1}{8} + \frac{1}{8} + \frac{1}{4} = 0.5, \qquad G_{44} = \frac{1}{8} + \frac{1}{2} + \frac{1}{1} = 1.625 -$$ - -The off-diagonal terms are - -$$ -G_{12} = -\frac{1}{5} = -0.2, \qquad G_{13} = G_{14} = 0 -$$ - -\n -$$ -G_{21} = -0.2, \qquad G_{23} = -\frac{1}{8} = -0.125, \qquad G_{24} = -\frac{1}{1} = -1 -$$ - -\n -$$ -G_{31} = 0, \qquad G_{32} = -0.125, \qquad G_{34} = -\frac{1}{8} = -0.125 -$$ - -\n -$$ -G_{41} = 0, \qquad G_{42} = -1, \qquad G_{43} = -0.125 -$$ - -The input current vector **i** has the following terms, in amperes: - -$$ -i_1 = 3 -$$ -, $i_2 = -1 - 2 = -3$ , $i_3 = 0$ , $i_4 = 2 + 4 = 6$ - -Thus, the node-voltage equations are - -$$ -\begin{bmatrix} 0.3 & -0.2 & 0 & 0 \ -0.2 & 1.325 & -0.125 & -1 \ 0 & -0.125 & 0.5 & -0.125 \ 0 & -1 & -0.125 & 1.625 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \ v_4 \end{bmatrix} = \begin{bmatrix} 3 \ -3 \ 0 \ 6 \end{bmatrix} -$$ - -which can be solved using *MATLAB* to obtain the node voltages *v*1, *v*2, *v*3, and *v*4. - -By inspection, obtain the node-voltage equations for the circuit in Fig. 3.28. - -**Answer:** - -| 1.25 | −0.2 | −1 | 0 | 0
v1 | | -|--------|------|-------|-------|------------------------------|--| -| −0.2 | 0.2 | 0 | 0 | 5
v2 | | -| −1 | 0 | 1.25 | −0.25 | =
−3
v3 | | -| [
0 | 0 | −0.25 | 1.25 | ]
[
]
[
2]
v4 | | - -By inspection, write the mesh-current equations for the circuit in Fig. 3.29. Example 3.9 - -# **Figure 3.29** - -For Example 3.9. - -# **Solution:** - -We have five meshes, so the resistance matrix is 5 by 5. The diagonal terms, in ohms, are: - -$$ -R_{11} = 5 + 2 + 2 = 9, \qquad R_{22} = 2 + 4 + 1 + 1 + 2 = 10, -$$ - -$$ -R_{33} = 2 + 3 + 4 = 9, \qquad R_{44} = 1 + 3 + 4 = 8, \qquad R_{55} = 1 + 3 = 4 -$$ - -The off-diagonal terms are: - -$$ -R_{12} = -2, \t R_{13} = -2, \t R_{14} = 0 = R_{15}, -$$ - -\n -$$ -R_{21} = -2, \t R_{23} = -4, \t R_{24} = -1, \t R_{25} = -1, -$$ - -\n -$$ -R_{31} = -2, \t R_{32} = -4, \t R_{34} = 0 = R_{35}, -$$ - -\n -$$ -R_{41} = 0, \t R_{42} = -1, \t R_{43} = 0, \t R_{45} = -3, -$$ - -\n -$$ -R_{51} = 0, \t R_{52} = -1, \t R_{53} = 0, \t R_{54} = -3 -$$ - -The input voltage vector **v** has the following terms in volts: - -$$ -v_1 = 4 -$$ -, $v_2 = 10 - 4 = 6$ , - $v_3 = -12 + 6 = -6$ , $v_4 = 0$ , $v_5 = -6$ - -Thus, the mesh-current equations are: - -| | 9 | −2 | −2 | 0 | 0 | | i1 | | 4 | | -|---|----|----|----|----|---------|---|---------|--------|--------|--| -| | −2 | 10 | −4 | −1 | −1 | | i2 | | 6 | | -| | −2 | −4 | 9 | 0 | 0 | | i3 | | −6 | | -| [ | 0 | −1 | 0 | 8 | −3
] | | i4 | =
[ | 0
] | | -| | 0 | −1 | 0 | −3 | 4 | [ | ]
i5 | | −6 | | - -From this, we can use *MATLAB* to obtain mesh currents *i*1, *i*2, *i*3, *i*4, and *i*5. - -# Practice Problem 3.9 - -By inspection, obtain the mesh-current equations for the circuit in Fig. 3.30. - -**Figure 3.30** For Practice Prob. 3.9. - -# **Answer:** - -| | 150 | −40 | 0 | −80 | 0 | i1 | | 30 | | -|---|-----|-----|-----|-----|--------|---------|-------------|---------|--| -| | −40 | 65 | −30 | −15 | 0 | i2 | | 0 | | -| | −0 | −30 | 50 | 0 | −20 | i3 | | −12 | | -| [ | 80 | −15 | 0 | 95 | 0
] | i4
[ | =
[
] | 20
] | | -| | 0 | 0 | −20 | 0 | 80 | i5 | | −20 | | - -# **3.7** Nodal Versus Mesh Analysis - -Both nodal and mesh analyses provide a systematic way of analyzing a complex network. Someone may ask: Gi ven a network to be analyzed, how do we know which method is better or more efficient? The choice of the better method is dictated by two factors. - -The first factor is the nature of the particular network. Networks that contain many series-connected elements, voltage sources, or supermeshes are more suitable for mesh analysis, whereas networks with parallelconnected elements, current sources, or supernodes are more suitable for nodal analysis. Also, a circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis. The key is to select the method that results in the smaller number of equations. - -The second factor is the information required. If node voltages are required, it may be expedient to apply nodal analysis. If branch or mesh currents are required, it may be better to use mesh analysis. - -It is helpful to be familiar with both methods of analysis, for at least two reasons. First, one method can be used to check the results from the other method, if possible. Second, since each method has its limitations, only one method may be suitable for a particular problem. For example, mesh analysis is the only method to use in analyzing transistor circuits, as we shall see in Section 3.9. But mesh analysis cannot easily be used to solve an op amp circuit, as we shall see in Chapter 5, because there is no direct way to obtain the voltage across the op amp itself. For nonplanar networks, nodal analysis is the only option, because mesh analysis only applies to planar networks. Also, nodal analysis is more amenable to solution by computer, as it is easy to program. This allows one to analyze complicated circuits that defy hand calculation. A computer software package based on nodal analysis is introduced next. - -# **3.8** Circuit Analysis with PSpice - -*PSpice* is a computer software circuit analysis program that we will gradually learn to use throughout the course of this text. This section illustrates how to use *PSpice for Windows* to analyze the dc circuits we have studied so far. - -The reader is expected to review the tutorial before proceeding in this section. It should be noted that *PSpice* is only helpful in determining branch voltages and currents when the numerical values of all the circuit components are known. - -A tutorial on using PSpice for Windows can be found in Connect. - -Use *PSpice* to find the node voltages in the circuit of Fig. 3.31. Example 3.10 - -# **Solution:** - -The first step is to draw the given circuit using Schematics. If one follows the instructions given in the PSpice tutorial found in Connect, the schematic in Fig. 3.32 is produced. Because this is a dc analysis, we use voltage source VDC and current source IDC. The pseudocomponent VIEWPOINTS are added to display the required node voltages. Once the circuit is drawn and saved as *exam310.sch*, we run *PSpice* by selecting **Analysis/Simulate**. The circuit is simulated and the results are displayed - -**Figure 3.32** For Example 3.10; the schematic of the circuit in Fig. 3.31. - -on VIEWPOINTS and also saved in output file *exam310.out*. The output file includes the following: - -| | NODE VOLTAGE | | NODE VOLTAGE NODE VOLTAGE | | | | -|-------------------------------------------------------------------|--------------|--|---------------------------|--|---------|--| -| (1) | 120.0000 (2) | | 81.2900 (3) | | 89.0320 | | -| indicating that V1
= 120 V, V2
= 81.29 V, V3
= 89.032 V. | | | | | | | - -Practice Problem 3.10 For the circuit in Fig. 3.33, use *PSpice* to find the node voltages. - -**Answer:** *V*1 = −10 V, *V*2 = 14.286 V, *V*3 = 50 V. - -Example 3.11 In the circuit of Fig. 3.34, determine the currents *i*1, *i*2, and *i*3. - -For Example 3.11. - -# **Solution:** - -The schematic is shown in Fig. 3.35. (The schematic in Fig. 3.35 includes the output results, implying that it is the schematic displayed on the screen *after* the simulation.) Notice that the voltage-controlled voltage source E1 in Fig. 3.35 is connected so that its input is the voltage across the 4- Ω resistor; its gain is set equal to 3. In order to dis play the required currents, we insert pseudocomponent IPROBES in the appropriate branches. The schematic is saved as *exam311.sch* and simulated by selecting **Analysis/Simulate**. The results are displayed on IPROBES as shown in Fig. 3.35 and saved in output file *exam311.out*. From the output file or the IPROBES, we obtain *i*1 = *i*2 = 1.333 A and *i*3 = 2.667 A. - -**Figure 3.35** The schematic of the circuit in Fig. 3.34. - -Use *PSpice* to determine currents *i*1, *i*2, and *i*3 in the circuit of Fig. 3.36. - -**Answer:** *i*1 = −428.6 mA, *i*2 = 2.286 A, *i*3 = 2 A. - -# **3.9** † - -# Applications: DC Transistor Circuits - -Most of us deal with electronic products on a routine basis and ha ve some e xperience with personal computers. A basic component for the integrated circuits found in these electronics and computers is the active, three-terminal device known as the *transistor*. Understanding the transistor is essential before an engineer can start an electronic circuit design. - -Figure 3.37 depicts various kinds of transistors commercially available. There are tw o basic types of transistors: *bipolar junction tr ansistors* (BJTs) and *field-effect transistors* (FETs). Here, we consider only the BJTs, which were the first of the two and are still used today . Our objective is to present enough detail about the BJT to enable us to apply the techniques developed in this chapter to analyze dc transistor circuits. - -Practice Problem 3.11 - -Courtesy of Lucent Technologies/Bell Labs - -# Historical - -**William Schockley** (1910–1989), **John Bardeen** (1908–1991), and **Walter Brattain** (1902–1987) co-invented the transistor. - -Nothing has had a greater impact on the transition from the "Industrial Age" to the "Age of the Engineer" than the transistor. I am sure that Dr. Shockley, Dr. Bardeen, and Dr. Brattain had no idea they would have this incredible effect on our history. While working at Bell Laboratories, they successfully demonstrated the point-contact transistor, invented by Bardeen and Brattain in 1947, and the junction transistor, which Shockley conceived in 1948 and successfully produced in 1951. - -It is interesting to note that the idea of the field-effect transistor, the most commonly used one today, was first conceived in 1925–1928 by J. E. Lilienfeld, a German immigrant to the United States. This is evident from his patents of what appears to be a field-effect transistor. Unfortunately, the technology to realize this device had to wait until 1954 when Shockley's field-effect transistor became a reality. Just think what today would be like if we had this transistor 30 years earlier! - -For their contributions to the creation of the transistor, Dr. Shockley, Dr. Bardeen, and Dr. Brattain received, in 1956, the Nobel Prize in physics. It should be noted that Dr. Bardeen is the only individual to win two Nobel prizes in physics; the second came later for work in superconductivity at the University of Illinois. - -**Figure 3.38** Two types of BJTs and their circuit symbols: (a) *npn*, (b) *pnp*. - -**Figure 3.37** Various types of transistors. *(© McGraw-Hill Education/Mark Dierker, photographer)* - -There are tw o types of BJTs: *npn* and *pnp*, with their circuit sym bols as shown in Fig. 3.38. Each type has three terminals, designated as emitter (E), base (B), and collector (C). For the *npn* transistor, the currents and voltages of the transistor are specified as in Fig. 3.39. Applying KCL to Fig. 3.39(a) gives - -$$ -I_E = I_B + I_C \tag{3.27} -$$ - -where *IE*, *IC*, and *IB* are emitter, collector, and base currents, respectively. Similarly, applying KVL to Fig. 3.39(b) gives - -$$ -V_{CE} + V_{EB} + V_{BC} = 0 \tag{3.28} -$$ - -where *VCE*, *VEB*, and *VBC* are collector -emitter, emitter-base, and basecollector voltages. The BJT can operate in one of three modes: acti ve, cutoff, and saturation. When transistors operate in the active mode, typically *VBE* ≃ 0.7 V, - -$$ -I_C = \alpha I_E \tag{3.29} -$$ - -where *α* is called the *common-base curr ent gain*. In Eq. (3.29), *α* denotes the fraction of electrons injected by the emitter that are col lected by the collector. Also, - -$$ -I_C = \beta I_B \tag{3.30} -$$ - -where *β* is known as the *common-emitter current gain*. The *α* and *β* are characteristic properties of a given transistor and assume constant values for that transistor. Typically, *α* takes values in the range of 0.98 to 0.999, while *β* takes values in the range of 50 to 1000. From Eqs. (3.27) to (3.30), it is evident that - -$$ -I_E = (1 + \beta)I_B \tag{3.31} -$$ - -and - -$$ -\beta = \frac{\alpha}{1 - \alpha} \tag{3.32} -$$ - -These equations show that, in the active mode, the BJT can be modeled as a dependent current-controlled current source. Thus, in circuit analysis, the dc equi valent model in Fig. 3.40(b) may be used to replace the *npn* transistor in Fig. 3.40(a). Since *β* in Eq. (3.32) is large, a small base current controls lar ge currents in the output circuit. Consequently , the bipolar transistor can serve as an amplifier, producing both current gain and voltage gain. Such amplifiers can be used to furnish a considerable amount of power to transducers such as loudspeakers or control motors. - -(a) An *npn* transistor, (b) its dc equivalent model. - -It should be observ ed in the follo wing e xamples that one cannot directly analyze transistor circuits using nodal analysis because of the potential difference between the terminals of the transistor . Only when the transistor is replaced by its equivalent model can we apply nodal analysis. - -# **Figure 3.39** - -The terminal variables of an *npn* transistor: (a) currents, (b) voltages. - -In fact, transistor circuits provide motivation to study dependent sources. - -Example 3.12 Find *IB*, *IC*, and *vo* in the transistor circuit of Fig. 3.41. Assume that the transistor operates in the active mode and that *β* = 50. - -**Figure 3.41** For Example 3.12. - -# **Solution:** - -For the input loop, KVL gives - -$$ --4 + I_B (20 \times 10^3) + V_{BE} = 0 -$$ - -Since *VBE* = 0.7 V in the active mode, - -$$ -I_B = \frac{4 - 0.7}{200 \times 10^3} = 16.5 \,\mu\text{A} -$$ - -But - -$$ -I_C = \beta I_B = 50 \times 16.5 \,\mu A = 0.825 \text{ mA} -$$ - -For the output loop, KVL gives - -$$ --v_o - 100I_C + 6 = 0 -$$ - -or - -+ - -$$ -v_o = 6 - 100I_C = 6 - 0.0825 = 5.917 -$$ - V - -Note that *vo* = *VCE* in this case. - -**Answer:** 4.691 V, 888.5 mV. - -Practice Problem 3.12 - -For Practice Prob. 3.12. - -For the BJT circuit in Fig. 3.43, *β* = 150 and *VBE* = 0.7 V. Find *vo*. - -# **Solution:** - -- 1. **Define.** The circuit is clearly defined and the problem is clearly stated. There appear to be no additional questions that need to be asked. -- 2. **Present.** We are to determine the output voltage of the circuit shown in Fig. 3.43. The circuit contains an ideal transistor with *β* = 150 and *VBE* = 0.7 V. -- 3. **Alternative.** We can use mesh analysis to solv e for *vo*. We can re place the transistor with its equivalent circuit and use nodal analysis. We can try both approaches and use them to check each other . As a third check, we can use the equi valent circuit and solv e it using *PSpice*. - -4. **Attempt.** - -■ **METHOD 1** Working with Fig. 3.44(a), we start with the first loop. - -−2 + 100k*I*1 + 200k(*I*1 − I2) = 0 or 3*I*1 − 2*I*2 = 2 × 10−5 **(3.13.1)** - -**Figure 3.44** Solution of the problem in Example 3.13: (a) Method 1, (b) Method 2, (c) Method 3. - -Example 3.13 - -1 kΩ - -Now for loop 2. - -$$ -200k(I_2 - I_1) + V_{BE} = 0 \qquad \text{or} \qquad -2I_1 + 2I_2 = -0.7 \times 10^{-5} -$$ -\n(3.13.2) - -Since we have two equations and two unknowns, we can solve for *I*1 and *I*2. Adding Eq. (3.13.1) to (3.13.2) we get; - -*I*1 = 1.3 × 10−5 A and *I*2 = (−0.7 + 2.6)10−5 ∕2 = 9.5 *µ*A Since *I*3 = −150*I*2 = −1.425 mA, we can now solve for *vo* using loop 3: −*vo* + 1 k*I*3 + 16 = 0 or *vo* = −1.425 + 16 = **14.575 V** - -■ **METHOD 2** Replacing the transistor with its equivalent circuit produces the circuit shown in Fig. 3.44(b). We can now use nodal analysis to solve for *vo*. - -At node number 1: *V*1 = 0.7 V - -$$ -(0.7 - 2)/100k + 0.7/200k + I_B = 0 \qquad \text{or} \qquad I_B = 9.5 \,\mu\text{A} -$$ - -At node number 2 we have: - -150*IB* + (*vo* − 16)∕1k = 0 or *vo* = 16 − 150 × 103 × 9.5 × 10−6 = **14.575 V** - -- 5. **Evaluate.** The answers check, b ut to further check we can use *PSpice* (Method 3), which gi ves us the solution sho wn in Fig. 3.44(c). -- 6. **Satisfactory?** Clearly, we have obtained the desired answer with a very high confidence level. We can now present our work as a solution to the problem. - -Practice Problem 3.13 - -The transistor circuit in Fig. 3.45 has *β* = 80 and *VBE* = 0.7 V. Find *vo* and *Io*. - -**Figure 3.45** For Practice Prob. 3.13. **Answer:** 12 V, 600 *µ* A. - -- 1. Nodal analysis is the application of Kirchhoff's current law at the nonreference nodes. (It is applicable to both planar and nonplanar circuits.) We express the result in terms of the node voltages. Solv ing the simultaneous equations yields the node voltages. -- 2. A supernode consists of two nonreference nodes connected by a (dependent or independent) voltage source. -- 3. Mesh analysis is the application of Kirchhoff's voltage law around meshes in a planar circuit. We express the result in terms of mesh currents. Solving the simultaneous equations yields the mesh currents. - -- 4. A supermesh consists of two meshes that have a (dependent or independent) current source in common. -- 5. Nodal analysis is normally used when a circuit has fewer node equations than mesh equations. Mesh analysis is normally used when a circuit has fewer mesh equations than node equations. -- 6. Circuit analysis can be carried out using *PSpice*. -- 7. DC transistor circuits can be analyzed using the techniques covered in this chapter. - -# Review Questions - -**3.1** At node 1 in the circuit of Fig. 3.46, applying KCL gives: - -(a) -$$ -2 + \frac{12 - v_1}{3} = \frac{v_1}{6} + \frac{v_1 - v_2}{4} -$$ - -\n(b) $2 + \frac{v_1 - 12}{3} = \frac{v_1}{6} + \frac{v_2 - v_1}{4}$ -\n(c) $2 + \frac{12 - v_1}{3} = \frac{0 - v_1}{6} + \frac{v_1 - v_2}{4}$ -\n(d) $2 + \frac{v_1 - 12}{3} = \frac{0 - v_1}{6} + \frac{v_2 - v_1}{4}$ - -# **Figure 3.46** - -- For Review Questions 3.1 and 3.2. -- **3.2** In the circuit of Fig. 3.46, applying KCL at node 2 gives: - -**3.3** For the circuit in Fig. 3.47, *v*1 and *v*2 are related as: - -(a) -$$ -v_1 = 6i + 8 + v_2 -$$ - -\n(b) $v_1 = 6i - 8 + v_2$ -\n(c) $v_1 = -6i + 8 + v_2$ -\n(d) $v_1 = -6i - 8 + v_2$ - -# **Figure 3.47** - -For Review Questions 3.3 and 3.4. - -**3.4** In the circuit of Fig. 3.47, the voltage *v*2 is: - -| (a) −8 V | (b) −1.6 V | -|-----------|------------| -| (c) 1.6 V | (d) 8 V | - -**3.5** The current *i* in the circuit of Fig. 3.48 is: - -| (a) −2.667 A | (b) −0.667 A | -|--------------|--------------| -| (c) 0.667 A | (d) 2.667 A | - -# **Figure 3.48** - -For Review Questions 3.5 and 3.6. - -- **3.6** The loop equation for the circuit in Fig. 3.48 is: - - (a) −10 + 4*i* + 6 + 2*i* = 0 (b) 10 + 4*i* + 6 + 2*i* = 0 (c) 10 + 4*i* − 6 + 2*i* = 0 (d) −10 + 4*i* − 6 + 2*i* = 0 - -- **3.7** In the circuit of Fig. 3.49, current *i*1 is: - - (a) 4 A (b) 3 A (c) 2 A (d) 1 A - -# **Figure 3.49** - -For Review Questions 3.7 and 3.8. - -**3.8** The voltage *v* across the current source in the circuit of Fig. 3.49 is: - -| (a) 20 V | (b) 15 V | (c) 10 V | (d) 5 V | -|----------|----------|----------|---------| -|----------|----------|----------|---------| - -**3.9** The *PSpice* part name for a current-controlled voltage source is: - -(a) EX (b) FX (c) HX (d) GX - -- **3.10** Which of the following statements are not true of the pseudocomponent IPROBE: - - (a) It must be connected in series. - - (b) It plots the branch current. - - (c) It displays the current through the branch in which it is connected. - - (d) It can be used to display voltage by connecting it in parallel. - - (e) It is used only for dc analysis. - - (f) It does not correspond to a particular circuit element. - -*Answers: 3.1a, 3.2c, 3.3a, 3.4c, 3.5c, 3.6a, 3.7d, 3.8b, 3.9c, 3.10b,d.* - -# Problems - -# Sections 3.2 and 3.3 Nodal Analysis - -**3.1** Using Fig. 3.50, design a problem to help other students better understand nodal analysis. - -# **Figure 3.50** - -For Prob. 3.1 and Prob. 3.39. - -**3.2** For the circuit in Fig. 3.51, obtain *v*1 and *v*2. - -**Figure 3.51** For Prob. 3.2. - -**3.3** Find the currents *I*1 through *I*4 and the voltage *vo* in the circuit of Fig. 3.52. - -**Figure 3.52** For Prob. 3.3. - -**3.4** Given the circuit in Fig. 3.53, calculate the currents *i*1 through *i*4. - -For Prob. 3.4. - -**3.5** Obtain *vo* in the circuit of Fig. 3.54. - -For Prob. 3.5. - -**3.6** Solve for *V*1 in the circuit of Fig. 3.55 using nodal analysis. - -**3.7** Apply nodal analysis to solve for *Vx* in the circuit of Fig. 3.56. - -For Prob. 3.7. - -**3.8** Using nodal analysis, find *vo* in the circuit of Fig. 3.57. - -**Figure 3.57** For Prob. 3.8 and Prob. 3.37. - -**3.9** Determine *Ib* in the circuit in Fig. 3.58 using nodal analysis. - -**Figure 3.58** For Prob. 3.9. - -**3.10** Find *Io* in the circuit of Fig. 3.59. - -**Figure 3.59** For Prob. 3.10. - -**3.11** Find *Vo* and the power dissipated in all the resistors in the circuit of Fig. 3.60. - -**Figure 3.60** For Prob. 3.11. - -**3.12** Using nodal analysis, determine *Vo* in the circuit in Fig. 3.61. - -**Figure 3.61** For Prob. 3.12. - -**3.13** Calculate *v*1 and *v*2 in the circuit of Fig. 3.62 using nodal analysis. - -**Figure 3.62** - -For Prob. 3.13. - -**3.14** Using nodal analysis, find *vo* in the circuit of Fig. 3.63. - -# **Figure 3.63** - -For Prob. 3.14. - -**3.15** Apply nodal analysis to find *io* and the power dissipated in each resistor in the circuit of Fig. 3.64. - -For Prob. 3.15. - -**3.16** Determine voltages *v*1 through *v*3 in the circuit of Fig. 3.65 using nodal analysis. - -For Prob. 3.16. - -**3.17** Using nodal analysis, find current *io* in the circuit of Fig. 3.66. - -**Figure 3.66** For Prob. 3.17. - -> **3.18** Determine the node voltages in the circuit in Fig. 3.67 using nodal analysis. - -For Prob. 3.18. - -**3.19** Use nodal analysis to find *v*1, *v*2, and *v*3 in the circuit of Fig. 3.68. - -**Figure 3.68** For Prob. 3.19. - -**3.20** For the circuit in Fig. 3.69, find *v*1, *v*2, and *v*3 using nodal analysis. - -# **Figure 3.69** - -For Prob. 3.20. - -**3.21** For the circuit in Fig. 3.70, find *v*1 and *v*2 using nodal analysis. - -# **Figure 3.70** - -For Prob. 3.21. - -**3.22** Determine *v*1 and *v*2 in the circuit of Fig. 3.71. - -# **Figure 3.71** For Prob. 3.22. - -**3.23** Use nodal analysis to find *Vo* in the circuit of Fig. 3.72. - -For Prob. 3.23. - -**3.24** Use nodal analysis and *MATLAB* to find *Vo* in the circuit of Fig. 3.73. - -# **Figure 3.73** For Prob. 3.24. - -- **3.25** Use nodal analysis along with *MATLAB* to determine - - the node voltages in Fig. 3.74. - -**3.26** Calculate the node voltages *v*1, *v*2, and *v*3 in the circuit of Fig. 3.75. - -**3.27** Use nodal analysis to determine voltages *v*1, *v*2, and *v*3 in the circuit of Fig. 3.76. \* - -# **Figure 3.76** - -For Prob. 3.27. - -**3.28** Use *MATLAB* to find the voltages at nodes *a*, *b*, *c*, and *d* in the circuit of Fig. 3.77. \* - -# **Figure 3.77** - -For Prob. 3.28. - -**3.29** Use *MATLAB* to solve for the node voltages in the circuit of Fig. 3.78. - -# **Figure 3.78** - -For Prob. 3.29. - -\* An asterisk indicates a challenging problem. - -**3.30** Using nodal analysis, find *vo* and *io* in the circuit of Fig. 3.79. - -**3.31** Find the node voltages for the circuit in Fig. 3.80. - -**Figure 3.80** For Prob. 3.31. - -**3.32** Obtain the node voltages *v*1, *v*2, and *v*3 in the circuit of Fig. 3.81. - -**Figure 3.81** For Prob. 3.32. - -# Sections 3.4 and 3.5 Mesh Analysis - -**3.33** Which of the circuits in Fig. 3.82 is planar? For the planar circuit, redraw the circuits with no crossing branches. - -**3.34** Determine which of the circuits in Fig. 3.83 is planar and redraw it with no crossing branches. - -# **Figure 3.83** - -For Prob. 3.34. - -- **3.35** Rework Prob. 3.5 using mesh analysis. -- **3.36** Use mesh analysis to obtain *ia*, *ib*, and *ic* in the circuit in Fig. 3.84. - -**Figure 3.84** For Prob. 3.36. - -**3.37** Solve Prob. 3.8 using mesh analysis. - -**3.38** Apply mesh analysis to the circuit in Fig. 3.85 and obtain *Io*. - -**Figure 3.85** For Prob. 3.38. - -**3.39** Using Fig. 3.50 from Prob. 3.1, design a problem to help other students better understand mesh analysis. - -**3.40** For the bridge network in Fig. 3.86, find *io* using mesh analysis. - -**3.41** Apply mesh analysis to find *i* in Fig. 3.87. - -**Figure 3.87** For Prob. 3.41. - -# **Figure 3.88** - -For Prob. 3.42. - -**3.43** Use mesh analysis to find *vab* and *io* in the circuit of Fig. 3.89. - -**Figure 3.89** For Prob. 3.43. - -**3.44** Use mesh analysis to obtain *io* in the circuit of Fig. 3.90. - -**Figure 3.90** For Prob. 3.44. - -**3.45** Find current *i* in the circuit of Fig. 3.91. - -For Prob. 3.45. - -**3.46** Calculate the mesh currents *i*1 and *i*2 in Fig. 3.92. - -For Prob. 3.46. - -**3.47** Rework Prob. 3.19 using mesh analysis. - -**3.48** Determine the current through the 10-kΩ resistor in the circuit of Fig. 3.93 using mesh analysis. - -**3.49** Find *vo* and *io* in the circuit of Fig. 3.94. - -**Figure 3.94** For Prob. 3.49. - -**3.50** Use mesh analysis to find the current *io* in the circuit of Fig. 3.95. - -For Prob. 3.50. - -**3.51** Apply mesh analysis to find *vo* in the circuit of Fig. 3.96. - -**Figure 3.96** For Prob. 3.51. - -**3.52** Use mesh analysis to find *i*1, *i*2, and *i*3 in the circuit of Fig. 3.97. - -**3.53** Find the mesh currents in the circuit of Fig. 3.98 using *MATLAB*. - -**Figure 3.98** For Prob. 3.53. - -**3.54** Find the mesh currents *i*1, *i*2, and *i*3 in the circuit in Fig. 3.99. - -**Figure 3.99** - -For Prob. 3.54. - -# **Figure 3.100** For Prob. 3.55. - -**3.56** Determine *v*1 and *v*2 in the circuit of Fig. 3.101. - -# **Figure 3.101** - -For Prob. 3.56. - -**3.57** In the circuit of Fig. 3.102, find the values of *R*, *V*1, and *V*2 given that *io* = 15 mA. - -**3.58** Find *i*1, *i*2, and *i*3 in the circuit of Fig. 3.103. - -# **Figure 3.103** - -For Prob. 3.58. - -**3.59** Rework Prob. 3.30 using mesh analysis. - -**3.60** Calculate the power dissipated in each resistor in the circuit of Fig. 3.104. - -# **Figure 3.104** - -For Prob. 3.60. - -**3.61** Calculate the current gain *io*∕*is* in the circuit of Fig. 3.105. - -**Figure 3.105** - -For Prob. 3.61. - -**3.62** Find the mesh currents *i*1, *i*2, and *i*3 in the network of Fig. 3.106. - -For Prob. 3.62. - -**3.63** Find *vx* and *ix* in the circuit shown in Fig. 3.107. - -**3.66** Write a set of mesh equations for the circuit in Fig. 3.110. Use *MATLAB* to determine the mesh currents. - -# **Figure 3.110** For Prob. 3.66. - -# Section 3.6 Nodal and Mesh Analyses by Inspection - -**3.67** Obtain the node-voltage equations for the circuit in Fig. 3.111 by inspection. Then solve for *Vo*. - -# **Figure 3.111** - -For Prob. 3.67. - -**3.68** Using Fig. 3.112, design a problem, to solve for *Vo*, to help other students better understand nodal analysis. Try your best to come up with values to make the calculations easier. - -**Figure 3.108** For Prob. 3.64. - -**3.65** Use *MATLAB* to solve for the mesh currents in the circuit of Fig. 3.109. 6 V - -**Figure 3.109** For Prob. 3.65. - -**3.69** For the circuit shown in Fig. 3.113, write the nodevoltage equations by inspection. - -For Prob. 3.69. - -**3.70** Write the node-voltage equations by inspection and then determine values of *V*1 and *V*2 in the circuit of Fig. 3.114. - -For Prob. 3.70. - -**3.71** Write the mesh-current equations for the circuit in Fig. 3.115. Next, determine the values of *i*1, *i*2, and *i*3. - -For Prob. 3.71. - -**3.72** By inspection, write the mesh-current equations for the circuit in Fig. 3.116. - -# **Figure 3.116** - -For Prob. 3.72. - -**3.73** Write the mesh-current equations for the circuit in Fig. 3.117. - -# **Figure 3.117** - -For Prob. 3.73. - -**3.74** By inspection, obtain the mesh-current equations for the circuit in Fig. 3.118. - -# **Figure 3.118** For Prob. 3.74. - -Section 3.8 Circuit Analysis with PSpice or MultiSim - -- **3.75** Use *PSpice* or *MultiSim* to solve Prob. 3.58. -- **3.76** Use *PSpice* or *MultiSim* to solve Prob. 3.27. - -**3.77** Solve for *V*1 and *V*2 in the circuit of Fig. 3.119 using *PSpice* or *MultiSim*. - -# **Figure 3.119** - -For Prob. 3.77. - -- **3.78** Solve Prob. 3.20 using *PSpice* or *MultiSim*. -- **3.79** Rework Prob. 3.28 using *PSpice* or *MultiSim*. -- **3.80** Find the nodal voltages *v*1 through *v*4 in the circuit of Fig. 3.120 using *PSpice* or *MultiSim*. - -# **Figure 3.120** - -For Prob. 3.80. - -- **3.81** Use *PSpice* or *MultiSim* to solve the problem in Example 3.4. -- **3.82** If the Schematics Netlist for a network is as follows, draw the network. - -| R_R1 | 1 | 2 | 2K | | -|---------|---|---|--------------|--------------| -| R_R2 | 2 | 0 | 4K | | -| R_R3 | 3 | 0 | 8K | | -| R R4 | 3 | 4 | 6K | | -| R_R5 | 1 | 3 | 3K | | -| V_VS | 4 | 0 | DC | 100 | -| I IS. | 0 | 1 | DC | 4 | -| $F_F1$ | 1 | 3 | VF F1 | $\mathsf{Z}$ | -| $VF_F1$ | 5 | 0 | 0V | | -| E E1 | 3 | 2 | $\mathbf{1}$ | 3 | -| | | | | | - -**3.83** The following program is the Schematics Netlist of a particular circuit. Draw the circuit and determine the voltage at node 2. - -| $\frac{1}{2}$ | | | | | -|---------------|--------------|---|-----|--| -| R R1 | 1 | 2 | 20 | | -| R R2 | $\mathsf{Z}$ | Ø | 50 | | -| R R3 | 2 | 3 | 70 | | -| R R4 | 3. | 0 | 30 | | -| v vs | 1 | a | 20V | | -| I IS | 2 | a | DC. | | - -# Section 3.9 Applications - -**3.84** Calculate *vo* and *Io* in the circuit of Fig. 3.121. - -- **3.85** An audio amplifier with a resistance of 9 Ω supplies power to a speaker. What should be the resistance of the speaker for maximum power to be delivered? -- **3.86** For the simplified transistor circuit of Fig. 3.122, calculate the voltage *vo*. - -**Figure 3.122** For Prob. 3.86. - -**3.87** For the circuit in Fig. 3.123, find the gain *vo*∕*vs*. - -**3.88** Determine the gain *vo*∕*vs* of the transistor amplifier circuit in Fig. 3.124. \* - -**Figure 3.124** - -- For Prob. 3.88. - - **3.89** For the transistor circuit shown in Fig. 3.125, find *IB* and *VCE*. Let *β* = 100, and *VBE* = 0.7 V. - -**Figure 3.125** - -For Prob. 3.89. - -**3.90** Calculate *vs* for the transistor in Fig. 3.126 given that *vo* = 6 V, *β* = 90, *VBE* = 0.7 V. - -**Figure 3.126** For Prob. 3.90. - -\***3.93** Rework Example 3.11 with hand calculation. - -**3.91** For the transistor circuit of Fig. 3.127, find *IB*, *VCE*, and *vo*. Take *β* = 150, *VBE* = 0.7 V. - -**3.92** Using Fig. 3.128, design a problem to help other students better understand transistors. Make sure you use reasonable numbers! - -**Figure 3.128** For Prob. 3.92. - -# **chapter** - -4 - -# Circuit Theorems - -*Your success as an engineer will be directly proportional to your ability to communicate!* - -—Charles K. Alexander - -# Enhancing Your Skills and Your Career - -# **Enhancing Your Communication Skills** - -Taking a course in circuit analysis is one step in preparing yourself for a career in electrical engineering. Enhancing your communication skills while in school should also be part of that preparation, as a large part of your time will be spent communicating. - -People in industry have complained again and again that graduating engineers are ill-prepared in written and oral communication. An engineer who communicates effectively becomes a valuable asset. - -You can probably speak or write easily and quickly. But how *effectively* do you communicate? The art of effective communication is of the utmost importance to your success as an engineer. - -For engineers in industry , communication is k ey to promotability . Consider the result of a survey of U.S. corporations that asked what factors influence managerial promotion. The survey includes a listing of 22 personal qualities and their importance in adv ancement. You may be surprised to note that "technical skill based on experience" placed fourth from the bottom. Attributes such as self-confidence, ambition, flexibility, maturity, ability to mak e sound decisions, getting things done with and through people, and capacity for hard work all ranked higher. At the top of the list w as "ability to communicate. " The higher your professional career progresses, the more you will need to communicate. Therefore, you should regard effective communication as an important tool in your engineering tool chest. - -Learning to communicate ef fectively is a lifelong task you should always work toward. The best time to begin is while still in school. Continually look for opportunities to de velop and strengthen your reading, writing, listening, and speaking skills. You can do this through classroom presentations, team projects, acti ve participation in student or ganizations, and enrollment in communication courses. The risks are less now than later in the workplace. - -*Ability to communicate effectively is regarded by many as the most important step to an executive promotion.* © IT Stock/PunchStock RF - -# Learning Objectives - -By using the information and exercises in this chapter you will be able to: - -- 1. Develop and enhance your skills in using nodal analysis and mesh analysis to analyze basic circuits. -- 2. Understand how linearity works with basic circuits. -- 3. Explain the principle of superposition and how it can be used to help analyze circuits. -- 4. Understand the value of source transformation and how it can be used to simplify circuits. -- 5. Recognize Thevenin's and Norton's theorems and know how they can lead to greatly simplified circuits. -- 6. Explain the maximum power transfer concept. - -# **4.1** Introduction - -A major adv antage of analyzing circuits using Kirchhof f's laws as we did in Chapter 3 is that we can analyze a circuit without tampering with its original configuration. A major disadvantage of this approach is that, for a large, complex circuit, tedious computation is involved. - -The growth in areas of application of electric circuits has led to an evolution from simple to complex circuits. To handle the complexity, engineers over the years have developed some theorems to simplify circuit analysis. Such theorems include Thevenin's and Norton's theorems. Since these theorems are applicable to *linear* circuits, we first discuss the concept of circuit linearity. In addition to circuit theorems, we discuss the concepts of superposition, source transformation, and maximum power transfer in this chapter. The concepts we de velop are applied in the last section to source modeling and resistance measurement. - -# **4.2** Linearity Property - -Linearity is the property of an element describing a linear relationship between cause and effect. Although the property applies to many circuit elements, we shall limit its applicability to resistors in this chapter. The property is a combination of both the homogeneity (scaling) property and the additivity property. - -The homogeneity property requires that if the input (also called the *excitation*) is multiplied by a constant, then the output (also called the *response*) is multiplied by the same constant. For a resistor, for example, Ohm's law relates the input *i* to the output *v*, - -$$ -v = iR \tag{4.1} -$$ - -If the current is increased by a constant *k*, then the voltage increases correspondingly by *k*; that is, - -$$ -kiR = kv \tag{4.2} -$$ - -The additivity property requires that the response to a sum of inputs is the sum of the responses to each input applied separately . Using the voltage-current relationship of a resistor, if - -$$ -v_1 = i_1 R \tag{4.3a} -$$ - -and - -$$ -v_2 = i_2 R \tag{4.3b} -$$ - -then applying (*i*1 + *i*2) gives - -$$ -v = (i_1 + i_2)R = i_1R + i_2R = v_1 + v_2 -$$ -\n(4.4) - -We say that a resistor is a linear element because the voltage-current relationship satisfies both the homogeneity and the additivity properties. - -In general, a circuit is linear if it is both additive and homogeneous. A linear circuit consists of only linear elements, linear dependent sources, and independent sources. - -A linear circuit is one whose output is linearly related (or directly proportional) to its input. - -Throughout this book we consider only linear circuits. Note that since *p* = *i* 2 *R* = *v* 2 ∕*R* (making it a quadratic function rather than a linear one), the relationship between power and voltage (or current) is nonlinear . Therefore, the theorems covered in this chapter are not applicable to power. - -To illustrate the linearity principle, consider the linear circuit shown in Fig. 4.1. The linear circuit has no independent sources inside it. It is excited by a v oltage source *vs*, which serv es as the input. The circuit is terminated by a load *R*. We may tak e the current *i* through *R* as the output. Suppose *vs* = 10 V gives *i* = 2 A. According to the linearity principle, *vs* = 1 V will give *i* = 0.2 A. By the same token, *i* = 1 mA must be due to *vs* = 5 mV. - -For example, when current i1 flows through resistor R, the power is p1 = Ri1 2 , and when current i2 flows through R, the power is p2 = Ri2 2 . If current i1 + i2 flows through R, the power absorbed is p3 = R (i1 + i2) 2 = Ri1 2 + Ri2 2 + 2Ri1i2 ≠ p1 + p2. Thus, the power relation is nonlinear. - -# **Figure 4.1** - -A linear circuit with input *vs* and output *i*. - -For the circuit in Fig. 4.2, find *Io* when *vs* = 12 V and *vs* = 24 V. Example 4.1 - -# **Solution:** - -Applying KVL to the two loops, we obtain - -$$ -12i1 - 4i2 + vs = 0 -$$ -\n(4.1.1) -\n -$$ --4i1 + 16i2 - 3vx - vs = 0 -$$ -\n(4.1.2) - -But *vx* = 2*i*1. Equation (4.1.2) becomes - -$$ --10i_1 + 16i_2 - v_s = 0 \tag{4.1.3} -$$ - -Adding Eqs. (4.1.1) and (4.1.3) yields - -$$ -2i_1 + 12i_2 = 0 \qquad \Rightarrow \qquad i_1 = -6i_2 -$$ - -Substituting this in Eq. (4.1.1), we get - -$$ --76i_2 + v_s = 0 \qquad \Rightarrow \qquad i_2 = \frac{v_s}{76} -$$ - -When *vs* = 12 V, - -When *vs* = 24 V, - -$$ -I_o = i_2 = \frac{24}{76} -$$ - A - -\_\_\_ 12 76 A - -*Io* = *i*2 = - -showing that when the source value is doubled, *Io* doubles. - -For Practice Prob. 4.1. - -Example 4.2 - -Assume *Io* = 1 A and use linearity to find the actual value of *Io* in the - -# **Figure 4.4** - -circuit of Fig. 4.4. - -**Answer:** 40 V, 60 V. - -# **Solution:** - -**Answer:** 16 V. - -If *Io* = 1 A, then *V*1 = (3 + 5)*Io* = 8 V and *I*1 = *V*1∕4 = 2 A. Applying KCL at node 1 gives - -$$ -I_2 = I_1 + I_o = 3 \text{ A} -$$ - -$$ -V_2 = V_1 + 2I_2 = 8 + 6 = 14 \text{ V}, \qquad I_3 = \frac{V_2}{7} = 2 \text{ A} -$$ - -Applying KCL at node 2 gives - -$$ -I_4 = I_3 + I_2 = 5 \text{ A} -$$ - -Therefore, *Is* = 5 A. This shows that assuming *Io* = 1 gives *Is* = 5 A, the actual source current of 15 A will give *Io* = 3 A as the actual value. - -Assume that *Vo* = 1 V and use linearity to calculate the actual value of *Vo* in the circuit of Fig. 4.5. - -Practice Problem 4.2 - -For Practice Prob. 4.2. - -# **4.3** Superposition - -If a circuit has tw o or more independent sources, one w ay to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis as in Chapter 3. Another way is to determine the contribution of each independent source to the v ariable and then add them up. The latter approach is known as the *superposition principle*. - -The idea of superposition rests on the linearity property. - -The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone. - -The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contrib ution of each independent source separately. However, to apply the superposition principle, we must keep two things in mind: - -- 1. We consider one independent source at a time while all other independent sources are *turned off*. This implies that we replace e very voltage source by 0 V (or a short circuit), and e very current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit. -- 2. Dependent sources are left intact because the y are controlled by circuit variables. - -With these in mind, we apply the superposition principle in three steps: - -# Steps to Apply Superposition Principle: - -- 1. Turn off all independent sources e xcept one source. Find the output (voltage or current) due to that active source using the techniques covered in Chapters 2 and 3. -- 2. Repeat step 1 for each of the other independent sources. -- 3. Find the total contribution by adding algebraically all the contributions due to the independent sources. - -Analyzing a circuit using superposition has one major disadvantage: It may very likely involve more work. If the circuit has three independent sources, we may ha ve to analyze three simpler circuits each pro viding the contribution due to the respective individual source. However, superposition does help reduce a comple x circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits. - -Keep in mind that superposition is based on linearity . For this reason, it is not applicable to the ef fect on po wer due to each source, be cause the po wer absorbed by a resistor depends on the square of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition. - - Superposition is not limited to circuit analysis but is applicable in many fields where cause and effect bear a linear relationship to one another. - -Other terms such as killed, made inactive, deadened, or set equal to zero are often used to convey the same idea. - -**Figure 4.6** For Example 4.3. - -8 Ω - -i 2 i 3 - -4 Ω - -(b) - -Example 4.3 Use the superposition theorem to find *v* in the circuit of Fig. 4.6. - -# **Solution:** - -Since there are two sources, let - -$$ -v = v_1 + v_2 -$$ - -where *v*1 and *v*2 are the contributions due to the 6-V v oltage source and the 3-A current source, respecti vely. To obtain *v*1, we set the current source to zero, as sho wn in Fig. 4.7(a). Applying KVL to the loop in Fig. 4.7(a) gives - -$$ -12i_1 - 6 = 0 \qquad \Rightarrow \qquad i_1 = 0.5 \text{ A} -$$ - -Thus, - -$$ -v_1 = 4i_1 = 2 \text{ V} -$$ - -We may also use voltage division to get *v*1 by writing - -$$ -v_1 = \frac{4}{4+8}(6) = 2 \text{ V} -$$ - -To get *v*2, we set the voltage source to zero, as in Fig. 4.7(b). Using current division, - -$$ -i_3 = \frac{8}{4+8}(3) = 2 \text{ A} -$$ - -Hence, - -3 A - -*v*2 - -+ – - -*v*2 = 4*i*3 = 8 V - -And we find - -**Answer:** 16 V. - -$$ -v = v_1 + v_2 = 2 + 8 = 10 \text{ V} -$$ - -**Figure 4.7** For Example 4.3: (a) calculating *v*1, (b) calculating *v*2. - -**Figure 4.8** For Practice Prob. 4.3. - -Find *io* in the circuit of Fig. 4.9 using superposition. - -# **Solution:** - -The circuit in Fig. 4.9 in volves a dependent source, which must be left intact. We let - -$$ -i_o = i'_o + i''_o \tag{4.4.1} -$$ - -where *i*′ *o* and *i o* ″ are due to the 4-A current source and 20-V voltage source respectively. To obtain *i*′ *o*, we turn off the 20-V source so that we have the circuit in Fig. 4.10(a). We apply mesh analysis in order to obtain *i*′ *o*. For loop 1, - -$$ -i_1 = 4 \text{ A} \tag{4.4.2} -$$ - -For loop 2, - -$$ -l_1 = 4 \,\mathrm{A} -$$ - -$$ --3i_1 + 6i_2 - 1i_3 - 5i'_o = 0 \tag{4.4.3} -$$ - -# **Figure 4.10** - -For Example 4.4: Applying superposition to (a) obtain *i*′ *o*, (b) obtain *i* ″ *o*. - -For loop 3, - -−5*i*1 − 1*i*2 + 10*i*3 + 5*i*′ *o* = 0 **(4.4.4)** But at node 0, *i*3 = *i*1 − *i*′ *o* = 4 − *i*′ *o* **(4.4.5)** - -Substituting Eqs. (4.4.2) and (4.4.5) into Eqs. (4.4.3) and (4.4.4) gi ves two simultaneous equations - -$$ -3i_2 - 2i'_o = 8 \tag{4.4.6} -$$ - -$$ -i_2 + 5i'_o = 20 \tag{4.4.7} -$$ - -which can be solved to get - -$$ -i'_{o} = \frac{52}{17} A -$$ - (4.4.8) - -To obtain *i*″ *o*, we turn of f the 4-A current source so that the circuit becomes that shown in Fig. 4.10(b). For loop 4, KVL gives - -$$ -6i_4 - i_5 - 5i''_o = 0 \tag{4.4.9} -$$ - -and for loop 5, - -$$ --i_4 + 10i_5 - 20 + 5i''_o = 0 \tag{4.4.10} -$$ - -But *i*5 = −*i*″ *o*. Substituting this in Eqs. (4.4.9) and (4.4.10) gives - -$$ -6i_4 - 4i''_o = 0 \tag{4.4.11} -$$ - -$$ -i_4 + 5i''_o = -20 \tag{4.4.12} -$$ - -which we solve to get - -$$ -i_o'' = -\frac{60}{17} \,\mathrm{A} \tag{4.4.13} -$$ - -Now substituting Eqs. (4.4.8) and (4.4.13) into Eq. (4.4.1) gives - -$$ -i_o = -\frac{8}{17} = -0.4706 \text{ A} -$$ - -Example 4.5 - -For the circuit in Fig. 4.12, use the superposition principle to find *i*. - -# **Solution:** - -In this case, we have three sources. Let - -$$ -i = i_1 + i_2 + i_3 -$$ - -where *i*1, *i*2, and *i*3 are due to the 12-V , 24-V, and 3-A sources respec tively. To get *i*1, consider the circuit in Fig. 4.13(a). Combining 4 Ω (on the right-hand side) in series with 8 Ω gives 12 Ω. The 12 Ω in parallel with 4 Ω gives 12 × 4∕16 = 3 Ω. Thus, - -$$ -i_1 = \frac{12}{6} = 2 \text{ A} -$$ - -To get *i*2, consider the circuit in Fig. 4.13(b). Applying mesh analysis gives - -$$ -16i_a - 4i_b + 24 = 0 \qquad \Rightarrow \qquad 4i_a - i_b = -6 \tag{4.5.1} -$$ - -$$ -\mathcal{T}_b - 4i_a = 0 \qquad \Rightarrow \qquad i_a = \frac{7}{4} i_b \tag{4.5.2} -$$ - -Substituting Eq. (4.5.2) into Eq. (4.5.1) gives - -$$ -i_2 = i_b = -1 -$$ - -To get *i*3, consider the circuit in Fig. 4.13(c). Using nodal analysis gives - -$$ -3 = \frac{v_2}{8} + \frac{v_2 - v_1}{4} \qquad \Rightarrow \qquad 24 = 3v_2 - 2v_1 \tag{4.5.3} -$$ - -$$ -\frac{v_2 - v_1}{4} = \frac{v_1}{4} + \frac{v_1}{3} \qquad \Rightarrow \qquad v_2 = \frac{10}{3}v_1 \tag{4.5.4} -$$ - -Substituting Eq. (4.5.4) into Eq. (4.5.3) leads to *v*1 = 3 and - -$$ -i_3 = \frac{v_1}{3} = 1 \text{ A} -$$ - -Thus, - -$$ -i = i_1 + i_2 + i_3 = 2 - 1 + 1 = 2 \text{ A} -$$ - -Find *I* in the circuit of Fig. 4.14 using the superposition principle. Practice Problem 4.5 - -2 Ω - -2 A - -– 6 V + - -8 Ω - -I - -– - -8 V - -+ - -**Figure 4.14** For Practice Prob. 4.5. - -# **4.4** Source Transformation - -6 Ω - -We have noticed that series-parallel combination and wye-delta transformation help simplify circuits. *Source transformation* is another tool for simplifying circuits. Basic to these tools is the concept of *equivalence*. We recall that an equivalent circuit is one whose *v*-*i* characteristics are identical with the original circuit. - -In Section 3.6, we sa w that node-v oltage (or mesh-current) equa tions can be obtained by mere inspection of a circuit when the sources are all independent current (or all independent v oltage) sources. It is therefore expedient in circuit analysis to be able to substitute a v oltage source in series with a resistor for a current source in parallel with a - -resistor, or vice versa, as shown in Fig. 4.15. Either substitution is known as a *source transformation*. - -**Figure 4.15** Transformation of independent sources. - -A source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa. - -The tw o circuits in Fig. 4.15 are equi valent—provided the y ha ve the same voltage-current relation at terminals *a*-*b*. It is easy to sho w that they are indeed equi valent. If the sources are turned off, the equivalent resistance at terminals *a*-*b* in both circuits is *R*. Also, when terminals *a*-*b* are short-circuited, the short-circuit current flowing from *a* to *b* is *isc* = *vs*∕*R* in the circuit on the left-hand side and *isc* = *is* for the circuit on the right-hand side. Thus, *vs*∕*R* = *is* in order for the two circuits to be equivalent. Hence, source transformation requires that - -$$ -v_s = i_s R \qquad \text{or} \qquad i_s = \frac{v_s}{R} \tag{4.5} -$$ - -Source transformation also applies to dependent sources, pro vided we carefully handle the dependent v ariable. As shown in Fig. 4.16, a dependent voltage source in series with a resistor can be transformed to a dependent current source in parallel with the resistor or vice versa where we make sure that Eq. (4.5) is satisfied. - -**Figure 4.16** Transformation of dependent sources. - -Like the wye-delta transformation we studied in Chapter 2, a source transformation does not af fect the remaining part of the circuit. When applicable, source transformation is a po werful tool that allo ws circuit manipulations to ease circuit analysis. However, we should keep the following points in mind when dealing with source transformation. - -- 1. Note from Fig. 4.15 (or Fig. 4.16) that the arrow of the current source is directed toward the positive terminal of the voltage source. -- 2. Note from Eq. (4.5) that source transformation is not possible when *R* = 0, which is the case with an ideal voltage source. However, for a practical, nonideal voltage source, *R* ≠ 0. Similarly, an ideal current source with *R* = ∞ cannot be replaced by a finite voltage source. More will be said on ideal and nonideal sources in Section 4.10.1. - -Use source transformation to find *vo* in the circuit of Fig. 4.17. - -# **Solution:** - -We first transform the current and voltage sources to obtain the cir cuit in Fig. 4.18(a). Combining the 4-Ω and 2-Ω resistors in series and transforming the 12-V v oltage source gi ves us Fig. 4.18(b). We now combine the 3-Ω and 6-Ω resistors in parallel to get 2-Ω. We also combine the 2-A and 4-A current sources to get a 2-A source. Thus, by repeatedly applying source transformations, we obtain the circuit in Fig. 4.18(c). - -Example 4.6 - -We use current division in Fig. 4.18(c) to get - -$$ -i = \frac{2}{2+8}(2) = 0.4 \text{ A} -$$ - -and - -$$ -v_o = 8i = 8(0.4) = 3.2 -$$ - V - -Alternatively, since the 8-Ω and 2-Ω resistors in Fig. 4.18(c) are in parallel, they have the same voltage *vo* across them. Hence, - -$$ -v_o = (8 \parallel 2)(2 \text{ A}) = \frac{8 \times 2}{10}(2) = 3.2 \text{ V} -$$ - -Find *io* in the circuit of Fig. 4.19 using source transformation. Practice Problem 4.6 - -**Answer:** 1.78 A. - -# Example 4.7 - -**Figure 4.20** For Example 4.7. Find *vx* in Fig. 4.20 using source transformation. - -# **Solution:** - -The circuit in Fig. 4.20 in volves a v oltage-controlled dependent cur rent source. We transform this dependent current source as well as the 6-V independent v oltage source as sho wn in Fig. 4.21(a). The 18-V voltage source is not transformed because it is not connected in series with any resistor. The two 2-Ω resistors in parallel combine to gi ve a 1-Ω resistor, which is in parallel with the 3-A current source. The current source is transformed to a v oltage source as shown in Fig. 4.21(b). Notice that the terminals for *vx* are intact. Applying KVL around the loop in Fig. 4.21(b) gives - -**Figure 4.21** For Example 4.7: Applying source transformation to the circuit in Fig. 4.20. - -Applying KVL to the loop containing only the 3-V v oltage source, the 1-Ω resistor, and *vx* yields - -$$ --3 + 1i + v_x = 0 \qquad \Rightarrow \qquad v_x = 3 - i \tag{4.7.2} -$$ - -Substituting this into Eq. (4.7.1), we obtain - -$$ -15 + 5i + 3 - i = 0 \qquad \Rightarrow \qquad i = -4.5 \text{ A} -$$ - -Alternatively, we may apply KVL to the loop containing *vx*, the 4-Ω resistor, the v oltage-controlled dependent v oltage source, and the 18-V voltage source in Fig. 4.21(b). We obtain - -$$ --v_x + 4i + v_x + 18 = 0 \qquad \Rightarrow \qquad i = -4.5 \text{ A} -$$ - -Thus, *vx* = 3 − *i* = 7.5 V. - -For Practice Prob. 4.7. - -# **4.5** Thevenin's Theorem - -It often occurs in practice that a particular element in a circuit is variable (usually called the *load*) while other elements are fixed. As a typical example, a household outlet terminal may be connected to dif ferent appliances constituting a v ariable load. Each time the v ariable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenin's theorem pro vides a technique by which the fixed part of the circuit is replaced by an equivalent circuit. - -According to Thevenin's theorem, the linear circuit in Fig. 4.23(a) can be replaced by that in Fig. 4.23(b). (The load in Fig. 4.23 may be a single resistor or another circuit.) The circuit to the left of the terminals *a*-*b* in Fig. 4.23(b) is kno wn as the *Thevenin equivalent cir cuit*; it w as developed in 1883 by M. Leon Thevenin (1857–1926), a French telegraph engineer. - -Thevenin's theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor RTh, where VTh is the open-circuit voltage at the terminals and RTh is the input or equivalent resistance at the terminals when the independent sources are turned off. - -The proof of the theorem will be gi ven later, in Section 4.7. Our major concern right now is how to find the Thevenin equivalent voltage *V*Th and resistance *R*Th. To do so, suppose the tw o circuits in Fig. 4.23 are equivalent. Two circuits are said to be *equivalent* if the y have the same voltage-current relation at their terminals. Let us find out what will make the tw o circuits in Fig. 4.23 equi valent. If the terminals *a*-*b* are made open-circuited (by remo ving the load), no current flows, so that the open-circuit voltage across the terminals *a*-*b* in Fig. 4.23(a) must be equal to the voltage source *V*Th in Fig. 4.23(b), since the two circuits are equivalent. Thus, *V*Th is the open-circuit v oltage across the terminals as shown in Fig. 4.24(a); that is, - -Finding *V*Th and *R*Th. - -Again, with the load disconnected and terminals *a*-*b* opencir cuited, we turn of f all independent sources. The input resistance (or equi valent resistance) of the dead circuit at the terminals *a*-*b* in Fig. 4.23(a) must be equal to *R*Th in Fig. 4.23(b) because the two circuits are equivalent. Thus, *R*Th is the input resistance at the terminals when the independent sources are turned off, as shown in Fig. 4.24(b); that is, - -$$ -R_{\rm Th} = R_{\rm in} \tag{4.7} -$$ - -# **Figure 4.23** Replacing a linear two-terminal circuit - -by its Thevenin equivalent: (a) original circuit, (b) the Thevenin equivalent circuit. - -# **Figure 4.25** - -Finding *R*Th when circuit has dependent sources. - -Later we will see that an alternative way of finding RTh is RTh = voc ∕isc. - -# **Figure 4.26** - -A circuit with a load: (a) original circuit, (b) Thevenin equivalent. - -To apply this idea in finding the Thevenin resistance *R*Th, we need to consider two cases. - -■ **CASE 1** If the network has no dependent sources, we turn of f all independent sources. *R*Th is the input resistance of the netw ork looking between terminals *a* and *b*, as shown in Fig. 4.24(b). - -■ **CASE 2** If the network has dependent sources, we turn of f all independent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables. We apply a voltage source *vo* at terminals *a* and *b* and determine the resulting current *io*. Then *R*Th = *vo*∕*io*, as shown in Fig. 4.25(a). Alternatively, we may insert a current source *io* at terminals *a*-*b* as shown in Fig. 4.25(b) and find the terminal voltage *vo*. Again *R*Th = *vo*∕*io*. Either of the two approaches will give the same result. In either approach we may assume an y value of *vo* and *io*. For example, we may use *vo* = 1 V or *io* = 1 A, or even use unspecified values of *vo* or *io*. - -It often occurs that *R*Th tak es a ne gative v alue. In this case, the negative resistance (*v* = −*iR*) implies that the circuit is supplying power. This is possible in a circuit with dependent sources; Example 4.10 will illustrate this. - -Thevenin's theorem is v ery important in circuit analysis. It helps simplify a circuit. A large circuit may be replaced by a single indepen dent voltage source and a single resistor. This replacement technique is a powerful tool in circuit design. - -As mentioned earlier, a linear circuit with a variable load can be replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same w ay externally as the original circuit. Con sider a linear circuit terminated by a load *RL*, as shown in Fig. 4.26(a). The current *IL* through the load and the voltage *VL* across the load are easily determined once the Thevenin equivalent of the circuit at the load' s terminals is obtained, as sho wn in Fig. 4.26(b). From Fig. 4.26(b), we obtain - -$$ -I_L = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_L} \tag{4.8a} -$$ - -$$ -V_{L} = R_{L}I_{L} = \frac{R_{L}}{R_{\text{Th}} + R_{L}}V_{\text{Th}} -$$ -(4.8b) - -Note from Fig. 4.26(b) that the Thevenin equivalent is a simple v oltage divider, yielding *VL* by mere inspection. - -Find the Thevenin equivalent circuit of the circuit shown in Fig. 4.27, to the left of the terminals *a*-*b*. Then find the current through *RL* = 6, 16, and 36 Ω. - -# **Solution:** - -We find *R*Th by turning of f the 32-V v oltage source (replacing it with a short circuit) and the 2-A current source (replacing it with an - -For Example 4.8. - -open circuit). The circuit becomes what is sho wn in Fig. 4.28(a). Thus, - -For Example 4.8: (a) finding *R*Th, (b) finding *V*Th. - -To find *V*Th, consider the circuit in Fig. 4.28(b). Applying mesh analysis to the two loops, we obtain - -$$ --32 + 4i_1 + 12(i_1 - i_2) = 0, \qquad i_2 = -2 \text{ A} -$$ - -Solving for *i*1, we get *i*1 = 0.5 A. Thus, - -$$ -V_{\text{Th}} = 12(i_1 - i_2) = 12(0.5 + 2.0) = 30 \text{ V} -$$ - -Alternatively, it is even easier to use nodal analysis. We ignore the 1-Ω resistor since no current flows through it. At the top node, KCL gives - -$$ -\frac{32 - V_{\text{Th}}}{4} + 2 = \frac{V_{\text{Th}}}{12} -$$ - -or - -$$ -96 - 3V_{\text{Th}} + 24 = V_{\text{Th}} \qquad \Rightarrow \qquad V_{\text{Th}} = 30 \text{ V} -$$ - -as obtained before. We could also use source transformation to find *V*Th. - -The Thevenin equivalent circuit is sho wn in Fig. 4.29. The current through *RL* is - -$$ -I_L = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_L} = \frac{30}{4 + R_L} -$$ - -When *RL* = 6, - -$$ -I_L = \frac{30}{10} = 3 \text{ A} -$$ - -When *RL* = 16, - -$$ -I_L = \frac{30}{20} = 1.5 \text{ A} -$$ - -When *RL* = 36, - -$$ -I_L = \frac{30}{40} = 0.75 \text{ A} -$$ - -**Figure 4.29** The Thevenin equivalent circuit for Example 4.8. - -# **Figure 4.30** - -For Practice Prob. 4.8. - -Example 4.9 Find the Thevenin equivalent of the circuit in Fig. 4.31 at terminals *a*-*b*. - -# **Solution:** - -This circuit contains a dependent source, unlik e the circuit in the pre vious e xample. To find *R*Th, we set the independent source equal to zero b ut lea ve the dependent source alone. Because of the presence of the dependent source, ho wever, we e xcite the netw ork with a v oltage source *vo* connected to the terminals as indicated in Fig. 4.32(a). We may set *vo* = 1 V to ease calculation, since the circuit is linear . Our goal is to find the current *io* through the terminals, and then obtain *R*Th = 1∕*io*. (Alternatively, we may insert a 1-A current source, find the corresponding voltage *vo*, and obtain *R*Th = *vo*∕1.) - -**Figure 4.31** For Example 4.9. - -**Figure 4.32** Finding *R*Th and *V*Th for Example 4.9. - -Applying mesh analysis to loop 1 in the circuit of Fig. 4.32(a) results in - -−2*vx* + 2(*i*1 − *i*2) = 0 or *vx* = *i*1 − *i*2 - -But −4*i*2 = *vx* = *i*1 − *i*2; hence, - -$$ -i_1 = -3i_2 \tag{4.9.1} -$$ - -For loops 2 and 3, applying KVL produces - -4*i*2 + 2(*i*2 − *i*1) + 6(*i*2 − *i*3) = 0 **(4.9.2)** - -$$ -6(i_3 - i_2) + 2i_3 + 1 = 0 \tag{4.9.3} -$$ - -Solving these equations gives - -$$ -i_3 = -\frac{1}{6} \,\mathrm{A} -$$ - -But *io* = −*i*3 = 1∕6 A. Hence, - -$$ -R_{\text{Th}} = \frac{1 \text{ V}}{i_o} = 6 \text{ }\Omega -$$ - -To get *V*Th, we find *voc* in the circuit of Fig. 4.32(b). Applying mesh analysis, we get - -$$ -i_1 = 5 \tag{4.9.4} -$$ - -$$ --2v_x + 2(i_3 - i_2) = 0 \Rightarrow v_x = i_3 - i_2 -$$ - (4.9.5) -$$ -4(i_2 - i_1) + 2(i_2 - i_3) + 6i_2 = 0 -$$ - -or - -$$ -12i_2 - 4i_1 - 2i_3 = 0 \tag{4.9.6} -$$ - -But 4(*i*1 − *i*2) = *vx*. Solving these equations leads to *i*2 = 10∕3. Hence, - -$$ -V_{\rm Th} = v_{oc} = 6i_2 = 20 \, \text{V} -$$ - -The Thevenin equivalent is as shown in Fig. 4.33. - -# **Figure 4.33** - -The Thevenin equivalent of the circuit in Fig. 4.31. - -Find the Thevenin equivalent circuit of the circuit in Fig. 4.34 to the left of the terminals. - -**Answer:** *V*Th = 5.333 V, *R*Th = 444.4 mΩ. - -6 V 5 Ω 3 Ω 4 Ω 1.5I x + – I x - -a - -b - -Practice Problem 4.9 - -Determine the Thevenin equi valent of the circuit in Fig. 4.35(a) at Example 4.10 terminals *a*-*b*. - -# **Solution:** - -- 1. **Define.** The problem is clearly defined; we are to determine the Thevenin equivalent of the circuit shown in Fig. 4.35(a). -- 2. **Present.** The circuit contains a 2-Ω resistor in parallel with a 4-Ω resistor. These are, in turn, in parallel with a dependent current source. It is important to note that there are no independent sources. -- 3. **Alternative.** The first thing to consider is that, since we have no independent sources in this circuit, we must excite the circuit externally. In addition, when you have no independent sources the value for *V*Th will be equal to zero so you will only have to find *R*Th. - -For Example 4.10. - -The simplest approach is to excite the circuit with either a 1-V voltage source or a 1-A current source. Because we will end up with an equivalent resistance (either positive or negative), I prefer to use the current source and nodal analysis which will yield a voltage at the output terminals equal to the resistance (with 1 A flowing in, *vo* is equal to 1 times the equivalent resistance). - - As an alternative, the circuit could also be excited by a 1-V voltage source and mesh analysis could be used to find the equivalent resistance. - -4. **Attempt.** We start by writing the nodal equation at *a* in Fig. 4.35(b) assuming *io* = 1 A. - -$$ -2i_x + (v_o - 0)/4 + (v_o - 0)/2 + (-1) = 0 \tag{4.10.1} -$$ - -Given that we have two unknowns and only one equation, we will need a constraint equation. - -$$ -i_x = (0 - v_o)/2 = -v_o/2 \tag{4.10.2} -$$ - -Substituting Eq. (4.10.2) into Eq. (4.10.1) yields - -$$ -2(-v_o/2) + (v_o - 0)/4 + (v_o - 0)/2 + (-1) = 0 -$$ - -= $(-1 + \frac{1}{4} + \frac{1}{2})v_o - 1$ or $v_o = -4$ V - -Since *vo* = 1 × *R*Th, then *R*Th = *vo* ∕1 = −4 Ω. - -The negative value of the resistance tells us that, according to the passive sign convention, the circuit in Fig. 4.35(a) is supplying power. Of course, the resistors in Fig. 4.35(a) cannot supply power (they absorb power); it is the dependent source that supplies the power. This is an example of how a dependent source and resistors could be used to simulate negative resistance. - -5. **Evaluate.** First of all, we note that the answer has a negative value. We know this is not possible in a passive circuit, but in this circuit we do have an active device (the dependent current source). Thus, the equivalent circuit is essentially an active circuit that can supply power. - - Now we must evaluate the solution. The best way to do this is to perform a check, using a different approach, and see if we obtain the same solution. Let us try connecting a 9-Ω resistor in series with a 10-V voltage source across the output terminals of the original circuit and then the Thevenin equivalent. To make the circuit easier to solve, we can take and change the parallel current source and 4-Ω resistor to a series voltage source and 4-Ω resistor by using source transformation. This, with the new load, gives us the circuit shown in Fig. 4.35(c). - -We can now write two mesh equations. - -$$ -8i_x + 4i_1 + 2(i_1 - i_2) = 0 -$$ - -$$ -2(i_2 - i_1) + 9i_2 + 10 = 0 -$$ - -Note, we only have two equations but have three unknowns, so we need a constraint equation. We can use - -$$ -i_x = i_2 - i_1 -$$ - -This leads to a new equation for loop 1. Simplifying leads to - -$$ -(4 + 2 - 8)i1 + (-2 + 8)i2 = 0 -$$ - -or - -$$ --2i_1 + 6i_2 = 0 \t or \t i_1 = 3i_2 -$$ - -$$ --2i_1 + 11i_2 = -10 -$$ - -Substituting the first equation into the second gives - -$$ --6i_2 + 11i_2 = -10 -$$ - or $i_2 = -10/5 = -2$ A - -Using the Thevenin equivalent is quite easy since we have only one loop, as shown in Fig. 4.35(d). - -$$ --4i + 9i + 10 = 0 -$$ - or $i = -10/5 = -2$ A - -5. **Satisfactory?** Clearly we have found the value of the equivalent circuit as required by the problem statement. Checking does validate that solution (we compared the answer we obtained by using the equivalent circuit with one obtained by using the load with the original circuit). We can present all this as a solution to the problem. - -Obtain the Thevenin equivalent of the circuit in Fig. 4.36. - -Answer: -$$ -V_{\text{Th}} = 0 \text{ V}, R_{\text{Th}} = -7.5 \Omega. -$$ - -b - -Practice Problem 4.10 - -# **4.6** Norton's Theorem - -In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an American engineer at Bell Telephone Laboratories, proposed a similar theorem. - -Norton's theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel with a resistor RN, where IN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the independent sources are turned off. - -Thus, the circuit in Fig. 4.37(a) can be replaced by the one in Fig. 4.37(b). - -The proof of Norton' s theorem will be gi ven in the ne xt section. For now, we are mainly concerned with ho w to get *RN* and *IN*. We find *RN* in the same w ay we find *R*Th. In f act, from what we kno w about source transformation, the Thevenin and Norton resistances are equal; that is, - -$$ -R_N = R_{\text{Th}} \tag{4.9} -$$ - -To find the Norton current *IN*, we determine the short-circuit current flowing from terminal *a* to *b* in both circuits in Fig. 4.37. It is e vident - -**Figure 4.36** For Practice Prob. 4.10. - -– - -# **Figure 4.37** (a) Original circuit, (b) Norton equivalent circuit. - -## **144** Chapter 4 Circuit Theorems - -Finding Norton current *IN*. - -The Thevenin and Norton equivalent circuits are related by a source transformation. - -that the short-circuit current in Fig. 4.37(b) is *IN*. This must be the same short-circuit current from terminal *a* to *b* in Fig. 4.37(a), since the tw o circuits are equivalent. Thus, - -$$ -I_N = i_{sc} \tag{4.10} -$$ - -shown in Fig. 4.38. Dependent and independent sources are treated the same way as in Thevenin's theorem. - -Observe the close relationship between Norton' s and Thevenin's theorems: *RN* = *R*Th as in Eq. (4.9), and - -$$ -I_N = \frac{V_{\text{Th}}}{R_{\text{Th}}} \tag{4.11} -$$ - -This is essentially source transformation. F or this reason, source trans formation is often called Thevenin-Norton transformation. - -Since *V*Th, *IN*, and *R*Th are related according to Eq. (4.11), to deter mine the Thevenin or Norton equivalent circuit requires that we find: - -- The open-circuit voltage *voc* across terminals *a* and *b*. -- The short-circuit current *isc* at terminals *a* and *b*. -- The equivalent or input resistance *R*in at terminals *a* and *b* when all independent sources are turned off. - -We can calculate an y two of the three using the method that tak es the least effort and use them to get the third using Ohm's law. Example 4.11 will illustrate this. Also, since - -$$ -V_{\rm Th} = v_{oc} \tag{4.12a} -$$ - -$$ -I_N = i_{sc} \tag{4.12b} -$$ - -$$ -R_{\text{Th}} = \frac{v_{oc}}{i_{sc}} = R_N \tag{4.12c} -$$ - -the open-circuit and short-circuit tests are sufficient to find any Thevenin or Norton equivalent, of a circuit which contains at least one independent source. - -Example 4.11 Find the Norton equi valent circuit of the circuit in Fig. 4.39 at terminals *a*-*b*. - -# **Solution:** - -We find *RN* in the same way we find *R*Th in the Thevenin equivalent circuit. Set the independent sources equal to zero. This leads to the circuit in Fig. 4.40(a), from which we find *RN*. Thus, - -$$ -R_N = 5 \parallel (8 + 4 + 8) = 5 \parallel 20 = \frac{20 \times 5}{25} = 4 \text{ }\Omega -$$ - -To find *IN*, we short-circuit terminals *a* and *b*, as shown in Fig. 4.40(b). We ignore the 5-Ω resistor because it has been short-circuited. Applying mesh analysis, we obtain - -$$ -i_1 = 2 \text{ A}, \qquad 20i_2 - 4i_1 - 12 = 0 -$$ - -From these equations, we obtain - -$$ -i_2=1 \; \mathrm{A}=i_{sc}=I_N -$$ - -**Figure 4.39** For Example 4.11. - -**Figure 4.40** For Example 4.11; finding: (a) *RN*, (b) *IN* = *isc*, (c) *V*Th = *voc*. - -Alternatively, we may determine *IN* from *V*Th ∕*R*Th. We obtain *V*Th as the open-circuit voltage across terminals *a* and *b* in Fig. 4.40(c). Using mesh analysis, we obtain - -$$ -i_3 = 2 \text{ A} -$$ - - $25i_4 - 4i_3 - 12 = 0 \Rightarrow i_4 = 0.8 \text{ A}$ - -and - -$$ -v_{oc} = V_{\text{Th}} = 5i_4 = 4 \text{ V} -$$ - -Hence, - -$$ -I_N = \frac{V_{\text{Th}}}{R_{\text{Th}}} = \frac{4}{4} = 1 \text{ A} -$$ - -as obtained pre viously. This also serv es to confirm Eq. (4.12c) that *R*Th = *voc*∕*isc* = 4 ∕1 = 4 Ω. Thus, the Norton equi valent circuit is as shown in Fig. 4.41. - -Find the Norton equi valent circuit for the circuit in Fig. 4.42, at Practice Problem 4.11 terminals *a*-*b*. - -**Answer:** *RN* = 90 Ω, *IN* = 4.5 A. - -For Practice Prob. 4.11. - -**Figure 4.43** For Example 4.12. - -Example 4.12 Using Norton's theorem, find *RN* and *IN* of the circuit in Fig. 4.43 at terminals *a*-*b*. - -# **Solution:** - -To find *RN*, we set the independent voltage source equal to zero and connect a voltage source of *vo* = 1 V (or any unspecified voltage *vo*) to the terminals. We obtain the circuit in Fig. 4.44(a). We ignore the 4-Ω resistor because it is short-circuited. Also due to the short circuit, the 5- Ω resistor, the voltage source, and the dependent current source are all in parallel. Hence, *ix* = 0. At node *a*, *io* = \_\_\_\_ 1 V 5Ω = 0.2 A, and - -$$ -R_N = \frac{v_o}{i_o} = \frac{1}{0.2} = 5 \ \Omega -$$ - -To find *IN*, we short-circuit terminals *a* and *b* and find the current *isc*, as indicated in Fig. 4.44(b). Note from this figure that the 4 Ω resistor, the 10-V voltage source, the 5-Ω resistor, and the dependent current source are all in parallel. Hence, - -$$ -i_x = \frac{10}{4} = 2.5 \text{ A} -$$ - -At node *a*, KCL gives - -$$ -i_{sc} = \frac{10}{5} + 2i_x = 2 + 2(2.5) = 7 \text{ A} -$$ - -Thus, - -$$ -I_N = 7 \,\mathrm{A} -$$ - -**Figure 4.44** For Example 4.12: (a) finding *RN*, (b) finding *IN*. - -For Practice Prob. 4.12. - -# **4.7** Derivations of Thevenin's and Norton's Theorems - -In this section, we will pro ve Thevenin's and Norton's theorems using the superposition principle. - -Consider the linear circuit in Fig. 4.46(a). It is assumed that the circuit contains resistors and dependent and independent sources. We have access to the circuit via terminals *a* and *b*, through which current from an external source is applied. Our objective is to ensure that the voltagecurrent relation at terminals *a* and *b* is identical to that of the Thevenin equivalent in Fig. 4.46(b). F or the sak e of simplicity , suppose the lin ear circuit in Fig. 4.46(a) contains tw o independent voltage sources *vs*1 and *vs*2 and two independent current sources *is*1 and *is*2. We may obtain any circuit variable, such as the terminal v oltage *v*, by applying super position. That is, we consider the contrib ution due to each independent source including the e xternal source *i*. By superposition, the terminal voltage *v* is - -$$ -v = A_0 i + A_1 v_{s1} + A_2 v_{s2} + A_3 i_{s1} + A_4 i_{s2} -$$ - (4.13) - -where *A*0, *A*1, *A*2, *A*3, and *A*4 are constants. Each term on the right-hand side of Eq. (4.13) is the contrib ution of the related independent source; that is, *A*0*i* is the contrib ution to *v* due to the e xternal current source *i*, *A*1*vs*1 is the contribution due to the voltage source *vs*1, and so on. We may collect terms for the internal independent sources together as *B*0, so that Eq. (4.13) becomes - -$$ -v = A_0 i + B_0 \tag{4.14} -$$ - -where *B*0 = *A*1*vs*1 + *A*2*vs*2 + *A*3*is*1 + *A*4*is*2. We now want to evaluate the values of constants *A*0 and *B*0. When the terminals *a* and *b* are opencircuited, *i* = 0 and *v* = *B*0. Thus, *B*0 is the open-circuit voltage *voc*, which is the same as *V*Th, so - -$$ -B_0 = V_{\text{Th}} \tag{4.15} -$$ - -When all the internal sources are turned off, *B*0 = 0. The circuit can then be replaced by an equivalent resistance *R*eq, which is the same as *R*Th, and Eq. (4.14) becomes - -$$ -v = A_0 i = R_{\text{Th}} i \qquad \Rightarrow \qquad A_0 = R_{\text{Th}} \tag{4.16} -$$ - -Substituting the values of *A*0 and *B*0 in Eq. (4.14) gives - -$$ -v = R_{\text{Th}} i + V_{\text{Th}} \tag{4.17} -$$ - -which expresses the voltage-current relation at terminals *a* and *b* of the circuit in Fig. 4.46(b). Thus, the two circuits in Fig. 4.46(a) and 4.46(b) are equivalent. - -When the same linear circuit is dri ven by a v oltage source *v* as shown in Fig. 4.47(a), the current flowing into the circuit can be obtained by superposition as - -$$ -i = C_0 v + D_0 \tag{4.18} -$$ - -where *C*0*v* is the contrib ution to *i* due to the e xternal voltage source *v* and *D*0 contains the contrib utions to *i* due to all internal independent sources. When the terminals *a*-*b* are short-circuited, *v* = 0 so that - -Derivation of Thevenin equivalent: (a) a current-driven circuit, (b) its Thevenin equivalent. - -**Figure 4.47** Derivation of Norton equivalent: (a) a voltage-driven circuit, (b) its Norton equivalent. - -## **148** Chapter 4 Circuit Theorems - -*i* = *D*0 = −*isc*, where *isc* is the short-circuit current flowing out of terminal *a*, which is the same as the Norton current *IN*, i.e., - -$$ -D_0 = -I_N \tag{4.19} -$$ - -When all the internal independent sources are turned of f, *D*0 = 0 and the circuit can be replaced by an equivalent resistance *R*eq (or an equivalent conductance *G*eq = 1∕*R*eq), which is the same as *R*Th or *RN*. Thus, Eq. (4.19) becomes - -$$ -i = \frac{v}{R_{\text{Th}}} - I_N \tag{4.20} -$$ - -This expresses the v oltage-current relation at terminals *a*-*b* of the cir cuit in Fig. 4.47(b), confirming that the two circuits in Fig. 4.47(a) and 4.47(b) are equivalent. - -# **4.8** Maximum Power Transfer - -In many practical situations, a circuit is designed to provide power to a load. There are applications in areas such as communications where it is desirable to maximize the power delivered to a load. We now address the problem of deli vering the maximum po wer to a load when gi ven a system with known internal losses. It should be noted that this will result in significant internal losses greater than or equal to the power delivered to the load. - -The Thevenin equivalent is useful in finding the maximum power a linear circuit can deliver to a load. We assume that we can adjust the load resistance *RL*. If the entire circuit is replaced by its Thevenin equivalent except for the load, as shown in Fig. 4.48, the power delivered to the load is - -$$ -p = i^2 R_L = \left(\frac{V_{\text{Th}}}{R_{\text{Th}} + R_L}\right)^2 R_L -$$ - (4.21) - -For a given circuit, *V*Th and *R*Th are fixed. By varying the load resistance *RL*, the power delivered to the load varies as sketched in Fig. 4.49. We notice from Fig. 4.49 that the power is small for small or large values of *RL* but maximum for some value of *RL* between 0 and ∞. We now want to show that this maximum power occurs when *RL* is equal to *R*Th. This is known as the *maximum power theorem.* - -Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load (RL = RTh). - -To prove the maximum power transfer theorem, we differentiate *p* in Eq. (4.21) with respect to *RL* and set the result equal to zero. We obtain - - *dp \_\_\_\_ dRL* = *V*2 Th [ (*R*Th + *RL*) 2 −2*RL*(*R*Th + *RL*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*R*Th + *RL*) 4 ] = *V*2 Th [ (*R*Th + *RL* − 2*RL*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*R*Th + *RL*) 3 ] =0 - -**Figure 4.48** - -The circuit used for maximum power transfer. - -**Figure 4.49** Power delivered to the load as a function of *RL*. - -This implies that - -$$ -0 = (R_{\text{Th}} + R_L - 2R_L) = (R_{\text{Th}} - R_L) -$$ -\n(4.22) - -which yields - -$$ -R_L = R_{\text{Th}} \tag{4.23} -$$ - -showing that the maximum power transfer takes place when the load resistance *RL* equals the Thevenin resistance *R*Th. We can readily confirm that Eq. (4.23) gives the maximum power by showing that *d*2 *p*∕*dR*2 *L*< 0. - -The maximum po wer transferred is obtained by substituting Eq. (4.23) into Eq. (4.21), for - -> *p*max = *V*2 \_\_\_\_Th 4*R*Th **(4.24)** - -Equation (4.24) applies only when *RL* = *R*Th. When *RL* ≠ *R*Th, we compute the power delivered to the load using Eq. (4.21). - -Find the v alue of *RL* for maximum po wer transfer in the circuit of Example 4.13 Fig. 4.50. Find the maximum power. - -The source and load are said to be - -matched when RL = RTh. - -**Solution:** - -We need to find the Thevenin resistance *R*Th and the Thevenin voltage *V*Th across the terminals *a*-*b*. To get *R*Th, we use the circuit in Fig. 4.51(a) and obtain - -$$ -R_{\text{Th}} = 2 + 3 + 6 \parallel 12 = 5 + \frac{6 \times 12}{18} = 9 \ \Omega -$$ - -For Example 4.13: (a) finding *R*Th, (b) finding *V*Th. - -To get *V*Th, we consider the circuit in Fig. 4.51(b). Applying mesh analysis gives - -$$ --12 + 18i_1 - 12i_2 = 0, \qquad i_2 = -2 \text{ A} -$$ - -Solving for *i*1, we get *i*1 = −2∕3. Applying KVL around the outer loop to get *V*Th across terminals *a*-*b*, we obtain - -−12 + 6*i*1 + 3*i*2 + 2(0) + *V*Th = 0 ⇒ *V*Th = 22 V - -For maximum power transfer, - -$$ -R_L = R_{\text{Th}} = 9 \ \Omega -$$ - -and the maximum power is - -$$ -p_{\text{max}} = \frac{V_{\text{Th}}^2}{4R_L} = \frac{22^2}{4 \times 9} = 13.44 \text{ W} -$$ - -Determine the value of *RL* that will draw the maximum power from the rest of the circuit in Fig. 4.52. Calculate the maximum power. - -**Answer:** 126.67 Ω, 96.71 mW. - -# **4.9** Verifying Circuit Theorems with PSpice - -In this section, we learn how to use *PSpice* to verify the theorems covered in this chapter. Specifically, we will consider using DC Sweep analysis to find the Thevenin or Norton equi valent at an y pair of nodes in a circuit and the maximum po wer transfer to a load. The reader is advised to read Section D.3 of Appendix D in preparation for this section. - -To find the Thevenin equivalent of a circuit at a pair of open terminals using *PSpice*, we use the schematic editor to dra w the circuit and insert an independent probing current source, say , Ip, at the terminals. The probing current source must have a part name ISRC. We then perform a DC Sweep on Ip, as discussed in Section D.3. Typically, we may let the current through Ip vary from 0 to 1 A in 0.1-A increments. After saving and simulating the circuit, we use Probe to display a plot of the voltage across Ip versus the current through Ip. The zero intercept of the plot gives us the Thevenin equivalent voltage, while the slope of the plot is equal to the Thevenin resistance. - -To find the Norton equivalent involves similar steps except that we insert a probing independent v oltage source (with a part name VSRC), say, Vp, at the terminals. We perform a DC Sweep on Vp and let Vp vary from 0 to 1 V in 0.1-V increments. A plot of the current through Vp versus the voltage across Vp is obtained using the Probe menu after simulation. The zero intercept is equal to the Norton current, while the slope of the plot is equal to the Norton conductance. - -To find the maximum power transfer to a load using *PSpice* in volves performing a DC parametric Sweep on the component v alue of *RL* in Fig. 4.48 and plotting the po wer delivered to the load as a function of *RL*. According to Fig. 4.49, the maximum po wer occurs - -**Figure 4.52** For Practice Prob. 4.13. - -when *RL* = *R*Th. This is best illustrated with an example, and Example 4.15 provides one. - -We use VSRC and ISRC as part names for the independent voltage and current sources, respectively. - -Consider the circuit in Fig. 4.31 (see Example 4.9). Use *PSpice* to find the Thevenin and Norton equivalent circuits. - -# **Solution:** - -(a) To find the Thevenin resistance *R*Th and Thevenin voltage *V*Th at the terminals *a*-*b* in the circuit in Fig. 4.31, we first use Schematics to draw the circuit as shown in Fig. 4.53(a). Notice that a probing current source I2 is inserted at the terminals. Under **Analysis**∕**Setput**, we select DC Sweep. In the DC Sweep dialog box, we select Linear for the *Sweep Type* and Current Source for the *Sweep Var*. *Type*. We enter I2 under the *Name* box, 0 as *Start Value*, 1 as *End Value*, and 0.1 as *Increment*. After simulation, we add trace V(I2:–) from the *PSpice* A∕D window and obtain the plot shown in Fig. 4.53(b). From the plot, we obtain *V*Th = Zero intercept = 20 V, *R*Th = Slope = \_\_\_\_\_\_\_ 26 − 20 1 = 6 Ω These agree with what we got analytically in Example 4.9. - -(b) To find the Norton equivalent, we modify the schematic in Fig. 4.53(a) by replaying the probing current source with a probing v oltage source V1. The result is the schematic in Fig. 4.54(a). Again, in the DC Sweep dialog box, we select Linear for the *Sweep Type* and Voltage Source for the *Sweep Var*. *Type*. We enter V1 under *Name* box, 0 as *Start Value*, 1 as *End Value*, and 0.1 as *Increment*. Under the *PSpice* A∕D Window, we add trace I (V1) and obtain the plot in Fig. 4.54(b). From the plot, we obtain - -$$ -I_N -$$ - = Zero intercept = 3.335 A - $G_N$ = Slope = $\frac{3.335 \text{ } \bigcirc \text{ } 3.165}{1} = 0.17 \text{ S}$ - -Example 4.14 - -**Figure 4.54** For Example 4.14: (a) schematic and (b) plot for finding *GN* and *IN*. - -| Practice Problem 4.14 | Rework Practice Prob. 4.9 using PSpice. | -|-----------------------|-------------------------------------------------| -| | Answer:
VTh
= 5.333 V, RTh
= 444.4 mΩ. | -| | | - -**Figure 4.55** - -For Example 4.15. - -# **Figure 4.56** - -Schematic for the circuit in Fig. 4.55. - -Example 4.15 Refer to the circuit in Fig. 4.55. Use *PSpice* to find the maximum power transfer to *RL*. - -# **Solution:** - -We need to perform a DC Sweep on *RL* to determine when the po wer across it is maximum. We first draw the circuit using Schematics as shown in Fig. 4.56. Once the circuit is dra wn, we tak e the follo wing three steps to further prepare the circuit for a DC Sweep. - -The first step involves defining the value of *RL* as a parameter, since we want to vary it. To do this: - -- 1. **DCLICKL** the value 1k of R2 (representing *RL*) to open up the *Set Attribute Value* dialog box. -- 2. Replace 1k with {RL} and click **OK** to accept the change. - -# Note that the curly brackets are necessary. - -The second step is to define parameter. To achieve this: - -- 1. Select **Draw**∕**Get New Part∕Libraries** ⋯∕**special.slb**. -- 2. Type PARAM in the *PartName* box and click **OK**. -- 3. **DRAG** the box to any position near the circuit. -- 4. **CLICKL** to end placement mode. -- 5. **DCLICKL** to open up the *PartName: PARAM* dialog box. -- 6. **CLICKL** on *NAME1* = and enter RL (with no curly brackets) in the *Value* box, and **CLICKL Save Attr** to accept change. -- 7. **CLICKL** on *VALUE1*= and enter 2k in the *Value* box, and **CLICKL Save Attr** to accept change. -- 8. Click **OK**. - -The value 2k in item 7 is necessary for a bias point calculation; it cannot be left blank. - -The third step is to set up the DC Sweep to sweep the parameter. To do this: - -- 1. Select **Analysis**∕**Setput** to bring up the DC Sweep dialog box. -- 2. For the *Sweep Type*, select Linear (or Octave for a wide range of *RL*). -- 3. For the *Sweep Var*. *Type*, select Global Parameter. -- 4. Under the *Name* box, enter RL. -- 5. In the *Start Value* box, enter 100. -- 6. In the *End Value* box, enter 5k. -- 7. In the *Increment* box, enter 100. -- 8. Click **OK** and **Close** to accept the parameters. - -After taking these steps and sa ving the circuit, we are ready to simulate. Select **Analysis/Simulate**. If there are no errors, we select **Add Trace** in the *PSpice* A/D windo w and type −V(R2:2)∗ I(R2) in the *Trace Command* box. [The ne gative sign is needed since I(R2) is negative.] This gives the plot of the power delivered to *RL* as *RL* varies from 100 Ω to 5 k Ω. We can also obtain the power absorbed by *RL* by typing V(R2:2)∗ V(R2:2)/RL in the *Trace Command* box. Either way, we obtain the plot in Fig. 4.57. It is evident from the plot that the maximum power is 250 *μ*W. Notice that the maximum occurs when *RL* = 1 kΩ, as expected analytically. - -150 uW 200 uW 100 uW 50 uW 0 2.0 K 4.0 K 6.0 K –V(R2:2)\*I(R2) RL - -# **Figure 4.57** - -250 uW - -For Example 4.15: the plot of power across *RL*. - -Find the maximum po wer transferred to *RL* if the 1-k Ω resistor in Practice Problem 4.15 Fig. 4.55 is replaced by a 2-kΩ resistor. - -**Answer:** 125 *μ*W. - -# **4.10** Applications - -In this section we will discuss tw o important practical applications of the concepts co vered in this chapter: source modeling and resistance measurement. - -# **4.10.1** Source Modeling - -Source modeling provides an example of the usefulness of the Thevenin or the Norton equivalent. An active source such as a battery is often characterized by its Thevenin or Norton equivalent circuit. An ideal voltage source provides a constant v oltage irrespective of the current dra wn by the load, while an ideal current source supplies a constant current regardless of the load voltage. As Fig. 4.58 shows, practical voltage and current sources are not ideal, due to their *internal resistances* or *source resistances Rs* and *Rp*. They become ideal as *Rs* → 0 and *Rp* → ∞. To show that this is the case, consider the effect of the load on voltage sources, - -## **154** Chapter 4 Circuit Theorems - -as shown in Fig. 4.59(a). By the voltage division principle, the load voltage is - -$$ -v_L = \frac{R_L}{R_s + R_L} v_s \tag{4.25} -$$ - -As *RL* increases, the load v oltage approaches a source v oltage *vs*, as illustrated in Fig. 4.59(b). From Eq. (4.25), we should note that: - -1. The load v oltage will be constant if the internal resistance *Rs* of the source is zero or , at least, *Rs* ≪ *RL*. In other w ords, the smaller *Rs* is compared with *RL*, the closer the v oltage source is to being ideal. - -(a) Practical voltage source connected to a load *RL*, (b) load voltage decreases as *RL* decreases. - -2. When the load is disconnected (i.e., the source is open-circuited so that *RL* → ∞), *voc* = *vs*. Thus, *vs* may be re garded as the *unloaded source* voltage. The connection of the load causes the terminal voltage to drop in magnitude; this is known as the *loading effect*. - -The same argument can be made for a practical current source when connected to a load as sho wn in Fig. 4.60(a). By the current di vision principle, - -$$ -i_L = \frac{R_p}{R_p + R_L} i_s \tag{4.26} -$$ - -Figure 4.60(b) shows the variation in the load current as the load resis tance increases. Again, we notice a drop in current due to the load (loading effect), and load current is constant (ideal current source) when the internal resistance is very large (i.e., *Rp* → ∞ or, at least, *Rp* ≫ *RL*). - -Sometimes, we need to know the unloaded source voltage *vs* and the internal resistance *Rs* of a v oltage source. To find *vs* and *Rs*, we follo w the procedure illustrated in Fig. 4.61. First, we measure the open-circuit voltage *voc* as in Fig. 4.61(a) and set - -$$ -v_s = v_{oc} \tag{4.27} -$$ - -Then, we connect a v ariable load *RL* across the terminals as in Fig. 4.61(b). We adjust the resistance *RL* until we measure a load voltage of exactly one-half of the open-circuit voltage, *vL* = *voc*∕2, be cause now *RL* = *R*Th = *Rs*. At that point, we disconnect *RL* and measure it. We set - -$$ -R_s = R_L \tag{4.28} -$$ - -For example, a car battery may have *vs* = 12 V and *Rs* = 0.05 Ω. - -# **Figure 4.60** - -(a) Practical current source connected to a load *RL*, (b) load current decreases as *RL* increases. - -(a) Measuring *voc*, (b) measuring *vL*. - -The terminal v oltage of a v oltage source is 12 V when connected to a Example 4.16 2-W load. When the load is disconnected, the terminal v oltage rises to 12.4 V. (a) Calculate the source v oltage *vs* and internal resistance *Rs*. (b) Determine the voltage when an 8-Ω load is connected to the source. - -# **Solution:** - -(a) We replace the source by its Thevenin equivalent. The terminal voltage when the load is disconnected is the open-circuit voltage, - -$$ -v_s = v_{oc} = 12.4 \text{ V} -$$ - -When the load is connected, as shown in Fig. 4.62(a), *vL* = 12 V and *pL* = 2 W. Hence, - -$$ -p_L = \frac{v_L^2}{R_L} -$$ - $\Rightarrow$ $R_L = \frac{v_L^2}{p_L} = \frac{12^2}{2} = 72 \Omega$ - -The load current is - -$$ -i_L = \frac{v_L}{R_L} = \frac{12}{72} = \frac{1}{6} -$$ - A - -The voltage across *Rs* is the difference between the source voltage *vs* and the load voltage *vL*, or - -$$ -12.4 - 12 = 0.4 = R_s i_L, \qquad R_s = \frac{0.4}{I_L} = 2.4 \ \Omega -$$ - -(b) Now that we have the Thevenin equivalent of the source, we connect the 8-Ω load across the Thevenin equivalent as shown in Fig. 4.62(b). Using voltage division, we obtain - -$$ -v = \frac{8}{8 + 2.4} (12.4) = 9.538 \text{ V} -$$ - -**Figure 4.62** For Example 4.16. - -The measured open-circuit voltage across a certain amplifier is 16 V. The voltage drops to 8 V when a 8-Ω loudspeaker is connected to the amplifier. Calculate the voltage when a 24-Ω loudspeaker is used instead. - -Practice Problem 4.16 - -**Answer:** 12 V. - -# **4.10.2** Resistance Measurement - -Although the ohmmeter method pro vides the simplest w ay to mea sure resistance, more accurate measurement may be obtained using the Wheatstone bridge. While ohmmeters are designed to measure resistance in low, mid, or high range, a Wheatstone bridge is used to mea sure resistance in the mid range, say, between 1 Ω and 1 MΩ. Very low values of resistances are measured with a *milliohmmeter,* while very high values are measured with a *Megger tester*. - -The Wheatstone bridge (or resistance bridge) circuit is used in a number of applications. Here we will use it to measure an unkno wn resistance. The unkno wn resistance *Rx* is connected to the bridge as shown in Fig. 4.63. The variable resistance is adjusted until no current flows through the galvanometer, which is essentially a d'Arsonval movement operating as a sensiti ve current-indicating device like an ammeter in the microamp range. Under this condition *v*1 = *v*2, and the bridge is said to be *balanced*. Since no current flows through the galvanometer, *R*1 and *R*2 behave as though the y were in series; so do *R*3 and *Rx*. The fact that no current flows through the galvanometer also implies that *v*1 = *v*2. Applying the voltage division principle, - -$$ -v_1 = \frac{R_2}{R_1 + R_2} v = v_2 = \frac{R_x}{R_3 + R_x} v -$$ -\n(4.29) - -Hence, no current flows through the galvanometer when - -$$ -\frac{R_2}{R_1 + R_2} = \frac{R_x}{R_3 + R_x} \qquad \Rightarrow \qquad R_2 R_3 = R_1 R_x -$$ - -$$ -R_x = \frac{R_3}{R_1} R_2 \tag{4.30} -$$ - -If *R*1 = *R*3, and *R*2 is adjusted until no current flows through the galvanometer, then *Rx* = *R*2. - -How do we find the current through the galvanometer when the Wheatstone bridge is *unbalanced*? We find the Thevenin equivalent (*V*Th and *R*Th) with respect to the g alvanometer terminals. If *Rm* is the resis tance of the g alvanometer, the current through it under the unbalanced condition is - -$$ -I = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_m} \tag{4.31} -$$ - -Example 4.18 will illustrate this. - -In Fig. 4.63, *R*1 = 500 Ω and *R*3 = 200 Ω. The bridge is balanced when *R*2 is adjusted to be 125 Ω. Determine the unknown resistance *Rx*. - -# **Solution:** - -or - -Using Eq. (4.30) gives - -$$ -R_x = \frac{R_3}{R_1} R_2 = \frac{200}{500} 125 = 50 \ \Omega -$$ - -Historical note: The bridge was invented by Charles Wheatstone (1802–1875), a British professor who also invented the telegraph, as Samuel Morse did independently in the - -United States. - -**Figure 4.63** The Wheatstone bridge; *Rx* is the resistance to be measured. - -A Wheatstone bridge has *R*1 = *R*3 = 2 kΩ. *R*2 is adjusted until no current flows through the galvanometer. At that point, *R*2 = 6.3 kΩ. What is the value of the unknown resistance? - -# **Answer:** 6.3 kΩ. - -The circuit in Fig. 4.64 represents an unbalanced bridge. If the galvanometer has a resistance of 40 Ω, find the current through the galvanometer. - -Unbalanced bridge of Example 4.18. - -# **Solution:** - -We first need to replace the circuit by its Thevenin equivalent at ter minals *a* and *b*. The Thevenin resistance is found using the circuit in Fig. 4.65(*a*). Notice that the 3-kΩ and 1-KΩ resistors are in parallel; so are the 400 and 600- Ω resistors. The two parallel combinations form a series combination with respect to terminals *a* and *b*. Hence, - -$$ -R_{\text{Th}} = 3000 \parallel 1000 + 400 \parallel 600 -$$ - -= -$$ -\frac{3000 \times 1000}{3000 + 1000} + \frac{400 \times 600}{400 + 600} = 750 + 240 = 990 \Omega -$$ - -To find the Thevenin voltage, we consider the circuit in Fig. 4.65(b). Using the voltage division principle gives - -sing the voltage division principle gives -$$ -v_1 = \frac{1000}{1000 + 3000} (220) = 55 \text{ V}, \qquad v_2 = \frac{600}{600 + 400} (220) = 132 \text{ V} -$$ - -Applying KVL around loop *ab* gives - -$$ --v_1 + V_{\text{Th}} + v_2 = 0 -$$ - or $V_{\text{Th}} = v_1 - v_2 = 55 - 132 = -77$ V - -Having determined the Thevenin equivalent, we find the current through the galvanometer using Fig. 4.65(c). - -$$ -I_G = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_m} = \frac{-77}{990 + 40} = -74.76 \text{ mA} -$$ - -The negative sign indicates that the current flows in the direction oppo site to the one assumed, that is, from terminal *b* to terminal *a*. - -Practice Problem 4.17 - -Example 4.18 - -For Example 4.18: (a) Finding *R*Th, (b) finding *V*Th, (c) determining the current through the galvanometer. - -Practice Problem 4.18 14 Ω 60 Ω 16 V 40 Ω 20 Ω 30 Ω G - -**Figure 4.66** For Practice Prob. 4.18. - -Obtain the current through the g alvanometer, ha ving a resistance of 14 Ω, in the Wheatstone bridge shown in Fig. 4.66. - -**Answer:** 64 mA. - -# **4.11** Summary - -- 1. A linear netw ork consists of linear elements, linear dependent sources, and linear independent sources. -- 2. Network theorems are used to reduce a complex circuit to a simpler one, thereby making circuit analysis much simpler. -- 3. The superposition principle states that for a circuit ha ving multiple independent sources, the voltage across (or current through) an element is equal to the algebraic sum of all the individual voltages (or currents) due to each independent source acting one at a time. -- 4. Source transformation is a procedure for transforming a v oltage source in series with a resistor to a current source in parallel with a resistor, or vice versa. -- 5. Thevenin's and Norton's theorems allow us to isolate a portion of a network while the remaining portion of the netw ork is replaced by an equivalent network. The Thevenin equivalent consists of a v oltage source *V*Th in series with a resistor *R*Th, while the Norton equivalent consists of a current source *IN* in parallel with a resistor *RN*. The two theorems are related by source transformation. - -$$ -R_N = R_{\text{Th}}, \qquad I_N = \frac{V_{\text{Th}}}{R_{\text{Th}}} -$$ - -- 6. For a gi ven Thevenin equivalent circuit, maximum po wer transfer occurs when *RL* = *R*Th; that is, when the load resistance is equal to the Thevenin resistance. -- 7. The maximum po wer transfer theorem states that the maximum power is delivered by a source to the load *RL* when *RL* is equal to *R*Th, the Thevenin resistance at the terminals of the load. -- 8. *PSpice* can be used to v erify the circuit theorems co vered in this chapter. -- 9. Source modeling and resistance measurement using the Wheatstone bridge provide applications for Thevenin's theorem. - -# Review Questions - -**4.1** The current through a branch in a linear network is 2 A when the input source voltage is 10 V. If the voltage is reduced to 1 V and the polarity is reversed, the current through the branch is: - -> (a) − 2 A (b) − 0.2 A (c) 0.2 A (d) 2 A (e) 20 A - -**4.2** For superposition, it is not required that only one independent source be considered at a time; any number of independent sources may be considered simultaneously. - -(a) True (b) False - -**4.3** The superposition principle applies to power calculation. - -(a) True (b) False - -- **4.4** Refer to Fig. 4.67. The Thevenin resistance at terminals *a* and *b* is: - - (a) 25 Ω (b) 20 Ω - -# **Figure 4.67** - -For Review Questions 4.4 to 4.6. - -**4.5** The Thevenin voltage across terminals *a* and *b* of the circuit in Fig. 4.67 is: - -| (a) 50 V | (b) 40 V | -|----------|----------| -| (c) 20 V | (d) 10 V | - -**4.6** The Norton current at terminals *a* and *b* of the circuit in Fig. 4.67 is: - -| (a) 10 A | (b) 2.5 A | -|----------|-----------| -| (c) 2 A | (d) 0 A | - -**4.7** The Norton resistance *RN* is exactly equal to the Thevenin resistance *R*Th. - -(a) True (b) False - -**4.8** Which pair of circuits in Fig. 4.68 are equivalent? - -| (a) a and b | (b) b and d | -|-------------|-------------| -| (c) a and c | (d) c and d | - -# **Figure 4.68** - -For Review Question 4.8. - -**4.9** A load is connected to a network. At the terminals to which the load is connected, *R*Th = 10 Ω and *V*Th = 40 V. The maximum possible power supplied to the load is: - -| (a) 160 W | (b) 80 W | -|-----------|----------| -| (c) 40 W | (d) 1 W | - -**4.10** The source is supplying the maximum power to the load when the load resistance equals the source resistance. - -(a) True (b) False - -*Answers: 4.1b, 4.2a, 4.3b, 4.4d, 4.5b, 4.6a, 4.7a, 4.8c, 4.9c, 4.10a.* - -# Problems - -# Section 4.2 Linearity Property - -**4.1** Calculate the current *io* in the circuit of Fig. 4.69. What value of input voltage is necessary to make *io* equal to 5 amps? - -# **Figure 4.69** - -For Prob. 4.1. - -- **4.3** (a) In the circuit of Fig. 4.71, calculate *vo* and *io* when *vs* = 1 V. - - (b) Find *vo* and *io* when *vs* = 10 V. - - (c) What are *vo* and *io* when each of the 1-Ω resistors is replaced by a 10-Ω resistor and *vs* = 10 V? - -For Prob. 4.3. - -**4.4** Use linearity to determine *io* in the circuit of Fig. 4.72. - -**4.5** For the circuit in Fig. 4.73, assume *vo* = 1 V, and use linearity to find the actual value of *vo*. - -# **Figure 4.73** - -For Prob. 4.5. - -**4.6** For the linear circuit shown in Fig. 4.74, use linearity to complete the following table. - -# **Figure 4.74** - -For Prob. 4.6. - -**4.7** Use linearity and the assumption that *Vo* = 1 V to find the actual value of *Vo* in Fig. 4.75. - -# Section 4.3 Superposition - -**4.8** Using superposition, find *Vo* in the circuit of Fig. 4.76. Check with *PSpice or MultiSim*. - -## Problems **161** - -**4.9** Given that *I* = 6 amps when *Vs* = 160 volts and *Is* = −10 amps and *I* = 5 amp when *Vs* = 200 volts and *Is* = 0, use superposition and linearity to determine the value of *I* when *Vs* = 120 volts and *Is* = 5 amps. - -**4.10** Using Fig. 4.78, design a problem to help other students better understand superposition. Note, the letter *k* is a gain you can specify to make the problem easier to solve but must not be zero. - -**Figure 4.78** For Prob. 4.10. - -**4.11** Use the superposition principle to find *io* and *vo* in the circuit of Fig. 4.79. - -For Prob. 4.11. - -**4.12** Determine *vo* in the circuit of Fig. 4.80 using the superposition principle. - -**4.13** Use superposition to find *vo* in the circuit of Fig. 4.81. - -**4.14** Apply the superposition principle to find *vo* in the circuit of Fig. 4.82. - -# **Figure 4.82** - -For Prob. 4.14. - -**4.15** For the circuit in Fig. 4.83, use superposition to find *i*. Calculate the power delivered to the 3-Ω resistor. - -# **Figure 4.83** For Probs. 4.15 and 4.56. - -**4.16** Given the circuit in Fig. 4.84, use superposition to obtain *io*. - -For Prob. 4.16. - -**Figure 4.85** For Prob. 4.17. - -**4.18** Use superposition to find *Vo* in the circuit of Fig. 4.86. - -**Figure 4.86** - -- For Prob. 4.18. -- **4.19** Use superposition to solve for *vx* in the circuit of Fig. 4.87. - -**Figure 4.87** - -For Prob. 4.19. - -# Section 4.4 Source Transformation - -**4.20** Use source transformation to reduce the circuit between terminals *a* and *b* shown in Fig. 4.88 to a single voltage source in series with a single resistor. - -**Figure 4.88** For Prob. 4.20. - -**4.21** Using Fig. 4.89, design a problem to help other students better understand source transformation. - -**Figure 4.89** For Prob. 4.21. - -**4.22** For the circuit in Fig. 4.90, use source transformation to find *i*. - -# **Figure 4.90** - -For Prob. 4.22. - -**4.23** Referring to Fig. 4.91, use source transformation to determine the current and power absorbed by the 8-Ω resistor. - -**Figure 4.91** For Prob. 4.23. - -**4.24** Use source transformation to find the voltage *Vx* in the circuit of Fig. 4.92. - -## Problems **163** - -**4.25** Obtain *vo* in the circuit of Fig. 4.93 using source transformation. Check your result using *PSpice or MultiSim*. - -# **Figure 4.93** - -For Prob. 4.25. - -**4.26** Use source transformation to find *io* in the circuit of Fig. 4.94. - -# **Figure 4.94** - -For Prob. 4.26. - -**4.27** Apply source transformation to find *vx* in the circuit of Fig. 4.95. - -**4.28** Use source transformation to find *Io* in Fig. 4.96. - -For Prob. 4.28. - -**4.29** Use source transformation to find *vo* in the circuit of Fig. 4.97. - -# **Figure 4.97** - -For Prob. 4.29. - -**4.30** Use source transformation on the circuit shown in Fig 4.98 to find *ix*. - -**4.31** Determine *vx* in the circuit of Fig. 4.99 using source transformation. - -**Figure 4.99** For Prob. 4.31. - -**4.32** Use source transformation to find *ix* in the circuit of Fig. 4.100. - -# Sections 4.5 and 4.6 Thevenin's and Norton's Theorems - -- **4.33** Determine the Thevenin equivalent circuit, shown in Fig. 4.101, as seen by the 7-ohm resistor. - - Then calculate the current flowing through the 7-ohm resistor. - -For Prob. 4.33. - -**4.34** Using Fig. 4.102, design a problem that will help other students better understand Thevenin equivalent circuits. - -# **Figure 4.104** - -For Prob. 4.37. - -**4.38** Apply Thevenin's theorem to find *Vo* in the circuit of Fig. 4.105. - -**Figure 4.105** For Prob. 4.38. - -**4.39** Obtain the Thevenin equivalent at terminals *a*-*b* of the circuit shown in Fig. 4.106. - -# **Figure 4.106** - -For Prob. 4.39. - -**4.40** Find the Thevenin equivalent at terminals *a*-*b* of the circuit in Fig. 4.107. - -For Prob. 4.40. - -**Figure 4.102** - -For Probs. 4.34 and 4.49. - -- **4.35** Use Thevenin's theorem to find *vo* in Prob. 4.12. -- **4.36** Solve for the current *i* in the circuit of Fig. 4.103 using Thevenin's theorem. (*Hint:* Find the Thevenin equivalent seen by the 12-Ω resistor.) - -**4.41** Find the Thevenin and Norton equivalents at terminals *a*-*b* of the circuit shown in Fig. 4.108. - -For Prob. 4.41. - -**4.42** For the circuit in Fig. 4.109, find the Thevenin equivalent between terminals *a* and *b*. **\*** - -# **Figure 4.109** - -For Prob. 4.42. - -**4.43** Find the Thevenin equivalent looking into terminals *a*-*b* of the circuit in Fig. 4.110 and solve for *ix*. - -# **Figure 4.110** - -For Prob. 4.43. - -**4.44** For the circuit in Fig. 4.111, obtain the Thevenin equivalent as seen from terminals: - -**Figure 4.111** - -For Prob. 4.44. - -\* An asterisk indicates a challenging problem. - -**4.45** Find the Thevenin equivalent of the circuit in Fig. 4.112 as seen by looking into terminals *a* and *b*. - -# **Figure 4.112** - -For Prob. 4.45. - -**4.46** Using Fig. 4.113, design a problem to help other students better understand Norton equivalent circuits. - -**Figure 4.113** For Prob. 4.46. - -**4.47** Obtain the Thevenin and Norton equivalent circuits of the circuit in Fig. 4.114 with respect to terminals *a* and *b*. - -# **Figure 4.114** - -For Prob. 4.47. - -**4.48** Determine the Norton equivalent at terminals *a*-*b* for the circuit in Fig. 4.115. - -**Figure 4.115** For Prob. 4.48. - -**4.49** Find the Norton equivalent looking into terminals *a*-*b* of the circuit in Fig. 4.102. Let *V* = 40 V, *I* = 3 A, *R*1 = 10 Ω, *R*2 = 40 Ω, and *R*3 = 20 Ω. - -**4.50** Obtain the Norton equivalent of the circuit in Fig. 4.116 to the left of terminals *a*-*b*. Use the result to find current *i*. - -# **Figure 4.116** - -For Prob. 4.50. - -**4.51** Given the circuit in Fig. 4.117, obtain the Norton equivalent as viewed from terminals: - -**Figure 4.117** - -For Prob. 4.51. - -**4.52** For the transistor model in Fig. 4.118, obtain the Thevenin equivalent at terminals *a*-*b*. - -**4.53** Find the Norton equivalent at terminals *a*-*b* of the circuit in Fig. 4.119. - -**4.54** Find the Thevenin equivalent between terminals *a*-*b* of the circuit in Fig. 4.120. - -For Prob. 4.54. - -**4.55** Obtain the Norton equivalent at terminals *a*-*b* of the circuit in Fig. 4.121. **\*** - -For Prob. 4.55. - -**4.56** Use Norton's theorem to find *Vo* in the circuit of Fig. 4.122. - -**Figure 4.122** For Prob. 4.56. - -**4.57** Obtain the Thevenin and Norton equivalent circuits at terminals *a*-*b* for the circuit in Fig. 4.123. - -**Figure 4.123** For Probs. 4.57 and 4.79. - -**4.58** The network in Fig. 4.124 models a bipolar transistor common-emitter amplifier connected to a load. Find the Thevenin resistance seen by the load. - -**Figure 4.124** - -For Prob. 4.58. - -**4.59** Determine the Thevenin and Norton equivalents at terminals *a*-*b* of the circuit in Fig. 4.125. - -# **Figure 4.125** - -For Probs. 4.59 and 4.80. - -**4.60** For the circuit in Fig. 4.126, find the Thevenin and Norton equivalent circuits at terminals *a*-*b*. **\*** - -# **Figure 4.126** - -For Probs. 4.60 and 4.81. - -**4.61** Obtain the Thevenin and Norton equivalent circuits at terminals *a*-*b* of the circuit in Fig. 4.127. **\*** - -For Prob. 4.61. - -**4.62** Find the Thevenin equivalent of the circuit in Fig. 4.128. **\*** - -# **Figure 4.128** For Prob. 4.62. - -**4.63** Find the Norton equivalent for the circuit in Fig. 4.129. - -# **Figure 4.129** - -For Prob. 4.63. - -**4.64** Obtain the Thevenin equivalent seen at terminals *a*-*b* of the circuit in Fig. 4.130. - -# **Figure 4.130** - -For Prob. 4.64. - -For Prob. 4.65. - -**4.65** For the circuit shown in Fig. 4.131, determine the relationship between *Vo* and *Io*. - -# Section 4.8 Maximum Power Transfer - -**4.66** Find the maximum power that can be delivered to the resistor *R* in the circuit of Fig. 4.132. - -# **Figure 4.132** - -For Prob. 4.66. - -- **4.67** The variable resistor *R* in Fig. 4.133 is adjusted until it absorbs the maximum power from the circuit. - - (a) Calculate the value of *R* for maximum power. (b) Determine the maximum power absorbed by *R*. - -# **Figure 4.133** - -For Prob. 4.67. - -**4.68** Consider the 30-Ω resistor in Fig. 4.134. First compute the Thevenin equivalent circuit as seen by the 30-Ω resistor. Compute the value of *R* that results in Thevenin equivalent resistance equal to the 30-Ω resistance and then calculate power delivered to the 30-Ω resistor. Now let *R* = 0 Ω, 110 Ω, and ∞, calculate the power delivered to the 30-Ω resistor in each case. What can you say about the value of *R* that will result in the maximum power that can be delivered to the 30-Ω resistor? **\*** - -**Figure 4.134** - -# For Prob. 4.68. - -**4.69** Find the maximum power transferred to resistor *R* in the circuit of Fig. 4.135. - -**Figure 4.135** For Prob. 4.69. - -**4.70** Determine the maximum power delivered to the variable resistor *R* shown in the circuit of Fig. 4.136. - -# **Figure 4.136** - -**4.71** For the circuit in Fig. 4.137, what resistor connected across terminals *a*-*b* will absorb maximum power from the circuit? What is that power? - -**Figure 4.137** For Prob. 4.71. - -- **4.72** (a) For the circuit in Fig. 4.138, obtain the Thevenin equivalent at terminals *a*-*b*. - - (b) Calculate the current in *RL* = 13 Ω. - - (c) Find *RL* for maximum power deliverable to *RL*. - - (d) Determine that maximum power. - -**Figure 4.138** For Prob. 4.72. - -**4.73** Determine the maximum power that can be delivered to the variable resistor *R* in the circuit of Fig. 4.139. - -# **Figure 4.139** - -- For Prob. 4.73. - - **4.74** For the bridge circuit shown in Fig. 4.140, find the load *RL* for maximum power transfer and the maximum power absorbed by the load. - -**Figure 4.140** For Prob. 4.74. - -- **4.75** For the circuit in Fig. 4.141, determine the value of *R* such that the maximum power delivered to the load is 12 mW. **\*** - -For Prob. 4.75. - -# Section 4.9 Verifying Circuit Theorems with PSpice - -- **4.76** Solve Prob. 4.34 using *PSpice or MultiSim*. Let *V* = 40 V, *I* = 3 A, *R*1 = 10 Ω, *R*2 = 40 Ω, and *R*3 = 20 Ω. -- **4.77** Use *PSpice or MultiSim* to solve Prob. 4.44. -- **4.78** Use *PSpice or MultiSim* to solve Prob. 4.52. -- **4.79** Obtain the Thevenin equivalent of the circuit in Fig. 4.123 using *PSpice or MultiSim*. - -- **4.80** Use *PSpice or MultiSim* to find the Thevenin equivalent circuit at terminals *a*-*b* of the circuit in Fig. 4.125. -- **4.81** For the circuit in Fig. 4.126, use *PSpice or MultiSim* to find the Thevenin equivalent at terminals *a*-*b*. - -# Section 4.10 Applications - -- **4.82** An automobile battery has an open circuit voltage of 14.7 V which drops to 12 V when connected to two 65-W headlights. What is the resistance of the headlights and the value of the internal resistance of the battery? -- **4.83** The following results were obtained from measurements taken between the two terminals of a resistive network. - -| Terminal Voltage | 72 V | 0 V | -|------------------|------|-----| -| Terminal Current | 0 A | 9 A | - -Find the Thevenin equivalent of the network. - -- **4.84** When connected to a 4- Ω resistor, a battery has a terminal voltage of 10.8 V but produces 12 V on an open circuit. Determine the Thevenin equivalent circuit for the battery. -- **4.85** The Thevenin equivalent at terminals *a*-*b* of the linear network shown in Fig. 4.142 is to be determined by measurement. When a 10-kΩ resistor is connected to terminals *a-b*, the voltage *Vab* is measured as 20 V. When a 30-kΩ resistor is connected to the terminals, *Vab* is measured as 40 V. Determine: (a) the Thevenin equivalent at terminals *a*-*b*, (b) *Vab* when a 20-kΩ resistor is connected to terminals *a*-*b*. - -# **Figure 4.142** - -For Prob. 4.85. - -**4.86** A black box with a circuit in it is connected to a variable resistor. An ideal ammeter (with zero resistance) and an ideal voltmeter (with infinite resistance) are used to measure current and voltage as shown in Fig. 4.143. The results are shown in the table on the next page. - -- (a) Find *i* when *R* = 12 Ω. -- (b) Determine the maximum power from the box. - -| R(Ω) | V(V) | i(A) | -|------|------|------| -| 2 | 6 | 3 | -| 8 | 16 | 2 | -| 14 | 21 | 1.5 | - -**4.87** A transducer is modeled with a current source *Is* and a parallel resistance *Rs*. The current at the terminals of the source is measured to be 9.975 mA when an ammeter with an internal resistance of 20 Ω is used. - -- (a) If adding a 2-kΩ resistor across the source terminals causes the ammeter reading to fall to 9.876 mA, calculate *Is* and *Rs*. -- (b) What will the ammeter reading be if the resistance between the source terminals is changed to 4 kΩ? -- **4.88** Consider the circuit in Fig. 4.144. An ammeter with internal resistance *Ri* is inserted between *A* and *B* to measure *Io*. Determine the reading of the ammeter if: (a) *Ri* = 500 Ω, (b) *Ri* = 0 Ω. (*Hint:* Find the Thevenin equivalent circuit at terminals *a*-*b*.) - -**Figure 4.144** - -For Prob. 4.88. - -**4.89** Consider the circuit in Fig. 4.145. (a) Replace the resistor *RL* by a zero resistance ammeter and determine the ammeter reading. (b) To verify the reciprocity theorem, interchange the ammeter and the 12-V source and determine the ammeter reading again. - -**Figure 4.145** For Prob. 4.89. - -**4.90** The Wheatstone bridge circuit shown in Fig. 4.146 is used to measure the resistance of a strain gauge. The adjustable resistor has a linear taper with a maximum value of 100 Ω. If the resistance of the strain gauge is found to be 42.6 Ω, what fraction of the full slider travel is the slider when the bridge is balanced? - -For Prob. 4.90. - -- can measure *Rx* in the range of 0–25 Ω. -- (b) Repeat for the range of 0–250 Ω. - -# **Figure 4.147** - -For Prob. 4.91. - -**4.92** Consider the bridge circuit of Fig. 4.148. Is the bridge balanced? If the 10-kΩ resistor is replaced by an 18-kΩ resistor, what resistor connected between terminals *a*-*b* absorbs the maximum power? What is this power? **\*** - -For Prob. 4.92. - -**\*** - -# Comprehensive Problems - -**4.93** The circuit in Fig. 4.149 models a common-emitter transistor amplifier. Find *ix* using source transformation. - -# **Figure 4.149** - -For Prob. 4.93. - -**4.94** An attenuator is an interface circuit that reduces the voltage level without changing the output resistance. - -> (a) By specifying *Rs* and *Rp* of the interface circuit in Fig. 4.150, design an attenuator that will meet the following requirements: - -$$ -\frac{V_o}{V_g} = 0.125, \qquad R_{\text{eq}} = R_{\text{Th}} = R_g = 100 \ \Omega -$$ - -(b) Using the interface designed in part (a), calculate the current through a load of *RL* = 50 Ω when *Vg* = 12 V. - -For Prob. 4.94. - -**4.95** A dc voltmeter with a sensitivity of 10 kΩ∕V is used to find the Thevenin equivalent of a linear network. Readings on two scales are as follows: **\*** - -(a) 0–10 V scale: 8 V (b) 0–50 V scale: 10 V - -Obtain the Thevenin voltage and the Thevenin resistance of the network. - -**4.96** A resistance array is connected to a load resistor *R* and a 9-V battery as shown in Fig. 4.151. - -- (a) Find the value of *R* such that *Vo* = 1.8 V. -- (b) Calculate the value of *R* that will draw the maximum current. What is the maximum current? - -# **Figure 4.151** For Prob. 4.96. - -**4.97** A common-emitter amplifier circuit is shown in Fig. 4.152. Obtain the Thevenin equivalent to the left of points *B* and *E*. - -# **Figure 4.152** - -For Prob. 4.97. - -**4.98** For Practice Prob. 4.18, determine the current through the 40-Ω resistor and the power dissipated by the resistor. **\*** - -*This page intentionally left blank* - -# **chapter** - -5 - -# Operational Amplifiers - -*He who will not reason is a bigot; he who cannot is a fool; and he who dares not is a slave.* - -—Lord Byron - -# Enhancing Your Career - -# **Career in Electronic Instrumentation** - -Engineering involves applying physical principles to design devices for the benefit of humanity. But ph ysical principles cannot be understood without measurement. In f act, ph ysicists often say that ph ysics is the science that measures reality. Just as measurements are a tool for understanding the physical world, instruments are tools for measurement. The operational amplifier introduced in this chapter is a building block of modern electronic instrumentation. Therefore, mastery of operational amplifier fundamentals is paramount to any practical application of electronic circuits. - -Electronic instruments are used in all fields of science and engineering. They have proliferated in science and technology to the e xtent that it would be ridiculous to have a scientific or technical education without exposure to electronic instruments. For example, physicists, physiologists, chemists, and biologists must learn to use electronic instruments. For electrical engineering students in particular, the skill in operating digital and analog electronic instruments is crucial. Such instruments include ammeters, v oltmeters, ohmmeters, oscilloscopes, spectrum analyzers, and signal generators. - -Beyond de veloping the skill for operating the instruments, some electrical engineers specialize in designing and constructing electronic instruments. These engineers deri ve pleasure in b uilding their o wn instruments. Most of them invent and patent their inventions. Specialists in electronic instruments find employment in medical schools, hospitals, research laboratories, aircraft industries, and thousands of other indus tries where electronic instruments are routinely used. - -Electronic Instrumentation used in medical research. © Royalty-Free/Corbis - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Comprehend how real operational amplifiers (op amps) function. -- 2. Understand that ideal op amps function nearly identically to real ones and that they can be used to model them effectively in a variety of circuit applications. -- 3. Realize how the basic inverting op amp is the workhorse of the op amp family. -- 4. Use the inverting op amp to create summers. -- 5. Use the op amp to create a difference amplifier. -- 6. Explain how to cascade a variety of op amp circuits. - -# **5.1** Introduction - -Having learned the basic la ws and theorems for circuit analysis, we are now ready to study an acti ve circuit element of paramount importance: the *operational amplifier,* or *op amp* for short. The op amp is a versatile circuit building block. - -The op amp is an electronic unit that behaves like a voltage-controlled voltage source. - -It can also be used in making a v oltage- or current -controlled current source. An op amp can sum signals, amplify a signal, inte grate it, or differentiate it. The ability of the op amp to perform these mathematical operations is the reason it is called an *operational amplifier*. It is also the reason for the widespread use of op amps in analog design. Op amps are popular in practical circuit designs because the y are versatile, inexpensive, easy to use, and fun to work with. - -We begin by discussing the ideal op amp and later consider the nonideal op amp. Using nodal analysis as a tool, we consider ideal op amp circuits such as the in verter, voltage follower, summer, and dif ference amplifier. We will also analyze op amp circuits with *PSpice*. Finally, we learn how an op amp is used in digital-to-analog converters and instrumentation amplifiers. - -# **5.2** Operational Amplifiers - -An operational amplifier is designed so that it performs some mathematical operations when e xternal components, such as resistors and capaci tors, are connected to its terminals. Thus, - -An op amp is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation, and integration. - -The op amp is an electronic device consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. A full discussion of what is inside the op amp is beyond the scope of this book. It will suffice - -The term operational amplifier was introduced in 1947 by John Ragazzini and his colleagues, in their work on analog computers for the National Defense Research Council after World War II. The first op amps used vacuum tubes rather than transistors. - -An op amp may also be regarded as a voltage amplifier with very high gain. - -to treat the op amp as a circuit b uilding block and simply study what takes place at its terminals. - -Op amps are commercially a vailable in integrated circuit packages in several forms. Figure 5.1 sho ws a typical op amp package. A typical one is the eight-pin dual in-line package (or DIP), shown in Fig. 5.2(a). Pin or terminal 8 is unused, and terminals 1 and 5 are of little concern to us. The five important terminals are: - -- 1. The inverting input, pin 2. -- 2. The noninverting input, pin 3. -- 3. The output, pin 6. -- 4. The positive power supply *V*+, pin 7. -- 5. The negative power supply *V*−, pin 4. - -The circuit symbol for the op amp is the triangle in Fig. 5.2(b); as shown, the op amp has tw o inputs and one output. The inputs are mark ed with minus (−) and plus (+) to specify *inverting* and *noninverting* inputs, respectively. An input applied to the noninverting terminal will appear with the same polarity at the output, while an input applied to the in verting terminal will appear inverted at the output. - -As an active element, the op amp must be powered by a voltage supply as typically shown in Fig. 5.3. Although the power supplies are often ignored in op amp circuit diagrams for the sake of simplicity, the power supply currents must not be overlooked. By KCL, - -$$ -i_o = i_1 + i_2 + i_+ + i_- \tag{5.1} -$$ - -The equivalent circuit model of an op amp is shown in Fig. 5.4. The output section consists of a v oltage-controlled source in series with the - -**Figure 5.2** A typical op amp: (a) pin configuration, (b) circuit symbol. - -Powering the op amp. - -**Figure 5.4** The equivalent circuit of the nonideal op amp. - -# **Figure 5.1** - -A typical operational amplifier. © McGraw-Hill Education/Mark Dierker, photographer - -The pin diagram in Fig. 5.2(a) corresponds to the 741 generalpurpose op amp made by Fairchild Semiconductor. - -output resistance *Ro*. It is evident from Fig. 5.4 that the input resis tance *Ri* is the Thevenin equivalent resistance seen at the input terminals, while the output resistance *Ro* is the Thevenin equivalent resistance seen at the output. The differential input voltage *vd* is given by - -$$ -v_d = v_2 - v_1 \tag{5.2} -$$ - -where *v*1 is the voltage between the inverting terminal and ground and *v*2 is the voltage between the noninverting terminal and ground. The op amp senses the difference between the two inputs, multiplies it by the gain *A*, and causes the resulting voltage to appear at the output. Thus, the output *vo* is given by - -$$ -v_o = Av_d = A(v_2 - v_1) -$$ - (5.3) - -Sometimes, voltage gain is expressed in decibels (dB), as discussed in Chapter 14. - -*A* dB = 20 log10 A - -*A* is called the *open-loop voltage gain* because it is the g ain of the op amp without any external feedback from output to input. Table 5.1 shows typical values of voltage gain *A*, input resistance *Ri*, output resistance *Ro*, and supply voltage *VCC*. - -The concept of feedback is crucial to our understanding of op amp circuits. A negative feedback is achieved when the output is fed back to the inverting terminal of the op amp. As Example 5.1 shows, when there is a feedback path from output to input, the ratio of the output voltage to the input voltage is called the *closed-loop gain*. As a result of the ne gative feedback, it can be shown that the closed-loop gain is almost insensitive to the open-loop gain *A* of the op amp. For this reason, op amps are used in circuits with feedback paths. - -A practical limitation of the op amp is that the magnitude of its output voltage cannot e xceed |*VCC*|. In other w ords, the output v oltage is dependent on and is limited by the po wer supply voltage. Figure 5.5 illustrates that the op amp can operate in three modes, depending on the differential input voltage *vd*: - -- 1. Positive saturation, *vo* = *VCC*. -- 2. Linear region, −*VCC* ≤ *vo* = *Avd* ≤ *VCC*. -- 3. Negative saturation, *vo* = −*VCC*. - -If we attempt to increase *vd* beyond the linear range, the op amp becomes saturated and yields *vo* = *VCC* or *vo* = −*VCC*. Throughout this book, we will assume that our op amps operate in the linear mode. This means that the output voltage is restricted by - -$$ --V_{CC} \le v_o \le V_{CC} \tag{5.4} -$$ - -# **TABLE 5.1** - -# Typical ranges for op amp parameters. - -| Parameter | Typical range | Ideal values | -|----------------------------------------------|--------------------------|--------------| -| Open-loop gain, A | 105
to 108 | ` | -| | 105 | `Ω | -| Output resistance, Ro
Supply voltage, VCC | 10 to 100 Ω
5 to 24 V | 0Ω | -| | Input resistance, Ri | to 1013 Ω | - -Op amp output voltage *vo* as a function of the differential input voltage *vd*. - -, input resis tance - -Although we shall al ways operate the op amp in the linear re gion, the possibility of saturation must be borne in mind when one designs with op amps, to a void designing op amp circuits that will not w ork in the laboratory. - -of 2 MΩ, and output resistance of 50 Ω. The op amp is used in the circuit of Fig. 5.6(a). Find the closed-loop gain *vo*∕*vs*. Determine current *i* when - -Throughout this book, we assume that an op amp operates in the linear range. Keep in mind the voltage constraint on the op amp in this mode. - -Example 5.1 - -**Figure 5.6** - -For Example 5.1: (a) original circuit, (b) the equivalent circuit. - -A 741 op amp has an open-loop voltage gain of 2 × 105 - -# **Solution:** - -Using the op amp model in Fig. 5.4, we obtain the equivalent circuit of Fig. 5.6(a) as shown in Fig. 5.6(b). We now solve the circuit in Fig. 5.6(b) by using nodal analysis. At node 1, KCL gives - -$$ -\frac{v_s - v_1}{10 \times 10^3} = \frac{v_1}{2000 \times 10^3} + \frac{v_1 - v_o}{20 \times 10^3} -$$ - -Multiplying through by 2000 × 103 , we obtain - -$$ -200v_s = 301v_1 - 100v_o -$$ - -or - -$$ -2v_s \simeq 3v_1 - v_o \quad \Rightarrow \quad v_1 = \frac{2v_s + v_o}{3} \tag{5.1.1} -$$ - -At node *O*, - -$$ -\frac{v_1 - v_o}{20 \times 10^3} = \frac{v_o - Av_d}{50} -$$ - -But *vd* = −*v*1 and *A* = 200,000. Then - -$$ -v_1 - v_o = 400(v_o + 200,000v_1) -$$ -\n(5.1.2) - -Substituting *v*1 from Eq. (5.1.1) into Eq. (5.1.2) gives - -$$ -0 \simeq 26,667,067v_o + 53,333,333v_s \quad \Rightarrow \quad \frac{v_o}{v_s} = -1.9999699 -$$ - -This is closed -loop gain, because the 20 -kΩ feedback resistor closes the loop between the output and input terminals. When *vs* = 2 V, *vo* = −3.9999398 V. From Eq. (5.1.1), we obtain *v*1 = 20.066667 *μ*V. Thus, - -$$ -i = \frac{v_1 - v_o}{20 \times 10^3} = 0.19999 \text{ mA} -$$ - -It is evident that working with a nonideal op amp is tedious, as we are dealing with very large numbers. - -# Practice Problem 5.1 - -**Figure 5.7** For Practice Prob. 5.1. - -If the same 741 op amp in Example 5.1 is used in the circuit of Fig. 5.7, calculate the closed-loop gain *v*o∕*v*s. Find *io* when *vs* = 1 V. - -**Answer:** 9.00041, 657 *μ*A. - -# **5.3** Ideal Op Amp - -To facilitate the understanding of op amp circuits, we will assume ideal op amps. An op amp is ideal if it has the following characteristics: - -- 1. Infinite open-loop gain, *A* ≃ ∞. -- 2. Infinite input resistance, *Ri* ≃ ∞. -- 3. Zero output resistance, *Ro* ≃ 0. - -An ideal op amp is an amplifier with infinite open-loop gain, infinite input resistance, and zero output resistance. - -Although assuming an ideal op amp provides only an approximate analysis, most modern amplifiers have such large gains and input im pedances that the approximate analysis is a good one. Unless stated otherwise, we will assume from now on that every op amp is ideal. - -For circuit analysis, the ideal op amp is illustrated in Fig. 5.8, which is derived from the nonideal model in Fig. 5.4. Two important properties of the ideal op amp are: - -1. The currents into both input terminals are zero: - -$$ -i_1 = 0, \quad i_2 = 0 -$$ - (5.5) - -This is due to infinite input resistance. An infinite resistance between the input terminals implies that an open circuit e xists there and current cannot enter the op amp. But the output current is not necessarily zero according to Eq. (5.1). - -2. The voltage across the input terminals is equal to zero; i.e., - -$$ -v_d = v_2 - v_1 = 0 \tag{5.6} -$$ - -or - -$$ -v_1 = v_2 \tag{5.7} -$$ - -The two characteristics can be exploited by noting that for voltage calculations the input port behaves as a short circuit, while for current calculations the input port behaves as an open circuit. - -Thus, an ideal op amp has zero current into its tw o input terminals and the v oltage between the tw o input terminals is equal to zero. Equations (5.5) and (5.7) are e xtremely important and should be regarded as the key handles to analyzing op amp circuits. - -Rework Practice Prob. 5.1 using the ideal op amp model. Example 5.2 - -# **Solution:** - -We may replace the op amp in Fig. 5.7 by its equivalent model in Fig. 5.9 as we did in Example 5.1. But we do not really need to do this. We just need to keep Eqs. (5.5) and (5.7) in mind as we analyze the circuit in Fig. 5.7. Thus, the Fig. 5.7 circuit is presented as in Fig. 5.9. Notice that - -$$ -v_2 = v_s \tag{5.2.1} -$$ - -Since *i*1 = 0, the 40 - and 5 -kΩ resistors are in series; the same current flows through them. *v*1 is the voltage across the 5 -kΩ resistor. Hence, using the voltage division principle, - -$$ -v_1 = \frac{5}{5+40} v_o = \frac{v_o}{9} -$$ - (5.2.2) - -According to Eq. (5.7), - -$$ -v_2 = v_1 \tag{5.2.3} -$$ - -Substituting Eqs. (5.2.1) and (5.2.2) into Eq. (5.2.3) yields the closed loop gain, - -$$ -v_s = \frac{v_o}{9} \quad \Rightarrow \quad \frac{v_o}{v_s} = 9 \tag{5.2.4} -$$ - -which is very close to the value of 9.00041 obtained with the nonideal model in Practice Prob. 5.1. This shows that negligibly small error re sults from assuming ideal op amp characteristics. - -At node *O*, - -$$ -i_o = \frac{v_o}{40 + 5} + \frac{v_o}{20} \text{ mA} -$$ - (5.2.5) - -From Eq. (5.2.4), when *vs* = 1 V, *vo* = 9 V. Substituting for *vo* = 9 V in Eq. (5.2.5) produces - -$$ -i_o = 0.2 + 0.45 = 0.65 \text{ mA} -$$ - -This, again, is close to the value of 0.657 mA obtained in Practice Prob. 5.1 with the nonideal model. - -Repeat Example 5.1 using the ideal op amp model. Practice Problem 5.2 - -**Answer:** −2, 200 *μ*A. - -# **5.4** Inverting Amplifier - -In this and the following sections, we consider some useful op amp cir cuits that often serv e as modules for designing more comple x circuits. The first of such op amp circuits is the inverting amplifier shown in Fig. 5.10. In this circuit, the nonin verting input is grounded, *vi* is con nected to the in verting input through *R*1, and the feedback resistor *Rf* is connected between the inverting input and output. Our goal is to obtain - -+ – *v*2 0 V + + - -– 1 - -i 2 - -**Figure 5.10** The inverting amplifier. - -*v*i - -R1 - -i 1 - -*v*1 0 A - -For Example 5.2. - -Rf - -– - -*v*o + - -– - -the relationship between the input v oltage *v*i and the output v oltage *vo*. Applying KCL at node 1, - -$$ -i_1 = i_2 \Rightarrow \frac{v_i - v_1}{R_1} = \frac{v_1 - v_o}{R_f} -$$ - (5.8) - -But *v*1 = *v*2 = 0 for an ideal op amp, since the nonin verting terminal is grounded. Hence, - -or - -$$ -\frac{v_i}{R_1} = -\frac{v_o}{R_f} -$$ -$$ -v_o = -\frac{R_f}{R_1}v_i -$$ -(5.9) - -The voltage gain is *Av* = *vo*∕*vi* = −*Rf*∕*R*1. The designation of the circuit in Fig. 5.10 as an *inverter* arises from the negative sign. Thus, - -An inverting amplifier reverses the polarity of the input signal while amplifying it. - -Notice that the gain is the feedback resistance di vided by the in put resistance which means that the gain depends only on the external elements connected to the op amp. In view of Eq. (5.9), an equivalent circuit for the inverting amplifier is shown in Fig. 5.11. The inverting amplifier is used, for example, in a current-to-voltage converter. - -Example 5.3 Refer to the op amp in Fig. 5.12. If *vi* = 0.5 V, calculate: (a) the output voltage *vo*, and (b) the current in the 10-kΩ resistor. - -# **Solution:** - -(a) Using Eq. (5.9), - -$$ -\frac{v_o}{v_i} = -\frac{R_f}{R_1} = -\frac{25}{10} = -2.5 -$$ - -$$ -v_o = -2.5v_i = -2.5(0.5) = -1.25 -$$ - V - -(b) The current through the 10-kΩ resistor is - -$$ -i = \frac{v_i - 0}{R_1} = \frac{0.5 - 0}{10 \times 10^3} = 50 \,\mu\text{A} -$$ - -Practice Problem 5.3 Find the output of the op amp circuit shown in Fig. 5.13. Calculate the current through the feedback resistor. - -**Answer:** −3.15 V, 11.25 *μ*A. - -A key feature of the inverting amplifier is that both the input signal and the feedback are applied at the inverting terminal of the op amp. - -Note there are two types of gains: The one here is the closed-loop voltage gain Av, while the op amp itself has an - -open-loop voltage gain A. - -# **Figure 5.11** - -An equivalent circuit for the inverter in Fig. 5.10. - -**Figure 5.13** For Practice Prob. 5.3. Determine *vo* in the op amp circuit shown in Fig. 5.14. - -# **Solution:** - -Applying KCL at node *a*, - -$$ -\frac{v_a - v_o}{40 \text{ k}\Omega} = \frac{6 - v_a}{20 \text{ k}\Omega} -$$ - -$$ -v_a - v_o = 12 - 2v_a \implies v_o = 3v_a - 12 -$$ - -But *va* = *vb* = 2 V for an ideal op amp, because of the zero voltage drop across the input terminals of the op amp. Hence, - -$$ -v_o = 6 - 12 = -6 -$$ - V - -Notice that if *vb* = 0 = *va*, then *vo* = −12, as expected from Eq. (5.9). - -For Example 5.4. - -Two kinds of current -to-voltage converters (also known as *transresis-* Practice Problem 5.4 *tance amplifiers*) are shown in Fig. 5.15. - -(a) Show that for the converter in Fig. 5.15(a), - -$$ -\frac{v_o}{i_s} = -R -$$ - -(b) Show that for the converter in Fig. 5.15(b), - -$$ -\frac{v_o}{i_s} = -R_1 \left( 1 + \frac{R_3}{R_1} + \frac{R_3}{R_2} \right) -$$ - -**Answer:** Proof. - -Another important application of the op amp is the nonin verting amplifier shown in Fig. 5.16. In this case, the input v oltage *vi* is applied directly at the nonin verting input terminal, and resistor *R*1 is connected - -**Figure 5.16** The noninverting amplifier. - -Example 5.4 - -between the ground and the in verting terminal. We are interested in the output voltage and the voltage gain. Application of KCL at the inverting terminal gives - -$$ -i_1 = i_2 \Rightarrow \frac{0 - v_1}{R_1} = \frac{v_1 - v_o}{R_f} -$$ - (5.10) - -But *v*1 = *v*2 = *vi*. Equation (5.10) becomes - -$$ -\frac{-v_i}{R_1} = \frac{v_i - v_o}{R_f} -$$ - -or - -$$ -v_o = \left(1 + \frac{R_f}{R_1}\right) v_i -$$ - (5.11) - -The voltage gain is *Av* = *vo*∕*vi* = 1 + *Rf*∕*R*1, which does not have a negative sign. Thus, the output has the same polarity as the input. - -A noninverting amplifier is an op amp circuit designed to provide a positive voltage gain. - -Again we notice that the gain depends only on the external resistors. - -Notice that if feedback resistor *Rf* = 0 (short circuit) or *R*1 = ∞ (open circuit) or both, the gain becomes 1. Under these conditions (*Rf* = 0 and *R*1 = ∞), the circuit in Fig. 5.16 becomes that shown in Fig. 5.17, which is called a *voltage follower* (or *unity gain amplifier*) because the output follows the input. Thus, for a voltage follower - -$$ -v_o = v_i \tag{5.12} -$$ - -Such a circuit has a ve ry high input impedance and is therefore use ful as an intermediate-stage (or buffer) amplifier to isolate one circuit from another, as portrayed in Fig. 5.18. The voltage follower minimizes interaction between the tw o stages and eliminates interstage loading. - -Example 5.3 For the op amp circuit in Fig. 5.19, calculate the output voltage *vo*. - -# **Solution:** - -We may solve this in two ways: using superposition and using nodal analysis. - -■ **METHOD 1** Using superposition, we let - -**Figure 5.17** The voltage follower. - -**Figure 5.18** A voltage follower used to isolate two cascaded stages of a circuit. - -where *vo*1 is due to the 6-V voltage source, and *vo*2 is due to the 4-V input. To get *vo*1, we set the 4-V source equal to zero. Under this condition, the circuit becomes an inverter. Hence Eq. (5.9) gives - -$$ -v_{o1} = -\frac{10}{4}(6) = -15 -$$ - V - -To get *vo*2, we set the 6 -V source equal to zero. The circuit becomes a noninverting amplifier so that Eq. (5.11) applies. - -$$ -v_{o2} = \left(1 + \frac{10}{4}\right)4 = 14 \text{ V} -$$ - -Thus, - -$$ -v_o = v_{o1} + v_{o2} = -15 + 14 = -1 -$$ - V - -■ **METHOD 2** Applying KCL at node *a*, - -$$ -\frac{6-v_a}{4} = \frac{v_a - v_o}{10} -$$ - -But *va*= *vb*=4, and so - -$$ -\frac{6-4}{4} = \frac{4-v_o}{10} \Rightarrow 5 = 4-v_o -$$ - -or *vo*= −1 V, as before. - -Calculate *vo* in the circuit of Fig. 5.20. Practice Problem 5.5 - -**Answer:** 21 V. - -# **5.6** Summing Amplifier - -Besides amplification, the op amp can perform addition and subtraction. The addition is performed by the summing amplifier covered in this section; the subtraction is performed by the difference amplifier covered in the next section. - -A summing amplifier is an op amp circuit that combines several inputs and produces an output that is the weighted sum of the inputs. - -The summing amplifier, shown in Fig. 5.21, is a v ariation of the inverting amplifier. It takes advantage of the fact that the inverting configuration can handle many inputs at the same time. We keep in mind - -**Figure 5.19** For Example 5.5. - -+ - -4 kΩ - -6 V *v*o - -– - -a b - -4 V - -– + - -+ - -# **Figure 5.20** - -4 kΩ - -For Practice Prob. 5.5. - -**Figure 5.21** The summing amplifier. - -+ - -– - -10 kΩ - -+ – that the current entering each op amp input is zero. Applying KCL at node *a* gives - -$$ -i = i_1 + i_2 + i_3 \tag{5.13} -$$ - -But - -$$ -i_1 = \frac{v_1 - v_a}{R_1}, \quad i_2 = \frac{v_2 - v_a}{R_2} -$$ - -\n -$$ -i_3 = \frac{v_3 - v_a}{R_3}, \quad i = \frac{v_a - v_o}{R_f} -$$ - (5.14) - -We note that *va* = 0 and substitute Eq. (5.14) into Eq. (5.13). We get - -$$ -v_o = -\left(\frac{R_f}{R_1}v_1 + \frac{R_f}{R_2}v_2 + \frac{R_f}{R_3}v_3\right) -$$ - (5.15) - -indicating that the output v oltage is a weighted sum of the inputs. F or this reason, the circuit in Fig. 5.21 is called a *summer*. Needless to say, the summer can have more than three inputs. - -# Example 5.6 - -Calculate *vo* and *io* in the op amp circuit in Fig. 5.22. - -**Figure 5.22** For Example 5.6. - -# **Solution:** - -This is a summer with two inputs. Using Eq. (5.15) gives - -$$ -v_o = -\left[\frac{10}{5}(2) + \frac{10}{2.5}(1)\right] = -(4 + 4) = -8 -$$ - V - -The current *io* is the sum of the currents through the 10- and 2-kΩ resistors. Both of these resistors have voltage *vo* = −8 V across them, since *va* = *vb* = 0. Hence, - -$$ -i_o = \frac{v_o - 0}{10} + \frac{v_o - 0}{2} -$$ - mA = -0.8 - 4 = -4.8 mA - -Find *vo* and *io* in the op amp circuit shown in Fig. 5.23. - -**Answer:** −3.8 V, −1.425 mA. - -or - -# **5.7** Difference Amplifier - -Difference (or dif ferential) amplifiers are used in various applications where there is a need to amplify the dif ference between tw o input sig nals. They are first cousins of the *instrumentation amplifier,* the most useful and popular amplifier, which we will discuss in Section 5.10. - -A difference amplifier is a device that amplifies the difference between two inputs but rejects any signals common to the two inputs. - -Consider the op amp circuit sho wn in Fig. 5.24. K eep in mind that zero currents enter the op amp terminals. Applying KCL to node *a*, - -> \_\_\_\_\_\_ *v*1 − *va R*1 = \_\_\_\_\_\_ *va* − *vo R*2 - -$$ -v_o = \left(\frac{R_2}{R_1} + 1\right) v_a - \frac{R_2}{R_1} v_1 \tag{5.16} -$$ - -Difference amplifier. - -The difference amplifier is also known as the subtractor, for reasons to be shown later. - -# Practice Problem 5.6 - -Applying KCL to node *b*, - -$$ -\frac{v_2 - v_b}{R_3} = \frac{v_b - 0}{R_4} -$$ - -or - -$$ -v_b = \frac{R_4}{R_3 + R_4} v_2 \tag{5.17} -$$ - -But *va* = *vb*. Substituting Eq. (5.17) into Eq. (5.16) yields - -$$ -v_o = \left(\frac{R_2}{R_1} + 1\right) \frac{R_4}{R_3 + R_4} v_2 - \frac{R_2}{R_1} v_1 -$$ - -or - -$$ -v_o = \frac{R_2(1 + R_1/R_2)}{R_1(1 + R_3/R_4)} v_2 - \frac{R_2}{R_1} v_1 -$$ - (5.18) - -Since a difference amplifier must reject a signal common to the two inputs, the amplifier must have the property that *vo* = 0 when *v*1 = *v*2. This property exists when - -$$ -\frac{R_1}{R_2} = \frac{R_3}{R_4} -$$ -\n(5.19) - -Thus, when the op amp circuit is a dif ference amplifier, Eq. (5.18) becomes - -$$ -v_o = \frac{R_2}{R_1}(v_2 - v_1) -$$ -\n(5.20) - -If *R*2 = *R*1 and *R*3 = *R*4, the difference amplifier becomes a *subtractor,* with the output - -$$ -v_o = v_2 - v_1 \tag{5.21} -$$ - -Example 5.7 Design an op amp circuit with inputs *v*1 and *v*2 such that *vo* = −5*v*1 + 3*v*2. - -# **Solution:** - -The circuit requires that - -$$ -v_o = 3v_2 - 5v_1 \tag{5.7.1} -$$ - -This circuit can be realized in two ways. - -**Design 1** If we desire to use only one op amp, we can use the op amp circuit of Fig. 5.24. Comparing Eq. (5.7.1) with Eq. (5.18), we see - -$$ -\frac{R_2}{R_1} = 5 \quad \Rightarrow \quad R_2 = 5R_1 \tag{5.7.2} -$$ - -Also, - -$$ -5\frac{(1+R_1/R_2)}{(1+R_3/R_4)} = 3 \quad \Rightarrow \quad \frac{\frac{6}{5}}{1+R_3/R_4} = \frac{3}{5} -$$ - -or - -$$ -2 = 1 + \frac{R_3}{R_4} \Rightarrow R_3 = R_4 \tag{5.7.3} -$$ - -If we choose *R*1 = 10 k Ω and *R*3 = 20 k Ω, then *R*2 = 50 k Ω and *R*4 = 20 kΩ. - -**Design 2** If we desire to use more than one op amp, we may cascade an inverting amplifier and a two-input inverting summer, as shown in Fig. 5.25. For the summer, - -$$ -v_o = -v_a - 5v_1 \tag{5.7.4} -$$ - -and for the inverter, - -$$ -v_a = -3v_2 \tag{5.7.5} -$$ - -Combining Eqs. (5.7.4) and (5.7.5) gives - -*vo* = 3*v*2 − 5*v*1 - -which is the desired result. In Fig. 5.25, we may select *R*1 = 10 kΩ and *R*3 = 20 kΩ or *R*1 = *R*3 = 10 kΩ. - -Design a difference amplifier with gain 7.5. Practice Problem 5.7 - -**Answer:** Typical: *R*1 = *R*3 = 20 kΩ, *R*2 = *R*4 = 150 kΩ. - -An *instrumentation amplifier* shown in Fig. 5.26 is an amplifier of lowlevel signals used in process control or measurement applications and commercially available in single-package units. Show that - -$$ -v_o = \frac{R_2}{R_1} \left( 1 + \frac{2R_3}{R_4} \right) (v_2 - v_1) -$$ - -# **Solution:** - -We recognize that the amplifier *A*3 in Fig. 5.26 is a difference amplifier. Thus, from Eq. (5.20), - -$$ -v_o = \frac{R_2}{R_1}(v_{o2} - v_{o1}) -$$ -\n(5.8.1) - -Since the op amps *A*1 and *A*2 draw no current, current *i* flows through the three resistors as though they were in series. Hence, - -$$ -v_{o1} - v_{o2} = i(R_3 + R_4 + R_3) = i(2R_3 + R_4) -$$ - (5.8.2) - -**Figure 5.25** For Example 5.7. - -Example 5.8 - -**Figure 5.26** Instrumentation amplifier; for Example 5.8. - -But - -$$ -i = \frac{v_a - v_b}{R_4} -$$ - -and *va* = *v*1, *vb* = *v*2. Therefore, - -$$ -i = \frac{v_1 - v_2}{R_4} -$$ - (5.8.3) - -Inserting Eqs. (5.8.2) and (5.8.3) into Eq. (5.8.1) gives - -$$ -v_o = \frac{R_2}{R_1} \left( 1 + \frac{2R_3}{R_4} \right) (v_2 - v_1) -$$ - -as required. We will discuss the instrumentation amplifier in detail in Section 5.10. - -Obtain *io* Practice Problem 5.8 in the instrumentation amplifier circuit of Fig. 5.27. - -# **Figure 5.27** - -Instrumentation amplifier; for Practice Prob. 5.8. - -**Answer:** 800 *μ*A. - -# **5.8** Cascaded Op Amp Circuits - -As we know, op amp circuits are modules or building blocks for designing complex circuits. It is often necessary in practical applications to connect op amp circuits in cascade (i.e., head to tail) to achie ve a large overall gain. In general, tw o circuits are cascaded when the y are con nected in tandem, one behind another in a single file. - -A cascade connection is a head-to-tail arrangement of two or more op amp circuits such that the output of one is the input of the next. - -When op amp circuits are cascaded, each circuit in the string is called a *stage;* the original input signal is increased by the g ain of the individual stage. Op amp circuits have the advantage that they can be cascaded without changing their input -output relationships. This is due to the fact that each (ideal) op amp circuit has infinite input resistance and zero output resistance. Figure 5.28 displays a block diagram represen tation of three op amp circuits in cascade. Since the output of one stage is the input to the next stage, the overall gain of the cascade connection is the product of the gains of the individual op amp circuits, or - -$$ -A = A_1 A_2 A_3 \tag{5.22} -$$ - -Although the cascade connection does not afect the op amp input-output relationships, care must be e xercised in the design of an actual op amp circuit to ensure that the load due to the ne xt stage in the cascade does not saturate the op amp. - -**Figure 5.28** - -A three-stage cascaded connection. - -Find *vo* and *io* in the circuit in Fig. 5.29. Example 5.9 - -# **Solution:** - -This circuit consists of two noninverting amplifiers cascaded. At the output of the first op amp, - -$$ -v_a = \left(1 + \frac{12}{3}\right)(20) = 100 \text{ mV} -$$ - -At the output of the second op amp, - -$$ -v_o = \left(1 + \frac{10}{4}\right) v_a = (1 + 2.5)100 = 350 -$$ - mV - -The required current *io* is the current through the 10-kΩ resistor. - -$$ -i_o = \frac{v_o - v_b}{10} -$$ - mA - -# **Figure 5.29** For Example 5.9. - -But -$$ -v_b = v_a = 100 -$$ - mV. Hence, - -mV. Hence, -$$ -i_o = \frac{(350 - 100) \times 10^{-3}}{10 \times 10^3} = 25 \,\mu\text{A} -$$ - -``` -For Practice Prob. 5.9. -``` - -| Example 5.10 | -|--------------| -|--------------| - -If *v*1 = 1 V and *v*2 = 2 V, find *vo* in the op amp circuit of Fig. 5.31. - -For Example 5.10. - -# **Solution:** - -- 1. **Define.** The problem is clearly defined. -- 2. **Present.** With an input of *v*1 of 1 V and of *v*2 of 2 V, determine the output voltage of the circuit shown in Figure 5.31. The op amp circuit is actually composed of three circuits. The first circuit acts as an amplifier of gain −3(−6 kΩ∕2 kΩ) for *v*1 and the second functions as an amplifier of gain −2(−8 kΩ∕4 kΩ) for *v*2. The last circuit serves as a summer of two different gains for the output of the other two circuits. -- 3. **Alternative.** There are different ways of working with this circuit. Because it involves ideal op amps, then a purely mathematical - -approach will work quite easily. A second approach would be to use *PSpice* as a confirmation of the math. - -4. **Attempt.** Let the output of the first op amp circuit be designated as *v*11 and the output of the second op amp circuit be designated as *v*22. Then we get - -$$ -v_{11} = -3v_1 = -3 \times 1 = -3 \text{ V}, -$$ - -$$ -v_{22} = -2v_2 = -2 \times 2 = -4 \text{ V} -$$ - -In the third circuit we have - -$$ -v_o = -(10 \text{ k}\Omega/5 \text{ k}\Omega)v_{11} + [-(10 \text{ k}\Omega/15 \text{ k}\Omega)v_{22}] -$$ - -= -2(-3) - (2/3)(-4) -= 6 + 2.667 = **8.667 V** - - 5. **Evaluate.** To properly evaluate our solution, we need to identify a reasonable check. Here we can easily use *PSpice* to provide that check. - - Now we can simulate this in *PSpice*. The results are shown in Fig. 5.32. - - We obtain the same results using two entirely different techniques (the first is to treat the op amp circuits as just gains and a summer and the second is to use circuit analysis with *PSpice*). This is a very good method of assuring that we have the correct answer. - - 6. **Satisfactory?** We are satisfied we have obtained the asked for results. We can now present our work as a solution to the problem. - -Practice Problem 5.10 - -If *v*1 = 5 V and *v*2 = 5 V, find *v*o in the op amp circuit of Fig. 5.33. - -For Practice Prob. 5.10. - -**Answer:** 35 V. - -# **5.9** Op Amp Circuit Analysis with PSpice - -*PSpice for Windows* does not have a model for an ideal op amp, although one may create one as a subcircuit using the *Create Subcircuit* line in the *Tools* menu. Rather than creating an ideal op amp, we will use one of the four nonideal, commercially a vailable op amps supplied in the *PSpice library eval.slb*. The op amp models have the part names LF411, LM111, LM324, and uA741, as shown in Fig. 5.34. Each of them can be obtained from **Draw/Get New Part/libraries . . . /eval.lib** or by simply selecting **Draw/Get New Part** and typing the part name in the *PartName* dialog box, as usual. Note that each of them requires dc supplies, without which the op amp will not work. The dc supplies should be connected as shown in Fig. 5.3. - -(a) JFET–input op amp subcircuit - -(b) Op amp subcircuit - -(c) Five– connection op amp subcircuit - -(d) Five–connection op amp subcircuit - -**Figure 5.34** - -Nonideal op amp model available in *PSpice*. - -Use *PSpice* to solve the op amp circuit for Example 5.1. - -# **Solution:** - -Using Schematics, we draw the circuit in Fig. 5.6(a) as shown in Fig.�5.35. Notice that the positive terminal of the voltage source *v*s is connected to the inverting terminal (pin 2) via the 10-kΩ resistor, while the noninverting terminal (pin 3) is grounded as required in Fig. 5.6(a). Also, notice how the op amp is powered; the positive power supply terminal V+ (pin 7) is connected to a 15 -V dc voltage source, while the negative power supply terminal V− (pin 4) is connected to −15 V. Pins 1 and 5 are left floating because they are used for offset null adjustment, which does not concern us in this chapter. Besides adding the dc power supplies to the original circuit in Fig. 5.6(a), we have also added pseudocomponents VIEWPOINT and IPROBE to respectively measure the output voltage *v*o at pin 6 and the required current *i* through the 20-kΩ resistor. - -Schematic for Example 5.11. - -After sa ving the schematic, we simulate the circuit by selecting **Analysis/Simulate** and ha ve the results displayed on VIEWPOINT and IPROBE. From the results, the closed-loop gain is - -$$ -\frac{v_o}{v_s} = \frac{-3.9983}{2} = -1.99915 -$$ - -and *i* = 0.1999 mA, in agreement with the results obtained analytically in Example 5.1. - -**Answer:** 9.0027, 650.2 *μ*A. - -Rework Practice Prob. 5.1 using *PSpice*. Practice Problem 5.11 - -## **5.10** † Applications - -The op amp is a fundamental b uilding block in modern electronic instrumentation. It is used e xtensively in man y de vices, along with resistors and other passi ve elements. Its numerous practical applica tions include instrumentation amplifiers, digital-to-analog converters, analog computers, level shifters, filters, calibration circuits, inverters, summers, integrators, differentiators, subtractors, log arithmic amplifiers, comparators, gyrators, oscillators, rectifiers, regulators, voltageto- current con verters, current -to-voltage converters, and clippers. Some of these we have already considered. We will consider two more applications here: the digital-to-analog converter and the instrumentation amplifier. - -# **5.10.1** Digital-to-Analog Converter - -The digital-to-analog converter (DAC) transforms digital signals into analog form. A typical e xample of a four -bit DAC is illustrated in Fig. 5.36(a). The four-bit DAC can be realized in many ways. A simple realization is the *binary weighted ladder*, shown in Fig. 5.36(b). The bits are weights according to the magnitude of their place v alue, by descending value of *Rf*∕*Rn* so that each lesser bit has half the weight of the ne xt higher. This is ob viously an in verting summing amplifier. The output is related to the inputs as sho wn in Eq. (5.15). Thus, - -$$ --V_o = \frac{R_f}{R_1}V_1 + \frac{R_f}{R_2}V_2 + \frac{R_f}{R_3}V_3 + \frac{R_f}{R_4}V_4 -$$ - (5.23) - -Input *V*1 is called the *most significant bit* (MSB), while input *V*4 is the *least significant bit* (LSB). Each of the four binary inputs *V*1, . . . , *V*4 can assume only two voltage levels: 0 or 1 V. By using the proper input and feedback resistor values, the DAC provides a single output that is pro portional to the inputs. - -# Example 5.12 - -In the op amp circuit of Fig. 5.36(b), let *Rf* = 10 k Ω, *R*1 = 10 k Ω, *R*2 = 20 kΩ, *R*3 = 40 kΩ, and *R*4 = 80 kΩ. Obtain the analog output for binary inputs [0000], [0001], [0010], . . . , [1111]. - -# **Solution:** - -Substituting the given values of the input and feedback resistors in Eq. (5.23) gives - -$$ --V_o = \frac{R_f}{R_1}V_1 + \frac{R_f}{R_2}V_2 + \frac{R_f}{R_3}V_3 + \frac{R_f}{R_4}V_4 -$$ -$$ -= V_1 + 0.5V_2 + 0.25V_3 + 0.125V_3 -$$ - -Using this equation, a digital input [ *V*1*V*2*V*3*V*4] = [0000] produces an analog output of −*Vo* = 0 V; [*V*1*V*2*V*3*V*4] = [0001] gives −*Vo* = 0.125 V. - -(b) binary weighted ladder type. - - In practice, the voltage levels may be typically 0 and ± 5 V. - -Similarly, - -$$ -[V_1 V_2 V_3 V_4] = [0010] \Rightarrow -V_o = 0.25 \text{ V} -$$ - -\n -$$ -[V_1 V_2 V_3 V_4] = [0011] \Rightarrow -V_o = 0.25 + 0.125 = 0.375 \text{ V} -$$ - -\n -$$ -[V_1 V_2 V_3 V_4] = [0100] \Rightarrow -V_o = 0.5 \text{ V} -$$ - -\n: -\n -$$ -[V_1 V_2 V_3 V_4] = [1111] \Rightarrow -V_o = 1 + 0.5 + 0.25 + 0.125 -$$ - -\n -$$ -= 1.875 \text{ V} -$$ - -Table 5.2 summarizes the result of the digital -to-analog conversion. Note that we have assumed that each bit has a value of 0.125 V. Thus, in this system, we cannot represent a voltage between 1.000 and 1.125, for example. This lack of resolution is a major limitation of digital-to-analog conversions. For greater accuracy, a word representation with a greater number of bits is required. Even then a digital representation of an ana log voltage is never exact. In spite of this inexact representation, digital representation has been used to accomplish remarkable things such as audio CDs and digital photography. - -# **TABLE 5.2** - -Input and output values of the four-bit DAC. - -| Binary input | | Output | -|--------------|---------------|--------| -| [V1V2V3V4] | Decimal value | −Vo | -| 0000 | 0 | 0 | -| 0001 | 1 | 0.125 | -| 0010 | 2 | 0.25 | -| 0011 | 3 | 0.375 | -| 0100 | 4 | 0.5 | -| 0101 | 5 | 0.625 | -| 0110 | 6 | 0.75 | -| 0111 | 7 | 0.875 | -| 1000 | 8 | 1.0 | -| 1001 | 9 | 1.125 | -| 1010 | 10 | 1.25 | -| 1011 | 11 | 1.375 | -| 1100 | 12 | 1.5 | -| 1101 | 13 | 1.625 | -| 1110 | 14 | 1.75 | -| 1111 | 15 | 1.875 | -| | | | - -A three-bit DAC is shown in Fig. 5.37. - -- (a) Determine |*Vo*| for [*V*1*V*2*V*3] = [010]. -- (b) Find |*Vo*| if [*V*1*V*2*V*3] = [110]. -- (c) If |*Vo*| = 1.25 V is desired, what should be [*V*1*V*2*V*3]? -- (d) To get |*Vo*| = 1.75 V, what should be [*V*1*V*2*V*3]? - -**Answer:** 0.5 V, 1.5 V, [101], [111]. - -Three-bit DAC; for Practice Prob. 5.12. - -# **5.10.2** Instrumentation Amplifiers - -One of the most useful and v ersatile op amp circuits for precision mea surement and process control is the *instrumentation amplifier* (IA), so called because of its widespread use in measurement systems. Typical applications of IAs include isolation amplifiers, thermocouple amplifiers, and data acquisition systems. - -The instrumentation amplifier is an extension of the difference amplifier in that it amplifies the difference between its input signals. As shown in Fig. 5.26 (see Example 5.8), an instrumentation amplifier typically consists of three op amps and seven resistors. For convenience, the amplifier is shown again in Fig. 5.38(a), where the resistors are made equal except for the external gain-setting resistor *RG*, connected between the gain set terminals. Figure 5.38(b) shows its schematic symbol. Example 5.8 showed that - -$$ -v_o = A_v(v_2 - v_1) -$$ - (5.24) - -where the voltage gain is - -$$ -A_v = 1 + \frac{2R}{R_G} \tag{5.25} -$$ - -As shown in Fig. 5.39, the instrumentation amplifier amplifies small differential signal v oltages superimposed on lar ger common -mode - -Instrumentation amplifier Amplified dierential signal, - -no common-mode signal - -# **Figure 5.39** - -The IA rejects common voltages but amplifies small signal voltages. - -voltages. Since the common-mode voltages are equal, they cancel each other. - -The IA has three major characteristics: - -- 1. The voltage gain is adjusted by *one* external resistor *RG*. -- 2. The input impedance of both inputs is v ery high and does not vary as the gain is adjusted. -- 3. The output *vo* depends on the dif ference between the inputs *v*1 and *v*2, not on the voltage common to them (common-mode voltage). - -Due to the widespread use of IAs, manuf acturers have developed these amplifiers on single-package units. A typical e xample is the LH0036, developed by National Semiconductor . The gain can be v aried from 1 to 1,000 by an e xternal resistor whose v alue may vary from 100 to 10 kΩ. - -In Fig. 5.38, let *R* = 10 kΩ, *v*1 = 2.011 V, and *v*2 = 2.017 V. If *RG* is Example 5.13 adjusted to 500 Ω, determine: (a) the voltage gain, (b) the output volt age *vo*. - -# **Solution:** - -(a) The voltage gain is - -$$ -A_v = 1 + \frac{2R}{R_G} = 1 + \frac{2 \times 10,000}{500} = 41 -$$ - -(b) The output voltage is - -$$ -v_o = A_v(v_2 - v_1) = 41(2.017 - 2.011) = 41(6) -$$ - mV = 246 mV - -Determine the value of the external gain-setting resistor *RG* required for the IA in Fig. 5.38 to produce a gain of 200 when *R* = 25 kΩ. - -Practice Problem 5.13 - -**Answer:** 251.3 Ω. - -**5.11** Summary - -- 1. The op amp is a high -gain amplifier that has high input resistance and low output resistance. -- 2. Table 5.3 summarizes the op amp circuits considered in this chapter. The expression for the gain of each amplifier circuit holds whether the inputs are dc, ac, or time-varying in general. - -# **TABLE 5.3** - -Summary of basic op amp circuits. - -- 3. An ideal op amp has an infinite input resistance, a zero output resistance, and an infinite gain. -- 4. For an ideal op amp, the current into each of its two input terminals is zero, and the voltage across its input terminals is negligibly small. -- 5. In an inverting amplifier, the output voltage is a negative multiple of the input. -- 6. In a noninverting amplifier, the output is a positive multiple of the input. -- 7. In a voltage follower, the output follows the input. -- 8. In a summing amplifier, the output is the weighted sum of the inputs. -- 9. In a difference amplifier, the output is proportional to the difference of the two inputs. -- 10. Op amp circuits may be cascaded without changing their inputoutput relationships. -- 11. *PSpice* can be used to analyze an op amp circuit. -- 12. Typical applications of the op amp considered in this chapter include the digital-to-analog converter and the instrumentation amplifier. - -# Review Questions - -- **5.1** The two input terminals of an op amp are labeled as: - - (a) high and low. - - (b) positive and negative. - - (c) inverting and noninverting. - - (d) differential and nondifferential. -- **5.2** For an ideal op amp, which of the following statements are not true? - - (a) The differential voltage across the input terminals is zero. - - (b) The current into the input terminals is zero. - - (c) The current from the output terminal is zero. - - (d) The input resistance is zero. - - (e) The output resistance is zero. -- **5.3** For the circuit in Fig. 5.40, voltage *vo* is: - -| (a) −6 V | (b) −5 V | -|------------|------------| -| (c) −1.2 V | (d) −0.2 V | - -# **Figure 5.40** - -For Review Questions 5.3 and 5.4. - -**5.4** For the circuit in Fig. 5.40, current *ix* is: - -| (a) 600 μA | (b) 500 μA | -|------------|-------------| -| (c) 200 μA | (d) 1∕12 μA | - -**5.5** If *vs* = 0 in the circuit of Fig. 5.41, current *io* is: - -| (a) −10 μA | (b) −2.5 μA | -|--------------|--------------| -| (c) 10∕12 μA | (d) 10∕14 μA | - -**5.6** If *vs* = 8 mV in the circuit of Fig. 5.41, the output - -**Figure 5.41** For Review Questions 5.5, 5.6, and 5.7. - -voltage is: - -| (a) −44 mV | (b) −8 mV | -|------------|-----------| -| (c) 4 mV | (d) 7 mV | - -**5.7** Refer to Fig. 5.41. If *vs* = 8 mV, voltage *va* is: - -| (a) −8 mV | (b) 0 mV | -|-------------|----------| -| (c) 10∕3 mV | (d) 8 mV | - -**5.8** The power absorbed by the 4-kΩ resistor in Fig. 5.42 is: - -| (a) 9 mW | (b) 4 mW | -|----------|----------| -| (c) 2 mW | (d) 1 mW | - -**Figure 5.42** - -For Review Questions 5.8. - -- **5.9** Which of these amplifiers is used in a digital-toanalog converter? - - (a) noninverter - - (b) voltage follower - - (c) summer - - (d) difference amplifier -- **5.10** Difference amplifiers are used in (please check all that apply): - - (a) instrumentation amplifiers - - (b) voltage followers - - (c) voltage regulators - - (d) buffers - - (e) summing amplifiers - - (f ) subtracting amplifiers - -*Answers: 5.1c, 5.2c,d, 5.3b, 5.4b, 5.5a, 5.6c, 5.7d, 5.8b, 5.9c, 5.10a, f.* - -# Problems - -# Section 5.2 Operational Amplifiers - -- **5.1** The equivalent model of a certain op amp is shown in Fig. 5.43. Determine: - - (a) the input resistance - - (b) the output resistance - - (c) the voltage gain in dB - -# **Figure 5.43** - -For Prob. 5.1. - -- **5.2** The open-loop gain of an op amp is 50,000. Calculate the output voltage when there are inputs of +10 *μ*V on the inverting terminal and +20 *μ*V on the noninverting terminal. -- **5.3** Determine the voltage input to the inverting terminal of an op amp when −40 *μ*V is applied to the noninverting terminal and the output through an openloop gain of 150,000 is 15 V. -- **5.4** The output voltage of an op amp is −4 V when the noninverting input is 1 mV. If the open-loop gain of the op amp is 2 × 106 , what is the inverting input? -- **5.5** For the op amp circuit of Fig. 5.44, the op amp has an open-loop gain of 100,000, an input resistance of 10 kΩ, and an output resistance of 100 Ω. Find the voltage gain *vo*∕*vi* using the nonideal model of the op amp. - -**5.6** Using the same parameters for the 741 op amp in Example 5.1, find *vo* in the op amp circuit of Fig. 5.45. - -# **Figure 5.45** - -For Prob. 5.6. - -**5.7** The op amp in Fig. 5.46 has *Ri* = 100 kΩ, *Ro* = 100 Ω, *A* = 100,000. Find the differential voltage *vd* and the output voltage *vo*. - -**Figure 5.46** For Prob. 5.7. - -# Section 5.3 Ideal Op Amp - -**5.8** Obtain *vo* for each of the op amp circuits in Fig. 5.47. - -For Prob. 5.8. - -**5.9** Determine *vo* for each of the op amp circuits in Fig. 5.48. - -**5.10** Find the gain *vo*∕*vs* of the circuit in Fig. 5.49. - -For Prob. 5.10. - -**5.11** Using Fig. 5.50, design a problem to help other students better understand how ideal op amps work. - -# **Figure 5.51** - -For Prob. 5.12. - -# **Figure 5.52** - -For Prob. 5.13. - -**5.14** Determine the output voltage *vo* in the circuit of Fig. 5.53. - -# **Figure 5.53** - -For Prob. 5.14. - -# Section 5.4 Inverting Amplifier - -- **5.15** (a) Determine the ratio *vo*∕*is* in the op amp circuit of Fig. 5.54. -- (b) Evaluate the ratio for *R*1 = 20 kΩ, *R*2 = 25 kΩ, *R*3 = 40 kΩ. - -**5.16** Using Fig. 5.55, design a problem to help students better understand inverting op amps. - -**Figure 5.55** - -- **5.17** Calculate the gain *vo*∕*vi* when the switch in Fig. 5.56 is in: - - (a) position 1 (b) position 2 (c) position 3. - -For Prob. 5.17. - -**5.18** For the circuit shown in Figure 5.57, solve for the Thevenin equivalent circuit looking into terminals A and B. \* - -For Prob. 5.18. - -\* An asterisk indicates a challenging problem. - -**5.19** Determine *io* in the circuit of Fig. 5.58. - -# **Figure 5.58** - -**5.20** In the circuit of Fig. 5.59, calculate *vo* of *vs* = 2 V. - -**Figure 5.59** - -For Prob. 5.20. - -**5.21** Calculate *vo* in the op amp circuit of Fig. 5.60. - -# **Figure 5.60** - -For Prob. 5.21. - -- **5.22** Design an inverting amplifier with a gain of −15. -- **5.23** For the op amp circuit in Fig. 5.61, find the voltage gain *vo*∕*vs*. - -**Figure 5.61** For Prob. 5.23. - -**5.24** In the circuit shown in Fig. 5.62, find *k* in the voltage transfer function *vo* = *kvs*. - -# **Figure 5.62** - -For Prob. 5.24. - -# Section 5.5 Noninverting Amplifier - -**Figure 5.63** For Prob. 5.25. - -**5.26** Using Fig. 5.64, design a problem to help other students better understand noninverting op amps. - -**Figure 5.64** For Prob. 5.26. - -**5.27** Find *vo* in the op amp circuit of Fig. 5.65. - -**Figure 5.65** For Prob. 5.27. - -**5.28** Find *io* in the op amp circuit of Fig. 5.66. - -# **Figure 5.66** - -For Prob. 5.28. - -**5.29** Determine the voltage gain *vo*∕*vi* of the op amp circuit in Fig. 5.67. - -# **Figure 5.67** For Prob. 5.29. - -**5.30** In the circuit shown in Fig. 5.68, find *ix* and the power absorbed by the 20-kΩ resistor. - -# **Figure 5.68** - -For Prob. 5.30. - -**5.31** For the circuit in Fig. 5.69, find *ix*. - -**Figure 5.69** For Prob. 5.31. - -**5.32** Calculate *ix* and *vo* in the circuit of Fig. 5.70. Find the power dissipated by the 60-kΩ resistor. - -**5.33** Refer to the op amp circuit in Fig. 5.71. Calculate *ix* and the power absorbed by the 3-kΩ resistor. - -**Figure 5.71** - -For Prob. 5.33. - -**5.34** Given the op amp circuit shown in Fig. 5.72, express *vo* in terms of *v*1 and *v*2. - -**5.35** Design a noninverting amplifier with a gain of 7.5. - -**5.36** For the circuit shown in Fig. 5.73, find the Thevenin equivalent at terminals *a*-*b*. (*Hint:* To find *R*Th, apply a current source *io* and calculate *vo*.) - -# **Figure 5.73** - -For Prob. 5.36. - -# Section 5.6 Summing Amplifier - -**5.37** Determine the output of the summing amplifier in Fig. 5.74. - -# **Figure 5.74** - -For Prob. 5.37. - -**5.38** Using Fig. 5.75, design a problem to help other students better understand summing amplifiers. - -# **Figure 5.75** - -For Prob. 5.38. - -**5.39** For the op amp circuit in Fig. 5.76, determine the value of *v*2 in order to make *vo* = −16.5 V. - -**5.40** Referring to the circuit shown in Fig. 5.77, determine *V*o in terms of *V*1 and *V*2. - -**Figure 5.77** For Prob. 5.40. - -**5.41** An *averaging amplifier* is a summer that provides an output equal to the average of the inputs. By using proper input and feedback resistor values, one can get - -$$ --v_{\text{out}} = \frac{1}{4}(v_1 + v_2 + v_3 + v_4) -$$ - - Using a feedback resistor of 10 kΩ, design an averaging amplifier with four inputs. - -- **5.42** The feedback resistor of a three-input averaging summing amplifier is 50 kΩ. What are the values of *R*1, *R*2, and *R*3? -- **5.43** The feedback resistor of a five-input averaging summing amplifier is 40 kΩ. What are the values of *R*1, *R*2, *R*3, *R*4, and *R*5? -- **5.44** Show that the output voltage *vo* of the circuit in Fig. 5.78 is - -$$ -v_o = \frac{(R_3 + R_4)}{R_3(R_1 + R_2)}(R_2v_1 + R_1v_2) -$$ - -**Figure 5.78** For Prob. 5.44. - -**5.45** Design an op amp circuit to perform the following operation: - -$$ -v_o = 3.5v_1 - 2.5v_2 -$$ - -All resistances must be ≤ 100 kΩ. - -**5.46** Using only two op amps, design a circuit to solve - -$$ --v_{\text{out}} = \frac{v_3 - v_1}{5} + \frac{v_1 - v_2}{2} -$$ - -# Section 5.7 Difference Amplifier - -**5.47** The circuit in Fig. 5.79 is for a difference amplifier. Find *vo* given that *v*1 = 1 V and *v*2 = 2 V. - -**5.48** The circuit in Fig. 5.80 is a differential amplifier driven by a bridge. Find *vo*. - -**5.49** Design a difference amplifier to have a gain of 4 and a common-mode input resistance of 20 kΩ at each input. - -**5.50** Design a circuit to amplify the difference between two inputs by 2.5. - -(a) Use only one op amp. - -(b) Use two op amps. - -**5.52** Design an op amp circuit such that \* - -$$ -v_o = 4v_1 + 6v_2 - 3v_3 - 5v_4 -$$ - -Let all the resistors be in the range of 20 to 200 kΩ. - -**5.53** The ordinary difference amplifier for fixed-gain operation is shown in Fig. 5.81(a). It is simple and reliable unless gain is made variable. One way of providing gain adjustment without losing simplicity and accuracy is to use the circuit in Fig. 5.81(b). Another way is to use the circuit in Fig. 5.81(c). Show that: \* - -(a) for the circuit in Fig. 5.81(a), - -$$ -\frac{v_o}{v_i} = \frac{R_2}{R_1} -$$ - -(b) for the circuit in Fig. 5.81(b), - -$$ -\frac{v_o}{v_i} = \frac{R_2}{R_1} \frac{1}{1 + \frac{R_1}{2R_G}} -$$ - -(c) for the circuit in Fig. 5.81(c), - -$$ -\frac{v_o}{v_i} = \frac{R_2}{R_1} \left( 1 + \frac{R_2}{2R_G} \right) -$$ - -# **Figure 5.81** - -For Prob. 5.53. - -# Section 5.8 Cascaded Op Amp Circuits - -**5.54** Determine the voltage transfer ratio *vo*∕*vs* in the op amp circuit of Fig. 5.82, where R = 10 kΩ. - -**Figure 5.82** - -For Prob. 5.54. - -- **5.55** In a certain electronic device, a three-stage amplifier is desired, whose overall voltage gain is 42 dB. The individual voltage gains of the first two stages are to be equal, while the gain of the third is to be onefourth of each of the first two. Calculate the voltage gain of each. - -**5.56** Using Fig. 5.83, design a problem to help other students better understand cascaded op amps. - -**5.57** Find *vo* in the op amp circuit of Fig. 5.84. - -# **Figure 5.84** - -For Prob. 5.57. - -**5.58** Calculate *io* in the op amp circuit of Fig. 5.85. - -**Figure 5.85** - -- For Prob. 5.58. - - **5.59** In the op amp circuit of Fig. 5.86, determine the voltage gain *vo*∕*vs*. Take *R* = 10 kΩ. - -**Figure 5.86** For Prob. 5.59. - -**5.60** Calculate *vo*∕*vi* in the op amp circuit of Fig. 5.87. - -For Prob. 5.60. - -**5.61** Determine *vo* in the circuit of Fig. 5.88. - -# **Figure 5.88** - -For Prob. 5.61. - -**5.62** Obtain the closed-loop voltage gain *vo*∕*vi* of the circuit in Fig. 5.89. - -**Figure 5.89** For Prob. 5.62. - -**5.63** Determine the gain *vo*∕*vi* of the circuit in Fig. 5.90. - -**Figure 5.90** For Prob. 5.63. - -**5.64** For the op amp circuit shown in Fig. 5.91, find *vo*∕*vs*. - -**Figure 5.91** For Prob. 5.64. - -**5.65** Find *vo* in the op amp circuit of Fig. 5.92. - -For Prob. 5.65. - -**5.66** For the circuit in Fig. 5.93, find *vo*. - -**Figure 5.93** For Prob. 5.66. - -**5.67** Obtain the output *vo* in the circuit of Fig. 5.94. - -**Figure 5.96** For Prob. 5.70. - -**5.68** Find *vo* in the circuit of Fig. 5.95, assuming that *Rf* = ∞ (open circuit). - -**Figure 5.95** For Probs. 5.68 and 5.69. - -- **5.69** Repeat the previous problem if *Rf* = 10 kΩ. -- **5.70** Determine *vo* in the op amp circuit of Fig. 5.96. - -30 kΩ - -40 kΩ - -**5.71** Determine *vo* in the op amp circuit of Fig. 5.97. - -- **5.72** Find the load voltage *vL* in the circuit of Fig. 5.98. - -**5.73** Determine the load voltage *vL* in the circuit of Fig. 5.99. - -**5.74** Find *io* in the op amp circuit of Fig. 5.100. - -# **Figure 5.100** For Prob. 5.74. - -# Section 5.9 Op Amp Circuit Analysis with PSpice - -- **5.75** Rework Example 5.11 using the nonideal op amp LM324 instead of uA741. -- **5.76** Solve Prob. 5.19 using *PSpice* or *MultiSim* and op amp uA741. -- **5.77** Solve Prob. 5.48 using *PSpice* or *MultiSim* and op amp LM324. -- **5.78** Use *PSpice* or *MultiSim* to obtain *vo* in the circuit of Fig. 5.101. - -**Figure 5.101** For Prob. 5.78. - -**5.79** Determine *vo* in the op amp circuit of Fig. 5.102, using *PSpice* or *MultiSim*. - -**Figure 5.102** For Prob. 5.79. - -- **5.80** Use *PSpice* or *MultiSim* to solve Prob. 5.70. -- **5.81** Use *PSpice* or *MultiSim* to verify the results in Example 5.9. Assume nonideal op amps LM324. - -# Section 5.10 Applications - -**5.82** A four-bit DAC covers a voltage range of 0 to 10 V. - -Calculate the resolution of the DAC in volts per discrete binary step. - -**5.83** Design a six-bit digital-to-analog converter. - -- (a) If |*Vo*| = 1.1875 V is desired, what should [*V*1*V*2*V*3*V*4*V*5*V*6] be? -- (b) Calculate |*Vo*| if [*V*1*V*2*V*3*V*4*V*5*V*6] = [011011]. -- (c) What is the maximum value |*Vo*| can assume? -- **5.84** A four-bit *R-2R ladder* DAC is presented in Fig. 5.103. \* - - (a) Show that the output voltage is given by - -$$ --V_o = R_f \left( \frac{V_1}{2R} + \frac{V_2}{4R} + \frac{V_3}{8R} + \frac{V_4}{16R} \right) -$$ - -(b) If *Rf* = 12 kΩ and *R* = 10 kΩ, find |*Vo*| for [*V*1*V*2*V*3*V*4] = [1011] and [*V*1*V*2*V*3*V*4] = [0101]. - -- For Prob. 5.84. - - **5.85** In the op amp circuit of Fig. 5.104, find the value of *R* so that the power absorbed by the 10-kΩ resistor is 10 mW. Determine the power gain. - -**Figure 5.104** For Prob. 5.85. - -- **5.86** Design a voltage controlled ideal current source (within the operating limits of the op amp) where the output current is equal to 200 *vs*(*t*) *μ*A. -- **5.87** Figure 5.105 displays a two-op-amp instrumentation amplifier. Derive an expression for *vo* in terms of *v*1 and *v*2. How can this amplifier be used as a subtractor? - -**Figure 5.105** For Prob. 5.87. - -**5.88** Figure 5.106 shows an instrumentation amplifier driven by a bridge. Obtain the gain *vo*∕*vi* of the amplifier. \* - -# Comprehensive Problems - -**5.89** Design a circuit that provides a relationship between output voltage *vo* and input voltage *vs* such that *vo* = 12*vs* − 10. Two op amps, a 6-V battery, and several resistors are available. - -**5.90** The op amp circuit in Fig. 5.107 is a *current amplifier*. Find the current gain *io*∕*is* of the amplifier. - -# **Figure 5.107** - -## For Prob. 5.90. - -**5.91** A noninverting current amplifier is portrayed in Fig. 5.108. Calculate the gain *io*∕*is*. Take *R*1 = 8 kΩ and *R*2 = 1 kΩ. - -**5.92** Refer to the *bridge amplifier* shown in Fig. 5.109. Determine the voltage gain *vo*∕*vi*. - -# **Figure 5.109** - -For Prob. 5.92. - -**5.93** A voltage-to-current converter is shown in Fig. 5.110, which means that *iL*= *Avi* if *R*1*R*2 = *R*3*R*4. Find the constant term *A*. \* - -*This page intentionally left blank* - -# **chapter** - -6 - -# Capacitors and Inductors - -*But in science the credit goes to the man who convinces the world, not to the man to whom the idea first occurs.* - -—Francis Darwin - -# Enhancing Your Skills and Your Career - -# **ABET EC 2000 criteria (3.c), "an ability to design a system, component, or process to meet desired needs."** - -The "ability to design a system, component, or process to meet desired needs" is why engineers are hired. That is why this is the most important *technical* skill that an engineer has. Interestingly, your success as an engineer is directly proportional to your ability to communicate but your being able to design is wh y you will be hired in the first place. - -Design takes place when you ha ve what is termed an open-ended problem that eventually is defined by the solution. Within the context of this course or te xtbook, we can only e xplore some of the elements of design. Pursuing all of the steps of our problem-solving technique teaches you se veral of the most important elements of the design process. - -Probably the most important part of design is clearly defining what the system, component, process, or , in our case, problem is. Rarely is an engineer given a perfectly clear assignment. Therefore, as a student, you can develop and enhance this skill by asking yourself, your colleagues, or your professors questions designed to clarify the problem statement. - -Exploring alternati ve solutions is another important part of the design process. Again, as a student, you can practice this part of the de sign process on almost every problem you work. - -Evaluating your solutions is critical to any engineering assignment. Again, this is a skill that you as a student can practice on every problem you work. - -Photo by Charles Alexander - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Fully understand the volt-amp characteristics of capacitors and inductors and their use in basic circuits. -- 2. Explain how capacitors behave when combined in parallel and in series. -- 3. Understand how inductors behave when combined in parallel and in series. -- 4. Know how to create integrators using capacitors and op amps. -- 5. Learn how to create differentiators and their limitations. -- 6. Learn how to create analog computers and to understand how they can be used to solve linear differential equations. - -In contrast to a resistor, which spends or dissipates energy irreversibly, an inductor or capacitor stores or releases energy (i.e., has a memory). - -# **6.1** Introduction - -So far we have limited our study to resistive circuits. In this chapter, we shall introduce two new and important passive linear circuit elements: the capacitor and the inductor. Unlike resistors, which dissipate ener gy, capacitors and inductors do not dissipate but store energy, which can be retrieved at a later time. F or this reason, capacitors and inductors are called *storage* elements. - -The application of resistive circuits is quite limited. With the introduction of capacitors and inductors in this chapter , we will be able to analyze more important and practical circuits. Be assured that the circuit analysis techniques covered in Chapters 3 and 4 are equally applicable to circuits with capacitors and inductors. - -We begin by introducing capacitors and describing how to combine them in series or in parallel. Later, we do the same for inductors. As typical applications, we explore how capacitors are combined with op amps to form integrators, differentiators, and analog computers. - -**Figure 6.1** A typical capacitor. - -# **6.2** Capacitors - -A capacitor is a passi ve element designed to store ener gy in its elec tric field. Besides resistors, capacitors are the most common electrical components. Capacitors are used extensively in electronics, communications, computers, and power systems. For example, they are used in the tuning circuits of radio recei vers and as dynamic memory elements in computer systems. - -A capacitor is typically constructed as depicted in Fig. 6.1. - -A capacitor consists of two conducting plates separated by an insulator (or dielectric). - -In many practical applications, the plates may be aluminum foil while the dielectric may be air, ceramic, paper, or mica. - -# Historical - -**Michael Faraday** (1791–1867), an Engl ish c hemist a nd phy sicist, was probably the greatest experimentalist who ever lived. - -Born near London, Faraday realized his boyhood dream by work ing with the great chemist Sir Humphry Davy at the Royal Institu tion, where he worked for 54 years. He made several contributions in all areas of physical science and coined such words as electrolysis, anode, and cathode. His discovery of electromagnetic induction in 1831 was a major breakthrough in engineering because it provided a way of generating electricity. The electric motor and generator operate on this principle. The unit of capacitance, the farad, was named in his honor. - -© Stock Montage/Getty Images - -When a voltage source *v* is connected to the capacitor, as in Fig. 6.2, the source deposits a positive charge *q* on one plate and a negative charge −*q* on the other . The capacitor is said to store the electric char ge. The amount of charge stored, represented by *q*, is directly proportional to the applied voltage *v* so that - -$$ -q = Cv \tag{6.1} -$$ - -where *C*, the constant of proportionality , is kno wn as the *capacitance* of the capacitor. The unit of capacitance is the farad (F), in honor of the English physicist Michael Faraday (1791–1867). From Eq. (6.1), we may derive the following definition. - -Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F). - -Note from Eq. (6.1) that 1 farad = 1 coulomb/volt. - -Although the capacitance *C* of a capacitor is the ratio of the charge *q* per plate to the applied voltage *v*, it does not depend on *q* or *v*. It depends on the physical dimensions of the capacitor. For example, for the parallelplate capacitor shown in Fig. 6.1, the capacitance is given by - -$$ -C = \frac{\epsilon A}{d} \tag{6.2} -$$ - -where *A* is the surf ace area of each plate, *d* is the distance between the plates, and *ϵ* is the permittivity of the dielectric material between the plates. Although Eq. (6.2) applies to only parallel-plate capacitors, we may infer from it that, in general, three f actors determine the v alue of the capacitance: - -- 1. The surface area of the plates—the lar ger the area, the greater the capacitance. -- 2. The spacing between the plates—the smaller the spacing, the greater the capacitance. - -**Figure 6.2** A capacitor with applied voltage *v*. - -Alternatively, capacitance is the amount of charge stored per plate for a unit voltage difference in a capacitor. - - Capacitor voltage rating and capacitance are typically inversely rated due to the relationships in Eqs. (6.1) and (6.2). Arcing occurs if d is small and V is high. - -**Figure 6.3** - -Circuit symbols for capacitors: (a) fixed capacitor, (b) variable capacitor. - -3. The permitti vity of the material—the higher the permitti vity, the greater the capacitance. - -Capacitors are commercially a vailable in different values and types. Typically, capacitors have values in the picofarad (pF) to microfarad (*μ*F) range. They are described by the dielectric material the y are made of and by whether they are of fixed or variable type. Figure 6.3 shows the circuit symbols for fixed and variable capacitors. Note that according to the passive sign convention, if *v* > 0 and *i* > 0 or if *v* < 0 and *i* < 0, the capacitor is being charged, and if *v* **·** *i* < 0, the capacitor is discharging. - -# **Figure 6.4** - -**Figure 6.5** Variable capacitors: (a) trimmer capacitor, (b) filmtrim capacitor. Courtesy of Johanson. - -According to Eq. (6.4), for a capacitor to carry current, its voltage must vary with time. Hence, for constant voltage, i = 0. - -Figure 6.4 shows common types of fixed-value capacitors. Polyester capacitors are light in weight, stable, and their change with tempera ture is predictable. Instead of polyester, other dielectric materials such as mica and polystyrene may be used. Film capacitors are rolled and housed in metal or plastic films. Electrolytic capacitors produce very high capacitance. Figure 6.5 shows the most common types of variable capacitors. The capacitance of a trimmer (or padder) capacitor is often placed in parallel with another capacitor so that the equi valent capacitance can be varied slightly. The capacitance of the variable air capacitor (meshed plates) is varied by turning the shaft. Variable capacitors are used in radio receivers allowing one to tune to various stations. In addition, capacitors are used to block dc, pass ac, shift phase, store energy, start motors, and suppress noise. - -To obtain the current-voltage relationship of the capacitor , we take the derivative of both sides of Eq. (6.1). Since - -$$ -i = \frac{dq}{dt} \tag{6.3} -$$ - -differentiating both sides of Eq. (6.1) gives - -$$ -i = C \frac{dv}{dt} -$$ - (6.4) - -This is the current-v oltage relationship for a capacitor , assuming the passive sign convention. The relationship is illustrated in Fig. 6.6 for a capacitor whose capacitance is independent of v oltage. Capacitors that satisfy Eq. (6.4) are said to be *linear*. For a *nonlinear capacitor*, the plot of the current-voltage relationship is not a straight line. Although some capacitors are nonlinear, most are linear. We will assume linear capaci tors in this book. - -The v oltage-current relation of the capacitor can be obtained by integrating both sides of Eq. (6.4). We get - -$$ -v(t) = \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau -$$ - (6.5) - -or - -$$ -v(t) = \frac{1}{C} \int_{t_0}^t i(\tau) \, d\tau + v(t_0) -$$ -\n(6.6) - -where *v*(*t*0) = *q*(*t*0)∕*C* is the voltage across the capacitor at time *t*0. Equation (6.6) shows that capacitor voltage depends on the past history of the capacitor current. Hence, the capacitor has memory—a property that is often exploited. - -The instantaneous power delivered to the capacitor is - -$$ -p = vi = Cv \frac{dv}{dt} -$$ - (6.7) - -The energy stored in the capacitor is therefore - -$$ -w = \int_{-\infty}^{t} p(\tau) d\tau = C \int_{-\infty}^{t} v \frac{dv}{d\tau} d\tau = C \int_{v(-\infty)}^{v(t)} v dv = \frac{1}{2} C v^{2} \Big|_{v(-\infty)}^{v(t)} \quad (6.8) -$$ - -We note that *v*(−∞) = 0, because the capacitor was uncharged at *t* = −∞. Thus, - -$$ -w = \frac{1}{2} C v^2 \tag{6.9} -$$ - -Using Eq. (6.1), we may rewrite Eq. (6.9) as - -$$ -w = \frac{q^2}{2C} \tag{6.10} -$$ - -Equation (6.9) or (6.10) represents the energy stored in the electric field that exists between the plates of the capacitor. This energy can be retrieved, since an ideal capacitor cannot dissipate energy. In fact, the word *capacitor* is derived from this element's capacity to store energy in an electric field. - -We should note the following important properties of a capacitor: - -1. Note from Eq. (6.4) that when the v oltage across a capacitor is not changing with time (i.e., dc voltage), the current through the capacitor is zero. Thus, - -A capacitor is an open circuit to dc. - -However, if a battery (dc voltage) is connected across a capacitor, the capacitor charges. - - An alternative way of looking at this is using Eq. (6.9), which indicates that energy is proportional to voltage squared. Since injecting or extracting energy can only be done over some finite time, voltage cannot change instantaneously across a capacitor. - -**Figure 6.6** Current-voltage relationship of a capacitor. - -Voltage across a capacitor: (a) allowed, (b) not allowable; an abrupt change is not possible. - -**Figure 6.8** Circuit model of a nonideal capacitor. - -Example 6.1 - -2. The voltage on the capacitor must be continuous. - -The voltage on a capacitor cannot change abruptly. - -The capacitor resists an abrupt change in the v oltage across it. According to Eq. (6.4), a discontinuous change in v oltage requires an infinite current, which is physically impossible. For example, the voltage across a capacitor may tak e the form shown in Fig. 6.7(a), whereas it is not physically possible for the capacitor voltage to take the form shown in Fig. 6.7(b) because of the abrupt changes. Con versely, the current through a capacitor can change instantaneously. - -- 3. The ideal capacitor does not dissipate ener gy. It tak es power from the circuit when storing ener gy in its field and returns previously stored energy when delivering power to the circuit. -- 4. A real, nonideal capacitor has a parallel-model leakage resistance, as shown in Fig. 6.8. The leakage resistance may be as high as 100 MΩ and can be neglected for most practical applications. For this reason, we will assume ideal capacitors in this book. - -(a) Calculate the charge stored on a 3-pF capacitor with 20 V across it. (b) Find the energy stored in the capacitor. - -# **Solution:** - -(a) Since *q* = *Cv*, - -$$ -q = 3 \times 10^{-12} \times 20 = 60 \text{ pC} -$$ - -(b) The energy stored is - -$$ -w = \frac{1}{2} Cv^2 = \frac{1}{2} \times 3 \times 10^{-12} \times 400 = 600 \text{ pJ} -$$ - -What is the voltage across a 4.5- *μ*F capacitor if the charge on one plate is 0.12 mC? How much energy is stored? Practice Problem 6.1 - -**Answer:** 26.67 V, 1.6 mJ. - -Example 6.2 The voltage across a 5-*μ*F capacitor is - -*v*(*t*) = 10 cos 6000*t* V - -Calculate the current through it. - -# **Solution:** - -By definition, the current is - -$$ -i(t) = C \frac{dv}{dt} = 5 \times 10^{-6} \frac{d}{dt} (10 \cos 6000t) -$$ - -= -5 × 10-6 × 6000 × 10 sin 6000t = -0.3 sin 6000t A - -If a 10-*μ*F capacitor is connected to a voltage source with - -*v*(*t*) = 20 cos(200*t*) V - -determine the current through the capacitor. - -**Answer:** −40 sin (200*t*) mA. - -Determine the voltage across a 2-*μ*F capacitor if the current through it is Example 6.3 - -$$ -i(t) = 6e^{-3000t} \text{ mA} -$$ - -Assume that the initial capacitor voltage is zero. - -# **Solution:** - -Since -$$ -v = \frac{1}{C} \int_0^t i \, d\tau + v(0) -$$ - and $v(0) = 0$ , -\n -$$ -v = \frac{1}{2 \times 10^{-6}} \int_0^t 6e^{-3000\tau} \, d\tau \cdot 10^{-3} -$$ -\n -$$ -= \frac{3 \times 10^3}{-3000} e^{-3000\tau} \Big|_0^t = (1 - e^{-3000t}) \text{ V} -$$ - -The current through a 100-*μ*F capacitor is *i*(*t*) = 50 sin 120*πt* mA. Calculate the voltage across it at *t* = 1 ms and *t* = 5 ms. Take *v*(0) = 0. - -**Answer:** 93.14 mV, 1.736 V. - -Determine the current through a 200- *μ*F capacitor whose voltage is Example 6.4 shown in Fig. 6.9. - -# **Solution:** - -The voltage waveform can be described mathematically as - -$$ -v(t) = \begin{cases} 50t \text{ V} & 0 < t < 1 \\ 100 - 50t \text{ V} & 1 < t < 3 \\ -200 + 50t \text{ V} & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases} -$$ - -Since *i* = *C dv*∕*dt* and *C* = 200 *μ*F, we take the derivative of *v* to obtain - -$$ -i(t) = 200 \times 10^{-6} \times \begin{cases} 50 & 0 < t < 1 \\ -50 & 1 < t < 3 \\ 50 & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases} -$$ -$$ -= \begin{cases} 10 \text{ mA} & 0 < t < 1 \\ -10 \text{ mA} & 1 < t < 3 \\ 10 \text{ mA} & 3 < t < 4 \\ 0 & \text{otherwise} \end{cases} -$$ - -Thus, the current waveform is as shown in Fig. 6.10. - -Practice Problem 6.3 - -**Figure 6.9** For Example 6.4. - -Practice Problem 6.2 - -For Practice Prob. 6.4. - -An initially uncharged 1-mF capacitor has the current shown in Fig. 6.11 across it. Calculate the voltage across it at *t* = 2 ms and *t* = 5 ms. - -**Answer:** 100 mV, 400 mV. - -Example 6.5 Obtain the ener gy stored in each capacitor in Fig. 6.12(a) under dc conditions. - -# **Solution:** - -Under dc conditions, we replace each capacitor with an open circuit, as shown in Fig. 6.12(b). The current through the series combination of the 2-kΩ and 4-kΩ resistors is obtained by current division as - -$$ -i = \frac{3}{3 + 2 + 4}(6 \text{ mA}) = 2 \text{ mA} -$$ - -Hence, the voltages *v*1 and *v*2 across the capacitors are - -$$ -v_1 = 2000i = 4 \text{ V} \qquad v_2 = 4000i = 8 \text{ V} -$$ - -and the energies stored in them are - -$$ -w_1 = \frac{1}{2}C_1v_1^2 = \frac{1}{2}(2 \times 10^{-3})(4)^2 = 16 \text{ mJ} -$$ - -$$ -w_2 = \frac{1}{2}C_2v_2^2 = \frac{1}{2}(4 \times 10^{-3})(8)^2 = 128 \text{ mJ} -$$ - -**Figure 6.12** - -For Example 6.5. - -# Practice Problem 6.5 - -**Figure 6.13** For Practice Prob. 6.5. - -Under dc conditions, find the energy stored in the capacitors in Fig. 6.13. - -**Answer:** 20.25 mJ, 3.375 mJ. - -# **6.3** Series and Parallel Capacitors - -We know from resisti ve circuits that the series-parallel combination is a powerful tool for reducing circuits. This technique can be e xtended to series-parallel connections of capacitors, which are sometimes encoun tered. We desire to replace these capacitors by a single equivalent capacitor *C*eq. - -In order to obtain the equi valent capacitor *C*eq of *N* capacitors in parallel, consider the circuit in Fig. 6.14(a). The equi valent circuit is in Fig. 6.14(b). Note that the capacitors ha ve the same voltage *v* across them. Applying KCL to Fig. 6.14(a), - -$$ -i = i_1 + i_2 + i_3 + \dots + i_N \tag{6.11} -$$ - -But *ik* = *Ck dv*∕*dt*. Hence, - -$$ -i = C_1 \frac{dv}{dt} + C_2 \frac{dv}{dt} + C_3 \frac{dv}{dt} + \dots + C_N \frac{dv}{dt} -$$ - -= -$$ -\left(\sum_{k=1}^{N} C_k\right) \frac{dv}{dt} = C_{eq} \frac{dv}{dt} -$$ - (6.12) - -where - -$$ -C_{\text{eq}} = C_1 + C_2 + C_3 + \dots + C_N \tag{6.13} -$$ - -The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances. - -We observe that capacitors in parallel combine in the same manner as resistors in series. - -We no w obtain *C*eq of *N* capacitors connected in series by comparing the circuit in Fig. 6.15(a) with the equi valent circuit in Fig. 6.15(b). Note that the same current *i* flows (and consequently the same charge) through the capacitors. Applying KVL to the loop in Fig. 6.15(a), - -$$ -v = v_1 + v_2 + v_3 + \dots + v_N \tag{6.14} -$$ - -But *vk*=\_\_\_1 *Ck* ∫ *t*0 *t i*(*τ*) *dτ*+*vk*(*t*0). Therefore, - -$$ -v = \frac{1}{C_1} \int_{t_0}^t i(\tau) d\tau + v_1(t_0) + \frac{1}{C_2} \int_{t_0}^t i(\tau) d\tau + v_2(t_0) -$$ - -+ $\cdots + \frac{1}{C_N} \int_{t_0}^t i(\tau) d\tau + v_N(t_0)$ -= $\left(\frac{1}{\tau_0} + \frac{1}{\tau_0} + \cdots + \frac{1}{\tau_N}\right) \int_0^t i(\tau) d\tau + v_1(t_0) + v_2$ - -$$ -= \left(\frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_N}\right) \int_{t_0}^t i(\tau) \, d\tau + v_1(t_0) + v_2(t_0) + \dots + v_N(t_0) -$$ - -$$ -=\frac{1}{C_{\text{eq}}}\int_{t_0}^t i(\tau)\,d\tau + v(t_0) -$$ - -where - -$$ -\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots + \frac{1}{C_N} -$$ - (6.16) - -# **Figure 6.14** - -(a) Parallel-connected *N* capacitors, (b) equivalent circuit for the parallel capacitors. - -(a) - -## **(6.15) Figure 6.15** - -(a) Series-connected *N* capacitors, (b) equivalent circuit for the series capacitor. - -The initial voltage *v*(*t*0) across *C*eq is required by KVL to be the sum of the capacitor voltages at *t*0. Or according to Eq. (6.15), - -$$ -v(t_0) = v_1(t_0) + v_2(t_0) + \dots + v_N(t_0) -$$ - -Thus, according to Eq. (6.16), - -The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances. - -Note that capacitors in series combine in the same manner as resistors in parallel. For *N* = 2 (i.e., two capacitors in series), Eq. (6.16) becomes - -> \_\_\_1 *C*eq = \_\_\_1 *C*1 + \_\_\_1 *C*2 - -or - -$$ -C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} \tag{6.17} -$$ - -Example 6.6 Find the equi valent capacitance seen between terminals *a* and *b* of the circuit in Fig. 6.16. - -# **Solution:** - -The 20- *μ*F and 5- *μ*F capacitors are in series; their equivalent capaci tance is - -$$ -\frac{20 \times 5}{20 + 5} = 4 \ \mu\text{F} -$$ - -This 4-*μ*F capacitor is in parallel with the 6- *μ*F and 20- *μ*F capacitors; their combined capacitance is - -$$ -4 + 6 + 20 = 30 \,\mu\text{F} -$$ - -This 30-*μ*F capacitor is in series with the 60- *μ*F capacitor. Hence, the equivalent capacitance for the entire circuit is - -$$ -C_{\text{eq}} = \frac{30 \times 60}{30 + 60} = 20 \,\mu\text{F} -$$ - -Find the equivalent capacitance seen at the terminals of the circuit in Practice Problem 6.6 Fig. 6.17. - -# **Answer:** 40 *μ*F. - -For Practice Prob. 6.6. - -For the circuit in Fig. 6.18, find the voltage across each capacitor. Example 6.7 - -# **Solution:** - -We first find the equivalent capacitance *C*eq, shown in Fig. 6.19. The two parallel capacitors in Fig. 6.18 can be combined to get 40 + 20 = 60 mF. This 60-mF capacitor is in series with the 20-mF and 30-mF capacitors. Thus, - -$$ -C_{\text{eq}} = \frac{1}{\frac{1}{60} + \frac{1}{30} + \frac{1}{20}} \text{ mF} = 10 \text{ mF} -$$ - -The total charge is - -$$ -q = C_{\text{eq}}v = 10 \times 10^{-3} \times 30 = 0.3 \text{ C} -$$ - -This is the charge on the 20-mF and 30-mF capacitors, because they are in series with the 30-V source. (A crude way to see this is to imagine that charge acts like current, since *i* = *dq*∕*dt*.) Therefore, - -$$ -v_1 = \frac{q}{C_1} = \frac{0.3}{20 \times 10^{-3}} = 15 \text{ V} -$$ - $v_2 = \frac{q}{C_2} = \frac{0.3}{30 \times 10^{-3}} = 10 \text{ V}$ - -Having determined *v*1 and *v*2, we now use KVL to determine *v*3 by - -$$ -v_3 = 30 - v_1 - v_2 = 5 -$$ - V - -Alternatively, since the 40-mF and 20-mF capacitors are in parallel, they have the same voltage *v*3 and their combined capacitance is 40 + 20 = 60 mF. This combined capacitance is in series with the 20-mF and 30-mF capacitors and consequently has the same charge on it. Hence, - -$$ -v_3 = \frac{q}{60 \text{ mF}} = \frac{0.3}{60 \times 10^{-3}} = 5 \text{ V} -$$ - -Find the voltage across each of the capacitors in Fig. 6.20. Practice Problem 6.7 - -**Answer:** -$$ -v_1 = 75 \text{ V}, v_2 = 75 \text{ V}, v_3 = 25 \text{ V}, v_4 = 50 \text{ V}. -$$ - -**Figure 6.18** For Example 6.7. - -**Figure 6.19** Equivalent circuit for Fig. 6.18. - -**Figure 6.20** For Practice Prob. 6.7. - -**Figure 6.21** Typical form of an inductor. - -In view of Eq. (6.18), for an inductor to have voltage across its terminals, its current must vary with time. Hence, v = 0 for constant current through the inductor. - -# **Figure 6.22** - -Various types of inductors: (a) solenoidal wound inductor, (b) toroidal inductor, (c) axial lead inductor. - -© McGraw-Hill Education/Mark Dierker, photographer - -# **6.4** Inductors - -An inductor is a passive element designed to store energy in its magnetic field. Inductors find numerous applications in electronic and power systems. They are used in power supplies, transformers, radios, TVs, radars, and electric motors. - -Any conductor of electric current has inductive properties and may be regarded as an inductor. But in order to enhance the inducti ve effect, a practical inductor is usually formed into a c ylindrical coil with man y turns of conducting wire, as shown in Fig. 6.21. - -An inductor consists of a coil of conducting wire. - -If current is allowed to pass through an inductor, it is found that the voltage across the inductor is directly proportional to the time rate of change of the current. Using the passive sign convention, - -$$ -v = L \frac{di}{dt} -$$ - (6.18) - -where *L* is the constant of proportionality called the *inductance* of the inductor. The unit of inductance is the henry (H), named in honor of the American inventor Joseph Henry (1797–1878). It is clear from Eq. (6.18) that 1 henry equals 1 volt-second per ampere. - -Inductance is the property whereby an inductor exhibits opposition to the change of current flowing through it, measured in henrys (H). - -The inductance of an inductor depends on its ph ysical dimension and construction. F ormulas for calculating the inductance of inductors of different shapes are deri ved from electromagnetic theory and can be found in standard electrical engineering handbooks. For example, for the inductor, (solenoid) shown in Fig. 6.21, - -$$ -L = \frac{N^2 \mu A}{\ell} \tag{6.19} -$$ - -where *N* is the number of turns, ℓ is the length, A is the cross-sectional area, and *μ* is the permeability of the core. We can see from Eq. (6.19) that inductance can be increased by increasing the number of turns of coil, using material with higher permeability as the core, increasing the cross-sectional area, or reducing the length of the coil. - -Like capacitors, commercially a vailable inductors come in dif ferent values and types. Typical practical inductors ha ve inductance values ranging from a few microhenrys (*μ*H), as in communication systems, to tens of henrys (H) as in po wer systems. Inductors may be fixed or variable. The core may be made of iron, steel, plastic, or air . The terms *coil* and *choke* are also used for inductors. Common inductors are sho wn in Fig. 6.22. The circuit symbols for inductors are shown in Fig. 6.23, following the passive sign convention. - -Equation (6.18) is the v oltage-current relationship for an inductor . Figure 6.24 sho ws this relationship graphically for an inductor whose - -# Historical - -**Joseph Henry** (1797–1878), an American physicist, discovered inductance and constructed an electric motor. - -Born in Albany, New York, Henry graduated from Albany Academy and taught philosophy at Princeton University from 1832 to 1846. He was the first secretary of the Smithsonian Institution. He conducted several experiments on electromagnetism and developed powerful electromagnets that could lift objects weighing thousands of pounds. Interestingly, Joseph Henry discovered electromagnetic induction before Faraday but failed to publish his findings. The unit of inductance, the henry, was named after him. - -inductance is independent of current. Such an inductor is kno wn as a *linear inductor*. For a *nonlinear inductor,* the plot of Eq. (6.18) will not be a straight line because its inductance v aries with current. We will assume linear inductors in this textbook unless stated otherwise. - -The current-voltage relationship is obtained from Eq. (6.18) as - -$$ -di = \frac{1}{L}v \, dt -$$ - -Integrating gives - -$$ -i = \frac{1}{L} \int_{-\infty}^{t} v(\tau) d\tau -$$ - (6.20) - -$$ -(6.20) -$$ - -or - -$$ -i = \frac{1}{L} \int_{t_0}^t v(\tau) d\tau + i(t_0) -$$ - (6.21) - -where *i*(*t*0) is the total current for −∞ < *t* < *t*0 and *i*(−∞) = 0. The idea of making *i*(−∞) = 0 is practical and reasonable, because there must be a time in the past when there was no current in the inductor. - -The inductor is designed to store ener gy in its magnetic field. The energy stored can be obtained from Eq. (6.18). The power delivered to the inductor is - -$$ -p = vi = \left(L\frac{di}{dt}\right)i\tag{6.22} -$$ - -The energy stored is - -$$ -w = \int_{-\infty}^{t} p(\tau) d\tau = L \int_{-\infty}^{t} \frac{di}{d\tau} i d\tau -$$ - -= $L \int_{-\infty}^{t} i di = \frac{1}{2} L i^{2}(t) - \frac{1}{2} L i^{2}(-\infty)$ (6.23) - -# **Figure 6.23** - -Circuit symbols for inductors: (a) air-core, (b) iron-core, (c) variable iron-core. - -**Figure 6.24** Voltage-current relationship of an inductor. - -Since *i*(−∞) = 0, - -$$ -w = \frac{1}{2}Li^2 \tag{6.24} -$$ - -We should note the following important properties of an inductor. - -1. Note from Eq. (6.18) that the v oltage across an inductor is zero when the current is constant. Thus, - -An inductor acts like a short circuit to dc. - -2. An important property of the inductor is its opposition to the change in current flowing through it. - -The current through an inductor cannot change instantaneously. - -According to Eq. (6.18), a discontinuous change in the current through an inductor requires an infinite voltage, which is not physically possible. Thus, an inductor opposes an abrupt change in the current through it. For example, the current through an inductor may take the form sho wn in Fig. 6.25(a), whereas the inductor current cannot take the form shown in Fig. 6.25(b) in real-life situations due to the discontinuities. Ho wever, the voltage across an inductor can change abruptly. - -- 3. Like the ideal capacitor , the ideal inductor does not dissipate energy. The energy stored in it can be retrie ved at a later time. The inductor tak es power from the circuit when storing ener gy and delivers po wer to the circuit when returning pre viously stored energy. -- 4. A practical, nonideal inductor has a significant resistive component, as shown in Fig. 6.26. This is due to the fact that the inductor is made of a conducting material such as copper , which has some resistance. This resistance is called the *winding resistance Rw*, and it appears in series with the inductance of the inductor . The presence of *Rw* makes it both an ener gy storage device and an energy dissipation device. Since *Rw* is usually v ery small, it is ignored in most cases. The nonideal inductor also has a *winding capacitance Cw* due to the capacitive coupling between the conducting coils. *Cw* is very small and can be ignored in most cases, except at high frequencies. We will assume ideal inductors in this book. - -Example 6.8 The current through a 0.1-H inductor is *i*(*t*) = 10*te*−5*t* A. Find the voltage across the inductor and the energy stored in it. - -# **Solution:** - -Since *v* = *L di*∕*dt* and *L* = 0.1 H, - -$$ -v = 0.1 \frac{d}{dt} (10te^{-5t}) = e^{-5t} + t(-5)e^{-5t} = e^{-5t}(1 - 5t) -$$ - V - -**Figure 6.25** - -Current through an inductor: (a) allowed, (b) not allowable; an abrupt change is not possible. - -Since an inductor is often made of a highly conducting wire, it has a very small resistance. - -**Figure 6.26** Circuit model for a practical inductor. - -The energy stored is - -$$ -w = \frac{1}{2}Li^2 = \frac{1}{2}(0.1)100t^2e^{-10t} = 5t^2e^{-10t} \text{ J} -$$ - -If the current through a 1-mH inductor is *i*(*t*) = 90 sin (200*t*) mA, find the Practice Problem 6.8 terminal voltage and the energy stored. - -**Answer:** 18 cos (200*t*) mV, 4.05 sin2 (200*t*) *μ*J. - -Find the current through a 5-H inductor if the voltage across it is Example 6.9 - -$$ -v(t) = \begin{cases} 30t^2, & t > 0 \\ 0, & t < 0 \end{cases} -$$ - -Also, find the energy stored at *t* = 5 s. Assume *i*(*v*) > 0. - -# **Solution:** - -Since *i* = \_\_1 *L* ∫ *t*0 *t v*(*τ*) *dτ* + *i*(*t*0) and *L* = 5 H, *i* = \_\_1 5 ∫ 0 *t* 30*τ* 2 *dτ* + 0 = 6 × *t* 3 \_\_ 3 = 2*t* 3 A - -The power *p* = *vi* = 60*t* 5 , and the energy stored is then - -$$ -w = \int p \, dt = \int_0^5 60t^5 \, dt = 60 \frac{t^6}{6} \Big|_0^5 = 156.25 \, \text{kJ} -$$ - -Alternatively, we can obtain the energy stored using Eq. (6.24), by writing - -$$ -w\Big|_0^5 = \frac{1}{2}Li^2(5) - \frac{1}{2}Li(0) = \frac{1}{2}(5)(2 \times 5^3)^2 - 0 = 156.25 \text{ kJ} -$$ - -as obtained before. - -The terminal voltage of a 2-H inductor is *v* = 10(1 − *t*) V. Find the Practice Problem 6.9 current flowing through it at *t* = 4 s and the energy stored in it at *t* = 4 s. Assume *i*(0) = 2 A. - -**Figure 6.27** For Example 6.10. - -Example 6.10 Consider the circuit in Fig. 6.27(a). Under dc conditions, find: (a) *i*, *vC*, and *iL*, (b) the energy stored in the capacitor and inductor. - -# **Solution:** - -(a) Under dc conditions, we replace the capacitor with an open circuit and the inductor with a short circuit, as in Fig. 6.27(b). It is evident from Fig. 6.27(b) that - -$$ -i = i_L = \frac{12}{1+5} = 2 -$$ - A - -The voltage *vC* is the same as the voltage across the 5-Ω resistor. Hence, - -$$ -v_C = 5i = 10 \text{ V} -$$ - -(b) The energy in the capacitor is - -$$ -w_C = \frac{1}{2}Cv_C^2 = \frac{1}{2}(1)(10^2) = 50 \text{ J} -$$ - -and that in the inductor is - -$$ -w_L = \frac{1}{2}Li_L^2 = \frac{1}{2}(2)(2^2) = 4 \text{ J} -$$ - -For Practice Prob. 6.10. - -**Figure 6.29** (a) A series connection of *N* inductors, (b) equivalent circuit for the series inductors. - -Practice Problem 6.10 Determine *vC*, *iL*, and the energy stored in the capacitor and inductor in the circuit of Fig. 6.28 under dc conditions. - -**Answer:** 15 V, 7.5 A, 450 J, 168.75 J. - -# **6.5** Series and Parallel Inductors - -Now that the inductor has been added to our list of passive elements, it is necessary to extend the powerful tool of series-parallel combination. We need to know how to find the equivalent inductance of a series-connected or parallel-connected set of inductors found in practical circuits. - -Consider a series connection of *N* inductors, as shown in Fig. 6.29(a), with the equivalent circuit shown in Fig. 6.29(b). The inductors have the same current through them. Applying KVL to the loop, - -$$ -v = v_1 + v_2 + v_3 + \dots + v_N \tag{6.25} -$$ - -Substituting *vk*=*Lk di*∕*dt* results in - -$$ -v = L_1 \frac{di}{dt} + L_2 \frac{di}{dt} + L_3 \frac{di}{dt} + \dots + L_N \frac{di}{dt} -$$ - -= $(L_1 + L_2 + L_3 + \dots + L_N) \frac{di}{dt}$ -= $\left(\sum_{k=1}^{N} L_k\right) \frac{di}{dt} = L_{eq} \frac{di}{dt}$ (6.26) - -where - -Thus, - -The equivalent inductance of series-connected inductors is the sum of the individual inductances. - -Inductors in series are combined in exactly the same way as resistors in series. - -We now consider a parallel connection of *N* inductors, as shown in Fig. 6.30(a), with the equi valent circuit in Fig. 6.30(b). The inductors have the same voltage across them. Using KCL, - -$$ -i = i_1 + i_2 + i_3 + \dots + i_N \tag{6.28} -$$ - -But *ik* = \_\_1 *Lk* ∫ *t*0 *t v dt* + *ik*(*t*0); hence, *i* = \_\_1 *L*1 ∫ *t*0 *t v dt* + *i*1(*t*0) + \_\_1 *L*2 ∫ *t*0 *t v dt* + *i*2(*t*0) + ⋯ + \_\_\_1 *LN* ∫ *t*0 *t v dt* + *iN* (*t*0) = ( \_\_1 *L*1 + \_\_1 *L*2 + ⋯ + \_\_\_1 *LN*)∫ *t*0 *t v dt* + *i*1(*t*0) + *i*2(*t*0) + ⋯ + *iN* (*t*0) *N t N t* - -$$ -= \left(\sum_{k=1}^{N} \frac{1}{L_k}\right) \int_{t_0}^{t} v \, dt + \sum_{k=1}^{N} i_k(t_0) = \frac{1}{L_{\text{eq}}} \int_{t_0}^{t} v \, dt + i(t_0) \tag{6.29} -$$ - -where - -$$ -\frac{1}{L_{\text{eq}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} + \dots + \frac{1}{L_N} -$$ - (6.30) - -The initial current *i*(*t*0) through *L*eq at *t*=*t*0 is expected by KCL to be the sum of the inductor currents at *t*0. Thus, according to Eq. (6.29), - -$$ -i(t_0) = i_1(t_0) + i_2(t_0) + \dots + i_N(t_0) -$$ - -According to Eq. (6.30), - -The equivalent inductance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances. - -Note that the inductors in parallel are combined in the same way as resistors in parallel. - -For two inductors in parallel (*N* = 2), Eq. (6.30) becomes - -$$ -\frac{1}{L_{\text{eq}}} = \frac{1}{L_1} + \frac{1}{L_2} \qquad \text{or} \qquad L_{\text{eq}} = \frac{L_1 L_2}{L_1 + L_2} \tag{6.31} -$$ - -As long as all the elements are of the same type, the ∆-Y transformations for resistors discussed in Section 2.7 can be e xtended to capacitors and inductors. - -# **Figure 6.30** - -(a) A parallel connection of *N* inductors, (b) equivalent circuit for the parallel inductors. - -# **TABLE 6.1** - -Important characteristics of the basic elements.† - -| Relation | Resistor (R) | Capacitor (C) | Inductor (L) | -|---------------------------------|----------------------------------------|-------------------------------------------------|----------------------------------------------------| -| v-i: | v = iR | t
__1
v =
C ∫
i(τ) dτ + v(t0)
t0 | v = L __di
dt | -| i-v: | i = v∕R | i = C ___ dv
dt | t
__1
i =
∫
v(τ) dτ + i(t0)
L
t0 | -| p or w: | v2
__
2
p = i
R =
R | __1
Cv2
w =
2 | __1
Li2
w =
2 | -| Series: | Req
= R1
+ R2 | _______ C1C2
Ceq
=
C1
+ C2 | Leq
= L1
+ L2 | -| Parallel: | _______ R1R2
Req
=
R1
+ R2 | Ceq
= C1
+ C2 | _______ L1L2
Leq
=
L1
+ L2 | -| At dc: | Same | Open circuit | Short circuit | -| Circuit variable
that cannot | | | | -| change abruptly: | Not applicable | v | i | - -† *Passive sign convention is assumed.* - -It is appropriate at this point to summarize the most important characteristics of the three basic circuit elements we have studied. The summary is given in Table 6.1. - -Example 6.11 Find the equivalent inductance of the circuit shown in Fig. 6.31. - -# **Solution:** - -The 10-H, 12-H, and 20-H inductors are in series; thus, combining them gives a 42-H inductance. This 42-H inductor is in parallel with the 7-H inductor so that they are combined, to give - -$$ -\frac{7 \times 42}{7 + 42} = 6 -$$ - -This 6-H inductor is in series with the 4-H and 8-H inductors. Hence, - -$$ -L_{\text{eq}} = 4 + 6 + 8 = 18 \text{ H} -$$ - -Practice Problem 6.11 Calculate the equivalent inductance for the inductive ladder network in Fig. 6.32. - -For the circuit in Fig. 6.33, *i*(*t*) = 4(2 − *e* Example 6.12 −10*t* ) mA. If *i*2(0) = −1 mA, find: (a) *i*1(0); (b) *v*(*t*), *v*1(*t*), and *v*2(*t*); (c) *i*1(*t*) and *i*2(*t*). - -# **Solution:** - -(a) From *i*(*t*) = 4(2 − *e*−10*t* ) mA, *i*(0) = 4(2 − 1) = 4 mA. Since *i* = *i*1 + *i*2, - -$$ -i_1(0) = i(0) - i_2(0) = 4 - (-1) = 5 \text{ mA} -$$ - -(b) The equivalent inductance is - -$$ -L_{\text{eq}} = 2 + 4 || 12 = 2 + 3 = 5 \text{ H} -$$ - -Thus, - -$$ -v(t) = L_{eq} \frac{di}{dt} = 5(4)(-1)(-10)e^{-10t} \text{ mV} = 200e^{-10t} \text{ mV} -$$ - -and - -$$ -v_1(t) = 2\frac{di}{dt} = 2(-4)(-10)e^{-10t} -$$ - mV = 80 $e^{-10t}$ mV - -Since *v* = *v*1 + *v*2, - -$$ -v_2(t) = v(t) - v_1(t) = 120e^{-10t} \text{ mV} -$$ - -(c) The current *i*1 is obtained as - -$$ -i_1(t) = \frac{1}{4} \int_0^t v_2 dt + i_1(0) = \frac{120}{4} \int_0^t e^{-10t} dt + 5 \text{ mA} -$$ - -= $-3e^{-10t} \Big|_0^t + 5 \text{ mA} = -3e^{-10t} + 3 + 5 = 8 - 3e^{-10t} \text{ mA}$ - -Similarly, - -$$ -i_2(t) = \frac{1}{12} \int_0^t v_2 dt + i_2(0) = \frac{120}{12} \int_0^t e^{-10t} dt - 1 \text{ mA} -$$ -$$ -= -e^{-10t} \Big|_0^t - 1 \text{ mA} = -e^{-10t} + 1 - 1 = -e^{-10t} \text{ mA} -$$ - -Note that *i*1(*t*) + *i*2(*t*) = *i*(*t*). - -In the circuit of Fig. 6.34, *i*1(*t*) = 3*e* Practice Problem 6.12 −2*t* A. If *i*(0) = 7 A, find: (a) *i*2(0); (b) *i*2(*t*) and *i*(*t*); (c) *v*1(*t*), *v*2(*t*), and *v*(*t*). - -**Answer:** (a) 4 A, (b) -$$ -(-2 + 6e^{-2t}) -$$ - A, $(-2 + 9e^{-2t})$ A, (c) $-36e^{-2t}$ V, $-144e^{-2t}$ V, $-180e^{-2t}$ V. - -**Figure 6.34** For Practice Prob. 6.12. - -# **6.6** Applications - -Circuit elements such as resistors and capacitors are commercially available in either discrete form or integrated-circuit (IC) form. Unlike capacitors and resistors, inductors with appreciable inductance are dif ficult to produce on IC substrates. Therefore, inductors (coils) usually - -**Figure 6.33** For Example 6.12. - -come in discrete form and tend to be more b ulky and e xpensive. For this reason, inductors are not as versatile as capacitors and resistors, and they are more limited in applications. However, there are several applications in which inductors have no practical substitute. They are routinely used in relays, delays, sensing devices, pick-up heads, telephone circuits, radio and TV receivers, power supplies, electric motors, microphones, and loudspeakers, to mention a few. - -Capacitors and inductors possess the following three special properties that make them very useful in electric circuits: - -- 1. The capacity to store ener gy makes them useful as temporary v oltage or current sources. Thus, they can be used for generating a large amount of current or voltage for a short period of time. -- 2. Capacitors oppose an y abrupt change in v oltage, while inductors oppose any abrupt change in current. This property makes inductors useful for spark or arc suppression and for converting pulsating dc voltage into relatively smooth dc voltage. -- 3. Capacitors and inductors are frequenc y sensiti ve. This property makes them useful for frequency discrimination. - -The first two properties are put to use in dc circuits, while the third one is taken advantage of in ac circuits. We will see how useful these properties are in later chapters. F or now, consider three applications involving capacitors and op amps: integrator, differentiator, and analog computer. - -# **6.6.1** Integrator - -Important op amp circuits that use energy-storage elements include integrators and dif ferentiators. These op amp circuits often in volve resistors and capacitors; inductors (coils) tend to be more b ulky and expensive. - -The op amp integrator is used in numerous applications, especially in analog computers, to be discussed in Section 6.6.3. - -An integrator is an op amp circuit whose output is proportional to the integral of the input signal. - -If the feedback resistor *Rf* in the f amiliar in verting amplifier of Fig. 6.35(a) is replaced by a capacitor , we obtain an ideal inte grator, as shown in Fig. 6.35(b). It is interesting that we can obtain a mathematical representation of integration this way. At node *a* in Fig. 6.35(b), - -$$ -i_R = i_C \tag{6.32} -$$ - -But - -$$ -i_R = \frac{v_i}{R}, \qquad i_C = -C \frac{dv_o}{dt} -$$ - -Substituting these in Eq. (6.32), we obtain - -$$ -\frac{v_i}{R} = -C \frac{dv_o}{dt} -$$ - (6.33a) - -$$ -dv_o = -\frac{1}{RC}v_i dt -$$ - (6.33b) - -Replacing the feedback resistor in the inverting amplifier in (a) produces an integrator in (b). - -Integrating both sides gives - -$$ -v_o(t) - v_o(0) = -\frac{1}{RC} \int_0^t v_i(\tau) \, d\tau \tag{6.34} -$$ - -To ensure that *vo*(0) = 0, it is always necessary to discharge the integrator's capacitor prior to the application of a signal. Assuming *vo*(0) = 0, - -$$ -v_o = -\frac{1}{RC} \int_0^t v_i(\tau) \, d\tau \tag{6.35} -$$ - -which shows that the circuit in Fig. 6.35(b) pro vides an output v oltage proportional to the integral of the input. In practice, the op amp integrator requires a feedback resistor to reduce dc gain and prevent saturation. Care must be tak en that the op amp operates within the linear range so that it does not saturate. - -If *v*1 = 10 cos 2 *t* mV and *v*2 = 0.5*t* mV, find *vo* in the op amp circuit in Example 6.13 Fig. 6.36. Assume that the voltage across the capacitor is initially zero. - -# **Solution:** - -This is a summing integrator, and - -$$ -v_o = -\frac{1}{R_1C} \int v_1 dt - \frac{1}{R_2C} \int v_2 dt -$$ - -= -$$ --\frac{1}{3 \times 10^6 \times 2 \times 10^{-6}} \int_0^t 10 \cos(2\tau) d\tau -$$ - -$$ --\frac{1}{100 \times 10^3 \times 2 \times 10^{-6}} \int_0^t 0.5\tau d\tau -$$ - -= -$$ --\frac{1}{6} \frac{10}{2} \sin 2t - \frac{1}{0.2} \frac{0.5t^2}{2} = -0.833 \sin 2t - 1.25t -$$ - -The integrator in Fig. 6.35(b) has *R* = 100 kΩ, *C* = 20 *μ*F. Determine the Practice Problem 6.13 output voltage when a dc voltage of 2.5 mV is applied at *t* = 0. Assume that the op amp is initially nulled. - -**Answer:** −1.25*t* m V. - -# **6.6.2** Differentiator - -A differentiator is an op amp circuit whose output is proportional to the rate of change of the input signal. - -In Fig. 6.35(a), if the input resistor is replaced by a capacitor , the resulting circuit is a dif ferentiator, shown in Fig. 6.37. Applying KCL at node *a*, - -$$ -i_R = i_C \tag{6.36} -$$ - -2 mV But - -$$ -i_R = -\frac{v_o}{R}, \qquad i_C = C \frac{dv_i}{dt} -$$ - -Substituting these in Eq. (6.36) yields - -$$ -v_o = -RC \frac{dv_i}{dt} \tag{6.37} -$$ - -showing that the output is the deri vative of the input. Differentiator circuits are electronically unstable because an y electrical noise within the circuit is exaggerated by the differentiator. For this reason, the differentiator circuit in Fig. 6.37 is not as useful and popular as the integrator. It is seldom used in practice. - -Example 6.14 Sketch the output voltage for the circuit in Fig. 6.38(a), given the input voltage in Fig. 6.38(b). Take *vo* = 0 at *t* = 0. - -# **Solution:** - -This is a differentiator with - -$$ -RC = 5 \times 10^3 \times 0.2 \times 10^{-6} = 10^{-3} \text{ s} -$$ - -For 0 < *t* < 4 ms, we can express the input voltage in Fig. 6.38(b) as - -*vi* = { 2000*t* 8 − 2000*t* 0 < *t* < 2 ms 2 < *t* < 4 ms - -This is repeated for 4 < *t* < 8 ms. Using Eq. (6.37), the output is ob tained as - -$$ -v_o = -RC \frac{dv_i}{dt} = \begin{cases} -2\text{ V} & 0 < t < 2\text{ ms} \\ 2\text{ V} & 2 < t < 4\text{ ms} \end{cases} -$$ - -Thus, the output is as sketched in Fig. 6.39. - -*v*o (V) - -8642 2 0 –2 t (ms) - -**Figure 6.39** Output of the circuit in Fig. 6.38(a). - -Practice Problem 6.14 The differentiator in Fig. 6.37 has *R* = 100 kΩ and *C* = 0.1 *μ*F. Given that *vi* = 5*t* V, determine the output *vo*. - -**Answer:** −50 mV. - -**Figure 6.37** An op amp differentiator. - -For Example 6.14. - -# **6.6.3** Analog Computer - -Op amps were initially de veloped for electronic analog computers. Analog computers can be programmed to solv e mathematical models of mechanical or electrical systems. These models are usually expressed in terms of differential equations. - -To solv e simple dif ferential equations using the analog computer requires cascading three types of op amp circuits: inte grator circuits, summing amplifiers, and inverting/noninverting amplifiers for negative/ positive scaling. The best way to illustrate how an analog computer solves a differential equation is with an example. - -Suppose we desire the solution *x*(*t*) of the equation - -$$ -a\frac{d^2x}{dt^2} + b\frac{dx}{dt} + cx = f(t), \qquad t > 0 -$$ -\n(6.38) - -where *a*, *b*, and *c* are constants, and *f*(*t*) is an arbitrary forcing func tion. The solution is obtained by first solving the highest-order derivative term. Solving for *d*2 *x*∕*dt*2 yields - -$$ -\frac{d^2x}{dt^2} = \frac{f(t)}{a} - \frac{b}{a}\frac{dx}{dt} - \frac{c}{a}x\tag{6.39} -$$ - -To obtain *dx*∕*dt*, the *d*2 *x*∕*dt*2 term is inte grated and inverted. Finally, to obtain *x*, the *dx*∕*dt* term is integrated and inverted. The forcing function is injected at the proper point. Thus, the analog computer for solving Eq. (6.38) is implemented by connecting the necessary summers, inverters, and inte grators. A plotter or oscilloscope may be used to vie w the output *x*, or *dx*∕*dt*, or *d*2 *x*∕*dt*2 , depending on where it is connected in the system. - -Although the above example is on a second-order differential equation, any differential equation can be simulated by an analog computer comprising integrators, inverters, and inverting summers. But care must be exercised in selecting the v alues of the resistors and capacitors, to ensure that the op amps do not saturate during the solution time interval. - -The analog computers with vacuum tubes were built in the 1950s and 1960s. Recently their use has declined. They ha ve been superseded by modern digital computers. Ho wever, we still study analog computers for two reasons. First, the a vailability of integrated op amps has made it pos sible to build analog computers easily and cheaply. Second, understanding analog computers helps with the appreciation of the digital computers. - -Design an analog computer circuit to solve the differential equation: Example 6.15 - -$$ -\frac{d^2v_o}{dt^2} + 2\frac{dv_o}{dt} + v_o = 10\sin 4t, \qquad t > 0 -$$ - -subject to *vo*(0) = −4, *vo* ′ (0) = 1, where the prime refers to the time derivative. - -# **Solution:** - -1. **Define.** We have a clearly defined problem and expected solution. I might remind the student that many times the problem is not so well defined and this portion of the problem-solving process could - -require much more effort. If this is so, then you should always keep in mind that time spent here will result in much less effort later and most likely save you a lot of frustration in the process. - -- 2. **Present.** Clearly, using the devices developed in Section 6.6.3 will allow us to create the desired analog computer circuit. We will need the integrator circuits (possibly combined with a summing capability) and one or more inverter circuits. -- 3. **Alternative.** The approach for solving this problem is straightforward. We will need to pick the correct values of resistances and capacitors to allow us to realize the equation we are representing. The final output of the circuit will give the desired result. -- 4. **Attempt.** There are an infinite number of possibilities for picking the resistors and capacitors, many of which will result in correct solutions. Extreme values of resistors and capacitors will result in incorrect outputs. For example, low values of resistors will overload the electronics. Picking values of resistors that are too large will cause the op amps to stop functioning as ideal devices. The limits can be determined from the characteristics of the real op amp. - -We first solve for the second derivative as - -$$ -\frac{d^2v_o}{dt^2} = 10\sin 4t - 2\frac{dv_o}{dt} - v_o -$$ - (6.15.1) - -Solving this requires some mathematical operations, includ ing summing, scaling, and integration. Integrating both sides of Eq. (6.15.1) gives - -$$ -\frac{dv_o}{dt} = -\int_0^t \left(-10\sin(4\tau) + 2\frac{dv_o(\tau)}{d\tau} + v_o(\tau)\right) d\tau + v'_o(0) \quad (6.15.2) -$$ - -where *v o*′ (0) = 1. We implement Eq. (6.15.2) using the summing integrator shown in Fig. 6.40(a). The values of the resistors and capacitors have been chosen so that *RC* = 1 for the term - -$$ --\frac{1}{RC}\int_0^t v_o(\tau)\,d\tau -$$ - -Other terms in the summing integrator of Eq. (6.15.2) are implemented accordingly. The initial condition *dv o*(0) ∕*dt* = 1 is imple mented by connecting a 1-V battery with a switch across the capacitor as shown in Fig. 6.40(a). - -The next step is to obtain *v o* by integrating *dv o* ∕*dt* and inverting the result, - -$$ -v_o = -\int_0^t \frac{dv_o(\tau)}{d\tau} d\tau + v(0) -$$ - (6.15.3) - -This is implemented with the circuit in Fig. 6.40(b) with the battery giving the initial condition of −4 V. We now combine the two circuits in Fig. 6.40(a) and (b) to obtain the complete circuit shown in Fig. 6.40(c). When the input signal 10 sin 4 *t* is applied, we open the switches at *t* = 0 to obtain the output waveform *v o*, which may be viewed on an oscilloscope. - -For Example 6.15. - -5. **Evaluate.** The answer looks correct, but is it? If an actual solution for *vo* is desired, then a good check would be to first find the solution by realizing the circuit in *PSpice*. This result could then be compared with a solution using the differential solution capability of *MATLAB.* - -Since all we need to do is check the circuit and confirm that it represents the equation, we have an easier technique to use. We just go through the circuit and see if it generates the desired equation. - -However, we still have choices to make. We could go through the circuit from left to right but that would involve differentiating the result to obtain the original equation. An easier approach would be to go from right to left. This is the approach we will use to check the answer. - - Starting with the output, *vo*, we see that the right-hand op amp is nothing more than an inverter with a gain of one. This means that the output of the middle circuit is −*vo*. The following represents the action of the middle circuit. - -$$ --v_o = -\left(\int_0^t \frac{dv_o}{d\tau} d\tau + v_o(0)\right) = -\left(v_o \Big|_0^t + v_o(0)\right) -$$ - -= -\left(v\_o(t) - v\_o(0) + v\_o(0)\right) - -where *vo*(0) = −4 V is the initial voltage across the capacitor. - -We check the circuit on the left the same way. - -$$ -\frac{dv_o}{dt} = -\left(\int_0^t -\frac{d^2v_o}{dt^2}d\tau - v'_o(0)\right) = -\left(-\frac{dv_o}{dt} + v'_o(0) - v'_o(0)\right) -$$ - -Now, all we need to verify is that the input to the first op amp is −*d*2 *vo*∕*dt*2 . - -Looking at the input we see that it is equal to - -$$ --10\sin(4t) + v_o + \frac{1/10^{-6}}{0.5\,\text{M}\Omega} \frac{dv_o}{dt} = -10\sin(4t) + v_o + 2\frac{dv_o}{dt} -$$ - -which does produce −*d*2 *vo*∕*dt*2 from the original equation. - -6. **Satisfactory?** The solution we have obtained is satisfactory. We can now present this work as a solution to the problem. - -Practice Problem 6.15 Design an analog computer circuit to solve the differential equation: - -$$ -\frac{d^2v_o}{dt^2} + 3\frac{dv_o}{dt} + 2v_o = 4\cos 10t, \qquad t > 0 -$$ - -subject to *vo*(0) = 2, *vo* ′ (0) = 0. - -**Answer:** See Fig. 6.41, where *RC* = 1 s. - -# **6.7** Summary - -1. The current through a capacitor is directly proportional to the time rate of change of the voltage across it. - -$$ -i = C \frac{dv}{dt} -$$ - -The current through a capacitor is zero unless the v oltage is changing. Thus, a capacitor acts like an open circuit to a dc source. - -2. The voltage across a capacitor is directly proportional to the time integral of the current through it. - -$$ -v = \frac{1}{C} \int_{-\infty}^{t} i \, d\tau = \frac{1}{C} \int_{t_0}^{t} i \, d\tau + v(t_0) -$$ - -The voltage across a capacitor cannot change instantly. - -- 3. Capacitors in series and in parallel are combined in the same w ay as conductances. -- 4. The voltage across an inductor is directly proportional to the time rate of change of the current through it. - -$$ -v = L\frac{di}{dt} -$$ - -The voltage across the inductor is zero unless the current is chang ing. Thus, an inductor acts like a short circuit to a dc source. - -5. The current through an inductor is directly proportional to the time integral of the voltage across it. - -$$ -i = \frac{1}{L} \int_{-\infty}^{t} v \, d\tau = \frac{1}{L} \int_{t_0}^{t} v \, d\tau + i(t_0) -$$ - -The current through an inductor cannot change instantly. - -- 6. Inductors in series and in parallel are combined in the same w ay resistors in series and in parallel are combined. -- 7. At any given time *t*, the energy stored in a capacitor is \_1 2 *Cv*2 , while the energy stored in an inductor is \_1 2 *Li*2 . -- 8. Three application circuits, the inte grator, the differentiator, and the analog computer , can be realized using resistors, capacitors, and op amps. - -# Review Questions - -**6.1** What charge is on a 5-F capacitor when it is connected across a 120-V source? - -| (a) 600 C | (b) 300 C | -|-----------|-----------| -| (c) 24 C | (d) 12 C | - -**6.2** Capacitance is measured in: - -| (a) coulombs | (b) joules | -|--------------|------------| -| (c) henrys | (d) farads | - -**6.3** When the total charge in a capacitor is doubled, the energy stored: - -| (a) remains the same | (b) is halved | -|----------------------|-------------------| -| (c) is doubled | (d) is quadrupled | - -**6.4** Can the voltage waveform in Fig. 6.42 be associated with a real capacitor? - -(a) Yes (b) No - -# **Figure 6.42** - -For Review Question 6.4. - -**6.5** The total capacitance of two 40-mF series-connected capacitors in parallel with a 4-mF capacitor is: - -| (a) 3.8 mF | (b) 5 mF | (c) 24 mF | -|------------|-----------|-----------| -| (d) 44 mF | (e) 84 mF | | - -**6.6** In Fig. 6.43, if *i* = cos 4*t* and *v* = sin 4*t*, the element is: - -(a) a resistor (b) a capacitor (c) an inductor - -# **Figure 6.43** - -For Review Question 6.6. - -**6.7** A 5-H inductor changes its current by 3 A in 0.2 s. The voltage produced at the terminals of the inductor is: - -| (a) 75 V | (b) 8.888 V | -|----------|-------------| -| (c) 3 V | (d) 1.2 V | - -**6.8** If the current through a 10-mH inductor increases from zero to 2 A, how much energy is stored in the inductor? - -| (a) 40 mJ | (b) 20 mJ | -|-----------|-----------| -| (c) 10 mJ | (d) 5 mJ | - -**6.9** Inductors in parallel can be combined just like resistors in parallel. - -$$ -(a) True \t\t (b) False -$$ - -**6.10** For the circuit in Fig. 6.44, the voltage divider formula is: - -(a) -$$ -v_1 = \frac{L_1 + L_2}{L_1} v_s -$$ - -\n(b) $v_1 = \frac{L_1 + L_2}{L_2} v_s$ -\n(c) $v_1 = \frac{L_2}{L_1 + L_2} v_s$ -\n(d) $v_1 = \frac{L_1}{L_1 + L_2} v_s$ -\n $v_s$ - -# **Figure 6.44** - -For Review Question 6.10. - -*Answers: 6.1a, 6.2d, 6.3d, 6.4b, 6.5c, 6.6b, 6.7a, 6.8b, 6.9a, 6.10d.* - -# Problems - -# Section 6.2 Capacitors - -- **6.1** If the voltage across a 7.5-F capacitor is 2*te*−3*t* V, find the current and the power. -- **6.2** A 50-*μ*F capacitor has energy *w*(*t*) = 10 cos2 377*t* J. Determine the current through the capacitor. -- **6.3** Design a problem to help other students better understand how capacitors work. -- **6.4** A voltage across a capacitor is equal to [2 – 2 cos(4*t*)] V and the current flowing through it is equal to 2 sin (4*t*) *μ*A. Determine the value of the capacitance. Calculate the power being stored by the capacitor. -- **6.5** The voltage across a 4-*μ*F capacitor is shown in Fig. 6.45. Find the current waveform. - -For Prob. 6.5. - -**6.6** The voltage waveform in Fig. 6.46 is applied across a 55-*μ*F capacitor. Draw the current waveform through it. - -# **Figure 6.46** - -For Prob. 6.6. - -- **6.7** At *t* = 0, the voltage across a 25-mF capacitor is 10 V. Calculate the voltage across the capacitor for *t* > 0 when current 5*t* mA flows through it. -- **6.8** A 4-mF capacitor has the terminal voltage - -$$ -v = \begin{cases} 50 \text{ V}, & t \le 0 \\ Ae^{-100t} + Be^{-600t} \text{ V}, & t \ge 0 \end{cases} -$$ - -If the capacitor has an initial current of 2 A, find: - -- (a) the constants *A* and *B*, -- (b) the energy stored in the capacitor at *t* = 0, -- (c) the capacitor current for *t* > 0. - -Problems **241** - -- **6.9** The current through a 0.5-F capacitor is 6(1 − *e* −*t* ) A. Determine the voltage and power at *t* = 2 s. Assume *v*(0) = 0. -- **6.10** The voltage across a 5-mF capacitor is shown in Fig. 6.47. Determine the current through the capacitor. - -# **Figure 6.47** - -For Prob. 6.10. - -**6.11** A 4-mF capacitor has the current waveform shown in Fig. 6.48. Assuming that *v*(0) = 10 V, sketch the voltage waveform *v*(*t*). - -# **Figure 6.48** - -For Prob. 6.11. - -- **6.12** A voltage of 45*e*−2000*t* V appears across a parallel combination of a 100-mF capacitor and a 12-Ω resistor. Calculate the power absorbed by the parallel combination. -- **6.13** Find the voltage across the capacitors in the circuit of Fig. 6.49 under dc conditions. - -**Figure 6.49** - -For Prob. 6.13. - -# Section 6.3 Series and Parallel Capacitors - -**6.14** Series-connected 20- and 60-pF capacitors are placed in parallel with series-connected 30- and 70-pF capacitors. Determine the equivalent capacitance. - -**6.15** Two capacitors (25 and 75 *μ*F) are connected to a 100-V source. Find the energy stored in each capacitor if they are connected in: - -(a) parallel (b) series - -**6.16** The equivalent capacitance at terminals *a*-*b* in the circuit of Fig. 6.50 is 20 *μ*F. Calculate the value of *C*. - -# **Figure 6.50** - -For Prob. 6.16. - -**6.17** Determine the equivalent capacitance for each of the circuits of Fig. 6.51. - -# **Figure 6.51** - -For Prob. 6.17. - -**6.18** Find *C*eq in the circuit of Fig. 6.52 if all capacitors are 4 *μ*F. - -**6.19** Find the equivalent capacitance between terminals *a* and *b* in the circuit of Fig. 6.53. All capacitances are in *μ*F. - -**Figure 6.53** For Prob. 6.19. - -> **6.20** Find the equivalent capacitance at terminals *a*-*b* of the circuit in Fig. 6.54. - -**6.21** Determine the equivalent capacitance at terminals *a*-*b* of the circuit in Fig. 6.55. - -**6.22** Obtain the equivalent capacitance of the circuit in Fig. 6.56. - -# **Figure 6.56** - -For Prob. 6.22. - -# **Figure 6.57** - -For Prob. 6.23. - -**6.24** In the circuit shown in Fig. 6.58 assume that the capacitors were initially uncharged and that the current source has been connected to the circuit long enough for all the capacitors to reach steady state (no current flowing through the capacitors). Determine the voltage across each capacitor and the energy - -# **Figure 6.58** For Prob. 6.24. - -**6.25** (a) Show that the voltage-division rule for two capacitors in series as in Fig. 6.59(a) is - -$$ -v_1 = \frac{C_2}{C_1 + C_2} v_s -$$ -, $v_2 = \frac{C_1}{C_1 + C_2} v_s$ - -assuming that the initial conditions are zero. - -**Figure 6.59** For Prob. 6.25. - -(b) For two capacitors in parallel as in Fig. 6.59(b), show that the current-division rule is - -$$ -i_1 = \frac{C_1}{C_1 + C_2} i_s -$$ -, $i_2 = \frac{C_2}{C_1 + C_2} i_s$ - -assuming that the initial conditions are zero. - -- **6.26** Three capacitors, *C*1 = 5 *μ*F, *C*2 = 10 *μ*F, and *C*3 = 20 *μ*F, are connected in parallel across a 200-V source. Determine: - - (a) the total capacitance, - - (b) the charge on each capacitor, - - (c) the total energy stored in the parallel combination. -- **6.27** Given that four 10-*μ*F capacitors can be connected -- in series and in parallel, find the minimum and maximum values that can be obtained by such series/parallel combinations. -- **6.28** Obtain the equivalent capacitance of the network shown in Fig. 6.60. **\*** - -**Figure 6.60** For Prob. 6.28. - -**6.29** Determine *C*eq for each circuit in Fig. 6.61. - -For Prob. 6.29. - -**6.30** Assuming that the capacitors are initially uncharged, find *vo*(*t*) in the circuit of Fig. 6.62. - -**6.31** If *v*(0) = 0, find *v*(*t*), *i*1(*t*), and *i*2(*t*) in the circuit of Fig. 6.63. - -**Figure 6.63** For Prob. 6.31. - -**6.32** In the circuit in Fig. 6.64, let *is* = 4.5*e*−2*t* mA and the voltage across each capacitor is equal to zero at *t* = 0. Determine *v*1 and *v*2 and the energy stored in each capacitor for all *t* > 0. - -For Prob. 6.32. - -\* An asterisk indicates a challenging problem. - -**6.33** Obtain the Thevenin equivalent at the terminals, *a*-*b*, of the circuit shown in Fig. 6.65. Please note that Thevenin equivalent circuits do not generally exist for circuits involving capacitors and resistors. This is a special case where the Thevenin equivalent circuit does exist. - -**Figure 6.65** - -For Prob. 6.33. - -# Section 6.4 Inductors - -- **6.34** The current through a 25-mH inductor is 10*e*−*t*∕2 A. Find the voltage and the power at *t* = 3 s. -- **6.35** An inductor has a linear change in current from 100 mA to 200 mA in 2 ms and induces a voltage of 160 mV. Calculate the value of the inductor. -- **6.36** Design a problem to help other students better understand how inductors work. -- **6.37** The current through a 12-mH inductor is 4 sin 100*t* A. Find the voltage, and the energy stored at *t* = \_\_\_*π* 200s. -- **6.38** The current through a 40-mH inductor is - -$$ -i(t) = \begin{cases} 0, & t < 0 \\ te^{-2t} \, \mathbf{A}, & t > 0 \end{cases} -$$ - -Find the voltage *v*(*t*). - -**6.39** The voltage across a 50-mH inductor is given by - -$$ -v(t) = [5e^{-2t} + 2t + 4] \text{ V} \qquad \text{for } t > 0. -$$ - - Determine the current *i*(*t*) through the inductor. Assume that *i*(0) = 0 A. - -**6.40** The current through a 5-mH inductor is shown in Fig. 6.66. Determine the voltage across the inductor at *t* = 1, 3, and 5 ms. - -**Figure 6.66** For Prob. 6.40. - -- **6.41** The voltage across a 2-H inductor is 20(1 − *e* −2*t* ) V. If the initial current through the inductor is 0.3 A, find the current and the energy stored in the inductor at *t* = 1 s. -- **6.42** If the voltage waveform in Fig. 6.67 is applied across the terminals of a 5-H inductor, calculate the current through the inductor. Assume *i*(0) = −1 A. - -# **Figure 6.67** - -For Prob. 6.42. - -- **6.43** The current in a 150-mH inductor increases from 0 to 60 mA (steady state). How much energy is stored in the inductor? -- **6.44** A 100-mH inductor is connected in parallel with a 2-kΩ resistor. The current through the inductor is *i*(*t*) = 35*e*−400*t* mA. (a) Find the voltage *vL* across the inductor. (b) Find the voltage *vR* across the resistor. (c) Does *vR*(*t*) + *vL*(*t*) = 0? (d) Calculate the energy stored in the inductor at *t* = 0. **\*** -- **6.45** If the voltage waveform in Fig. 6.68 is applied to a 25-mH inductor, find the inductor current *i*(*t*) for 0 < *t* < 2 seconds. Assume *i*(0) = 0. - -# **Figure 6.68** - -For Prob. 6.45. - -**6.46** Find *vC*, *iL*, and the energy stored in the capacitor and inductor in the circuit of Fig. 6.69 under dc conditions. - -For Prob. 6.46. - -# **Figure 6.70** - -For Prob. 6.47. - -**6.48** Under steady-state dc conditions, find *i* and *v* in the circuit in Fig. 6.71. - -For Prob. 6.48. - -# Section 6.5 Series and Parallel Inductors - -**6.49** Find the equivalent inductance of the circuit in Fig. 6.72. Assume all inductors are 40 mH. - -# **Figure 6.72** - -For Prob. 6.49. - -**6.50** An energy-storage network consists of seriesconnected 16- and 14-mH inductors in parallel with series-connected 24- and 36-mH inductors. Calculate the equivalent inductance. - -**6.51** Determine *L*eq at terminals *a*-*b* of the circuit in Fig. 6.73. - -**Figure 6.73** For Prob. 6.51. - -**6.53** Find *L*eq at the terminals of the circuit in Fig. 6.75. - -**6.54** Find the equivalent inductance looking into the terminals of the circuit in Fig. 6.76. - -**6.55** Find *L*eq in each of the circuits in Fig. 6.77. - -**6.56** Find *L*eq in the circuit of Fig. 6.78. - -For Prob. 6.56. - -**6.57** Determine *L*eq that may be used to represent the inductive network of Fig. 6.79 at the terminals. **\*** - -For Prob. 6.57. - -**6.58** The current waveform in Fig. 6.80 flows through a 3-H inductor. Sketch the voltage across the inductor over the interval 0 < *t* < 6 s. - -**Figure 6.80** For Prob. 6.58. - -**6.59** (a) For two inductors in series as in Fig. 6.81(a), show that the voltage division principle is - -$$ -v_1 = \frac{L_1}{L_1 + L_2} v_s -$$ -, $v_2 = \frac{L_2}{L_1 + L_2} v_s$ - -assuming that the initial conditions are zero. - -(b) For two inductors in parallel as in Fig. 6.81(b), show that the current-division principle is - -$$ -i_1 = \frac{L_2}{L_1 + L_2} i_s -$$ -, $i_2 = \frac{L_1}{L_1 + L_2} i_s$ - -assuming that the initial conditions are zero. - -For Prob. 6.59. - -**6.60** In the circuit of Fig. 6.82, *io*(0) = 2 A. Determine *io*(*t*) and *vo*(*t*) for *t* > 0. - -For Prob. 6.60. - -**6.61** Consider the circuit in Fig. 6.83. Find: (a) *L*eq, *i*1(*t*), and *i*2(*t*) if *is* = 3*e*−*t* mA, (b) *vo*(*t*), (c) energy stored in the 20-mH inductor at *t* = 1 s. - -**6.62** Consider the circuit in Fig. 6.84. Given that *v*(*t*) = 12*e*−3*t* mV for *t* > 0 and *i*1(0) = −30 mA, find: (a) *i*2(0), (b) *i*1(*t*) and *i*2(*t*). - -For Prob. 6.62. - -**6.63** In the circuit of Fig. 6.85, sketch *vo*. - -**6.64** The switch in Fig. 6.86 has been in position *A* for a long time. At *t* = 0, the switch moves from position *A* to *B*. The switch is a make-before-break type so that there is no interruption in the inductor current. Find: - -(a) *i*(*t*) for *t* > 0, - -(b) *v just after* the switch has been moved to position *B*, (c) *v*(*t*) long after the switch is in position *B*. - -**6.65** The inductors in Fig. 6.87 are initially charged and are connected to the black box at *t* = 0. If *i*1(0) = 4 A, *i*2(0) = −2 A, and *v*(*t*) = 50*e*−200*t* mV, *t* ≥ 0, find: - -(a) the energy initially stored in each inductor, - -- (b) the total energy delivered to the black box from *t* = 0 to *t* = ∞, -- (c) *i*1(*t*) and *i*2(*t*), *t* ≥ 0, (d) *i*(*t*), *t* ≥ 0. - -**6.66** The current *i*(*t*) through a 20-mH inductor is equal, in magnitude, to the voltage across it for all values of time. If *i*(0) = 2 A, find *i*(*t*). - -# Section 6.6 Applications - -**6.67** An op amp integrator has *R* = 50 kΩ and *C* = 0.04 *μ*F. If the input voltage is *vi* = 10 sin 50*t* mV, obtain the output voltage. Assume that at *t* equal to zero, the output is equal to zero. - -- **6.68** A 6-V dc voltage is applied to an integrator with *R* = 50 kΩ, *C* = 100 *μ*F at *t* = 0. How long will it take for the op amp to saturate if the saturation voltages are +12 V and −12 V? Assume that the initial capacitor voltage was zero. -- **6.69** An op amp integrator with *R* = 4 MΩ and *C* = 1 *μ*F has the input waveform shown in Fig. 6.88. Plot the output waveform. - -**6.70** Using a single op amp, a capacitor, and resistors of 100 kΩ or less, design a circuit to implement - -$$ -v_o = -2 \int_0^t v_i(\tau) \, d\tau -$$ - -Assume *vo*=0 at *t*=0. - -**6.71** Show how you would use a single op amp to generate - -$$ -v_o = -\int_0^t (v_1 + 4v_2 + 10v_3) \, dt -$$ - -If the inte grating capacitor is *C* = 5 *μ*F, obtain the other component values. - -**6.72** At *t* = 1.5 ms, calculate *vo* due to the cascaded integrators in Fig. 6.89. Assume that the integrators are reset to 0 V at *t* = 0. - -For Prob. 6.72. - -**6.73** Show that the circuit in Fig. 6.90 is a noninverting integrator. - -# **Figure 6.90** For Prob. 6.73. - -**6.74** The triangular waveform in Fig. 6.91(a) is applied to the input of the op amp differentiator in Fig. 6.91(b). Plot the output. - -For Prob. 6.74. - -- **6.75** An op amp differentiator has *R* = 250 kΩ and *C* = 10 *μ*F. The input voltage is a ramp *r*(*t*) = 7*t* mV. Find the output voltage. -- **6.76** A voltage waveform has the following characteristics: a positive slope of 20 V/s for 5 ms followed by a negative slope of 10 V/s for 10 ms. If the waveform is applied to a differentiator with *R* = 50 kΩ, *C* = 10 *μ*F, sketch the output voltage waveform. - -**6.77** The output *vo* of the op amp circuit in Fig. 6.92(a) is shown in Fig. 6.92(b). Let *Ri* = *Rf* = 1 MΩ and *C* = 1 *μ*F. Determine the input voltage waveform and sketch it. **\*** - -**6.79** Design an analog computer circuit to solve for *v*(*t*), given the following equation and a value for *f*(*t*) and that *v*(0) = 0 V. - -$$ -\left(\frac{dv(t)}{dt}\right) + 3vt = f t dt -$$ - -**6.80** Figure 6.93 presents an analog computer designed to solve a differential equation. Assuming *f*(*t*) is known, set up the equation for *f*(*t*). - -**Figure 6.93** For Prob. 6.80. - -**6.81** Design an analog computer to simulate the following equation to solve for *v*(*t*) (assume the initial conditions are zero): - -$$ -(d^{3}v(t)/dt^{3}) + 3(dv(t)/dt) = 4f(t) -$$ - -**6.82** Design an op amp circuit such that - -$$ -v_o = 10v_s + 2 \int v_s dt -$$ - - where *vs* and *vo* are the input v oltage and output voltage, respectively. - -**6.83** Your laboratory has available a large number of 5-*μ*F capacitors rated at 150 V. To design a capacitor bank of 10 *μ*F rated at 600 V, how many 5-*μ*F capacitors are needed and how would you connect them? - -**6.84** An 8-mH inductor is used in a fusion power experiment. If the current through the inductor is *i*(*t*) = 10 cos2 (*πt*) mA, for all *t* > 0 s, find the power being delivered to the inductor and the energy stored in it at *t* = 0.5 s. - -–4 - -**6.78** Design an analog computer to simulate *d*2 \_\_\_\_*vo dvo* - -$$ -\frac{d v_o}{dt^2} + 2\frac{d v_o}{dt} + v_o = 10 \sin 2t -$$ - -(b) - -where *v*0(0)=−6 V and *v*0 ′ (0)=0. **6.85** A square-wave generator produces the voltage waveform shown in Fig. 6.94(a). What kind of a circuit component is needed to convert the voltage waveform to the triangular current waveform shown in Fig. 6.94(b)? Calculate the value of the component, assuming that it is initially uncharged. - -**6.86** An electric motor can be modeled as a series combination of a 12-Ω resistor and 200-mH inductor. If a current *i*(*t*) = 2*te*−10*t* A flows through the series combination, find the voltage across the combination. - -# **chapter** - -7 - -# First-Order Circuits - -*We live in deeds, not years; in thoughts, not breaths; in feelings, not in figures on a dial. We should count time in heart-throbs. He most lives who thinks most, feels the noblest, acts the best.* - -—P. J. Bailey - -# Enhancing Your Career - -# **Careers in Computer Engineering** - -Electrical engineering education has gone through drastic changes in recent decades. Most departments have come to be known as Department of Electrical and Computer Engineering, emphasizing the rapid changes due to computers. Computers occup y a prominent place in modern so ciety and education. They have become commonplace and are helping to change the f ace of research, de velopment, production, business, and entertainment. The scientist, engineer , doctor, attorney, teacher, airline pilot, businessperson—almost anyone benefits from a computer's abilities to store large amounts of information and to process that information in very short periods of time. The internet, a computer communication network, is essential in b usiness, education, and library science. Com puter usage continues to grow by leaps and bounds. - -An education in computer engineering should provide breadth in software, hardware design, and basic modeling techniques. It should include courses in data structures, digital systems, computer architecture, micro processors, interfacing, software engineering, and operating systems. - -Electrical engineers who specialize in computer engineering find jobs in computer industries and in numerous fields where computers are being used. Companies that produce software are growing rapidly in number and size and providing employment for those who are skilled in programming. An excellent way to advance one's knowledge of computers is to join the IEEE Computer Society, which sponsors diverse magazines, journals, and conferences. - -Computer design of very large scale integrated (VLSI) circuits. Courtesy Brian Fast, Cleveland State University - -# Learning Objectives - -By using the information and exercises in this chapter you will be able to: - -- 1. Understand solutions to unforced, first-order linear differential equations. -- 2. Comprehend singularity equations and know their importance in solving linear differential equations. -- 3. Understand the effect of unit step sources on first-order linear differential equations. -- 4. Explain how dependent sources and op amps influence simple first-order linear differential equations. -- 5. Use *PSpice* to solve simple transient circuits with an inductor or a capacitor. - -# **7.1** Introduction - -Now that we ha ve considered the three passi ve elements (resistors, ca pacitors, and inductors) and one active element (the op amp)individually, we are prepared to consider circuits that contain various combinations of two or three of the passi ve elements. In this chapter , we shall e xamine two types of simple circuits: a circuit comprising a resistor and capacitor and a circuit comprising a resistor and an inductor . These are called *RC* and *RL* circuits, respectively. As simple as these circuits are, the y find continual applications in electronics, communications, and control sys tems, as we shall see. - -We carry out the analysis of *RC* and *RL* circuits by applying Kirchhoff's laws, as we did for resisti ve circuits. The only dif ference is that applying Kirchhoff's laws to purely resistive circuits results in algebraic equations, while applying the la ws to *RC* and *RL* circuits produces differential equations, which are more difficult to solve than algebraic equations. The differential equations resulting from analyzing *RC* and *RL* circuits are of the first order. Hence, the circuits are collectively known as *first-order* circuits. - -A first-order circuit is characterized by a first-order differential equation. - -In addition to there being tw o types of first-order circuits *(RC* and *RL)*, there are tw o ways to e xcite the circuits. The first way is by ini tial conditions of the storage elements in the circuits. In these so-called *source-free circuits*, we assume that ener gy is initially stored in the ca pacitive or inductive element. The energy causes current to flow in the circuit and is gradually dissipated in the resistors. Although source-free circuits are by definition free of independent sources, they may ha ve dependent sources. The second way of exciting first-order circuits is by independent sources. In this chapter, the independent sources we will consider are dc sources. (In later chapters, we shall consider sinusoidal and exponential sources.) The two types of first-order circuits and the two ways of exciting them add up to the four possible situations we will study in this chapter. - -Finally, we consider four typical applications of *RC* and *RL* circuits: delay and relay circuits, a photoflash unit, and an automobile ignition circuit. - -# **7.2** The Source-Free RC Circuit - -A source-free *RC* circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor is released to the resistors. - -Consider a series combination of a resistor and an initially char ged capacitor, as shown in Fig. 7.1. (The resistor and capacitor may be the equivalent resistance and equivalent capacitance of combinations of re sistors and capacitors.) Our objective is to determine the circuit response, which, for pedagogic reasons, we assume to be the voltage *v*(*t*) across the capacitor. Since the capacitor is initially charged, we can assume that at time *t* = 0, the initial voltage is - -$$ -v(0) = V_0 \tag{7.1} -$$ - -with the corresponding value of the energy stored as - -$$ -w(0) = \frac{1}{2}CV_0^2 -$$ - (7.2) - -Applying KCL at the top node of the circuit in Fig. 7.1 yields - -$$ -i_C + i_R = 0 \tag{7.3} -$$ - -By definition, *iC* = *C dv*∕*dt* and *iR* = *v*∕*R*. Thus, - -$$ -C\frac{dv}{dt} + \frac{v}{R} = 0 -$$ -\n(7.4a) - -or - -$$ -\frac{dv}{dt} + \frac{v}{RC} = 0\tag{7.4b} -$$ - -This is a *first-order differential equation,* since only the first derivative of *v* is involved. To solve it, we rearrange the terms as - -$$ -\frac{dv}{v} = -\frac{1}{RC}dt\tag{7.5} -$$ - -Integrating both sides, we get - -$$ -\ln v = -\frac{t}{RC} + \ln A -$$ - -where ln *A* is the integration constant. Thus, - -$$ -\ln\frac{v}{A} = -\frac{t}{RC} -$$ - (7.6) - -Taking powers of *e* produces - -$$ -v(t) = Ae^{-t/RC} -$$ - -But from the initial conditions, *v*(0) = *A* = *V*0. Hence, - -$$ -v(t) = V_0 e^{-t/RC} -$$ -\n -$$ -(7.7) -$$ - -This shows that the voltage response of the *RC* circuit is an exponential decay of the initial voltage. Since the response is due to the initial energy stored and the physical characteristics of the circuit and not due to some e xternal voltage or current source, it is called the *natural response* of the circuit. - -**Figure 7.1** A source-free *RC* circuit. - -A circuit response is the manner in which the circuit reacts to an excitation. - -The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation. - -The natural response depends on the nature of the circuit alone, with no external sources. In fact, the circuit has a response only because of the energy initially stored in the capacitor. - -| (t)∕V0 = e−t∕τ
Values of v
t
τ
2τ
3τ
4τ
5τ | TABLE 7.1 | | -|-----------------------------------------------------------------|-----------|---------| -| | | | -| | | v(t)∕V0 | -| | | 0.36788 | -| | | 0.13534 | -| | | 0.04979 | -| | | 0.01832 | -| | | 0.00674 | - -Graphical determination of the time constant *τ* from the response curve. - -The natural response is illustrated graphically in Fig. 7.2. Note that at *t* = 0, we have the correct initial condition as in Eq. (7.1). As *t* increases, the voltage decreases toward zero. The rapidity with which the v oltage decreases is e xpressed in terms of the *time constant*, denoted by *τ*, the lowercase Greek letter tau. - -The time constant of a circuit is the time required for the response to decay to a factor of 1∕e or 36.8 percent of its initial value.1 - -This implies that at *t* = *τ*, Eq. (7.7) becomes - -$$ -V_0 e^{-\tau/RC} = V_0 e^{-1} = 0.368 V_0 -$$ - -or - -$$ -\tau = RC \tag{7.8} -$$ - -In terms of the time constant, Eq. (7.7) can be written as - -$$ -v(t) = V_0 e^{-t/\tau} \tag{7.9} -$$ - -With a calculator it is easy to sho w that the v alue of *v*(*t*)∕*V*0 is as shown in Table 7.1. It is evident from Table 7.1 that the voltage *v*(*t*) is less than 1 percent of *V*0 after 5*τ* (five time constants). Thus, it is customary to assume that the capacitor is fully dischar ged (or charged) after five time constants. In other words, it takes 5*τ* for the circuit to reach its final state or steady state when no changes take place with time. Notice that for every time interval of *τ*, the voltage is reduced by 36.8 percent of its pre vious value, *v*(*t* + *τ*) = *v*(*t*)∕*e* = 0.368*v*(*t*), regardless of the value of *t*. - -Observe from Eq. (7.8) that the smaller the time constant, the more rapidly the v oltage decreases, that is, the f aster the response. This is illustrated in Fig. 7.4. A circuit with a small time constant gi ves a f ast response in that it reaches the steady state (or final state) quickly due to quick dissipation of ener gy stored, whereas a circuit with a lar ge time constant gives a slo w response because it tak es longer to reach steady state. At any rate, whether the time constant is small or large, the circuit reaches steady state in five time constants. - -With the voltage *v*(*t*) in Eq. (7.9), we can find the current *iR*(*t*), - -$$ -i_R(t) = \frac{v(t)}{R} = \frac{V_0}{R} e^{-t/\tau} -$$ -\n(7.10) - -1 The time constant may be viewed from another perspective. Evaluating the derivative of *v*(*t*) in Eq. (7.7) at *t* = 0, we obtain - -$$ -\frac{d}{dt}\left(\frac{v}{V_0}\right)\Big|_{t=0} = -\frac{1}{\tau}e^{-t/\tau}\Big|_{t=0} = -\frac{1}{\tau} -$$ - -Thus, the time constant is the initial rate of decay, or the time taken for *v*∕*V*0 to decay from unity to zero, assuming a constant rate of decay. This initial slope interpretation of the time constant is often used in the laboratory to find *τ* graphically from the response curve displayed on an oscilloscope. To find *τ* from the response curve, draw the tangent to the curve at *t* = 0, as shown in Fig. 7.3. The tangent intercepts with the time axis at *t* = *τ*. - -Plot of *v*∕*V*0 = *e* −*t*∕*τ* for various values of the time constant. - -The power dissipated in the resistor is - -$$ -p(t) = vi_R = \frac{V_0^2}{R}e^{-2t/\tau} -$$ -\n(7.11) - -The energy absorbed by the resistor up to time *t* is - -$$ -w_R(t) = \int_0^t p(\lambda) d\lambda = \int_0^t \frac{V_0^2}{R} e^{-2\lambda/\tau} d\lambda -$$ - -= $-\frac{\tau V_0^2}{2R} e^{-2\lambda/\tau} \Big|_0^t = \frac{1}{2} C V_0^2 (1 - e^{-2t/\tau}), \qquad \tau = RC$ (7.12) - -Notice that as *t* → ∞, *wR* (∞) → \_1 2*CV* 0 2 , which is the same as *wC* (0), the energy initially stored in the capacitor. The energy that was initially stored in the capacitor is eventually dissipated in the resistor. - -In summary: - -The Key to Working with a Source-Free RC Circuit Is Finding: - -- 1. The initial voltage *v*(0) = *V*0 across the capacitor. -- 2. The time constant *τ*. - -With these tw o items, we obtain the response as the capacitor v oltage *vC* (*t*) = *v*(*t*) = *v*(0) *e*−*t*∕*τ* . Once the capacitor voltage is first obtained, other variables (capacitor current *iC*, resistor voltage *vR*, and resistor current *iR*) can be determined. In finding the time constant *τ* = *RC*, *R* is often the Thevenin equivalent resistance at the terminals of the capacitor; that is, we take out the capacitor *C* and find *R* = *R*Th at its terminals. - -In Fig. 7.5, let *vC* (0) = 15 V. Find *vC*, *vx*, and *ix* for *t* > 0. - -# **Solution:** - -We first need to make the circuit in Fig. 7.5 conform with the standard *RC* circuit in Fig. 7.1. We find the equivalent resistance or the Thevenin The time constant is the same regardless of what the output is defined to be. - -When a circuit contains a single capacitor and several resistors and dependent sources, the Thevenin equivalent can be found at the terminals of the capacitor to form a simple RC circuit. Also, one can use Thevenin's theorem when several capacitors can be combined to form a single equivalent capacitor. - -# Example 7.1 - -**Figure 7.5** - -For Example 7.1. - -**Figure 7.6** Equivalent circuit for the circuit in Fig. 7.5. - -The 8-Ω and 12- Ω resistors in series can be combined to give a 20-Ω resistor. This 20-Ω resistor in parallel with the 5-Ω resistor can be combined so that the equivalent resistance is - -$$ -R_{\text{eq}} = \frac{20 \times 5}{20 + 5} = 4 \ \Omega -$$ - -Hence, the equivalent circuit is as shown in Fig. 7.6, which is analogous to Fig. 7.1. The time constant is - -Thus, - -$$ -\tau = R_{\text{eq}}C = 4(0.1) = 0.4 \text{ s} -$$ - -$$ -^{as,} -$$ - -$$ -v = v(0)e^{-t/\tau} = 15e^{-t/0.4} \text{ V}, \qquad v_C = v = 15e^{-2.5t} \text{ V} -$$ - -From Fig. 7.5, we can use voltage division to get *vx*; so - -$$ -v_x = \frac{12}{12 + 8}v = 0.6(15e^{-2.5t}) = 9e^{-2.5t} -$$ - V - -Finally, - -$$ -i_x = \frac{v_x}{12} = 0.75e^{-2.5t} \text{ A} -$$ - -# Practice Problem 7.1 - -Refer to the circuit in Fig. 7.7. Let *vC* (0) = 60 V. Determine *vC*, *vx*, and *io* for *t* ≥ 0. - -For Practice Prob. 7.1. - -**Answer:** 60*e*−0.25*t* V, 20*e*−0.25*t*V, −5*e*−0.25*t*A. - -**Figure 7.8** For Example 7.2. - -Example 7.2 The switch in the circuit in Fig. 7.8 has been closed for a long time, and it is opened at *t* = 0. Find *v*(*t*) for *t* ≥ 0. Calculate the initial energy stored in the capacitor. - -# **Solution:** - -For *t* < 0, the switch is closed; the capacitor is an open circuit to dc, as represented in Fig. 7.9(a). Using voltage division - -$$ -v_C(t) = \frac{9}{9+3}(20) = 15 \text{ V}, \qquad t < 0 -$$ - -Since the voltage across a capacitor cannot change instantaneously, the voltage across the capacitor at *t* = 0− is the same at *t* = 0, or - -$$ -v_C(0) = V_0 = 15 \text{ V} -$$ - -For *t* > 0, the switch is opened, and we have the *RC* circuit shown in Fig. 7.9(b). [Notice that the *RC* circuit in Fig. 7.9(b) is source free; the independent source in Fig. 7.8 is needed to provide *V*0 or the initial energy in the capacitor.] The 1-Ω and 9-Ω resistors in series give - -$$ -R_{\text{eq}} = 1 + 9 = 10 \Omega -$$ - -The time constant is - -$$ -\tau = R_{\text{eq}}C = 10 \times 20 \times 10^{-3} = 0.2 \text{ s} -$$ - -Thus, the voltage across the capacitor for *t* ≥ 0 is - -$$ -v(t) = v_C(0)e^{-t/\tau} = 15e^{-t/0.2} \text{ V} -$$ - -or - -$$ -v(t) = 15e^{-5t} \,\mathrm{V} -$$ - -The initial energy stored in the capacitor is - -$$ -w_C(0) = \frac{1}{2}Cv_C^2(0) = \frac{1}{2} \times 20 \times 10^{-3} \times 15^2 = 2.25 \text{ J} -$$ - -If the switch in Fig. 7.10 opens at *t* = 0, find *v*(*t*) for *t* ≥ 0 and *wC* (0). Practice Problem 7.2 - -**Answer:** 8 *e*−2*t* V, 5.333 J. - -9 Ω 1 Ω *v*C(0) 3 Ω + 20 V (a) 9 Ω 1 Ω (b) + ‒ Vo = 15 V 20 mF + ‒ - -**Figure 7.9** For Example 7.2: (a) *t* < 0, (b) *t* > 0. - -**Figure 7.10** For Practice Prob. 7.2. - -# **7.3** The Source-Free RL Circuit - -Consider the series connection of a resistor and an inductor, as shown in Fig. 7.11. Our goal is to determine the circuit response, which we will assume to be the current *i*(*t*) through the inductor. We select the induc tor current as the response in order to take advantage of the idea that the inductor current cannot change instantaneously. At *t* = 0, we assume that the inductor has an initial current *I*0, or - -$$ -i(0) = I_0 \tag{7.13} -$$ - -with the corresponding energy stored in the inductor as - -$$ -w(0) = \frac{1}{2} L I_0^2 \tag{7.14} -$$ - -Applying KVL around the loop in Fig. 7.11, - -$$ -v_L + v_R = 0 \tag{7.15} -$$ - -But *vL* = *L di*∕*dt* and *vR* = *iR*. Thus, - -$$ -L\frac{di}{dt} + Ri = 0 -$$ - -**Figure 7.11** A source-free *RL* circuit. - -$$ -\frac{di}{dt} + \frac{R}{L}i = 0\tag{7.16} -$$ - -Rearranging terms and integrating gives - -$$ -\int_{I_0}^{i(t)} \frac{di}{i} = -\int_0^t \frac{R}{L} dt -$$ - -\n -$$ -\ln i \Big|_{I_0}^{i(t)} = -\frac{Rt}{L} \Big|_0^t \implies \ln i(t) - \ln I_0 = -\frac{Rt}{L} + 0 -$$ - -or - -$$ -\ln \frac{i(t)}{I_0} = -\frac{Rt}{L} \tag{7.17} -$$ - -Taking the powers of *e*, we have - -$$ -i(t) = I_0 e^{-Rt/L} -$$ -\n(7.18) - -This shows that the natural response of the *RL* circuit is an e xponential decay of the initial current. The current response is shown in Fig. 7.12. It is evident from Eq. (7.18) that the time constant for the *RL* circuit is - -$$ -\tau = \frac{L}{R} \tag{7.19} -$$ - -with *τ* ag ain ha ving the unit of seconds. Thus, Eq. (7.18) may be written as - -$$ -i(t) = I_0 e^{-t/\tau} -$$ -\n(7.20) - -With the current in Eq. (7.20), we can find the voltage across the resistor as - -$$ -v_R(t) = iR = I_0 Re^{-t/\tau} -$$ - (7.21) - -The power dissipated in the resistor is - -$$ -p = v_R i = I_0^2 Re^{-2t/\tau} -$$ - (7.22) - -The energy absorbed by the resistor is - -$$ -w_R(t) = \int_0^t p(\lambda) d\lambda = \int_0^t I_0^2 Re^{-2\lambda/\tau} d\lambda = -\frac{\tau}{2} I_0^2 Re^{-2\lambda/\tau} \Big|_0^t, \qquad \tau = \frac{L}{R} -$$ - -or - -$$ -w_R(t) = \frac{1}{2} L \, I_0^2 (1 - e^{-2t/\tau}) \tag{7.23} -$$ - -Figure 7.12 shows an initial slope interpretation may be given to *τ*. - -Note that as *t* → ∞,*wR*(∞) → \_1 2 *L I*2 0, which is the same as *wL*(0), the initial energy stored in the inductor as in Eq. (7.14). Again, the energy initially stored in the inductor is eventually dissipated in the resistor. - -The smaller the time constant *τ* of a circuit, the faster the rate of decay of the response. The larger the time constant, the slower the rate of decay of the response. At any rate, the response decays to less than 1 percent of its initial value (i.e., reaches steady - -state) after 5*τ*. - -In summary: - -# The Key to Working with a Source-Free RL Circuit Is to Find: - -- 1. The initial current *i*(0) = *I*0 through the inductor. -- 2. The time constant *τ* of the circuit. - -With the tw o items, we obtain the response as the inductor current *iL*(*t*) = *i*(*t*) = *i*(0)*e*−*t*∕*τ* . Once we determine the inductor current *iL*, other variables (inductor voltage *vL*, resistor voltage *vR*, and resistor current *iR*) can be obtained. Note that in general, *R* in Eq. (7.19) is the Thevenin resistance at the terminals of the inductor. - -Assuming that *i*(0) = 10 A, calculate *i*(*t*) and *ix*(*t*) in the circuit of Example 7.3 Fig. 7.13. - -# **Solution:** - -There are two ways we can solve this problem. One way is to obtain the equivalent resistance at the inductor terminals and then use Eq. (7.20). The other way is to start from scratch by using Kirchhoff's voltage law. Whichever approach is taken, it is always better to first obtain the inductor current. - -■ **METHOD 1** The equivalent resistance is the same as the Thevenin resistance at the inductor terminals. Because of the dependent source, we insert a voltage source with *vo* = 1 V at the inductor terminals *a*-*b*, as in Fig. 7.14(a). (We could also insert a 1-A current source at the terminals.) Applying KVL to the two loops results in - -$$ -2(i_1 - i_2) + 1 = 0 \qquad \Rightarrow \qquad i_1 - i_2 = -\frac{1}{2} \tag{7.3.1} -$$ - -$$ -6i_2 - 2i_1 - 3i_1 = 0 \qquad \Rightarrow \qquad i_2 = \frac{5}{6}i_1 \tag{7.3.2} -$$ - -Substituting Eq. (7.3.2) into Eq. (7.3.1) gives - -$$ -i_1 = -3 -$$ - A, $i_o = -i_1 = 3$ A - -**Figure 7.14** Solving the circuit in Fig. 7.13. When a circuit has a single inductor and several resistors and dependent sources, the Thevenin equivalent can be found at the terminals of the inductor to form a simple RL circuit. Also, one can use Thevenin's theorem when several inductors can be combined to form a single equivalent inductor. - -Hence, - -The time constant is - -$$ -R_{\text{eq}} = R_{\text{Th}} = \frac{v_o}{i_o} = \frac{1}{3} \Omega -$$ - -*τ* = \_\_\_*L R*eq = \_1 2 \_ \_1 3 =\_\_3 2 s - -Thus, the current through the inductor is - -$$ -i(t) = i(0)e^{-t/\tau} = 10e^{-(2/3)t} \text{ A}, \qquad t > 0 -$$ - -■ **METHOD 2** We may directly apply KVL to the circuit as in Fig. 7.14(b). For loop 1, - -$$ -\frac{1}{2}\frac{di_1}{dt} + 2(i_1 - i_2) = 0 -$$ - -or - -$$ -\frac{di_1}{dt} + 4i_1 - 4i_2 = 0 \tag{7.3.3} -$$ - -For loop 2, - -$$ -6i_2 - 2i_1 - 3i_1 = 0 \qquad \Rightarrow \qquad i_2 = \frac{5}{6}i_1 \tag{7.3.4} -$$ - -Substituting Eq. (7.3.4) into Eq. (7.3.3) gives - -$$ -\frac{di_1}{dt} + \frac{2}{3}i_1 = 0 -$$ - -Rearranging terms, - -$$ -\frac{di_1}{i_1} = -\frac{2}{3}dt -$$ - -Since *i*1 = *i*, we may replace *i*1 with *i* and integrate: - -$$ -\ln i \Big|_{i(0)}^{i(t)} = -\frac{2}{3}t \Big|_0^t -$$ - -or - -$$ -\ln \frac{i(t)}{i(0)} = -\frac{2}{3}t -$$ - -Taking the powers of *e*, we finally obtain - -$$ -i(t) = i(0)e^{-(2/3)t} = 10e^{-(2/3)t} \text{ A}, \qquad t > 0 -$$ - -which is the same as by Method 1. The voltage across the inductor is - -$$ -v = L \frac{di}{dt} = 0.5(10) \left( -\frac{2}{3} \right) e^{-(2/3)t} = -\frac{10}{3} e^{-(2/3)t} -$$ - V - -Since the inductor and the 2-Ω resistor are in parallel, - -$$ -i_x(t) = \frac{v}{2} = -1.6667e^{-(2/3)t} \text{A}, \qquad t > 0 -$$ - -Find *i* and -$$ -v_x -$$ - in the circuit of Fig. 7.15. Let $i(0) = 7$ A. - -**Answer:** 7 *e*−2*t* A, −7 *e*−2*t* V, *t* > 0. - -Practice Problem 7.3 - -The switch in the circuit of Fig. 7.16 has been closed for a long time. At Example 7.4 *t* = 0, the switch is opened. Calculate *i*(*t*) for *t* > 0. - -# **Solution:** - -When *t* < 0, the switch is closed, and the inductor acts as a short circuit to dc. The 16-Ω resistor is short-circuited; the resulting circuit is shown in Fig. 7.17(a). To get *i*1 in Fig. 7.17(a), we combine the 4- Ω and 12-Ω resistors in parallel to get - -$$ -\frac{4 \times 12}{4 + 12} = 3 \ \Omega -$$ - -Hence, - -$$ -i_1 = \frac{40}{2+3} = 8 \text{ A} -$$ - -We obtain *i*(*t*) from *i*1 in Fig. 7.17(a) using current division, by writing - -$$ -i(t) = \frac{12}{12 + 4} i_1 = 6 \text{ A}, \qquad t < 0 -$$ - -Since the current through an inductor cannot change instantaneously, - -$$ -i(0) = i(0^-) = 6\,\mathrm{A} -$$ - - When *t* > 0, the switch is open and the voltage source is disconnected. We now have the source-free *RL* circuit in Fig. 7.17(b). Combining the resistors, we have - -$$ -R_{\text{eq}} = (12 + 4) || 16 = 8 \Omega -$$ - -The time constant is - -$$ -\tau = \frac{L}{R_{\text{eq}}} = \frac{2}{8} = \frac{1}{4} \,\text{s} -$$ - -Thus, - -$$ -i(t) = i(0)e^{-t/\tau} = 6e^{-4t} A -$$ - -**Figure 7.16** For Example 7.4. - -Solving the circuit of Fig. 7.16: (a) for *t* < 0, (b) for *t* > 0. - -For Practice Prob. 7.4. - -**Figure 7.19** For Example 7.5. - -**Figure 7.20** - -The circuit in Fig. 7.19 for: (a) *t* < 0, (b) *t* > 0. - -Example 7.5 In the circuit shown in Fig. 7.19, find *io*, *vo*, and *i* for all time, assuming that the switch was open for a long time. - -# **Solution:** - -It is better to first find the inductor current *i* and then obtain other quantities from it. - -For *t* < 0, the switch is open. Since the inductor acts like a short circuit to dc, the 6-Ω resistor is short-circuited, so that we ha ve the circuit shown in Fig. 7.20(a). Hence, *io* = 0, and - -$$ -i(t) = \frac{10}{2+3} = 2 \text{ A}, \qquad t < 0 -$$ -$$ -v_o(t) = 3i(t) = 6 \text{ V}, \qquad t < 0 -$$ - -Thus, *i*(0) = 2. - - For *t* > 0, the switch is closed, so that the v oltage source is shortcircuited. We now have a source-free *RL* circuit as shown in Fig. 7.20(b). At the inductor terminals, - -$$ -R_{\text{Th}} = 3 \parallel 6 = 2 \text{ }\Omega -$$ - -so that the time constant is - -$$ -\tau = \frac{L}{R_{\text{Th}}} = 1 \text{ s} -$$ - -Hence, - -$$ -i(t) = i(0)e^{-t/\tau} = 2e^{-t}A, \qquad t > 0 -$$ - -Because the inductor is in parallel with the 6- and 3-Ω resistors, - -$$ -v_o(t) = -v_L = -L\frac{di}{dt} = -2(-2e^{-t}) = 4e^{-t} \text{ V}, \qquad t > 0 -$$ - -and - -$$ -i_o(t) = \frac{v_L}{6} = -\frac{2}{3}e^{-t}A, \qquad t > 0 -$$ - -Thus, for all time, - -$$ -i_o(t) = \begin{cases} 0 \text{ A}, & t < 0 \\ -\frac{2}{3}e^{-t} \text{ A}, & t > 0 \end{cases}, \quad v_o(t) = \begin{cases} 6 \text{ V}, & t < 0 \\ 4e^{-t} \text{ V}, & t > 0 \end{cases} -$$ -$$ -i(t) = \begin{cases} 2 \text{ A}, & t < 0 \\ 2e^{-t} \text{ A}, & t \ge 0 \end{cases} -$$ - -o We notice that the inductor current is continuous at *t* = 0, while the (t) current through the 6-Ω resistor drops from 0 to −2∕3 at *t* = 0, and the voltage across the 3-Ω resistor drops from 6 to 4 at *t* = 0. We also notice that the time constant is the same regardless of what the output is defined to be. Figure 7.21 plots *i* and *io*. - -Determine *i*, *io*, and *vo* for all *t* in the circuit shown in Fig. 7.22. Assume that the switch was closed for a long time. It should be noted that opening a switch in series with an ideal current source creates an infinite voltage at the current source terminals. Clearly this is impossible. For the purposes of problem solving, we can place a shunt resistor in parallel with the source (which now makes it a voltage source in series with a resistor). In more practical circuits, devices that act like current sources are, for the most part, electronic circuits. These circuits will allow the source to act like an ideal current source over its operating range but voltage-limit it when the load resistor becomes too large (as in an open circuit). - -# **Answer:** - -$$ -i = \begin{cases} 16 \text{ A}, & t < 0 \\ 16e^{-2t} \text{ A}, & t \ge 0 \end{cases}, \quad i_o = \begin{cases} 8 \text{ A}, & t < 0 \\ -5.333e^{-2t} \text{ A}, & t > 0 \end{cases}, -$$ -$$ -v_o = \begin{cases} 32 \text{ V}, & t < 0 \\ 10.667e^{-2t} \text{ V}, & t > 0 \end{cases} -$$ - -# **7.4** Singularity Functions - -Before going on with the second half of this chapter, we need to digress and consider some mathematical concepts that will aid our understand ing of transient analysis. A basic understanding of singularity functions will help us make sense of the response of first-order circuits to a sudden application of an independent dc voltage or current source. - -Singularity functions (also called *switching functions*) are very useful in circuit analysis. They serve as good approximations to the switching signals that arise in circuits with switching operations. They are helpful in the neat, compact description of some circuit phenomena, especially the step response of *RC* or *RL* circuits to be discussed in the next sections. By definition, - -Singularity functions are functions that either are discontinuous or have discontinuous derivatives. - -**Figure 7.21** A plot of *i* and *io*. - -**Figure 7.22** For Practice Prob. 7.5. - -The three most widely used singularity functions in circuit analysis are the *unit step*, the *unit impulse*, and the *unit ramp* functions. - -The unit step function u(t) is 0 for negative values of t and 1 for positive values of t. - -In mathematical terms, - -$$ -u(t) = \begin{cases} 0, & t < 0 \\ 1, & t > 0 \end{cases} -$$ - (7.24) - -The unit step function is undefined at *t* = 0, where it changes abrupt ly from 0 to 1. It is dimensionless, lik e other mathematical functions such as sine and cosine. Figure 7.23 depicts the unit step function. If the abrupt change occurs at *t* = *t*0 (where *t*0 > 0) instead of *t* = 0, the unit step function becomes - -$$ -u(t - t_0) = \begin{cases} 0, & t < t_0 \\ 1, & t > t_0 \end{cases} -$$ - (7.25) - -which is the same as saying that *u*(*t*) is delayed by *t*0 seconds, as shown in Fig. 7.24(a). To get Eq. (7.25) from Eq. (7.24), we simply replace every *t* by *t* − *t*0. If the change is at *t* = −*t*0, the unit step function becomes - -$$ -u(t+t_0) = \begin{cases} 0, & t < -t_0 \\ 1, & t > -t_0 \end{cases} \tag{7.26} -$$ - -meaning that *u*(*t*) is advanced by *t*0 seconds, as shown in Fig. 7.24(b). - -We use the step function to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control systems and digital computers. For example, the voltage - -$$ -v(t) = \begin{cases} 0, & t < t_0 \\ V_0, & t > t_0 \end{cases} \tag{7.27} -$$ - -may be expressed in terms of the unit step function as - -$$ -v(t) = V_0 u(t - t_0) -$$ -\n(7.28) - -(b) - -t = 0 - -b - -a - -If we let *t*0 = 0, then *v*(*t*) is simply the step v oltage *V*0 *u*(*t*). A voltage source of *V*0 *u*(*t*) is shown in Fig. 7.25(a); its equivalent circuit is shown in Fig. 7.25(b). It is e vident in Fig. 7.25(b) that terminals *a-b* are shortcircuited (*v* = 0) for *t* < 0 and that *v* = *V*0 appears at the terminals for - -V0 - -= + - -**Figure 7.25** (a) Voltage source of *V*0 *u*(*t*), (b) its equivalent circuit. - -‒ + - -b - -a - -(a) - -‒ - -V0u(t) - -Alternatively, we may derive Eqs. (7.25) and (7.26) from Eq. (7.24) by writing u [f (t)] = 1, f (t) > 0, where f(t) may be t − t0 or t + t0. - -**Figure 7.24** - -The unit step function. - -(a) The unit step function delayed by *t*0, (b) the unit step advanced by *t*0. - -*t* > 0. Similarly, a current source of *I*0 *u*(*t*) is shown in Fig. 7.26(a), while its equivalent circuit is in Fig. 7.26(b). Notice that for *t* < 0, there is an open circuit (*i* = 0), and that *i* = *I*0 flows for *t* > 0. - -**Figure 7.26** - -(a) Current source of *I*0*u*(*t*), (b) its equivalent circuit. - -The derivative of the unit step function *u*(*t*) is the *unit impulse function δ*(*t*), which we write as - -$$ -\delta(t) = \frac{d}{dt}u(t) = \begin{cases} 0, & t < 0 \\ \text{Undefined}, & t = 0 \\ 0, & t > 0 \end{cases} \tag{7.29} -$$ - -The unit impulse function—also known as the *delta* function—is shown in Fig. 7.27. - -**Figure 7.27** 0 t *δ*(t) (∞) - -The unit impulse function. - -The unit impulse function *δ*(t) is zero everywhere except at t = 0, where it is undefined. - -Impulsive currents and v oltages occur in electric circuits as a result of switching operations or impulsi ve sources. Although the unit im pulse function is not ph ysically realizable ( just like ideal sources, ideal resistors, etc.), it is a very useful mathematical tool. - -The unit impulse may be regarded as an applied or resulting shock. It may be visualized as a very short duration pulse of unit area. This may be expressed mathematically as - -$$ -\int_{0^{-}}^{0^{+}} \delta(t) dt = 1 \tag{7.30} -$$ - -where *t* = 0− denotes the time just before *t* = 0 and *t* = 0 + is the time just after *t* = 0. For this reason, it is customary to write 1 (denoting unit area) beside the arrow that is used to symbolize the unit impulse function, as in Fig. 7.27. The unit area is known as the *strength* of the impulse function. When an impulse function has a strength other than unity , the area of the impulse is equal to its strength. F or example, an impulse func tion 10*δ*(*t*) has an area of 10. Figure 7.28 sho ws the impulse functions 5*δ*(*t* + 2),10*δ*(*t*), and −4*δ*(*t* − 3). - -To illustrate how the impulse function affects other functions, let us evaluate the integral - -$$ -\int_{a}^{b} f(t)\delta(t - t_0)dt -$$ -\n(7.31) - -Three impulse functions. - -where *a* < *t*0 < *b*. Since δ(*t* − *t*0) = 0 except at *t* = *t*0, the integrand is zero except at *t*0. Thus, - -$$ -\int_{a}^{b} f(t)\delta(t - t_0) dt = \int_{a}^{b} f(t_0)\delta(t - t_0) dt -$$ -$$ -= f(t_0) \int_{a}^{b} \delta(t - t_0) dt = f(t_0) -$$ - -or - -$$ -\int_{a}^{b} f(t)\delta(t - t_0) dt = f(t_0) -$$ -\n(7.32) - -This shows that when a function is integrated with the impulse function, we obtain the value of the function at the point where the impulse occurs. This is a highly useful property of the impulse function kno wn as the *sampling* or *sifting* property. The special case of Eq. (7.31) is for *t*0 = 0. Then Eq. (7.32) becomes - -$$ -\int_{0^{-}}^{0^{+}} f(t) \delta(t) dt = f(0) -$$ -\n(7.33) - -Integrating the unit step function *u*(*t*) results in the *unit ramp function r*(*t*); we write - -$$ -r(t) = \int_{-\infty}^{t} u(\lambda)d\lambda = tu(t) -$$ - (7.34) - -$$ -\overline{\text{or}} -$$ - -$$ -r(t) = \begin{cases} 0, & t \le 0 \\ t, & t \ge 0 \end{cases} \tag{7.35} -$$ - -The unit ramp function is zero for negative values of t and has a unit slope for positive values of t. - -Figure 7.29 shows the unit ramp function. In general, a ramp is a func tion that changes at a constant rate. - -The unit ramp function may be delayed or adv anced as sho wn in Fig. 7.30. For the delayed unit ramp function, - -$$ -r(t - t_0) = \begin{cases} 0, & t \le t_0 \\ t - t_0, & t \ge t_0 \end{cases} -$$ - (7.36) - -and for the advanced unit ramp function, - -$$ -r(t + t_0) = \begin{cases} 0, & t \le -t_0 \\ t + t_0, & t \ge -t_0 \end{cases} -$$ -(7.37) - -We should keep in mind that the three singularity functions (impulse, step, and ramp) are related by differentiation as - -$$ -\delta(t) = \frac{du(t)}{dt}, \qquad u(t) = \frac{dr(t)}{dt} -$$ - (7.38) - -The unit ramp function: (a) delayed by *t*0, (b) advanced by *t*0. - -1 - -r(t ‒ t0) - -or by integration as - -$$ -u(t) = \int_{-\infty}^{t} \delta(\lambda) d\lambda, \quad r(t) = \int_{-\infty}^{t} u(\lambda) d\lambda -$$ - (7.39) - -Although there are man y more singularity functions, we are only inter ested in these three (the impulse function, the unit step function, and the ramp function) at this point. - -Express the voltage pulse in Fig. 7.31 in terms of the unit step. Calculate Example 7.6 its derivative and sketch it. - -# **Solution:** - -The type of pulse in Fig. 7.31 is called the *gate function*. It may be regarded as a step function that switches on at one value of *t* and switches off at another value of *t*. The gate function shown in Fig. 7.31 switches on at *t* = 2 s and switches off at *t* = 5 s. It consists of the sum of two unit step functions as shown in Fig. 7.32(a). From the figure, it is evident that - -$$ -v(t) = 10u(t - 2) - 10u(t - 5) = 10[u(t - 2) - u(t - 5)] -$$ - -Taking the derivative of this gives - -$$ -\frac{dv}{dt} = 10[\delta(t-2) - \delta(t-5)] -$$ - -which is shown in Fig. 7.32(b). We can obtain Fig. 7.32(b) directly from Fig. 7.31 by simply observing that there is a sudden increase by 10 V at *t* = 2 s leading to 10 *δ*(*t* − 2). At *t* = 5 s, there is a sudden decrease by 10 V leading to −10V *δ*(*t* − 5). - -Gate functions are used along with switches to pass or block another signal. - -For Example 7.6. - -# Practice Problem 7.6 - -Express the current pulse in Fig. 7.33 in terms of the unit step. Find its integral and sketch it. - -**Answer:** 10[*u* (*t*) − 2*u* (*t* − 2) + *u* (*t* − 4)] A, 10[*r* (*t*) − 2*r* (*t* − 2) + *r*(*t* − 4)] amp-sec. See Fig. 7.34. - -Example 7.7 Express the *sawtooth* function shown in Fig. 7.35 in terms of singularity functions. - -# **Solution:** - -There are three ways of solving this problem. The first method is by mere observation of the given function, while the other methods involve some graphical manipulations of the function. - -■ **METHOD 1** By looking at the sketch of *v*(*t*) in Fig. 7.35, it is not hard to notice that the given function *v*(*t*) is a combination of singularity functions. So we let - -$$ -v(t) = v_1(t) + v_2(t) + \cdots -$$ - (7.7.1) - -The function *v*1(*t*) is the ramp function of slope 5, shown in Fig. 7.36(a); that is, - -$$ -v_1(t) = 5r(t) \tag{7.7.2} -$$ - -Partial decomposition of *v*(*t*) in Fig. 7.35. - -Since *v*1(*t*) goes to infinity, we need another function at *t* = 2s in order to get *v*(*t*). We let this function be *v*2, which is a ramp function of slope −5, as shown in Fig. 7.36(b); that is, - -$$ -v_2(t) = -5r(t-2) -$$ -\n(7.7.3) - -Adding *v*1 and *v*2 gives us the signal in Fig. 7.36(c). Obviously, this is not the same as *v*(*t*) in Fig. 7.35. But the difference is simply a constant 10 units for *t* > 2 s. By adding a third signal *v*3, where - -$$ -v_3 = -10u(t - 2) \tag{7.7.4} -$$ - -we get *v*(*t*), as shown in Fig. 7.37. Substituting Eqs. (7.7.2) through (7.7.4) into Eq. (7.7.1) gives - -Complete decomposition of *v*(*t*) in Fig. 7.35. - -■ **METHOD 2** A close observation of Fig. 7.35 reveals that *v*(*t*) is a multiplication of two functions: a ramp function and a gate function. Thus, - -$$ -v(t) = 5t[u(t) - u(t-2)] -$$ - -= 5tu(t) - 5tu(t - 2) -= 5r(t) - 5(t - 2 + 2)u(t - 2) -= 5r(t) - 5(t - 2)u(t - 2) - 10u(t - 2) -= 5r(t) - 5r(t - 2) - 10u(t - 2) - -the same as before. - -■ **METHOD 3** This method is similar to Method 2. We observe from Fig. 7.35 that *v*(*t*) is a multiplication of a ramp function and a unit step function, as shown in Fig. 7.38. Thus, - -$$ -v(t) = 5r(t)u(-t+2) -$$ - -If we replace *u*(−*t*) by 1 −*u*(*t*), then we can replace *u*(−*t* + 2) by 1 −*u*(*t* − 2). Hence, - -$$ -v(t) = 5r(t)[1 - u(t - 2)] -$$ - -which can be simplified as in Method 2 to get the same result. - -Decomposition of *v*(*t*) in Fig. 7.35. - -For Practice Prob. 7.7. - -Example 7.8 Given the signal - -| | 3, | t < 0 | -|--------|-----------------|--------------| -| g(t) = | −2, | 0 <
t < 1 | -| | {
2t
− 4, | t
> 1 | - -express *g*(*t*) in terms of step and ramp functions. - -# **Solution:** - -The signal *g*(*t*) may be regarded as the sum of three functions specified within the three intervals *t* < 0, 0 < *t* < 1, and *t* > 1. - - For *t* < 0, *g*(*t*) may be regarded as 3 multiplied by *u*(−*t*), where *u*(−*t*) = 1 for *t* < 0 and 0 for *t* > 0. Within the time interval 0 < *t* < 1, the function may be considered as −2 multiplied by a gated function [*u*(*t*) − *u*(*t* − 1)]. For *t* > 1, the function may be regarded as 2*t* − 4 multiplied by the unit step function *u*(*t* − 1). Thus, - -$$ -g(t) = 3u(-t) - 2[u(t) - u(t-1)] + (2t - 4)u(t-1) -$$ - -= 3u(-t) - 2u(t) + (2t - 4 + 2)u(t-1) -= 3u(-t) - 2u(t) + 2(t-1)u(t-1) -= 3u(-t) - 2u(t) + 2r(t-1) - -One may avoid the trouble of using *u*(−*t*) by replacing it with 1 − *u*(*t*). Then - -$$ -g(t) = 3[1 - u(t)] - 2u(t) + 2r(t - 1) = 3 - 5u(t) + 2r(t - 1) -$$ - -Alternatively, we may plot *g*(*t*) and apply Method 1 from Example 7.7. - -If -$$ -h(t) = \begin{cases} 0, & t < 0 \\ -4, & 0 < t < 2 \\ 3t - 8, & 2 < t < 6 \\ 0, & t > 6 \end{cases} -$$ -Practice Problem 7.8 - -express *h*(*t*) in terms of the singularity functions. - -**Answer:** -$$ --4u(t) + 2u(t-2) + 3r(t-2) - 10u(t-6) - 3r(t-6) -$$ -. - -Evaluate the following integrals involving the impulse function: Example 7.9 - -$$ -\int_0^{10} (t^2 + 4t - 2)\delta(t - 2) dt -$$ - -$$ -\int_{-\infty}^{\infty} [\delta(t - 1)e^{-t} \cos t + \delta(t + 1)e^{-t} \sin t] dt -$$ - -# **Solution:** - -For the first integral, we apply the sifting property in Eq. (7.32). - -$$ -\int_0^{10} (t^2 + 4t - 2) \delta(t - 2) dt = (t^2 + 4t - 2)|_{t=2} = 4 + 8 - 2 = 10 -$$ - -Similarly, for the second integral, - -$$ -\int_{-\infty}^{\infty} \left[ \delta(t-1)e^{-t} \cos t + \delta(t+1)e^{-t} \sin t \right] dt -$$ - -= $e^{-t} \cos t|_{t=1} + e^{-t} \sin t|_{t=-1}$ -= $e^{-t} \cos 1 + e^{t} \sin (-1) = 0.1988 - 2.2873 = -2.0885$ - -Evaluate the following integrals: - -$$ -\int_{-\infty}^{\infty} (t^3 + 5t^2 + 10)\delta(t+3) \, dt, \qquad \int_{0}^{10} \delta(t-\pi) \cos 3t \, dt -$$ - -**Answer:** 28, −1. - -# **7.5** Step Response of an RC Circuit - -When the dc source of an *RC* circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a *step response*. - -The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. - -Practice Problem 7.9 - -The step response is the response of the circuit due to a sudden applica tion of a dc voltage or current source. - -Consider the *RC* circuit in Fig. 7.40(a) which can be replaced by the circuit in Fig. 7.40(b), where *Vs* is a constant dc v oltage source. Again, we select the capacitor voltage as the circuit response to be determined. We assume an initial v oltage *V*0 on the capacitor , although this is not necessary for the step response. Since the v oltage of a capacitor cannot change instantaneously, - -$$ -v(0^{-}) = v(0^{+}) = V_0 \tag{7.40} -$$ - -where *v*(0−) is the voltage across the capacitor just before switching and *v*( 0 +) is its voltage immediately after switching. Applying KCL, we have - -$$ -C\frac{dv}{dt} + \frac{v - V_s u(t)}{R} = 0 -$$ - -$$ -\frac{dv}{dt} + \frac{v}{RC} = \frac{V_s}{RC}u(t) -$$ -\n(7.41) - -where *v* is the voltage across the capacitor. For *t* > 0, Eq. (7.41) becomes - -$$ -\frac{dv}{dt} + \frac{v}{RC} = \frac{V_s}{RC} -$$ - (7.42) - -Rearranging terms gives - -$$ -\frac{dv}{dt} = -\frac{v - V_s}{RC} -$$ - -\_\_\_\_\_\_ *dv v*− *Vs* = − \_\_\_ *dt RC* **(7.43)** - -Integrating both sides and introducing the initial conditions, - -$$ -\ln(v - V_s) \Big|_{V_0}^{v(t)} = -\frac{t}{RC} \Big|_{0}^{t} -$$ -$$ -\ln(v(t) - V_s) - \ln(V_0 - V_s) = -\frac{t}{RC} + 0 -$$ - -or - -or - -or - -$$ -\ln \frac{v - V_s}{V_0 - V_s} = -\frac{t}{RC} -$$ -\n(7.44) - -Taking the exponential of both sides - -$$ -\frac{v - V_s}{V_0 - V_s} = e^{-t/\tau}, \qquad \tau = RC -$$ -$$ -v - V_s = (V_0 - V_s)e^{-t/\tau} -$$ - -or - -$$ -v(t) = V_s + (V_0 - V_s)e^{-t/\tau}, \qquad t > 0 \tag{7.45} -$$ - -Thus, - -$$ -v(t) = \begin{cases} V_0, & t < 0 \\ V_s + (V_0 - V_s)e^{-t/\tau}, & t > 0 \end{cases} -$$ - (7.46) - -This is known as the *complete response* (or total response) of the *RC* circuit to a sudden application of a dc voltage source, assuming the capacitor is initially charged. The reason for the term "complete" will become evident a little later . Assuming that *Vs* > *V*0, a plot of *v*(*t*) is sho wn in Fig. 7.41. - -If we assume that the capacitor is uncharged initially, we set *V*0 = 0 in Eq. (7.46) so that - -$$ -v(t) = \begin{cases} 0, & t < 0 \\ V_s(1 - e^{-t/\tau}), & t > 0 \end{cases} -$$ -(7.47) - -which can be written alternatively as - -$$ -v(t) = V_s(1 - e^{-t/\tau})u(t) -$$ -\n(7.48) - -This is the complete step response of the *RC* circuit when the capacitor is initially uncharged. The current through the capacitor is obtained from Eq. (7.47) using *i*(*t*) = *C dv ∕ dt*. We get - -$$ -i(t) = C\frac{dv}{dt} = \frac{C}{\tau}V_s e^{-t/\tau}, \qquad \tau = RC, \qquad t > 0 -$$ -$$ -i(t) = \frac{V_s}{R} e^{-t/\tau} u(t) \tag{7.49} -$$ - -Figure 7.42 shows the plots of capacitor v oltage *v*(*t*) and capacitor current *i*(*t*). - -*R* - -Rather than going through the derivations above, there is a systematic approach—or rather, a shortcut method—for finding the step response of an *RC* or *RL* circuit. Let us reexamine Eq. (7.45), which is more general than Eq. (7.48). It is evident that *v*(*t*) has two components. Classically there are two ways of decomposing this into tw o components. The first is to break it into a "natural response and a forced response'' and the second is to break it into a "transient response and a steady-state response.'' Starting with the natural response and forced response, we write the total or complete response as - -> Complete response = natural response + forced response stored energy independent source - -> > *v* = *vn* + *vf* **(7.50)** - -or - -or - -where - -and - -$$ -v_f = V_s(1 - e^{-t/\tau}) -$$ - -*vn* = *Vo e*−*t*∕*τ* - -We are familiar with the natural response *vn* of the circuit, as discussed in Section 7.2. *vf* is known as the *forced* response because it is produced by the circuit when an external "force'' (a voltage source in this case) is applied. It represents what the circuit is forced to do by the input excitation. The natural response e ventually dies out along with the transient component of the forced response, leaving only the steady-state component of the forced response. - -Step response of an *RC* circuit with initially uncharged capacitor: (a) voltage response, (b) current response. - -Response of an *RC* circuit with initially charged capacitor. - -## **274** Chapter 7 First-Order Circuits - -Another way of looking at the complete response is to break into two components—one temporary and the other permanent, that is - -Complete response = transient response + steady-state response temporary part permanent part - -or - -$$ -v = v_t + v_{ss} \tag{7.51} -$$ - -where - -$$ -v_t = (V_o - V_s)e^{-t/\tau} -$$ - (7.52a) - -and - -$$ -v_{ss} = V_s \tag{7.52b} -$$ - -The *transient response vt* is temporary; it is the portion of the complete response that decays to zero as time approaches infinity. Thus, - -The transient response is the circuit's temporary response that will die out with time. - -The *steady-state response vss* is the portion of the complete response that remains after the transient reponse has died out. Thus, - -The steady-state response is the behavior of the circuit a long time after an external excitation is applied. - -The first decomposition of the complete response is in terms of the source of the responses, while the second decomposition is in terms of the permanency of the responses. Under certain conditions, the natural response and transient response are the same. The same can be said about the forced response and steady-state response. - -Whichever way we look at it, the complete response in Eq. (7.45) may be written as - -$$ -v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau} -$$ - (7.53) - -where *v*(0) is the initial voltage at *t* = 0 + and *v*(∞) is the final or steadystate value. Thus, to find the step response of an *RC* circuit requires three things: - -- 1. The initial capacitor voltage *v*(0). -- 2. The final capacitor voltage *v*(∞). -- 3. The time constant *τ*. - -Once we know x (0), x (∞), and *τ*, almost all the circuit problems in this chapter can be solved using the formula - -*x*(*t*) = *x*(∞) + [*x*(0) − *x*(∞)] *e*−*t*∕*τ* - -We obtain item 1 from the given circuit for *t* < 0 and items 2 and 3 from the circuit for *t* > 0. Once these items are determined, we obtain the - -This is the same as saying that the complete response is the sum of the transient response and the steady-state response. - -response using Eq. (7.53). This technique equally applies to *RL* circuits, as we shall see in the next section. - -Note that if the switch changes position at time *t* = *t*0 instead of at *t* = 0, there is a time delay in the response so that Eq. (7.53) becomes - -$$ -v(t) = v(\infty) + [v(t_0) - v(\infty)]e^{-(t - t_0)/\tau} -$$ - (7.54) - -where *v*(*t*0) is the initial value at *t* = *t*0 + . Keep in mind that Eq. (7.53) or (7.54) applies only to step responses, that is, when the input e xcitation is constant. - -The switch in Fig. 7.43 has been in position *A* for a long time. At *t* = 0, the switch moves to *B*. Determine *v*(*t*) for *t* > 0 and calculate its value at *t* = 1 and 4 s. - -# **Solution:** - -For *t* < 0, the switch is at position *A*. The capacitor acts like an open circuit to dc, but *v* is the same as the voltage across the 5-k Ω resistor. Hence, the voltage across the capacitor just before *t* = 0 is obtained by voltage division as - -$$ -v(0^{-}) = \frac{5}{5+3}(24) = 15 \text{ V} -$$ - -Using the fact that the capacitor voltage cannot change instantaneously, - -$$ -v(0) = v(0^-) = v(0^+) = 15 -$$ - V - -For *t* > 0, the switch is in position *B*. The Thevenin resistance connected to the capacitor is *R*Th = 4 kΩ, and the time constant is - -$$ -\tau = R_{\text{Th}} C = 4 \times 10^3 \times 0.5 \times 10^{-3} = 2 \text{ s} -$$ - -Since the capacitor acts like an open circuit to dc at steady state, *v*(∞) = 30 V. Thus, - -$$ -v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau} -$$ - -= 30 + (15 - 30)e^{-t/2} = (30 - 15e^{-0.5t}) V - -At *t* = 1, - -$$ -v(1) = 30 - 15e^{-0.5} = 20.9 \text{ V} -$$ - -At *t* = 4, - -$$ -v(4) = 30 - 15e^{-2} = 27.97 -$$ - V - -Example 7.10 - -# Practice Problem 7.10 - -Find *v*(*t*) for *t* > 0 in the circuit of Fig. 7.44. Assume the switch has been open for a long time and is closed at *t* = 0. Calculate *v*(*t*) at *t* = 0.5. - -**Answer:** (9.375 + 5.625*e*−2*t* ) V for all *t* > 0, 11.444 V. - -Example 7.11 In Fig. 7.45, the switch has been closed for a long time and is opened at *t* = 0. Find *i* and *v* for all time. - -# **Solution:** - -The resistor current *i* can be discontinuous at *t* = 0, while the capacitor voltage *v* cannot. Hence, it is always better to find *v* and then obtain *i* from *v*. - -By definition of the unit step function, - -$$ -30u(t) = \begin{cases} 0, & t < 0\\ 30, & t > 0 \end{cases} -$$ - -For *t* < 0, the switch is closed and 30 *u*(*t*) = 0, so that the 30*u*(*t*) voltage source is replaced by a short circuit and should be regarded as contributing nothing to *v*. Since the switch has been closed for a long time, the capacitor voltage has reached steady state and the capacitor acts like an open circuit. Hence, the circuit becomes that shown in Fig. 7.46(a) for *t* < 0. From this circuit we obtain - -$$ -v = 10 \text{ V}, \qquad i = -\frac{v}{10} = -1 \text{ A} -$$ - -Since the capacitor voltage cannot change instantaneously, - -$$ -v(0) = v(0^-) = 10 \text{ V} -$$ - - For *t* > 0, the switch is opened and the 10-V voltage source is disconnected from the circuit. The 30 *u*(*t*) voltage source is now operative, so the circuit becomes that shown in Fig. 7.46(b). After a long time, the circuit reaches steady state and the capacitor acts like an open circuit again. We obtain *v*(∞) by using voltage division, writing - -$$ -v(\infty) = \frac{20}{20 + 10}(30) = 20 \text{ V} -$$ - -**Figure 7.46** Solution of Example 7.11: (a) for *t* < 0, (b) for *t* > 0. - -The Thevenin resistance at the capacitor terminals is - -$$ -R_{\text{Th}} = 10||20 = \frac{10 \times 20}{30} = \frac{20}{3}\Omega -$$ - -and the time constant is - -$$ -\tau = R_{\text{Th}} C = \frac{20}{3} \cdot \frac{1}{4} = \frac{5}{3} \text{ s} -$$ - -Thus, - -$$ -v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau} -$$ - -= 20 + (10 - 20)e^{-(3/5)t} = (20 - 10e^{-0.6t}) V - -To obtain *i*, we notice from Fig. 7.46(b) that *i* is the sum of the currents through the 20-Ω resistor and the capacitor; that is, - -$$ -i = \frac{v}{20} + C\frac{dv}{dt} -$$ - -= 1 - 0.5e-0.6t + 0.25(-0.6)(-10)e-0.6t = (1 + e-0.6t) A - -Notice from Fig. 7.46(b) that *v* + 10*i* = 30 is satisfied, as expected. Hence, - -$$ -v = \begin{cases} 10 \text{ V}, & t < 0 \\ (20 - 10e^{-0.6t}) \text{ V}, & t \ge 0 \end{cases} -$$ -$$ -i = \begin{cases} -1 \text{ A}, & t < 0 \\ (1 + e^{-0.6t}) \text{ A}, & t > 0 \end{cases} -$$ - -Notice that the capacitor voltage is continuous while the resistor current is not. - -The switch in Fig. 7.47 is closed at *t* = 0. Find *i*(*t*) and *v*(*t*) for all time. Note that *u*(−*t*) = 1 for *t* < 0 and 0 for *t* > 0. Also, *u*(−*t*) = 1 − *u*(*t*). Practice Problem 7.11 - - - -# **7.6** Step Response of an RL Circuit - -Consider the *RL* circuit in Fig. 7.48(a), which may be replaced by the circuit in Fig. 7.48(b). Again, our goal is to find the inductor current *i* as the circuit response. Rather than apply Kirchhoff's laws, we will use the simple technique in Eqs. (7.50) through (7.53). Let the response be the sum of the transient response and the steady-state response, - -$$ -i = i_t + i_{ss} \tag{7.55} -$$ - -We know that the transient response is al ways a decaying e xponential, that is, - -$$ -i_t = Ae^{-t/\tau}, \qquad \tau = \frac{L}{R} \tag{7.56} -$$ - -where *A* is a constant to be determined. - -The steady-state response is the value of the current a long time after the switch in Fig. 7.48(a) is closed. We know that the transient response essentially dies out after five time constants. At that time, the inductor becomes a short circuit, and the v oltage across it is zero. The entire source v oltage *Vs* appears across *R*. Thus, the steady-state response is - -$$ -i_{ss} = \frac{V_s}{R} -$$ - (7.57) - -Substituting Eqs. (7.56) and (7.57) into Eq. (7.55) gives - -$$ -i = Ae^{-t/\tau} + \frac{V_s}{R} -$$ -\n -$$ -(7.58) -$$ - -We now determine the constant *A* from the initial v alue of *i*. Let *I*0 be the initial current through the inductor, which may come from a source other than *Vs*. Since the current through the inductor cannot change instantaneously, - -$$ -i(0^+) = i(0^-) = I_0 \tag{7.59} -$$ - -Thus, at *t* = 0, Eq. (7.58) becomes - -$$ -I_0 = A + \frac{V_s}{R} -$$ - -From this, we obtain *A* as - -$$ -A = I_0 - \frac{V_s}{R} -$$ - -Substituting for *A* in Eq. (7.58), we get - -$$ -i(t) = \frac{V_s}{R} + \left(I_0 - \frac{V_s}{R}\right)e^{-t/\tau} -$$ -\n(7.60) - -This is the complete response of the *RL* circuit. It is illustrated in Fig. 7.49. The response in Eq. (7.60) may be written as - -$$ -i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau} -$$ - (7.61) - -**Figure 7.49** Total response of the *RL* circuit with initial inductor current *I*0. - -where *i*(0) and *i*(∞) are the initial and final values of *i*, respectively. Thus, to find the step response of an *RL* circuit requires three things: - -- 1. The initial inductor current *i*(0) at *t* = 0. 2. The final inductor current *i*(∞). -- 3. The time constant *τ*. - -We obtain item 1 from the given circuit for *t* < 0 and items 2 and 3 from the circuit for *t* > 0. Once these items are determined, we obtain the response using Eq. (7.61). Keep in mind that this technique applies only for step responses. - -Again, if the switching tak es place at time *t* = *t*0 instead of *t* = 0, Eq. (7.61) becomes - -$$ -i(t) = i(\infty) + [i(t_0) - i(\infty)]e^{-(t-t_0)/\tau} -$$ -\n(7.62) - -If *I*0 = 0, then - -$$ -i(t) = \begin{cases} 0, & t < 0 \\ \frac{V_s}{R}(1 - e^{-t/\tau}), & t > 0 \end{cases} -$$ -(7.63a) - -or - -$$ -i(t) = \frac{V_s}{R} (1 - e^{-t/\tau}) u(t) -$$ -\n(7.63b) - -This is the step response of the *RL* circuit with no initial inductorcurrent. The v oltage across the inductor is obtained from Eq. (7.63) using *v* = *L di*∕*dt*. We get - -$$ -v(t) = L\frac{di}{dt} = V_s \frac{L}{\tau R} e^{-t/\tau}, \qquad \tau = \frac{L}{R}, \qquad t > 0 -$$ - -or - -$$ -v(t) = V_s e^{-t/\tau} u(t) \tag{7.64} -$$ - -Figure 7.50 shows the step responses in Eqs. (7.63) and (7.64). - -Step responses of an *RL* circuit with no initial inductor current: (a) current response, (b) voltage response. - -For Example 7.12. - -Find *i*(*t*) in the circuit of Fig. 7.51 for *t* > 0. Assume that the switch has been closed for a long time. - -# **Solution:** - -When *t* < 0, the 3-Ω resistor is short-circuited, and the inductor acts like a short circuit. The current through the inductor at *t* = 0− (i.e., just before *t* = 0) is - -$$ -i(0^{-}) = \frac{10}{2} = 5 \text{ A} -$$ - -Because the inductor current cannot change instantaneously, - -$$ -i(0) = i(0^+) = i(0^-) = 5 \text{ A} -$$ - -When *t* > 0, the switch is open. The 2- and 3- Ω resistors are in series, so that - -$$ -i(\infty) = \frac{10}{2+3} = 2 \text{ A} -$$ - -The Thevenin resistance across the inductor terminals is - -$$ -R_{\rm Th} = 2 + 3 = 5 \ \Omega -$$ - -For the time constant, - -*τ* = \_\_\_*L R*Th = \_1 3 \_\_ 5 =\_\_\_1 15 s - -Thus, - -$$ -i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau} -$$ - -= 2 + (5 - 2)e-15t = 2 + 3e-15t A, t > 0 - -*Check:* In Fig. 7.51, for *t* > 0, KVL must be satisfied; that is, - -$$ -10 = 5i + L\frac{di}{dt} -$$ - -$$ -5i + L\frac{di}{dt} = [10 + 15e^{-15t}] + \left[\frac{1}{3}(3)(-15)e^{-15t}\right] = 10 -$$ - -This confirms the result. - -At *t* = 0, switch 1 in Fig. 7.53 is closed, and switch 2 is closed 4 s later. Example 7.13 Find *i*(*t*) for *t* > 0. Calculate *i* for *t* = 2 s and *t* = 5 s. - -**Figure 7.53** For Example 7.13. - -# **Solution:** - -We need to consider the three time intervals *t* ≤ 0, 0 ≤ *t* ≤ 4, and *t* ≥ 4 separately. For *t* < 0, switches *S*1 and *S*2 are open so that *i* = 0. Since the inductor current cannot change instantly, - -$$ -i(0^-) = i(0) = i(0^+) = 0 -$$ - -For 0 ≤ *t* ≤ 4, *S*1 is closed so that the 4- and 6- Ω resistors are in series. (Remember, at this time, *S*2 is still open.) Hence, assuming for now that *S*1 is closed forever, - -$$ -i(\infty) = \frac{40}{4+6} = 4 \text{ A}, \qquad R_{\text{Th}} = 4+6 = 10 \text{ }\Omega -$$ -$$ -\tau = \frac{L}{R_{\text{Th}}} = \frac{5}{10} = \frac{1}{2} \text{ s} -$$ - -Thus, - -$$ -i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau} -$$ - -= 4 + (0 - 4)e-2t = 4(1 - e-2t) A, 0 \le t \le 4 - - For *t* ≥ 4, *S*2 is closed; the 10-V voltage source is connected, and the circuit changes. This sudden change does not affect the inductor current because the current cannot change abruptly. Thus, the initial current is - -$$ -i(4) = i(4^-) = 4(1 - e^{-8}) \simeq 4 \text{ A} -$$ - -To find *i*(∞), let *v* be the voltage at node *P* in Fig. 7.53. Using KCL, - -$$ -\frac{40 - v}{4} + \frac{10 - v}{2} = \frac{v}{6} \implies v = \frac{180}{11} \text{ V} -$$ -$$ -i(\infty) = \frac{v}{6} = \frac{30}{11} = 2.727 \text{ A} -$$ - -The Thevenin resistance at the inductor terminals is - -$$ -R_{\text{Th}} = 4||2 + 6 = \frac{4 \times 2}{6} + 6 = \frac{22}{3} \,\Omega -$$ - -and - -$$ -\tau = \frac{L}{R_{\text{Th}}} = \frac{5}{\frac{22}{3}} = \frac{15}{22} \text{ s} -$$ - -Hence, - -$$ -i(t) = i(\infty) + [i(4) - i(\infty)]e^{-(t-4)/\tau}, \qquad t \ge 4 -$$ - -We need (*t* − 4) in the exponential because of the time delay. Thus, - -$$ -i(t) = 2.727 + (4 - 2.727)e^{-(t-4)/\tau}, \qquad \tau = \frac{15}{22} -$$ -$$ -= 2.727 + 1.273e^{-1.4667(t-4)}, \qquad t \ge 4 -$$ - -Putting all this together, - -$$ -i(t) = \begin{cases} 0, & t \le 0 \\ 4(1 - e^{-2t}), & 0 \le t \le 4 \\ 2.727 + 1.273e^{-1.4667(t-4)}, & t \ge 4 \end{cases} -$$ - -At *t* = 2, - -At *t* = 5, - -$$ -i(2) = 4(1 - e^{-4}) = 3.93 \text{ A} -$$ - -$$ -i(5) = 2.727 + 1.273e^{-1.4667} = 3.02 -$$ - A - -Switch *S*1 in Fig. 7.54 is closed at *t* = 0, and switch *S*2 is closed at *t* = 2s. - -**Answer:** *i*(*t*) = { 0, 4(1 − *e*−9*t* ), 7.2 − 3.2 *e*−5(*t*−2) , *t* < 0 0 < *t* < 2 *t* > 2 *i*(1) = 4 A, *i*(3) = 7.178 A. - -Calculate *i*(*t*) for all *t*. Find *i*(1) and *i*(3). - -# **Figure 7.54** For Practice Prob. 7.13. - -# **7.7** † First-Order Op Amp Circuits - -An op amp circuit containing a storage element will exhibit first-order behavior. Differentiators and integrators treated in Section 6.6 are examples of first-order op amp circuits. Again, for practical reasons, inductors are hardly ever used in op amp circuits; therefore, the op amp circuits we consider here are of the *RC* type. - -As usual, we analyze op amp circuits using nodal analysis. Some times, the Thevenin equivalent circuit is used to reduce the op amp circuit to one that we can easily handle. The following three examples illustrate the concepts. The first one deals with a source-free op amp circuit, while the other two involve step responses. The three examples have been carefully selected to co ver all possible *RC* types of op amp circuits, depending on the location of the capacitor with respect to the op amp; that is, the capacitor can be located in the input, the output, or the feedback loop. - -For the op amp circuit in Fig. 7.55(a), find *vo* for *t* > 0, gi ven that Example 7.14 *v*(0) = 3 V. Let *Rf* = 80 kΩ, *R*1 = 20 kΩ, and *C* = 5 *µ*F. - -For Example 7.14. - -# **Solution:** - -This problem can be solved in two ways: - -■ **METHOD 1** Consider the circuit in Fig. 7.55(a). Let us derive the appropriate differential equation using nodal analysis. If *v*1 is the voltage at node 1, at that node, KCL gives - -$$ -\frac{0 - v_1}{R_1} = C \frac{dv}{dt} -$$ - (7.14.1) - -Because nodes 2 and 3 must be at the same potential, the potential at node 2 is zero. Thus, *v*1 − 0 = *v* or *v*1 = *v* and Eq. (7.14.1) becomes - -$$ -\frac{dv}{dt} + \frac{v}{CR_1} = 0 -$$ - (7.14.2) - -This is similar to Eq. (7.4b) so that the solution is obtained the same way as in Section 7.2, i.e., - -$$ -v(t) = V_0 e^{-t/\tau}, \qquad \tau = R_1 C \tag{7.14.3} -$$ - -where *V*0 is the initial voltage across the capacitor. But *v*(0) = 3 = *V*0 and *τ* = 20 × 103 × 5 × 10*−*6 = 0*.*1*.* Hence, - -$$ -v(t) = 3e^{-10t} -$$ - (7.14.4) - -Applying KCL at node 2 gives - -$$ -C\frac{dv}{dt} = \frac{0 - v_o}{R_f} -$$ - -or - -$$ -v_o = -R_f C \frac{dv}{dt} \tag{7.14.5} -$$ - -Now we can find *v*0 as - -$$ -v_o = -80 \times 10^3 \times 5 \times 10^{-6} (-30e^{-10t}) = 12e^{-10t} \text{ V}, \qquad t > 0 -$$ - -■ **METHOD 2** Let us apply the shortcut method from Eq. (7.53). We need to find *vo*( 0+)*, vo*(∞)*,* and *τ.* Since *v*( 0+) = *v*(0*−*) = 3 V*,* we apply KCL at node 2 in the circuit of Fig. 7.55(b) to obtain - -$$ -\frac{3}{20,000} + \frac{0 - v_o(0^+)}{80,000} = 0 -$$ - -or *vo*( 0+) = 12 V*.* Since the circuit is source free, *v*(∞) = 0 V*.* To find *τ,* we need the equivalent resistance *Req* across the capacitor terminals. If we remove the capacitor and replace it by a 1-A current source, we have the circuit shown in Fig. 7.55(c). Applying KVL to the input loop yields - -$$ -20,000(1) - v = 0 \qquad \Rightarrow \qquad v = 20 \text{ kV} -$$ - -Then - -$$ -R_{\text{eq}} = \frac{v}{1} = 20 \text{ k}\Omega -$$ - -and *τ* = *R*eq*C* = 0*.*1*.* Thus, - -$$ -v_o(t) = v_o(\infty) + [v_o(0) - v_o(\infty)]e^{-t/\tau} -$$ - -= 0 + (12 - 0)e^{-10t} = 12e^{-10t} V, \qquad t > 0 - -as before. - -For the op amp circuit in Fig. 7.56, find *vo* for *t >* 0 if *v*(0) = 4 V*.* Assume that *Rf* = 50 kΩ*, R*1 = 10 kΩ*,* and *C* = 10 *µ*F*.* Practice Problem 7.14 - -> **Answer:** −4 *e* −2*t* V*, t >* 0*.* - -**Figure 7.56** For Practice Prob. 7.14. - -Example 7.15 Determine *v*(*t*) and *vo*(*t*) in the circuit of Fig. 7.57. - -# **Solution:** - -This problem can be solved in two ways, just like the previous example. However, we will apply only the second method. Since what we are looking for is the step response, we can apply Eq. (7.53) and write - -$$ -v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}, \qquad t > 0 \tag{7.15.1} -$$ - -*v*o + - -where we need only find the time constant *τ,* the initial value *v*(0)*,* and the final value *v*(∞). Notice that this applies strictly to the capacitor voltage due a step input. Since no current enters the input terminals of the op amp, the elements on the feedback loop of the op amp constitute an *RC* circuit, with - -$$ -\tau = RC = 50 \times 10^3 \times 10^{-6} = 0.05 \tag{7.15.2} -$$ - -For *t* < 0*,* the switch is open and there is no voltage across the capacitor. Hence, *v*(0) = 0*.* For *t* > 0*,* we obtain the voltage at node 1 by voltage division as - -$$ -v_1 = \frac{20}{20 + 10}3 = 2 \text{ V} -$$ - (7.15.3) - -Since there is no storage element in the input loop, *v*1 remains constant for all *t*. At steady state, the capacitor acts like an open circuit so that the op amp circuit is a noninverting amplifier. Thus, - -$$ -v_o(\infty) = \left(1 + \frac{50}{20}\right)v_1 = 3.5 \times 2 = 7 \text{ V} -$$ - (7.15.4) - -But - -$$ -v_1 - v_o = v \tag{7.15.5} -$$ - -so that - -$$ -v(\infty) = 2 - 7 = -5 -$$ - V - -Substituting *τ, v*(0)*,* and *v*(∞) into Eq. (7.15.1) gives - -$$ -v(t) = -5 + [0 - (-5)]e^{-20t} = 5(e^{-20t} - 1) \text{ V}, \qquad t > 0 \tag{7.15.6} -$$ - -From Eqs. (7.15.3), (7.15.5), and (7.15.6), we obtain - -$$ -v_o(t) = v_1(t) - v(t) = 7 - 5e^{-20t} \text{ V}, \qquad t > 0 \tag{7.15.7} -$$ - -Find *v*(*t*) and *vo*(*t*) in the op amp circuit of Fig. 7.58. - -**Answer:** (Note, the v oltage across the capacitor and the output v oltage must be both equal to zero, for *t* < 0, since the input w as zero for all *t* < 0.) 40(1 − *e*−10*t* ) *u*(*t*) mV, 40( *e*−10*t* − 1) *u*(*t*) mV. - -Practice Problem 7.15 - -*v*o + - -1 *μ*F - -100 kΩ - -‒ - -**Figure 7.58** For Practice Prob. 7.15. - -**Figure 7.57** For Example 7.15. - -Example 7.16 Find the step response *vo*(*t*) for *t* > 0 in the op amp circuit of Fig. 7.59. Let *vi* = 2*u*(*t*) V, *R*1 = 20 kΩ, *Rf* = 50 kΩ, *R*2 = *R*3 = 10 kΩ, *C* = 2 *µ*F. - -# **Solution:** - -Notice that the capacitor in Example 7.14 is located in the input loop, while the capacitor in Example 7.15 is located in the feedback loop. In this example, the capacitor is located in the output of the op amp. Again, we can solve this problem directly using nodal analysis. However, using the Thevenin equivalent circuit may simplify the problem. - -We temporarily remove the capacitor and find the Thevenin equivalent at its terminals. To obtain *V*Th, consider the circuit in Fig. 7.60(a). Since the circuit is an inverting amplifier, - -$$ -V_{ab} = -\frac{R_f}{R_1} v_i -$$ - -By voltage division, - -# **Figure 7.60** - -Obtaining *V*Th and *R*Th across the capacitor in Fig. 7.59. - -To obtain *R*Th, consider the circuit in Fig. 7.60(b), where *Ro* is the output resistance of the op amp. Since we are assuming an ideal op amp, *Ro* = 0, and - -$$ -R_{\text{Th}} = R_2 \parallel R_3 = \frac{R_2 R_3}{R_2 + R_3} -$$ - -Substituting the given numerical values, - -$$ -V_{\text{Th}} = -\frac{R_3}{R_2 + R_3} \frac{R_f}{R_1} v_i = -\frac{10}{20} \frac{50}{20} 2u(t) = -2.5u(t) -$$ -$$ -R_{\text{Th}} = \frac{R_2 R_3}{R_2 + R_3} = 5k\Omega -$$ - -$$ -v_o(t) = -2.5(1 - e^{-t/\tau})u(t) -$$ - -where *τ* = *R*Th*C* = 5 × 103 × 2 × 10−6 = 0.01. Thus, the step response for *t* > 0 is - -$$ -v_o(t) = 2.5(e^{-100t} - 1)u(t) -$$ - V - -**Figure 7.61** Thevenin equivalent circuit of the circuit in Fig. 7.59. - -**Answer:** 27(1 − *e*−50*t* )*u*(*t*) V. - -# **7.8** Transient Analysis with PSpice - -As we discussed in Section 7.5, the transient response is the temporary response of the circuit that soon disappears. *PSpice* can be used to obtain the transient response of a circuit with storage elements. Section D.4 in Appendix D provides a review of transient analysis using *PSpice for Windows*. It is recommended that you read Section D.4 before continuing with this section. - -If necessary, dc *PSpice* analysis is first carried out to determine the initial conditions. Then the initial conditions are used in the transient *PSpice* analysis to obtain the transient responses. It is rec om mended but not necessary that during this dc analysis, all capac itors should be opencircuited while all inductors should be short-circuited. - -Practice Problem 7.16 - -**Figure 7.62** For Practice Prob. 7.16. - -PSpice uses "transient" to mean "function of time." Therefore, the transient response in PSpice may not actually die out as expected. - -Use *PSpice* to find the response *i*(*t*) for *t* > 0 in the circuit of Fig. 7.63. Example 7.17 - -# **Solution:** - -Solving this problem by hand gives *i*(0) = 0, *i*(∞) = 2A, *R*Th = 6, *τ* = 3∕6 = 0.5 s, so that - -$$ -i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau} = 2(1 - e^{-2t}), \qquad t > 0 -$$ - -To use *PSpice,* we first draw the schematic as shown in Fig. 7.64. We recall from Appendix D that the part name for a closed switch is Sw\_tclose. We do not need to specify the initial condition of the induc tor because *PSpice* will determine that from the circuit. By selecting **Analysis/Setup/Transient**, we set *Print Step* to 25 ms and *Final Step* to 5*τ* = 2.5 s. After saving the circuit, we simulate by selecting **Analysis/ Simulate**. In the *PSpice* A/D window, we select **Trace/Add** and display –I(L1) as the current through the inductor. Figure 7.65 shows the plot of *i*(*t*), which agrees with that obtained by hand calculation. - -**Figure 7.64** The schematic of the circuit in Fig. 7.63. - -**Figure 7.63** For Example 7.17. - -**Figure 7.65** For Example 7.17; the response of the circuit in Fig. 7.63. - -**Answer:** *v*(*t*) = 8(1 − *e*−*t* - -that in Fig. 7.65. - -Note that the negative sign on I(L1) is needed because the current enters through the upper terminal of the inductor, which happens to be the negative terminal after one counterclock wise rotation. A way to avoid the negative sign is to ensure that current enters pin 1 of the inductor. To obtain this desired direction of positive current flow, the initially horizontal inductor symbol should be rotated counterclockwise 270° and placed in the desired location. - -) V, *t* > 0. The response is similar in shape to - -For the circuit in Fig. 7.66, use *Pspice* to find *v*(*t*) for *t* > 0. - -Practice Problem 7.17 - -**Figure 7.66** For Practice Prob. 7.17. - -12 Ω 30 V 6 Ω 6 Ω 3 Ω 0.1 F 4 A + ‒ t=0 t=0 (a) *v*(t) + ‒ 6 Ω 6 Ω 12 Ω 0.1 F + *v*(t) ‒ 30 V (b) 10 Ω 0.1 F + *v*(t) ‒ 10 V (c) + ‒ + ‒ Example 7.18 In the circuit of Fig. 7.67(a), determine the response *v*(*t*). - -**Figure 7.67** For Example 7.18. Original circuit (a), circuit *for t* > 0 (b), and reduced circuit for *t* > 0 (c). - -# **Solution:** - -- 1. **Define.** The problem is clearly stated and the circuit is clearly labeled. -- 2. **Present.** Given the circuit shown in Fig. 7.67(a), determine the response *v* ( *t*). -- 3. **Alternative.** We can solve this circuit using circuit analysis techniques, nodal analysis, mesh analysis, or *PSpice*. Let us solve the problem using circuit analysis techniques (this time Thevenin equivalent circuits) and then check the answer using two methods of *PSpice* . -- 4. **Attempt.** For time < 0, the switch on the left is open and the switch on the right is closed. Assume that the switch on the right has been closed long enough for the circuit to reach steady state; then the capacitor acts like an open circuit and the current from the 4-A source flows through the parallel combination of the 6- Ω and 3-Ω resistors (6 ∥ 3 = 18 ∕ 9 = 2), producing a voltage equal to 2 × 4 = 8 V = − *v*(0). - - At *t* = 0, the switch on the left closes and the switch on the right opens, producing the circuit shown in Fig. 7.67(b). - - The easiest way to complete the solution is to find the Thevenin equivalent circuit as seen by the capacitor. The opencircuit voltage (with the capacitor removed) is equal to the voltage drop across the 6-Ω resistor on the left, or 10 V (the voltage drops uniformly across the 12- Ω resistor, 20 V, and across the 6- Ω resistor, 10 V). This is *V*Th . The resistance looking in where the capacitor was is equal to 12 ∥ 6 + 6 = 72 ∕18 + 6 = 10 Ω, which is *R*eq. This produces the Thevenin equivalent circuit shown in Fig. 7.67(c). Matching up the boundary conditions ( *v*(0) = −8 V and *v* ( ∞ ) = 10 V) and *τ* = *RC* = 1, we get - -$$ -v(t) = 10 - 18e^{-t} V -$$ - -## 5. **Evaluate.** There are two ways of solving the problem using *PSpice* . - -■ **METHOD 1** One way is to first do the dc *PSpice* analysis to determine the initial capacitor voltage. The schematic of the revelant circuit is in Fig. 7.68(a). Two pseudocomponent VIEWPOINTs are inserted to measure the voltages at nodes 1 and 2. When the circuit is simulated, we obtain the displayed values in Fig. 7.68(a) as *V* 1 = 0 V and *V*2 = 8 V. Thus, the initial capacitor voltage is *v*(0) = *V* 1 − *V*2 = −8 V. The *PSpice* transient analysis uses this value along with the schematic in Fig. 7.68(b). Once the circuit in Fig. 7.68(b) is drawn, we insert the capacitor initial voltage as IC = −8. We select **Analysis/ Setup/Transient** and set *Print Step* to 0.1 s and *Final Step* to 4 *τ* = 4 s. After saving the circuit, we select **Analysis/Simulate** to simulate the circuit. In the *PSpice* A/D window, we select **Trace/Add** and display V(R2:2) – V(R3:2) or V(C1:1) − V(C1:2) as the capacitor voltage *v* ( *t*). The plot of *v* ( *t*) is shown in Fig. 7.69. This agrees with the result ob tained by hand calculation, *v* ( *t* ) = 10 − 18 *e* − *t* V. - -**Figure 7.69** Response *v*(*t*) for the circuit in Fig. 7.67. - -**Figure 7.68** - -(a) Schematic for dc analysis to get *v*(0), (b) schematic for transient analysis used in getting the response *v*(*t*). - -■ **METHOD 2** We can simulate the circuit in Fig. 7.67 directly, since *PSpice* can handle the open and closed switches and determine the initial conditions automatically. Using this approach, the schemat ic is drawn as shown in Fig. 7.70. After drawing the circuit, we select **Analysis/Setup/Transient** and set *Print Step* to 0.1 s and *Final Step* to 4*τ* = 4 s. We save the circuit, then select **Analysis/Simulate** to simulate the circuit. In the *PSpice* A/D window, we select **Trace/Add** and display V(R2:2) − V(R3:2) as the capacitor voltage *v*(*t*). The plot of *v*(*t*) is the same as that shown in Fig. 7.69. - -6. **Satisfactory?** Clearly, we have found the value of the output response *v*(*t*), as required by the problem statement. Checking does validate that solution. We can present all this as a complete solution to the problem. - -The switch in Fig. 7.71 was open for a long time but closed at *t* = 0. If *i*(0) = 10A, find *i*(*t*) for *t* > 0 by hand and also by *PSpice*. - -# **Answer:** *i*(*t*) = 6 + 4 *e*−5*t* A. The plot of *i*(*t*) obtained by *PSpice* analysis is shown in Fig. 7.72. - -# Practice Problem 7.18 - -# **7.9** †Applications - -The various devices in which *RC* and *RL* circuits find applications include filtering in dc power supplies, smoothing circuits in digital communications, differentiators, integrators, delay circuits, and relay circuits. Some of these applications take advantage of the short or long time constants of the *RC* or *RL* circuits. We will consider four simple applications here. The first two are *RC* circuits, the last two are *RL* circuits. - -# **7.9.1** Delay Circuits - -An *RC* circuit can be used to pro vide various time delays. Figure 7.73 shows such a circuit. It basically consists of an *RC* circuit with the ca pacitor connected in parallel with a neon lamp. The voltage source can provide enough voltage to fire the lamp. When the switch is closed, the capacitor voltage increases gradually toward 110 V at a rate determined by the circuit's time constant, (*R*1 + *R*2)*C*. The lamp will act as an open - -An *RC* delay circuit. - -circuit and not emit light until the voltage across it exceeds a particular level, say 70 V. When the voltage level is reached, the lamp fires (goes on), and the capacitor dischar ges through it. Due to the lo w resistance of the lamp when on, the capacitor voltage drops fast and the lamp turns off. The lamp acts again as an open circuit and the capacitor recharges. By adjusting *R*2, we can introduce either short or long time delays into the circuit and mak e the lamp fire, recharge, and fire repeatedly every time constant *τ* = (*R*1 + *R*2)*C*, because it takes a time period *τ* to get the capacitor voltage high enough to fire or low enough to turn off. - -The warning blinkers commonly found on road construction sites are one example of the usefulness of such an *RC* delay circuit. - -Example 7.19 Consider the circuit in Fig. 7.73, and assume that *R*1 = 1.5 MΩ, 0 &lt; *R*2 < 2.5 MΩ. (a) Calculate the extreme limits of the time constant of the circuit. (b) How long does it take for the lamp to glow for the first time after the switch is closed? Let *R*2 assume its largest value. - -# **Solution:** - -(a) The smallest value for *R*2 is 0 Ω, and the corresponding time constant for the circuit is - -$$ -\tau = (R_1 + R_2)C = (1.5 \times 10^6 + 0) \times 0.1 \times 10^{-6} = 0.15 \text{ s} -$$ - -The largest value for *R*2 is 2.5 MΩ, and the corresponding time constant for the circuit is - -$$ -\tau = (R_1 + R_2)C = (1.5 + 2.5) \times 10^6 \times 0.1 \times 10^{-6} = 0.4 \text{ s} -$$ - -Thus, by proper circuit design, the time constant can be adjusted to introduce a proper time delay in the circuit. - -(b) Assuming that the capacitor is initially uncharged, *vC*(0) = 0, while *vC*(∞) = 110. But - -$$ -v_C(t) = v_C(\infty) + [v_C(0) - v_C(\infty)]e^{-t/\tau} = 110[1 - e^{-t/\tau}] -$$ - -where *τ* = 0.4 s, as calculated in part (a). The lamp glows when *vC* = 70 V. If *vC*(*t*) = 70 V at *t* = *t*0, then - -$$ -70 = 110[1 - e^{-t_0/\tau}] \qquad \Rightarrow \qquad \frac{7}{11} = 1 - e^{-t_0/\tau} -$$ - -or - -$$ -e^{-t_0/\tau} = \frac{4}{11} \qquad \Rightarrow \qquad e^{t_0/\tau} = \frac{11}{4} -$$ - -Taking the natural logarithm of both sides gives - -$$ -t_0 = \tau \ln \frac{11}{4} = 0.4 \ln 2.75 = 0.4046 \text{ s} -$$ - -A more general formula for finding *t*0 is - -$$ -t_0 = \tau \ln \frac{-v(\infty)}{v(t_0) - v(\infty)} -$$ - -The lamp will fire repeatedly every *t*0 seconds if and only if *v*(*t*0) < *v*(∞). - -The *RC* circuit in Fig. 7.74 is designed to operate an alarm which acti - Practice Problem 7.19 vates when the current through it exceeds 90 *μ*A. If 0 ≤ *R* ≤ 6 kΩ, find the range of the time delay that the variable resistor can create. - -**Answer:** Between 34.47 and 89.26 ms. - -# **7.9.2** Photoflash Unit - -An electronic flash unit pro vides a common e xample of an *RC* circuit. This application exploits the ability of the capacitor to oppose any abrupt change in voltage. Figure 7.75 shows a simplified circuit. It consists essentially of a high-voltage dc supply, a current-limiting large resistor *R*1, and a capacitor *C* in parallel with the flashlamp of low resistance *R*2. When the switch is in position 1, the capacitor charges slowly due to the large time constant ( *τ*1 = *R*1*C*). As shown in Fig. 7.76(a), the capacitor voltage rises gradually from zero to *Vs*, while its current decreases gradually from *I*1 = *Vs*∕*R*1 to zero. The charging time is approximately five times the time constant, - -$$ -t_{\text{charge}} = 5R_1C \tag{7.65} -$$ - -With the switch in position 2, the capacitor v oltage is dischar ged. The low resistance *R*2 of the photolamp permits a high discharge current with peak *I*2 = *Vs*∕*R*2 in a short duration, as depicted in Fig. 7.76(b). Discharging takes place in approximately five times the time constant, - -$$ -t_{\text{discharge}} = 5R_2C \tag{7.66} -$$ - -(a) Capacitor voltage showing slow charge and fast discharge, (b) capacitor current showing low charging current *I*1 = *Vs*∕*R*1 and high discharge current *I*2 = *Vs*∕*R*2. - -Thus, the simple *RC* circuit of Fig. 7.75 provides a short-duration, highcurrent pulse. Such a circuit also finds applications in electric spot welding and the radar transmitter tube. - -# **Figure 7.75** - -Circuit for a flash unit providing slow charge in position 1 and fast discharge in position 2. - -Example 7.20 An electronic flashgun has a current-limiting 6-kΩ resistor and 2000-*μ*F electrolytic capacitor char ged to 240 V. If the lamp resistance is 12 Ω, find: (a) the peak charging current, (b) the time required for the capaci tor to fully charge, (c) the peak discharging current, (d) the total energy stored in the capacitor, and the average power dissipated by the lamp. - -# **Solution:** - -(a) The peak charging current is - -$$ -I_1 = \frac{V_s}{R_1} = \frac{240}{6 \times 10^3} = 40 \text{ mA} -$$ - -(b) From Eq. (7.65), - -$$ -t_{\text{charge}} = 5R_1C = 5 \times 6 \times 10^3 \times 2000 \times 10^{-6} = 60 \text{ s} = 1 \text{ minute} -$$ - -(c) The peak discharging current is - -$$ -I_2 = \frac{V_s}{R_2} = \frac{240}{12} = 20 \text{ A} -$$ - -(d) The energy stored is - -$$ -W = \frac{1}{2}CV_s^2 = \frac{1}{2} \times 2000 \times 10^{-6} \times 240^2 = 57.6 \text{ J} -$$ - - The energy stored in the capacitor is dissipated across the lamp during the discharging period. From Eq. (7.66), - -$$ -t_{\text{discharge}} = 5R_2C = 5 \times 12 \times 2000 \times 10^{-6} = 0.12 \text{ s} -$$ - -Thus, the average power dissipated is - -$$ -p = \frac{W}{t_{\text{discharge}}} = \frac{57.6}{0.12} = 480 \text{ watts} -$$ - -The flash unit of a camera has a 2-mF capacitor charged to 40 V. Practice Problem 7.20 - -- (a) How much charge is on the capacitor? -- (b) What is the energy stored in the capacitor? -- (c) If the flash fires in 0.8 ms, what is the average current through the flashtube? -- (d) How much power is delivered to the flashtube? After a picture has been taken, the capacitor needs to be recharged by a power unit that supplies a maximum of 5 mA. How much time does it take to charge the capacitor? - -**Answer:** (a) 80 mC, (b) 1.6 J, (c) 100 A, (d) 2 kW, 16 s. - -# **7.9.3** Relay Circuits - -A magnetically controlled switch is called a *relay*. A relay is essentially an electromagnetic de vice used to open or close a switch that controls another circuit. Figure 7.77(a) sho ws a typical relay circuit. The coil - -circuit is an *RL* circuit like that in Fig. 7.77(b), where *R* and *L* are the resistance and inductance of the coil. When switch *S*1 in Fig. 7.77(a) is closed, the coil circuit is energized. The coil current gradually increases and produces a magnetic field. Eventually the magnetic field is sufficiently strong to pull the mo vable contact in the other circuit and close switch *S*2. At this point, the relay is said to be *pulled in*. The time interval *td* between the closure of switches *S*1 and *S*2 is called the *relay delay time*. - -Relays were used in the earliest digital circuits and are still used for switching high-power circuits. - -The coil of a certain relay is operated by a 12-V battery. If the coil has a Example 7.21 resistance of 150 Ω and an inductance of 30 mH and the current needed to pull in is 50 mA, calculate the relay delay time. - -# **Solution:** - -The current through the coil is given by - -$$ -i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau} -$$ - -where - -$$ -i(0) = 0, -$$ - $i(\infty) = \frac{12}{150} = 80 \text{ mA}$ - -$$ -\tau = \frac{L}{R} = \frac{30 \times 10^{-3}}{150} = 0.2 \text{ ms} -$$ - -Thus, - -$$ -i(t) = 80[1 - e^{-t/\tau}] \text{ mA} -$$ - -If *i*(*td*) = 50 mA, then - -50 = 80[1 − *e*−*td*∕*τ* ] ⇒ \_\_5 8 = 1 − *e*−*td*∕*τ* - -or - -$$ -e^{-t_d/\tau} = \frac{3}{8} \qquad \Rightarrow \qquad e^{t_d/\tau} = \frac{8}{3} -$$ - -By taking the natural logarithm of both sides, we get - -$$ -t_d = \tau \ln \frac{8}{3} = 0.2 \ln \frac{8}{3} \text{ ms} = 0.1962 \text{ ms} -$$ - -Alternatively, we may find *td* using - -$$ -t_d = \tau \ln \frac{i(0) - i(\infty)}{i(t_d) - i(\infty)} -$$ - -# Practice Problem 7.21 - -A relay has a resistance of 200 Ω and an inductance of 500 mH. The relay contacts close when the current through the coil reaches 175 *μ*A. What time elapses between the application of 110 V to the coil and contact closure? - -**Answer:** 957.5 *μ*s. - -# **7.9.4** Automobile Ignition Circuit - -The ability of inductors to oppose rapid change in current mak es them useful for arc or spark generation. An automobile ignition system tak es advantage of this feature. - -The g asoline engine of an automobile requires that the fuel-air mixture in each c ylinder be ignited at proper times. This is achie ved by means of a spark plug (Fig. 7.78), which essentially consists of a pair of electrodes separated by an air g ap. By creating a lar ge voltage (thousands of v olts) between the electrodes, a spark is formed across the air g ap, thereby igniting the fuel. But ho w can such a lar ge voltage be obtained from the car battery, which supplies only 12 V? This is achieved by means of an inductor (the spark coil) *L*. Since the v oltage across the inductor is *v* = *L di*∕*dt*, we can make *di*∕*dt* large by creating a large change in current in a very short time. When the ignition switch in Fig. 7.78 is closed, the current through the inductor increases gradu ally and reaches the final value of *i* = *Vs*∕*R*, where *Vs* = 12 V. Again, the time taken for the inductor to char ge is five times the *time constant* of the circuit (*τ* = *L*∕*R*), - -$$ -t_{\text{charge}} = 5 \frac{L}{R} \tag{7.67} -$$ - -Since at steady state, *i* is constant, *di*∕*dt* = 0, and the inductor v oltage *v* = 0. When the switch suddenly opens, a lar ge voltage is de veloped across the inductor (due to the rapidly collapsing field) causing a spark or arc in the air g ap. The spark continues until the ener gy stored in the inductor is dissipated in the spark dischar ge. In laboratories, when one is working with inductive circuits, this same ef fect causes a v ery nasty shock, and one must exercise caution. - -A solenoid with resistance 4 Ω and inductance 6 mH is used in an auto- Example 7.22 mobile ignition circuit similar to that in Fig. 7.78. If the battery supplies 12 V, determine: the final current through the solenoid when the switch is closed, the energy stored in the coil, and the voltage across the air gap, assuming that the switch takes 1 *μ*s to open. - -# **Solution:** - -The final current through the coil is - -*I* = \_\_ *Vs R* =\_\_\_ 12 4 = 3 A - -The energy stored in the coil is - -$$ -W = \frac{1}{2}L\ l^2 = \frac{1}{2} \times 6 \times 10^{-3} \times 3^2 = 27 \text{ mJ} -$$ - -The voltage across the gap is - -$$ -V = L\frac{\Delta I}{\Delta t} = 6 \times 10^{-3} \times \frac{3}{1 \times 10^{-6}} = 18 \text{ kV} -$$ - -The spark coil of an automobile ignition system has a 20-mH induc - Practice Problem 7.22 tance and a 5-Ω resistance. With a supply voltage of 12 V, calculate: the time needed for the coil to fully charge, the energy stored in the coil, and the voltage developed at the spark gap if the switch opens in 2 *μ*s. - -**Answer:** 20 ms, 57.6 mJ, and 24 kV. - -# **7.10** Summary - -- 1. The analysis in this chapter is applicable to an y circuit that can be reduced to an equi valent circuit comprising a resistor and a single energy-storage element (inductor or capacitor). Such a circuit is first-order because its behavior is described by a first-order differential equation. When analyzing *RC* and *RL* circuits, one must always keep in mind that the capacitor is an open circuit to steady-state dc conditions while the inductor is a short circuit to steady-state dc conditions. -- 2. The natural response is obtained when no independent source is present. It has the general form - -$$ -x(t) = x(0)e^{-t/\tau} -$$ - - where *x* represents current through (or v oltage across) a resistor, a capacitor, or an inductor, and *x*(0) is the initial value of *x*. Be cause most practical resistors, capacitors, and inductors always have losses, the natural response is a transient response, i.e. it dies out with time. - -3. The time constant *τ* is the time required for a response to decay to 1∕e of its initial value. For *RC* circuits, *τ* = *RC* and for *RL* circuits, *τ* = *L*∕*R*. - -4. The singularity functions include the unit step, the unit ramp func tion, and the unit impulse functions. The unit step function *u*(*t*) is - -$$ -u(t) = \begin{cases} 0, & t < 0 \\ 1, & t > 0 \end{cases} -$$ - -The unit impulse function is - -$$ -\delta(t) = \begin{cases}\n0, & t < 0 \\ -\text{Undefined}, & t = 0 \\ -0, & t > 0\n\end{cases} -$$ - -The unit ramp function is - -$$ -r(t) = \begin{cases} 0, & t \le 0 \\ t, & t \ge 0 \end{cases} -$$ - -- 5. The steady-state response is the beha vior of the circuit after an in dependent source has been applied for a long time. The transient response is the component of the complete response that dies out with time. -- 6. The total or complete response consists of the steady-state response and the transient response. -- 7. The step response is the response of the circuit to a sudden application of a dc current or v oltage. Finding the step response of a firstorder circuit requires the initial v alue *x*( 0 +), the final value *x*(∞), and the time constant *τ*. With these three items, we obtain the step response as - -$$ -x(t) = x(\infty) + [x(0^+) - x(\infty)]e^{-t/\tau} -$$ - -A more general form of this equation is - -$$ -x(t) = x(\infty) + [x(t_0^+) - x(\infty)]e^{-(t-t_0)/\tau} -$$ - -Or we may write it as - -Instantaneous value = Final + [Initial − Final] *e*−(*t*−*t*0)∕*τ* - -- 8. *PSpice* is very useful for obtaining the transient response of a circuit. - - 9. Four practical applications of *RC* and *RL* circuits are: a delay circuit, a photoflash unit, a relay circuit, and an automobile ignition circuit. - -# Review Questions - -**7.1** An *RC* circuit has *R* = 2 Ω and *C* = 4 F. The time constant is: - -> (a) 0.5 s (b) 2 s (c) 4 s (d) 8 s 15 s - -**7.2** The time constant for an *RL* circuit with *R* = 2 Ω and *L* = 4 H is: - -> (a) 0.5 s (b) 2 s (c) 4 s (d) 8 s 15 s - -**7.3** A capacitor in an *RC* circuit with *R* = 2 Ω and *C* = 4 F is being charged. The time required for the capacitor voltage to reach 63.2 percent of its steadystate value is: - -(a) 2 s (b) 4 s (c) 8 s (d) 16 s none of the above - -**7.4** An *RL* circuit has *R* = 2 Ω and *L* = 4 H. The time needed for the inductor current to reach 40 percent of its steady-state value is: - -> (a) 0.5 s (b) 1 s (c) 2 s (d) 4 s none of the above - -Problems **299** - -- **7.5** In the circuit of Fig. 7.79, the capacitor voltage just before *t* = 0 is: - - (a) 10 V (b) 7 V (c) 6 V (d) 4 V 0 V - -# **Figure 7.79** - -For Review Questions 7.5 and 7.6. - -**7.6** In the circuit in Fig. 7.79, *v*(∞) is: - -(a) -$$ -10 \text{ V} -$$ - (b) $7 \text{ V}$ (c) $6 \text{ V}$ -(d) $4 \text{ V}$ 0 V - -**7.7** For the circuit in Fig. 7.80, the inductor current just before *t* = 0 is: - -> (a) 8 A (b) 6 A (c) 4 A (d) 2 A 0 A - -# **Figure 7.80** - -For Review Questions 7.7 and 7.8. - -**7.8** In the circuit of Fig. 7.80, *i*(∞) is: - -(a) 10 A (b) 6 A (c) 4 A (d) 2 A 0 A - -**7.9** If *vs* changes from 2 V to 4 V at *t* = 0, we may express *vs* as: - -> (a) *δ*(*t*) V (b) 2*u*(*t*) V (c) 2*u*(−*t*) + 4*u*(*t*) V (d) 2 + 2*u*(*t*) V 4*u*(*t*) − 2 V - -**7.10** The pulse in Fig. 7.116(a) can be expressed in terms of singularity functions as: - -| (a) 2u(t) + 2u(t − 1) V | (b) 2u(t) − 2u(t − 1) V | -|-------------------------|-------------------------| -| (c) 2u(t) − 4u(t − 1) V | (d) 2u(t) + 4u(t − 1) V | - -*Answers: 7.1d, 7.2b, 7.3c, 7.4b, 7.5d, 7.6a, 7.7c, 7.8e , 7.9c,d, 7.10b.* - -# Problems - -# Section 7.2 The Source-Free RC Circuit - -**7.1** In the circuit shown in Fig. 7.81 - -$$ -v(t) = 56e^{-200t} \,\text{V}, \quad t > 0 -$$ - -$$ -i(t) = 8e^{-200t} \text{ mA}, \quad t > 0 -$$ - -- (a) Find the values of *R* and *C*. -- (b) Calculate the time constant *τ*. -- (c) Determine the time required for the voltage to decay half its initial value at *t* = 0. - -**7.2** Find the time constant for the *RC* circuit in Fig. 7.82. - -# **Figure 7.82** - -For Prob. 7.2. - -For Prob. 7.3. - -**7.3** Determine the time constant for the circuit in Fig. 7.83. - -**7.4** The switch in Fig. 7.84 has been in position *A* for a long time. Assume the switch moves instantaneously from *A* to *B* at *t* = 0. Find *v* for *t* > 0. - -For Prob. 7.4. - -**7.5** Using Fig. 7.85, design a problem to help other students better understand source-free *RC* circuits. - -**Figure 7.85** - -For Prob. 7.5. - -**7.6** The switch in Fig. 7.86 has been closed for a long time, and it opens at *t* = 0. Find *v*(*t*) for *t* ≥ 0. - -For Prob. 7.6. - -**7.7** Assuming that the switch in Fig. 7.87 has been in position *A* for a long time and is moved to position *B* at *t* = 0, Then at *t* = 1 second, the switch moves from *B* to *C*. Find *vC*(*t*) for *t* ≥ 0. - -**Figure 7.87** For Prob. 7.7. - -**7.8** For the circuit in Fig. 7.88, if - -$$ -v = 10e^{-4t} -$$ - V and $i = 0.2e^{-4t}$ A, $t > 0$ - -- (a) Find *R* and *C*. -- (b) Determine the time constant. -- (c) Calculate the initial energy in the capacitor. -- (d) Obtain the time it takes to dissipate 50 percent of the initial energy. - -# **Figure 7.88** - -For Prob. 7.8. - -**7.9** The switch in Fig. 7.89 opens at *t* = 0. Find *vo* for *t* > 0. - -# **Figure 7.89** - -For Prob. 7.9. - -**7.10** For the circuit in Fig. 7.90, find *vo*(*t*) for *t* > 0. Determine the time necessary for the capacitor voltage to decay to one-third of its value at *t* = 0. - -# **Figure 7.90** - -For Prob. 7.10. - -# Section 7.3 The Source-Free RL Circuit - -**7.11** For the circuit in Fig. 7.91, find *i*o for *t* > 0. - -**Figure 7.91** For Prob. 7.11. - -# **Figure 7.92** - -For Prob. 7.12. - -**7.13** In the circuit of Fig. 7.93, - -$$ -v(t) = 80e^{-10^{3}t} \text{V}, \quad t > 0 -$$ -$$ -i(t) = 5e^{-10^{3}t} \text{mA}, \quad t > 0 -$$ - -- (a) Find *R*, *L*, and *τ*. -- (b) Calculate the energy dissipated in the resistance for 0 < *t* < 0.5 ms. - -# **Figure 7.93** - -For Prob. 7.13. - -# **Figure 7.94** - -- For Prob. 7.14. -- **7.15** Find the time constant for each of the circuits in Fig. 7.95. - -For Prob. 7.15. - -**7.16** Determine the time constant for each of the circuits in Fig. 7.96. - -For Prob. 7.16. - -**7.17** Consider the circuit of Fig. 7.97. Find *vo*(*t*) if *i*(0) = 15 A and *v*(*t*) = 0. - -**Figure 7.97** For Prob. 7.17. - -**7.18** For the circuit in Fig. 7.98, determine *vo*(*t*) when *i*(0) = 5 A and *v*(*t*) = 0. - -# **Figure 7.98** - -For Prob. 7.18. - -**7.19** In the circuit of Fig. 7.99, find *i*(*t*) for *t* > 0 if *i*(0) = 5 A. - -**Figure 7.99** For Prob. 7.19. - -**7.20** For the circuit in Fig. 7.100, - -and - -$$ -i = 30e^{-50t}A, \qquad t > 0 -$$ - -*v* = 90 *e*−50*t* - -V - -- (a) Find *L* and *R*. -- (b) Determine the time constant. -- (c) Calculate the initial energy in the inductor. -- (d) What fraction of the initial energy is dissipated in 10 ms? - -# **Figure 7.100** - -For Prob. 7.20. - -**7.21** In the circuit of Fig. 7.101, find the value of *R* for which the steady-state energy stored in the inductor will be 2 J. - -# **Figure 7.101** - -For Prob. 7.21. - -**7.22** Find *i*(*t*) and *v*(*t*) for *t* > 0 in the circuit of Fig. 7.102 if *i*(0) = 10 A. - -For Prob. 7.22. - -**7.23** Consider the circuit in Fig. 7.103. Given that *vo*(0) = 10 V, find *vo* and *vx* for *t* > 0. - -**Figure 7.103** For Prob. 7.23. - -# Section 7.4 Singularity Functions - -**7.24** Express the following signals in terms of singularity functions. - -**7.25** Design a problem to help other students better understand singularity functions. - -**7.26** Express the signals in Fig. 7.104 in terms of singularity functions. - -**Figure 7.104** For Prob. 7.26. - -**7.27** Express *v*(*t*) in Fig. 7.105 in terms of step functions. - -Problems **303** - -‒1 0 3 21 15 10 5 ‒10 ‒5 t *v*(t) - -# **Figure 7.105** - -For Prob. 7.27. - -**7.28** Sketch the waveform represented by - -$$ -i(t) = [r(t) - r(t-1) - u(t-2) - r(t-2) -$$ - -+ -$$ -r(t-3) + u(t)(t-4)] -$$ - A - -**7.29** Sketch the following functions: - -(a) -$$ -x(t) = 10e^{-t}u(t - 1) -$$ -, -\n(b) $y(t) = 10e^{-(t-1)}u(t)$ , -\n(c) $z(t) = \cos 4t\delta(t - 1)$ - -**7.30** Evaluate the following integrals involving the impulse functions: - -(a) -$$ -\int_{-\infty}^{\infty} 4t^2 \delta(t-1) dt -$$ - -(b) -$$ -\int_{-\infty}^{\infty} 4t^2 \cos 2\pi t \delta(t-0.5) dt -$$ - -**7.31** Evaluate the following integrals: - -(a) -$$ -\int_{-\infty}^{\infty} e^{-4t^2} \delta(t - 2) dt -$$ - -(b) -$$ -\int_{-\infty}^{\infty} [5\delta(t) + e^{-t} \delta(t) + \cos 2\pi t \delta(t)] dt -$$ - -**7.32** Evaluate the following integrals: - -(a) -$$ -\int_{1}^{t} u(\lambda) d\lambda -$$ - -\n(b) -$$ -\int_{0}^{4} r(t-1) dt -$$ - -\n(c) -$$ -\int_{1}^{5} (t-6)^{2} \delta(t-2) dt -$$ - -- **7.33** The voltage across a 10-mH inductor is 45*δ*(*t* − 2)mV. Find the inductor current, assuming that the inductor is initially uncharged. -- **7.34** Evaluate the following derivatives: - -(a) -$$ -\frac{d}{dt}[u(t-1)u(t+1)] -$$ - -\n(b) -$$ -\frac{d}{dt}[r(t-6)u(t-2)] -$$ - -\n(c) -$$ -\frac{d}{dt}[\sin 4tu(t-3)] -$$ - -**7.35** Find the solution to the following differential equations: - -(a) -$$ -\frac{dv}{dt} + 2v = 0 -$$ -, $v(0) = -1$ V -(b) $2\frac{di}{dt} - 3i = 0$ , $i(0) = 2$ - -**7.36** Solve for *v* in the following differential equations, subject to the stated initial condition. - -(a) -$$ -dv/dt + v = u(t) -$$ -, $v(0) = 0$ -(b) $2 dv/dt - v = 3u(t)$ , $v(0) = -6$ - -**7.37** A circuit is described by - -$$ -4\frac{dv}{dt} + v = 10 -$$ - -- (a) What is the time constant of the circuit? -- (b) What is *v*(∞), the final value of *v*? -- (c) If *v*(0) = 2, find *v*(*t*) for *t* ≥ 0. -- **7.38** A circuit is described by - -$$ -\frac{di}{dt} + 3i = 2u(t) -$$ - -Find *i*(*t*) for *t* > 0 given that *i*(0) = 0. - -# Section 7.5 Step Response of an RC Circuit - -**7.39** Calculate the capacitor voltage for *t* < 0 and *t* > 0 for each of the circuits in Fig. 7.106. - -**Figure 7.106** For Prob. 7.39. - -**7.40** Find the capacitor voltage for *t* < 0 and *t* > 0 for each of the circuits in Fig. 7.107. - -For Prob. 7.40. - -**7.41** Using Fig. 7.108, design a problem to help other students better understand the step response of an *RC* circuit. - -**Figure 7.108** - -- For Prob. 7.41. - - **7.42** (a) If the switch in Fig. 7.109 has been open for a long time and is closed at *t* = 0, find *vo*(*t*). - - (b) Suppose that the switch has been closed for a long time and is opened at *t* = 0. Find *vo*(*t*). - -For Prob. 7.42. - -**7.43** Consider the circuit in Fig. 7.110. Find *i*(*t*) for *t* < 0 and *t* > 0. - -For Prob. 7.43. - -**7.44** The switch in Fig. 7.111 has been in position *a* for a long time. At *t* = 0, it moves to position *b*. Calculate *i*(*t*) for all *t* > 0. - -**Figure 7.111** For Prob. 7.44. - -**7.45** Find *vo* in the circuit of Fig. 7.112 when *vs* = 30*u*(*t*) V. Assume that *vo*(0) = 5 V. - -**Figure 7.112** For Prob. 7.45. - -**7.46** For the circuit in Fig. 7.113, *is*(*t*) = 5*u*(*t*). Find *v*(*t*). - -**Figure 7.113** For Prob. 7.46. - -**7.47** Determine *v*(*t*) for *t* > 0 in the circuit of Fig. 7.114 if *v*(0) = 0. - -For Prob. 7.47. - -Problems **305** - -**7.48** Find *v*(*t*) and *i*(*t*) in the circuit of Fig. 7.115. - -# **Figure 7.115** - -For Prob. 7.48. - -**7.49** If the waveform in Fig. 7.116(a) is applied to the circuit of Fig. 7.116(b), find *v*(*t*). Assume *v*(0) = 0. - -# **Figure 7.116** - -For Prob. 7.49 and Review Question 7.10. - -\***7.50** In the circuit of Fig. 7.117, find *ix* for *t* > 0. Let *R*1 = *R*2 = 1 kΩ, *R*3 = 2 kΩ, and *C* = 0.25 mF. - -# **Figure 7.117** For Prob. 7.50. - -# Section 7.6 Step Response of an RL Circuit - -- **7.51** Rather than applying the shortcut technique used in Section 7.6, use KVL to obtain Eq. (7.60). -- **7.52** Using Fig. 7.118, design a problem to help other students better understand the step response of an *RL* circuit. - -**Figure 7.118** For Prob. 7.52. - -**7.53** Determine the inductor current *i*(*t*) for both *t* < 0 and *t* > 0 for each of the circuits in Fig. 7.119. - -**Figure 7.119** For Prob. 7.53. - -**7.54** Obtain the inductor current for both *t* < 0 and *t* > 0 in each of the circuits in Fig. 7.120. - -For Prob. 7.54. - -\* An asterisk indicates a challenging problem. - -**7.55** Find *v*(*t*) for *t* < 0 and *t* > 0 in the circuit of Fig. 7.121. - -# **Figure 7.121** - -For Prob. 7.55. - -**7.56** For the network shown in Fig. 7.122, find *v*(*t*) for *t* > 0. - -**Figure 7.122** - -For Prob. 7.56. - -\***7.57** Find *i*1(*t*) and *i*2(*t*) for *t* > 0 in the circuit of Fig. 7.123. - -**Figure 7.123** - -For Prob. 7.57. - -- **7.58** Rework Prob. 7.17 if *i*(0) = 10 A and *v*(*t*) = 20*u*(*t*) V. -- **7.59** Determine the step response *vo*(*t*) to *is* = 6*u*(*t*) A in the circuit of Fig. 7.124. - -**Figure 7.124** For Prob. 7.59. - -**7.60** Find *v*(*t*) for *t* > 0 in the circuit of Fig. 7.125 if the initial current in the inductor is zero. - -# **Figure 7.125** - -For Prob. 7.60. - -**7.61** In the circuit in Fig. 7.126, *is* changes from 5 A to 10 A at *t* = 0; that is, *is* = 5*u*(−*t*) + 10*u*(*t*). Find *v* and *i*. - -# **Figure 7.126** - -For Prob. 7.61. - -**7.62** For the circuit in Fig. 7.127, calculate *i*(*t*) if *i*(0) = 0. - -**Figure 7.127** - -For Prob. 7.62. - -**7.63** Obtain *v*(*t*) and *i*(*t*) in the circuit of Fig. 7.128. - -# **Figure 7.128** - -For Prob. 7.63. - -**7.64** Determine the value of *iL*(*t*) and the total energy dissipated by the circuit from *t* = 0 sec to *t* = ∞ sec. The value of *iin*(*t*) is equal to [6 – 6*u*(*t*)] A. - -**7.65** If the input pulse in Fig. 7.130(a) is applied to the circuit in Fig. 7.130(b), determine the response *i*(*t*). - -# Section 7.7 First-order Op Amp Circuits - -**7.66** Using Fig. 7.131, design a problem to help other students better understand first-order op amp circuits. - -For Prob. 7.66. - -**7.67** If *v*(0) = 10 V, find *vo*(*t*) for *t* > 0 in the op amp circuit in Fig. 7.132. Let *R* = 100 kΩ and *C* = 20 *µ*F. - -# **Figure 7.133** - -For Prob. 7.68. - -**7.69** For the op amp circuit in Fig. 7.134, find *vo*(*t*) for *t* > 0. - -# **Figure 7.134** - -For Prob. 7.69. - -**7.70** Determine *vo* for *t* > 0 when *vs* = 20 mV in the op amp circuit of Fig. 7.135. - -# **Figure 7.135** For Prob. 7.70. - -- **7.71** For the op amp circuit in Fig. 7.136, suppose *vs* = 10*u*(*t*) V. Find *v*(*t*) for *t* > 0. - -**7.72** Find *io* in the op amp circuit in Fig. 7.137. Assume that *v*(0) = −2 V, *R* = 10 kΩ, and *C* = 10 *µ*F. - -# **Figure 7.137** - -For Prob. 7.72. - -**7.73** For the op amp circuit of Fig. 7.138, let *R*1 = 10 kΩ, *Rf* = 30 kΩ, *C* = 20 *μ*F, and *v*(0) = 1 V. Find *v*0. - -# **Figure 7.138** - -For Prob. 7.73. - -**7.74** Determine *vo*(*t*) for *t* > 0 in the circuit of Fig. 7.139. Let *is* = 10*u*(*t*) *μ*A and assume that the capacitor is initially uncharged. - -# **Figure 7.139** - -For Prob. 7.74. - -**7.75** In the circuit of Fig. 7.140, find *vo* and *io*, given that *vs* = 10[1 − *e*−*t* ]*u*(*t*) V. - -**Figure 7.140** For Prob. 7.75. - -# Section 7.8 Transient Analysis with PSpice - -- **7.76** Repeat Prob. 7.49 using *PSpice or MultiSim*. -- **7.77** The switch in Fig. 7.141 opens at *t* = 0. Use *PSpice or MultiSim* to determine *v*(*t*) for *t* > 0. - -# **Figure 7.141** - -For Prob. 7.77. - -**7.78** The switch in Fig. 7.142 moves from position *a* to *b* at *t* = 0. Use *PSpice or MultiSim* to find *i*(*t*) for *t* > 0. - -# **Figure 7.142** For Prob. 7.78. - -**7.79** In the circuit of Fig. 7.143, determine *io*(*t*). - -# **Figure 7.143** - -- For Prob. 7.79. -- **7.80** In the circuit of Fig. 7.144, find the value of *io* for all values of 0 < *t*. - -For Prob. 7.80. - -**7.81** Repeat Prob. 7.65 using *PSpice or MultiSim*. - -# Section 7.9 Applications - -- **7.82** In designing a signal-switching circuit, it was found that a 100-*µ*F capacitor was needed for a time constant of 3 ms. What value resistor is necessary for the circuit? -- **7.83** An *RC* circuit consists of a series connection of a 120-V source, a switch, a 34-MΩ resistor, and a 15-*µ*F capacitor. The circuit is used in estimating the speed of a horse running a 4-km racetrack. The switch closes when the horse begins and opens when the horse crosses the finish line. Assuming that the capacitor charges to 85.6 V, calculate the speed of the horse. -- **7.84** A capacitor with a value of 10 mF has a leakage resistance of 2 MΩ. How long does it take the voltage across the capacitor to decay to 40% of the initial voltage to which the capacitor is charged? Assume that the capacitor is charged and then set aside by itself. -- **7.85** A simple relaxation oscillator circuit is shown in Fig. 7.145. The neon lamp fires when its voltage reaches 75 V and turns off when its voltage drops to 30 V. Its resistance is 120 Ω when on and infinitely high when off. - - (a) For how long is the lamp on each time the capacitor discharges? - - (b) What is the time interval between light flashes? - -**Figure 7.145** For Prob. 7.85. - -**7.86** Figure 7.146 shows a circuit for setting the length of time voltage is applied to the electrodes of a welding machine. The time is taken as how long it takes the capacitor to charge from 0 to 8 V. What is the time range covered by the variable resistor? - -# **Figure 7.146** For Prob. 7.86. - -# Comprehensive Problems - -- **7.88** The circuit in Fig. 7.148(a) can be designed as an approximate differentiator or an integrator, depending on whether the output is taken across the resistor or the capacitor, and also on the time constant *τ* = *RC* of the circuit and the width *T* of the input pulse in Fig. 7.148(b). The circuit is a differentiator if *τ*≪ *T*, say *τ*< 0.1*T*, or an integrator if *τ*≫ *T*, say *τ*> 10*T*. -- (a) What is the minimum pulse width that will allow a differentiator output to appear across the capacitor? -- (b) If the output is to be an integrated form of the input, what is the maximum value the pulse width can assume? - -For Prob. 7.88. - -**7.89** An *RL* circuit may be used as a differentiator if the output is taken across the inductor and *τ*≪ *T* (say *τ*< 0.1*T*), where *T* is the width of the input pulse. If *R* is fixed at 200 kΩ, determine the maximum value of *L* required to differentiate a pulse with *T* = 10 *µ*s. - -**7.90** An attenuator probe employed with oscilloscopes was designed to reduce the magnitude of the input voltage *vi* by a factor of 10. As shown in Fig. 7.149, the oscilloscope has internal resistance *Rs* and capacitance *Cs*, while the probe has an internal resistance *Rp*. If *Rp* is fixed at 6 MΩ, find *Rs* and *Cs* for the circuit to have a time constant of 15 *µ*s. - -**Figure 7.149** For Prob. 7.90. - -**7.91** The circuit in Fig. 7.150 is used by a biology student to study "frog kick." She noticed that the frog kicked a little when the switch was closed but kicked violently for 5 s when the switch was opened. Model the frog as a resistor and calculate its resistance. Assume that it takes 10 mA for the frog to kick violently. - -# **Figure 7.150** For Prob. 7.91. - -**7.92** To move a spot of a cathode-ray tube across the screen requires a linear increase in the voltage across the deflection plates, as shown in Fig. 7.151. Given that the capacitance of the plates is 4 nF, sketch the current flowing through the plates. - -**Figure 7.151** For Prob. 7.92. - -# **chapter** - -8 - -# Second-Order Circuits - -*Everyone who can earn a masters degree in engineering must earn a masters degree in engineering in order to maximize the success of their career! If you want to do research, state-of-the-art engineering, teach in a university, or start your own business, you really need to earn a doctoral degree!* - -—Charles K. Alexander - -# Enhancing Your Career - -To increase your engineering career opportunities after graduation, develop a strong fundamental understanding in a broad set of engineer ing areas. When possible, this might best be accomplished by working toward a graduate degree immediately upon receiving your undergraduate degree. - -Each de gree in engineering represents certain skills the student acquires. At the Bachelor de gree level, you learn the language of engi neering and the fundamentals of engineering and design. At the Master's level, you acquire the ability to do advanced engineering projects and to communicate your work effectively both orally and in writing. The Ph.D. represents a thorough understanding of the fundamentals of electrical engineering and a mastery of the skills necessary both for w orking at the frontiers of an engineering area and for communicating one' s effort to others. - -If you have no idea what career you should pursue after graduation, a graduate de gree program will enhance your ability to e xplore career options. Since your undergraduate degree will only provide you with the fundamentals of engineering, a Master' s degree in engineering supple mented by business courses benefits more engineering students than does getting a Master's of Business Administration (MBA). The best time to get your MB A is after you ha ve been a practicing engineer for some years and decide your career path would be enhanced by strengthening your business skills. - -Engineers should constantly educate themselv es, formally and informally, taking advantage of all means of education. Perhaps there is no better way to enhance your career than to join a professional society such as IEEE and be an active member. - -Enhancing your career involves understanding your goals, adapting to changes, anticipating opportunities, and planning your own niche. - -© 2005 Institute of Electrical and Electronics Engineers (IEEE), from *IEEE Potentials* cover, April/May 2005 - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Develop a better understanding of the solution of generalsecond order differential equations. -- 2. Learn how to determine initial and final values. -- 3. Understand the response in source-free series *RLC* circuits. -- 4. Understand the response in source-free parallel *RLC* circuits. -- 5. Understand the step response of series *RLC* circuits. -- 6. Understand the step response of parallel *RLC* circuits. -- 7. Understand general second-order circuits. -- 8. Understand general second-order circuits with op amps. - -# **8.1** Introduction - -In the previous chapter we considered circuits with a single storage element (a capacitor or an inductor). Such circuits are first-order because the differential equations describing them are first-order. In this chap ter we will consider circuits containing two storage elements. These are known as *second-order* circuits because their responses are described by differential equations that contain second derivatives. - -Typical e xamples of second-order circuits are *RLC* circuits, in which the three kinds of passive elements are present. Examples of such circuits are shown in Fig. 8.1(a) and (b). Other examples are *RL* and *RC* circuits, as shown in Fig. 8.1(c) and (d). It is apparent from Fig. 8.1 that a second-order circuit may have two storage elements of different type or the same type (provided elements of the same type cannot be represented by an equi valent single element). An op amp circuit with tw o storage elements may also be a second-order circuit. As with first-order circuits, a second-order circuit may contain se veral resistors and dependent and independent sources. - -A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements. - -Our analysis of second-order circuits will be similar to that used for first-order. We will first consider circuits that are excited by the initial conditions of the storage elements. Although these circuits may contain dependent sources, they are free of independent sources. These sourcefree circuits will give natural responses as expected. Later we will consider circuits that are e xcited by independent sources. These circuits will give both the transient response and the steady-state response. We consider only dc independent sources in this chapter. The case of sinusoidal and exponential sources is deferred to later chapters. - -We begin by learning ho w to obtain the initial conditions for the circuit variables and their deri vatives, as this is crucial to analyzing second-order circuits. Then we consider series and parallel *RLC* circuits such as shown in Fig. 8.1 for the two cases of excitation: by initial - - - -# **Figure 8.1** - -Typical examples of second-order circuits: (a) series *RLC* circuit, (b) parallel *RLC* circuit, (c) *RL* circuit, (d) *RC* circuit. - -conditions of the ener gy storage elements and by step inputs. Later we e xamine other types of second-order circuits, including op amp circuits. We will consider *PSpice* analysis of second-order circuits. Finally, we will consider the automobile ignition system and smooth ing circuits as typical applications of the circuits treated in this chapter. Other applications such as resonant circuits and filters will be covered in Chapter 14. - -# **8.2** Finding Initial and Final Values - -Perhaps the major problem students f ace in handling second-order circuits is finding the initial and final conditions on circuit variables. Students are usually comfortable getting the initial and final values of *v* and *i* but often have difficulty finding the initial values of their derivatives: *dv*∕*dt* and *di*∕*dt*. For this reason, this section is e xplicitly devoted to the subtleties of getting *v*(0), *i*(0), *dv*(0)∕*dt*, *di*(0)∕*dt*, *i*(∞), and *v*(∞). Unless otherwise stated in this chapter , *v* denotes capacitor v oltage, while *i* is the inductor current. - -There are two key points to keep in mind in determining the initial conditions. - -First—as always in circuit analysis—we must carefully handle the polarity of voltage *v*(*t*) across the capacitor and the direction of the current *i*(*t*) through the inductor. Keep in mind that *v* and *i* are defined strictly according to the passive sign convention (see Figs. 6.3 and 6.23). One should carefully observe how these are defined and apply them accordingly. - -Second, keep in mind that the capacitor voltage is always continuous so that - -$$ -v(0^+) = v(0^-) \tag{8.1a} -$$ - -and the inductor current is always continuous so that - -$$ -i(0^+) = i(0^-) \tag{8.1b} -$$ - -where *t* = 0− denotes the time just before a switching event and *t* = 0+ is the time just after the switching event, assuming that the switching event takes place at *t* = 0. - -Thus, in finding initial conditions, we first focus on those variables that cannot change abruptly, capacitor voltage and inductor current, by applying Eq. (8.1). The following examples illustrate these ideas. - -The switch in Fig. 8.2 has been closed for a long time. It is open at *t* = 0. Example 8.1 Find: (a) *i*(0+), *v*(0+), (b) *di*(0+)∕*dt*, *dv*(0+)∕*dt*, (c) *i*(∞), *v*(∞). - -# **Solution:** - -(a) If the switch is closed a long time before *t* = 0, it means that the circuit has reached dc steady state at *t* = 0. At dc steady state, the inductor acts like a short circuit, while the capacitor acts like an open circuit, so we have the circuit in Fig. 8.3(a) at *t* = 0−. Thus, - -$$ -i(0^{-}) = \frac{12}{4+2} = 2 \text{ A}, \qquad v(0^{-}) = 2i(0^{-}) = 4 \text{ V} -$$ - -As the inductor current and the capacitor voltage cannot change abruptly, - -$$ -i(0^+) = i(0^-) = 2 -$$ - A, $v(0^+) = v(0^-) = 4$ V - -(b) At *t* = 0+, the switch is open; the equivalent circuit is as shown in Fig. 8.3(b). The same current flows through both the inductor and capacitor. Hence, - -$$ -i_C(0^+) = i(0^+) = 2 \text{ A} -$$ - -Since *C dv*∕*dt* = *iC*, *dv*∕*dt* = *iC*∕*C*, and - -$$ -\frac{dv(0^{+})}{dt} = \frac{i_C(0^{+})}{C} = \frac{2}{0.1} = 20 \text{ V/s} -$$ - -Similarly, since *L di*∕*dt* = *vL*, *di*∕*dt* = *vL*∕*L*. We now obtain *vL* by applying KVL to the loop in Fig. 8.3(b). The result is - -$$ --12 + 4i(0^+) + v_L(0^+) + v(0^+) = 0 -$$ - -or - -$$ -v_L(0^+) = 12 - 8 - 4 = 0 -$$ - -Thus, - -$$ -\frac{di(0^{+})}{dt} = \frac{v_L(0^{+})}{L} = \frac{0}{0.25} = 0 \text{ A/s} -$$ - -(c) For *t* > 0, the circuit undergoes transience. But as *t* → ∞, the circuit reaches steady state again. The inductor acts like a short circuit and the capacitor like an open circuit, so that the circuit in Fig. 8.3(b) becomes that shown in Fig. 8.3(c), from which we have - -$$ -i(\infty) = 0 \text{ A}, \qquad v(\infty) = 12 \text{ V} -$$ - -Practice Problem 8.1 The switch in Fig. 8.4 was open for a long time but closed at *t* = 0. Determine: (a) *i*(0+), *v*(0+), (b) *di*(0+)∕*dt*, *dv*(0+)∕*dt*, (c) *i*(∞), *v*(∞). - -**Answer:** (a) 3.5 A, 7 V, (b) 87.5 A/s, 0 V/s, (c) 21 A, 42 V. - -In the circuit of Fig. 8.5, calculate: (a) *iL*(0+), *vC* (0+), *vR*(0+), (b) *diL*(0+)∕*dt*, *dvC*(0+)∕*dt*, *dvR*(0+)∕*dt*, (c) *iL*(∞), *vC*(∞), *vR*(∞). Example 8.2 - -# **Solution:** - -(a) For *t* < 0, 3*u*(*t*) = 0. At *t* = 0−, since the circuit has reached steady state, the inductor can be replaced by a short circuit, while the capacitor is replaced by an open circuit as shown in Fig. 8.6(a). From this figure we obtain - -$$ -i_L(0^-) = 0 -$$ -, $v_R(0^-) = 0$ , $v_C(0^-) = -20$ V (8.2.1) - -Although the derivatives of these quantities at *t* = 0− are not required, it is evident that they are all zero, since the circuit has reached steady state and nothing changes. - -# **Figure 8.6** - -The circuit in Fig. 8.5 for: (a) *t* = 0−, (b) *t* = 0+. - -For *t* > 0, 3*u*(*t*) = 3, so that the circuit is no w equivalent to that in Fig. 8.6(b). Since the inductor current and capacitor v oltage cannot change abruptly, - -$$ -i_L(0^+) = i_L(0^-) = 0, -$$ - $v_C(0^+) = v_C(0^-) = -20$ V (8.2.2) - -Although the voltage across the 4-Ω resistor is not required, we will use it to apply KVL and KCL; let it be called *vo*. Applying KCL at node *a* in Fig. 8.6(b) gives - -$$ -3 = \frac{v_R(0^+)}{2} + \frac{v_o(0^+)}{4} -$$ - (8.2.3) - -Applying KVL to the middle mesh in Fig. 8.6(b) yields - -$$ --v_R(0^+) + v_o(0^+) + v_C(0^+) + 20 = 0 \tag{8.2.4} -$$ - -Since *vC* (0+) = −20 V from Eq. (8.2.2), Eq. (8.2.4) implies that - -$$ -v_R(0^+) = v_o(0^+) \tag{8.2.5} -$$ - -From Eqs. (8.2.3) and (8.2.5), we obtain - -$$ -v_R(0^+) = v_o(0^+) = 4 \text{ V} -$$ - (8.2.6) - -(b) Since *L diL*∕*dt* = *vL*, - -$$ -\frac{di_L(0^+)}{dt} = \frac{v_L(0^+)}{L} -$$ - -But applying KVL to the right mesh in Fig. 8.6(b) gives - -$$ -v_L(0^+) = v_C(0^+) + 20 = 0 -$$ - -Hence, - -$$ -\frac{di_L(0^+)}{dt} = 0\tag{8.2.7} -$$ - -Similarly, since *C dvC*∕*dt* = *iC*, then *dvC*∕*dt* = *iC*∕*C*. We apply KCL at node *b* in Fig. 8.6(b) to get *iC*: - -$$ -\frac{v_o(0^+)}{4} = i_C(0^+) + i_L(0^+) \tag{8.2.8} -$$ - -Since *vo*(0+) = 4 and *iL*(0+) = 0, *iC*(0+) = 4∕4 = 1 A. Then - -$$ -\frac{dv_C(0^+)}{dt} = \frac{i_C(0^+)}{C} = \frac{1}{0.5} = 2 \text{ V/s} -$$ - (8.2.9) - -To get *dvR*(0+)∕*dt*, we apply KCL to node *a* and obtain - -$$ -3 = \frac{v_R}{2} + \frac{v_o}{4} -$$ - -Taking the derivative of each term and setting *t* = 0+ gives - -$$ -0 = 2\frac{dv_R(0^+)}{dt} + \frac{dv_o(0^+)}{dt} -$$ - (8.2.10) - -We also apply KVL to the middle mesh in Fig. 8.6(b) and obtain - -$$ --v_R + v_C + 20 + v_o = 0 -$$ - -Again, taking the derivative of each term and setting *t* = 0+ yields - -$$ --\frac{dv_R(0^+)}{dt} + \frac{dv_C(0^+)}{dt} + \frac{dv_o(0^+)}{dt} = 0 -$$ - -Substituting for *dvC* (0+)∕*dt* = 2 gives - -$$ -\frac{dv_R(0^+)}{dt} = 2 + \frac{dv_o(0^+)}{dt} -$$ - (8.2.11) - -From Eqs. (8.2.10) and (8.2.11), we get - -$$ -\frac{dv_R(0^+)}{dt} = \frac{2}{3} \text{ V/s} -$$ - -We can find *diR*(0+)∕*dt* although it is not required. Since *vR* = 2*iR*, - -$$ -\frac{di_R(0^+)}{dt} = \frac{1}{2}\frac{dv_R(0^+)}{dt} = \frac{1}{2}\frac{2}{3} = \frac{1}{3} \text{ A/s} -$$ - -(c) As *t* → ∞, the circuit reaches steady state. We have the equivalent circuit in Fig. 8.6(a) except that the 3-A current source is now operative. By current division principle, - -$$ -i_L(\infty) = \frac{2}{2+4} \cdot 3 \text{ A} = 1 \text{ A} -$$ -\n -$$ -v_R(\infty) = \frac{4}{2+4} \cdot 3 \text{ A} \times 2 = 4 \text{ V}, \qquad v_C(\infty) = -20 \text{ V} -$$ -\n(8.2.12) - -For the circuit in Fig. 8.7, find: (a) *iL*(0+), *vC*(0+), *vR*(0+), Practice Problem 8.2 (b) *diL*(0+)∕*dt*, *dvC*(0+)∕*dt*, *dvR*(0+)∕*dt*, (c) *iL*(∞), *vC*(∞), *vR*(∞). - -**Answer:** (a) −6 A, 0, 0, (b) 0, 20 V/s, 0, (c) −2 A, 20 V, 20 V. - -# **8.3** The Source-Free Series RLC Circuit - -An understanding of the natural response of the series *RLC* circuit is a necessary background for future studies in filter design and communications networks. - -Consider the series *RLC* circuit sho wn in Fig. 8.8. The circuit is being excited by the energy initially stored in the capacitor and inductor. The energy is represented by the initial capacitor v oltage *V*0 and initial inductor current *I*0. Thus, at *t* = 0, - -$$ -v(0) = \frac{1}{C} \int_{-\infty}^{0} i \, dt = V_0 \tag{8.2a} -$$ - -$$ -i(0) = I_0 \tag{8.2b} -$$ - -Applying KVL around the loop in Fig. 8.8, - -$$ -Ri + L\frac{di}{dt} + \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau = 0 -$$ - (8.3) - -A source-free series *RLC* circuit. - -To eliminate the integral, we differentiate with respect to *t* and rearrange terms. We get - -$$ -\frac{d^2i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{i}{LC} = 0 -$$ -\n(8.4) - -This is a *second-order differential equation* and is the reason for calling the *RLC* circuits in this chapter second-order circuits. Our goal is to solve Eq. (8.4). To solv e such a second-order dif ferential equation requires that we have two initial conditions, such as the initial v alue of *i* and its first derivative or initial values of some *i* and *v*. The initial value of *i* is given in Eq. (8.2b). We get the initial v alue of the deri vative of *i* from Eqs. (8.2a) and (8.3); that is, - -$$ -Ri(0) + L\frac{di(0)}{dt} + V_0 = 0 -$$ - -or - -$$ -\frac{di(0)}{dt} = -\frac{1}{L} (RI_0 + V_0) -$$ -\n(8.5) - -With the two initial conditions in Eqs. (8.2b) and (8.5), we can no w solve Eq. (8.4). Our e xperience in the preceding chapter on first-order circuits suggests that the solution is of exponential form. So we let - -$$ -i = Ae^{st} \tag{8.6} -$$ - -where *A* and *s* are constants to be determined. Substituting Eq. (8.6) into Eq. (8.4) and carrying out the necessary differentiations, we obtain - -$$ -As2est + \frac{AR}{L}sest + \frac{A}{LC}est = 0 -$$ - -or - -$$ -Ae^{st}(s^2 + \frac{R}{L}s + \frac{1}{LC}) = 0 -$$ - (8.7) - -Since *i* = *Aest* is the assumed solution we are trying to find, only the expression in parentheses can be zero: - -$$ -s^2 + \frac{R}{L}s + \frac{1}{LC} = 0 -$$ - (8.8) - -This quadratic equation is kno wn as the *characteristic equation* of the differential Eq. (8.4), since the roots of the equation dictate the character of *i*. The two roots of Eq. (8.8) are - -$$ -s_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 - \frac{1}{LC}} -$$ - (8.9a) - -$$ -s_2 = -\frac{R}{2L} - \sqrt{\left(\frac{R}{2L}\right)^2 - \frac{1}{LC}} -$$ - (8.9b) - -A more compact way of expressing the roots is - -$$ -s_1 = -\alpha + \sqrt{\alpha^2 - \omega_0^2} -$$ -, $s_2 = -\alpha - \sqrt{\alpha^2 - \omega_0^2}$ (8.10) - -See Appendix C.1 for the formula to find the roots of a quadratic equation. where - -$$ -\alpha = \frac{R}{2L}, \qquad \omega_0 = \frac{1}{\sqrt{LC}} \tag{8.11} -$$ - -The roots *s*1 and *s*2 are called *natural frequencies*, measured in nepers per second (Np/s), because they are associated with the natural response of the circuit; *ω*0 is kno wn as the *resonant frequency* or strictly as the *undamped natural frequency*, expressed in radians per second (rad/s); and *α* is the *neper frequency* expressed in nepers per second. In terms of *α* and *ω*0, Eq. (8.8) can be written as - -$$ -x^2 + 2\alpha s + \omega_0^2 = 0 -$$ - (8.8a) - -The variables *s* and *ω*0 are important quantities we will be discussing throughout the rest of the text. - -*s* - -The two values of *s* in Eq. (8.10) indicate that there are two possible solutions for *i*, each of which is of the form of the assumed solution in Eq. (8.6); that is, - -$$ -i_1 = A_1 e^{s_1 t}, \qquad i_2 = A_2 e^{s_2 t} \tag{8.12} -$$ - -Since Eq. (8.4) is a linear equation, an y linear combination of the tw o distinct solutions *i*1 and *i*2 is also a solution of Eq. (8.4). A complete or total solution of Eq. (8.4) would therefore require a linear combination of *i*1 and *i*2. Thus, the natural response of the series *RLC* circuit is - -$$ -i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} -$$ - (8.13) - -where the constants *A*1 and *A*2 are determined from the initial values *i*(0) and *di*(0)∕*dt* in Eqs. (8.2b) and (8.5). - -From Eq. (8.10), we can infer that there are three types of solutions: - -1. If *α* > *ω*0, we have the *overdamped* case. - -- 2. If *α* = *ω*0, we have the *critically damped* case. -- 3. If *α* < *ω*0, we have the *underdamped* case. - -We will consider each of these cases separately. - -# **Overdamped Case (***α* > *ω***0)** - -From Eqs. (8.9) and (8.10), *α* > *ω*0 implies *C* > 4*L*∕*R*2 . When this happens, both roots *s*1 and *s*2 are negative and real. The response is - -$$ -i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} -$$ - (8.14) - -which decays and approaches zero as *t* increases. Figure 8.9(a) illustrates a typical overdamped response. - -# **Critically Damped Case (***α* **=** *ω***0)** - -When *α* = *ω*0, *C* = 4*L*∕*R*2 and - -$$ -s_1 = s_2 = -\alpha = -\frac{R}{2L} -$$ - (8.15) - -The response is overdamped when the roots of the circuit's characteristic equation are unequal and real, critically damped when the roots are equal and real, and underdamped when the roots are complex. - -The neper (Np) is a dimensionless unit named after John Napier (1550–1617), a Scottish mathematician. - -The ratio *α*/*ω*0 is known as the damping - -ratio ζ. - -For this case, Eq. (8.13) yields - -$$ -i(t) = A_1 e^{-\alpha t} + A_2 e^{-\alpha t} = A_3 e^{-\alpha t} -$$ - -where *A*3 = *A*1 + *A*2 . This cannot be the solution, because the two initial conditions cannot be satisfied with the single constant *A*3. What then could be wrong? Our assumption of an e xponential solution is incor rect for the special case of critical damping. Let us go back to Eq. (8.4). When *α* = *ω*0 = *R*∕2*L*, Eq. (8.4) becomes - -$$ -\frac{d^2i}{dt^2} + 2\alpha \frac{di}{dt} + \alpha^2 i = 0 -$$ - -$$ -\frac{d}{dt}\left(\frac{di}{dt} + \alpha i\right) + \alpha \left(\frac{di}{dt} + \alpha i\right) = 0 -$$ -\n(8.16) - -If we let - -or - -$$ -f = \frac{di}{dt} + \alpha i \tag{8.17} -$$ - -then Eq. (8.16) becomes - -$$ -\frac{df}{dt} + \alpha f = 0 -$$ - -which is a first-order differential equation with solution *f* = *A*1*e*−*αt* , where *A*1 is a constant. Equation (8.17) then becomes - -$$ -\frac{di}{dt} + \alpha i = A_1 e^{-\alpha t} -$$ - -$$ -e^{\alpha t} \frac{di}{dt} + e^{\alpha t} \alpha i = A_1 \tag{8.18} -$$ - -This can be written as - -$$ -\frac{d}{dt}(e^{at}i) = A_1 \tag{8.19} -$$ - -Integrating both sides yields - -$$ -e^{\alpha t}i = A_1t + A_2 -$$ - -or - -or - -$$ -i = (A_1 t + A_2)e^{-\alpha t} -$$ - (8.20) - -where *A*2 is another constant. Hence, the natural response of the critically damped circuit is a sum of two terms: a negative exponential and a negative exponential multiplied by a linear term, or - -$$ -i(t) = (A_2 + A_1 t)e^{-\alpha t} -$$ - (8.21) - -A typical critically damped response is sho wn in Fig. 8.9(b). In f act, Fig. 8.9(b) is a sk etch of *i*(*t*) = *te*−*αt* , which reaches a maximum v alue of *e* −1 ∕*α* at *t* = 1∕*α*, one time constant, and then decays all the way to zero. - -(a) Overdamped response, (b) critically damped response, (c) underdamped response. - -(c) - -# **Underdamped Case (***α* < *ω***0)** - -For *α* < *ω*0, *C* < 4*L*∕*R*2 . The roots may be written as - -$$ -s_1 = -\alpha + \sqrt{-(\omega_0^2 - \alpha^2)} = -\alpha + j\omega_d -$$ - (8.22a) - -$$ -s_2 = -\alpha - \sqrt{-(\omega_0^2 - \alpha^2)} = -\alpha - j\omega_d -$$ - (8.22b) - -where *j* = √ \_\_\_ −1 and *ωd* = √ \_\_\_\_\_\_\_ *ω* 0 2− *α*2 ,which is called the *damped frequency*. Both *ω*0 and *ωd* are natural frequencies because they help determine the natural response; while *ω*0 is often called the *undamped natur al fr equency, ωd* is called the *damped natural frequency*. The natural response is - -$$ -i(t) = A_1 e^{-(\alpha - j\omega_d)t} + A_2 e^{-(\alpha + j\omega_d)t} -$$ - -= $e^{-\alpha t} (A_1 e^{-j\omega_d t} + A_2 e^{-j\omega_d t})$ (8.23) - -Using Euler's identities, - -$$ -e^{j\theta} = \cos\theta + j\sin\theta, \qquad e^{-j\theta} = \cos\theta - j\sin\theta \qquad (8.24) -$$ - -we get - -$$ -i(t) = e^{-\alpha t} [A_1(\cos \omega_d t + j \sin \omega_d t) + A_2(\cos \omega_d t - j \sin \omega_d t)] -$$ - -= $e^{-\alpha t} [(A_1 + A_2) \cos \omega_d t + j(A_1 - A_2) \sin \omega_d t]$ (8.25) - -Replacing constants ( *A*1 + *A*2) and *j*(*A*1 − *A*2) with constants *B*1 and *B*2, we write - -$$ -i(t) = e^{-\alpha t} (B_1 \cos \omega_d t + B_2 \sin \omega_d t) -$$ - (8.26) - -With the presence of sine and cosine functions, it is clear that the natural response for this case is e xponentially damped and oscillatory in nature. The response has a time constant of 1 ∕*α* and a period of *T* = 2*π*∕*ωd*. Figure 8.9(c) depicts a typical underdamped response. Part (a) and (b) of Fig. 8.9 assume for each case that *i*(0) = 0. - -Once the inductor current *i*(*t*) is found for the *RLC* series circuit as shown above, other circuit quantities such as indi vidual element voltages can easily be found. F or example, the resistor voltage is *vR* = *Ri*, and the inductor voltage is *vL* = *L di*∕*dt*. The inductor current *i*(*t*) is selected as the key variable to be determined first in order to take advantage of Eq. (8.1b). - -We conclude this section by noting the follo wing interesting, peculiar properties of an *RLC* network: - -- 1. The behavior of such a network is captured by the idea of *damping,* which is the gradual loss of the initial stored ener gy, as evidenced by the continuous decrease in the amplitude of the response. The damping effect is due to the presence of resistance *R*. The neper frequency *α* determines the rate at which the response is damped. If *R* = 0, then *α* = 0, and we have an *LC* circuit with 1∕√ \_\_\_ *LC* as the undamped natural frequency. Since *α* < *ω*0 in this case, the response is not only undamped b ut also oscillatory. The circuit is said to be *loss-less,* because the dissipating or damping element (*R*) is absent. By adjusting the value of *R*, the response may be made undamped, overdamped, critically damped, or underdamped. -- 2. Oscillatory response is possible due to the presence of the tw o types of storage elements. Ha ving both *L* and *C* allows the flow - -R = 0 produces a perfectly sinusoidal response. This response cannot be practically accomplished with L and C because of the inherent losses in them. See Figs 6.8 and 6.26. An electronic device called an oscillator can produce a perfectly sinusoidal response. - -Examples 8.5 and 8.7 demonstrate the effect of varying R. - -The response of a second-order circuit with two storage elements of the same type, as in Fig. 8.1(c) and (d), cannot be oscillatory. - -of energy back and forth between the tw o. The damped oscillation exhibited by the underdamped response is kno wn as *ringing*. It stems from the ability of the storage elements *L* and *C* to transfer energy back and forth between them. - -3. Observe from Fig. 8.9 that the w aveforms of the responses dif fer. In general, it is dif ficult to tell from the waveforms the dif ference between the overdamped and critically damped responses. The critically damped case is the borderline between the underdamped and overdamped cases and it decays the fastest. With the same initial conditions, the overdamped case has the longest settling time, because it tak es the longest time to dissipate the initial stored energy. If we desire the response that approaches the final value most rapidly without oscillation or ringing, the critically damped circuit is the right choice. - -circuit that is as close as possible to a critically damped circuit. - -What this means in most practical circuits is that we seek an overdamped - -Example 8.3 - -In Fig. 8.8, *R* = 40 Ω, *L* = 4 H, and *C* = 1∕4 F. Calculate the charac teristic roots of the circuit. Is the natural response o verdamped, underdamped, or critically damped? - -# **Solution:** - -We first calculate - -$$ -\alpha = \frac{R}{2L} = \frac{40}{2(4)} = 5 -$$ -, $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{4 \times \frac{1}{4}}} = 1$ - -The roots are - -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -5 \pm \sqrt{25 - 1} -$$ - -or - -$$ -s_1 = -0.101, \qquad s_2 = -9.899 -$$ - -Since *α* > *ω*0, we conclude that the response is overdamped. This is also evident from the fact that the roots are real and negative. - -| Practice Problem 8.3 | If R = 10 Ω, L = 5 H, and C = 2 mF in Fig. 8.8, find α, ω0, s1,
and s2. What | -|----------------------|---------------------------------------------------------------------------------| -| | type of natural response will the circuit have? | - -**Answer:** 1, 10, −1 ± *j*9.95, underdamped. - -| Example 8.4 | | Find i(t) in the circuit of Fig. 8.10. Assume that the circuit has reached | -|-------------|-------------------------|----------------------------------------------------------------------------| -| | steady state at t = 0−. | | - -# **Solution:** - -For *t* < 0, the switch is closed. The capacitor acts like an open circuit while the inductor acts like a shunted circuit. The equivalent circuit is shown in Fig. 8.11(a). Thus, at *t* = 0, - -$$ -i(0) = \frac{10}{4+6} = 1 \text{ A}, \qquad v(0) = 6i(0) = 6 \text{ V} -$$ - -The circuit in Fig. 8.10: (a) for *t* < 0, (b) for *t* > 0. - -where *i*(0) is the initial current through the inductor and *v*(0) is the initial voltage across the capacitor. - -For *t* > 0, the switch is opened and the v oltage source is discon nected. The equivalent circuit is shown in Fig. 8.11(b), which is a sourcefree series *RLC* circuit. Notice that the 3-Ω and 6-Ω resistors, which are in series in Fig. 8.10 when the switch is opened, have been combined to give *R* = 9 Ω in Fig. 8.11(b). The roots are calculated as follows: - -$$ -\alpha = \frac{R}{2L} = \frac{9}{2\left(\frac{1}{2}\right)} = 9, \qquad \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\frac{1}{2} \times \frac{1}{50}}} = 10 -$$ -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -9 \pm \sqrt{81 - 100} -$$ - -or - -*s*1,2 = −9 ± *j*4.359 - -Hence, the response is underdamped (*α* < *ω*); that is, - -$$ -i(t) = e^{-9t} (A_1 \cos 4.359t + A_2 \sin 4.359t) -$$ - (8.4.1) - -We now obtain *A*1 and *A*2 using the initial conditions. At *t* = 0, - -$$ -i(0) = 1 = A_1 \tag{8.4.2} -$$ - -From Eq. (8.5), - -$$ -\left. \frac{di}{dt} \right|_{t=0} = -\frac{1}{L} [Ri(0) + v(0)] = -2[9(1) - 6] = -6 \text{ A/s} -$$ - (8.4.3) - -Note that *v*(0) = *V*0 = −6 V is used, because the polarity of *v* in Fig. 8.11(b) is opposite that in Fig. 8.8. Taking the derivative of *i*(*t*) in Eq. (8.4.1), - -$$ -\frac{di}{dt} = -9e^{-9t}(A_1 \cos 4.359t + A_2 \sin 4.359t) \n+ e^{-9t}(4.359)(-A_1 \sin 4.359t + A_2 \cos 4.359t) -$$ - -Imposing the condition in Eq. (8.4.3) at *t* = 0 gives - -$$ --6 = -9(A_1 + 0) + 4.359(-0 + A_2) -$$ - -But *A*1 = 1 from Eq. (8.4.2). Then - -$$ --6 = -9 + 4.359A_2 -$$ - $\Rightarrow$ $A_2 = 0.6882$ - -Substituting the values of *A*1 and *A*2 in Eq. (8.4.1) yields the com plete solution as - -$$ -i(t) = e^{-9t}(\cos 4.359t + 0.6882 \sin 4.359t) A -$$ - -# Practice Problem 8.4 - -**Figure 8.12** For Practice Prob. 8.4. - -**Figure 8.13** A source-free parallel *RLC* circuit. - -The circuit in Fig. 8.12 has reached steady state at *t* = 0−. If the makebefore-break switch moves to position *b* at *t* = 0, calculate *i*(*t*) for *t* > 0. - -**Answer:** -$$ -e^{-2.5t} (10 \cos 1.6583t - 15.076 \sin 1.6583t) -$$ - A. - -# **8.4** The Source-Free Parallel RLC Circuit - -Parallel *RLC* circuits find many practical applications, notably in communications networks and filter designs. - -Consider the parallel *RLC* circuit shown in Fig. 8.13. Assume initial inductor current *I*0 and initial capacitor voltage *V*0, - -$$ -i(0) = I_0 = \frac{1}{L} \int_{-\infty}^{0} v(t) dt -$$ - (8.27a) - -$$ -v(0) = V_0 \tag{8.27b} -$$ - -Because the three elements are in parallel, they have the same voltage *v* across them. According to passive sign convention, the current is entering each element; that is, the current through each element is leaving the top node. Thus, applying KCL at the top node gives - -$$ -\frac{v}{R} + \frac{1}{L} \int_{-\infty}^{t} v(\tau) \, d\tau + C \frac{dv}{dt} = 0 \tag{8.28} -$$ - -Taking the derivative with respect to *t* and dividing by *C* results in - -$$ -\frac{d^2v}{dt^2} + \frac{1}{RC}\frac{dv}{dt} + \frac{1}{LC}v = 0 -$$ - (8.29) - -We obtain the characteristic equation by replacing the first derivative by *s* and the second derivative by *s* 2 . By following the same reasoning used in establishing Eqs. (8.4) through (8.8), the characteristic equation is obtained as - -$$ -s^2 + \frac{1}{RC} s + \frac{1}{LC} = 0 -$$ - (8.30) - -The roots of the characteristic equation are - -$$ -s_{1,2} = -\frac{1}{2RC} \pm \sqrt{\left(\frac{1}{2RC}\right)^2 - \frac{1}{LC}} -$$ - -or - -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} -$$ - (8.31) - -where - -$$ -\alpha = \frac{1}{2RC}, \qquad \omega_0 = \frac{1}{\sqrt{LC}} \tag{8.32} -$$ - -The names of these terms remain the same as in the preceding section, as they play the same role in the solution. Again, there are three possible solutions, depending on whether *α* > *ω*0, *α* = *ω*0, or *α* < *ω*0. Let us consider these cases separately. - -# **Overdamped Case (***α* **>** *ω***0)** - -From Eq. (8.32), *α* > *ω*0 when *L* > 4*R*2 *C*. The roots of the characteristic equation are real and negative. The response is - -$$ -v(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} -$$ - (8.33) - -# **Critically Damped Case (***α* **=** *ω***0)** - -For *α* = *ω*0, *L* = 4*R*2 *C*. The roots are real and equal so that the response is - -$$ -v(t) = (A_1 + A_2 t)e^{-\alpha t} -$$ - (8.34) - -# **Underdamped Case (***α* **<** *ω***0)** - -When *α* < *ω*0, *L* < 4*R*2 *C*. In this case the roots are complex and may be expressed as - -$$ -s_{1,2} = -\alpha \pm j\omega_d \tag{8.35} -$$ - -where - -$$ -\omega_d = \sqrt{\omega_0^2 - \alpha^2} \tag{8.36} -$$ - -The response is - -$$ -v(t) = e^{-at}(A_1 \cos \omega_d t + A_2 \sin \omega_d t) -$$ - (8.37) - -The constants *A*1 and *A*2 in each case can be determined from the initial conditions. We need *v*(0) and *dv*(0)∕*dt*. The first term is known from Eq. (8.27b). We find the second term by combining Eqs. (8.27) and (8.28), as - -> *dv*(0) \_\_\_\_\_ *dt* =0 - -+ *I*0 + *C* - - \_\_\_ *V*0 *R* - -or - -$$ -\frac{dv(0)}{dt} = -\frac{(V_0 + RI_0)}{RC} -$$ - (8.38) - -The voltage waveforms are similar to those sho wn in Fig. 8.9 and will depend on whether the circuit is overdamped, underdamped, or critically damped. - -Having found the capacitor voltage *v*(*t*) for the parallel *RLC* circuit as shown above, we can readily obtain other circuit quantities such as indi vidual element currents. F or e xample, the resistor current is *iR* = *v*∕*R* and the capacitor current is *iC* = *C dv*∕*dt*. We have selected the capacitor voltage *v*(*t*) as the key variable to be determined first in order to take advantage of Eq. (8.1a). Notice that we first found the inductor current *i*(*t*) for the *RLC* series circuit, whereas we first found the capacitor voltage *v*(*t*) for the parallel *RLC* circuit. - -Example 8.5 - -In the parallel circuit of Fig. 8.13, find *v*(*t*) for *t* > 0, assuming *v*(0) = 5 V, *i*(0) = 0, *L* = 1 H, and *C* = 10 mF. Consider these cases: *R* = 1.923 Ω, *R* = 5 Ω, and *R* = 6.25 Ω. - -# **Solution:** - -Solution: -\n**CASE 1** If -$$ -R = 1.923 \Omega -$$ -, -\n -$$ -\alpha = \frac{1}{2RC} = \frac{1}{2 \times 1.923 \times 10 \times 10^{-3}} = 26 -$$ -\n -$$ -\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 10 \times 10^{-3}}} = 10 -$$ - -Since *α* > *ω*0 in this case, the response is overdamped. The roots of the characteristic equation are - -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -2, -50 -$$ - -and the corresponding response is - -$$ -v(t) = A_1 e^{-2t} + A_2 e^{-50t} -$$ - (8.5.1) - -We now apply the initial conditions to get *A*1 and *A*2. - -$$ -v(0) = 5 = A_1 + A_2 \tag{8.5.2} -$$ - -$$ -v(0) = 5 = A_1 + A_2 -$$ -$$ -\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{1.923 \times 10 \times 10^{-3}} = -260 -$$ - -But differentiating Eq. (8.5.1), - -$$ -\frac{dv}{dt} = -2A_1e^{-2t} - 50A_2e^{-50t} -$$ - -$$ -At t = 0, -$$ - -$$ --260 = -2A_1 - 50A_2 \tag{8.5.3} -$$ - -From Eqs. (8.5.2) and (8.5.3), we obtain *A*1 = −0.2083 and *A*2 = 5.208. Substituting *A*1 and *A*2 in Eq. (8.5.1) yields - -$$ -v(t) = -0.2083e^{-2t} + 5.208e^{-50t} -$$ - (8.5.4) - -■ **CASE 2** When *R* = 5 Ω, - -en -$$ -R = 5 \Omega -$$ -, -\n -$$ -\alpha = \frac{1}{2RC} = \frac{1}{2 \times 5 \times 10 \times 10^{-3}} = 10 -$$ - -while *ω*0 = 10 remains the same. Since *α* = *ω*0 = 10, the response is critically damped. Hence, *s*1 = *s*2 = −10, and - -$$ -v(t) = (A_1 + A_2 t)e^{-10t} -$$ -\n(8.5.5) - -To get *A*1 and *A*2, we apply the initial conditions - -$$ -v(0) = 5 = A_1 \tag{8.5.6} -$$ - -$$ -v(0) = 5 = A_1 -$$ -$$ -\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{5 \times 10 \times 10^{-3}} = -100 -$$ - -But differentiating Eq. (8.5.5), - -$$ -\frac{dv}{dt} = (-10A_1 - 10A_2t + A_2)e^{-10t} -$$ - -At *t* = 0, - -$$ --100 = -10A_1 + A_2 \tag{8.5.7} -$$ - -From Eqs. (8.5.6) and (8.5.7), *A*1 = 5 and *A*2 = −50. Thus, - -$$ -v(t) = (5 - 50t)e^{-10t} \text{ V} -$$ - (8.5.8) - -■ **CASE 3** When *R* = 6.25 Ω, - -$$ -\text{hen } R = 6.25 \, \Omega, -$$ -\n -$$ -\alpha = \frac{1}{2RC} = \frac{1}{2 \times 6.25 \times 10 \times 10^{-3}} = 8 -$$ - -while *ω*0 = 10 remains the same. As *α* < *ω*0 in this case, the response is underdamped. The roots of the characteristic equation are - -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -8 \pm j6 -$$ - -Hence, - -$$ -v(t) = (A_1 \cos 6t + A_2 \sin 6t)e^{-8t} -$$ - (8.5.9) - -We now obtain *A*1 and *A*2, as - -$$ -v(0) = 5 = A_1 \tag{8.5.10} -$$ - -$$ -\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{5 + 0}{6.25 \times 10 \times 10^{-3}} = -80 -$$ - -But differentiating Eq. (8.5.9), - -$$ -\frac{dv}{dt} = (-8A_1 \cos 6t - 8A_2 \sin 6t - 6A_1 \sin 6t + 6A_2 \cos 6t)e^{-8t} -$$ - -At $t = 0$ , -$$ --80 = -8A_1 + 6A_2 -$$ - (8.5.11) - -From Eqs. (8.5.10) and (8.5.11), *A*1 = 5 and *A*2 = −6.667. Thus, - -$$ -v(t) = (5 \cos 6t - 6.667 \sin 6t)e^{-8t} -$$ - (8.5.12) - -Notice that by increasing the value of *R*, the degree of damping decreases and the responses differ. Figure 8.14 plots the three cases. - -For Example 8.5: responses for three degrees of damping. - -Practice Problem 8.5 - -In Fig. 8.13, let *R* = 2 Ω, *L* = 0.4 H, *C* = 25 mF, *v*(0) = 0, *i*(0) = 50 mA. Find *v*(*t*) for *t* > 0. - -**Answer:** −2*te*−10*t* V. - -Example 8.6 Find *v*(*t*) for *t* > 0 in the *RLC* circuit of Fig. 8.15. - -# **Solution:** - -When *t* < 0, the switch is open; the inductor acts like a short circuit while the capacitor behaves like an open circuit. The initial voltage across the capacitor is the same as the voltage across the 50-Ω resistor; that is, - -$$ -v(0) = \frac{50}{30 + 50}(40) = \frac{5}{8} \times 40 = 25 \text{ V} -$$ - (8.6.1) - -The initial current through the inductor is - -$$ -i(0) = -\frac{40}{30 + 50} = -0.5 \text{ A} -$$ - -The direction of *i* is as indicated in Fig. 8.15 to conform with the direc tion of *I*0 in Fig. 8.13, which is in agreement with the convention t hat current flows into the positive terminal of an inductor (see Fig. 6.23). We need to express this in terms of *dv*∕*dt*, since we are looking for *v*. - -need to express this in terms of -$$ -dv/dt -$$ -, since we are looking for *v*. -\n -$$ -\frac{dv(0)}{dt} = -\frac{v(0) + Ri(0)}{RC} = -\frac{25 - 50 \times 0.5}{50 \times 20 \times 10^{-6}} = 0 -$$ -\n(8.6.2) - -When *t* > 0, the switch is closed. The voltage source along with the 30-Ω resistor is separated from the rest of the circuit. The parallel *RLC* circuit acts independently of the voltage source, as illustrated in Fig. 8.16. Next, we determine that the roots of the characteristic equation are - -circuit acts independently of the voltage source, as illustrated in -\nNext, we determine that the roots of the characteristic equation -\n -$$ -\alpha = \frac{1}{2RC} = \frac{1}{2 \times 50 \times 20 \times 10^{-6}} = 500 -$$ -\n -$$ -\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{0.4 \times 20 \times 10^{-6}} = 354 -$$ -\n -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} -$$ -\n -$$ -= -500 \pm \sqrt{250,000 - 124,997.6} = -500 \pm 354 -$$ - -or - -$$ -s_1 = -854, \qquad s_2 = -146 -$$ - -The circuit in Fig. 8.15 when *t* > 0. The parallel *RLC* circuit on the right-hand side acts independently of the circuit on the left-hand side of the junction. - -Since *α* > *ω*0, we have the overdamped response - -$$ -v(t) = A_1 e^{-854t} + A_2 e^{-146t} -$$ - (8.6.3) - -At *t* = 0, we impose the condition in Eq. (8.6.1), - -$$ -v(0) = 25 = A_1 + A_2 -$$ - $\Rightarrow$ $A_2 = 25 - A_1$ (8.6.4) - -Taking the derivative of *v*(*t*) in Eq. (8.6.3), - -$$ -\frac{dv}{dt} = -854A_1e^{-854t} - 146A_2e^{-146t} -$$ - -Imposing the condition in Eq. (8.6.2), - -$$ -\frac{dv(0)}{dt} = 0 = -854A_1 - 146A_2 -$$ - -or - -$$ -0 = 854A_1 + 146A_2 \tag{8.6.5} -$$ - -Solving Eqs. (8.6.4) and (8.6.5) gives - -$$ -A_1 = -5.156, \qquad A_2 = 30.16 -$$ - -Thus, the complete solution in Eq. (8.6.3) becomes - -$$ -v(t) = -5.156e^{-854t} + 30.16e^{-146t} -$$ - V - -Refer to the circuit in Fig. 8.17. Find *v*(*t*) for *t* > 0. - -**Answer:** 50(*e*−10*t* − *e*−2.5*t* ) V. - -# **8.5** Step Response of a Series RLC Circuit - -As we learned in the preceding chapter, the step response is obtained by the sudden application of a dc source. Consider the series *RLC* circuit shown in Fig. 8.18. Applying KVL around the loop for *t* > 0, - -$$ -L\frac{di}{dt} + Ri + v = V_s \tag{8.39} -$$ - -# **Figure 8.17** For Practice Prob. 8.6. - -Vs R L C i t = 0 *v* + ‒ + ‒ - -**Figure 8.18** Step voltage applied to a series *RLC* circuit. - -But - -$$ -i = C \frac{dv}{dt} -$$ - -Substituting for *i* in Eq. (8.39) and rearranging terms, - -$$ -\frac{d^2v}{dt^2} + \frac{R}{L}\frac{dv}{dt} + \frac{v}{LC} = \frac{V_s}{LC} -$$ - (8.40) - -which has the same form as Eq. (8.4). More specifically, the coefficients are the same (and that is important in determining the frequency parameters) but the variable is different. (Likewise, see Eq. (8.47).) Hence, the characteristic equation for the series *RLC* circuit is not af fected by the presence of the dc source. - -The solution to Eq. (8.40) has two components: the transient re sponse *vt*(*t*) and the steady-state response *vss*(*t*); that is, - -$$ -v(t) = v_t(t) + v_{ss}(t) -$$ - (8.41) - -The transient response *vt*(*t*) is the component of the total response that dies out with time. The form of the transient response is the same as the form of the solution obtained in Section 8.3 for the source-free circuit, given by Eqs. (8.14), (8.21), and (8.26). Therefore, the transient response *vt*(*t*) for the overdamped, underdamped, and critically damped cases are: - -$$ -v_t(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} \qquad \text{(Overdamped)} \tag{8.42a} -$$ - -$$ -v_t(t) = (A_1 + A_2 t)e^{-\alpha t} -$$ - (Critically damped) (8.42b) - -$$ -v_t(t) = (A_1 \cos \omega_d t + A_2 \sin \omega_d t)e^{-\alpha t} -$$ - (Underdamped) (8.42c) - -The steady-state response is the final value of *v*(*t*). In the circuit in Fig. 8.18, the final value of the capacitor voltage is the same as the source voltage *Vs*. Hence, - -$$ -v_{ss}(t) = v(\infty) = V_s \tag{8.43} -$$ - -Thus, the complete solutions for the o verdamped, underdamped, and critically damped cases are: - -| v(t) = Vs + A1es1t
+ A2es2t
(Overdamped) | (8.44a) | -|------------------------------------------------------------|---------| -| v(t) = Vs + (A1 + A2t)e−αt
(Critically damped) | (8.44b) | -| v(t) = Vs + (A1 cos ωdt + A2 sin ωdt)e−αt
(Underdamped) | (8.44c) | - -The values of the constants *A*1 and *A*2 are obtained from the initial conditions: *v*(0) and *dv*(0)∕*dt*. Keep in mind that *v* and *i* are, respectively, the voltage across the capacitor and the current through the inductor. Therefore, Eq. (8.44) only applies for finding *v*. But once the capacitor voltage *vC* = *v* is known, we can determine *i* = *C dv*∕*dt*, which is the same current through the capacitor , inductor, and resistor. Hence, the v oltage across the resistor is *vR* = *iR*, while the inductor voltage is *vL* = *L di*∕*dt*. - -Alternatively, the complete response for any v ariable *x*(*t*) can be found directly, because it has the general form - -$$ -x(t) = x_{ss}(t) + x_t(t) -$$ - (8.45) - -where the *xss* = *x*(∞) is the final value and *xt*(*t*) is the transient response. The final value is found as in Section 8.2. The transient response has the same form as in Eq. (8.42), and the associated constants are determined from Eq. (8.44) based on the values of *x*(0) and *dx*(0)∕*dt*. - -# **Solution:** - -■ **CASE 1** When *R* = 5 Ω. For *t* < 0, the switch is closed for a long time. The capacitor behaves like an open circuit while the inductor acts like a short circuit. The initial current through the inductor is - -$$ -i(0) = \frac{24}{5+1} = 4 \text{ A} -$$ - -and the initial voltage across the capacitor is the same as the voltage across the 1-Ω resistor; that is, - -$$ -v(0) = 1i(0) = 4 \text{ V} -$$ - -For *t* > 0, the switch is opened, so that we ha ve the 1- Ω resistor disconnected. What remains is the series *RLC* circuit with the v oltage source. The characteristic roots are determined as follows: - -$$ -\alpha = \frac{R}{2L} = \frac{5}{2 \times 1} = 2.5, \qquad \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 0.25}} = 2 -$$ -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -1, -4 -$$ - -Since *α* > *ω*0, we have the overdamped natural response. The total response is therefore - -$$ -v(t) = v_{ss} + (A_1 e^{-t} + A_2 e^{-4t}) -$$ - -where *vss* is the steady-state response. It is the final value of the capacitor voltage. In Fig. 8.19, *vf* = 24 V. Thus, - -$$ -v(t) = 24 + (A_1 e^{-t} + A_2 e^{-4t}) -$$ - (8.7.1) - -We now need to find *A*1 and *A*2 using the initial conditions. - -$$ -v(0) = 4 = 24 + A_1 + A_2 -$$ - -or - -$$ --20 = A_1 + A_2 \tag{8.7.2} -$$ - -The current through the inductor cannot change abruptly and is the same current through the capacitor at *t* = 0+ because the inductor and capacitor are now in series. Hence, - -$$ -i(0) = C \frac{dv(0)}{dt} = 4 \qquad \Rightarrow \qquad \frac{dv(0)}{dt} = \frac{4}{C} = \frac{4}{0.25} = 16 -$$ - -Before we use this condition, we need to take the derivative of *v* in Eq. (8.7.1). - -$$ -\frac{dv}{dt} = -A_1 e^{-t} - 4A_2 e^{-4t} -$$ - (8.7.3) - -At *t* = 0, - -$$ -\frac{dv(0)}{dt} = 16 = -A_1 - 4A_2 \tag{8.7.4} -$$ - -R 1 H - -i - -24 V - -+ ‒ - -Example 8.7 - -t = 0 - -0.25 F 1 Ω - -*v* + ‒ - -From Eqs. (8.7.2) and (8.7.4), *A*1 = −64∕3 and *A*2 = 4∕3. Substituting *A*1 and *A*2 in Eq. (8.7.1), we get - -$$ -v(t) = 24 + \frac{4}{3}(-16e^{-t} + e^{-4t}) \text{ V} -$$ - (8.7.5) - -Since the inductor and capacitor are in series for *t* > 0, the inductor current is the same as the capacitor current. Hence, - -$$ -i(t) = C \frac{dv}{dt} -$$ - -Multiplying Eq. (8.7.3) by *C* = 0.25 and substituting the values of *A*1 and *A*2 gives - -$$ -i(t) = \frac{4}{3}(4e^{-t} - e^{-4t}) \text{ A} -$$ - (8.7.6) - -Note that *i*(0) = 4 A, as expected. - -■ **CASE 2** When *R* = 4 Ω. Again, the initial current through the inductor is - -$$ -i(0) = \frac{24}{4+1} = 4.8 \text{ A} -$$ - -and the initial capacitor voltage is - -$$ -v(0) = 1i(0) = 4.8 \text{ V} -$$ - -For the characteristic roots, - -$$ -\alpha = \frac{R}{2L} = \frac{4}{2 \times 1} = 2 -$$ - -while *ω*0 = 2 remains the same. In this case, *s*1 = *s*2 = −*α* = −2, and we have the critically damped natural response. The total response is therefore - -$$ -v(t) = v_{ss} + (A_1 + A_2 t)e^{-2t} -$$ - -and, as before *vss* = 24 V, - -$$ -v(t) = 24 + (A_1 + A_2 t)e^{-2t} -$$ - (8.7.7) - -To find *A*1 and *A*2, we use the initial conditions. We write - -$$ -v(0) = 4.8 = 24 + A_1 -$$ - $\Rightarrow$ $A_1 = -19.2$ (8.7.8) - -Since *i*(0) = *C dv*(0)∕*dt* = 4.8 or - -$$ -\frac{dv(0)}{dt} = \frac{4.8}{C} = 19.2 -$$ - -From Eq. (8.7.7), - -$$ -\frac{dv}{dt} = (-2A_1 - 2tA_2 + A_2)e^{-2t} -$$ -\n(8.7.9) - -At *t* = 0, - -$$ -\frac{dv(0)}{dt} = 19.2 = -2A_1 + A_2 \tag{8.7.10} -$$ - -From Eqs. (8.7.8) and (8.7.10), *A*1 = −19.2 and *A*2 = −19.2. Thus, Eq. (8.7.7) becomes - -$$ -v(t) = 24 - 19.2(1 + t)e^{-2t} -$$ - V (8.7.11) - -The inductor current is the same as the capacitor current; that is, - -$$ -i(t) = C \frac{dv}{dt} -$$ - -Multiplying Eq. (8.7.9) by *C* = 0.25 and substituting the values of *A*1 and *A*2 gives - -$$ -i(t) = (4.8 + 9.6t)e^{-2t} A -$$ - (8.7.12) - -Note that *i*(0) = 4.8 A, as expected. - -■ **CASE 3** When *R* = 1 Ω. The initial inductor current is - -$$ -i(0) = \frac{24}{1+1} = 12 \text{ A} -$$ - -and the initial voltage across the capacitor is the same as the voltage across the 1-Ω resistor, - -$$ -v(0) = 1i(0) = 12 \text{ V} -$$ -$$ -\alpha = \frac{R}{2L} = \frac{1}{2 \times 1} = 0.5 -$$ - -Since *α* = 0.5 < *ω*0 = 2, we have the underdamped response \_\_\_\_\_\_\_ - -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -0.5 \pm j1.936 -$$ - -The total response is therefore - -$$ -v(t) = 24 + (A_1 \cos 1.936t + A_2 \sin 1.936t)e^{-0.5t} -$$ - (8.7.13) - -We now determine *A*1 and *A*2. We write - -$$ -v(0) = 12 = 24 + A_1 \qquad \Rightarrow \qquad A_1 = -12 \tag{8.7.14} -$$ - -Since *i*(0) = *C dv*(0)∕*dt* = 12, - -$$ -\frac{dv(0)}{dt} = \frac{12}{C} = 48\tag{8.7.15} -$$ - -But - -$$ -\frac{dv}{dt} = e^{-0.5t}(-1.936A_1 \sin 1.936t + 1.936A_2 \cos 1.936t) -$$ -$$ --0.5e^{-0.5t}(A_1 \cos 1.936t + A_2 \sin 1.936t) -$$ -(8.7.16) - -At *t* = 0, - -$$ -\frac{dv(0)}{dt} = 48 = (-0 + 1.936A_2) - 0.5(A_1 + 0) -$$ - -Substituting *A*1 = −12 gives *A*2 = 21.694, and Eq. (8.7.13) becomes - -$$ -v(t) = 24 + (21.694 \sin 1.936t - 12 \cos 1.936t)e^{-0.5t} \text{ V} -$$ - (8.7.17) - -The inductor current is - -$$ -i(t) = C \frac{dv}{dt} -$$ - -Multiplying Eq. (8.7.16) by *C* = 0.25 and substituting the values of *A*1 and *A*2 gives - -$$ -i(t) = (3.1 \sin 1.936t + 12 \cos 1.936t)e^{-0.5t} \text{ A} -$$ - (8.7.18) - -Note that *i*(0) = 12 A, as expected. - -Figure 8.20 plots the responses for the three cases. From this figure, we observe that the critically damped response approaches the step input of 24 V the fastest. - -For Example 8.7: response for three degrees of damping. - -Practice Problem 8.7 Having been in position *a* for a long time, the switch in Fig. 8.21 is moved to position *b* at *t* = 0. Find *v*(*t*) and *vR*(*t*) for *t* > 0. - -**Figure 8.21** For Practice Prob. 8.7. - -**Answer:** 15 − (1.7321 sin 3.464*t* + 3 cos 3.464*t*)*e*−2*t* V, 3.464*e*−2*t* sin 3.464*t* V. - -**Figure 8.22** Parallel *RLC* circuit with an applied current. - -# **8.6** Step Response of a Parallel RLC Circuit - -Consider the parallel *RLC* circuit sho wn in Fig. 8.22. We want to find *i* due to a sudden application of a dc current. Applying KCL at the top node for *t* > 0, - -$$ -\frac{v}{R} + i + C\frac{dv}{dt} = I_s \tag{8.46} -$$ - -$$ -v = L \frac{di}{dt} -$$ - -Substituting for *v* in Eq. (8.46) and dividing by *LC*, we get - -$$ -\frac{d^2i}{dt^2} + \frac{1}{RC}\frac{di}{dt} + \frac{i}{LC} = \frac{I_s}{LC} -$$ - (8.47) - -which has the same characteristic equation as Eq. (8.29). - -The complete solution to Eq. (8.47) consists of the transient response *it*(*t*) and the steady-state response *iss*; that is, - -$$ -i(t) = it(t) + iss(t) -$$ - (8.48) - -The transient response is the same as what we had in Section 8.4. The steady-state response is the final value of *i*. In the circuit in Fig. 8.22, the final value of the current through the inductor is the same as the source current *Is*. Thus, - -$$ -i(t) = I_s + A_1 e^{s_1 t} + A_2 e^{s_2 t} -$$ - (Overdamped) -\n -$$ -i(t) = I_s + (A_1 + A_2 t)e^{-\alpha t} -$$ - (Critically damped) -\n -$$ -i(t) = I_s + (A_1 \cos \omega_d t + A_2 \sin \omega_d t)e^{-\alpha t} -$$ - (Underdamped) (8.49) - -The constants *A*1 and *A*2 in each case can be determined from the initial conditions for *i* and *di*∕*dt*. Again, we should keep in mind that Eq. (8.49) only applies for finding the inductor current *i*. But once the inductor current *iL* = *i* is known, we can find *v* = *L di*∕*dt*, which is the same v oltage across inductor, capacitor, and resistor . Hence, the current through the resistor is *iR* = *v*∕*R*, while the capacitor current is *iC* = *C dv*∕*dt*. Alternatively, the complete response for any variable *x*(*t*) may be found directly, using - -$$ -x(t) = x_{ss}(t) + x_t(t) -$$ - (8.50) - -where *xss* and *xt* are its final value and transient response, respectively. - -# **Solution:** - -For *t* < 0, the switch is open, and the circuit is partitioned into two inde pendent subcircuits. The 4-A current flows through the inductor, so that - -$$ -i(0) = 4 \text{ A} -$$ - -Since 30*u*(− *t*) = 30 when *t* < 0 and 0 when *t* > 0, the voltage source is operative for *t* < 0. The capacitor acts like an open circuit and the voltage across it is the same as the voltage across the 20-Ω resistor connected in parallel with it. By voltage division, the initial capacitor voltage is - -$$ -v(0) = \frac{20}{20 + 20}(30) = 15 \text{ V} -$$ - -For *t* > 0, the switch is closed, and we ha ve a parallel *RLC* circuit with a current source. The voltage source is zero which means it acts like a short-circuit. The tw o 20-Ω resistors are no w in parallel. They are combined to gi ve *R* = 20 ‖ 20 = 10 Ω. The characteristic roots are determined as follows: - -$$ -\alpha = \frac{1}{2RC} = \frac{1}{20 \times 10 \times 8 \times 10^{-3}} = 6.25 -$$ - -\n -$$ -\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{20 \times 8 \times 10^{-3}}} = 2.5 -$$ - -\n -$$ -s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -6.25 \pm \sqrt{39.0625 - 6.25} -$$ - -\n -$$ -= -6.25 \pm 5.7282 -$$ - -or - -$$ -s_1 = -11.978, \qquad s_2 = -0.5218 -$$ - -Since *α* > *ω*0, we have the overdamped case. Hence, - -$$ -i(t) = I_s + A_1 e^{-11.978t} + A_2 e^{-0.5218t} -$$ - (8.8.1) - -where *Is* = 4 is the final value of *i*(*t*). We now use the initial conditions to determine *A*1 and *A*2. At *t* = 0, - -*i*(0) = 4 = 4 + *A*1 + *A*2 ⇒ *A*2 = −*A*1 **(8.8.2)** - -Taking the derivative of *i*(*t*) in Eq. (8.8.1), - -$$ -\frac{di}{dt} = -11.978A_1e^{-11.978t} - 0.5218A_2e^{-0.5218t} -$$ - -so that at *t* = 0, - -$$ -\frac{di(0)}{dt} = -11.978A_1 - 0.5218A_2 \tag{8.8.3} -$$ - -But - -$$ -L\frac{di(0)}{dt} = v(0) = 15 -$$ - $\Rightarrow$ $\frac{di(0)}{dt} = \frac{15}{L} = \frac{15}{20} = 0.75$ - -Substituting this into Eq. (8.8.3) and incorporating Eq. (8.8.2), we get - -$$ -0.75 = (11.978 - 0.5218)A_2 \Rightarrow A_2 = 0.0655 -$$ - -Thus, *A*1 = −0.0655 and *A*2 = 0.0655. Inserting *A*1 and *A*2 in Eq. (8.8.1) gives the complete solution as - -$$ -i(t) = 4 + 0.0655(e^{-0.5218t} - e^{-11.978t}) -$$ - A - -From *i*(*t*), we obtain *v*(*t*) = *L di*∕*dt* and - -$$ -i_R(t) = \frac{v(t)}{20} = \frac{L}{20} \frac{di}{dt} = 0.785e^{-11.978t} - 0.0342e^{-0.5218t} -$$ - A - -Find *i*(*t*) and *v*(*t*) for *t* > 0 in the circuit of Fig. 8.24. - -# **Answer:** 10(1− cos(0.5*t*)) A, 100 sin(0.5*t*) V. - -Practice Problem 8.8 - -‒ - -# **8.7** General Second-Order Circuits - -Now that we have mastered series and parallel *RLC* circuits, we are prepared to apply the ideas to an y second-order circuit having one or more independent sources with constant values. Although the series and parallel *RLC* circuits are the second-order circuits of greatest interest, other second-order circuits including op amps are also useful. Given a secondorder circuit, we determine its step response *x*(*t*) (which may be voltage or current) by taking the following four steps: - -- 1. We first determine the initial conditions *x*(0) and *dx*(0)∕*dt* and the final value *x*(∞), as discussed in Section 8.2. -- 2. We turn off the independent sources and find the form of the transient response *xt* (*t*) by applying KCL and KVL. Once a second-order differential equation is obtained, we determine its characteristic roots. Depending on whether the response is overdamped, critically damped, or underdamped, we obtain *xt*(*t*) with two unknown constants as we did in the previous sections. -- 3. We obtain the steady-state response as - -$$ -x_{ss}(t) = x(\infty) \tag{8.51} -$$ - -where *x*(∞) is the final value of *x*, obtained in step 1. - -4. The total response is now found as the sum of the transient response and steady-state response - -$$ -x(t) = x_t(t) + x_{ss}(t) -$$ - (8.52) - -We finally determine the constants associated with the transient response by imposing the initial conditions *x*(0) and *dx*(0)∕*dt*, determined in step 1. - -We can apply this general procedure to find the step response of any second-order circuit, including those with op amps. The following examples illustrate the four steps. - -Find the complete response *v* and then *i* for *t* > 0 in the circuit of Fig. 8.25. - -# **Solution:** - -We first find the initial and final values. At *t* = 0−, the circuit is at steady state. The switch is open; the equivalent circuit is shown in Fig. 8.26(a). It is evident from the figure that - -$$ -v(0^-) = 12 \text{ V}, \qquad i(0^-) = 0 -$$ - -At *t* = 0+, the switch is closed; the equivalent circuit is in Fig. 8.26(b). By the continuity of capacitor voltage and inductor current, we know that - -$$ -v(0^+) = v(0^-) = 12 \text{ V}, \qquad i(0^+) = i(0^-) = 0 -$$ - (8.9.1) - -# **Figure 8.24** For Practice Prob. 8.8. - - A circuit may look complicated at first. But once the sources are turned off in an attempt to find the form of the transient response, it may be reducible to a first-order circuit, when the storage elements can be combined, or to a parallel/series RLC circuit. If it is reducible to a first-order circuit, the solution becomes simply what we had in Chapter 7. If it is reducible to a parallel or series RLC circuit, we apply the techniques of previous sections in this chapter. - - Problems in this chapter can also be solved by using Laplace transforms, which are covered in Chapters 15 and 16. - -# Example 8.9 - -Equivalent circuit of the circuit in Fig. 8.25 for: (a) *t* < 0, (b) *t* > 0. - -Obtaining the form of the transient response for Example 8.9. - -To get *dv*(0+)∕*dt*, we use *C dv*∕*dt* = *iC* or *dv*∕*dt* = *iC*∕*C*. Applying KCL at node *a* in Fig. 8.26(b), - -$$ -i(0^{+}) = i_{C}(0^{+}) + \frac{v(0^{+})}{2} -$$ -$$ -0 = i_{C}(0^{+}) + \frac{12}{2} \implies i_{C}(0^{+}) = -6 \text{ A} -$$ - -Hence, - -$$ -\frac{dv(0^{+})}{dt} = \frac{-6}{0.5} = -12 \text{ V/s} -$$ - (8.9.2) - -The final values are obtained when the inductor is replaced by a short circuit and the capacitor by an open circuit in Fig. 8.26(b), giving - -$$ -i(\infty) = \frac{12}{4+2} = 2 \text{ A}, \qquad v(\infty) = 2i(\infty) = 4 \text{ V} -$$ - (8.9.3) - -Next, we obtain the form of the transient response for *t* > 0. By turning off the 12-V voltage source, we have the circuit in Fig. 8.27. Applying KCL at node *a* in Fig. 8.27 gives - -$$ -i = \frac{v}{2} + \frac{1}{2} \frac{dv}{dt} -$$ - (8.9.4) - -Applying KVL to the left mesh results in - -$$ -4i + 1\frac{di}{dt} + v = 0 -$$ - (8.9.5) - -Since we are interested in *v* for the moment, we substitute *i* from Eq. (8.9.4) into Eq. (8.9.5). We obtain - -$$ -2v + 2\frac{dv}{dt} + \frac{1}{2}\frac{dv}{dt} + \frac{1}{2}\frac{d^2v}{dt^2} + v = 0 -$$ - -or - -$$ -\frac{d^2v}{dt^2} + 5\frac{dv}{dt} + 6v = 0 -$$ - -From this, we obtain the characteristic equation as - -$$ -s^2 + 5s + 6 = 0 -$$ - -with roots *s* = −2 and *s* = −3. Thus, the natural response is - -$$ -v_n(t) = Ae^{-2t} + Be^{-3t} -$$ - (8.9.6) - -where *A* and *B* are unknown constants to be determined later. The steadystate response is - -$$ -v_{ss}(t) = v(\infty) = 4 \tag{8.9.7} -$$ - -The complete response is - -$$ -v(t) = v_t + v_{ss} = 4 + Ae^{-2t} + Be^{-3t} -$$ - (8.9.8) - -We now determine *A* and *B* using the initial values. From Eq. (8.9.1), *v*(0) = 12. Substituting this into Eq. (8.9.8) at *t* = 0 gives - -$$ -12 = 4 + A + B \qquad \Rightarrow \qquad A + B = 8 \tag{8.9.9} -$$ - -Taking the derivative of *v* in Eq. (8.9.8), - -$$ -\frac{dv}{dt} = -2Ae^{-2t} - 3Be^{-3t} -$$ - (8.9.10) - -Substituting Eq. (8.9.2) into Eq. (8.9.10) at *t* = 0 gives - -$$ --12 = -2A - 3B \qquad \Rightarrow \qquad 2A + 3B = 12 \tag{8.9.11} -$$ - -From Eqs. (8.9.9) and (8.9.11), we obtain - -$$ -A = 12, \qquad B = -4 -$$ - -so that Eq. (8.9.8) becomes - -$$ -v(t) = 4 + 12e^{-2t} - 4e^{-3t} \text{ V}, \qquad t > 0 \tag{8.9.12} -$$ - -From *v*, we can obtain other quantities of interest by referring to Fig. 8.26(b). To obtain *i*, for example, - -$$ -i = \frac{v}{2} + \frac{1}{2} \frac{dv}{dt} = 2 + 6e^{-2t} - 2e^{-3t} - 12e^{-2t} + 6e^{-3t} -$$ - -= 2 - 6e^{-2t} + 4e^{-3t} A, \t t > 0 (8.9.13) - -Notice that *i*(0) = 0, in agreement with Eq. (8.9.1). - -Determine *v* and *i* for *t* > 0 in the circuit of Fig. 8.28. (See comments Practice Problem 8.9 about current sources in Practice Prob. 7.5.) - -**Answer:** 20(1 − *e*−5*t* ) V, 5(1 − *e*−5*t* ) A. - -**Figure 8.28** For Practice Prob. 8.9. - -Find *vo*(*t*) for *t* > 0 in the circuit of Fig. 8.29. - -# **Solution:** - -This is an example of a second-order circuit with two inductors. We first obtain the mesh currents *i*1 and *i*2, which happen to be the currents through the inductors. We need to obtain the initial and final values of these currents. - -For *t* < 0, 7*u*(*t*) = 0, so that *i*1(0− ) = 0 = *i*2(0− ). For *t* > 0, 7*u*(*t*) = 7, so that the equivalent circuit is as shown in Fig. 8.30(a). Due to the continuity of inductor current, - -$$ -i_1(0^+) = i_1(0^-) = 0, -$$ - $i_2(0^+) = i_2(0^-) = 0$ (8.10.1) - -$$ -v_{L_2}(0^+) = v_o(0^+) = 1[(i_1(0^+) - i_2(0^+)] = 0 \tag{8.10.2} -$$ - -Applying KVL to the left loop in Fig. 8.30(a) at *t* = 0+, - -$$ -7 = 3i_1(0^+) + v_{L_1}(0^+) + v_o(0^+) -$$ - -**Figure 8.29** For Example 8.10. - -Example 8.10 - -**Figure 8.30** Equivalent circuit of that in Fig. 8.29 for: (a) *t* > 0, (b) *t* → ∞. - -$$ -v_{L_1}(0^+) = 7 \, \text{V} -$$ - -Since *L*1 *di*1∕*dt* = *vL*1, - -$$ -\frac{di_1(0^+)}{dt} = \frac{v_{L1}}{L_1} = \frac{7}{\frac{1}{2}} = 14 \text{ A/s} -$$ - (8.10.3) - -Similarly, since *L*2 *di*2∕*dt* = *vL*2, - -$$ -\frac{di_2(0^+)}{dt} = \frac{v_{L2}}{L_2} = 0 -$$ -\n(8.10.4) - -As *t* → ∞, the circuit reaches steady state, and the inductors can be replaced by short circuits, as shown in Fig. 8.30(b). From this figure, - -$$ -i_1(\infty) = i_2(\infty) = \frac{7}{3} A -$$ - (8.10.5) - -Next, we obtain the form of the transient responses by removing the voltage source, as shown in Fig. 8.31. Applying KVL to the two meshes yields - -$$ -4i_1 - i_2 + \frac{1}{2}\frac{di_1}{dt} = 0 -$$ - (8.10.6) - -and - -Obtaining the form of the transient response for Example 8.10. - -$$ -i_2 + \frac{1}{5} \frac{di_2}{dt} - i_1 = 0 \tag{8.10.7} -$$ - -From Eq. (8.10.6), - -$$ -i_2 = 4i_1 + \frac{1}{2} \frac{di_1}{dt} -$$ - (8.10.8) - -Substituting Eq. (8.10.8) into Eq. (8.10.7) gives - -$$ -4i_1 + \frac{1}{2}\frac{di_1}{dt} + \frac{4}{5}\frac{di_1}{dt} + \frac{1}{10}\frac{d^2i_1}{dt^2} - i_1 = 0 -$$ -$$ -\frac{d^2i_1}{dt^2} + 13\frac{di_1}{dt} + 30i_1 = 0 -$$ - -From this we obtain the characteristic equation as - -$$ -s^2 + 13s + 30 = 0 -$$ - -which has roots *s* = −3 and *s* = −10. Hence, the form of the transient response is - -$$ -i_{1n} = Ae^{-3t} + Be^{-10t} -$$ - (8.10.9) - -**Figure 8.31** - -where *A* and *B* are constants. The steady-state response is - -$$ -i_{1ss} = i_1(\infty) = \frac{7}{3} \text{ A} -$$ - (8.10.10) - -From Eqs. (8.10.9) and (8.10.10), we obtain the complete response as - -$$ -i_1(t) = \frac{7}{3} + Ae^{-3t} + Be^{-10t} -$$ - (8.10.11) - -We finally obtain *A* and *B* from the initial values. From Eqs. (8.10.1) and (8.10.11), - -$$ -0 = \frac{7}{3} + A + B \tag{8.10.12} -$$ - -Taking the derivative of Eq. (8.10.11), setting *t* = 0 in the derivative, and enforcing Eq. (8.10.3), we obtain - -$$ -14 = -3A - 10B \tag{8.10.13} -$$ - -From Eqs. (8.10.12) and (8.10.13), *A* = −4∕3 and *B* = −1. Thus, - -$$ -i_1(t) = \frac{7}{3} - \frac{4}{3} e^{-3t} - e^{-10t} -$$ - (8.10.14) - -We no w obtain *i*2 from *i*1. Applying KVL to the left loop in Fig. 8.30(a) gives - -$$ -7 = 4i_1 - i_2 + \frac{1}{2} \frac{di_1}{dt} \qquad \Rightarrow \qquad i_2 = -7 + 4i_1 + \frac{1}{2} \frac{di_1}{dt} -$$ - -Substituting for *i*1 in Eq. (8.10.14) gives - -$$ -i_2(t) = -7 + \frac{28}{3} - \frac{16}{3}e^{-3t} - 4e^{-10t} + 2e^{-3t} + 5e^{-10t} -$$ - -= $\frac{7}{3} - \frac{10}{3}e^{-3t} + e^{-10t}$ (8.10.15) - -From Fig. 8.29, - -**Answer:** 14(*e*−*t* - -$$ -v_o(t) = 1[i_1(t) - i_2(t)] -$$ - (8.10.16) - -Substituting Eqs. (8.10.14) and (8.10.15) into Eq. (8.10.16) yields - -$$ -v_o(t) = 2(e^{-3t} - e^{-10t}) -$$ -\n(8.10.17) - -Note that *vo*(0) = 0, as expected from Eq. (8.10.2). - -) V, *t* > 0. - -− *e*−6*t* - -For *t* > 0, obtain *vo*(*t*) in the circuit of Fig. 8.32. ( *Hint:* First find *v*1 Practice Problem 8.10 and *v*2.) - -**Figure 8.32** For Practice Prob. 8.10. - -The use of op amps in second-order circuits avoids the use of inductors, which are undesirable in some applications. - -# **8.8** Second-Order Op Amp Circuits - -An op amp circuit with tw o storage elements that cannot be combined into a single equi valent element is second-order. Because inductors are bulky and hea vy, they are rarely used in practical op amp circuits. F or this reason, we will only consider *RC* second-order op amp circuits here. Such circuits find a wide range of applications in devices such as filters and oscillators. - -The analysis of a second-order op amp circuit follows the same four steps given and demonstrated in the previous section. - -Example 8.11 In the op amp circuit of Fig. 8.33, find *vo*(*t*) for *t* > 0 when *vs* = 10*u*(*t*) mV. Let *R*1 = *R*2 = 10 kΩ, *C*1 = 20 *μ*F, and *C*2 = 100 *μ*F. - -**Figure 8.33** For Example 8.11. - -# **Solution:** - -Although we could follow the same four steps given in the previous section to solve this problem, we will solve it a little differently. Due to the voltage follower configuration, the voltage across *C*1 is *vo*. Applying KCL at node 1, - -$$ -\frac{v_s - v_1}{R_1} = C_2 \frac{dv_2}{dt} + \frac{v_1 - v_o}{R_2} -$$ - (8.11.1) - -At node 2, KCL gives - -$$ -\frac{v_1 - v_o}{R_2} = C_1 \frac{dv_o}{dt} -$$ - (8.11.2) - -But - -$$ -v_2 = v_1 - v_o \tag{8.11.3} -$$ - -We now try to eliminate *v*1 and *v*2 in Eqs. (8.11.1) to (8.11.3). Substituting Eqs. (8.11.2) and (8.11.3) into Eq. (8.11.1) yields - -$$ -\frac{v_s - v_1}{R_1} = C_2 \frac{dv_1}{dt} - C_2 \frac{dv_o}{dt} + C_1 \frac{dv_o}{dt} -$$ - (8.11.4) - -From Eq. (8.11.2), - -$$ -v_1 = v_o + R_2 C_1 \frac{dv_o}{dt} -$$ - (8.11.5) - -Substituting Eq. (8.11.5) into Eq. (8.11.4), we obtain - -$$ -\frac{v_s}{R_1} = \frac{v_o}{R_1} + \frac{R_2 C_1}{R_1} \frac{dv_o}{dt} + C_2 \frac{dv_o}{dt} + R_2 C_1 C_2 \frac{d^2 v_o}{dt^2} - C_2 \frac{dv_o}{dt} + C_1 \frac{dv_o}{dt} -$$ - -or - -$$ -\frac{d^2v_o}{dt^2} + \left(\frac{1}{R_1C_2} + \frac{1}{R_2C_2}\right)\frac{dv_o}{dt} + \frac{v_o}{R_1R_2C_1C_2} = \frac{v_s}{R_1R_2C_1C_2} -$$ -(8.11.6) - -With the given values of *R*1, *R*2, *C*1, and *C*2, Eq. (8.11.6) becomes - -$$ -\frac{d^2v_o}{dt^2} + 2\frac{dv_o}{dt} + 5v_o = 5v_s -$$ - (8.11.7) - -To obtain the form of the transient response, set *vs* = 0 in Eq. (8.11.7), which is the same as turning off the source. The characteristic equation is - -$$ -s^2 + 2s + 5 = 0 -$$ - -which has complex roots *s*1,2 = −1 ± *j*2. Hence, the form of the transient response is - -$$ -v_{ot} = e^{-t}(A \cos 2t + B \sin 2t) -$$ - (8.11.8) - -where *A* and *B* are unknown constants to be determined. - -As *t* → ∞, the circuit reaches the steady-state condition, and the capacitors can be replaced by open circuits. Since no current flows through *C*1 and *C*2 under steady-state conditions and no current can enter the input terminals of the ideal op amp, current does not flow through *R*1 and *R*2. - -Thus, - -$$ -v_o(\infty) = v_1(\infty) = v_s -$$ - -The steady-state response is then - -$$ -v_{\text{oss}} = v_o(\infty) = v_s = 10 \text{ mV}, \qquad t > 0 \tag{8.11.9} -$$ - -The complete response is - -$$ -v_o(t) = v_{ot} + v_{oss} = 10 + e^{-t}(A \cos 2t + B \sin 2t) \text{ mV} \qquad (8.11.10) -$$ - -To determine *A* and *B*, we need the initial conditions. For *t* < 0, *vs* = 0, so that - -$$ -v_o(0^-) = v_2(0^-) = 0 -$$ - -For *t* > 0, the source is operative. However, due to capacitor voltage continuity, - -$$ -v_o(0^+) = v_2(0^+) = 0 \tag{8.11.11} -$$ - -From Eq. (8.11.3), - -$$ -v_1(0^+) = v_2(0^+) + v_o(0^+) = 0 -$$ - -and, hence, from Eq. (8.11.2), - -$$ -\frac{dv_o(0^+)}{dt} = \frac{v_1 - v_o}{R_2 C_1} = 0 -$$ -\n(8.11.12) - -We now impose Eq. (8.11.11) on the complete response in Eq. (8.11.10) at *t* = 0, for - -$$ -0 = 10 + A \qquad \Rightarrow \qquad A = -10 \tag{8.11.13} -$$ - -Taking the derivative of Eq. (8.11.10), - -\_\_\_ *dvo dt* = *e*−*t* (−*A* cos 2*t* − *B* sin 2*t* − 2*A* sin 2*t* + 2*B* cos 2*t*) - -Setting *t* = 0 and incorporating Eq. (8.11.12), we obtain - -$$ -0 = -A + 2B \tag{8.11.14} -$$ - -From Eqs. (8.11.13) and (8.11.14), *A* = −10 and *B* = −5. Thus, the step response becomes - -$$ -v_o(t) = 10 - e^{-t}(10\cos 2t + 5\sin 2t) \text{ mV}, \qquad t > 0 -$$ - -In the op amp circuit shown in Fig. 8.34, *vs* = 25*u*(*t*) V, find *vo*(*t*) for *t* > 0. Assume that *R*1 = *R*2 = 10 kΩ, *C*1 = 20 *μ*F, and *C*2 = 100 *μ*F. - -**Answer:** -$$ -(25 - 31.25e^{-t} + 6.25e^{-5t}) -$$ - V, $t > 0$ . - -**8.9** PSpice Analysis of RLC Circuits *RLC* circuits can be analyzed with great ease using *PSpice*, just like the - -*RC* or *RL* circuits of Chapter 7. The following two examples will illustrate this. The reader may review Section D.4 in Appendix D on *PSpice* for transient analysis. - -# Example 8.12 - -The input voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.35(b). Use *PSpice* to plot *v*(*t*) for 0 < *t* < 4 s. - -# **Solution:** - -- 1. **Define.** As true with most textbook problems, the problem is clearly defined. -- 2. **Present.** The input is equal to a single square wave of amplitude 12 V with a period of 2 s. We are asked to plot the output, using *PSpice*. -- 3. **Alternative.** Since we are required to use *PSpice*, that is the only alternative for a solution. However, we can check it using the technique illustrated in Section 8.5 (a step response for a series *RLC* circuit). -- 4. **Attempt.** The given circuit is drawn using Schematics as in Fig. 8.36. The pulse is specified using VPWL voltage source, but VPULSE could be used instead. Using the piecewise linear function, we set the attributes of VPWL as T1 = 0, V1 = 0, T2 = 0.001, V2 = 12, and so forth, as shown in Fig. 8.36. Two voltage markers are inserted to plot the input and output voltages. Once the circuit is drawn and the attributes are set, we select **Analysis/Setup/Transient** to open up the *Transient Analysis* dialog box. As a parallel *RLC* circuit, the roots of the characteristic equation are −1 and −9. Thus, we may set *Final Time* as 4 s (four times the magnitude of the lower root). When the schematic - -is saved, we select **Analysis/Simulate** and obtain the plots for the input and output voltages under the *PSpice* A/D window as shown in Fig. 8.37. - - Now we check using the technique from Section 8.5. We can start by realizing the Thevenin equivalent for the resistor-source combination is *V*Th = 12∕2 (the open circuit voltage divides equally across both resistors) = 6 V. The equivalent resistance is 30 Ω (60 ‖ 60). Thus, we can now solve for the response using *R* = 30 Ω, *L* = 3 H, and *C* = (1∕27) F. - -We first need to solve for *α* and *ω*0: - -$$ -\alpha = R/(2L) = 30/6 = 5 -$$ - and $\omega_0 = \frac{1}{\sqrt{3\frac{1}{27}}} = 3$ - - Since 5 is greater than 3, we have the overdamped case \_\_\_\_\_\_ - -*i*(*t*) = *C* - -$$ -s_{1,2} = -5 \pm \sqrt{5^2 - 9} = -1, -9, \qquad v(0) = 0, -$$ - - $v(\infty) = 6 \text{ V}, \qquad i(0) = 0$ - -*dv*(*t*) \_\_\_\_ *dt* , - -where - -where -\n -$$ -v(t) = A_1 e^{-t} + A_2 e^{-9t} + 6 -$$ -\n -$$ -v(0) = 0 = A_1 + A_2 + 6 -$$ -\n -$$ -i(0) = 0 = C(-A_1 - 9A_2) -$$ - -which yields *A*1 = −9*A*2. Substituting this into the above, we get 0 = 9*A*2 − *A*2 + 6, or *A*2 = 0.75 an*d A*1 = −6.75. - -$$ -v(t) = (-6.75e^{-t} + 0.75e^{-9t} + 6)u(t) V -$$ - for all $0 < t < 2$ s. - - At *t* = 1 s, *v*(1) = −6.75*e*−1 + 0.75*e*−9 + 6 = −2.483 + 0.0001 + 6 = 3.552 V. At *t* = 2 s, *v*(2) = −6.75*e* −2 + 0 + 6 = 5.086 V. - - Note that from 2 < *t* < 4 s, *V*Th = 0, which implies that *v*(∞) = 0. Therefore, *v*(*t*) = (*A*3*e*−(*t* − 2) + *A*4*e*−9(*t* − 2))*u*(*t* − 2) V. At *t* = 2 s, *A*3 + *A*4 = 5.086. - -$$ -i(t) = (A_3e^{-(t-2)} - 9A_4e^{-9(t-2)}) -$$ - -\n -$$ -i(t) = \frac{(-A_3e^{-(t-2)} - 9A_4e^{-9(t-2)})}{27} -$$ - -and - -$$ -i(2) = \frac{(6.75e^{-2} - 6.75e^{-18})}{27} = 33.83 \text{ mA} -$$ - -Therefore, −*A*3 − 9*A*4 = 0.9135. - - Combining the two equations, we get −*A*3 − 9(5.086 − *A*3) = 0.9135, which leads to *A*3 = 5.835 and *A*4 = −0.749. - -$$ -v(t) = (5.835e^{-(t-2)} - 0.749e^{-9(t-2)}) u(t-2) -$$ - V - - At *t* = 3 s, *v*(3) = (2.147 − 0) = 2.147 V. At *t* = 4 s, *v*(4) = 0.7897 V. - -- 5. **Evaluate.** A check between the values calculated above and the plot shown in Figure 8.37 shows good agreement within the obvious level of accuracy. -- 6. **Satisfactory?** Yes, we have agreement and the results can be presented as a solution to the problem. - -Practice Problem 8.12 Find *i*(*t*) using *PSpice* for 0 < *t* < 4 s if the pulse voltage in Fig. 8.35(a) is applied to the circuit in Fig. 8.38. - -# **Solution:** - -When the switch is in position *a*, the 6- Ω resistor is redundant. The schematic for this case is shown in Fig. 8.41(a). To ensure that current - -**Figure 8.41** For Example 8.13: (a) for dc analysis, (b) for transient analysis. - -*i*(*t*) enters pin 1, the inductor is rotated three times before it is placed in the circuit. The same applies for the capacitor. We insert pseudo -components VIEWPOINT and IPROBE to determine the initial capacitor voltage and initial inductor current. We carry out a dc *PSpice* analysis by select ing **Analysis/Simulate.** As shown in Fig. 8.41(a), we obtain the initial capacitor voltage as 0 V and the initial inductor current *i*(0) as 4 A from the dc analysis. These initial values will be used in the transient analysis. - -When the switch is moved to position *b*, the circuit becomes a sourcefree parallel *RLC* circuit with the schematic in Fig. 8.41(b). We set the initial condition IC = 0 for the capacitor and IC = 4 A for the inductor. A current marker is inserted at pin 1 of the inductor. We select **Analysis/ Setup/Transient** to open up the *Transient Analysis* dialog box and set *Final Time* to 3 s. After saving the schematic, we select **Analysis/ Transient**. Figure 8.42 sho ws the plot of *i*(*t*). The plot agrees with *i*(*t*) = 4.8*e*−*t* − 0.8*e*−6*t* A, which is the solution by hand calculation. - -Plot of *i*(*t*) for Example 8.13. - -Refer to the circuit in Fig. 8.21 (see Practice Prob. 8.7). Use *PSpice* to Practice Problem 8.13 obtain *v*(*t*) for 0 < *t* < 2. - -**Answer:** See Fig. 8.43. - -# **8.10** Duality - -The concept of duality is a time-sa ving, ef fort-effective measure of solving circuit problems. Consider the similarity between Eqs. (8.4) and (8.29). The two equations are the same, e xcept that we must interchange the following quantities: (1) voltage and current, (2) resistance and conductance, (3) capacitance and inductance. Thus, it sometimes occurs in circuit analysis that two different circuits have the same equations and solutions, except that the roles of certain complementary elements are interchanged. This interchangeability is kno wn as the principle of *duality*. - -The duality principle asserts a parallelism between pairs of characterizing equations and theorems of electric circuits. - -Dual pairs are sho wn in Table 8.1. Note that po wer does not appear in Table 8.1, because power has no dual. The reason for this is the principle of linearity; since power is not linear, duality does not apply. Also notice from Table 8.1 that the principle of duality extends to circuit elements, configurations, and theorems. - -Two circuits that are described by equations of the same form, b ut in which the variables are interchanged, are said to be dual to each other. - -Two circuits are said to be duals of one another if they are described by the same characterizing equations with dual quantities interchanged. - -The usefulness of the duality principle is self-e vident. Once we know the solution to one circuit, we automatically ha ve the solution for the dual circuit. It is ob vious that the circuits in Figs. 8.8 and 8.13 are dual. Consequently , the result in Eq. (8.32) is the dual of that in Eq. (8.11). We must k eep in mind that the method described here for finding a dual is limited to planar circuits. Finding a dual for a nonplanar circuit is be yond the scope of this te xtbook because nonplanar circuits cannot be described by a system of mesh equations. - -To find the dual of a given circuit, we do not need to write do wn the mesh or node equations. We can use a graphical technique. Gi ven a planar circuit, we construct the dual circuit by taking the following three steps: - -- 1. Place a node at the center of each mesh of the given circuit. Place the reference node (the ground) of the dual circuit outside the given circuit. -- 2. Draw lines between the nodes such that each line crosses an ele ment. Replace that element by its dual (see Table 8.1). -- 3. To determine the polarity of voltage sources and direction of current sources, follow this rule: A voltage source that produces a positi ve (clockwise) mesh current has as its dual a current source whose reference direction is from the ground to the nonreference node. - -In case of doubt, one may verify the dual circuit by writing the nodal or mesh equations. The mesh (or nodal) equations of the original circuit are similar to the nodal (or mesh) equations of the dual circuit. The duality principle is illustrated with the following two examples. - -# **TABLE 8.1** - -# Dual pairs. - -| Resistance R | Conductance G | -|----------------|----------------| -| Inductance L | Capacitance C | -| Voltage v | Current i | -| Voltage source | Current source | -| Node | Mesh | -| Series path | Parallel path | -| Open circuit | Short circuit | -| KVL | KCL | -| Thevenin | Norton | -| | | - -Even when the principle of linearity applies, a circuit element or variable may not have a dual. For example, mutual inductance (to be covered in Chapter 13) has no dual. - -Construct the dual of the circuit in Fig. 8.44. - -# **Solution:** - -As shown in Fig. 8.45(a), we first locate nodes 1 and 2 in the two meshes and also the ground node 0 for the dual circuit. We draw a line between one node and another crossing an element. We replace the line joining the nodes by the duals of the elements which it crosses. For example, a line between nodes 1 and 2 crosses a 2-H inductor, and we place a 2-F capacitor (an inductor's dual) on the line. A line between nodes 1 and 0 crossing the 6-V voltage source will contain a 6-A current source. By drawing lines crossing all the elements, we construct the dual circuit on the given circuit as in Fig. 8.45(a). The dual circuit is redrawn in Fig. 8.45(b) for clarity. - -**Figure 8.44** For Example 8.14. - -# **Figure 8.45** - -(a) Construction of the dual circuit of Fig. 8.44, (b) dual circuit redrawn. - -Draw the dual circuit of the one in Fig. 8.46. - -**Answer:** See Fig. 8.47. - -Obtain the dual of the circuit in Fig. 8.48. Example 8.15 - -# **Solution:** - -The dual circuit is constructed on the original circuit as in Fig. 8.49(a). We first locate nodes 1 to 3 and the reference node 0. Joining nodes 1 and 2, we cross the 2-F capacitor, which is replaced by a 2-H inductor. - -Practice Problem 8.14 - -For Example 8.15. - -Joining nodes 2 and 3, we cross the 20- Ω resistor, which is replaced by a \_\_\_1 20 -Ω resistor. We keep doing this until all the elements are crossed. The result is in Fig. 8.49(a). The dual circuit is redrawn in Fig. 8.49(b). - -# **Figure 8.49** - -For Example 8.15: (a) construction of the dual circuit of Fig. 8.48, (b) dual circuit redrawn. - -To v erify the polarity of the v oltage source and the direction of the current source, we may apply mesh currents *i*1, *i*2, and *i*3 (all in the clockwise direction) in the original circuit in Fig. 8.48. The 10-V voltage source produces positive mesh current *i*1, so that its dual is a 10-A current source directed from 0 to 1. Also, *i*3 = −3 A in Fig. 8.48 has as its dual *v*3 = −3 V in Fig. 8.49(b). - -**Answer:** See Fig. 8.51. - -For Practice Prob. 8.15. - -**Figure 8.51** Dual of the circuit in Fig. 8.50. - -# **8.11** Applications - -Practical applications of *RLC* circuits are found in control and communications circuits such as ringing circuits, peaking circuits, resonant circuits, smoothing circuits, and filters. Most of these circuits cannot be covered until we treat ac sources. For now, we will limit ourselves to two simple applications: automobile ignition and smoothing circuits. - -# **8.11.1** Automobile Ignition System - -In Section 7.9.4, we considered the automobile ignition system as a charging system. That was only a part of the system. Here, we consider another part—the voltage generating system. The system is modeled by the circuit shown in Fig. 8.52. The 12-V source is due to the battery and alternator. The 4-Ω resistor represents the resistance of the wiring. The ignition coil is modeled by the 8-mH inductor . The 1-*μ*F capacitor (known as the *condenser* to automechanics) is in parallel with the switch (known as the *breaking points* or *electronic ignition*). In the follo wing example, we determine how the *RLC* circuit in Fig. 8.52 is used in gen erating high voltage. - -Automobile ignition circuit. - -Assuming that the switch in Fig. 8.52 is closed prior to *t* = 0−, find the inductor voltage *vL* for *t* > 0. - -# **Solution:** - -If the switch is closed prior to *t* = 0− and the circuit is in steady state, then - -$$ -i(0^{-}) = \frac{12}{4} = 3 \text{ A}, \qquad v_C(0^{-}) = 0 -$$ - -At *t* = 0+, the switch is opened. The continuity conditions require that - -$$ -i(0^+) = 3 \text{ A}, \qquad v_C(0^+) = 0 -$$ - (8.16.1) - -We obtain *di*(0+)/*dt* from *vL*(0+). Applying KVL to the mesh at *t* = 0+ yields - -$$ --12 + 4i(0^{+}) + v_L(0^{+}) + v_C(0^{+}) = 0 -$$ - -$$ --12 + 4 \times 3 + v_L(0^{+}) + 0 = 0 \implies v_L(0^{+}) = 0 -$$ - -# Example 8.16 - -Hence, - -$$ -\frac{di(0^{+})}{dt} = \frac{v_L(0^{+})}{L} = 0 -$$ -\n(8.16.2) - -As *t* → ∞, the system reaches steady state, so that the capacitor acts like an open circuit. Then - -$$ -i(\infty) = 0 \tag{8.16.3} -$$ - -If we apply KVL to the mesh for *t* > 0, we obtain - -$$ -12 = Ri + L\frac{di}{dt} + \frac{1}{C} \int_0^t i \, dt + v_C(0) -$$ - -Taking the derivative of each term yields - -$$ -\frac{d^2i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{i}{LC} = 0 -$$ - (8.16.4) - -We obtain the form of the transient response by following the procedure in Section 8.3. Substituting *R* = 4 Ω, *L* = 8 mH, and *C* = 1 *μ*F, we get - -$$ -\alpha = \frac{R}{2L} = 250 -$$ -, $\omega_0 = \frac{1}{\sqrt{LC}} = 1.118 \times 10^4$ - -Since *α* < *ω*0, the response is underdamped. The damped natural fre quency is - -$$ -\omega_d = \sqrt{\omega_0^2 - \alpha^2} \simeq \omega_0 = 1.118 \times 10^4 -$$ - -The form of the transient response is - -$$ -it(t) = e^{-\alpha} (A \cos \omega_d t + B \sin \omega_d t) -$$ - (8.16.5) - -where *A* and *B* are constants. The steady-state response is - -$$ -i_{ss}(t) = i(\infty) = 0 \tag{8.16.6} -$$ - -so that the complete response is - -$$ -i(t) = it(t) + iss(t) = e^{-250t}(A \cos 11,180t + B \sin 11,180t) -$$ - (8.16.7) - -We now determine *A* and *B*. - -$$ -i(0) = 3 = A + 0 \qquad \Rightarrow \qquad A = 3 -$$ - -Taking the derivative of Eq. (8.16.7), - -$$ -\frac{di}{dt} = -250e^{-250t}(A \cos 11,180t + B \sin 11,180t) \n+ e^{-250t}(-11,180A \sin 11,180t + 11,180B \cos 11,180t) -$$ - -Setting *t* = 0 and incorporating Eq. (8.16.2), - -$$ -0 = -250A + 11{,}180B \qquad \Rightarrow \qquad B = 0.0671 -$$ - -Thus, - -$$ -i(t) = e^{-250t}(3 \cos 11,180t + 0.0671 \sin 11,180t) -$$ - (8.16.8) - -The voltage across the inductor is then - -$$ -v_L(t) = L\frac{di}{dt} = -268e^{-250t}\sin 11{,}180t -$$ - (8.16.9) - -This has a maximum value when sine is unity, that is, at 11,180*t*0 = *π*∕2 or *t*0 = 140.5 *μ*s. At *ti*me = *t*0, the inductor voltage reaches its peak, which is - -$$ -v_L(t_0) = -268e^{-250t_0} = -259 \text{ V} \tag{8.16.10} -$$ - -Although this is far less than the voltage range of 6000 to 10,000 V required to fire the spark plug in a typical automobile, a device known as a *transformer* (to be discussed in Chapter 13) is used to step up the inductor voltage to the required level. - -In Fig. 8.52, find the capacitor voltage *vC* for *t* > 0. - -**Answer:** 12 − 12*e*−250*t* cos 11,180*t* + 267.7*e*−250*t* sin 11,180*t* V. - -# **8.11.2** Smoothing Circuits - -In a typical digital communication system, the signal to be transmitted is first sampled. Sampling refers to the procedure of selecting samples of a signal for processing, as opposed to processing the entire signal. Each sample is con verted into a binary number represented by a series of pulses. The pulses are transmitted by a transmission line such as a coaxial cable, twisted pair, or optical fiber. At the receiving end, the signal is applied to a digital-to-analog (D/A) con verter whose output is a "staircase" function, that is, constant at each time interv al. In order to recover the transmitted analog signal, the output is smoothed by letting it pass through a "smoothing" circuit, as illustrated in Fig. 8.53. An *RLC* circuit may be used as the smoothing circuit. - -The output of a D/A converter is shown in Fig. 8.54(a). If the *RLC* circuit in Fig. 8.54(b) is used as the smoothing circuit, determine the output - -Practice Problem 8.16 - -# **Figure 8.53** - -A series of pulses is applied to the digitalto-analog (D/A) converter, whose output is applied to the smoothing circuit. - -# Example 8.17 - -**Figure 8.54** For Example 8.17: (a) output of a D/A converter, (b) an *RLC* smoothing circuit. - -# **Solution:** - -voltage *vo*(*t*). - -This problem is best solved using *PSpice*. The schematic is shown in Fig. 8.55(a). The pulse in Fig. 8.54(a) is specified using the piecewise - -**Figure 8.55** For Example 8.17: (a) schematic, (b) input and output voltages. - -linear function. The attributes of V1 are set as T1 = 0, V1 = 0, T2 = 0.001, V2 = 4, T3 = 1, V3 = 4, and so on. To be able to plot both input and output voltages, we insert two voltage markers as shown. We select **Analysis/Setup/Transient** to open up the *Transient Analysis* dialog box and set *Final Time* as 6 s. Once the schematic is saved, we select **Analysis/Simulate** to run and obtain the plots shown in Fig. 8.55(b). - -Practice Problem 8.17 Rework Example 8.17 if the output of the D/A converter is as shown in Fig. 8.56. - -**Answer:** See Fig. 8.57. - -# **8.12** Summary - -- 1. The determination of the initial v alues *x*(0) and *dx*(0)∕*dt* and final value *x*(∞) is crucial to analyzing second-order circuits. -- 2. The *RLC* circuit is second-order because it is described by a second-order dif ferential equation. Its characteristic equation is - -*s* 2 + 2*αs* + *ω*0 2 = 0, where *α* is the neper frequenc y and *ω*0 is the undamped natural frequency. For a series circuit, *α* = *R*∕2*L*, for a parallel circuit *α* = 1∕2*RC*, and for both cases *ω*0 = 1∕ √ \_\_\_ *LC* . - -- 3. If there are no independent sources in the circuit after switching (or sudden change), we regard the circuit as source-free. The complete solution is the natural response. -- 4. The natural response of an *RLC* circuit is o verdamped, under damped, or critically damped, depending on the roots of the characteristic equation. The response is critically damped when the roots are equal (*s*1 = *s*2 or *α* = *ω*0), overdamped when the roots are real and unequal (*s*1 ≠ *s*2 or *α* > *ω*0), or underdamped when the roots are complex conjugate (*s*1 = *s*2 \* or *α* < *ω*0). -- 5. If independent sources are present in the circuit after switching, the complete response is the sum of the transient response and the steady-state response. -- 6. *PSpice* is used to analyze *RLC* circuits in the same w ay as for *RC* or *RL* circuits. -- 7. Two circuits are dual if the mesh equations that describe one circuit have the same form as the nodal equations that describe the other. The analysis of one circuit gives the analysis of its dual circuit. -- 8. The automobile ignition circuit and the smoothing circuit are typical applications of the material covered in this chapter. - -# Review Questions - -**8.1** For the circuit in Fig. 8.58, the capacitor voltage at *t* = 0− (just before the switch is closed) is: - -# **Figure 8.58** - -For Review Questions 8.1 and 8.2. - -**8.2** For the circuit in Fig. 8.58, the initial inductor current (at *t* = 0) is: - -(a) 0 A (b) 2 A (c) 6 A (d) 12 A - -- **8.3** When a step input is applied to a second-order circuit, the final values of the circuit variables are found by: - - (a) Replacing capacitors with closed circuits and inductors with open circuits. - - (b) Replacing capacitors with open circuits and inductors with closed circuits. - - (c) Doing neither of the above. - -**8.4** If the roots of the characteristic equation of an *RLC* circuit are −2 and −3, the response is: - -> (a) (*A* cos 2*t* + *B* sin 2*t*)*e*−3*t* (b) (*A* + 2*Bt*)*e*−3*t* (c) *Ae*−2*t* + *Bte*−3*t* (d) *Ae*−2*t* + *Be*−3*t* - -where *A* and *B* are constants. - -- **8.5** In a series *RLC* circuit, setting *R* = 0 will produce: - - (a) an overdamped response - - (b) a critically damped response - - (c) an underdamped response - - (d) an undamped response - - (e) none of the above -- **8.6** A parallel *RLC* circuit has *L* = 2 H and *C* = 0.25 F. The value of *R* that will produce a unity neper frequency is: - -(a) 0.5 Ω (b) 1 Ω (c) 2 Ω (d) 4 Ω - -- **8.7** Refer to the series *RLC* circuit in Fig. 8.59. What kind of response will it produce? - - (a) overdamped - - (b) underdamped - - (c) critically damped - - (d) none of the above - -C1 - -i - - - -# **Figure 8.59** - -For Review Question 8.7. - -- **8.8** Consider the parallel *RLC* circuit in Fig. 8.60. What type of response will it produce? - - (a) overdamped - - (b) underdamped - - (c) critically damped - - (d) none of the above - -For Review Question 8.8. - -- **8.9** Match the circuits in Fig. 8.61 with the following items: - - (i) first-order circuit - - (ii) second-order series circuit - - (iii) second-order parallel circuit - - (iv) none of the above - -## *v*s R (c) i s C2 C1 L R1 (d) C2 R2 + ‒ R1 R2 - -# **Figure 8.61** - -For Review Question 8.9. - -**8.10** In an electric circuit, the dual of resistance is: - -| (a) conductance | (b) inductance | -|-------------------|------------------| -| (c) capacitance | (d) open circuit | -| (e) short circuit | | - -*Answers: 8.1a, 8.2c, 8.3b, 8.4d, 8.5d, 8.6c, 8.7b, 8.8b, 8.9 (i)-c, (ii)-b, e, (iii)-a, (iv)-d, f, 8.10a.* - -# Problems - -# Section 8.2 Finding Initial and Final Values - -**8.1** For the circuit in Fig. 8.62, find: - -(a) *i*(0+) and *v*(0+), (b) *di*(0+)∕*dt* and *dv*(0+)∕*dt*, - -(c) *i*(∞) and *v*(∞). - -**Figure 8.62** For Prob. 8.1. - -**Figure 8.63** For Prob. 8.2. - -**8.3** Refer to the circuit shown in Fig. 8.64. Calculate: - -(a) *iL*(0+), *vC*(0+), and *vR*(0+), (b) *diL*(0+)∕*dt*, *dvC*(0+)∕*dt*, and *dvR*(0+)∕*dt*, (c) *iL*(∞), *vC*(∞), and *vR*(∞). - -**Figure 8.64** - -- For Prob. 8.3. - - **8.4** In the circuit of Fig. 8.65, find: (a) *v*(0+) and *i*(0+), (b) *dv*(0+)∕*dt* and *di*(0+)∕*dt*, (c) *v*(∞) and *i*(∞). - -**Figure 8.65** - -For Prob. 8.4. - -- **8.5** Refer to the circuit in Fig. 8.66. Determine: - - (a) *i*(0+) and *v*(0+), (b) *di*∕(0+)*dt* and *dv*(0+)∕*dt*, - - (c) *i*(∞) and *v*(∞). - -For Prob. 8.5. - -**8.6** In the circuit of Fig. 8.67, find: - -(a) *vR*(0+) and *vL*(0+), (b) *dvR*(0+)∕*dt* and *dvL*(0+)∕*dt*, (c) *vR*(∞) and *vL*(∞). - -# **Figure 8.67** - -For Prob. 8.6. - -# Section 8.3 Source-Free Series RLC Circuit - -- **8.7** A series *RLC* circuit has *R* = 20 kΩ, *L* = 0.2 mH, and *C* = 5 *μ*F. What type of damping is exhibited by the circuit? -- **8.8** Design a problem to help other students better understand source-free *RLC* circuits. -- **8.9** The current in an *RLC* circuit is described by - -$$ -\frac{d^2i}{dt^2} + 10\frac{di}{dt} + 25i = 0 -$$ - -If *i*(0) = 10 A and *di*(0)∕*dt* = 0, find *i*(*t*) for *t* > 0. - -**8.10** The differential equation that describes the current in an *RLC* network is - -$$ -3\frac{d^2i}{dt^2} + 15\frac{di}{dt} + 12i = 0 -$$ - -Given that *i*(0) = 0, *di*(0)∕*dt* = 6 mA/s, obtain *i*(*t*). - -**8.11** The natural response of an *RLC* circuit is described by the differential equation - -$$ -\frac{d^2v}{dt^2} + 2\frac{dv}{dt} + v = 0 -$$ - - for which the initial conditions are *v*(0) = 10 V and *dv*(0)∕*dt* = 0. Solve for *v*(*t*). - -- **8.12** If *R* = 50 Ω, *L* = 1.5 H, what value of *C* will make an *RLC* series circuit: - - (a) overdamped, - - (b) critically damped, - - (c) underdamped? -- **8.13** For the circuit in Fig. 8.68, calculate the value of *R* needed to have a critically damped response. - -**Figure 8.68** For Prob. 8.13. - -**8.14** The switch in Fig. 8.69 moves from position *A* to position *B* at *t* = 0 (please note that the switch must connect to point *B* before it breaks the connection at *A*, a make-before-break switch). Let *v*(0) = 0, find *v*(*t*) for *t* > 0. - -**8.15** The responses of a series *RLC* circuit are - -$$ -v_C(t) = 30 - 10e^{-20t} + 30e^{-10t} -$$ -V -$$ -i_L(t) = 40e^{-20t} - 60e^{-10t} -$$ - mA - - where *vC* and *iL* are the capacitor voltage and inductor current, respectively. Determine the values of *R*, *L*, and *C*. - -**8.16** Find *i*(*t*) for *t* > 0 in the circuit of Fig. 8.70. - -**Figure 8.70** - -For Prob. 8.16. - -**8.17** In the circuit of Fig. 8.71, the switch instantaneously moves from position *A* to *B* at *t* = 0. Find *v*(*t*) for all *t* ≥ 0. - -**8.18** Find the voltage across the capacitor as a function of time for *t* > 0 for the circuit in Fig. 8.72. Assume steady-state conditions exist at *t* = 0−. - -**Figure 8.72** - -For Prob. 8.18. - -**8.19** Obtain *v*(*t*) for *t* > 0 in the circuit of Fig. 8.73. - -**Figure 8.73** - -For Prob. 8.19. - -**8.20** The switch in the circuit of Fig. 8.74 has been closed for a long time but is opened at *t* = 0. Determine *i*(*t*) for *t* > 0. - -# **Figure 8.74** - -For Prob. 8.20. - -\***8.21** Calculate *v*(*t*) for *t* > 0 in the circuit of Fig. 8.75. - -**Figure 8.75** For Prob. 8.21. - -\* An asterisk indicates a challenging problem. - -# Section 8.4 Source-Free Parallel RLC Circuit - -**8.22** Assuming *R* = 2 kΩ, design a parallel *RLC* circuit that has the characteristic equation - -$$ -s^2 + 100s + 10^6 = 0. -$$ - -**8.23** For the network in Fig. 8.76, what value of *C* is needed to make the response underdamped with unity neper frequency (*α* = 1)? - -# **Figure 8.76** For Prob. 8.23. - -**8.24** The switch in Fig. 8.77 moves from position A to position *B* at *t* = 0 (please note that the switch must connect to point *B* before it breaks the connection at *A*, a make-before-break switch). Determine *i*(*t*) for *t* > 0. - -**Figure 8.77** For Prob. 8.24. - -**8.25** Using Fig. 8.78, design a problem to help other students better understand source-free *RLC* circuits. - -**Figure 8.78** For Prob. 8.25. - -# Section 8.5 Step Response of a Series RLC Circuit - -**8.26** The step response of an *RLC* circuit is given by - -$$ -\frac{d^2i}{dt^2} + 2\frac{di}{dt} + 5i = 10 -$$ - -Given that *i*(0) = 2 and *di*(0)∕*dt* = 4, solve for *i*(*t*). - -**8.27** A branch voltage in an *RLC* circuit is described by - -$$ -\frac{d^2v}{dt^2} + 4\frac{dv}{dt} + 8v = 24 -$$ - - If the initial conditions are *v*(0) = 0 = *dv*(0)∕*dt*, find *v*(*t*). - -**8.28** A series *RLC* circuit is described by - -$$ -L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{i}{C} = 10 -$$ - - Find the response when *L* = 0.5 H, *R* = 4 Ω, and *C* = 0.2 F. Let *i*(0) = 1, *di*(0)∕*dt* = 0. - -**8.29** Solve the following differential equations subject to the specified initial conditions - -(a) -$$ -d^2v/dt^2 + 4v = 12 -$$ -, $v(0) = 0$ , $dv(0)/dt = 2$ -\n(b) $d^2i/dt^2 + 5 \frac{di}{dt} + 4i = 8$ , $i(0) = -1$ , -\n $\frac{di(0)}{dt} = 0$ -\n(c) $\frac{d^2v}{dt^2} + 2 \frac{dv}{dt} + v = 3$ , $v(0) = 5$ , -\n $\frac{dv(0)}{dt} = 1$ -\n(d) $\frac{d^2i}{dt^2} + 2 \frac{di}{dt} + 5i = 10$ , $i(0) = 4$ , -\n $\frac{di(0)}{dt} = -2$ - -**8.30** The step responses of a series *RLC* circuit are - -$$ -v_C = 40 - 10e^{-2000t} - 10e^{-4000t} \text{ V}, \qquad t > 0 -$$ - -$$ -i_L(t) = 3e^{-2000t} + 6e^{-4000t} \text{ mA}, \qquad t > 0 -$$ - -- (a) Find *C*. (b) Determine what type of damping is exhibited by the circuit. -- **8.31** Consider the circuit in Fig. 8.79. Find *vL*(0+) and *vC*(0+). - -For Prob. 8.31. - -**Figure 8.80** For Prob. 8.32. - -**Figure 8.81** - -For Prob. 8.33. - -**8.34** Calculate *i*(*t*) for *t* > 0 in the circuit of Fig. 8.82. - -For Prob. 8.34. - -**8.35** Using Fig. 8.83, design a problem to help other students better understand the step response of series *RLC* circuits. - -# **Figure 8.83** - -For Prob. 8.35. - -**8.36** Obtain *v*(*t*) and *i*(*t*) for *t* > 0 in the circuit of Fig. 8.84. - -**Figure 8.84** For Prob. 8.36. - -\***8.37** For the network in Fig. 8.85, solve for *i*(*t*) for *t* > 0. - -# **Figure 8.85** - -For Prob. 8.37. - -**8.38** Refer to the circuit in Fig. 8.86. Calculate *i*(*t*) for *t* > 0. - -# **Figure 8.86** For Prob. 8.38. - -# **Figure 8.87** - -For Prob. 8.39. - -**8.40** The switch in the circuit of Fig. 8.88 is moved from position *a* to *b* at *t* = 0. Assume that the voltage across the capacitor is equal to zero at *t* = 0 and that the switch is a make before break switch. Determine *i*(*t*) for all *t* > 0. - -**\*8.41** For the network in Fig. 8.89, find *i*(*t*) for *t* > 0. - -**Figure 8.89** For Prob. 8.41. - -**\*8.42** Given the network in Fig. 8.90, find *v*(*t*) for *t* > 0. - -- For Prob. 8.42. - - **8.43** The switch in Fig. 8.91 is opened at *t* = 0 after the circuit has reached steady state. Choose *R* and *C* such that *α* = 8 Np/s and *ωd* = 30 rad/s. - -**Figure 8.91** - -For Prob. 8.43. - -**8.44** A series *RLC* circuit has the following parameters: *R* = 1 kΩ, *L* = 1 H, and *C* = 10 nF. What type of damping does this circuit exhibit? - -# Section 8.6 Step Response of a Parallel RLC Circuit - -**8.45** In the circuit of Fig. 8.92, find *v*(*t*) and *i*(*t*) for *t* > 0. - -For Prob. 8.45. - -**8.46** Using Fig. 8.93, design a problem to help other students better understand the step response of a parallel *RLC* circuit. - -# **Figure 8.93** - -For Prob. 8.46. - -**8.47** Find the output voltage *vo*(*t*) in the circuit of Fig. 8.94. - -# **Figure 8.94** - -For Prob. 8.47. - -**8.48** Given the circuit in Fig. 8.95, find *i*(*t*) and *v*(*t*) for *t* > 0. - -For Prob. 8.48. - -**8.49** Determine *i*(*t*) for *t* > 0 in the circuit of Fig. 8.96. - -**Figure 8.97** - -For Prob. 8.50. - -**8.51** Find *v*(*t*) for *t* > 0 in the circuit of Fig. 8.98. - -# **Figure 8.98** - -For Prob. 8.51. - -**8.52** The step response of a parallel *RLC* circuit is *v* = 10 + 20*e*−300*t* (cos 400*t* − 2 sin 400*t*) V, *t* ≥ 0 when the inductor is 25 mH. Find *R* and *C*. - -# Section 8.7 General Second-Order Circuits - -**8.53** After being open for a day, the switch in the circuit of Fig. 8.99 is closed at *t* = 0. Find the differential equation describing *i*(*t*), *t* > 0. - -# **Figure 8.99** - -For Prob. 8.53. - -**8.55** For the circuit in Fig. 8.101, find *v*(*t*) for *t* > 0. Assume that *i*(0+) = 2 A. - -# **Figure 8.101** - -For Prob. 8.55. - -**8.56** In the circuit of Fig. 8.102, find *i*(*t*) for *t* > 0. - -# **Figure 8.102** - -For Prob. 8.56. - -**8.57** Given the circuit shown in Fig. 8.103, determine the characteristic equation of the circuit and the values for *i*(*t*) and *v*(*t*) for all *t* > 0. - -# **Figure 8.103** - -For Prob. 8.57. - -- **8.58** In the circuit of Fig. 8.104, the switch has been in position 1 for a long time but moved to position 2 at *t* = 0. Find: - - (a) *v*(0+), *dv*(0+)∕*dt*, (b) *v*(*t*) for *t* ≥ 0. - -**Figure 8.104** For Prob. 8.58. - -**8.59** The switch in Fig. 8.105 has been in position 1 for *t* < 0. At *t* = 0, it is moved from position 1 to the top of the capacitor at *t* = 0. Please note that the switch is a make before break switch, it stays in contact with position 1 until it makes contact with the top of the capacitor and then breaks the contact at position 1. Given that the initial voltage across the capacitor is equal to zero, determine *v*(*t*). - -# **Figure 8.105** - -For Prob. 8.59. - -# **Figure 8.106** - -For Prob. 8.60. - -- **8.61** For the circuit in Prob. 8.5, find *i* and *v* for *t* > 0. -- **8.62** Find the response *vR*(*t*) for *t* > 0 in the circuit of Fig. 8.107. Let *R* = 8 Ω, *L* = 2 H, and *C* = 125 mF. - -# **Figure 8.107** - -For Prob. 8.62. - -# Section 8.8 Second-Order Op Amp Circuits - -**Figure 8.108** For Prob. 8.63. - -- **8.64** Using Fig. 8.109, design a problem to help other students better understand second-order op amp circuits. - -**8.65** Determine the differential equation for the op amp circuit in Fig. 8.110. If *v*1(0+) = 2 V and *v*2(0+) = 0 V, find *vo* for *t* > 0. Let *R* = 100 kΩ and *C* = 1 *μ*F. - -**Figure 8.110** For Prob. 8.65. - -**8.66** Obtain the differential equations for *vo*(*t*) in the op amp circuit of Fig. 8.111. - -**\*8.67** In the op amp circuit of Fig. 8.112, determine *vo*(*t*) for *t* > 0. Let *v*in = *u*(*t*) V, *R*1 = *R*2 = 10 kΩ, *C*1 = *C*2 = 100 *μ*F. - -# **Figure 8.112** - -**8.68** For the step function *vs* = *u*(*t*), use *PSpice* or *MultiSim* to find the response *v*(*t*) for 0 < *t* < 6 s in the circuit of Fig. 8.113. - -# **Figure 8.113** - -For Prob. 8.68. - -**8.69** Given the source-free circuit in Fig. 8.114, use *PSpice* or *MultiSim* to get *i*(*t*) for 0 < *t* < 20 s. Take *v*(0) = 30 V and *i*(0) = 2 A. - -# **Figure 8.114** - -For Prob. 8.69. - -**8.70** For the circuit in Fig. 8.115, use *PSpice* or *MultiSim* to obtain *v*(*t*) for 0 < *t* < 4 s. Assume that the capacitor voltage and inductor current at *t* = 0 are both zero. - -For Prob. 8.70. - -**8.71** Obtain *v*(*t*) for 0 < *t* < 4 s in the circuit of Fig. 8.116 using *PSpice* or *MultiSim*. - -# **Figure 8.116** - -For Prob. 8.71. - -**8.72** The switch in Fig. 8.117 has been in position 1 for a long time. At *t* = 0, it is switched to position 2. Use *PSpice* or *MultiSim* to find *i*(*t*) for 0 < *t* < 0.2 s. - -# **Figure 8.117** - -For Prob. 8.72. - -**8.73** Design a problem, to be solved using *PSpice* or *MultiSim*, to help other students better understand source-free *RLC* circuits. - -# Section 8.10 Duality - -**8.74** Draw the dual of the circuit shown in Fig. 8.118. - -**Figure 8.118** For Prob. 8.74. - -**8.75** Obtain the dual of the circuit in Fig. 8.119. - -**8.76** Find the dual of the circuit in Fig. 8.120. - -# **Figure 8.120** - -For Prob. 8.76. - -**8.77** Draw the dual of the circuit in Fig. 8.121. - -# **Figure 8.121** - -For Prob. 8.77. - -# Comprehensive Problems - -- **8.80** A mechanical system is modeled by a series *RLC* circuit. It is desired to produce an overdamped response with time constants 0.1 and 0.5 ms. If a series 50-kΩ resistor is used, find the values of *L* and *C*. -- **8.81** An oscillogram can be adequately modeled by a second-order system in the form of a parallel *RLC* circuit. It is desired to give an underdamped voltage across a 200-Ω resistor. If the damped frequency is 4 kHz and the time constant of the envelope is 0.25 s, find the necessary values of *L* and *C*. -- **8.82** The circuit in Fig. 8.123 is the electrical analog of body functions used in medical schools to study convulsions. The analog is as follows: - - *C*1 = Volume of fluid in a drug - - *C*2 = Volume of blood stream in a specified region - - *R*1 = Resistance in the passage of the drug from the input to the blood stream - - *R*2 = Resistance of the excretion mechanism, such as kidney, etc. - - *v*0 = Initial concentration of the drug dosage - - *v*(*t*) = Percentage of the drug in the blood stream - - Find *v*(*t*) for *t* > 0 given that *C*1 = 0.5 *μ*F, *C*2 = 5 *μ*F, *R*1 = 5 MΩ, *R*2 = 2.5 MΩ, and *v*0 = 60*u*(*t*) V. - -# Section 8.11 Applications - -**8.78** An automobile airbag igniter is modeled by the circuit in Fig. 8.122. Determine the time it takes the voltage across the igniter to reach its first peak after switching from *A* to *B*. Let *R* = 3 Ω, *C* = 1∕30 F, and *L* = 60 mH. - -**8.79** A load is modeled as a 100-mH inductor in parallel with a 12-Ω resistor. A capacitor is needed to be connected to the load so that the network is critically damped at 60 Hz. Calculate the size of the capacitor. - -8.83 Figure 8.124 shows a typical tunnel-diode oscillator circuit. The diode is modeled as a nonlinear resistor with *iD* = *f* (*vD*), i.e., the diode current is a nonlinear function of the voltage across the diode. Derive the differential equation for the circuit in - -terms of *v* and *iD*. - -# **PART TWO** - -# AC Circuits - -# OUTLINE - -- 9 Sinusoids and Phasors -- 10 Sinusoidal Steady-State Analysis -- 11 AC Power Analysis -- 12 Three-Phase Circuits -- 13 Magnetically Coupled Circuits -- 14 Frequency Response - -# **chapter** - -9 - -# Sinusoids and Phasors - -*He who knows not, and knows not that he knows not, is a fool—shun him. He who knows not, and knows that he knows not, is a c hild—teach him. He who knows, and knows not that he knows, is asleep—wake him up. He who knows, and knows that he knows, is wise—follow him.* - -—Persian Proverb - -# Enhancing Your Skills and Your Career - -# **ABET EC 2000 criteria (3.d), "an ability to function on multi‑disciplinary teams."** - -The "ability to function on multidisciplinary teams" is inherently critical for the working engineer. Engineers rarely, if ever, work by themselves. Engineers will always be part of some team. One of the things I lik e to remind students is that you do not ha ve to like everyone on a team; you just have to be a successful part of that team. - -Most frequently, these teams include indi viduals from a v ariety of engineering disciplines, as well as individuals from nonengineering disciplines such as marketing and finance. - -Students can easily de velop and enhance this skill by w orking in study groups in every course they take. Clearly, working in study groups in nonengineering courses, as well as engineering courses outside your discipline, will also give you experience with multidisciplinary teams. - -Photo by Charles Alexander - -# Historical - -George Westinghouse. Photo © Bettmann/Corbis - -**Nikola Tesla** (1856–1943) and **George Westinghouse** (1846–1914) helped establish alternating current as the primary mode of electricity transmission and distribution. - -Today it is obvious that ac generation is well established as the form of electric power that makes widespread distribution of electric power efficient and economical. However, at the end of the 19th century, which was the better—ac or dc—w as hotly debated and had e xtremely outspoken supporters on both sides. The dc side was led by Thomas Edison, who had earned a lot of respect for his many contributions. Power generation using ac really be gan to b uild after the successful contrib utions of Tesla. The real commercial success in ac came from Geor ge Westinghouse and the outstanding team, including Tesla, he assembled. In addition, tw o other big names were C. F. Scott and B. G. Lamme. - -The most significant contribution to the early success of ac w as the patenting of the polyphase ac motor by Tesla in 1888. The induction motor and polyphase generation and distrib ution systems doomed the use of dc as the prime energy source. - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Better understand sinusoids. -- 2. Understand phasors. -- 3. Understand the phasor relationships for circuit elements. -- 4. Know and understand the concepts of impedance and admittance. -- 5. Understand Kirchhoff's laws in the frequency domain. -- 6. Comprehend the concept of phase-shift. -- 7. Understand the concept of AC bridges. - -# **9.1** Introduction - -Thus far our analysis has been limited for the most part to dc circuits: those circuits e xcited by constant or time-in variant sources. We ha ve restricted the forcing function to dc sources for the sake of simplicity, for pedagogic reasons, and also for historic reasons. Historically, dc sources were the main means of providing electric power up until the late 1800s. At the end of that century, the battle of direct current versus alternating current began. Both had their advocates among the electrical engineers of the time. Because ac is more efficient and economical to transmit over long distances, ac systems ended up the winner . Thus, it is in k eeping with the historical sequence of events that we considered dc sources first. - -We now begin the analysis of circuits in which the source v oltage or current is time-v arying. In this chapter , we are particularly interested in sinusoidally time-varying excitation, or simply, excitation by a *sinusoid*. - -## A sinusoid is a signal that has the form of the sine or cosine function. - -A sinusoidal current is usually referred to as *alternating current* (*ac*). Such a current reverses at regular time intervals and has alternately positive and negative values. Circuits driven by sinusoidal current or voltage sources are called *ac circuits*. - -We are interested in sinusoids for a number of reasons. First, nature itself is characteristically sinusoidal. We e xperience sinusoidal v ariation in the motion of a pendulum, the vibration of a string, the ripples on the ocean surface, and the natural response of underdamped secondorder systems, to mention but a few. Second, a sinusoidal signal is easy to generate and transmit. It is the form of v oltage generated throughout the world and supplied to homes, factories, laboratories, and so on. It is the dominant form of signal in the communications and electric power industries. Third, through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids, therefore, play an important role in the analysis of periodic signals. Lastly , a sinu soid is easy to handle mathematically . The derivative and integral of a sinusoid are themselve s sinusoids. For these and other reasons, the sinusoid is an extremely important function in circuit analysis. - -A sinusoidal forcing function produces both a transient response and a steady-state response, much like the step function, which we studied in Chapters 7 and 8. The transient response dies out with time so that only the steady-state response remains. When the transient response has become negligibly small compared with the steady-state response, we say that the circuit is operating at sinusoidal steady state. It is this *sinusoidal steady-state response* that is of main interest to us in this chapter. - - We begin with a basic discussion of sinusoids and phasors. We then introduce the concepts of impedance and admittance. The basic circuit laws, Kirchhoff's and Ohm's, introduced for dc circuits, will be applied to ac circuits. Finally , we consider applications of ac circuits in phaseshifters and bridges. - -# **9.2** Sinusoids - -Consider the sinusoidal voltage - -$$ -v(t) = V_m \sin \omega t \tag{9.1} -$$ - -where - -*Vm* = the *amplitude* of the sinusoid - -*ω* = the *angular frequency* in radians/s - -*ωt* = the *argument* of the sinusoid - -The sinusoid is shown in Fig. 9.1(a) as a function of its argument and in Fig. 9.1(b) as a function of time. It is e vident that the sinusoid repeats itself every *T* seconds; thus, *T* is called the *period* of the sinusoid. From the two plots in Fig. 9.1, we observe that *ωT* = 2*π*, - -$$ -T = \frac{2\pi}{\omega} \tag{9.2} -$$ - -# Historical - -© Hulton Archives/Getty Images - -**Heinrich Rudorf Hertz** (1857–1894), a German experimental physicist, demonstrated that electromagnetic waves obey the same fundamental laws as light. His work confirmed James Clerk Maxwell's celebrated 1864 theory and prediction that such waves existed. - -Hertz w as born into a prosperous f amily in Hamb urg, German y. He attended the Uni versity of Berlin and did his doctorate under the prominent physicist Hermann von Helmholtz. He became a professor at Karlsruhe, where he be gan his quest for electromagnetic w aves. Hertz successfully generated and detected electromagnetic w aves; he was the first to show that light is electromagnetic ener gy. In 1887, Hertz noted for the first time the photoelectric effect of electrons in a molecular structure. Although Hertz only lived to the age of 37, his discovery of electromagnetic waves paved the way for the practical use of such waves in radio, television, and other communication systems. The unit of frequency, the hertz, bears his name. - -The fact that *v*(*t*) repeats itself every *T* seconds is shown by replacing *t* by *t* + *T* in Eq. (9.1). We get - -$$ -v(t+T) = V_m \sin \omega(t+T) = V_m \sin \omega \left(t + \frac{2\pi}{\omega}\right) -$$ - -= $V_m \sin (\omega t + 2\pi) = V_m \sin \omega t = v(t)$ (9.3) - -Hence, - -$$ -v(t+T) = v(t) \tag{9.4} -$$ - -that is, *v* has the same value at *t* + *T* as it does at *t* and *v*(*t*) is said to be *periodic*. In general, - -A periodic function is one that satisfies f (t) = f (t + nT ), for all t and for all integers n. - -As mentioned, the *period T* of the periodic function is the time of one complete cycle or the number of seconds per c ycle. The reciprocal of this quantity is the number of c ycles per second, kno wn as the *cyclic frequency f* of the sinusoid. Thus, - -$$ -f = \frac{1}{T} -$$ - (9.5) - -From Eqs. (9.2) and (9.5), it is clear that - -$$ -\omega = 2 \pi f \tag{9.6} -$$ - -While *ω* is in radians per second (rad/s), *f* is in hertz (Hz). - -Let us now consider a more general expression for the sinusoid, - -$$ -v(t) = V_m \sin(\omega t + \phi) -$$ -\n(9.7) - -where (*ωt* + *ϕ*) is the argument and *ϕ* is the *phase*. Both argument and phase can be in radians or degrees. - -Let us examine the two sinusoids - -$$ -v_1(t) = V_m \sin \omega t \qquad \text{and} \qquad v_2(t) = V_m \sin(\omega t + \phi) \tag{9.8} -$$ - -shown in Fig. 9.2. The starting point of *v*2 in Fig. 9.2 occurs first in time. Therefore, we say that *v*2 *leads v*1 by *ϕ* or that *v*1 *lags v*2 by *ϕ*. If *ϕ* ≠ 0, we also say that *v*1 and *v*2 are *out of phase*. If *ϕ* = 0, then *v*1 and *v*2 are said to be *in phase;* they reach their minima and maxima at e xactly the same time. We can compare *v*1 and *v*2 in this manner because they operate at the same frequency; they do not need to have the same amplitude. - -A sinusoid can be e xpressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. This is achieved by using the following trigonometric identities: - -$$ -sin(A \pm B) = sin A cos B \pm cos A sin B -$$ - -\n -$$ -cos(A \pm B) = cos A cos B \mp sin A sin B -$$ -\n(9.9) - -With these identities, it is easy to show that - -$$ -sin(\omega t \pm 180^\circ) = -sin \omega t -$$ - -\n -$$ -cos(\omega t \pm 180^\circ) = -cos \omega t -$$ - -\n -$$ -sin(\omega t \pm 90^\circ) = \pm cos \omega t -$$ - -\n -$$ -cos(\omega t \pm 90^\circ) = \mp sin \omega t -$$ -\n(9.10) - -Using these relationships, we can transform a sinusoid from sine form to cosine form or vice versa. - -The unit of f is named after the German physicist Heinrich R. Hertz (1857–1894). - -# **Figure 9.3** - -A graphical means of relating cosine and sine: (a) cos(*ωt* − 90°) = sin *ωt*, (b) sin(*ωt* + 180°) = −sin *ωt*. - -A graphical approach may be used to relate or compare sinusoids as an alternati ve to using the trigonometric identities in Eqs. (9.9) and (9.10). Consider the set of axes shown in Fig. 9.3(a). The horizontal axis represents the magnitude of cosine, while the vertical axis (pointing down) denotes the magnitude of sine. Angles are measured positi vely counterclockwise from the horizontal, as usual in polar coordinates. This graphical technique can be used to relate tw o sinusoids. F or example, we see in Fig. 9.3(a) that subtracting 90° from the ar gument of cos *ωt* gives sin *ωt*, o r cos(*ωt* − 90°) = sin *ωt*. Similarly, adding 1 80° to the argument of sin*ωt* gives −sin *ωt*, or sin(*ωt* + 180°) = −sin *ωt*, as shown in Fig. 9.3(b). - -The graphical technique can also be used to add tw o sinusoids of the same frequency when one is in sine form and the other is in cosine form. To add *A* cos *ωt* and *B* sin *ωt*, we note that *A* is the magnitude of cos *ωt* while *B* is the magnitude of sin*ωt*, as shown in Fig. 9.4(a). The magnitude and argument of the resultant sinusoid in cosine form is readily obtained from the triangle. Thus, - -$$ -A\cos\omega t + B\sin\omega t = C\cos(\omega t - \theta) -$$ -\n(9.11) - -where - -$$ -C = \sqrt{A^2 + B^2} -$$ -, $\theta = \tan^{-1} \frac{B}{A}$ (9.12) - -For example, we may add 3 cos *ωt* and −4 sin*ωt* as shown in Fig. 9.4(b) and obtain - -$$ -3\cos \omega t - 4\sin \omega t = 5\cos(\omega t + 53.1^{\circ}) -$$ - (9.13) - -Compared with the trigonometric identities in Eqs. (9.9) and (9.10), the graphical approach eliminates memorization. Ho wever, we must not confuse the sine and cosine ax es with the ax es for comple x numbers to be discussed in the ne xt section. Something else to note in Figs. 9.3 and 9.4 is that although the natural tendenc y is to ha ve the vertical axis point up, the positive direction of the sine function is down in the present case. - -(a) Adding *A* cos *ωt* and *B* sin *ωt*, (b) adding 3 cos *ωt* and −4 sin *ωt*. - -*v*(*t*) = 12 cos(50*t* + 10°) V. - -# **Solution:** - -The amplitude is *Vm* = 12 V. The phase is *ϕ* = 10°. The angular frequency is *ω* = 50 rad/s. The period *T* = \_\_\_ 2*π ω* = \_\_\_ 2*π* 50 = 0.1257 s. The frequency is *f* = \_\_1 *T* = 7.958 Hz. - -Given the sinusoid 45 cos(5 *πt* + 36°), calculate its amplitude, phase, angular frequency, period, and frequency. - -**Answer:** 45, 36°, 15.708 rad/s, 400 ms, 2.5 Hz. - -Calculate the phase angle between *v*1 = −10 cos(*ωt* + 50°) and *v*2 = Example 9.2 12 sin(*ωt* − 10°). State which sinusoid is leading. - -# **Solution:** - -Let us calculate the phase in three ways. The first two methods use trigonometric identities, while the third method uses the graphical approach. - -■ **METHOD 1** In order to compare *v*1 and *v*2, we must e xpress them in the same form. If we e xpress them in cosine form with posi tive amplitudes, - -*v*1 = −10 cos(*ωt* + 50°) = 10 cos(*ωt* + 50° − 180°) *v*1 = 10 cos(*ωt* − 130°) or *v*1 = 10 cos(*ωt* + 230°) **(9.2.1)** and *v*2 = 12 sin(*ωt* − 10°) = 12 cos(*ωt* − 10° − 90°) *v*2 = 12 cos(*ωt* − 100°) **(9.2.2)** - -It can be deduced from Eqs. (9.2.1) and (9.2.2) that the phase difference between *v*1 and *v*2 is 30°. We can write *v*2 as - -*v*2 = 12 cos(*ωt* − 130° + 30°) or *v*2 = 12 cos(*ωt* + 260°) **(9.2.3)** - -Comparing Eqs. (9.2.1) and (9.2.3) shows clearly that *v*2 leads *v*1 by 30°. - -■ **METHOD 2** Alternatively, we may express *v*1 in sine form: - -*v*1 = −10 cos(*ωt* + 50°) = 10 sin(*ωt* + 50° − 90°) = 10 sin(*ωt* − 40°) = 10 sin(*ωt* − 10° − 30°) - -Practice Problem 9.1 - -But *v*2 = 12 sin(*ωt* − 10°). Comparing the tw o shows that *v*1 lags *v*2 by 30°. This is the same as saying that *v*2 leads *v*1 by 30°. - -■ **METHOD 3** We may regard *v*1 as simply −10 cos*ωt* with a phase shift of +50°. Hence, *v*1 is as shown in Fig. 9.5. Similarly, *v*2 is 12 sin*ωt* with a phase shift of −10°, as shown in Fig. 9.5. It is easy to see from Fig. 9.5 that *v*2 leads *v*1 by 30°, that is, 90° − 50° − 10°. - -Find the phase angle between - -*i*1 = −4 sin(377*t* + 55°) and *i*2 = 5 cos(377*t* − 65°) - -Does *i*1 lead or lag *i*2? - -**Answer:** 210°, *i*1 leads *i*2. - -# **9.3** Phasors - -Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions. - -A phasor is a complex number that represents the amplitude and phase of a sinusoid. - -Phasors provide a simple means of analyzing linear circuits e xcited by sinusoidal sources; solutions of such circuits would be intractable otherwise. The notion of solving ac circuits using phasors was first introduced by Charles Steinmetz in 1893. Before we completely define phasors and apply them to circuit analysis, we need to be thoroughly f amiliar with complex numbers. - -A complex number *z* can be written in rectangular form as - -$$ -z = x + jy \tag{9.14a} -$$ - -where *j* = √ \_\_\_ −1 ; *x* is the real part of *z*; *y* is the imaginary part of *z*. In this context, the variables *x* and *y* do not represent a location as in tw odimensional vector analysis but rather the real and imaginary parts of *z* in the complex plane. Nevertheless, we note that there are some resemblances between manipulating complex numbers and manipulating twodimensional vectors. - -The complex number *z* can also be written in polar or e xponential form as - -$$ -z = r/\phi = re^{j\phi} \tag{9.14b} -$$ - -Charles Proteus Steinmetz (1865–1923) was a German-Austrian mathematician and electrical engineer. - -Appendix B presents a short tutorial on complex numbers. - -# Historical - -**Charles Proteus Steinmetz** (1865–1923), a German-Austrian mathematician and engineer, introduced the phasor method (covered in this chapter) in ac circuit analysis. He is also noted for his w ork on the theory of hysteresis. - -Steinmetz was born in Breslau, Germany, and lost his mother at the age of one. As a youth, he was forced to leave Germany because of his political activities just as he w as about to complete his doctoral dis sertation in mathematics at the Uni versity of Breslau. He migrated to Switzerland and later to the United States, where he w as employed by General Electric in 1893. That same year, he published a paper in which complex numbers were used to analyze ac circuits for the first time. This led to one of his man y textbooks, *Theory and Calculation of ac Phenomena,* published by McGra w-Hill in 1897. In 1901, he became the president of the American Institute of Electrical Engineers, which later became the IEEE. - -where *r* is the magnitude of *z*, and *ϕ* is the phase of *z*. We notice that *z* can be represented in three ways: - -| z = x + jy | Rectangular form | | -|------------|------------------|--------| -| z = r⧸ϕ | Polar form | (9.15) | -| z = rejϕ | Exponential form | | - -The relationship between the rectangular form and the polar form is shown in Fig. 9.6, where the *x* axis represents the real part and the *y* axis represents the imaginary part of a comple x number. Given *x* and *y*, we can get *r* and *ϕ* as - -$$ -r = \sqrt{x^2 + y^2} -$$ -, $\phi = \tan^{-1} \frac{y}{x}$ (9.16a) - -On the other hand, if we know *r* and *ϕ*, we can obtain *x* and *y* as - -$$ -x = r \cos \phi, \qquad y = r \sin \phi \tag{9.16b} -$$ - -Thus, *z* may be written as - -$$ -z = x + jy = r/\underline{\phi} = r(\cos\phi + j\sin\phi) -$$ - (9.17) - -Addition and subtraction of complex numbers are better performed in rectangular form; multiplication and division are better done in polar form. Given the complex numbers - -$$ -z = x + jy = r/\underline{\phi}, \qquad z_1 = x_1 + jy_1 = r_1/\underline{\phi}_1 -$$ -$$ -z_2 = x_2 + jy_2 = r_2/\underline{\phi}_2 -$$ - -the following operations are important. **Addition:** - -$$ -z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2) -$$ -\n(9.18a) - -# **Subtraction:** - -$$ -z_1 - z_2 = (x_1 - x_2) + j(y_1 - y_2) -$$ -\n(9.18b) - -**Multiplication:** - -$$ -z_1 z_2 = r_1 r_2 / \phi_1 + \phi_2 \tag{9.18c} -$$ - -**Division:** - -$$ -\frac{z_1}{z_2} = \frac{r_1}{r_2} / \phi_1 - \phi_2 \tag{9.18d} -$$ - -**Reciprocal:** - -$$ -\frac{1}{z} = \frac{1}{r} \angle -\phi \tag{9.18e} -$$ - -**Square Root:** - -$$ -\sqrt{z} = \sqrt{r} \sqrt{\phi/2} -$$ - (9.18f) - -# **Complex Conjugate:** - -$$ -z^* = x - jy = r'_\text{p} = re^{-j\phi} \tag{9.18g} -$$ - -Note that from Eq. (9.18e), - -$$ -\frac{1}{j} = -j \tag{9.18h} -$$ - -These are the basic properties of complex numbers we need. Other properties of complex numbers can be found in Appendix B. - -The idea of phasor representation is based on Euler' s identity. In general, - -$$ -e^{\pm i\phi} = \cos\phi \pm j\sin\phi \qquad (9.19) -$$ - -which shows that we may re gard cos *ϕ* and sin *ϕ* as the real and imagi nary parts of *e jϕ* ; we may write - -$$ -\cos \phi = \text{Re}(e^{j\phi}) \tag{9.20a} -$$ - -$$ -\sin \phi = \text{Im}(e^{j\phi})\tag{9.20b} -$$ - -where Re and Im stand for the *real part of* and the *imaginary part of*. Given a sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), we use Eq. (9.20a) to e xpress *v*(*t*) as - -$$ -v(t) = V_m \cos(\omega t + \phi) = \text{Re}(V_m e^{j(\omega t + \phi)}) -$$ - (9.21) - -or - -$$ -v(t) = \text{Re}(V_m e^{j\phi} e^{j\omega t}) -$$ -\n(9.22) - -Thus, - -$$ -v(t) = \text{Re}(\mathbf{V}e^{j\omega t}) -$$ - (9.23) - -where - -$$ -\mathbf{V} = V_m e^{j\phi} = V_m / \phi \tag{9.24} -$$ - -**V** is thus the *phasor representation* of the sinusoid *v*(*t*), as we saidearlier. In other words, a phasor is a complex representation of the magnitude and phase of a sinusoid. Either Eq. (9.20a) or Eq. (9.20b) can be used to develop the phasor, but the standard convention is to use Eq. (9.20a). - -One way of looking at Eqs. (9.23) and (9.24) is to consider the plot of the *sinor* **V***e jωt* = *Vme j*(*ωt*+*ϕ*) on the comple x plane. As time increases, the sinor rotates on a circle of radius *Vm* at an angular velocity *ω* in the counterclockwise direction, as shown in Fig. 9.7(a). We may regard *v*(*t*) as the projection of the sinor **V***e jωt* on the real axis, as shown in Fig. 9.7(b). The value of the sinor at time *t* = 0 is the phasor **V** of the sinusoid *v*(*t*). The sinor may be regarded as a rotating phasor. Thus, whenever a sinusoid is expressed as a phasor, the term *e jωt* is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency *ω* of the phasor; otherwise we can make serious mistakes. - - A phasor may be regarded as a mathematical equivalent of a sinusoid with the time dependence dropped. - - If we use sine for the phasor instead of cosine, then v (t) = V m sin(*ω*t + *ϕ*) = Im(Vmej(*ω*t+*ϕ*) ) and the corresponding phasor is the same as that in Eq. (9.24). - -Equation (9.23) states that to obtain the sinusoid corresponding to a given phasor **V**, multiply the phasor by the time f actor *ejωt* and tak e the real part. As a complex quantity, a phasor may be expressed in rectangular form, polar form, or exponential form. Because a phasor has magnitude and phase ("direction"), it behaves as a vector and is printed in boldface. For example, phasors **V** = *Vm*⧸*ϕ* and **I** = *Im*⧸−*θ* are graphically represented in Fig. 9.8. Such a graphical representation of phasors is known as a *phasor diagram*. - -Equations (9.21) through (9.23) re veal that to get the phasor cor responding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we tak e out the time f actor *ejωt* , and whate ver is left is the pha sor corresponding to the sinusoid. By suppressing the time f actor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows: - -$$ -v(t) = V_m \cos(\omega t + \phi) \qquad \Leftrightarrow \qquad \mathbf{V} = V_m / \underline{\phi} -$$ - (9.25) -\n(Time-domain representation) (Phasor-domain representation) - -We use lightface italic letters such as z to represent complex numbers but boldface letters such as **V** to represent phasors, because phasors are vectorlike quantities. - -A phasor diagram showing **V** = *Vm*⧸*ϕ* and **I** = *Im*⧸−*θ*. - -Given a sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), we obtain the corresponding phasor as **V** = *Vm* ⧸*ϕ*. Equation (9.25) is also demonstrated in Table 9.1, where the sine function is considered in addition to the cosine function. From Eq. (9.25), we see that to get the phasor representation of a sinu soid, we e xpress it in cosine form and tak e the magnitude and phase. Given a phasor, we obtain the time domain representation as the cosine function with the same magnitude as the phasor and the ar gument as *ωt* plus the phase of the phasor. The idea of expressing information in alternate domains is fundamental to all areas of engineering. - -## **TABLE 9.1** - -Sinusoid-phasor transformation. - -| Phasor domain representation | -|------------------------------| -| Vm⧸ϕ | -| Vm⧸ϕ − 90° | -| Im⧸θ | -| Im⧸θ − 90° | -| | - -Note that in Eq. (9.25) the frequenc y (or time) f actor *ejωt* is sup pressed, and the frequency is not explicitly shown in the phasor domain representation because *ω* is constant. However, the response depends on *ω*. For this reason, the phasor domain is also known as the *frequency domain.* - -From Eqs. (9.23) and (9.24), *v*(*t*) = Re(**V***e jωt* ) = *Vm* cos(*ωt* + *ϕ*), so that - -$$ -\frac{dv}{dt} = -\omega V_m \sin(\omega t + \phi) = \omega V_m \cos(\omega t + \phi + 90^\circ) -$$ - -= Re( $\omega V_m e^{j\omega t} e^{j\phi} e^{j90^\circ}$ ) = Re( $j\omega V e^{j\omega t}$ ) (9.26) - -This shows that the deri vative *v*(*t*) is transformed to the phasor domain as *jω***V** - -$$ -\frac{dv}{dt} \qquad \Leftrightarrow \qquad j\omega V \qquad (9.27) -$$ -\n(Time domain) - -\n(Phasor domain) - -Similarly, the inte gral of *v*(*t*) is transformed to the phasor domain as **V**∕*jω* - -$$ -\int v \, dt \qquad \Leftrightarrow \qquad \frac{V}{j\omega} \qquad (9.28) -$$ -\n(Time domain) - -\n(Phasor domain) - -Equation (9.27) allows the replacement of a derivative with respect to time with multiplication of *jω* in the phasor domain, whereas Eq. (9.28) allows the replacement of an inte gral with respect to time with di vision by *jω* in the phasor domain. Equations (9.27) and (9.28) are useful in finding the steady-state solution, which does not require knowing the initial values of the variable involved. This is one of the important applications of phasors. - -Besides time differentiation and integration, another important use of phasors is found in summing sinusoids of the same frequency. This is best illustrated with an example, and Example 9.6 provides one. - -The differences between *v*(*t*) and **V** should be emphasized: - -- 1. *v*(*t*) is the *instantaneous or time domain* representation, while **V** is the *frequency or phasor domain* representation. -- 2. *v*(*t*) is time dependent, while **V** is not. (This f act is often for gotten by students.) -- 3. *v*(*t*) is al ways real with no comple x term, while **V** is generally complex. - -Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of the same frequency. - - Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by j*ω*. - - Integrating a sinusoid is equivalent to dividing its corresponding phasor by j*ω*. - - Adding sinusoids of the same frequency is equivalent to adding their corresponding phasors. - -Evaluate these complex numbers: Example 9.3 - -(a) -$$ -(40/50^{\circ} + 20/-30^{\circ})^{1/2} -$$ - -\n(b) -$$ -\frac{10/-30^{\circ} + (3-j4)}{(2+j4)(3-j5)^{*}} -$$ - -# **Solution:** - -(a) Using polar to rectangular transformation, - -$$ -40/50^{\circ} = 40(\cos 50^{\circ} + j \sin 50^{\circ}) = 25.71 + j30.64 -$$ - -$$ -20\angle -30^{\circ} = 20[\cos(-30^{\circ}) + j\sin(-30^{\circ})] = 17.32 - j10 -$$ - -Adding them up gives - -$$ -40/50^{\circ} + 20/-30^{\circ} = 43.03 + j20.64 = 47.72/25.63^{\circ} -$$ - -Taking the square root of this, - -$$ -(40/50^{\circ} + 20/-30^{\circ})^{1/2} = 6.91/12.81^{\circ} -$$ - -(b) Using polar-rectangular transformation, addition, multiplication, and division, - -(b) Using polar-rectangular transformation, a -division, -$$ -\frac{10(-30^\circ + (3 - j4))}{(2 + j4)(3 - j5)^*} = \frac{8.66 - j5 + (3 - j4)}{(2 + j4)(3 + j5)} -$$ -$$ -= \frac{11.66 - j9}{-14 + j22} = \frac{14.73(-37.66^\circ)}{26.08(122.47^\circ)} -$$ -$$ -= 0.565(-160.13^\circ) -$$ - -| Practice Problem 9.3 | Evaluate the following complex numbers: | -|----------------------|------------------------------------------------------------------------| -| | (a) [(5 + j2)(−1 + j4) − 5⧸ 60°]* | -| | 10 + j5 + 3⧸ 40°
______________
(b)
+ 10⧸ 30° + j5
−3 + j4 | -| | Answer: (a) −15.5 −
j13.67, (b) 8.293 + j7.2. | -| | | - -Example 9.4 Transform these sinusoids to phasors: - -(a) *i* = 6 cos(50*t* − 40°) A (b) *v* = −4 sin(30*t* + 50°) V - -# **Solution:** - -(a) *i* = 6 cos(50*t* − 40°) has the phasor - -$$ -I = 6/–40^{\circ} A -$$ - -(b) Since -$$ --\sin A = \cos(A + 90^\circ) -$$ -, - $v = -4 \sin(30t + 50^\circ) = 4 \cos(30t + 50^\circ + 90^\circ)$ - -$$ -= 4\cos(30t + 140^\circ) \text{ V} -$$ - -The phasor form of *v* is - -$$ -V = 4/140^{\circ} V -$$ - -Practice Problem 9.4 Express these sinusoids as phasors: - -(a) *v* = −14 sin(5*t* − 22°) V (b) *i* = −8 cos(16*t* + 15°) A - -**Answer:** (a) **V** = 14⧸ 68° V, (b) **I** = 8⧸−165° A. - -Find the sinusoids represented by these phasors: Example 9.5 - -(a) -$$ -\mathbf{I} = -3 + j4 \text{ A} -$$ - -\n(b) $\mathbf{V} = j8e^{-j20^{\circ}} \text{ V}$ - -# **Solution:** - -(a) **I** = −3 + *j*4 = 5⧸ 126.87°. Transforming this to the time domain gives - -$$ -i(t) = 5 \cos(\omega t + 126.87^{\circ}) -$$ - A - -(b) Because *j* = 1⧸ 90°, - -$$ -\mathbf{V} = j8 \underline{/ -20^{\circ}} = (1 \underline{/ 90^{\circ}})(8 \underline{/ -20^{\circ}}) -$$ -$$ -= 8 \underline{/ 90^{\circ} - 20^{\circ}} = 8 \underline{/ 70^{\circ}} \text{ V} -$$ - -Converting this to the time domain gives - -$$ -v(t) = 8\cos(\omega t + 70^\circ)\text{V} -$$ - -Find the sinusoids corresponding to these phasors: Practice Problem 9.5 - -(a) **V** = −25⧸ 40° V (b) **I** = *j*(12 − *j*5) A **Answer:** (a) *v*(*t*) = 25 cos(*ωt* − 140°) V or 25 cos(*ωt* + 220°) V, (b) *i*(*t*) = 13 cos(*ωt* + 67.38°) A. - -Given *i*1(*t*) = 4 cos(*ωt* + 30°) A and *i*2(*t*) = 5 sin(*ωt* − 20°) A, find Example 9.6 their sum. - -# **Solution:** - -Here is an important use of phasors—for summing sinusoids of the same frequency. Current *i*1(*t*) is in the standard form. Its phasor is - -$$ -\mathbf{I}_1 = 4/30^\circ -$$ - -We need to express *i*2(*t*) in cosine form. The rule for converting sine to cosine is to subtract 90°. Hence, - -$$ -i_2 = 5\cos(\omega t - 20^\circ - 90^\circ) = 5\cos(\omega t - 110^\circ) -$$ - -and its phasor is - -**I**2 = 5⧸−110° - -If we let *i* = *i*1 + *i*2, then - -$$ -\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 = 4/30^\circ + 5/-110^\circ -$$ - -= 3.464 + j2 - 1.71 - j4.698 = 1.754 - j2.698 -= 3.218/-56.97° A - -Transforming this to the time domain, we get - -$$ -i(t) = 3.218 \cos(\omega t - 56.97^{\circ}) \text{ A} -$$ - -Of course, we can find *i*1 + *i*2 using Eq. (9.9), but that is the hard way. - -Practice Problem 9.6 If *v*1 = −10 sin(*ωt* − 30°) V and *v*2 = 20 cos(*ωt* + 45°) V, find *v*= *v*1 + *v*2. - -**Answer:** *v*(*t*) = 29.77 cos(*ωt* + 49.98°) V. - -Example 9.7 Using the phasor approach, determine the current *i*(*t*) in a circuit described by the integrodifferential equation - -$$ -4i + 8\int i\,dt - 3\frac{di}{dt} = 50\cos(2t + 75^\circ) -$$ - -# **Solution:** - -We transform each term in the equation from time domain to phasor domain. Keeping Eqs. (9.27) and (9.28) in mind, we obtain the phasor form of the given equation as - -$$ -4\mathbf{I} + \frac{8\mathbf{I}}{j\omega} - 3j\omega\mathbf{I} = 50/75^{\circ} -$$ - -But *ω* = 2, so - -$$ -I(4 - j4 - j6) = 50 / 75^{\circ} -$$ - -$$ -I = \frac{50/75^{\circ}}{4 - j10} = \frac{50/75^{\circ}}{10.77/-68.2^{\circ}} = 4.642/143.2^{\circ} -$$ - A - -Converting this to the time domain, - -$$ -i(t) = 4.642 \cos(2t + 143.2^{\circ}) \text{ A} -$$ - -Keep in mind that this is only the steady-state solution, and it does not require knowing the initial values. - -Practice Problem 9.7 Find the v oltage *v*(*t*) in a circuit described by the inte grodifferential equation - -$$ -2\frac{dv}{dt} + 5v + 10 \int v \, dt = 50 \cos(5t - 30^{\circ}) -$$ - -using the phasor approach. - -**Answer:** -$$ -v(t) = 5.3 \cos(5t - 88^\circ) -$$ - V. - -# **9.4** Phasor Relationships for Circuit Elements - -Now that we know how to represent a voltage or current in the phasor or frequency domain, one may le gitimately ask ho w we apply this to cir cuits involving the passive elements *R*, *L*, and *C*. What we need to do is to transform the voltage-current relationship from the time domain to the frequency domain for each element. Again, we will assume the passi ve sign convention. - -We be gin with the resistor . If the current through a resistor *R* is *i* = *Im* cos(*ωt* + *ϕ*), the voltage across it is given by Ohm's law as - -$$ -v = iR = RIm \cos(\omega t + \phi) -$$ - (9.29) - -The phasor form of this voltage is - -$$ -\mathbf{V} = R I_m / \underline{\phi} \tag{9.30} -$$ - -But the phasor representation of the current is **I** = I*m*⧸*ϕ*. Hence, - -$$ -V = RI \tag{9.31} -$$ - -showing that the v oltage-current relation for the resistor in the phasor domain continues to be Ohm's law, as in the time domain. Figure 9.9 illustrates the voltage-current relations of a resistor. We should note from Eq. (9.31) that voltage and current are in phase, as illustrated in the phasor diagram in Fig. 9.10. - -For the inductor *L*, assume the current through it is *i* = *Im* cos(*ωt* + *ϕ*). The voltage across the inductor is - -$$ -v = L\frac{di}{dt} = -\omega L I_m \sin(\omega t + \phi) -$$ -\n(9.32) - -Recall from Eq. (9.10) that −sin *A* = cos(*A* + 90°). We can write the voltage as - -$$ -v = \omega L I_m \cos(\omega t + \phi + 90^\circ) \tag{9.33} -$$ - -which transforms to the phasor - -$$ -\mathbf{V} = \omega L I_m e^{j(\phi + 90^\circ)} = \omega L I_m e^{j\phi} e^{j90^\circ} = \omega L I_m / \phi + 90^\circ \tag{9.34} -$$ - -But *Im*⧸*ϕ* = **I**, and from Eq. (9.19), *ej*90° = *j*. Thus, - -$$ -V = j\omega L I \tag{9.35} -$$ - -showing that the v oltage has a magnitude of *ωLIm* and a phase of *ϕ* + 90°. The voltage and current are 90° out of phase. Specifically, the current lags the v oltage by 90°. Figure 9.11 sho ws the v oltage-current relations for the inductor. Figure 9.12 shows the phasor diagram. - -For the capacitor *C*, assume the voltage across it is *v* = *Vm* cos(*ωt* + *ϕ*). The current through the capacitor is - -$$ -i = C \frac{dv}{dt} -$$ - (9.36) - -By following the same steps as we took for the inductor or by applying Eq. (9.27) on Eq. (9.36), we obtain - -$$ -\mathbf{I} = j\omega C \mathbf{V} \qquad \Rightarrow \qquad \mathbf{V} = \frac{\mathbf{I}}{j\omega C} \tag{9.37} -$$ - -# **Figure 9.9** - -Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain. - -Phasor diagram for the resistor. - -# **Figure 9.11** - -Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain. - -**Figure 9.12** Phasor diagram for the inductor; **I** lags **V**. - -Although it is equally correct to say that the inductor voltage leads the current by 90°, convention gives the current phase relative to the voltage. - -## **384** Chapter 9 Sinusoids and Phasors - -showing that the current and voltage are 90° out of phase. To be specific, the current leads the voltage by 90°. Figure 9.13 shows the voltage-current relations for the capacitor; Fig. 9.14 gives the phasor diagram. Table 9.2 summarizes the time domain and phasor domain representations of the circuit elements. - -# **TABLE 9.2** - -Summary of voltage-current relationships. - -| Element | Time domain | Frequency domain | -|---------|--------------------|-------------------| -| R | v = Ri | V = RI | -| L | v = L __di
dt | V = jωLI | -| C | i = C ___ dv
dt | V = ____ I
jωC | - -Example 9.8 The voltage *v* = 12cos(60*t* + 45°) is applied to a 0.1-H inductor. Find the steady-state current through the inductor. - -# **Solution:** - -For the inductor , **V** = *jωL***I**, where *ω* = 60 rad/s and **V** = 12 ⧸45° V. Hence, - -$$ -I = \frac{V}{j\omega L} = \frac{12/45^{\circ}}{j60 \times 0.1} = \frac{12/45^{\circ}}{6/90^{\circ}} = 2/45^{\circ} -$$ - A - -Converting this to the time domain, - -$$ -i(t) = 2\cos(60t - 45^\circ) \,\mathrm{A} -$$ - -Practice Problem 9.8 If voltage *v* = 25 sin(100*t* − 15°) *V* is applied to a 50 *μ*F capacitor, calculate the current through the capacitor. - -# **Answer:** 125 sin(100*t* + 75°) mA. - -$$ -0.125 \sim -$$ - -# **9.5** Impedance and Admittance - -In the preceding section, we obtained the v oltage-current relations for the three passive elements as - -$$ -V = RI, \t V = j\omega LI, \t V = \frac{I}{j\omega C} -$$ -(9.38) - -These equations may be written in terms of the ratio of the phasor v oltage to the phasor current as - -$$ -\frac{V}{I} = R, \qquad \frac{V}{I} = j\omega L, \qquad \frac{V}{I} = \frac{1}{j\omega C} -$$ - (9.39) - -From these three e xpressions, we obtain Ohm's law in phasor form for any type of element as - -$$ -Z = \frac{V}{I} \qquad \text{or} \qquad V = ZI \tag{9.40} -$$ - -where **Z** is a frequenc y-dependent quantity known as *impedance*, measured in ohms. - -The impedance **Z** of a circuit is the ratio of the phasor voltage **V** to the phasor current **I**, measured in ohms (Ω). - -The impedance represents the opposition that the circuit e xhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity. - -The impedances of resistors, inductors, and capacitors can be readily obtained from Eq. (9.39). Table 9.3 summarizes their impedances. From the table we notice that **Z***L* = *jωL* and **Z***C* = −*j*∕*ωC*. Consider two extreme cases of angular frequenc y. When *ω* = 0 (i.e., for dc sources), **Z***L* = 0 and **Z***C* → ∞, confirming what we already know—that the inductor acts like a short circuit, while the capacitor acts like an open circuit. When *ω* → ∞ (i.e., for high frequencies), **Z***L* → ∞ and **Z***C* = 0, indicating that the inductor is an open circuit to high frequencies, while the capacitor is a short circuit. Figure 9.15 illustrates this. - -As a complex quantity, the impedence may be e xpressed in rectangular form as - -$$ -Z = R \pm jX \tag{9.41} -$$ - -where *R* = Re **Z** is the *resistance* and *X* = Im **Z** is the *reactance*. The reactance, *X,* is just a magnitude, a positi ve value, but when used as a vector, a *j* is associated with inductance and a −*j* is associated with capacitance. Thus, impedance **Z** = *R* + *jX* is said to be *inductive* o r lagging since current lags v oltage, while impedance **Z** = *R* − *jX* i s capacitive or leading because current leads v oltage. The impedance, resistance, and reactance are all measured in ohms. The impedance may also be expressed in polar form as - -$$ -\mathbf{Z} = |\mathbf{Z}| \; / \theta \tag{9.42} -$$ - -**TABLE 9.3** - -Impedances and admittances of passive elements. - -| Impedance | Admittance | -|--------------------|------------------------------------------------------------| -| Z = R | Y = __1
R | -| Z = jωL | Y = ____ 1
jωL | -| Z = ____ 1
jω C | Y = jωC | -| | Short circuit at dc
Open circuit at
high frequencies | -| (a) | | -| | Open circuit at dc | -| (b) | Short circuit at
high frequencies | -| | | - -# **Figure 9.15** - -Equivalent circuits at dc and high frequencies: (a) inductor, (b) capacitor. Comparing Eqs. (9.41) and (9.42), we infer that - -$$ -Z = R \pm jX = |Z|/\theta \tag{9.43} -$$ - -where - -$$ -|Z| = \sqrt{R^2 + X^2}, \qquad \theta = \tan^{-1} \frac{\pm X}{R} -$$ - (9.44) - -and - -$$ -R = |\mathbf{Z}| \cos \theta, \qquad X = |\mathbf{Z}| \sin \theta \tag{9.45} -$$ - -It is sometimes con venient to w ork with the reciprocal of imped ance, known as *admittance*. - -The admittance **Y** is the reciprocal of impedance, measured in siemens (S). - -The admittance **Y** of an element (or a circuit) is the ratio of the phasor current through it to the phasor voltage across it, or - -$$ -Y = \frac{1}{Z} = \frac{I}{V} -$$ - (9.46) - -The admittances of resistors, inductors, and capacitors can be obtained from Eq. (9.39). They are also summarized in Table 9.3. - -As a complex quantity, we may write **Y** as - -$$ -Y = G + jB \tag{9.47} -$$ - -where *G* = Re **Y** is called the *conductance* and *B* = Im **Y** is called the *susceptance*. Admittance, conductance, and susceptance are all expressed in the unit of siemens (or mhos). From Eqs. (9.41) and (9.47), - -$$ -G + jB = \frac{1}{R + jX} \tag{9.48} -$$ - -By rationalization, - -$$ -G + jB = \frac{1}{R + jX} \cdot \frac{R - jX}{R - jX} = \frac{R - jX}{R^2 + X^2} -$$ -(9.49) - -Equating the real and imaginary parts gives - -$$ -G = \frac{R}{R^2 + X^2}, \qquad B = -\frac{X}{R^2 + X^2} -$$ -(9.50) - -showing that *G* ≠ 1∕*R* as it is in resisti ve circuits. Of course, if *X* = 0, then *G* = 1∕*R*. - -Find *v*(*t*) and *i*(*t*) in the circuit shown in Fig. 9.16. Example 9.9 - -# **Solution:** - -From the voltage source 10 cos 4*t*, *ω* = 4, - -$$ -V_s = 10/0^{\circ} V -$$ - -The impedance is - -$$ -\mathbf{Z} = 5 + \frac{1}{j\omega C} = 5 + \frac{1}{j4 \times 0.1} = 5 - j2.5 \ \Omega -$$ - -Hence the current - -$$ -\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{10/0^{\circ}}{5 - j2.5} = \frac{10(5 + j2.5)}{5^2 + 2.5^2} -$$ -(9.9.1) -= 1.6 + j0.8 = 1.789/26.57° A - -The voltage across the capacitor is - -The voltage across the capacitor is -\n -$$ -\mathbf{V} = IZ_C = \frac{I}{j\omega C} = \frac{1.789/26.57^\circ}{j4 \times 0.1} = \frac{1.789/26.57^\circ}{0.4/90^\circ} = 4.47/-63.43^\circ \text{ V} -$$ -\n(9.9.2) - -Converting **I** and **V** in Eqs. (9.9.1) and (9.9.2) to the time domain, we get - -$$ -i(t) = 1.789 \cos(4t + 26.57^{\circ}) \text{ A} -$$ - -$$ -v(t) = 4.47 \cos(4t - 63.43^{\circ}) \text{ V} -$$ - -Notice that *i*(*t*) leads *v*(*t*) by 90° as expected. - -Refer to Fig. 9.17. Determine *v*(*t*) and *i*(*t*). Practice Problem 9.9 - -**Answer:** 8.944 sin (10*t* + 93.43°) V, 4.472 sin(10*t* + 3.43°) A. - -**9.6** Kirchhoff's Laws in the Frequency Domain - -We cannot do circuit analysis in the frequenc y domain without Kirch hoff's current and v oltage laws. Therefore, we need to e xpress them in the frequency domain. - -For KVL, let *v*1,*v*2, … , *vn* be the voltages around a closed loop. Then - -$$ -v_1 + v_2 + \dots + v_n = 0 \tag{9.51} -$$ - -In the sinusoidal steady state, each v oltage may be written in cosine form, so that Eq. (9.51) becomes - -$$ -V_{m1}\cos(\omega t + \theta_1) + V_{m2}\cos(\omega t + \theta_2) -$$ - -+ ... + $V_{mn}\cos(\omega t + \theta_n) = 0$ (9.52) - -# **Figure 9.17** - -For Practice Prob. 9.9. - -This can be written as - -$$ -\text{Re}(V_{m1}e^{j\theta_1}e^{j\omega t}) + \text{Re}(V_{m2}e^{j\theta_2}e^{j\omega t}) + \dots + \text{Re}(V_{mn}e^{j\theta_n}e^{j\omega t}) = 0 -$$ - -or - -$$ -\text{Re}[(V_{m1}e^{j\theta_1} + V_{m2}e^{j\theta_2} + \dots + V_{mn}e^{j\theta_n})e^{j\omega t}] = 0 \tag{9.53} -$$ - -If we let **V***k* = *Vmkejθk*, then - -$$ -Re[(V_1 + V_2 + \dots + V_n)e^{j\omega t}] = 0 -$$ -\n(9.54) - -Because *ejωt* ≠ 0, - -$$ -V_1 + V_2 + \dots + V_n = 0 \tag{9.55} -$$ - -indicating that Kirchhoff's voltage law holds for phasors. - -By following a similar procedure, we can show that Kirchhoff's current law holds for phasors. If we let *i*1, *i*2, … , *in* be the current leaving or entering a closed surface in a network at time *t*, then - -$$ -i_1 + i_2 + \dots + i_n = 0 \tag{9.56} -$$ - -If **I**1, **I**2, … , **I***n* are the phasor forms of the sinusoids *i*1,*i*2, … ,*in*, then - -$$ -\mathbf{I}_1 + \mathbf{I}_2 + \dots + \mathbf{I}_n = 0 \tag{9.57} -$$ - -which is Kirchhoff's current law in the frequency domain. - -Once we have shown that both KVL and KCL hold in the frequency domain, it is easy to do man y things, such as impedance combination, nodal and mesh analyses, superposition, and source transformation. - -# **9.7** Impedance Combinations - -Consider the *N* series-connected impedances sho wn in Fig. 9.18. The same current **I** flows through the impedances. Applying KVL around the loop gives - -$$ -V = V_1 + V_2 + \dots + V_N = I(Z_1 + Z_2 + \dots + Z_N) -$$ -(9.58) - -*N* impedances in series. - -The equivalent impedance at the input terminals is - -$$ -\mathbf{Z}_{\text{eq}} = \frac{\mathbf{V}}{\mathbf{I}} = \mathbf{Z}_1 + \mathbf{Z}_2 + \dots + \mathbf{Z}_N -$$ - -$$ -Z_{eq} = Z_1 + Z_2 + \dots + Z_N \tag{9.59} -$$ - -or - -showing that the total or equi valent impedance of series-connected impedances is the sum of the indi vidual impedances. This is similar to the series connection of resistances. - -If *N* = 2, as shown in Fig. 9.19, the current through the imped ances is - -$$ -\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}_1 + \mathbf{Z}_2} \tag{9.60} -$$ - -Because **V**1 = **Z**1**I** and **V**2 = **Z**2**I**, then - -$$ -V_1 = \frac{Z_1}{Z_1 + Z_2} V, \qquad V_2 = \frac{Z_2}{Z_1 + Z_2} V -$$ - (9.61) - -which is the *voltage-division* relationship. - -In the same manner , we can obtain the equi valent impedance or admittance of the *N* parallel-connected impedances shown in Fig. 9.20. The voltage across each impedance is the same. Applying KCL at the top node, - -$$ -\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 + \dots + \mathbf{I}_N = \mathbf{V} \left( \frac{1}{\mathbf{Z}_1} + \frac{1}{\mathbf{Z}_2} + \dots + \frac{1}{\mathbf{Z}_N} \right) \tag{9.62} -$$ - -*N* impedances in parallel. - -The equivalent impedance is - -$$ -\frac{1}{Z_{\text{eq}}} = \frac{I}{V} = \frac{1}{Z_1} + \frac{1}{Z_2} + \dots + \frac{1}{Z_N} -$$ -(9.63) - -and the equivalent admittance is - -$$ -\mathbf{Y}_{\text{eq}} = \mathbf{Y}_1 + \mathbf{Y}_2 + \dots + \mathbf{Y}_N \tag{9.64} -$$ - -This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances. - -When *N* = 2, as sho wn in Fig. 9.21, the equi valent impedance becomes - -$$ -Z_{\text{eq}} = \frac{1}{Y_{\text{eq}}} = \frac{1}{Y_1 + Y_2} = \frac{1}{1/Z_1 + 1/Z_2} = \frac{Z_1 Z_2}{Z_1 + Z_2} -$$ -(9.65) - -**Figure 9.21** Current division. - -Voltage division. - -Also, since - -$$ -\mathbf{V} = \mathbf{I}\mathbf{Z}_{\text{eq}} = \mathbf{I}_1\mathbf{Z}_1 = \mathbf{I}_2\mathbf{Z}_2 -$$ - -the currents in the impedances are - -$$ -\mathbf{I}_1 = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{I}, \qquad \mathbf{I}_2 = \frac{\mathbf{Z}_1}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{I} -$$ - (9.66) - -which is the *current-division* principle. - -The delta-to-wye and wye-to-delta transformations that we applied to resisti ve circuits are also v alid for impedances. With reference to Fig. 9.22, the conversion formulas are as follows. - -**Figure 9.22** Superimposed *Y* and ∆ networks. - -*Y*-∆ *Conversion:* - -$$ -Z_a = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_1} -$$ - -\n -$$ -Z_b = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_2} -$$ - -\n -$$ -Z_c = \frac{Z_1 Z_2 + Z_2 Z_3 + Z_3 Z_1}{Z_3} -$$ - -\n(9.67) - -∆-*Y Conversion:* - -$$ -\mathbf{Z}_{1} = \frac{\mathbf{Z}_{b}\mathbf{Z}_{c}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}} -$$ -\n -$$ -\mathbf{Z}_{2} = \frac{\mathbf{Z}_{c}\mathbf{Z}_{a}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}} -$$ -\n -$$ -\mathbf{Z}_{3} = \frac{\mathbf{Z}_{a}\mathbf{Z}_{b}}{\mathbf{Z}_{a} + \mathbf{Z}_{b} + \mathbf{Z}_{c}} -$$ -\n(9.68) - -A delta or wye circuit is said to be balanced if it has equal impedances in all three branches. - -When a ∆-*Y* circuit is balanced, Eqs. (9.67) and (9.68) become - -$$ -\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y} \qquad \text{or} \qquad \mathbf{Z}_{Y} = \frac{1}{3}\mathbf{Z}_{\Delta} \qquad (9.69) -$$ - -where **Z***Y* = **Z**1 = **Z**2 = **Z**3 and **Z**∆ = **Z***a* = **Z***b* = **Z***c*. - -As you see in this section, the principles of voltage division, current division, circuit reduction, impedance equi valence, and *Y*-∆ transformation all apply to ac circuits. Chapter 10 will sho w that other c ircuit techniques—such as superposition, nodal analysis, mesh analysis, source transformation, the Thevenin theorem, and the Norton theorem are all applied to ac circuits in a manner similar to their application in dc circuits. - -Find the input impedance of the circuit in Fig. 9.23. Assume that the Example 9.10 circuit operates at *ω* = 50 rad/s. - -# **Solution:** - -Let - -- **Z**1 = Impedance of the 2-mF capacitor -- **Z**2 = Impedance of the 3-Ω resistor in series with the10-mF capacitor -- **Z**3 = Impedance of the 0.2-H inductor in series with the 8-Ω resistor - -Then - -$$ -\mathbf{Z}_1 = \frac{1}{j\omega C} = \frac{1}{j50 \times 2 \times 10^{-3}} = -j10 \text{ }\Omega -$$ - -$$ -\mathbf{Z}_2 = 3 + \frac{1}{j\omega C} = 3 + \frac{1}{j50 \times 10 \times 10^{-3}} = (3 - j2) \text{ }\Omega -$$ - -$$ -\mathbf{Z}_3 = 8 + j\omega L = 8 + j50 \times 0.2 = (8 + j10) \text{ }\Omega -$$ - -The input impedance is - -at impedance is -\n -$$ -\mathbf{Z}_{in} = \mathbf{Z}_1 + \mathbf{Z}_2 \parallel \mathbf{Z}_3 = -j10 + \frac{(3 - j2)(8 + j10)}{11 + j8} -$$ -\n -$$ -= -j10 + \frac{(44 + j14)(11 - j8)}{11^2 + 8^2} = -j10 + 3.22 - j1.07 \Omega -$$ - -Thus, - -$$ -\mathbf{Z}_{\text{in}} = 3.22 - j11.07 \ \Omega -$$ - -For Example 9.10. - -**Figure 9.25** - -**Figure 9.26** - -circuit in Fig. 9.25. - -For Example 9.11. - -The frequency domain equivalent of the - -*vs* = 20 cos(4*t* − 15°) ⇒ **V***s* = 20⧸−15° V, *ω* = 4 10 mF ⇒ \_\_\_\_ 1 *jωC* = \_\_\_\_\_\_\_\_\_\_\_\_ 1 *j*4 × 10 × 10−3 = −*j*25 Ω 5 H ⇒ *jωL* = *j*4 × 5 = *j*20 Ω - - **Z**1 = Impedance of the 60-Ω resistor - - **Z**2 = Impedance of the parallel combination of the 10-mF capacitor and the 5-H inductor - -Then **Z**1 = 60 Ω and - -$$ -\mathbf{Z}_2 = -j25 \parallel j20 = \frac{-j25 \times j20}{-j25 + j20} = j100 \text{ }\Omega -$$ - -By the voltage-division principle, - -$$ -\mathbf{V}_o = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{V}_s = \frac{j100}{60 + j100} (20/15^\circ) -$$ - -= (0.8575/30.96°)(20/15°) = 17.15/15.96° V - -We convert this to the time domain and obtain - -*vo*(*t*) = 17.15 cos(4*t* + 15.96°) V - -For Practice Prob. 9.11. - -# Find current **I** in the circuit of Fig. 9.28. Example 9.12 - -# **Solution:** - -The delta netw ork connected to nodes *a*, *b*, and *c* can be con verted to the *Y* network of Fig. 9.29. We obtain the *Y* impedances as follows using Eq. (9.68): - -68): -\n -$$ -\mathbf{Z}_{an} = \frac{j4(2-j4)}{j4+2-j4+8} = \frac{4(4+j2)}{10} = (1.6+j0.8) \ \Omega -$$ -\n -$$ -\mathbf{Z}_{bn} = \frac{j4(8)}{10} = j3.2 \ \Omega, \qquad \mathbf{Z}_{cn} = \frac{8(2-j4)}{10} = (1.6-j3.2) \ \Omega -$$ - -The total impedance at the source terminals is - -$$ -\mathbf{Z} = 12 + \mathbf{Z}_{an} + (\mathbf{Z}_{bn} - j3) \parallel (\mathbf{Z}_{cn} + j6 + 8) -$$ - -= 12 + 1.6 + j0.8 + (j0.2) || (9.6 + j2.8) -= 13.6 + j0.8 + $\frac{j0.2(9.6 + j2.8)}{9.6 + j3}$ -= 13.6 + j1 = 13.64/4.204° Ω - -The desired current is - -$$ -I = \frac{V}{Z} = \frac{50/0^{\circ}}{13.64/4.204^{\circ}} = 3.666/-4.204^{\circ} A -$$ - -# Practice Problem 9.12 Find **I** in the circuit of Fig. 9.30. - -**Figure 9.30** For Practice Prob. 9.12. - -# **Figure 9.31** - -Series *RC* shift circuits: (a) leading output, (b) lagging output. - -**Answer:** 12.728 ⧸ 63.8° A. - -# **9.8** Applications - -In Chapters 7 and 8, we sa w certain uses of *RC*, *RL*, and *RLC* circuits in dc applications. These circuits also have ac applications; among them are coupling circuits, phase-shifting circuits, filters, resonant circuits, ac bridge circuits, and transformers. This list of applications is ine xhaustive. We will consider some of them later. It will suffice here to observe two simple ones: *RC* phase-shifting circuits, and ac bridge circuits. - -# **9.8.1** Phase-Shifters - -A phase-shifting circuit is often emplo yed to correct an undesirable phase shift already present in a circuit or to produce special desired effects. An *RC* circuit is suitable for this purpose because its capacitor causes the circuit current to lead the applied v oltage. Two commonly used *RC* circuits are shown in Fig. 9.31. (*RL* circuits or any reactive circuits could also serve the same purpose.) - -In Fig. 9.31(a), the circuit current **I** leads the applied voltage **V***i* by some phase angle *θ*, where 0 < *θ*< 90°, depending on the values of *R* and *C*. If *XC* = −1∕*ωC*, then the total impedance is **Z** = *R* + *jXC*, and the phase shift is given by - -$$ -\theta = \tan^{-1} \frac{X_C}{R} -$$ - (9.70) - -This shows that the amount of phase shift depends on the v alues of *R*, *C*, and the operating frequenc y. Since the output v oltage **V***o* across the resistor is in phase with the current, **V***o* leads (positive phase shift) **V***i* as shown in Fig. 9.32(a). - -In Fig. 9.31(b), the output is taken across the capacitor. The current **I** leads the input v oltage **V***i* by *θ*, but the output voltage *vo*(*t*) across the capacitor lags (negative phase shift) the input v oltage *vi*(*t*) as illustrated in Fig. 9.32(b). - -**Figure 9.32** Phase shift in *RC* circuits: (a) leading output, (b) lagging output. - -We should keep in mind that the simple *RC* circuits in Fig. 9.31 also act as voltage dividers. Therefore, as the phase shift *θ* approaches 90°, the output vo ltage **V***o* approaches zero. F or this reason, these simple *RC* circuits are used only when small amounts of phase shift are required. If it is desired to ha ve phase shifts greater than 60°, simple *RC* networks are cascaded, thereby providing a total phase shift equal to the sum of the indi vidual phase shifts. In practice, the phase shifts due to the stages are not equal, because the succeeding stages load down the earlier stages unless op amps are used to sepa rate the stages. - -# **Solution:** - -If we select circuit components of equal ohmic v alue, say *R* = ∣*XC*∣ = 20 Ω, at a particular frequenc y, according to Eq. (9.70), the phase shift is exactly 45°. By cascading two similar *RC* circuits in Fig. 9.31(a), we obtain the circuit in Fig. 9.33, providing a positive or leading phase shift of 90°, as we shall soon show. Using the series-parallel combination technique, **Z** in Fig. 9.33 is obtained as - -$$ -\mathbf{Z} = 20 \| (20 - j20) = \frac{20(20 - j20)}{40 - j20} = 12 - j4 \ \Omega -$$ - (9.13.1) - -Using voltage division, - -$$ -\mathbf{V}_1 = \frac{\mathbf{Z}}{\mathbf{Z} - j20} \mathbf{V}_i = \frac{12 - j4}{12 - j24} \mathbf{V}_i = \frac{\sqrt{2}}{3} \angle 45^\circ \mathbf{V}_i \quad (9.13.2) -$$ - -and - -$$ -\mathbf{V}_o = \frac{20}{20 - j20} \mathbf{V}_1 = \frac{\sqrt{2}}{2} \underline{45^\circ} \mathbf{V}_1 -$$ - (9.13.3) - -Substituting Eq. (9.13.2) into Eq. (9.13.3) yields - -$$ -\mathbf{V}_o = \left(\frac{\sqrt{2}}{2} \angle 45^\circ\right) \left(\frac{\sqrt{2}}{3} \angle 45^\circ \mathbf{V}_i\right) = \frac{1}{3} \angle 90^\circ \mathbf{V}_i -$$ - -Thus, the output leads the input by 90° b ut its magnitude is only about 33 percent of the input. - -Design an *RC* circuit to pro vide a 90° lagging phase shift of the out - Practice Problem 9.13 put voltage relative to the input v oltage. If an ac v oltage of 60 V rms is applied, what is the output voltage? - -**Answer:** Figure 9.34 shows a typical design; 20 V rms. - -# **Figure 9.33** - -An *RC* phase shift circuit with 90° leading phase shift; for Example 9.13. - -**Figure 9.35** For Example 9.14. - -Example 9.14 For the *RL* circuit shown in Fig. 9.35(a), calculate the amount of phase shift produced at 2 kHz. - -# **Solution:** - -At 2 kHz, we transform the 10- and 5-mH inductances to the corresponding impedances. - -$$ -10 \text{ mH} \Rightarrow X_L = \omega L = 2\pi \times 2 \times 10^3 \times 10 \times 10^{-3} -$$ -$$ -= 40\pi = 125.7 \text{ }\Omega -$$ -$$ -5 \text{ mH} \Rightarrow X_L = \omega L = 2\pi \times 2 \times 10^3 \times 5 \times 10^{-3} -$$ -$$ -= 20\pi = 62.83 \text{ }\Omega -$$ - -Consider the circuit in Fig. 9.35(b). The impedance **Z** is the parallel combination of *j*125.7 Ω and 100 + *j*62.83 Ω. Hence, - -$$ -\mathbf{Z} = j125.7 \parallel (100 + j62.83) -$$ - -\n -$$ -\mathbf{Z} = j125.7 \parallel (100 + j62.83) -$$ - -\n -$$ -= \frac{j125.7(100 + j62.83)}{100 + j188.5} = 69.56 / 60.1^{\circ} \ \Omega -$$ - (9.14.1) - -Using voltage division, - -$$ -\mathbf{V}_1 = \frac{\mathbf{Z}}{\mathbf{Z} + 150} \mathbf{V}_i = \frac{69.56 / 60.1^{\circ}}{184.7 + j60.3} \mathbf{V}_i -$$ - -= 0.3582 / 42.02° $\mathbf{V}_i$ (9.14.2) - -and - -$$ -\mathbf{V}_o = \frac{j62.832}{100 + j62.832} \mathbf{V}_1 = 0.532 \, \text{/} 57.86^{\circ} \, \mathbf{V}_1 \tag{9.14.3} -$$ - -Combining Eqs. (9.14.2) and (9.14.3), - -**V***o* = (0.532 ⧸ 57.86°)(0.3582 ⧸ 42.02°)**V***i* = 0.1906 ⧸ 100° **V***i* - -showing that the output is about 19 percent of the input in magnitude but leading the input by 100°. If the circuit is terminated by a load, the load will affect the phase shift. - -Practice Problem 9.14 Refer to the *RL* circuit in Fig. 9.36. If 10 V is applied to the input, find the magnitude and the phase shift produced at 5 kHz. Specify whether the phase shift is leading or lagging. - -**Answer:** 1.7161 V, 120.39°, lagging. - -# **9.8.2** AC Bridges - -An ac bridge circuit is used in measuring the inductance *L* of an inductor or the capacitance *C* of a capacitor. It is similar in form to the Wheatstone bridge for measuring an unknown resistance (discussed in Section 4.10) and follows the same principle. To measure *L* and *C*, however, an ac - -source is needed as well as an ac meter instead of the galvanometer. The ac meter may be a sensitive ac ammeter or voltmeter. - -Consider the general ac bridge circuit displayed in Fig. 9.37. The bridge is *balanced* when no current flows through the meter. This means that **V**1 = **V**2. Applying the voltage division principle, - -$$ -\mathbf{V}_1 = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{V}_s = \mathbf{V}_2 = \frac{\mathbf{Z}_x}{\mathbf{Z}_3 + \mathbf{Z}_x} \mathbf{V}_s -$$ -(9.71) - -Thus, - -$$ -\frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} = \frac{\mathbf{Z}_x}{\mathbf{Z}_3 + \mathbf{Z}_x} \qquad \Rightarrow \qquad \mathbf{Z}_2 \mathbf{Z}_3 = \mathbf{Z}_1 \mathbf{Z}_x \tag{9.72} -$$ - -or - -$$ -Z_x = \frac{Z_3}{Z_1} Z_2 \tag{9.73} -$$ - -This is the balanced equation for the ac bridge and is similar to Eq. (4.30) for the resistance bridge except that the *R*'s are replaced by **Z**'s. - -Specific ac bridges for measuring *L* and *C* are shown in Fig. 9.38, where *Lx* and *Cx* are the unknown inductance and capacitance to be measured while *Ls* and *Cs* are a standard inductance and capacitance (the values of which are kno wn to great precision). In each case, tw o resistors, *R*1 and *R*2, are varied until the ac meter reads zero. Then the bridge is balanced. From Eq. (9.73), we obtain - -$$ -L_x = \frac{R_2}{R_1} L_s \tag{9.74} -$$ - -and - -$$ -C_x = \frac{R_1}{R_2} C_s \tag{9.75} -$$ - -Notice that the balancing of the ac bridges in Fig. 9.38 does not depend on the frequency *f* of the ac source, since *f* does not appear in the relationships in Eqs. (9.74) and (9.75). - -Specific ac bridges: (a) for measuring *L*, (b) for measuring *C*. - -**Figure 9.37** - -A general ac bridge. - -Example 9.15 The ac bridge circuit of Fig. 9.37 balances when **Z**1 is a 1-kΩ resistor, **Z**2 is a 4.2-kΩ resistor, **Z**3 is a parallel combination of a 1.5-MΩ resistor and a 12-pF capacitor, and *f* = 2 kHz. Find: (a) the series com ponents that make up **Z***x*, and (b) the parallel components that make up **Z***x*. - -# **Solution:** - -- 1. **Define.** The problem is clearly stated. -- 2. **Present.** We are to determine the unknown components subject to the fact that they balance the given quantities. Given that a parallel and series equivalent exists for this circuit, we need to find both. -- 3. **Alternative.** Although there are alternative techniques that can be used to find the unknown values, a straightforward equality works best. Once we have answers, we can check them by using hand techniques such as nodal analysis or just using *PSpice*. -- 4. **Attempt.** From Eq. (9.73), - -$$ -Z_{x} = \frac{Z_{3}}{Z_{1}} Z_{2} -$$ - (9.15.1) - -where **Z***x* = *Rx* + *jXx*, - -$$ -\mathbf{Z}_1 = 1000 \,\Omega, \qquad \mathbf{Z}_2 = 4200 \,\Omega \tag{9.15.2} -$$ - -and - -$$ -\mathbf{Z}_3 = R_3 \parallel \frac{1}{j\omega C_3} = \frac{\frac{R_3}{j\omega C_3}}{R_3 + 1/j\omega C_3} = \frac{R_3}{1 + j\omega R_3 C_3} -$$ - -Since *R*3 = 1.5 MΩ and *C*3 = 12 pF, - -Since -$$ -R_3 = 1.5 \text{ M}\Omega -$$ - and $C_3 = 12 \text{ pF}$ , -\n -$$ -\mathbf{Z}_3 = \frac{1.5 \times 10^6}{1 + j2\pi \times 2 \times 10^3 \times 1.5 \times 10^6 \times 12 \times 10^{-12}} = \frac{1.5 \times 10^6}{1 + j0.2262} -$$ - -or - -$$ -Z_3 = 1.427 - j0.3228 M\Omega -$$ - (9.15.3) - -(a) Assuming that **Z***x* is made up of series components, we substitute Eqs. (9.15.2) and (9.15.3) in Eq. (9.15.1) and obtain - -$$ -R_x + jX_x = \frac{4200}{1000}(1.427 - j0.3228) \times 10^6 -$$ - -= (5.993 - j1.356) MΩ (9.15.4) - -Equating the real and imaginary parts yields *Rx* = 5.993 MΩ and a capacitive reactance - -$$ -X_x = \frac{1}{\omega C} = 1.356 \times 10^6 -$$ - -or - -$$ -C = \frac{1}{\omega X_x} = \frac{1}{2\pi \times 2 \times 10^3 \times 1.356 \times 10^6} = 58.69 \text{ pF} -$$ - -(b) *Zx* remains the same as in Eq. (9.15.4) but *Rx* and *Xx* are in parallel. Assuming an *RC* parallel combination, - -$$ -\mathbf{Z}_x = (5.993 - j1.356) \text{ M}\Omega -$$ -$$ -= R_x \parallel \frac{1}{j\omega C_x} = \frac{R_x}{1 + j\omega R_x C_x} -$$ - -By equating the real and imaginary parts, we obtain - -By equating the real and imaginary parts, we obtain -\n -$$ -R_x = \frac{\text{Real}(\mathbf{Z}_x)^2 + \text{Imag}(\mathbf{Z}_x)^2}{\text{Real}(\mathbf{Z}_x)} = \frac{5.993^2 + 1.356^2}{5.993} = 6.3 \text{ M}\Omega -$$ -\n -$$ -C_x = -\frac{\text{Imag}(\mathbf{Z}_x)}{\omega[\text{Real}(\mathbf{Z}_x)^2 + \text{Imag}(\mathbf{Z}_x)^2]} -$$ -\n -$$ -= -\frac{-1.356}{2\pi(2000)(5.917^2 + 1.356^2)} = 2.852 \mu\text{F} -$$ - -We have assumed a parallel *RC* combination which works in this case. - -5. **Evaluate.** Let us now use *PSpice* to see if we indeed have the correct equalities. Running *PSpice* with the equivalent circuits, an open circuit between the "bridge" portion of the circuit, and a 10-volt input voltage yields the following voltages at the ends of the "bridge" relative to a reference at the bottom of the circuit: - -| FREQ | VM(\$N_0002) | VP(\$N_0002) | -|-------------|--------------|--------------| -| 2.000E + 03 | 9.993E + 00 | -8.634E - 03 | -| 2.000E + 03 | 9.993E + 00 | -8.637E - 03 | - -Because the voltages are essentially the same, then no measurable current can flow through the "bridge" portion of the circuit for any element that connects the two points together and we have a balanced bridge, which is to be expected. This indicates we have properly determined the unknowns. - -There is a very important problem with what we have done! Do you know what that is? We have what can be called an ideal, "theoretical" answer, but one that really is not very good in the real world. The difference between the magnitudes of the upper impedances and the lower impedances is much too large and would never be accepted in a real bridge circuit. For greatest accuracy, the overall magnitude of the impedances must at least be within the same relative order. To increase the accuracy of the solution of this problem, I would recommend increasing the magnitude of the top impedances to be in the range of 500 kΩ to 1.5 MΩ. One additional real-world comment: The size of these impedances also creates serious problems in making actual measurements, so the appropriate instruments must be used in order to minimize their loading (which would change the actual voltage readings) on the circuit. - -6. **Satisfactory?** Because we solved for the unknown terms and then tested to see if they worked, we validated the results. They can now be presented as a solution to the problem. - -Practice Problem 9.15 In the ac bridge circuit of Fig. 9.37, suppose that balance is achie ved when **Z**1 is a 4.8-k Ω resistor , **Z**2 is a 10- Ω resistor in series with a 0.25-*μ*H inductor, **Z**3 is a 12-kΩ resistor, and *f* = 6 MHz. Determine the series components that make up **Z***x*. - -**Answer:** A 25-Ω resistor in series with a 0.625-*μ*H inductor. - -# **9.9** Summary - -1. A sinusoid is a signal in the form of the sine or cosine function. It has the general form - -$$ -v(t) = V_m \cos(\omega t + \phi) -$$ - -where *Vm* is the amplitude, *ω* = 2*πf* is the angular frequency, (*ωt* + *ϕ*) is the argument, and *ϕ* is the phase. - -2. A phasor is a complex quantity that represents both the magnitude and the phase of a sinusoid. Given the sinusoid *v*(*t*) = *Vm* cos(*ωt* + *ϕ*), its phasor **V** is - -$$ -\mathbf{V}=V_m\big/\phi -$$ - -- 3. In ac circuits, v oltage and current phasors al ways ha ve a fixed relation to one another at an y moment of time. If *v*(*t*) = *Vm* cos(*ωt* + *ϕv*) represents the v oltage through an element and *i*(*t*) = *Im* cos(*ωt* + *ϕi*) represents the current through the element, then *ϕi* = *ϕv* if the element is a resistor , *ϕi* leads *ϕv* by 90° if the element is a capacitor , and *ϕi* lags *ϕv* by 90° if the element is an inductor. -- 4. The impedance **Z** of a circuit is the ratio of the phasor voltage across it to the phasor current through it: - -$$ -\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = R(\omega) + jX(\omega) -$$ - -The admittance **Y** is the reciprocal of impedance: - -$$ -\mathbf{Y} = \frac{1}{\mathbf{Z}} = G(\omega) + jB(\omega) -$$ - -Impedances are combined in series or in parallel the same w ay as resistances in series or parallel; that is, impedances in series add while admittances in parallel add. - -- 5. For a resistor **Z** = *R*, for an inductor **Z** = *jX* = *jωL*, and for a capacitor **Z** = −*jX* = 1∕*jωC*. -- 6. Basic circuit la ws (Ohm's and Kirchhof f's) apply to ac circuits in the same manner as they do for dc circuits; that is, - -$$ -\mathbf{V} = \mathbf{Z}\mathbf{I} -$$ -$$ -\Sigma \mathbf{I}_k = 0 \quad \text{(KCL)} -$$ -$$ -\Sigma \mathbf{V}_k = 0 \quad \text{(KVL)} -$$ - -- 7. The techniques of voltage/current division, series/parallel combination of impedance/admittance, circuit reduction, and *Y*-∆ transformation all apply to ac circuit analysis. -- 8. AC circuits are applied in phase-shifters and bridges. - -# Review Questions - -**9.1** Which of the following is *not* a right way to express the sinusoid *A* cos *ωt*? - -> (a) *A* cos 2*π ft* (b) *A* cos(2*πt*∕*T*) (c) *A* cos *ω*(*t* − *T*) (d) *A* sin(*ωt* − 90°) - -**9.2** A function that repeats itself after fixed intervals is said to be: - -| (a) a phasor | (b) harmonic | -|--------------|--------------| -| (c) periodic | (d) reactive | - -**9.3** Which of these frequencies has the shorter period? - -(a) 1 krad/s (b) 1 kHz - -- **9.4** If *v*1 = 30 sin(*ωt* + 10°) and *v*2 = 20 sin(*ωt* + 50°), which of these statements are true? - - (a) *v*1 leads *v*2 (b) *v*2 leads *v*1 (c) *v*2 lags *v*1 (d) *v*1 lags *v*2 - - (e) *v*1 and *v*2 are in phase -- **9.5** The voltage across an inductor leads the current through it by 90°. - -(a) True (b) False - -- **9.6** The imaginary part of impedance is called: - - (a) resistance (b) admittance (c) susceptance (d) conductance - - (e) reactance -- **9.7** The impedance of a capacitor increases with increasing frequency. - -(a) True (b) False - -**9.8** At what frequency will the output voltage *vo*(*t*) in Fig. 9.39 be equal to the input voltage *v*(*t*) ? - -| (a) 0 rad/s | (b) 1 rad/s | (c) 4 rad/s | -|-------------|-----------------------|-------------| -| (d) ∞ rad/s | (e) none of the above | | - -# **Figure 9.39** - -For Review Question 9.8. - -**9.9** A series *RC* circuit has ∣*VR*∣ = 12 V and ∣*VC*∣ = 5 V. The magnitude of the supply voltage is: - -(a) −7 V (b) 7 V (c) 13 V (d) 17 V - -**9.10** A series *RCL* circuit has *R* = 30 Ω, *XC* = 50 Ω, and *XL* = 90 Ω. The impedance of the circuit is: - -> (a) 30 + *j*140 Ω (b) 30 + *j*40 Ω (c) 30 − *j*40 Ω (d) −30 − *j*40 Ω (e) −30 + *j*40 Ω - -*Answers: 9.1d, 9.2c, 9.3b, 9.4b,d, 9.5a, 9.6e, 9.7b, 9.8d, 9.9c, 9.10b.* - -# Problems - -# Section 9.2 Sinusoids - -- **9.1** Given the sinusoidal voltage *v*(*t*) = 50 cos (30*t* + 10°) V, find: (a) the amplitude *Vm*, (b) the period *T*, (c) the frequency *f*, and (d) *v*(*t*) at *t* = 10 ms. -- **9.2** A current source in a linear circuit has - -*is* = 15 cos (25 *π t* + 25°) A - -- (a) What is the amplitude of the current? -- (b) What is the angular frequency? -- (c) Find the frequency of the current. -- (d) Calculate *is* at *t* = 2 ms. -- **9.3** Express the following functions in cosine form: - -(a) 10 sin(*ωt* + 30°) (b) −9 sin (8*t*) (c) −20 sin(*ωt* + 45°) - -- **9.4** Design a problem to help other students better understand sinusoids. -- **9.5** Given *v*1 = 45 sin(*ωt* + 30°) V and *v*2 = 50 cos(*ωt* − 30°) V, determine the phase angle between the two sinusoids and which one lags the other. -- **9.6** For the following pairs of sinusoids, determine which one leads and by how much. - -(a) -$$ -v(t) = 10 \cos(4t - 60^{\circ}) -$$ - and - $i(t) = 4 \sin(4t + 50^{\circ})$ - -- (b) *v*1(*t*) = 4 cos(377*t* + 10°) and *v*2(*t*) = −20 cos 377*t* -- (c) *x*(*t*) = 13 cos 2*t* + 5 sin 2*t* and *y*(*t*) = 15 cos(2*t* − 11.8°) - -# Section 9.3 Phasors - -- **9.7** If *f*(*ϕ*) = cos *ϕ* + *j* sin *ϕ*, show that *f*(*ϕ*) = *e jϕ* . -- **9.8** Calculate these complex numbers and express your results in rectangular form: - -(a) -$$ -\frac{60/45^{\circ}}{7.5 - j10} + j2 -$$ - -\n(b) -$$ -\frac{32/20^{\circ}}{(6 - j8)(4 + j2)} + \frac{20}{-10 + j24} -$$ - -\n(c) -$$ -20 + (16/-50^{\circ})(5 + j12) -$$ - -**9.9** Evaluate the following complex numbers and leave your results in polar form: - -(a) -$$ -5/30^{\circ} -$$ - $\left(6 - j8 + \frac{3/60^{\circ}}{2 + j}\right)$ -(b) $\frac{(10/60^{\circ})(35/ -50^{\circ})}{(2 + j6) - (5 + j)}$ - -**9.10** Design a problem to help other students better understand phasors. - -**9.11** Find the phasors corresponding to the following signals: - -(a) -$$ -v(t) = 21 \cos(4t - 15^\circ) \text{V} -$$ - -(b) -$$ -i(t) = -8 \sin(10t + 70^{\circ}) -$$ - mA - -(c) -$$ -v(t) = 120 \sin(10t - 50^{\circ}) -$$ - V - -(d) -$$ -i(t) = -60 \cos(30t + 10^{\circ}) -$$ - mA - -**9.12** Let **X** = 4⧸ 40° and **Y** = 20⧸−30°. Evaluate the following quantities and express your results in polar form: - -$$ -\text{(a)}\,(X+Y)X^* -$$ - -$$ -(b) (X - Y)^* -$$ - -(c) (**X** + **Y**)∕**X** - -**9.13** Evaluate the following complex numbers: - -(a) -$$ -\frac{2+j3}{1-j6} + \frac{7-j8}{-5+j11} -$$ - -\n(b) -$$ -\frac{(5/10°)(10/-40°)}{(4/-80°)(-6/50°)} -$$ - -\n(c) -$$ -\begin{vmatrix} 2+j3 & -j2 \\ -j2 & 8-j5 \end{vmatrix} -$$ - -**9.14** Simplify the following expressions: - -Simplify the following expressions: -\n(a) -$$ -\frac{(5 - j6) - (2 + j8)}{(-3 + j4)(5 - j) + (4 - j6)} -$$ -\n(b) -$$ -\frac{(240/75^\circ + 160/–30^\circ)(60 - j80)}{(67 + j84)(20/32^\circ)} -$$ -\n(c) -$$ -\left(\frac{10 + j20}{3 + j4}\right)^2 \sqrt{(10 + j5)(16 - j20)} -$$ - -**9.15** Evaluate these determinants: - -(a) -$$ -\begin{vmatrix} 10 + j6 & 2 - j3 \ -5 & -1 + j \end{vmatrix} -$$ - -\n(b) -$$ -\begin{vmatrix} 20 \underline{/-30^{\circ}} & -4 \underline{/-10^{\circ}} \\ 16 \underline{/-0^{\circ}} & 3 \underline{/-40^{\circ}} \end{vmatrix} -$$ - -\n(c) -$$ -\begin{vmatrix} 1 - j & -j & 0 \\ j & 1 & -j \\ 1 & j & 1 + j \end{vmatrix} -$$ - -**9.16** Transform the following sinusoids to phasors: - -(a) −20 cos(4*t* + 135°) (b) 8 sin(20*t* + 30°) (c) 20 cos (2*t*) + 15 sin (2*t*) - -- **9.17** Two voltages *v*1 and *v*2 appear in series so that their sum is *v* = *v*1 + *v*2. If *v*1 = 10 cos(50*t* − *π*∕3) V and *v*2 = 12 cos(50*t* + 30°) V, find *v*. -- **9.18** Obtain the sinusoids corresponding to each of the following phasors: - -(a) -$$ -V_1 = 60/15^{\circ} -$$ - V, $\omega = 1$ -\n(b) $V_2 = 6 + j8$ V, $\omega = 40$ -\n(c) $I_1 = 2.8e^{-j\pi/3}$ A, $\omega = 377$ -\n(d) $I_2 = -0.5 - j1.2$ A, $\omega = 10^3$ - -**9.19** Using phasors, find: - -(a) 3 cos(20*t* + 10°) − 5 cos(20*t* − 30°) - -- (b) 40 sin 50*t* + 30 cos(50*t* − 45°) -- (c) 20 sin 400*t* + 10 cos(400*t* + 60°) - -$$ --5\sin(400t-20^\circ) -$$ - -**9.20** A linear network has a current input 7.5 cos(10*t* + 30°) A and a voltage output 120 cos(10*t* + 75°) V. Determine the associated impedance. - -**9.21** Simplify the following: - -(a) -$$ -f(t) = 5 \cos(2t + 15^\circ) - 4 \sin(2t - 30^\circ) -$$ - -(b) $g(t) = 8 \sin t + 4 \cos(t + 50^\circ)$ - -- (c) *h*(*t*) = ∫ 0 (10 cos 40*t* + 50 sin 40*t*) *dt* -- **9.22** An alternating voltage is given by *v*(*t*) = 55 cos(5*t* + 45°) V. Use phasors to find - -$$ -10v(t) + 4\frac{dv}{dt} - 2\int_{-\infty}^{t} v(t) dt -$$ - -Assume that the v alue of the inte gral is zero at *t* = −∞. - -**9.23** Apply phasor analysis to evaluate the following: - -(a) -$$ -v = [110 \sin(20t + 30^\circ) + 220 \cos(20t - 90^\circ)] -$$ - V -(b) $i = [30 \cos(5t + 60^\circ) - 20 \sin(5t + 60^\circ)]$ A - -**9.24** Find *v*(*t*) in the following integrodifferential equations using the phasor approach: - -(a) -$$ -v(t) + \int v dt = 10 \cos t -$$ - -\n(b) $\frac{dv}{dt} + 5v(t) + 4 \int v dt = 20 \sin(4t + 10^{\circ})$ - -**9.25** Using phasors, determine *i*(*t*) in the following equations: - -(a) -$$ -2\frac{di}{dt} + 3i(t) = 4\cos(2t - 45^{\circ}) -$$ - -\n(b) $10 \int i \, dt + \frac{di}{dt} + 6i(t) = 5\cos(5t + 22^{\circ})$ A - -**9.26** The loop equation for a series *RLC* circuit gives - -$$ -\frac{di}{dt} + 2i + \int_{-\infty}^{t} i \, dt = \cos 2t \, A -$$ - -Assuming that the v alue of the inte gral at *t* = −∞ is zero, find *i*(*t*) using the phasor method. - -**9.27** A parallel *RLC* circuit has the node equation - -$$ -\frac{dv}{dt} + 50v + 100 \int v \, dt = 110 \cos(377t - 10^{\circ}) \, \text{V} -$$ - -Determine *v*(*t*) using the phasor method. You may assume that the value of the integral at *t* = −∞ is zero. - -# Section 9.4 Phasor Relationships for Circuit Elements - -- **9.28** Determine the current that flows through an 20-Ω resistor connected to a voltage source *v*s = 120 cos (377*t* + 37°) V. -- **9.29** Given that *vc*(0) = 2 cos(155°) V, what is the instantaneous voltage across a 2-*μ*F capacitor when the current through it is *i* = 4 sin(106 *t* + 25°) A? - -- **9.30** A voltage *v*(*t*) = 100 cos(60*t* + 20°) V is applied to a parallel combination of a 40-kΩ resistor and a 50-*μ*F capacitor. Find the steady-state currents through the resistor and the capacitor. -- **9.31** A series *RLC* circuit has *R* = 80 Ω, *L* = 240 mH, and *C* = 5 mF. If the input voltage is *v*(*t*) = 115 cos 2*t*, find the current flowing through the circuit. -- **9.32** Using Fig. 9.40, design a problem to help other students better understand phasor relationships for circuit elements. - -- **9.33** A series *RL* circuit is connected to a 220-V ac source. If the voltage across the resistor is 170 V, find the voltage across the inductor. -- **9.34** What value of *ω* will cause the forced response, *vo*, in Fig. 9.41 to be zero? - -**Figure 9.41** For Prob. 9.34. - -# Section 9.5 Impedance and Admittance - -**9.35** Find the steady-state current *i* in the circuit of Fig. 9.42, when *vs*(*t*) = 115 cos 200*t* V. - -**Figure 9.42** For Prob. 9.35. - -# **9.36** Using Fig. 9.43, design a problem to help other students better understand impedance. - -**9.37** Determine the admittance **Y** for the circuit in Fig. 9.44. - -# **Figure 9.45** - -For Prob. 9.38. - -**9.39** For the circuit shown in Fig. 9.46, find *Z*eq and use that to find current **I**. Let *ω* = 10 rad/s. - -**Figure 9.46** For Prob. 9.39. - -**9.40** In the circuit of Fig. 9.47, find *io* when: - -(a) -$$ -\omega = 1 -$$ - rad/s (b) $\omega = 5$ rad/s -(c) $\omega = 10$ rad/s - -# **Figure 9.47** For Prob. 9.40. - -**9.41** Find *v*(*t*) in the *RLC* circuit of Fig. 9.48. - -# **Figure 9.48** - -For Prob. 9.41. - -**9.42** Calculate *vo*(*t*) in the circuit of Fig. 9.49. - -# **Figure 9.49** - -For Prob. 9.42. - -**9.43** Find current **I***o* in the circuit shown in Fig. 9.50. - -**Figure 9.50** For Prob. 9.43. - -**9.44** Calculate *i*(*t*) in the circuit of Fig. 9.51. - -**Figure 9.51** For prob. 9.44. - -**Figure 9.52** For Prob. 9.45. - -200 mH 100 mF 2 Ω 2Ω i o *v*s + ‒ - -# **Figure 9.53** - -For Prob. 9.46. - -**9.47** In the circuit of Fig. 9.54, determine the value of *is*(*t*). - -# **Figure 9.54** - -For Prob. 9.47. - -**9.48** Given that *vs*(*t*) = 20 sin(100*t* − 40°) in Fig. 9.55, determine *ix*(*t*). - -# **Figure 9.55** - -For Prob. 9.48. - -**9.49** Find *v s* (*t*) in the circuit of Fig. 9.56 if the current *ix* through the 1-Ω resistor is 8 sin 200*t* A. - -**Figure 9.56** For Prob. 9.49. - -**9.50** Determine *vx* in the circuit of Fig. 9.57. Let *is*(*t*) = 5 cos(100*t* + 40°) A. - -# **Figure 9.57** - -For Prob. 9.50. - -**9.51** If the voltage *vo* across the 2-Ω resistor in the circuit of Fig. 9.58 is 90 cos 2*t* V, obtain *is*. - -# **Figure 9.58** - -For Prob. 9.51. - -**9.52** If **V***o* = 8⧸ 30° V in the circuit of Fig. 9.59, find **I***s*. - -# **Figure 9.59** - -For Prob. 9.52. - -**Figure 9.60** - -For Prob. 9.53. - -**9.54** In the circuit of Fig. 9.61, find **V***s* if **I***o* = 30⧸ 0° A. - -**Figure 9.61** For Prob. 9.54. - -# **Figure 9.62** - -For Prob. 9.55. - -# Section 9.7 Impedance Combinations - -**9.56** At *ω* = 377 rad/s, find the input impedance of the circuit shown in Fig. 9.63. - -**Figure 9.63** For Prob. 9.56. - -**9.57** At *ω* = 1 rad/s, obtain the input admittance in the circuit of Fig. 9.64. - -# **Figure 9.64** - -For Prob. 9.57. - -**9.58** Using Fig. 9.65, design a problem to help other students better understand impedance combinations. - -\* An asterisk indicates a challenging problem. - -**\* 9.59** For the network in Fig. 9.66, find **Z**in. Let *ω* = 100 rad/s. - -**Figure 9.66** - -For Prob. 9.59. - -**9.60** Obtain **Z**in for the circuit in Fig. 9.67. - -**Figure 9.67** - -For Prob. 9.60. - -**9.61** Find **Z**eq in the circuit of Fig. 9.68. - -# **Figure 9.68** - -For Prob. 9.61. - -**9.62** For the circuit in Fig. 9.69, find the input impedance **Z**in at 10 krad/s. - -**Figure 9.69** For Prob. 9.62. - -**9.63** For the circuit in Fig. 9.70, find the value of **Z***T*. - -**Figure 9.70** For Prob. 9.63. - -**9.64** Find **Z***T* and Vo in the circuit in Fig. 9.71. Let the value of the inductance equal *j*20 Ω. - -**9.65** Determine **Z***T* and **I** for the circuit in Fig. 9.72. - -# **Figure 9.72** - -For Prob. 9.65. - -**9.66** For the circuit in Fig. 9.73, calculate **Z***T* and **V***ab*. - -**Figure 9.73** For Prob. 9.66. - -**9.67** At *ω* = 103 rad/s, find the input admittance of each of the circuits in Fig. 9.74. - -# **Figure 9.74** - -For Prob. 9.67. - -**9.68** Determine **Y**eq for the circuit in Fig. 9.75. - -# **Figure 9.75** - -For Prob. 9.68. - -**9.69** Find the equivalent admittance **Y**eq of the circuit in Fig. 9.76. - -# **Figure 9.76** - -For Prob. 9.69. - -**9.70** Find the equivalent impedance of the circuit in Fig. 9.77. - -**Figure 9.77** For Prob. 9.70. - -**9.71** Obtain the equivalent impedance of the circuit in Fig. 9.78. - -# **Figure 9.78** - -For Prob. 9.71. - -**Figure 9.79** - -For Prob. 9.72. - -**9.73** Determine the equivalent impedance of the circuit in Fig. 9.80. - -# Section 9.8 Applications - -- **9.74** Design an *RL* circuit to provide a 90° leading phase shift. -- **9.75** Design a circuit that will transform a sinusoidal -- voltage input to a cosinusoidal voltage output. -- **9.76** For the following pairs of signals, determine if *v*1 leads or lags *v*2 and by how much. - -(a) *v*1 = 10 cos(5*t* − 20°), *v*2 = 8 sin 5*t* - -(b) -$$ -v_1 = 19 \cos(2t + 90^\circ) -$$ -, $v_2 = 6 \sin 2t$ - -(c) -$$ -v_1 = -4 \cos 10t -$$ -, $v_2 = 15 \sin 10t$ - -- **9.77** Refer to the *RC* circuit in Fig. 9.81. - - (a) Calculate the phase shift at 2 MHz. - - (b) Find the frequency where the phase shift is 45°. - -# **Figure 9.81** - -For Prob. 9.77. - -- **9.78** A coil with impedance 8 + *j*6 Ω is connected in series with a capacitive reactance *X*. The series combination is connected in parallel with a resistor *R*. Given that the equivalent impedance of the resulting circuit is 5⧸ 0° Ω, find the value of *R* and *X*. -- **9.79** (a) Calculate the phase shift of the circuit in Fig. 9.82. (b) State whether the phase shift is leading or lagging (output with respect to input). - - (c) Determine the magnitude of the output when the input is 120 V. - -# **Figure 9.82** - -For Prob. 9.79. - -- **9.80** Consider the phase-shifting circuit in Fig. 9.83. Let **V***i* = 120 V operating at 60 Hz. Find: - - (a) **V***o* when *R* is maximum - - (b) **V***o* when *R* is minimum - - (c) the value of *R* that will produce a phase shift of 45° - -# **Figure 9.83** - -For Prob. 9.80. - -- **9.81** The ac bridge in Fig. 9.37 is balanced when *R*1 = 400 Ω, *R*2 = 600 Ω, *R*3 = 1.2 kΩ, and *C*2 = 0.3 *μ*F. Find *Rx* and *Cx*. Assume *R*2 and *C*2 are in series. -- **9.82** A capacitance bridge balances when *R*1 = 100 Ω, *R*2 = 2 kΩ, and *Cs* = 40 *μ*F. What is *Cx*, the capacitance of the capacitor under test? -- **9.83** An inductive bridge balances when *R*1 = 1.2 kΩ, *R*2 = 500 Ω, and *Ls* = 250 mH. What is the value of *Lx*, the inductance of the inductor under test? - -**9.84** The ac bridge shown in Fig. 9.84 is known as a *Maxwell bridge* and is used for accurate measurement of inductance and resistance of a coil in terms of a standard capacitance *Cs*. Show that when the bridge is balanced, - -$$ -L_x = R_2 R_3 C_s \qquad \text{and} \qquad R_x = \frac{R_2}{R_1} R_3 -$$ - -Find *Lx* and *Rx* for *R*1 = 40 k Ω, *R*2 = 1.6 k Ω, *R*3 = 4 kΩ, and *Cs* = 0.45 *μ*F. - -**Figure 9.84** Maxwell bridge; For Prob. 9.84. - -# *f* = \_\_\_\_\_\_\_\_\_\_\_\_ 1 2*π*√ - -**9.86** The circuit shown in Fig. 9.86 is used in a television receiver. What is the total impedance of this circuit? - -Comprehensive Problems - -For Prob. 9.86. - -**9.87** The network in Fig. 9.87 is part of the schematic describing an industrial electronic sensing device. What is the total impedance of the circuit at 4 kHz? - -**Figure 9.87** For Prob. 9.87. - -- (a) What is the impedance of the circuit? -- (b) If the frequency were halved, what would be the impedance of the circuit? - -# **Figure 9.88** - -For Prob. 9.88. - -**9.89** An industrial load is modeled as a series combination of an inductor and a resistance as shown in Fig. 9.89. Calculate the value of a capacitor *C* across the series combination so that the net impedance is resistive at a frequency of 2 kHz. - -For Prob. 9.89. - -**9.90** An industrial coil is modeled as a series combination of an inductance *L* and resistance *R*, as shown in Fig. 9.90. Since an ac voltmeter measures only the magnitude of a sinusoid, the following - -**9.85** The ac bridge circuit of Fig. 9.85 is called a *Wien bridge*. It is used for measuring the frequency of a source. Show that when the bridge is balanced, - -> *\_\_\_\_\_\_\_\_ R*2*R*4*C*2*C*4 - -measurements are taken at 60 Hz when the circuit operates in the steady state: - -$$ -|\mathbf{V}_s| = 145 \text{ V}, \qquad |\mathbf{V}_1| = 50 \text{ V}, \qquad |\mathbf{V}_o| = 110 \text{ V} -$$ - -Use these measurements to determine the values of *L* and *R*. - -# **Figure 9.90** - -For Prob. 9.90. - -**9.91** Figure 9.91 shows a series combination of an inductance and a resistance. If it is desired to connect a capacitor in parallel with the series combination such that the net impedance is resistive at 10 kHz, what is the required value of *C*? - -**Figure 9.91** For Prob. 9.91. - -- **9.92** A transmission line has a series impedance of **Z** = 100⧸ 75° Ω and a shunt admittance of **Y** = 450⧸ 48° *μ*S. Find: (a) the characteristic impedance **Z***o* = √ \_\_\_\_\_ **Z**∕**Y** , (b) the propagation constant *γ* = √ \_\_\_ **ZY** . -- **9.93** A power transmission system is modeled as shown in Fig. 9.92. Given the source voltage and circuit elements - -| Vs = 115⧸ 0° V, | source impedance | -|------------------------|---------------------------| -| Zs = (1 + j0.5) Ω, | line impedance | -| Zt = (0.4 + j0.3) Ω, | and load impedance | -| ZL = (23.2 + j18.9) Ω, | find the load current IL. | - -**Figure 9.92** For Prob. 9.93. - -# **chapter** - -# 10 - -# Sinusoidal Steady-State Analysis - -*Three men are my friends—he that loves me, he that hates me, he that is indifferent to me. Who loves me, teaches me tenderness; who hates me, teaches me caution; who is indifferent to me, teaches me self-reliance.* —J. E. Dinger - -# Enhancing Your Career - -# **Career in Software Engineering** - -Software engineering is that aspect of engineering that deals with the practical application of scientific knowledge in the design, construction, and v alidation of computer programs and the associated documenta ‑ tion required to de velop, operate, and maintain them. It is a branch of electrical engineering that is becoming increasingly important as more and more disciplines require one form of software package or another to perform routine tasks and as programmable microelectronic systems are used in more and more applications. - -The role of a softw are engineer should not be confused with that of a computer scientist; the softw are engineer is a practitioner , not a theoretician. A softw are engineer should ha ve good computer ‑ pro gramming skills and be familiar with programming languages, in par‑ ticular C++, which is becoming increasingly popular. Because hardware and software are closely interlinked, it is essential that a softw are engi‑ neer have a thorough understanding of hardware design. Most important, the software engineer should ha ve some specialized kno wledge of the area in which the software development skill is to be applied. - -All in all, the field of software engineering offers a great career to those who enjo y programming and de veloping softw are packages. The higher rewards will go to those having the best preparation, with the most interesting and challenging opportunities going to those with graduate education. - -A three‑dimensional printing of the output of an AutoCAD model of a NASA flywheel. © Ansoft Corporation - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Analyze electrical circuits in the frequency domain using nodal analysis. -- 2. Analyze electrical circuits in the frequency domain using mesh analysis. -- 3. Apply the superposition principle to frequency domain electrical circuits. -- 4. Apply source transformation in frequency domain circuits. -- 5. Understand how Thevenin and Norton equivalent circuits can be used in the frequency domain. -- 6. Analyze electrical circuits with op amps. - -# **10.1** Introduction - -In Chapter 9, we learned that the forced or steady ‑state response of cir‑ cuits to sinusoidal inputs can be obtained by using phasors. We also know that Ohm's and Kirchhoff's laws are applicable to ac circuits. In this chapter, we want to see ho w nodal analysis, mesh analysis, Thevenin's theorem, Norton's theorem, superposition, and source transformations are applied in analyzing ac circuits. Since these techniques were already introduced for dc circuits, our major effort here will be to illustrate with examples. - -Analyzing ac circuits usually requires three steps. - -# Steps to Analyze AC Circuits: - -- 1. Transform the circuit to the phasor or frequency domain. -- 2. Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.). -- 3. Transform the resulting phasor to the time domain. - -Step 1 is not necessary if the problem is specified in the frequency domain. In step 2, the analysis is performed in the same manner as dc circuit analysis except that complex numbers are involved. Having read Chapter 9, we are adept at handling step 3. - -Toward the end of the chapter , we learn ho w to apply *PSpice* in solving ac circuit problems. We finally apply ac circuit analysis to two practical ac circuits: oscillators and ac transistor circuits. - -# **10.2** Nodal Analysis - -The basis of nodal analysis is Kirchhof f's current la w. Since KCL is valid for phasors, as demonstrated in Section 9.6, we can analyze ac cir‑ cuits by nodal analysis. The following examples illustrate this. - -Frequency domain analysis of an ac circuit via phasors is much easier than analysis of the circuit in the time domain. - -Find *ix* in the circuit of Fig. 10.1 using nodal analysis. Example 10.1 - -For Example 10.1. - -# **Solution:** - -We first convert the circuit to the frequency domain: - -$$ -20 \cos 4t \Rightarrow 20 \underline{/0^{\circ}}, \qquad \Omega = 4 \text{ rad/s} -$$ - -\n -$$ -1 \text{ H} \Rightarrow j \Omega L = j4 -$$ - -\n -$$ -0.5 \text{ H} \Rightarrow j \Omega L = j2 -$$ - -\n -$$ -0.1 \text{ F} \Rightarrow \frac{1}{j \Omega C} = -j2.5 -$$ - -Thus, the frequency domain equivalent circuit is as shown in Fig. 10.2. - -# **Figure 10.2** - -Frequency domain equivalent of the circuit in Fig. 10.1. - -Applying KCL at node 1, - -$$ -\frac{20 - V_1}{10} = \frac{V_1}{-j2.5} + \frac{V_1 - V_2}{j4} -$$ - -or - -(1 + *j*1.5)**V**1 + *j*2.5**V**2 = 20 **(10.1.1)** - -At node 2, - -$$ -2\mathbf{I}_x + \frac{\mathbf{V}_1 - \mathbf{V}_2}{j4} = \frac{\mathbf{V}_2}{j2} -$$ - -But **I***x* = **V**1∕−*j*2.5. Substituting this gives - -$$ -\frac{2V_1}{-j2.5} + \frac{V_1 - V_2}{j4} = \frac{V_2}{j2} -$$ - -By simplifying, we get - -$$ -11V_1 + 15V_2 = 0 \tag{10.1.2} -$$ - -Equations (10.1.1) and (10.1.2) can be put in matrix form as - -$$ -\begin{bmatrix} 1+j1.5 & j2.5 \\ 11 & 15 \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} 20 \\ 0 \end{bmatrix} -$$ - -We obtain the determinants as - -$$ -\Delta = \begin{vmatrix} 1+j1.5 & j2.5 \\ 11 & 15 \end{vmatrix} = 15 - j5 -$$ - -$$ -\Delta_1 = \begin{vmatrix} 20 & j2.5 \\ 0 & 15 \end{vmatrix} = 300, \qquad \Delta_2 = \begin{vmatrix} 1+j1.5 & 20 \\ 11 & 0 \end{vmatrix} = -220 -$$ -$$ -\mathbf{V}_1 = \frac{\Delta_1}{\Delta} = \frac{300}{15 - j5} = 18.97 \underline{/18.43^\circ} \text{ V} -$$ -$$ -\mathbf{V}_2 = \frac{\Delta_2}{\Delta} = \frac{-220}{15 - j5} = 13.91 \underline{/198.3^\circ} \text{ V} -$$ - -The current **I***x* is given by - -$$ -\mathbf{I}_x \text{ is given by} -$$ -\n -$$ -\mathbf{I}_x = \frac{\mathbf{V}_1}{-j2.5} = \frac{18.97 \; / 18.43^\circ}{2.5 \; / -90^\circ} = 7.59 \; / 108.4^\circ \; \text{A} -$$ - -Transforming this to the time domain, - -$$ -i_x = 7.59 \cos(4t + 108.4^\circ) -$$ - A - -Using nodal analysis, find *v*1 and *v*2 Practice Problem 10.1 in the circuit of Fig. 10.3. - -**Figure 10.3** For Practice Prob. 10.1. - -**Answer:** *v*1(*t*) = 28.31 cos(2*t* + 60.01°) V, *v*2(*t*) = 82.56 cos(2*t* + 57.12°) V. - -Compute **V**1 and **V**2 Example 10.2 in the circuit of Fig. 10.4. - -For Example 10.2. - -# **Solution:** - -Nodes 1 and 2 form a supernode as shown in Fig. 10.5. Applying KCL at the supernode gives - -$$ -3 = \frac{\mathbf{V}_1}{-j3} + \frac{\mathbf{V}_2}{j6} + \frac{\mathbf{V}_2}{12} -$$ - -or - -$$ -36 = j4V_1 + (1 - j2)V_2 \tag{10.2.1} -$$ - - - -# **Figure 10.5** - -A supernode in the circuit of Fig. 10.4. - -But a voltage source is connected between nodes 1 and 2, so that - -$$ -V_1 = V_2 + 10 \frac{\angle 45^{\circ}}{\angle 0.2.2} -$$ - -Substituting Eq. (10.2.2) in Eq. (10.2.1) results in - -$$ -36 - 40 \underline{1135^\circ} = (1 + i2) \mathbf{V}_2 \implies \mathbf{V}_2 = 31.41 \underline{18^\circ} \text{ V} -$$ - -From Eq. (10.2.2), - -$$ -V_1 = V_2 + 10 \underline{745^\circ} = 25.78 \underline{770.48^\circ} \text{ V} -$$ - -Calculate -$$ -V_1 -$$ - and $V_2$ in the circuit shown in Fig. 10.6. **Practice Problem 10.2** - -For Practice Prob. 10.2. - -**Answer: V**1 = 96.8 ∕69.66° V, **V**2 = 16.88∕165.72° V. - -# **10.3** Mesh Analysis - -Kirchhoff's voltage law (KVL) forms the basis of mesh analysis. The validity of KVL for ac circuits was shown in Section 9.6 and is illustrated in the follo wing examples. Keep in mind that the v ery nature of using mesh analysis is that it is to be applied to planar circuits. - -Determine current **I***o* in the circuit of Fig. 10.7 using mesh analysis. Example 10.3 - -# **Solution:** - -Applying KVL to mesh 1, we obtain - -$$ -(8+j10-j2)\mathbf{I}_1 - (-j2)\mathbf{I}_2 - j10\mathbf{I}_3 = 0 \tag{10.3.1} -$$ - -**Figure 10.7** For Example 10.3. - -For mesh 2, - -$$ -(4 - j2 - j2)I2 - (-j2)I1 - (-j2)I3 + 20 / 90o = 0 -$$ - (10.3.2) - -For mesh 3, **I**3 = 5. Substituting this in Eqs. (10.3.1) and (10.3.2), we get - -$$ -(8+j8)\mathbf{I}_1 + j2\mathbf{I}_2 = j50 \tag{10.3.3} -$$ - -$$ -j2\mathbf{I}_1 + (4 - j4)\mathbf{I}_2 = -j20 - j10 -$$ - (10.3.4) - -Equations (10.3.3) and (10.3.4) can be put in matrix form as - -[ 8 + *j*8 *j*2 *j*2 4 − *j*4 ] [ **I**1 **I**2 ] = [ *j*50 −*j*30] - -from which we obtain the determinants - -$$ -\Delta = \begin{vmatrix} 8 + j8 & j2 \\ j2 & 4 - j4 \end{vmatrix} = 32(1 + j)(1 - j) + 4 = 68 -$$ - -$$ -\Delta_2 = \begin{vmatrix} 8 + j8 & j50 \\ j2 & -j30 \end{vmatrix} = 340 - j240 = 416.17 \underline{/-35.22^{\circ}} -$$ - -$$ -\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{416.17 \underline{/-35.22^{\circ}}}{68} = 6.12 \underline{/-35.22^{\circ}} A -$$ - -The desired current is - -$$ -I_o = -I_2 = 6.12 \underline{144.78}^{\circ} A -$$ - -For Practice Prob. 10.3. - -# **Solution:** - -As shown in Fig. 10.10, meshes 3 and 4 form a supermesh due to the current source between the meshes. For mesh 1, KVL gives - -$$ --10 + (8 - j2)\mathbf{I}_1 - (-j2)\mathbf{I}_2 - 8\mathbf{I}_3 = 0 -$$ - -or - -$$ -(8 - j2)\mathbf{I}_1 + j2\mathbf{I}_2 - 8\mathbf{I}_3 = 10 \tag{10.4.1} -$$ - -For mesh 2, - -$$ -I_2 = -3 \tag{10.4.2} -$$ - -For the supermesh, - -$$ -(8 - j4)\mathbf{I}_3 - 8\mathbf{I}_1 + (6 + j5)\mathbf{I}_4 - j5\mathbf{I}_2 = 0 \tag{10.4.3} -$$ - -Due to the current source between meshes 3 and 4, at node A, - -$$ -\mathbf{I}_4 = \mathbf{I}_3 + 4 \tag{10.4.4} -$$ - -■ **METHOD 1** Instead of solving the above four equations, we re ‑ duce them to two by elimination. - -Combining Eqs. (10.4.1) and (10.4.2), - -$$ -(8 - j2)\mathbf{I}_1 - 8\mathbf{I}_3 = 10 + j6 \tag{10.4.5} -$$ - -Combining Eqs. (10.4.2) to (10.4.4), - -$$ --8I1 + (14 + j)I3 = -24 - j35 -$$ - (10.4.6) - -**Figure 10.10** Analysis of the circuit in Fig. 10.9. - -From Eqs. (10.4.5) and (10.4.6), we obtain the matrix equation - -$$ -\begin{bmatrix} 8-j2 & -8 \ -8 & 14+j \end{bmatrix} \begin{bmatrix} I_1 \ I_3 \end{bmatrix} = \begin{bmatrix} 10+j6 \ -24-j35 \end{bmatrix} -$$ - -We obtain the following determinants - -$$ -\Delta = \begin{vmatrix} 8 - j2 & -8 \\ -8 & 14 + j \end{vmatrix} = 112 + j8 - j28 + 2 - 64 = 50 - j20 -$$ - -$$ -\Delta_1 = \begin{vmatrix} 10 + j6 & -8 \\ -24 - j35 & 14 + j \end{vmatrix} = 140 + j10 + j84 - 6 - 192 - j280 -$$ - -$$ -= -58 - j186 -$$ - -Current **I**1 is obtained as - -$$ -\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{-58 - j186}{50 - j20} = 3.618 \; \underline{/274.5^\circ} \, \text{A} -$$ - -The required voltage **V***o* is - -$$ -\mathbf{V}_o = -j2(\mathbf{I}_1 - \mathbf{I}_2) = -j2(3.618 \angle 274.5^\circ + 3) -$$ - -= -7.2134 - j6.568 = 9.756 \angle 222.32^\circ \text{V} - -■ **METHOD 2** We can use *MATLAB* to solve Eqs. (10.4.1) to (10.4.4). We first cast the equations as - -$$ -\begin{bmatrix} 8-j2 & j2 & -8 & 0 \ 0 & 1 & 0 & 0 \ -8 & -j5 & 8-j4 & 6+j5 \ 0 & 0 & -1 & 1 \ \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \\ I_3 \\ I_4 \end{bmatrix} = \begin{bmatrix} 10 \\ -3 \\ 0 \\ 4 \end{bmatrix} -$$ - (10.4.7a) - -or - -$$ -AI = B -$$ - -By inverting **A**, we can obtain **I** as - -$$ -\mathbf{I} = \mathbf{A}^{-1} \mathbf{B} \tag{10.4.7b} -$$ - -We now apply *MATLAB* as follows: - -``` ->> A = [(8-j*2) j*2 -8 0; - 0 1 0 0; - -8 -j*5 (8-j*4) (6+j*5); - 0 0 -1 1]; ->> B = [10 -3 0 4]'; ->> I = inv(A)*B -I = - 0.2828 - 3.6069i - -3.0000 - -1.8690 - 4.4276i - 2.1310 - 4.4276i ->> Vo = -2*j*(I(1) - I(2)) -Vo = - -7.2138 - 6.5655i -``` - -as obtained previously. - -Calculate current **I***o* in the circuit of Fig. 10.11. - -**Answer:** 6.089∕5.94° A. - -# **10.4** Superposition Theorem - -Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to dc circuits. The theorem becomes important if the circuit has sources operating at *different* frequencies. In this case, since the impedances depend on frequency, we must have a different frequency domain circuit for each frequency. The total re ‑ sponse must be obtained by adding the individual responses in the *time* domain. It is incorrect to try to add the responses in the phasor or fre‑ quency domain. Why? Because the exponential factor *ejωt* is implicit in sinusoidal analysis, and that factor would change for every angular frequency *ω*. It would therefore not make sense to add responses at different frequencies in the phasor domain. Thus, when a circuit has sources operating at different frequencies, one must add the responses due to the individual frequencies in the time domain. - -Use the superposition theorem to find **I***o* in the circuit in Fig. 10.7. - -# **Solution:** - -Let - -$$ -\mathbf{I}_o = \mathbf{I}_o' + \mathbf{I}_o'' \tag{10.5.1} -$$ - -where **I**′ *o* and **I**″ *o* are due to the voltage and current sources, respectively. To find **I**′ *o*, consider the circuit in Fig. 10.12(a). If we let **Z** be the parallel combination of −*j*2 and 8 + *j*10, then - -2 and 8 + j10, then -\n -$$ -\mathbf{Z} = \frac{-j2(8+j10)}{-2j+8+j10} = 0.25 - j2.25 -$$ - -and current **I**′ *o* is - -$$ -\mathbf{I}'_o = \frac{j20}{4 - j2 + \mathbf{Z}} = \frac{j20}{4.25 - j4.25} -$$ - -or - -$$ -\mathbf{I}'_o = -2.353 + j2.353\tag{10.5.2} -$$ - -To get **I**″ *o*, consider the circuit in Fig. 10.12(b). For mesh 1, - -$$ -(8 + j8)\mathbf{I}_1 - j10\mathbf{I}_3 + j2\mathbf{I}_2 = 0 \tag{10.5.3} -$$ - -For mesh 2, - -$$ -(4 - j4)\mathbf{I}_2 + j2\mathbf{I}_1 + j2\mathbf{I}_3 = 0 \tag{10.5.4} -$$ - -For mesh 3, - -**I**3 = 5 **(10.5.5)** - -**Figure 10.12** Solution of Example 10.5. - -# Example 10.5 - -(a) - -From Eqs. (10.5.4) and (10.5.5), - -$$ -(4 - j4)\mathbf{I}_2 + j2\mathbf{I}_1 + j10 = 0 -$$ - -Expressing **I**1 in terms of **I**2 gives - -$$ -\mathbf{I}_1 = (2 + j2)\mathbf{I}_2 - 5\tag{10.5.6} -$$ - -Substituting Eqs. (10.5.5) and (10.5.6) into Eq. (10.5.3), we get - -$$ -(8+j8)[(2+j2)I2 - 5] - j50 + j2I2 = 0 -$$ - -or - -$$ -I_2 = \frac{90 - j40}{34} = 2.647 - j1.176 -$$ - -Current **I**″ *o* is obtained as - -$$ -\mathbf{I}_{o}'' = -\mathbf{I}_{2} = -2.647 + j1.176 -$$ - (10.5.7) - -From Eqs. (10.5.2) and (10.5.7), we write - -$$ -I_o = I'_o + I''_o = -5 + j3.529 = 6.12 \underline{144.78^\circ} A -$$ - -which agrees with what we got in Example 10.3. It should be noted that applying the superposition theorem is not the best way to solve this prob‑ lem. It seems that we have made the problem twice as hard as the origi ‑ nal one by using superposition. However, in Example 10.6, superposi ‑ tion is clearly the easiest approach. - -| Practice Problem 10.5 | Find current | Io in the circuit of Fig. 10.8 using the superposition | -|-----------------------|--------------|--------------------------------------------------------| -| | theorem. | | -| | | | - -**Answer:** 5.97∕65.45° A. - -Because the circuit operates at three different frequencies (*ω* = 0 for the dc voltage source), one way to obtain a solution is to use superposition, which breaks the problem into single‑frequency problems. So we let - -$$ -v_o = v_1 + v_2 + v_3 \tag{10.6.1} -$$ - -where *v*1 is due to the 5‑V dc voltage source, *v*2 is due to the 10 cos 2*t* V voltage source, and *v*3 is due to the 2 sin 5*t* A current source. - -To find *v*1, we set to zero all sources except the 5‑V dc source. We recall that at steady state, a capacitor is an open circuit to dc while an inductor is a short circuit to dc. There is an alternative way of looking at this. Because *ω* = 0, *jωL* = 0, 1∕*jωC* = ∞. Either way, the equivalent circuit is as shown in Fig. 10.14(a). By voltage division, - -$$ --v_1 = \frac{1}{1+4} (5) = 1 \text{ V} -$$ - (10.6.2) - -To find *v*2, we set to zero both the 5‑V source and the 2 sin 5*t* current source and transform the circuit to the frequency domain. - -10 cos 2t -$$ -\Rightarrow -$$ - $10\underline{/0}^{\circ}$ , $\omega = 2 \text{ rad/s}$ -2 H $\Rightarrow$ $j\omega L = j4 \Omega$ -0.1 F $\Rightarrow$ $\frac{1}{j\omega C} = -j5 \Omega$ - -The equivalent circuit is now as shown in Fig. 10.14(b). Let - -$$ -\mathbf{Z} = -j5 \| 4 = \frac{-j5 \times 4}{4 - j5} = 2.439 - j1.951 -$$ - -**Figure 10.14** - -Solution of Example 10.6: (a) setting all sources to zero except the 5‑V dc source, (b) setting all sources to zero except the ac voltage source, (c) setting all sources to zero except the ac current source. - -By voltage division, - -oltage division, -\n -$$ -\mathbf{V}_2 = \frac{1}{1 + j4 + \mathbf{Z}} (10/0^\circ) = \frac{10}{3.439 + j2.049} = 2.498 \underline{\text{ } 20.79^\circ} -$$ - -In the time domain, - -$$ -v_2 = 2.498 \cos(2t - 30.79^\circ) \tag{10.6.3} -$$ - -To obtain *v*3, we set the voltage sources to zero and transform what is left to the frequency domain. - -$$ -2 \sin 5t \Rightarrow 2(-90^\circ, \omega = 5 \text{ rad/s}) -$$ - -$$ -2 \text{ H} \Rightarrow j\omega L = j10 \Omega -$$ - -$$ -0.1 \text{ F} \Rightarrow \frac{1}{j\omega C} = -j2 \Omega -$$ - -The equivalent circuit is in Fig. 10.14(c). Let - -$$ -\mathbf{Z}_1 = -j2 \parallel 4 = \frac{-j2 \times 4}{4 - j2} = 0.8 - j1.6 \ \Omega -$$ - -By current division, - -$$ -\mathbf{I}_1 = \frac{j10}{j10 + 1 + \mathbf{Z}_1} (2 \angle -90^\circ) \text{ A} -$$ -$$ -\mathbf{V}_3 = \mathbf{I}_1 \times 1 = \frac{j10}{1.8 + j8.4} (-j2) = 2.328 \angle -80^\circ \text{ V} -$$ - -In the time domain, - -$$ -v_3 = 2.33 \cos(5t - 80^\circ) = 2.33 \sin(5t + 10^\circ) \text{ V} \qquad (10.6.4) -$$ - -Substituting Eqs. (10.6.2) to (10.6.4) into Eq. (10.6.1), we have - -$$ -v_o(t) = -1 + 2.498 \cos(2t - 30.79^\circ) + 2.33 \sin(5t + 10^\circ) \text{ V} -$$ - -Practice Problem 10.6 Calculate *vo* in the circuit of Fig. 10.15 using the superposition theorem. - -**Answer:** 11.577 sin(5*t* − 81.12°) + 3.154 cos(10*t* − 86.24°) V. - -# **10.5** Source Transformation - -As Fig. 10.16 shows, source transformation in the frequency domain involves transforming a voltage source in series with an impedance to a current source in parallel with an impedance, or vice versa. As we go from one source type to another, we must keep the following relationship in mind: - -$$ -\mathbf{V}_s = \mathbf{Z}_s \mathbf{I}_s \qquad \Leftrightarrow \qquad \mathbf{I}_s = \frac{\mathbf{V}_s}{\mathbf{Z}_s} \qquad (10.1) -$$ - -Calculate **V***x* in the circuit of Fig. 10.17 using the method of source transformation. - -For Example 10.7. - -# **Solution:** - -We transform the voltage source to a current source and obtain the cir ‑ cuit in Fig. 10.18(a), where - -$$ -I_s = \frac{20/-90^{\circ}}{5} = 4/-90^{\circ} = -j4 \text{ A} -$$ - -The parallel combination of 5‑Ω resistance and (3 + *j*4) impedance gives - -$$ -\mathbf{Z}_1 = \frac{5(3+j4)}{8+j4} = 2.5 + j1.25 \ \Omega -$$ - -Converting the current source to a voltage source yields the circuit in Fig. 10.18(b), where - -$$ -\mathbf{V}_s = \mathbf{I}_s \mathbf{Z}_1 = -j4(2.5 + j1.25) = 5 - j10 \text{ V} -$$ - -**Figure 10.18** Solution of the circuit in Fig. 10.17. - -By voltage division, - -oltage division, -$$ -\mathbf{V}_x = \frac{10}{10 + 2.5 + j1.25 + 4 - j13} (5 - j10) = 5.519 \underline{/-28}^\circ \text{ V} -$$ - -Example 10.7 - -Practice Problem 10.7 Find **I***o* in the circuit of Fig. 10.19 using the concept of source transformation. - -**Answer:** 1.9727∕99.46° A. - -**10.6** Thevenin and Norton Equivalent Circuits - -Thevenin's and Norton's theorems are applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The frequency domain version of a Thevenin equivalent circuit is depicted in Fig. 10.20, where a linear circuit is replaced by a voltage source in series with an impedance. The Norton equivalent circuit is illustrated in Fig. 10.21, where a linear cir ‑ cuit is replaced by a current source in parallel with an impedance. Keep in mind that the two equivalent circuits are related as - -$$ -\mathbf{V}_{\mathrm{Th}} = \mathbf{Z}_N \mathbf{I}_N, \qquad \mathbf{Z}_{\mathrm{Th}} = \mathbf{Z}_N \tag{10.2} -$$ - -just as in source transformation. **V**Th is the open‑circuit voltage while **I***N* is the short‑circuit current. - -If the circuit has sources operating at dif ferent frequencies (see Example 10.6, for example), the Thevenin or Norton equivalent circuit must be determined at each frequenc y. This leads to entirely dif ferent equivalent circuits, one for each frequenc y, not one equi valent circuit with equivalent sources and equivalent impedances. - -Example 10.8 Obtain the Thevenin equivalent at terminals *a*‑*b* of the circuit in Fig. 10.22. - -**Figure 10.22** For Example 10.8. - -**Figure 10.21** Norton equivalent. - -# **Solution:** - -We find **Z**Th by setting the voltage source to zero. As shown in Fig. 10.23(a), the 8 ‑Ω resistance is now in parallel with the −*j*6 reac‑ tance, so that their combination gives - -$$ -\mathbf{Z}_1 = -j6 \| 8 = \frac{-j6 \times 8}{8 - j6} = 2.88 - j3.84 \ \Omega -$$ - -Similarly, the 4 ‑Ω resistance is in parallel with the *j*12 reactance, and their combination gives - -$$ -\mathbf{Z}_2 = 4 || j12 = \frac{j12 \times 4}{4 + j12} = 3.6 + j1.2 \ \Omega -$$ - -The Thevenin impedance is the series combination of **Z**1 and **Z**2; that is, - -$$ -\mathbf{Z}_{\text{Th}} = \mathbf{Z}_1 + \mathbf{Z}_2 = 6.48 - j2.64 \ \Omega -$$ - -To find **V**Th, consider the circuit in Fig. 10.23(b). Currents **I**1 and **I**2 are obtained as - -$$ -\mathbf{I}_1 = \frac{120/75^{\circ}}{8 - j6} \text{ A}, \qquad \mathbf{I}_2 = \frac{120/75^{\circ}}{4 + j12} \text{ A} -$$ - -Applying KVL around loop *bcdeab* in Fig. 10.23(b) gives - -$$ -\mathbf{V}_{\mathrm{Th}} - 4\mathbf{I}_2 + (-j6)\mathbf{I}_1 = 0 -$$ - -or - -$$ -\mathbf{V}_{\text{Th}} = 4\mathbf{I}_2 + j6\mathbf{I}_1 = \frac{480/75^\circ}{4 + j12} + \frac{720/75^\circ + 90^\circ}{8 - j6} -$$ -$$ -= 37.95/3.43^\circ + 72/201.87^\circ -$$ -$$ -= -28.936 - j24.55 = 37.95/220.31^\circ \text{ V} -$$ - -Practice Problem 10.8 Find the Thevenin equivalent at terminals *a*‑*b* of the circuit in Fig. 10.24. - -For Practice Prob. 10.8. - -**Answer: Z**Th = 12.4 − *j*3.2 Ω, **V**Th = 47.43∕−51.57° V. - -# Example 10.9 - -Find the Thevenin equivalent of the circuit in Fig. 10.25 as seen from terminals *a*‑*b*. - -# **Figure 10.25** - -For Example 10.9. - -# **Solution:** - -To find **V**Th, we apply KCL at node 1 in Fig. 10.26(a). - -$$ -15 = Io + 0.5Io \Rightarrow Io = 10 A -$$ - -Applying KVL to the loop on the right ‑hand side in Fig. 10.26(a), we obtain - -$$ --I_o(2-j4) + 0.5I_o(4+j3) + V_{Th} = 0 -$$ - -or - -$$ -\mathbf{V}_{\text{Th}} = 10(2 - j4) - 5(4 + j3) = -j55 -$$ - -Thus, the Thevenin voltage is - -$$ -V_{\text{Th}} = 55 \angle -90^{\circ} \text{ V} -$$ - -**Figure 10.26** Solution of the problem in Fig. 10.25: (a) finding **V**Th, (b) finding **Z**Th. - -To obtain **Z**Th, we remove the independent source. Due to the presence of the dependent current source, we connect a 3-A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals *a*-*b* as shown in Fig. 10.26(b). At the node, KCL gives - -$$ -3 = I_o + 0.5I_o \qquad \Rightarrow \qquad I_o = 2A -$$ - -Applying KVL to the outer loop in Fig. 10.26(b) gives - -$$ -\mathbf{V}_s = \mathbf{I}_o(4 + j3 + 2 - j4) = 2(6 - j) -$$ - -The Thevenin impedance is - -$$ -\mathbf{Z}_{\text{Th}} = \frac{\mathbf{V}_s}{\mathbf{I}_s} = \frac{2(6-j)}{3} = 4 - j0.6667 \ \Omega -$$ - -Determine the Thevenin equivalent of the circuit in Fig. 10.27 as seen from the terminals *a*-*b*. - -**Answer:** -$$ -\mathbb{Z}_{\text{Th}} = 4.473 \underline{\smash{\big)}\, -7.64^{\circ}} \,\Omega, \, \mathbf{V}_{\text{Th}} = 11.763 \underline{\smash{\big)}\, 72.9^{\circ}} \text{ volts.} -$$ - -**Figure 10.27** For Practice Prob. 10.9. - -Obtain current **I***o* in Fig. 10.28 using Norton's theorem. Example 10.10 - -For Example 10.10. - -# **Solution:** - -Our first objective is to find the Norton equivalent at terminals *a*-*b*. **Z***N* is found in the same way as **Z**Th. We set the sources to zero as shown in Fig. 10.29(a). As evident from the figure, the (8 − *j*2) and (10 + *j*4) impedances are short-circuited, so that - -$$ -\mathbf{Z}_N = 5 \ \Omega -$$ - -To get **I***N*, we short-circuit terminals *a*-*b* as in Fig. 10.29(b) and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1, - -$$ --j40 + (18 + j2)\mathbf{I}_1 - (8 - j2)\mathbf{I}_2 - (10 + j4)\mathbf{I}_3 = 0 \tag{10.10.1} -$$ - -Solution of the circuit in Fig. 10.28: (a) finding **Z***N*, (b) finding **V***N*, (c) calculating **I***o*. - -For the supermesh, - -$$ -(13 - j2)I2 + (10 + j4)I3 - (18 + j2)I1 = 0 \t(10.10.2) -$$ - -At node *a*, due to the current source between meshes 2 and 3, - -$$ -I_3 = I_2 + 3 \tag{10.10.3} -$$ - -Adding Eqs. (10.10.1) and (10.10.2) gives - -$$ --j40 + 5\mathbf{I}_2 = 0 \qquad \Rightarrow \qquad \mathbf{I}_2 = j8 -$$ - -From Eq. (10.10.3), - -$$ -\mathbf{I}_3 = \mathbf{I}_2 + 3 = 3 + j8 -$$ - -The Norton current is - -$$ -\mathbf{I}_N = \mathbf{I}_3 = (3 + j8) \text{ A} -$$ - -Figure 10.29(c) shows the Norton equivalent circuit along with the im ‑ pedance at terminals *a*‑*b*. By current division, - -$$ -\mathbf{I}_o = \frac{5}{5 + 20 + j15} \mathbf{I}_N = \frac{3 + j8}{5 + j3} = 1.465 / 38.48^{\circ} \text{ A} -$$ - -# Practice Problem 10.10 - -Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from terminals *a*‑*b*. Use the equivalent to find **I***o*. - -# **Figure 10.30** - -For Practice Prob. 10.10 and Prob. 10.35. - -**Answer: Z***N* = 3.176 + *j*0.706 Ω, **I***N* = 8.396⧸−32.68° A, **I***o* = 1.9714⧸−2.10° A. - -# **10.7** Op Amp AC Circuits - -The three steps stated in Section 10.1 also apply to op amp circuits, as long as the op amp is operating in the linear region. As usual, we will assume ideal op amps. (See Section 5.2.) As discussed in Chapter 5, the key to analyzing op amp circuits is to keep two important properties of an ideal op amp in mind: - -- 1. No current enters either of its input terminals. -- 2. The voltage across its input terminals is zero. - -The following examples will illustrate these ideas. - -# **Figure 10.31** - -For Example 10.11: (a) the original circuit in the time domain, (b) its frequency domain equivalent. - -# **Solution:** - -We first transform the circuit to the frequency domain, as shown in Fig. 10.31(b), where **V***s* = 3⧸ 0°, *ω* = 1000 rad/s. Applying KCL at node 1, we obtain - -$$ -\frac{3/0^{\circ} - V_1}{10} = \frac{V_1}{-j5} + \frac{V_1 - 0}{10} + \frac{V_1 - V_o}{20} -$$ - -or - -$$ -6 = (5 + j4)\mathbf{V}_1 - \mathbf{V}_o \tag{10.11.1} -$$ - -At node 2, KCL gives - -$$ -\frac{\mathbf{V}_1 - 0}{10} = \frac{0 - \mathbf{V}_o}{-j10} -$$ - -which leads to - -$$ -\mathbf{V}_1 = -j\mathbf{V}_o \tag{10.11.2} -$$ - -Substituting Eq. (10.11.2) into Eq. (10.11.1) yields - -$$ -6 = -j(5 + j4)\mathbf{V}_o - \mathbf{V}_o = (3 - j5)\mathbf{V}_o -$$ -$$ -\mathbf{V}_o = \frac{6}{3 - j5} = 1.029 \angle 59.04^\circ -$$ - -Hence, - -$$ -v_o(t) = 1.029 \cos(1000t + 59.04^{\circ}) \text{ V} -$$ - -Practice Problem 10.11 Find *vo* and *io* in the op amp circuit of Fig. 10.32. Let *vs*= 12 cos 5000*t* V. - -**Answer:** 4 sin 5,000*t* V, 400 sin 5,000*t μ*A. - -**Figure 10.33** For Example 10.12. - -Example 10.12 Compute the closed ‑loop g ain and phase shift for the circuit in Fig. 10.33. Assume that *R*1 = *R*2 = 10 kΩ, *C*1 = 2 *μ*F, *C*2 = 1 *μ*F, and *ω* = 200 rad/s. - -# **Solution:** - -The feedback and input impedances are calculated as - -$$ -\mathbf{Z}_f = R_2 \left\| \frac{1}{j\omega C_2} = \frac{R_2}{1 + j\omega R_2 C_2} -$$ -$$ -\mathbf{Z}_i = R_1 + \frac{1}{j\omega C_1} = \frac{1 + j\omega R_1 C_1}{j\omega C_1} -$$ - -Since the circuit in Fig. 10.33 is an inverting amplifier, the closed‑loop gain is given by - -$$ -G = \frac{V_o}{V_s} = \frac{Z_f}{Z_i} = \frac{-j\omega C_1 R_2}{(1 + j\omega R_1 C_1)(1 + j\omega R_2 C_2)} -$$ - -Substituting the given values of *R*1, *R*2, *C*1, *C*2, and *ω*, we obtain - -given values of -$$ -R_1 -$$ -, $R_2$ , $C_1$ , $C_2$ , and $\omega$ , -\n -$$ -G = \frac{-j4}{(1+j4)(1+j2)} = 0.434 / 130.6^{\circ} -$$ - -Thus, the closed‑loop gain is 0.434 and the phase shift is 130.6°. - -Practice Problem 10.12 Obtain the closed‑loop gain and phase shift for the circuit in Fig. 10.34. Let *R* = 10 kΩ, *C* = 1 *μ*F, and *ω* = 1000 rad/s. - -**Answer:** 1.0147, −5.6°. - -# **10.8** AC Analysis Using PSpice - -*PSpice* affords a big relief from the tedious task of manipulating com‑ plex numbers in ac circuit analysis. The procedure for using *PSpice* for ac analysis is quite similar to that required for dc analysis. The reader should read Section D.5 in Appendix D for a review of *PSpice* concepts for ac analysis. AC circuit analysis is done in the phasor or frequency domain, and all sources must have the same frequency. Although ac analysis with *PSpice* involves using AC Sweep, our analysis in this chapter requires a single frequency *f* = *ω*∕2*π*. The out‑ put file of *PSpice* contains voltage and current phasors. If necessary, the impedances can be calculated using the voltages and currents in the output file. - -Obtain *vo* and *io* in the circuit of Fig. 10.35 using *PSpice*. Example 10.13 - -# **Solution:** - -We first convert the sine function to cosine. - -$$ -8 \sin(1000t + 50^{\circ}) = 8 \cos(1000t + 50^{\circ} - 90^{\circ}) -$$ -$$ -= 8 \cos(1000t - 40^{\circ}) -$$ - -The frequency *f* is obtained from *ω* as - -$$ -f = \frac{\omega}{2\pi} = \frac{1000}{2\pi} = 159.155 -$$ - Hz - -The schematic for the circuit is shown in Fig. 10.36. Notice that the current‑controlled current source F1 is connected such that its current flows from node 0 to node 3 in conformity with the original circuit in Fig. 10.35. Since we only want the magnitude and phase of *vo* and *io*, we set the attributes of IPRINT and VPRINT1 each to *AC* = *yes*, *MAG* = *yes*, *PHASE* = *yes*. As a single ‑frequency analysis, we select **Analysis/ Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 159.155, and *Final Freq* = 159.155. After saving the schematic, we simulate it by selecting **Analysis/Simulate.** The output file includes the source fre‑ quency in addition to the attributes checked for the pseudocomponents IPRINT and VPRINT1, - -| FREQ | | IM(V_PRINT3) IP(V_PRINT3) | -|-----------|-----------|---------------------------| -| 1.592E+02 | 3.264E–03 | –3.743E+01 | -| FREQ | VM(3) | VP(3) | -| 1.592E+02 | 1.550E+00 | –9.518E+01 | - -**Figure 10.36** - -The schematic of the circuit in Fig. 10.35. - -From this output file, we obtain - -**V***o* = 1.55⧸−95.18° V, **I***o* = 3.264⧸−37.43° mA - -which are the phasors for - -*vo* = 1.55 cos(1000*t* − 95.18°) = 1.55 sin(1000*t* − 5.18°) V - -and - -*io* = 3.264 cos(1000*t* − 37.43°) mA - -For Practice Prob. 10.13. - -**Answer:** 3.219 cos(3,000*t* − 154.6°) V, 6.527 cos(3,000*t* − 55.12°) mA. - -Example 10.14 - -Find **V**1 and **V**2 in the circuit of Fig. 10.38. - -# **Solution:** - -1. **Define.** In its present form, the problem is clearly stated. Again, we must emphasize that time spent here will save lots of time and expense later on! One thing that might have created a problem for you is that, if the reference was missing for this problem, you would then need to ask the individual assigning the problem where - -it is to be located. If you could not do that, then you would need to assume where it should be and then clearly state what you did and why you did it. - -- 2. **Present.** The given circuit is a frequency domain circuit and the unknown node voltages **V**1 and **V**2 are also frequency domain values. Clearly, we need a process to solve for these unknowns in the frequency domain. -- 3. **Alternative.** We have two direct alternative solution techniques that we can easily use. We can do a straightforward nodal analysis approach or use *PSpice*. Since this example is in a section dedicated to using *PSpice* to solve problems, we will use *PSpice* to find **V**1 and **V**2. We can then use nodal analysis to check the answer. -- 4. **Attempt.** The circuit in Fig. 10.35 is in the time domain, whereas the one in Fig. 10.38 is in the frequency domain. Since we are not given a particular frequency and *PSpice* requires one, we select any frequency consistent with the given impedances. For example, if we select *ω* = 1 rad/s, the corresponding frequency is *f* = *ω*∕2*π* = 0.15916 Hz. We obtain the values of the capacitance (*C* = 1∕*ωXC*) and inductances (*L* = *XL*∕*ω*). Making these changes results in the schematic in Fig. 10.39. To ease wiring, we have exchanged the positions of the voltage‑controlled current source - -# **Figure 10.39** Schematic for the circuit in the Fig. 10.38. - -G1 and the 2 + *j*2 Ω impedance. Notice that the current of G1 flows from node 1 to node 3, while the controlling voltage is across the capacitor C2, as required in Fig. 10.38. The attributes of pseudo components VPRINT1 are set as shown. As a single‑ frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 0.15916, and *Final Freq* = 0.15916. After saving the schematic, we select **Analysis/Simulate** to simulate the circuit. When this is done, the output file includes - -| FREQ | VM(1) | VP(1) | -|-----------|-----------|------------| -| 1.592E–01 | 2.708E+00 | –5.673E+01 | -| | | | -| FREQ | VM(3) | VP(3) | -| 1.592E-01 | 4.468E+00 | –1.026E+02 | - -from which we obtain, - -**V**1 = **2.708**⧸**−56.74° V** and **V**2 = **6.911**⧸**−80.72° V** - -5. **Evaluate.** One of the most important lessons to be learned is that when using programs such as *PSpice* you still need to validate the answer. There are many opportunities for making a mistake, including coming across an unknown "bug" in *PSpice* that yields incorrect results. - -So, how can we validate this solution? Obviously, we can rework the entire problem with nodal analysis, and perhaps using *MATLAB*, to see if we obtain the same results. There is another way we will use here: Write the nodal equations and substitute the answers obtained in the *PSpice* solution, and see if the nodal equations are satisfied. - -The nodal equations for this circuit are given below. Note we have substituted **V**1 = **V***x* into the dependent source. - -$$ --3 + \frac{\mathbf{V}_1 - 0}{1} + \frac{\mathbf{V}_1 - 0}{-j1} + \frac{\mathbf{V}_1 - \mathbf{V}_2}{2 + j2} + 0.2\mathbf{V}_1 + \frac{\mathbf{V}_1 - \mathbf{V}_2}{-j2} = 0 -$$ - -(1 + j + 0.25 - j0.25 + 0.2 + j0.5) $\mathbf{V}_1$ --(0.25 - j0.25 + j0.5) $\mathbf{V}_2$ = 3 -(1.45 + j1.25) $\mathbf{V}_1$ - (0.25 + j0.25) $\mathbf{V}_2$ = 3 -1.9144/40.76° $\mathbf{V}_1$ - 0.3536/45° $\mathbf{V}_2$ = 3 - -Now, to check the answer, we substitute the *PSpice* answers into this. - -$$ -1.9144 \underline{/40.76^{\circ}} \times 2.708 \underline{/-56.74^{\circ}} - 0.3536 \underline{/45^{\circ}} \times 6.911 \underline{/-80.72^{\circ}} -$$ - -= 5.184 \underline{/-15.98^{\circ}} - 2.444 \underline{/-35.72^{\circ}} -= 4.984 - j1.4272 - 1.9842 + j1.4269 -= 3 - j0.0003 [Answer checks] - -6. **Satisfactory?** Although we used only the equation from node 1 to check the answer, this is more than satisfactory to validate the answer from the *PSpice* solution. We can now present our work as a solution to the problem. - -Obtain **V***x* and **I***x* in the circuit depicted in Fig. 10.40. Practice Problem 10.14 - -For Practice Prob. 10.14. - -**Answer:** 39.37⧸ 44.78° V, 10.336⧸158° A. - -# **10.9** Applications - -The concepts learned in this chapter will be applied in later chapters to calculate electric power and determine frequency response. The con ‑ cepts are also used in analyzing coupled circuits, three‑phase circuits, ac transistor circuits, filters, oscillators, and other ac circuits. In this section, we apply the concepts to develop two practical ac circuits: the capaci ‑ tance multiplier and the sine wave oscillators. - -# **10.9.1** Capacitance Multiplier - -The op amp circuit in Fig. 10.41 is known as a *capacitance multiplier*, for reasons that will become obvious. Such a circuit is used in integrated‑ circuit technology to produce a multiple of a small physical capacitance *C* when a large capacitance is needed. The circuit in Fig. 10.41 can be used to multiply capacitance values by a factor up to 1,000. For exam‑ ple, a 10‑pF capacitor can be made to behave like a 100‑nF capacitor. - -In Fig. 10.41, the first op amp operates as a voltage follower, while the second one is an inverting amplifier. The voltage follower iso‑ lates the capacitance formed by the circuit from the loading imposed by the inverting amplifier. Since no current enters the input terminals of the op amp, the input current **I***i* flows through the feedback capaci‑ tor. Hence, at node 1, - -$$ -\mathbf{I}_{i} = \frac{\mathbf{V}_{i} - \mathbf{V}_{o}}{1/j\omega C} = j\omega C(\mathbf{V}_{i} - \mathbf{V}_{o}) -$$ -\n(10.3) - -Applying KCL at node 2 gives - -$$ -\frac{\mathbf{V}_i - \mathbf{0}}{R_1} = \frac{\mathbf{0} - \mathbf{V}_o}{R_2} -$$ - -or - -$$ -\mathbf{V}_o = -\frac{R_2}{R_1} \mathbf{V}_i \tag{10.4} -$$ - -Substituting Eq. (10.4) into (10.3) gives - -$$ -\mathbf{I}_i = j\omega C \bigg( 1 + \frac{R_2}{R_1} \bigg) \mathbf{V}_i -$$ - -or - -$$ -\frac{\mathbf{I}_i}{\mathbf{V}_i} = j\omega \left( 1 + \frac{R_2}{R_1} \right) C \tag{10.5} -$$ - -The input impedance is - -$$ -\mathbf{Z}_{i} = \frac{\mathbf{V}_{i}}{\mathbf{I}_{i}} = \frac{1}{j\omega C_{\text{eq}}} -$$ -(10.6) - -where - -$$ -C_{\text{eq}} = \left(1 + \frac{R_2}{R_1}\right)C\tag{10.7} -$$ - -Thus, by a proper selection of the values of *R*1 and *R*2, the op amp circuit in Fig. 10.41 can be made to produce an effective capacitance between the input terminal and ground, which is a multiple of the physical capaci‑ tance *C*. The size of the effective capacitance is practically limited by the inverted output voltage limitation. Thus, the larger the capacitance multiplication, the smaller is the allowable input voltage to prevent the op amps from reaching saturation. - -A similar op amp circuit can be designed to simulate inductance. (See Prob. 10.89.) There is also an op amp circuit configuration to create a resistance multiplier. - -Example 10.15 Calculate *C*eq in Fig. 10.41 when *R*1 = 10 kΩ, *R*2 = 1 MΩ, and *C* = 1 nF. - -# **Solution:** - -From Eq. (10.7) *C*eq = (1 + \_\_\_ *R*2 *R*1 )*C* = (1 + 1 × 106 \_\_\_\_\_\_\_\_ 10 × 103 ) 1 nF = 101 nF Determine the equivalent capacitance of the op amp circuit in Fig. 10.41 if *R*1 = 10 kΩ, *R*2 = 10 MΩ, and *C* = 10 nF. - -**Answer:** 10 *μ*F. - -# **10.9.2** Oscillators - -We know that dc is produced by batteries. But how do we produce ac? One way is using *oscillators,* which are circuits that convert dc to ac. - -An oscillator is a circuit that produces an ac waveform as output when powered by a dc input. - -The only external source an oscillator needs is the dc power supply. Ironically, the dc power supply is usually obtained by con verting the ac supplied by the electric utility company to dc. Having gone through the trouble of conversion, one may wonder why we need to use the oscillator to convert the dc to ac again. The problem is that the ac supplied by the utility company operates at a preset frequenc y of 60 Hz in the United States (50 Hz in some other nations), whereas man y applications such as electronic circuits, communication systems, and micro wave devices require internally generated frequencies that range from 0 to 10 GHz or higher. Oscillators are used for generating these frequencies. - -In order for sine w ave oscillators to sustain oscillations, the y must meet the *Barkhausen criteria*: - -- 1. The overall gain of the oscillator must be unity or greater. Therefore, losses must be compensated for by an amplifying device. -- 2. The overall phase shift (from input to output and back to the input) must be zero. - -Three common types of sine wave oscillators are phase ‑shift, twin *T*, and Wien ‑bridge oscillators. Here we consider only the Wien ‑bridge oscillator. - -The *Wien-bridge oscillator* is widely used for generating sinusoids in the frequency range below 1 MHz. It is an *RC* op amp circuit with only a few components, easily tunable and easy to design. As shown in Fig. 10.42, the oscillator essentially consists of a noninverting amplifier with two feedback paths: The positive feedback path to the noninverting input creates oscillations, while the ne gative feedback path to the in verting input controls the gain. If we define the impedances of the *RC* series and parallel combinations as **Z***s* and **Z***p*, then - -$$ -Z_s = R_1 + \frac{1}{j\omega C_1} = R_1 - \frac{j}{\omega C_1} -$$ -(10.8) -$$ -Z_p = R_2 \Big| \frac{1}{j\omega C_2} = \frac{R_2}{1 + j\omega R_2 C_2} -$$ -(10.9) - -The feedback ratio is - -$$ -\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{\mathbf{Z}_p}{\mathbf{Z}_s + \mathbf{Z}_p} -$$ - -Practice Problem 10.15 - -Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives - -Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives -\n -$$ -\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{R_2}{R_2 + \left(R_1 - \frac{j}{\omega C_1}\right)(1 + j\omega R_2 C_2)} -$$ -\n -$$ -= \frac{\omega R_2 C_1}{\omega (R_2 C_1 + R_1 C_1 + R_2 C_2) + j(\omega^2 R_1 C_1 R_2 C_2 - 1)} -$$ -\n(10.11) - -To satisfy the second Barkhausen criterion, **V**2 must be in phase with **V***o*, which implies that the ratio in Eq. (10.11) must be purely real. Hence, the imaginary part must be zero. Setting the imaginary part equal to zero gives the oscillation frequency *ωo* as - -$$ -\omega_o^2 R_1 C_1 R_2 C_2 - 1 = 0 -$$ - -or - -$$ -\omega_o = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}\tag{10.12} -$$ - -In most practical applications, *R*1 = *R*2 = *R* and *C*1 = *C*2 = *C*, so that - -$$ -\omega_o = \frac{1}{RC} = 2\pi f_o \tag{10.13} -$$ - -or - -$$ -f_o = \frac{1}{2\pi RC} -$$ - (10.14) - -Substituting Eq. (10.13) and *R*1 = *R*2 = *R*, *C*1 = *C*2 = *C* into Eq. (10.11) yields - -$$ -\frac{V_2}{V_o} = \frac{1}{3} -$$ - (10.15) - -Thus, in order to satisfy the first Barkhausen criterion, the op amp must compensate by providing a gain of 3 or greater so that the overall gain is at least 1 or unity. We recall that for a noninverting amplifier, - -$$ -\frac{\mathbf{V}_o}{\mathbf{V}_2} = 1 + \frac{R_f}{R_g} = 3\tag{10.16} -$$ - -or - -$$ -R_f = 2R_g \tag{10.17} -$$ - -Due to the inherent delay caused by the op amp, Wien‑bridge oscil‑ lators are limited to operating in the frequency range of 1 MHz or less. - -# Example 10.16 - -Design a Wien‑bridge circuit to oscillate at 100 kHz. - -# **Solution:** - -Using Eq. (10.14), we obtain the time constant of the circuit as - -14), we obtain the time constant of the circuit as -$$ -RC = \frac{1}{2\pi f_o} = \frac{1}{2\pi \times 100 \times 10^3} = 1.59 \times 10^{-6} -$$ - (10.16.1) - -If we select *R* = 10 k Ω, then we can select *C* = 159 pF to satisfy Eq. (10.16.1). Since the gain must be 3, *Rf*∕*Rg* = 2. We could select *Rf* = 20 kΩ while *Rg* = 10 kΩ. - -In the Wien‑bridge oscillator circuit in Fig. 10.42, let *R*1 = *R*2 = 2.5 kΩ, *C*1 = *C*2 = 1 nF. Determine the frequency *fo* of the oscillator. Practice Problem 10.16 - -**Answer:** 63.66 kHz. - -# **10.10** Summary - -- 1. We apply nodal and mesh analysis to ac circuits by applying KCL and KVL to the phasor form of the circuits. -- 2. In solving for the steady ‑state response of a circuit that has inde ‑ pendent sources with different frequencies, each independent source *must* be considered separately. The most natural approach to analyz‑ ing such circuits is to apply the superposition theorem. A separate phasor circuit for each frequency *must* be solved independently, and the corresponding response should be obtained in the time domain. The overall response is the sum of the time domain responses of all the individual phasor circuits. -- 3. The concept of source transformation is also applicable in the fre ‑ quency domain. -- 4. The Thevenin equivalent of an ac circuit consists of a voltage source **V**Th in series with the Thevenin impedance **Z**Th. -- 5. The Norton equivalent of an ac circuit consists of a current source **I***N* in parallel with the Norton impedance **Z***N* (=**Z**Th). -- 6. *PSpice* is a simple and powerful tool for solving ac circuit problems. It relieves us of the tedious task of working with the complex num‑ bers involved in steady‑state analysis. -- 7. The capacitance multiplier and the ac oscillator provide two typical applications for the concepts presented in this chapter . A capaci‑ tance multiplier is an op amp circuit used in producing a multiple of a physical capacitance. An oscillator is a device that uses a dc input to generate an ac output. - -‒ - -# Review Questions - -10 0° V ‒j1 Ω **V**o + - -**10.1** The voltage **V***o* across the capacitor in Fig. 10.43 is: - -‒ - -For Review Question 10.1. - -**10.2** The value of the current **I***o* in the circuit of Fig. 10.44 is: - -(a) -$$ -4\angle 0^{\circ} -$$ - A -(b) $2.4\angle -90^{\circ}$ A -(c) $0.6\angle 0^{\circ}$ A -(d) $-1$ A - -**Figure 10.44** For Review Question 10.2. - -**10.3** Using nodal analysis, the value of **V***o* in the circuit of Fig. 10.45 is: - -(a) -$$ --24 \text{ V} -$$ - (b) $-8 \text{ V}$ - -$$ -(c) 8 V \t\t (d) 24 V -$$ - -# **10.6** For the circuit in Fig. 10.48, the Thevenin impedance at terminals *a*‑*b* is: - -| (a) 1 Ω | (b) 0.5 − j0.5 Ω | -|------------------|------------------| -| (c) 0.5 + j0.5 Ω | (d) 1 + j2 Ω | -| (e) 1 − j2 Ω | | - -# **Figure 10.48** - -For Review Questions 10.6 and 10.7. - -- - (a) 10 cos *t* A (b) 10 sin *t* A (c) 5 cos *t* A - -**10.4** In the circuit of Fig. 10.46, current *i*(*t*) is: - -(d) 5 sin *t* A (e) 4.472 cos(*t* − 63.43°) A - -**Figure 10.46** - -**Figure 10.45** - -For Review Question 10.3. - -For Review Question 10.4. - -- **10.5** Refer to the circuit in Fig. 10.47 and observe that the two sources do not have the same frequency. The current *ix*(*t*) can be obtained by: - - (a) source transformation - - (b) the superposition theorem - - (c) *PSpice* - -**Figure 10.47** For Review Question 10.5. **10.7** In the circuit of Fig. 10.48, the Thevenin voltage at terminals *a*‑*b* is: - -(a) -$$ -3.535 \angle -45^{\circ} \text{ V} -$$ - (b) $3.535 \angle 45^{\circ} \text{ V}$ -(c) $7.071 \angle -45^{\circ} \text{ V}$ (d) $7.071 \angle 45^{\circ} \text{ V}$ - -**10.8** Refer to the circuit in Fig. 10.49. The Norton equivalent impedance at terminals *a*‑*b* is: - -| (a) −j4 Ω | (b) −j2 Ω | -|-----------|-----------| -| (c) j2 Ω | (d) j4 Ω | - -# **Figure 10.49** - -For Review Questions 10.8 and 10.9. - -**10.9** The Norton current at terminals *a*‑*b* in the circuit of Fig. 10.49 is: - -(a) -$$ -1/\underline{0^{\circ}} -$$ - A -(b) $1.5/\underline{-90^{\circ}}$ A -(c) $1.5/90^{\circ}$ A -(d) $3/90^{\circ}$ A - -- **10.10** *PSpice* can handle a circuit with two independent sources of different frequencies. - - (a) True (b) False - -*Answers: 10.1c, 10.2a, 10.3d, 10.4a, 10.5b, 10.6c, 10.7a, 10.8a, 10.9d, 10.10b.* - -# Problems - -# Section 10.2 Nodal Analysis - -**10.1** Determine *i* in the circuit of Fig. 10.50. - -# **Figure 10.50** - -For Prob. 10.1. - -# **Figure 10.51** - -For Prob. 10.2. - -**10.3** Determine *vo* in the circuit of Fig. 10.52. - -# **Figure 10.52** For Prob. 10.3. - -# **Figure 10.53** - -For Prob. 10.4. - -For Prob. 10.5. - -**10.6** Determine **V***x* in Fig. 10.55. - -# **Figure 10.55** For Prob. 10.6. - -**10.7** Use nodal analysis to find **V** in the circuit of Fig. 10.56. - -# **Figure 10.56** For Prob. 10.7. - -**10.8** Use nodal analysis to find current *io* in the circuit of Fig. 10.57. Let *is* = 6 cos(200*t* + 15°) A. - -# **Figure 10.57** - -For Prob. 10.8. - -For Prob. 10.9. - -‒ - -**Figure 10.59** - -**10.11** Using nodal analysis, find *io*(*t*) in the circuit in Fig. 10.60. - -For Prob. 10.11. - -**10.12** Using Fig. 10.61, design a problem to help other students better understand nodal analysis. - -**Figure 10.61** For Prob. 10.12. - -**10.13** Determine **V***x* in the circuit of Fig. 10.62 using any method of your choice. - -For Prob. 10.13. - -**10.14** Calculate the voltage at nodes 1 and 2 in the circuit of Fig. 10.63 using nodal analysis. - -# **Figure 10.63** - -For Prob. 10.14. - -**10.15** Solve for the current **I** in the circuit of Fig. 10.64 using nodal analysis. - -**Figure 10.64** - -For Prob. 10.15. - -# **Figure 10.65** - -For Prob. 10.16. - -**10.17** By nodal analysis, obtain current **I***o* in the circuit of Fig. 10.66. - -**Figure 10.66** For Prob. 10.17. - -**10.19** Obtain **V***o* in Fig. 10.68 using nodal analysis. - -**10.20** Refer to Fig. 10.69. If *vs*(*t*) = *Vm* sin *ωt* and *vo*(*t*) = *A* sin(*ωt* + *ϕ*), derive the expressions for *A* and *ϕ*. - -**Figure 10.69** For Prob. 10.20. - -**10.21** For each of the circuits in Fig. 10.70, find **V***o*∕**V***i* for *ω* = 0, *ω* → ∞, and *ω*2 = 1∕*LC*. - -**Figure 10.70** For Prob. 10.21. - -**10.22** For the circuit in Fig. 10.71, determine **V***o*∕**V***s*. - -**Figure 10.71** For Prob. 10.22. - -**10.23** Using nodal analysis obtain **V** in the circuit of Fig. 10.72. - -**Figure 10.72** - -For Prob. 10.23. - -# Section 10.3 Mesh Analysis - -**10.24** Design a problem to help other students better understand mesh analysis. - -**10.25** Solve for *io* in Fig. 10.73 using mesh analysis. - -For Prob. 10.25. - -**10.26** Use mesh analysis to find current *io* in the circuit of Fig. 10.74. - -**10.29** Using Fig. 10.77, design a problem to help other students better understand mesh analysis. - -R3 - -For Prob. 10.26. - -**10.27** Using mesh analysis, find **I**1 and **I**2 in the circuit of Fig. 10.75. - -jXL1 - -**Figure 10.77** For Prob. 10.29. - -For Prob. 10.30. - -**Figure 10.76** For Prob. 10.28. - -**10.32** Determine **V***o* and **I***o* in the circuit of Fig. 10.80 using mesh analysis. - -**Figure 10.80** For Prob. 10.32. - -**10.33** Compute **I** in Prob. 10.15 using mesh analysis. - -**10.34** Use mesh analysis to find **I***o* in Fig. 10.28 (for Example 10.10). - -**10.35** Calculate **I***o* in Fig. 10.30 (for Practice Prob. 10.10) using mesh analysis. - -**10.36** Compute **V***o* in the circuit of Fig. 10.81 using mesh analysis. - -**Figure 10.81** For Prob. 10.36. - -**10.37** Use mesh analysis to find currents **I**1, **I**2, and **I**3 in the circuit of Fig. 10.82. - -**10.38** Using mesh analysis, obtain **I***o* in the circuit shown in Fig. 10.83. - -For Prob. 10.38. - -**10.39** Find **I**1, **I**2, **I**3, and **I***x* in the circuit of Fig. 10.84. - -**Figure 10.84** For Prob. 10.39. - -# Section 10.4 Superposition Theorem - -**10.40** Find *io* in the circuit shown in Fig. 10.85 using superposition. - -# **Figure 10.85** - -For Prob. 10.40. - -**10.41** Find *vo* for the circuit in Fig. 10.86, assuming that *is*(*t*) = 2 sin (2*t*) + 3 cos (4*t*) A. - -**Figure 10.86** For Prob. 10.41. - -**10.42** Using Fig. 10.87, design a problem to help other students better understand the superposition theorem. - -**Figure 10.87** For Prob. 10.42. - -For Prob. 10.43. - -**10.43** Using the superposition principle, find *ix* in the circuit of Fig. 10.88. - -For Prob. 10.46. - -**10.47** Determine *io* in the circuit of Fig. 10.92, using the superposition principle. - -For Prob. 10.47. - -**10.44** Use the superposition principle to obtain *vx* in the circuit of Fig. 10.89. Let *vs* = 50 sin 2*t* V and *is* = 12 cos(6*t* + 10°) A. - -**10.45** Use superposition to find *i*(*t*) in the circuit of Fig. 10.90. - -**Figure 10.90** For Prob. 10.45. - -**10.48** Find *io* in the circuit of Fig. 10.93 using superposition. - -**Figure 10.93** - -For Prob. 10.48. - -# Section 10.5 Source Transformation - -**10.49** Using source transformation, find *i* in the circuit of Fig. 10.94. - -For Prob. 10.49. - -**Figure 10.95** - -For Prob. 10.50. - -- **10.51** Use source transformation to find **I***o* in the circuit of Prob. 10.42. -- **10.52** Use the method of source transformation to find **I***x* in the circuit of Fig. 10.96. - -**Figure 10.96** For Prob. 10.52. - -**10.53** Use the concept of source transformation to find **V***o* in the circuit of Fig. 10.97. - -For Prob. 10.53. - -**10.54** Rework Prob. 10.7 using source transformation. - -# Section 10.6 Thevenin and Norton Equivalent Circuits - -**10.55** Find the Thevenin and Norton equivalent circuits at terminals *a*‑*b* for each of the circuits in Fig. 10.98. - -**Figure 10.98** For Prob. 10.55. - -**10.56** For each of the circuits in Fig. 10.99, obtain Thevenin and Norton equivalent circuits at terminals *a*‑*b*. - -# **Figure 10.99** - -For Prob. 10.56. - -**10.57** Using Fig. 10.100, design a problem to help other students better understand Thevenin and Norton equivalent circuits. - -# **Figure 10.100** For Prob. 10.57. - -**10.58** For the circuit depicted in Fig. 10.101, find the Thevenin equivalent circuit at terminals *a*‑*b*. - -**Figure 10.101** For Prob. 10.58. - -**10.59** Calculate the output impedance of the circuit shown in Fig. 10.102. - -For Prob. 10.59. - -**10.60** Find the Thevenin equivalent of the circuit in Fig. 10.103 as seen from: - -**10.61** Find the Thevenin equivalent at terminals *a*-*b* of the circuit in Fig. 10.104. - -**Figure 10.104** For Prob. 10.61. - -**10.62** Using Thevenin's theorem, find *vo* in the circuit of Fig. 10.105. - -For Prob. 10.62. - -**10.63** Obtain the Norton equivalent of the circuit depicted in Fig. 10.106 at terminals *a*-*b*. - -For Prob. 10.63. - -**10.64** For the circuit shown in Fig. 10.107, find the Norton equivalent circuit at terminals *a*-*b*. - -# **Figure 10.107** - -For Prob. 10.64. - -**10.65** Using Fig. 10.108, design a problem to help other students better understand Norton's theorem. - -Problems **449** - -For Prob. 10.70. - -**10.71** Find *vo* in the op amp circuit of Fig. 10.114. - -**Figure 10.114** - -For Prob. 10.71. - -**10.72** Compute *io*(*t*) in the op amp circuit in Fig. 10.115 if *vs* = 4 cos(104 *t*) V. - -# **Figure 10.115** - -For Prob. 10.72. - -**10.73** If the input impedance is defined as **Z**in = **V***s*∕**I***s*, find the input impedance of the op amp circuit in Fig. 10.116 when *R*1 = 10 kΩ, *R*2 = 20 kΩ, *C*1 = 10 nF, *C*2 = 20 nF, and *ω* = 5000 rad/s. - -**Figure 10.116** For Prob. 10.73. - -**Figure 10.111** For Prob. 10.68. - -**Figure 10.110** For Prob. 10.67. - -# Section 10.7 Op Amp AC Circuits - -**10.69** For the integrator shown in Fig. 10.112, obtain **V***o*∕**V***s*. Find *vo*(*t*) when *vs*(*t*) = **V***m* sin *ωt* and *ω* = 1∕*RC*. - -**Figure 10.112** For Prob. 10.69. - -**10.70** Using Fig. 10.113, design a problem to help other students better understand op amps in AC circuits. **10.74** Evaluate the voltage gain **A***v* = **V***o*∕**V***s* in the op amp circuit of Fig. 10.117. Find **A***v* at *ω* = 0, *ω* → ∞, *ω* = 1∕*R*1*C*1, and *ω* = 1∕*R*2*C*2. - -10 kΩ - -**V**o + - -20 kΩ - -i o - -‒ + - -+ - -‒ - -‒ - -6 30° V ‒j2 kΩ - -‒j4 kΩ - -**10.75** In the op amp circuit of Fig. 10.118, find the closed‑ loop gain and phase shift of the output voltage with respect to the input voltage if *C*1 = *C*2 = 1 nF, *R*1 = *R*2 = 100 kΩ, *R*3 = 20 kΩ, *R*4 = 40 kΩ, and *ω* = 2000 rad/s. - -*v*s *v*o - -R2 - -C2 - -R1 - -R4 - -+ - -‒ - -R3 - -‒ + - -C1 - -+ ‒ - -**Figure 10.118** For Prob. 10.75. - -**10.77** Compute the closed‑loop gain **V***o*∕**V***s* for the op amp circuit of Fig. 10.120. - -**10.78** Determine *vo*(*t*) in the op amp circuit in Fig. 10.121 below. - -**10.79** For the op amp circuit in Fig. 10.122, obtain **V***o*. - -For Prob. 10.79. - -**10.80** Obtain *vo*(*t*) for the op amp circuit in Fig. 10.123 if *vs* = 12 cos(1000*t* − 60°) V. - -**Figure 10.123** For Prob. 10.80. - -**10.81** Use *PSpice or MultiSim* to determine **V***o* in the circuit of Fig. 10.124. Assume *ω* = 1 rad/s. - -**Figure 10.124** For Prob. 10.81. - -**10.82** Solve Prob. 10.19 using *PSpice or MultiSim*. - -**10.83** Use *PSpice or MultiSim* to find *vo*(*t*) in the circuit of Fig. 10.125. Let *is* = 2 cos(103 *t*) A. - -# **Figure 10.125** - -For Prob. 10.83. - -i - -**10.84** Obtain **V***o* in the circuit of Fig. 10.126 using *PSpice or MultiSim*. - -# **Figure 10.126** - -For Prob. 10.84. - -**10.85** Using Fig. 10.127, design a problem to help other students better understand performing AC analysis with *PSpice or MultiSim*. - -# **Figure 10.127** For Prob. 10.85. - -**10.86** Use *PSpice or MultiSim* to find **V**1, **V**2, and **V**3 in the network of Fig. 10.128. - -**Figure 10.128** For Prob. 10.86. - -C - -+ + **V**o ‒ ‒ - -R R1 - -R2 R - -C - -**10.88** Use *PSpice or MultiSim* to find *vo* and *io* in the circuit of Fig. 10.130 below. - -**Figure 10.132** For Prob. 10.90. - -**V**i - -# Section 10.9 Applications - -**10.89** The op amp circuit in Fig. 10.131 is called an *inductance simulator*. Show that the input impedance is given by - -$$ -\mathbf{Z}_{in} = \frac{\mathbf{V}_{in}}{\mathbf{I}_{in}} = j\omega L_{eq} -$$ - -where - -$$ -L_{\text{eq}} = \frac{R_1 R_3 R_4}{R_2 C} -$$ - -**Figure 10.131** For Prob. 10.89. - -- **10.91** Consider the oscillator in Fig. 10.133. - - (a) Determine the oscillation frequency. - - (b) Obtain the minimum value of *R* for which oscillation takes place. - -**Figure 10.133** For Prob. 10.91. - -**10.87** Determine **V**1, **V**2, and **V**3 in the circuit of Fig. 10.129 using *PSpice or MultiSim*. - -- **10.92** The oscillator circuit in Fig. 10.134 uses an ideal op amp. - - (a) Calculate the minimum value of *Ro* that will cause oscillation to occur. - - (b) Find the frequency of oscillation. - -# **Figure 10.134** - -**10.93** Figure 10.135 shows a *Colpitts oscillator*. Show that the oscillation frequency is - -$$ -f_o = \frac{1}{2\pi\sqrt{LC_T}} -$$ - -where *CT* = *C*1*C*2∕(*C*1 + C2). Assume *Ri* ≫ *XC*2 . - -# **Figure 10.135** - -A Colpitts oscillator; for Prob. 10.93. - -(*Hint:* Set the imaginary part of the impedance in the feedback circuit equal to zero.) - -**10.94** Design a Colpitts oscillator that will operate at 50 kHz. - -**10.95** Figure 10.136 shows a *Hartley oscillator*. Show that the frequency of oscillation is - -# **Figure 10.136** A Hartley oscillator; for Prob. 10.95. - -**10.96** Refer to the oscillator in Fig. 10.137. - -(a) Show that - -$$ -\mathbf{v} \text{ that} -$$ -\n -$$ -\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{1}{3 + j(oL/R - R/oL)} -$$ - -- (b) Determine the oscillation frequency *fo*. -- (c) Obtain the relationship between *R*1 and *R*2 in order for oscillation to occur. - -**Figure 10.137** For Prob. 10.96. - -*This page intentionally left blank* - -# **chapter** - -11 - -# AC Power Analysis - -*Four things come not back: the spoken word; the sped arrow; time past; the neglected opportunity.* - -—Al Halif Omar Ibn - -# Enhancing Your Career - -# **Career in Power Systems** - -The discovery of the principle of an ac generator by Michael Faraday in 1831 was a major breakthrough in engineering; it provided a convenient way of generating the electric po wer that is needed in e very electronic, electrical, or electromechanical device we use now. - -Electric power is obtained by converting energy from sources such as fossil fuels (gas, oil, and coal), nuclear fuel (uranium), h ydro energy (water falling through a head), geothermal energy (hot water, steam), wind energy, tidal ener gy, and biomass ener gy (wastes). These various ways of generating electric po wer are studied in detail in the field of power engineering, which has become an indispensable subdiscipline of electrical engineering. An electrical engineer should be f amiliar with the analysis, generation, transmission, distrib ution, and cost of electric power. - -The electric po wer industry is a v ery large employer of electrical engineers. The industry includes thousands of electric utility systems ranging from large, interconnected systems serving large regional areas to small power companies serving indi vidual communities or f actories. Due to the comple xity of the po wer industry, there are numerous elec trical engineering jobs in dif ferent areas of the industry: po wer plant (generation), transmission and distribution, maintenance, research, data acquisition and flow control, and management. Since electric po wer is used e verywhere, electric utility companies are e verywhere, of fering exciting training and steady emplo yment for men and w omen in thou sands of communities throughout the world. - -A pole-type transformer with a lowvoltage, three-wire distribution system. © Dennis Wise/Getty Images RF - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Fully understand instantaneous and average power. -- 2. Understand the basics of maximum average power. -- 3. Understand effective or rms values and how to calculate them and to understand their importance. -- 4. Understand apparent power (complex power), power, and reactive power and power factor. -- 5. Understand power factor correction and the importance of its use. - -# **11.1** Introduction - -Our effort in ac circuit analysis so f ar has been focused mainly on cal culating voltage and current. Our major concern in this chapter is power analysis. - -Power analysis is of paramount importance. Po wer is the most important quantity in electric utilities, electronic, and communication systems, because such systems in volve transmission of po wer from one point to another . Also, e very industrial and household electrical device—every fan, motor, lamp, pressing iron, TV, personal computer has a po wer rating that indicates ho w much po wer the equipment re quires; e xceeding the po wer rating can do permanent damage to an appliance. The most common form of electric po wer is 50- or 60-Hz ac power. The choice of ac o ver dc allowed high-voltage power transmission from the power generating plant to the consumer. - -We will be gin by defining and deriving *instantaneous power* and *average power*. We will then introduce other power concepts. As practical applications of these concepts, we will discuss how power is measured and reconsider how electric utility companies charge their customers. - -# **11.2** Instantaneous and Average Power - -As mentioned in Chapter 2, the *instantaneous power p*(*t*) absorbed by an element is the product of the instantaneous v oltage *v*(*t*) across the ele ment and the instantaneous current *i*(*t*) through it. Assuming the passive sign convention, - -$$ -p(t) = v(t)i(t) -$$ - (11.1) - -The instantaneous power (in watts) is the power at any instant of time. - -It is the rate at which an element absorbs energy. - -Consider the general case of instantaneous po wer absorbed by an arbitrary combination of circuit elements under sinusoidal excitation, as - -We can also think of the instantaneous power as the power absorbed by the element at a specific instant of time. Instantaneous quantities are denoted by lowercase letters. - -shown in Fig. 11.1. Let the v oltage and current at the terminals of the circuit be - -$$ -v(t) = V_m \cos(\omega t + \theta_v) -$$ - (11.2a) - -$$ -i(t) = I_m \cos(\omega t + \theta_i) -$$ - (11.2b) - -where *Vm* and *Im* are the amplitudes (or peak values), and *θv* and *θi* are the phase angles of the voltage and current, respectively. The instantaneous power absorbed by the circuit is - -$$ -p(t) = v(t)i(t) = V_m I_m \cos(\omega t + \theta_v) \cos(\omega t + \theta_i) -$$ - (11.3) - -We apply the trigonometric identity - -$$ -\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)] \tag{11.4} -$$ - -and express Eq. (11.3) as - -$$ -p(t) = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) + \frac{1}{2}V_m I_m \cos(2\omega t + \theta_v + \theta_i) -$$ - (11.5) - -This shows us that the instantaneous power has two parts. The first part is constant or time independent. Its value depends on the phase dif ference between the voltage and the current. The second part is a sinusoidal function whose frequency is 2*ω*, which is twice the angular frequency of the voltage or current. - -A sketch of *p*(*t*) in Eq. (11.5) is shown in Fig. 11.2, where *T* = 2*π*∕*ω* is the period of v oltage or current. We observ e that *p*(*t*) is periodic, *p*(*t*) = *p*(*t* + *T*0), and has a period of *T*0 = *T*∕2, since its frequency is twice that of voltage or current. We also observe that *p*(*t*) is positive for some part of each c ycle and ne gative for the rest of the c ycle. When *p*(*t*) is positive, power is absorbed by the circuit. When *p*(*t*) is negative, power is absorbed by the source; that is, power is transferred from the circuit to the source. This is possible because of the storage elements (capacitors and inductors) in the circuit. - -**Figure 11.2** The instantaneous power *p*(*t*) entering a circuit. - -The instantaneous power changes with time and is therefore difficult to measure. The *average* power is more convenient to measure. In f act, the wattmeter, the instrument for measuring power, responds to average power. - -The average power, in watts, is the average of the instantaneous power over one period. - -# **Figure 11.1** - -Sinusoidal source and passive linear circuit. - -Thus, the average power is given by - -$$ -P = \frac{1}{T} \int_0^T p(t) \, dt \tag{11.6} -$$ - -Although Eq. (11.6) shows the averaging done over *T*, we would get the same result if we performed the integration over the actual period of *p* ( *t*) which is *T*0 = *T*∕2. - - Substituting *p* ( *t*) in Eq. (11.5) into Eq. (11.6) gives - -$$ -P = \frac{1}{T} \int_0^T \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) dt -$$ - -+ $\frac{1}{T} \int_0^T \frac{1}{2} V_m I_m \cos(2\omega t + \theta_v + \theta_i) dt$ -= $\frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) \frac{1}{T} \int_0^T dt$ -+ $\frac{1}{2} V_m I_m \frac{1}{T} \int_0^T \cos(2\omega t + \theta_v + \theta_i) dt$ (11.7) - -The first integrand is constant, and the average of a constant is the same constant. The second integrand is a sinusoid. We know that the average of a sinusoid over its period is zero because the area under the sinusoid during a positi ve half-cycle is canceled by the area under it during the following negative half-cycle. Thus, the second term in Eq. (11.7) v an ishes and the average power becomes - -$$ -P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) -$$ - (11.8) - -Since cos( *θv* − *θ i* ) = cos( *θi* − *θv*), what is important is the diference in the phases of the voltage and current. - -Note that *p* ( *t*) is time-varying while *P* does not depend on time. To find the instantaneous power, we must necessarily have *v* ( *t*) and *i* ( *t*) in the time domain. But we can find the average power when voltage and cur rent are expressed in the time domain, as in Eq. (11.8), or when they are expressed in the frequency domain. The phasor forms of *v* ( *t*) and *i* ( *t*) in Eq. (11.2) are **V** = *Vm* ⧸ *θv* and **I** = *Im* ⧸ *θ i*, respectively. *P* is calculated using Eq. (11.8) or using phasors **V** and **I**. To use phasors, we notice that - -$$ -\frac{1}{2}\mathbf{V}\mathbf{I}^* = \frac{1}{2}V_m I_m / \theta_v - \theta_i -$$ - -= -$$ -\frac{1}{2}V_m I_m [\cos(\theta_v - \theta_i) + j \sin(\theta_v - \theta_i)] -$$ - (11.9) - -We recognize the real part of this e xpression as the a verage power *P* according to Eq. (11.8). Thus, - -$$ -P = \frac{1}{2} \text{Re}[\mathbf{V} \mathbf{I}^*] = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) -$$ - (11.10) - -Consider two special cases of Eq. (11.10). When *θ v* = *θ i*, the voltage and current are in phase. This implies a purely resistive circuit or resis tive load *R*, and - -$$ -P = \frac{1}{2} V_m I_m = \frac{1}{2} I_m^2 R = \frac{1}{2} |\mathbf{I}|^2 R -$$ - (11.11) - -where ∣**I**∣ 2 = **I** × **I**\*. Equation (11.11) shows that a purely resistive circuit absorbs power at all times. When *θv* − *θi* = ±90°, we have a purely reactive circuit, and - -$$ -P = \frac{1}{2} V_m I_m \cos 90^\circ = 0 \tag{11.12} -$$ - -showing that a purely reacti ve circuit absorbs no a verage po wer. In summary, - -A resistive load (R) absorbs power at all times, while a reactive load (L or C ) absorbs zero average power. - -$$ -v(t) = 120 \cos(377t + 45^{\circ}) -$$ - V and $i(t) = 10 \cos(377t - 10^{\circ})$ A - -find the instantaneous power and the a verage po wer absorbed by the passive linear network of Fig. 11.1. - -# **Solution:** - -The instantaneous power is given by - -$$ -p = vi = 1200 \cos(377t + 45^{\circ}) \cos(377t - 10^{\circ}) -$$ - -Applying the trigonometric identity - -$$ -\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] -$$ - -gives - -$$ -p = 600[\cos(754t + 35^\circ) + \cos 55^\circ] -$$ - -or - -$$ -p(t) = 344.2 + 600 \cos(754t + 35^\circ) -$$ - W - -The average power is - -$$ -P = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) = \frac{1}{2}120(10) \cos[45^\circ - (-10^\circ)] -$$ - -= 600 cos 55° = 344.2 W - -which is the constant part of *p*(*t*) above. - -Calculate the instantaneous power and average power absorbed by the passive linear network of Fig. 11.1 if Practice Problem 11.1 - -*v*(*t*) = 330 cos(10*t* + 20°) V and *i*(*t*) = 33 sin(10*t* + 60°) A - -**Answer:** 3.5 + 5.445 cos(20*t* − 10°) kW, 3.5 kW. - -Calculate the average power absorbed by an impedance **Z** = 30 − *j*70 Ω Example 11.2 when a voltage **V** = 120⧸ 0**°** is applied across it. - -**Solution:** - -The current through the impedance is - -ent through the impedance is -\n -$$ -I = \frac{V}{Z} = \frac{120/0^{\circ}}{30 - j70} = \frac{120/0^{\circ}}{76.16/-66.8^{\circ}} = 1.576/66.8^{\circ} A -$$ - -Given that Example 11.1 - -The average power is - -$$ -P = \frac{1}{2}V_m I_m \cos(\theta_v - \theta_i) = \frac{1}{2}(120)(1.576)\cos(0 - 66.8^\circ) = 37.24 \text{ W} -$$ - -# Practice Problem 11.2 - -A current **I** = 33⧸ 30**°** A flows through an impedance **Z** = 40⧸ −22° Ω. Find the average power delivered to the impedance. - -**Answer:** 20.19 kW. - -Example 11.3 For the circuit shown in Fig. 11.3, find the average power supplied by the source and the average power absorbed by the resistor. - -# **Solution:** - -The current **I** is given by - -1 is given by -\n -$$ -I = \frac{5/30^{\circ}}{4 - j2} = \frac{5/30^{\circ}}{4.472/-26.57^{\circ}} = 1.118/56.57^{\circ} A -$$ - -The average power supplied by the voltage source is - -$$ -P = \frac{1}{2}(5)(1.118)\cos(30^\circ - 56.57^\circ) = 2.5 \text{ W} -$$ - -The current through the resistor is - -$$ -I_R = I = 1.118 / 56.57^{\circ} -$$ - A - -and the voltage across it is - -$$ -V_R = 4I_R = 4.472/56.57^\circ -$$ - V - -The average power absorbed by the resistor is - -$$ -P = \frac{1}{2}(4.472)(1.118) = 2.5 -$$ - W - -which is the same as the average power supplied. Zero average power is absorbed by the capacitor. - -Practice Problem 11.3 - -For Practice Prob. 11.3. - -In the circuit of Fig. 11.4, calculate the average power absorbed by the resistor and inductor. Find the average power supplied by the voltage source. - -**Answer:** 29.04 kW, 0 W, 29.04 kW. - -Determine the average power generated by each source and the average Example 11.4 power absorbed by each passive element in the circuit of Fig. 11.5(a). - -For Example 11.4. - -# **Solution:** - -We apply mesh analysis as shown in Fig. 11.5(b). For mesh 1, - -$$ -\mathbf{I}_1 = 4 \text{ A} -$$ - -For mesh 2, - -$$ -(j10 - j5) -$$ -**I**2 - $j10$ **I**1 + 60/ $\cancel{30^{\circ}}$ = 0, **I**1 = 4 A - -or - -$$ -j5I_2 = -60/30^{\circ} + j40 -$$ - $\Rightarrow$ $I_2 = -12/-60^{\circ} + 8$ -= 10.58/79.1° A - -For the voltage source, the current flowing from it is **I**2 = 10.58⧸ 79.1**°** A and the voltage across it is 60⧸ 30**°** V, so that the average power is - -$$ -P_5 = \frac{1}{2} (60)(10.58) \cos(30^\circ - 79.1^\circ) = 207.8 \text{ W} -$$ - -Following the passive sign convention (see Fig. 1.8), this average power is absorbed by the source, in view of the direction of **I**2 and the polarity of the voltage source. That is, the circuit is delivering average power to the voltage source. - -For the current source, the current through it is **I**1 = 4⧸ 0**°** and the voltage across it is - -$$ -\mathbf{V}_1 = 20\mathbf{I}_1 + j10(\mathbf{I}_1 - \mathbf{I}_2) = 80 + j10(4 - 2 - j10.39) -$$ - -= 183.9 + j20 = 184.984/(6.21° V) - -The average power supplied by the current source is - -$$ -P_1 = \frac{1}{2} (184.984)(4) \cos(6.21^\circ - 0) = -367.8 \text{ W} -$$ - -It is negative according to the passive sign convention, meaning that the current source is supplying power to the circuit. - -For the resistor, the current through it is **I**1 = 4⧸ 0**°** and the voltage across it is 20**I**1 = 80⧸ 0**°**, so that the power absorbed by the resistor is - -$$ -P_2 = \frac{1}{2} (80)(4 \pm 160 \text{ W}) -$$ - -For the capacitor, the current through it is **I**2 = 10.58⧸ 79.1**°** and the volt age across it is −*j*5**I**2 = (5⧸ −90**°**)(10.58⧸ 79.1**°**) = 52.9⧸ 79.1**°**− 90°. The average power absorbed by the capacitor is - -$$ -P_4 = \frac{1}{2} (52.9)(10.58)\cos(-90^\circ) = 0 -$$ - -For the inductor, the current through it is **I**1 − **I**2 = 2 − *j*10.39 = 10.58⧸ −79.1**°**. The voltage across it is *j*10(**I**1 − **I**2) = 105.8⧸ −79.1**°** + 90°. Hence, the average power absorbed by the inductor is - -$$ -P_3 = \frac{1}{2} (105.8)(10.58) \text{ as } 90^\circ = 0 -$$ - -Notice that the inductor and the capacitor absorb zero average power and that the total power supplied by the current source equals the power absorbed by the resistor and the voltage source, or - -*P*1 + *P*2 + *P*3 + *P*4 + *P*5 = −367.8 + 160 + 0 + 0 + 207.8 = 0 - -indicating that power is conserved. - -Calculate the average power absorbed by each of the five elements in the circuit of Fig. 11.6. Practice Problem 11.4 - -For Practice Prob. 11.4. - -**Answer:** 40-V Voltage source: −60 W; *j*20-V Voltage source: −40 W; resistor: 100 W; others: 0 W. - -# **11.3** Maximum Average Power Transfer - -In Section 4.8 we solv ed the problem of maximizing the po wer delivered by a power-supplying resistive network to a load *RL*. Representing the circuit by its Thevenin equi valent, we pro ved that the maximum power would be delivered to the load if the load resistance is equal to the Thevenin resistance *RL* = *R*Th. We now extend that result to ac circuits. - -Consider the circuit in Fig. 11.7, where an ac circuit is connected to a load **Z***L* and is represented by its Thevenin equivalent. The load is usually represented by an impedance, which may model an electric motor, an antenna, a TV, and so forth. In rectangular form, the Thevenin impedance **Z**Th and the load impedance **Z***L* are - -$$ -\mathbf{Z}_{\mathrm{Th}} = R_{\mathrm{Th}} + jX_{\mathrm{Th}} \tag{11.13a} -$$ - -$$ -\mathbf{Z}_L = R_L + jX_L \tag{11.13b} -$$ - -The current through the load is - -ough the load is -\n -$$ -I = \frac{V_{\text{Th}}}{Z_{\text{Th}} + Z_L} = \frac{V_{\text{Th}}}{(R_{\text{Th}} + jX_{\text{Th}}) + (R_L + jX_L)} -$$ -\n(11.14) - -From Eq. (11.11), the average power delivered to the load is - -1), the average power delivered to the load is -\n -$$ -P = \frac{1}{2} |\mathbf{I}|^2 R_L = \frac{|\mathbf{V}_{\text{Th}}|^2 R_L/2}{(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2} -$$ -\n(11.15) - -Our objective is to adjust the load parameters *RL* and *XL* so that *P* is maximum. To do this we set *∂P*∕*∂RL* and *∂P*∕*∂XL* equal to zero. From Eq. (11.15), we obtain - -num. To do this we set -$$ -\partial P/\partial R_L -$$ - and $\partial P/\partial X_L$ equal to zero. From -1.15), we obtain -$$ -\frac{\partial P}{\partial X_L} = -\frac{|\mathbf{V}_{\text{Th}}|^2 R_L (X_{\text{Th}} + X_L)}{[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2} -$$ -(11.16a) - -$$ -\frac{\partial P}{\partial X_L} = -\frac{|\mathbf{V}_{\text{Th}}| \mathbf{r}_L(X_{\text{Th}} + X_L)}{[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2} -$$ -(11.16a) -$$ -\frac{\partial P}{\partial R_L} = \frac{|\mathbf{V}_{\text{Th}}|^2 [(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2 - 2R_L(R_{\text{Th}} + R_L)]}{2[(R_{\text{Th}} + R_L)^2 + (X_{\text{Th}} + X_L)^2]^2} -$$ -(11.16b) - -Setting *∂P*∕*∂XL* to zero gives - -$$ -X_L = -X_{\text{Th}} \tag{11.17} -$$ - -and setting *∂P*∕*∂RL* to zero results in - -gives us the maximum average power as - -ero results in -\n -$$ -R_L = \sqrt{R_{\text{Th}}^2 + (X_{\text{Th}} + X_L)^2} -$$ -\n(11.18) - -Combining Eqs. (11.17) and (11.18) leads to the conclusion that for maximum average power transfer, **Z***L* must be selected so that *XL* = −*X*Th and *RL* = *R*Th, i.e., - -$$ -Z_L = R_L + jX_L = R_{Th} - jX_{Th} = Z_{Th}^* -$$ - (11.19) - -For maximum average power transfer, the load impedance **Z**L must be equal to the complex conjugate of the Thevenin impedance **Z**Th. - -This result is known as the *maximum average power transfer theorem* for the sinusoidal steady state. Setting *RL* = *R*Th and *XL* = −*X*Th in Eq. (11.15) When **Z**L = **Z**\*Th, we say that the load is matched to the source. - -In a situation in which the load is purely real, the condition for maximum power transfer is obtained from Eq. (11.18) by setting *XL* = 0; that is, - -*P*max = ∣**V**Th∣ - -2 \_\_\_\_\_ 8*R*Th - -$$ -R_L = \sqrt{R_{\text{Th}}^2 + X_{\text{Th}}^2} = |\mathbf{Z}_{\text{Th}}| -$$ - (11.21) - -**(11.20)** - -# **Figure 11.7** Finding the maximum average power - -transfer: (a) circuit with a load, (b) the Thevenin equivalent. - -This means that for maximum a verage power transfer to a purely resis tive load, the load impedance (or resistance) is equal to the magnitude of the Thevenin impedance. - -**Figure 11.8** For Example 11.5. - -Example 11.5 Determine the load impedance **Z***L* that maximizes the a verage po wer drawn from the circuit of Fig. 11.8. What is the maximum a verage power? - -# **Solution:** - -First we obtain the Thevenin equivalent at the load terminals. To get **Z**Th, consider the circuit shown in Fig. 11.9(a). We find - -$$ -\mathbf{Z}_{\text{Th}} = j5 + 4 || (8 - j6) = j5 + \frac{4(8 - j6)}{4 + 8 - j6} = 2.933 + j4.467 \ \Omega -$$ - -# **Figure 11.9** - -Finding the Thevenin equivalent of the circuit in Fig. 11.8. - -To find **V**Th, consider the circuit in Fig. 11.8(b). By voltage division, - -$$ -\mathbf{V}_{\text{Th}} = \frac{8 - j6}{4 + 8 - j6} (10) = 7.454 \underline{\text{/}} - 10.3^{\circ} \text{ V} -$$ - -The load impedance draws the maximum power from the circuit when - -$$ -\mathbf{Z}_L = \mathbf{Z}_{\text{Th}}^* = 2.933 - j4.467 \ \Omega -$$ - -According to Eq. (11.20), the maximum average power is - -$$ -P_{\text{max}} = \frac{|\mathbf{V}_{\text{Th}}|^2}{8R_{\text{Th}}} = \frac{(7.454)^2}{8(2.933)} = 2.368 \text{ W} -$$ - -For the circuit shown in Fig. 11.10, find the load impedance **Z***L* that absorbs the maximum average power. Calculate that maximum average power. - -8 Ω 5 Ω ‒j4 Ω j10 Ω **Z**L 12 A **Figure 11.10** - -Practice Problem 11.5 - -For Practice Prob. 11.5. - -**Answer:** 3.415 − *j*0.7317 Ω, 51.47 W. - -In the circuit in Fig. 11.11, find the value of *RL* that will absorb the Example 11.6 maximum average power. Calculate that power. - -# **Solution:** - -We first find the Thevenin equivalent at the terminals of *RL*. - -rst find the Thevenin equivalent at the terminals of *RL*. -**Z**Th = (40 − *j*30) -$$ -||j20 = \frac{j20(40 - j30)}{j20 + 40 - j30} = 9.412 + j22.35 Ω -$$ - -By voltage division, - -division, -\n -$$ -\mathbf{V}_{\text{Th}} = \frac{j20}{j20 + 40 - j30} (150/30^{\circ}) = 72.76/134^{\circ} \text{ V} -$$ - -The value of *RL* that will absorb the maximum average power is - -$$ -V_L \text{ that will absorb the maximum average po} -$$ -\n -$$ -R_L = |\mathbf{Z}_{\text{Th}}| = \sqrt{9.412^2 + 22.35^2} = 24.25 \ \Omega -$$ - -The current through the load is - -t through the load is -\n -$$ -I = \frac{V_{\text{Th}}}{Z_{\text{Th}} + R_L} = \frac{72.76/134^{\circ}}{33.66 + j22.35} = 1.8/100.42^{\circ} \text{ A} -$$ - -The maximum average power absorbed by *RL* is - -$$ -P_{\text{max}} = \frac{1}{2} |\mathbf{I}|^2 R_L = \frac{1}{2} (1.8)^2 (24.25) = 39.29 \text{ W} -$$ - -In Fig. 11.12, the resistor *RL* is adjusted until it absorbs the maximum average power. Calculate *RL* and the maximum average power absorbed by it. - -**Answer:** 30 Ω, 23.06 W. - -# **11.4** Effective or RMS Value - -The idea of *effective value* arises from the need to measure the efectiveness of a voltage or current source in delivering power to a resistive load. - -The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current. - -Practice Problem 11.6 - -For Example 11.6. - -Finding the effective current: (a) ac circuit, (b) dc circuit. - -In Fig. 11.13, the circuit in (a) is ac while that of (b) is dc. Our objecti ve is to find *I*eff that will transfer the same po wer to resistor *R* as the sinusoid *i*. The average power absorbed by the resistor in the ac circuit is - -$$ -P = \frac{1}{T} \int_0^T i^2 R \, dt = \frac{R}{T} \int_0^T i^2 \, dt \tag{11.22} -$$ - -while the power absorbed by the resistor in the dc circuit is - -$$ -P = I_{\text{eff}}^2 R \tag{11.23} -$$ - -Equating the expressions in Eqs. (11.22) and (11.23) and solving for *I*eff, we obtain - -$$ -I_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T i^2 dt} -$$ - (11.24) - -The effective value of the v oltage is found in the same w ay as current; that is, - -$$ -V_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T v^2 dt} -$$ - (11.25) - -This indicates that the effective value is the (square) *root* of the *mean* (or average) of the *square* of the periodic signal. Thus, the effective value is often known as the *root-mean-square* value, or *rms* value for short; and we write - -$$ -I_{\rm eff} = I_{\rm rms}, \qquad V_{\rm eff} = V_{\rm rms} \tag{11.26} -$$ - -For any periodic function *x*(*t*) in general, the rms value is given by - -$$ -X_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T x^2 dt} -$$ - (11.27) - -The effective value of a periodic signal is its root mean square (rms) value. - -Equation 11.27 states that to find the rms value of *x*(*t*), we first find its *square x*2 and then find the *mean* of that, or - -$$ -\frac{1}{T} \int_0^T x^2 dt -$$ - -and then the square *root* ( √ \_\_\_\_\_\_ ) of that mean. The rms value of a constant is the constant itself. For the sinusoid *i*(*t*) = *Im* cos *ωt*, the effective or rms value is - -$$ -I_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T I_m^2 \cos^2 \omega t \, dt} -$$ -$$ -= \sqrt{\frac{I_m^2}{T} \int 0 \, T \, \frac{1}{2} (1 + \cos 2\omega t) \, dt} = \frac{I_m}{\sqrt{2}} \tag{11.28} -$$ - -Similarly, for *v*(*t*) = *Vm* cos *ωt*, - -$$ -V_{\rm rms} = \frac{V_m}{\sqrt{2}}\tag{11.29} -$$ - -Keep in mind that Eqs. (11.28) and (11.29) are only valid for sinusoidal signals. - -The average power in Eq. (11.8) can be written in terms of the rms values. - -$$ -P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) = \frac{V_m}{\sqrt{2}} \frac{I_m}{\sqrt{2}} \cos(\theta_v - \theta_i) -$$ - -= $V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i)$ (11.30) - -Similarly, the average power absorbed by a resistor *R* in Eq. (11.11) can be written as - -$$ -P = I_{\rm rms}^2 R = \frac{V_{\rm rms}^2}{R} -$$ - (11.31) - -When a sinusoidal voltage or current is specified, it is often in terms of its maximum (or peak) v alue or its rms v alue, since its a verage value is zero. The power industries specify phasor magnitudes in terms of their rms values rather than peak v alues. For instance, the 110 V available at every household is the rms value of the voltage from the power company. It is convenient in power analysis to express voltage and current in their rms values. Also, analog v oltmeters and ammeters are designed to read directly the rms value of voltage and current, respectively. - -Determine the rms v alue of the current w aveform in Fig. 11.14. If the Example 11.7 current is passed through a 2- Ω resistor, find the average power absorbed by the resistor. - -# **Solution:** - -The period of the waveform is *T* = 4. Over a period, we can write the current waveform as - -$$ -i(t) = \begin{cases} 5t, & 0 < t < 2 \\ -10, & 2 < t < 4 \end{cases} -$$ - -The rms value is - -$$ -I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T i^2 dt} = \sqrt{\frac{1}{4} \left[ \int_0^2 (5t)^2 dt + \int_2^4 (-10)^2 dt \right]} -$$ -$$ -= \sqrt{\frac{1}{4} \left[ 25 \frac{t^3}{3} \right]_0^2 + 100t \Big|_2^4} = \sqrt{\frac{1}{4} \left( \frac{200}{3} + 200 \right)} = 8.165 \text{ A} -$$ - -The power absorbed by a 2-Ω resistor is - -$$ -P = I_{\text{rms}}^2 R = (8.165)^2 (2) = 133.3 \text{ W} -$$ - -Find the rms value of the current waveform of Fig. 11.15. If the current flows through a 9-Ω resistor, calculate the average power absorbed by the resistor. - -**Answer:** 9.238 A, 768 W. - -**Figure 11.14** For Example 11.7. - -For Example 11.8. - -Example 11.8 The waveform shown in Fig. 11.16 is a half-w ave rectified sine wave. Find the rms value and the amount of average power dissipated in a 10-Ω resistor. - -# **Solution:** - -The period of the voltage waveform is *T* = 2*π*, and - -$$ -v(t) = \begin{cases} 10 \sin t, & 0 < t < \pi \\ 0, & \pi < t < 2\pi \end{cases} -$$ - -The rms value is obtained as - -$$ -V_{\text{rms}}^2 = \frac{1}{T} \int_0^T v^2(t) \, dt = \frac{1}{2\pi} \left[ \int_0^{\pi} (10 \sin t)^2 \, dt + \int_{\pi}^{2\pi} 0^2 \, dt \right] -$$ - -But sin2 *t* = \_\_1 2 (1 − cos 2*t*). Hence, - -$$ -V_{\text{rms}}^2 = \frac{1}{2\pi} \int_0^{\pi} \frac{100}{2} (1 - \cos 2t) dt = \frac{50}{2\pi} \left( t - \frac{\sin 2t}{2} \right) \Big|_0^{\pi} -$$ -$$ -= \frac{50}{2\pi} \left( \pi - \frac{1}{2} \frac{\sin 2\pi - 0}{2} \right) = 25, \qquad V_{\text{rms}} = 5 \text{ V} -$$ -The average power absorbed is -$$ -\sqrt{\pi} \alpha \int_0^{\pi} \alpha \sqrt{1 - \frac{1}{2} \sin 2\pi} dt -$$ - -$$ -P = \frac{V_{\text{rms}}^2}{R} = \frac{5^2}{10} = 2.5 \text{ W} -$$ - -# Practice Problem 11.8 - -Find the rms value of the full-wave rectified sine wave in Fig. 11.17. Calculate the average power dissipated in a 6-Ω resistor. - -**Answer:** 70.71 V, 833.3 W. - -# For Practice Prob. 11.8. **11.5** Apparent Power and Power Factor - -In Section 11.2 we saw that if the voltage and current at the terminals of a circuit are - -$$ -v(t) = V_m \cos(\omega t + \theta_v) -$$ - and $i(t) = I_m \cos(\omega t + \theta_i)$ (11.32) - -or, in phasor form, **V** = *Vm*⧸*θv* and **I** = *Im*⧸*θi* , the average power is - -$$ -P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) -$$ - (11.33) - -In Section 11.4, we saw that - -$$ -P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}) = S \cos(\theta_{\nu} - \theta_{i}) -$$ - (11.34) - -We have added a new term to the equation: - -$$ -S = V_{\rm rms} I_{\rm rms} -$$ - (11.35) - -The average power is a product of two terms. The product *V*rms*I*rms is known as the *apparent power S* . The factor cos( *θv* − *θi*) is called the *power factor* (pf). - -The apparent power (in VA) is the product of the rms values of voltage and current. - -The apparent power is so called because it seems apparent that the power should be the v oltage-current product, by analogy with dc resisti ve circuits. It is measured in volt-amperes or VA to distinguish it from the average or real power, which is measured in watts. The power factor is dimensionless, since it is the ratio of the average power to the apparent power, - -$$ -pf = \frac{P}{S} = \cos(\theta_v - \theta_i) -$$ - (11.36) - -The angle *θv* − *θi* is called the *power factor angle,* because it is the angle whose cosine is the power factor. The power factor angle is equal to the angle of the load impedance if **V** is the voltage across the load and **I** is the current through it. This is evident from the fact that - -$$ -Z = \frac{V}{I} = \frac{V_m/\theta_v}{I_m/\theta_i} = \frac{V_m}{I_m}/\theta_v - \theta_i -$$ - (11.37) - -Alternatively, since - -$$ -\mathbf{V}_{\rm rms} = \frac{\mathbf{V}}{\sqrt{2}} = V_{\rm rms} \underline{\theta_v} \tag{11.38a} -$$ - -and - -$$ -\mathbf{I}_{\rm rms} = \frac{\mathbf{I}}{\sqrt{2}} = I_{\rm rms} / \theta_i \tag{11.38b} -$$ - -the impedance is - -$$ -Z = \frac{V}{I} = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{V_{\text{rms}}}{I_{\text{rms}}} \underbrace{\beta_{\nu} - \theta_{i}} \tag{11.39} -$$ - -The power factor is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of the load impedance. - -From Eq. (11.36), the power factor may be seen as that f actor by which the apparent power must be multiplied to obtain the real or average power. The value of pf ranges between zero and unity . For a purely resisti ve load, the voltage and current are in phase, so that *θv* − *θi* = 0 and pf = 1. This implies that the apparent po wer is equal to the a verage power. For a purely reactive load, *θv* − *θi* = ±90° and pf = 0. In this case the a verage power is zero. In between these tw o extreme cases, pf is said to be *leading* or *lagging*. Leading power factor means that current leads v oltage, which implies a capaciti ve load. Lagging po wer factor means that current lags voltage, implying an inductive load. Power factor affects the From Eq. (11.36), the power factor may also be regarded as the ratio of the real power dissipated in the load to the apparent power of the load. - -electric bills consumers pay the electric utility companies, as we will see in Section 11.9.2. - -Example 11.9 A series-connected load dra ws a current *i*(*t*) = 4 cos(100 *πt* + 10°) A when the applied v oltage is *v*(*t*) = 120 cos(100*πt* − 20°) V. Find the apparent power and the power factor of the load. Determine the element values that form the series-connected load. - -# **Solution:** - -The apparent power is - -$$ -S = V_{\text{rms}} I_{\text{rms}} = \frac{120}{\sqrt{2}} \frac{4}{\sqrt{2}} = 240 \text{ VA} -$$ - -The power factor is - -$$ -pf = \cos(\theta_v - \theta_i) = \cos(-20^\circ - 10^\circ) = 0.866 \quad \text{(leading)} -$$ - -The pf is leading because the current leads the voltage. The pf may also be obtained from the load impedance. - -$$ -\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = \frac{120/-20^{\circ}}{4/10^{\circ}} = 30/-30^{\circ} = 25.98 - j15 \ \Omega -$$ -\n -$$ -\text{pf} = \cos(-30^{\circ}) = 0.866 \qquad \text{(leading)} -$$ - -The load impedance **Z** can be modeled by a 25.98-Ω resistor in series with a capacitor with - -$$ -X_C = -15 = -\frac{1}{\omega C} -$$ - -or - -$$ -C = \frac{1}{15\omega} = \frac{1}{15 \times 100\pi} = 212.2 \,\mu\text{F} -$$ - -| Practice Problem 11.9 | Obtain the power factor and the apparent power of a load whose | -|-----------------------|------------------------------------------------------------------------------------------------| -| | impedance is Z = 60 + j40 Ω when the applied voltage
is v(t) =
155.56 cos(377t + 10°) V. | -| | | - -**Answer:** 0.8321 lagging, 167.69⧸ 33.69° VA. - -# **Solution:** - -The total impedance is - -$$ -\mathbf{Z} = 6 + 4 \left( (-j2) \right) = 6 + \frac{-j2 \times 4}{4 - j2} = 6.8 - j1.6 = 7 \underline{\text{/} -13.24^{\circ}} \,\Omega -$$ - -The power factor is - -$$ -pf = \cos(-13.24) = 0.9734 \text{ (leading)} -$$ - -since the impedance is capacitive. The rms value of the current is - -$$ -\mathbf{I}_{\rm rms} = \frac{\mathbf{V}_{\rm rms}}{\mathbf{Z}} = \frac{30/0^{\circ}}{7/-13.24^{\circ}} = 4.286/13.24^{\circ} \text{A} -$$ - -The average power supplied by the source is - -*P* = *V*rms*I*rmspf = (30)(4.286)0.9734 = 125 W - -or - -$$ -P = I_{\text{rms}}^2 R = (4.286)^2 (6.8) = 125 \text{ W} -$$ - -where *R* is the resistive part of **Z**. - -Calculate the power factor of the entire circuit of Fig. 11.19 as seen by the source. What is the average power supplied by the source? - -**Answer:** 0.936 lagging, 2.008 kW. - -# For Practice Prob. 11.10. **11.6** Complex Power - -Considerable effort has been expended over the years to express power relations as simply as possible. Po wer engineers have coined the term *complex power,* which they use to find the total effect of parallel loads. Complex power is important in po wer analysis because it contains *all* the information pertaining to the power absorbed by a given load. - -Consider the ac load in Fig. 11.20. Gi ven the phasor form **V** = *Vm*⧸*θv* and **I** = *Im*⧸*θi* of voltage *v*(*t*) and current *i*(*t*), the *complex power* **S** absorbed by the ac load is the product of the v oltage and the comple x conjugate of the current, or - -$$ -S = \frac{1}{2}VI^* -$$ -\n(11.40) - -assuming the passi ve sign con vention (see Fig. 11.20). In terms of the rms values, - -$$ -S = V_{\rm rms}I_{\rm rms}^* \tag{11.41} -$$ - -where - -$$ -\mathbf{V}_{\rm rms} = \frac{\mathbf{V}}{\sqrt{2}} = V_{\rm rms} \underline{\theta_v} \tag{11.42} -$$ - -and - -$$ -\mathbf{I}_{\rm rms} = \frac{\mathbf{I}}{\sqrt{2}} = I_{\rm rms} / \theta_i \tag{11.43} -$$ - -10 Ω 8 Ω 165 0° V rms + j4 Ω ‒j6 Ω ‒ Practice Problem 11.10 - -# **Figure 11.19** - -# **Figure 11.20** - -The voltage and current phasors associated with a load. - -When working with the rms values of currents or voltages, we may drop the subscript rms if no confusion will be caused by doing so. - -Thus, we may write Eq. (11.41) as - -$$ -\mathbf{S} = V_{\text{rms}} I_{\text{rms}} \underline{\beta_v - \theta_i} -$$ - -= $V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i) + j V_{\text{rms}} I_{\text{rms}} \sin(\theta_v - \theta_i)$ (11.44) - -This equation can also be obtained from Eq. (11.9). We notice from Eq. (11.44) that the magnitude of the complex power is the apparent power; hence, the comple x power is measured in v olt-amperes (VA). Also, we notice that the angle of the complex power is the power factor angle. - -The complex power may be expressed in terms of the load impedance **Z**. From Eq. (11.37), the load impedance **Z** may be written as - -$$ -Z = \frac{V}{I} = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{V_{\text{rms}}}{I_{\text{rms}}} \frac{\beta_v - \theta_i}{\beta} -$$ -(11.45) - -Thus, **V**rms = **ZI**rms. Substituting this into Eq. (11.41) gives - -$$ -S = I_{\rm rms}^2 Z = \frac{V_{\rm rms}^2}{Z^*} = V_{\rm rms}I_{\rm rms}^* -$$ - (11.46) - -Since **Z** = *R* + *jX*, Eq. (11.46) becomes - -$$ -S = I_{\text{rms}}^2(R + jX) = P + jQ \tag{11.47} -$$ - -where *P* and *Q* are the real and imaginary parts of the complex power; that is, - -$$ -P = \text{Re}(\mathbf{S}) = I_{\text{rms}}^2 R \tag{11.48} -$$ - -$$ -Q = \text{Im}(\mathbf{S}) = I_{\text{rms}}^2 X \tag{11.49} -$$ - -*P* is the a verage or real po wer and it depends on the load' s resistance *R*. *Q* depends on the load' s reactance *X* and is called the *reactive* (or quadrature) power. - -Comparing Eq. (11.44) with Eq. (11.47), we notice that - -$$ -P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}), \qquad Q = V_{\text{rms}} I_{\text{rms}} \sin(\theta_{\nu} - \theta_{i}) \tag{11.50} -$$ - -The real power *P* is the a verage power in watts delivered to a load; it is the only useful po wer. It is the actual po wer dissipated by the load. The reactive power *Q* is a measure of the energy exchange between the source and the reactive part of the load. The unit of *Q* is the *volt-ampere reactive* (VAR) to distinguish it from the real po wer, whose unit is the w att. We know from Chapter 6 that ener gy storage elements neither dissipate nor supply power, but exchange power back and forth with the rest of the network. In the same w ay, the reactive power is being transferred back and forth between the load and the source. It represents a lossless interchange between the load and the source. Notice that: - -- 1. *Q* = 0 for resistive loads (unity pf). -- 2. *Q* < 0 for capacitive loads (leading pf). -- 3. *Q* > 0 for inductive loads (lagging pf). - -Thus, - -Complex power (in VA) is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. As a complex quantity, its real part is real power P and its imaginary part is reactive power Q. - -Introducing the complex power enables us to obtain the real and reactive powers directly from voltage and current phasors. - -Complex Power = -$$ -\mathbf{S} = P + jQ = \mathbf{V}_{\text{rms}}(\mathbf{I}_{\text{rms}})^* -$$ - -\n= $|\mathbf{V}_{\text{rms}}| |\mathbf{I}_{\text{rms}}| / \theta_v - \theta_i$ -\nApparent Power = $S = |\mathbf{S}| = |\mathbf{V}_{\text{rms}}| |\mathbf{I}_{\text{rms}}| = \sqrt{P^2 + Q^2}$ -\nReal Power = $P = \text{Re}(\mathbf{S}) = S \cos(\theta_v - \theta_i)$ -\nReactive Power = $Q = \text{Im}(\mathbf{S}) = S \sin(\theta_v - \theta_i)$ -\nPower Factor = $\frac{P}{S} = \cos(\theta_v - \theta_i)$ - -This sho ws ho w the comple x po wer contains *all* the rele vant po wer information in a given load. - -It is a standard practice to represent **S**, *P*, and *Q* in the form of a triangle, known as the *power triangle,* shown in Fig. 11.21(a). This is similar to the impedance triangle sho wing the relationship between **Z**, *R*, and *X*, illustrated in Fig. 11.21(b). The power triangle has four items—the apparent/complex power, real power, reactive power, and the power factor angle. Given two of these items, the other two can easily be obtained from the triangle. As shown in Fig. 11.22, when **S** lies in the first quadrant, we have an inductive load and a lagging pf. When **S** lies in the fourth quadrant, the load is capacitive and the pf is leading. It is also possible for the comple x power to lie in the second or third quadrant. This requires that the load impedance ha ve a negative resistance, which is possible with active circuits. - -**S** contains all power information of a load. The real part of **S** is the real power P; its imaginary part is the reactive power Q; its magnitude is the apparent power S; and the cosine of its phase angle is the power factor pf. - -The voltage across a load is *v*(*t*) = 60 cos(*ωt* − 10°) V and the cur - Example 11.11 rent through the element in the direction of the v oltage drop is *i*(*t*) = 1.5 cos( *ωt* + 50°) A. Find: (a) the comple x and apparent po wers, (b) the real and reacti ve powers, and (c) the po wer factor and the load impedance. - -# **Solution:** - -(a) For the rms values of the voltage and current, we write - -$$ -\mathbf{V}_{\rm rms} = \frac{60}{\sqrt{2}} \angle 10^{\circ}, \qquad \mathbf{I}_{\rm rms} = \frac{1.5}{\sqrt{2}} \angle 50^{\circ} -$$ - -The complex power is - -$$ -\mathbf{S} = \mathbf{V}_{\rm rms} \mathbf{I}_{\rm rms}^* = \left(\frac{60}{\sqrt{2}} \angle 10^\circ \right) \left(\frac{1.5}{\sqrt{2}} \angle 50^\circ \right) = 45 \angle 60^\circ \text{ VA} -$$ - -The apparent power is - -$$ -S = |\mathbf{S}| = 45 \text{ VA} -$$ - -(b) We can express the complex power in rectangular form as - -$$ -S = 45 \underline{/-60^{\circ}} = 45 [\cos(-60^{\circ}) + j \sin(-60^{\circ})] = 22.5 - j38.97 -$$ - -Since **S** = *P* + *jQ*, the real power is - -$$ -P = 22.5 \, \mathrm{W} -$$ - -while the reactive power is - -$$ -Q = -38.97 -$$ - **VAR** - -(c) The power factor is - -$$ -pf = \cos(-60^\circ) = 0.5 \text{ (leading)} -$$ - -It is leading, because the reactive power is negative. The load impedance is - -$$ -Z = \frac{V}{I} = \frac{60/-10^{\circ}}{1.5/+50^{\circ}} = 40/-60^{\circ} \,\Omega -$$ - -which is a capacitive impedance. - -For a load, **V**rms = 110⧸ 85° V, **I**rms = 3⧸ 15° A. Determine: (a) the complex and apparent powers, (b) the real and reactive powers, and (c) the power factor and the load impedance. Practice Problem 11.11 - -> **Answer:** (a) 330 ⧸ 70° VA, 44 VA, (b) 112.87 W, 310.1 VAR, (c) 0.342 lagging, (12.541 + *j*34.46) Ω. - -Example 11.12 A load **Z** dra ws 12 kV A at a po wer f actor of 0.856 lagging from a 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance. - -# **Solution:** - -(a) Given that pf = cos *θ* = 0.856, we obtain the power angle as *θ* = cos−1 0.856 = 31.13°. If the apparent power is *S* = 12,000 VA, then the average or real power is - -*P* = *S* cos *θ* = 12,000 × 0.856 = 10.272 kW - -while the reactive power is - -$$ -Q = S \sin \theta = 12,000 \times 0.517 = 6.204 -$$ -kVA - -(b) Since the pf is lagging, the complex power is - -$$ -S = P + jQ = 10.272 + j6.204 -$$ - kVA - -From **S** = **V**rms**I**\*rms, we obtain - -$$ -\text{Im } \mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* \text{, we obtain} -$$ -\n -$$ -\mathbf{I}_{\text{rms}}^* = \frac{\mathbf{S}}{\mathbf{V}_{\text{rms}}} = \frac{10,272 + j6204}{120/0^{\circ}} = 85.6 + j51.7 \text{ A} = 100/31.13^{\circ} \text{ A} -$$ - -Thus **I**rms = 100⧸ −31.13° and the peak current is - -$$ -I_m = \sqrt{2}I_{\text{rms}} = \sqrt{2}(100) = 141.4 \text{ A} -$$ - -(c) The load impedance - -update - -\n -$$ -\mathbf{Z} = \frac{\mathbf{V}_{\text{rms}}}{\mathbf{I}_{\text{rms}}} = \frac{120/0^{\circ}}{100/-31.13^{\circ}} = 1.2/31.13^{\circ} \ \Omega -$$ - -which is an inductive impedance. - -A sinusoidal source supplies 100 kVAR reactive power to load **Z** = 250⧸ −75° Ω. Determine: (a) the power factor, (b) the apparent power delivered to the load, and (c) the rms voltage. - -**Answer:** (a) 0.2588 leading, (b) 103.53 kVA, (c) 5.087 kV. - -# **11.7** Conservation of AC Power - -The principle of conservation of power applies to ac circuits as well as to dc circuits (see Section 1.5). - -To see this, consider the circuit in Fig. 11.23(a), where two load impedances **Z**1 and **Z**2 are connected in parallel across an ac source **V**. KCL gives - -$$ -\mathbf{I} = \mathbf{I}_1 + \mathbf{I}_2 \tag{11.52} -$$ - -The complex power supplied by the source is (from now on, unless otherwise specified, all values of voltages and currents will be assumed to be rms values) - -$$ -S = VI^* = V(I_1^* + I_2^*) = VI_1^* + VI_2^* = S_1 + S_2 \qquad (11.53) -$$ - -11.3 and 11.4 that average power is conserved in ac circuits. - -In fact, we already saw in Examples - -Practice Problem 11.12 - -An ac voltage source supplied loads connected in: (a) parallel, (b) series. - -where **S**1 and **S**2 denote the comple x powers delivered to loads **Z**1 and **Z**2, respectively. - -If the loads are connected in series with the voltage source, as shown in Fig. 11.23(b), KVL yields - -$$ -\mathbf{V} = \mathbf{V}_1 + \mathbf{V}_2 \tag{11.54} -$$ - -The complex power supplied by the source is - -$$ -S = VI^* = (V_1 + V_2)I^* = V_1I^* + V_2I^* = S_1 + S_2 \quad (11.55) -$$ - -where **S**1 and **S**2 denote the comple x powers delivered to loads **Z**1 and **Z**2, respectively. - -We conclude from Eqs. (11.53) and (11.55) that whether the loads are connected in series or in parallel (or in general), the total po wer *supplied* by the source equals the total power *delivered* to the load. Thus, in general, for a source connected to *N* loads, - -$$ -S = S_1 + S_2 + \dots + S_N \tag{11.56} -$$ - -This means that the total comple x power in a network is the sum of the complex powers of the individual components. (This is also true of real power and reactive power, but not true of apparent power.) This expresses the principle of conservation of ac power: - -The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads. - -From this we imply that the real (or reactive) power flow from sources in a network equals the real (or reactive) power flow into the other elements in the network. - -Example 11.13 Figure 11.24 sho ws a load being fed by a v oltage source through a transmission line. The impedance of the line is represented by the (4 + *j*2) Ω impedance and a return path. Find the real power and reactive power absorbed by: (a) the source, (b) the line, and (c) the load. - -# **Solution:** - -The total impedance is - -$$ -\mathbf{Z} = (4+j2) + (15-j10) = 19 - j8 = 20.62 \underline{\smash{\big)}\,22.83^\circ} -$$ - $\Omega$ - -In fact, all forms of ac power are conserved: instantaneous, real, reactive, and complex. - -The current through the circuit is - -through the circuit is -\n -$$ -\mathbf{I} = \frac{\mathbf{V}_s}{\mathbf{Z}} = \frac{220/0^{\circ}}{20.62/-22.83^{\circ}} = 10.67/22.83^{\circ} \text{ A rms} -$$ - -(a) For the source, the complex power is - -$$ -S_s = V_s I^* = (220/0^\circ)(10.67/-22.83^\circ) -$$ - -= 2347.4/-22.83° = (2163.5 - j910.8) VA - -From this, we obtain the real power as 2163.5 W and the reactive power as 910.8 VAR (leading). - -(b) For the line, the voltage is - -$$ -\mathbf{V}_{\text{line}} = (4 + j2)\mathbf{I} = (4.472 \underline{/ 26.57^{\circ}})(10.67 \underline{/ 22.83^{\circ}}) -$$ -$$ -= 47.72 \underline{/ 49.4^{\circ}} \text{ V rms} -$$ - -The complex power absorbed by the line is - -$$ -S_{line} = V_{line}I^* = (47.72/49.4^{\circ})(10.67/-22.83^{\circ}) -$$ - -= 509.2/26.57° = 455.4 + j227.7 VA - -or - -$$ -S_{\text{line}} = |I|^2 Z_{\text{line}} = (10.67)^2 (4 + j2) = 455.4 + j227.7 VA -$$ - -That is, the real power is 455.4 W and the reactive power is 227.76 VAR (lagging). - -(c) For the load, the voltage is - -$$ -\mathbf{V}_L = (15 - j10)\mathbf{I} = (18.03 \text{/} - 33.7^{\circ})(10.67 \text{/} 22.83^{\circ}) -$$ - -= 192.38 \text{/} - 10.87° V rms - -The complex power absorbed by the load is - -$$ -\mathbf{S}_L = \mathbf{V}_L \mathbf{I}^* = (192.38 \text{/} - 10.87^\circ)(10.67 \text{/} - 22.83^\circ) -$$ - -= 2053 \text{/} - 33.7^\circ = (1708 - j1139) VA - -The real power is 1708 W and the reactive power is 1139 VAR (leading). Note that **S***s* = **S**line + **S***L*, as expected. We have used the rms values of voltages and currents. - -In the circuit in Fig. 11.25, the 60- Ω resistor absorbs an average power of 240 W. Find **V** and the complex power of each branch of the circuit. What is the overall complex power of the circuit? (Assume the current through the 60-Ω resistor has no phase shift.) - -**Answer:** 240.7 ⧸ 21.45° V (rms); the 20- Ω resistor: 656 VA; the (30 − *j*10) Ω impedance: 480 − *j*160 VA; the (60 + *j*20) Ω impedance: 240 + *j*80 VA; overall: 1376 − *j*80 VA. - -Practice Problem 11.13 - -For Practice Prob. 11.13. - -For Example 11.14. - -# **Solution:** - -The current through **Z**1 is - -$$ -\mathbf{I}_1 = \frac{\mathbf{V}}{\mathbf{Z}_1} = \frac{120/10^{\circ}}{60/-30^{\circ}} = 2/40^{\circ} \text{ A rms} -$$ - -while the current through **Z**2 is - -$$ -I_2 = \frac{V}{Z_2} = \frac{120/10^{\circ}}{40/45^{\circ}} = 3/-35^{\circ} -$$ - A rms - -The complex powers absorbed by the impedances are - -$$ -\mathbf{S}_1 = \frac{V_{\text{rms}}^2}{\mathbf{Z}_1^*} = \frac{(120)^2}{60/30^\circ} = 240/-30^\circ = 207.85 - j120 \text{ VA} -$$ -\n -$$ -\mathbf{S}_2 = \frac{V_{\text{rms}}^2}{\mathbf{Z}_2^*} = \frac{(120)^2}{40/-45^\circ} = 360/45^\circ = 254.6 + j254.6 \text{ VA} -$$ - -The total complex power is - -$$ -S_t = S_1 + S_2 = 462.4 + j134.6 VA -$$ - -(a) The total apparent power is - -arent power is -\n -$$ -|\mathbf{S}_t| = \sqrt{462.4^2 + 134.6^2} = 481.6 \text{ VA}. -$$ - -(b) The total real power is - -$$ -P_t = \text{Re}(S_t) = 462.4 \text{ W or } P_t = P_1 + P_2. -$$ - -(c) The total reactive power is - -$$ -Q_t = \text{Im}(S_t) = 134.6 \text{ VAR or } Q_t = Q_1 + Q_2. -$$ - -(d) The pf = *Pt*∕∣**S***t*∣ = 462.4∕481.6 = 0.96 (lagging). - -We may cross check the result by finding the complex power **S***s* supplied by the source. - -$$ -\mathbf{I}_t = \mathbf{I}_1 + \mathbf{I}_2 = (1.532 + j1.286) + (2.457 - j1.721) -$$ - -= 4 - j0.435 = 4.024 $\underline{/$ -6.21° A rms -$$ -\mathbf{S}_s = \mathbf{V}\mathbf{I}_t^* = (120/10°)(4.024/6.21°) -$$ - -= 482.88/16.21° = 463 + j135 VA - -which is the same as before. - -Two loads connected in parallel are respectively 3 kW at a pf of 0.75 leading and 6 kW at a pf of 0.95 lagging. Calculate the pf of the com bined two loads. Find the complex power supplied by the source. Practice Problem 11.14 - -**Answer:** 0.9972 (leading), 9 − *j*0.6742 kVA. - -# **11.8** Power Factor Correction - -Most domestic loads (such as w ashing machines, air conditioners, and refrigerators) and industrial loads (such as induction motors) are induc tive and operate at a lo w lagging po wer factor. Although the inducti ve nature of the load cannot be changed, we can increase its power factor. - -The process of increasing the power factor without altering the voltage or current to the original load is known as power factor correction. - -Since most loads are inducti ve, as shown in Fig. 11.27(a), a load' s power factor is improved or corrected by deliberately installing a capacitor in parallel with the load, as shown in Fig. 11.27(b). The effect of adding the capacitor can be illustrated using either the power triangle or the phasor diagram of the currents in volved. Figure 11.28 sho ws the latter, where it is assumed that the circuit in Fig. 11.27(a) has a power factor of cos *θ*1, while the one in Fig. 11.27(b) has a power factor of cos *θ*2. It is evident from Fig. 11.28 that adding the capacitor has caused the phase angle between the supplied v oltage and current to reduce from *θ*1 to *θ*2, thereby increasing the power factor. We also notice from the magnitudes of the vectors in Fig. 11.28 that with the same supplied v oltage, the circuit in Fig. 11.27(a) dra ws larger current *IL* than the current *I* drawn by the circuit in Fig. 11.27(b). Power companies charge more for larger currents, because they result in increased power losses (by a squared factor, since *P* = *IL* 2 *R*). Therefore, it is beneficial to both the power company and the consumer that every effort is made to minimize current level or keep the power factor as close to unity as possible. By choosing a suitable size for the capacitor, the current can be made to be completely in phase with the voltage, implying unity power factor. - -Alternatively, power factor correction may be viewed as the addition of a reactive element (usually a capacitor) in parallel with the load in order to make the power factor closer to unity. - - An inductive load is modeled as a series combination of an inductor and a resistor. - -**Figure 11.27** Power factor correction: (a) original inductive load, (b) inductive load with improved power factor. - -We can look at the power factor correction from another perspective. Consider the power triangle in Fig. 11.29. If the original inducti ve load has apparent power *S*1, then - -$$ -P = S_1 \cos \theta_1 -$$ -, $Q_1 = S_1 \sin \theta_1 = P \tan \theta_1$ (11.57) - -**Figure 11.29** Power triangle illustrating power factor correction. - -If we desire to increase the po wer factor from cos *θ*1 to cos *θ*2 without altering the real power (i.e., *P* = *S*2 cos *θ*2), then the new reactive power is - -$$ -Q_2 = P \tan \theta_2 \tag{11.58} -$$ - -The reduction in the reacti ve power is caused by the shunt capacitor; that is, - -$$ -Q_C = Q_1 - Q_2 = P(\tan \theta_1 - \tan \theta_2) -$$ - (11.59) - -But from Eq. (11.46), *QC* = *V*2 rms∕*XC* = *ωCV* 2 rms. The value of the required shunt capacitance *C* is determined as - -$$ -C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{P(\tan \theta_1 - \tan \theta_2)}{\omega V_{\text{rms}}^2} -$$ - (11.60) - -Note that the real po wer *P* dissipated by the load is not af fected by the power factor correction because the a verage power due to the capaci tance is zero. - -Although the most common situation in practice is that of an inductive load, it is also possible that the load is capaciti ve; that is, the load is operating at a leading power factor. In this case, an inductor should be connected across the load for po wer factor correction. The required shunt inductance *L* can be calculated from - -$$ -Q_L = \frac{V_{\text{rms}}^2}{X_L} = \frac{V_{\text{rms}}^2}{\omega L} \qquad \Rightarrow \qquad L = \frac{V_{\text{rms}}^2}{\omega Q_L} \tag{11.61} -$$ - -where *QL* = *Q*1 − *Q*2, the dif ference between the ne w and old reacti ve powers. - -Example 11.15 When connected to a 120-V (rms), 60-Hz po wer line, a load absorbs 4 kW at a lagging po wer factor of 0.8. Find the v alue of capacitance necessary to raise the pf to 0.95. - -# **Solution:** - -If the pf = 0.8, then - -cos *θ*1 = 0.8 ⇒ *θ*1 = 36.87° - -where *θ*1 is the phase difference between voltage and current. We obtain the apparent power from the real power and the pf as - -$$ -S_1 = \frac{P}{\cos \theta_1} = \frac{4000}{0.8} = 5000 \text{ VA} -$$ - -The reactive power is - -$$ -Q_1 = S_1 \sin \theta = 5000 \sin 36.87 = 3000 -$$ - VAR - -When the pf is raised to 0.95, - -$$ -\cos \theta_2 = 0.95 \qquad \Rightarrow \qquad \theta_2 = 18.19^\circ -$$ - -The real power *P* has not changed. But the apparent power has changed; its new value is - -$$ -S_2 = \frac{P}{\cos \theta_2} = \frac{4000}{0.95} = 4210.5 \text{ VA} -$$ - -The new reactive power is - -$$ -Q_2 = S_2 \sin \theta_2 = 1314.4 \text{ VAR} -$$ - -The difference between the new and old reactive powers is due to the parallel addition of the capacitor to the load. The reactive power due to the capacitor is - -$$ -Q_C = Q_1 - Q_2 = 3000 - 1314.4 = 1685.6 \text{ VAR} -$$ - -and - -$$ -C = \frac{Q_C}{\omega V_{\text{rms}}^2} = \frac{1685.6}{2\pi \times 60 \times 120^2} = 310.5 \,\mu\text{F} -$$ - -*Note:* Capacitors are normally purchased for voltages they expect to see. In this case, the maximum voltage this capacitor will see is about 170 V peak. We would suggest purchasing a capacitor with a voltage rating equal to, say, 200 V. - -Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume that the load is sup plied by a 220-V (rms), 60-Hz line. - -**Answer:** 7.673 mF. - -# **11.9** Applications - -In this section, we consider two important application areas: how power is measured and how electric utility companies determine the cost of electricity consumption. - -# **11.9.1** Power Measurement - -The average power absorbed by a load is measured by an instrument called the *wattmeter*. - -The wattmeter is the instrument used for measuring the average power. - -Figure 11.30 sho ws a w attmeter that consists essentially of tw o coils: the current coil and the voltage coil. A current coil with very low impedance (ideally zero) is connected in series with the load (Fig. 11.31) and responds to the load current. The voltage coil with very high impedance (ideally infinite) is connected in parallel with the load as shown in Fig. 11.31 and responds to the load v oltage. The current coil acts lik e a short circuit because of its low impedance; the voltage coil behaves like - -Reactive power is measured by an instrument called the varmeter. The varmeter is often connected to the load in the same way as the wattmeter. - - Some wattmeters do not have coils; the wattmeter considered here is the electromagnetic type. - -Practice Problem 11.15 - -A wattmeter. - -**Figure 11.31** The wattmeter connected to the load. - -an open circuit because of its high impedance. As a result, the presence of the w attmeter does not disturb the circuit or ha ve an ef fect on the power measurement. - -When the two coils are energized, the mechanical inertia of the moving system produces a deflection angle that is proportional to the average value of the product *v*(*t*)*i*(*t*). If the current and voltage of the load are *v*(*t*) = *Vm* cos(*ωt* + *θv*) and *i*(*t*) = *Im* cos(*ωt* + *θi*), their corresponding rms phasors are - -$$ -\mathbf{V}_{\rm rms} = \frac{V_m}{\sqrt{2}} \underline{\theta_v} \quad \text{and} \quad \mathbf{I}_{\rm rms} = \frac{I_m}{\sqrt{2}} \underline{\theta_i} \quad (11.62) -$$ - -and the wattmeter measures the average power given by - -$$ -P = |\mathbf{V}_{\text{rms}}||\mathbf{I}_{\text{rms}}| \cos(\theta_{\nu} - \theta_{i}) = V_{\text{rms}} I_{\text{rms}} \cos(\theta_{\nu} - \theta_{i}) \qquad (11.63) -$$ - -As shown in Fig. 11.31, each wattmeter coil has two terminals with one marked ±. To ensure upscale deflection, the ± terminal of the current coil is toward the source, while the ± terminal of the voltage coil is connected to the same line as the current coil. Re versing both coil connections still results in upscale deflection. However, reversing one coil and not the other results in downscale deflection and no wattmeter reading. - -# **Solution:** - -1. **Define.** The problem is clearly defined. Interestingly, this is a problem where the student could actually v alidate the results by doing the problem in the laboratory with a real wattmeter. - -- 2. **Present.** This problem consists of finding the average power delivered to a load by an external source with a series impedance. -- 3. **Alternative.** This is a straightforward circuit problem where all we need to do is find the magnitude and phase of the current through the load and the magnitude and the phase of the voltage across the load. These quantities could also be found by using *PSpice*, which we will use as a check. -- 4. **Attempt.** In Fig. 11.32, the w attmeter reads the average power absorbed by the (8 − *j*6) Ω impedance because the current coil is in series with the impedance while the v oltage coil is in parallel with it. The current through the circuit is - -The impedance while the voltage coil is in -int through the circuit is -$$ -I_{\rm rms} = \frac{150/0^{\circ}}{(12 + j10) + (8 - j6)} = \frac{150}{20 + j4} A -$$ - -The voltage across the (8 − *j*6) Ω impedance is - -$$ -\mathbf{V}_{\rm rms} = \mathbf{I}_{\rm rms}(8 - j6) = \frac{150(8 - j6)}{20 + j4} \text{ V} -$$ - -The complex power is - -$$ -\mathbf{S} = \mathbf{V}_{\text{rms}} \mathbf{I}_{\text{rms}}^* = \frac{150(8 - j6)}{20 + j4} \cdot \frac{150}{20 - j4} = \frac{150^2(8 - j6)}{20^2 + 4^2} -$$ -$$ -= 423.7 - j324.6 \text{ VA} -$$ - -The wattmeter reads - -$$ -P = \text{Re}(S) = 432.7 \text{ W} -$$ - -5. **Evaluate.** We can check our results by using *PSpice*. - -To check our answer , all we need is the magnitude of the current (7.354 A) flowing through the load resistor: - -$$ -P = (I_L)^2 R = (7.354)^2 8 = 432.7 \text{ W} -$$ - -As expected, the answer does check! - -6. **Satisfactory?** We ha ve satisf actorily solv ed the problem and the results can now be presented as a solution to the problem. - -For Practice Prob. 11.16. - -**Answer:** 1.437 kW. - -# **11.9.2** Electricity Consumption Cost - -In Section 1.7, we considered a simplified model of the way the cost of electricity consumption is determined. But the concept of po wer factor was not included in the calculations. Now we consider the importance of power factor in electricity consumption cost. - -Loads with low power factors are costly to serve because they require large currents, as explained in Section 11.8. The ideal situation would be to draw minimum current from a supply so that *S* = *P*, *Q* = 0, and pf = 1. A load with nonzero *Q* means that energy flows back and forth between the load and the source, gi ving rise to additional po wer losses. In vie w of this, power companies often encourage their customers to have power factors as close to unity as possible and penalize some customers who do not improve their load power factors. - -Utility companies divide their customers into categories: as residential (domestic), commercial, and industrial, or as small po wer, medium power, and large power. They have different rate structures for each category. The amount of energy consumed in units of kilowatt-hours (kWh) is measured using a kilo watt-hour meter installed at the customer' s premises. - -Although utility companies use different methods for charging customers, the tarif f or char ge to a consumer is often tw o-part. The first part is fixed and corresponds to the cost of generation, transmission, and distribution of electricity to meet the load requirements of the con sumers. This part of the tarif f is generally e xpressed as a certain price - -per kW of maximum demand. Or it may be based on kVA of maximum demand, to account for the power factor (pf) of the consumer. A pf penalty charge may be imposed on the consumer whereby a certain percentage of kW or kVA maximum demand is charged for every 0.01 fall in pf below a prescribed value, say 0.85 or 0.9. On the other hand, a pf credit may be given for every 0.01 that the pf exceeds the prescribed value. - -The second part is proportional to the ener gy consumed in kWh; i t may be in graded form, for example, the first 100 kWh at 16 cents/kWh, the next 200 kWh at 10 cents/kWh and so forth. Thus, the bill is determined based on the following equation: - -Total Cost = Fixed Cost + Cost of Energy **(11.64)** - -A manufacturing industry consumes 200 MWh in one month. If the Example 11.17 maximum demand is 1,600 kW, calculate the electricity bill based on the following two-part rate: - -Demand charge: \$5.00 per month per kW of billing demand. Energy charge: 8 cents per kWh for the first 50,000 kWh, 5 cents per kWh for the remaining energy. - -# **Solution:** - -The demand charge is - -\$5.00 × 1,600 = \$8,000 **(11.17.1)** - -The energy charge for the first 50,000 kWh is - -$$ -$0.08 \times 50,000 = $4,000 \tag{11.17.2} -$$ - -The remaining energy is 200,000 kWh− 50,000 kWh = 150,000 kWh, and the corresponding energy charge is - -\$0.05 × 150,000 = \$7,500 **(11.17.3)** - -Adding the results of Eqs. (11.17.1) to (11.17.3) gives - -Total bill for the month = \$8,000 + \$4,000 + \$7,500 = \$19,500 - -It may appear that the cost of electricity is too high. But this is often a small fraction of the overall cost of production of the goods manufactured or the selling price of the finished product. - -The monthly reading of a paper mill's meter is as follows: - -Maximum demand: 48,000 kW Energy consumed: 750 MWh - -Using the two-part rate in Example 11.17, calculate the monthly bill for the paper mill. - -**Answer:** \$279,000. - -Practice Problem 11.17 - -Example 11.18 A 300-kW load supplied at 13 kV (rms) operates 520 hours a month at 80 percent power factor. Calculate the average cost per month based on this simplified tariff: - -Energy charge: 6 cents per kWh - -Power-factor penalty: 0.1 percent of energy charge for every 0.01 that pf falls below 0.85. - -Power-factor credit: 0.1 percent of energy charge for every 0.01 that pf exceeds 0.85. - -# **Solution:** - -The energy consumed is - -$$ -W = 300 \text{ kW} \times 520 \text{ h} = 156,000 \text{ kWh} -$$ - -The operating power factor pf = 80% = 0.8 is 5 × 0.01 below the prescribed power factor of 0.85. Since there is 0.1 percent energy charge for every 0.01, there is a power-factor penalty charge of 0.5 percent. This amounts to an energy charge of - -$$ -\Delta W = 156,000 \times \frac{5 \times 0.1}{100} = 780 \text{ kWh} -$$ - -The total energy is - -$$ -W_t = W + \Delta W = 156,000 + 780 = 156,780 \text{ kWh} -$$ - -The cost per month is given by - -Cost = 6 cents × *Wt* = \$0.06 × 156,780 = \$9,406.80 - -An 500-kW induction furnace at 0.88 power factor operates 20 hours per day for 26 days in a month. Determine the electricity bill per month based on the tariff in Example 11.18. Practice Problem 11.18 - -**Answer:** \$15,553.20. - -# **11.10** Summary - -1. The instantaneous power absorbed by an element is the product of the element's terminal voltage and the current through the element: - -$$ -p = vi. -$$ - -2. Average or real po wer *P* (in w atts) is the a verage of instantaneous power *p*: - -$$ -P = \frac{1}{T} \int_0^T p \, dt -$$ - - If *v*(*t*) = *Vm*cos(*ωt* + *θv*) and *i*(*t*) = *Im* cos (*ωt* + *θi* ), then *V*rms = *Vm*∕√ 2 , *I*rms = *Im*∕ √ \_\_ 2 , and - -$$ -P = \frac{1}{2} V_m I_m \cos(\theta_v - \theta_i) = V_{\text{rms}} I_{\text{rms}} \cos(\theta_v - \theta_i) -$$ - - Inductors and capacitors absorb no a verage power, while the a verage power absorbed by a resistor is (1∕2)*Im* 2*R* = *I*rms 2 *R*. - -- 3. Maximum average power is transferred to a load when the load impedance is the comple x conjug ate of the Thevenin impedance as seen from the load terminals, **Z***L* = *Z*Th \* . -- 4. The effective value of a periodic signal x(*t*) is its root-mean-square (rms) value. - -$$ -X_{\rm eff} = X_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T x^2 dt} -$$ - - For a sinusoid, the ef fective or rms v alue is its amplitude di vided by √ \_\_ 2 . - -5. The power factor is the cosine of the phase difference between voltage and current: - -$$ -\mathrm{pf} = \cos(\theta_v - \theta_i) -$$ - - It is also the cosine of the angle of the load impedance or the ratio of real power to apparent power. The pf is lagging if the current lags voltage (inductive load) and is leading when the current leads v oltage (capacitive load). - -6. Apparent power *S* (in VA) is the product of the rms values of voltage and current: - -$$ -S = V_{\rm rms} I_{\rm rms} -$$ - - It is also given by *S* = ∣**S**∣ = √ \_\_\_\_\_\_\_ *P*2 + *Q*2 , where *P* is the real power and *Q* is reactive power. - -7. Reactive power (in VAR) is: - -$$ -Q = \frac{1}{2} V_m I_m \sin(\theta_v - \theta_i) = V_{\text{rms}} I_{\text{rms}} \sin(\theta_v - \theta_i) -$$ - -8. Complex power **S** (in VA) is the product of the rms v oltage phasor and the complex conjugate of the rms current phasor . It is also the complex sum of real power *P* and reactive power *Q*. - -$$ -\mathbf{S} = \mathbf{V}_{\rm rms} \, \mathbf{I}_{\rm rms}^* = V_{\rm rms} I_{\rm rms} / \theta_{\rm v} - \theta_{\rm i} = P + jQ -$$ - -Also, - -$$ -\mathbf{S} = I_{\text{rms}}^2 \, \mathbf{Z} = \frac{V_{\text{rms}}^2}{\mathbf{Z}^*} -$$ - -- 9. The total comple x power in a netw ork is the sum of the comple x powers of the individual components. Total real power and reactive power are also, respectively, the sums of the individual real powers and the reactive powers, but the total apparent po wer is not calcu lated by the process. -- 10. Power factor correction is necessary for economic reasons; it is the process of impro ving the po wer f actor of a load by reducing the overall reactive power. -- 11. The wattmeter is the instrument for measuring the average power. Energy consumed is measured with a kilowatt-hour meter. - -# Review Questions - -**11.1** The average power absorbed by an inductor is zero. - -(a) True (b) False - -**11.2** The Thevenin impedance of a network seen from the load terminals is 80 + *j*55 Ω. For maximum power transfer, the load impedance must be: - -| (a) −80 + j55 Ω | (b) −80 − j55 Ω | | | -|-----------------|-----------------|--|--| -| (c) 80 − j55 Ω | (d) 80 + j55 Ω | | | - -**11.3** The amplitude of the voltage available in the 60-Hz, 120-V power outlet in your home is: - -| (a) 110 V | (b) 120 V | | -|-----------|-----------|--| -| (c) 170 V | (d) 210 V | | - -- **11.4** If the load impedance is 20 − *j*20, the power factor is - - (a) ⧸−45*°* (b) 0 (c) 1 - - (d) 0.7071 (e) none of these -- **11.5** A quantity that contains all the power information in a given load is the - - (a) power factor (b) apparent power (c) average power (d) reactive power - - (e) complex power -- **11.6** Reactive power is measured in: - - (a) watts (b) VA - - (c) VAR (d) none of these -- **11.7** In the power triangle shown in Fig. 11.34(a), the reactive power is: - - (a) 1000 VAR leading (b) 1000 VAR lagging (c) 866 VAR leading (d) 866 VAR lagging - -# **Figure 11.34** - -For Review Questions 11.7 and 11.8. - -**11.8** For the power triangle in Fig. 11.34(b), the apparent power is: (a) 2000 VA (b) 1000 VAR - -| (c) 866 VAR | (d) 500 VAR | -|-------------|-------------| - -**11.9** A source is connected to three loads **Z**1, **Z**2, and **Z**3 in parallel. Which of these is not true? - -| (a) P = P1 + P2 + P3 | (b) Q = Q1 + Q2 + Q3 | -|----------------------|----------------------| -| (c) S = S1 + S2 + S3 | (d) S = S1 + S2 + S3 | - -**11.10** The instrument for measuring average power is the: - -| (a) voltmeter | (b) ammeter | -|-------------------------|--------------| -| (c) wattmeter | (d) varmeter | -| (e) kilowatt-hour meter | | - -*Answers: 11.1a, 11.2c, 11.3c, 11.4d, 11.5e, 11.6c, 11.7d, 11.8a, 11.9c, 11.10c.* - -# Problems1 - -# Section 11.2 Instantaneous and Average Power - -- **11.1** If *v*(*t*) = 160 cos 50*t* V and *i*(*t*) = −33 sin (50*t* − 30°)A, calculate the instantaneous power and the average power. -- **11.2** Given the circuit in Fig. 11.35, find the average power supplied or absorbed by each element. - -- **11.3** A load consists of a 60-Ω resistor in parallel with a 90-*μ*F capacitor. If the load is connected to a voltage source *vs*(*t*) = 160 cos 2000*t*, find the average power delivered to the load. -- **11.4** Using Fig. 11.36, design a problem to help other students better understand instantaneous and average power. - -1Starting with problem 11.22, unless otherwise specified, assume that all values of currents and voltages are rms. - -## Problems **489** - -**11.5** ssuming that *vs* = 8 cos(2*t* − 40°) V in the circuit of Fig. 11.37, find the average power delivered to each of the passive elements. - -**Figure 11.37** For Prob. 11.5. - -**11.6** For the circuit in Fig. 11.38, *is* = 6 cos 103 *t* A. Find the average power absorbed by the 50-Ω resistor. - -**Figure 11.38** For Prob. 11.6. - -**11.7** Given the circuit of Fig. 11.39, find the average power absorbed by the 10-Ω resistor. - -For Prob. 11.7. - -**11.8** In the circuit of Fig. 11.40, determine the average power absorbed by the 40-Ω resistor. - -**11.9** For the op amp circuit in Fig. 11.41, **V***s* = 2⧸30*° V*. Find the average power absorbed by the 20-kΩ resistor. - -- For Prob. 11.9. -- **11.10** In the op amp circuit in Fig. 11.42, find the total average power absorbed by the resistors. - -**Figure 11.42** For Prob. 11.10. - -**11.11** For the network in Fig. 11.43, assume that the port impedance is - -$$ -\mathbf{Z}_{ab} = \frac{R}{\sqrt{1 + \omega^2 R^2 C^2}} \sqrt{-\tan^{-1} \omega RC} -$$ - - Find the average power consumed by the network when *R* = 10 kΩ, *C* = 200 nF, and *i* = 33 sin(377*t* + 22°) mA. - -# **Figure 11.43** For Prob. 11.11. - -# Section 11.3 Maximum Average Power Transfer - -**11.12** For the circuit shown in Fig. 11.44, determine the load impedance *ZL* for maximum power transfer (to *ZL*). Calculate the maximum power absorbed by the load. - -# **Figure 11.44** - -For Prob. 11.12. - -- **11.13** The Thevenin impedance of a source is **Z**Th = 120 + *j*60 Ω, while the peak Thevenin voltage is **V**Th = 165 + *j*0 V. Determine the maximum available average power from the source. -- **11.14** Using Fig. 11.45, design a problem to help other students better understand maximum average power transfer to a load *Z*. - -**Figure 11.45** For Prob. 11.14. - -**11.15** In the circuit of Fig. 11.46, find the value of **Z***L* that will absorb the maximum power and the value of the maximum power. - -For Prob. 11.15. - -**11.16** For the circuit in Fig. 11.47, find the value of **Z***L* that will receive the maximum power from the circuit. Then calculate the power delivered to the load **Z***L*. - -**11.17** Calculate the value of **Z***L* in the circuit of Fig. 11.48 in order for **Z***L* to receive maximum average power. ‒j3 Ω What is the maximum average power received by **Z***L*? 4Ω - -**Figure 11.48** - -For Prob. 11.17. - -**11.18** Find the value of **Z***L* in the circuit of Fig. 11.49 for maximum power transfer. - -**Figure 11.49** - -For Prob. 11.18. - -**11.19** The variable resistor *R* in the circuit of Fig. 11.50 is adjusted until it absorbs the maximum average power. Find *R* and the maximum average power absorbed. - -**Figure 11.50** For Prob. 11.19. - -**11.20** The load resistance *RL* in Fig. 11.51 is adjusted until it absorbs the maximum average power. Calculate the value of *RL* and the maximum average power. - -For Prob. 11.20. - -**11.21** Assuming that the load impedance is to be purely resistive, what load should be connected to terminals *a*-*b* of the circuits in Fig. 11.52 so that the maximum power is transferred to the load? - -**Figure 11.52** For Prob. 11.21. - -# Section 11.4 Effective or RMS Value - -**11.22** Find the rms value of the offset sine wave shown in Fig. 11.53. - -For Prob. 11.22. - -**11.23** Using Fig. 11.54, design a problem to help other students better understand how to find the rms value of a waveshape. - -- For Prob. 11.23. -- **11.24** Determine the rms value of the waveform in Fig. 11.55. - -**Figure 11.55** For Prob. 11.24. - -**Figure 11.56** For Prob. 11.25. - -**11.26** Find the effective value of the voltage waveform in Fig. 11.57. - -For Prob. 11.26. - -**11.27** Calculate the rms value of the current waveform of Fig. 11.58. - -**11.28** Find the rms value of the voltage waveform of Fig. 11.59 as well as the average power absorbed by a 2-Ω resistor when the voltage is applied across the resistor. - -**11.29** Calculate the effective value of the current waveform in Fig. 11.60 and the average power delivered to a 12-Ω resistor when the current runs through the resistor. - -**Figure 11.60** For Prob. 11.29. - -**11.30** Compute the rms value of the waveform depicted in Fig. 11.61. - -**Figure 11.61** - -For Prob. 11.30. - -**11.31** Find the rms value of the signal shown in Fig. 11.62. - -**Figure 11.62** For Prob. 11.31. - -**11.32** Obtain the rms value of the current waveform shown in Fig. 11.63. - -**Figure 11.63** For Prob. 11.32. - -**11.33** Determine the rms value for the waveform in Fig. 11.64. - -**11.34** Find the effective value of *f*(*t*) defined in Fig. 11.65. - -**Figure 11.65** For Prob. 11.34. - -**11.35** One cycle of a periodic voltage waveform is depicted in Fig. 11.66. Find the effective value of the voltage. Note that the cycle starts at *t* = 0 and ends at *t* = 6 s. - -**Figure 11.66** For Prob. 11.35. - -**11.36** Calculate the rms value for each of the following functions: - -(a) *i*(*t*) = 10 A (b) *v*(*t*) = 4 + 3 cos 5*t* V (c) *i*(*t*) = 8 − 6 sin 2*t* A (d) *v*(*t*) = 5 sin *t* + 4 cos*t* V - -**11.37** Design a problem to help other students better understand how to determine the rms value of the sum of multiple currents. - -# Section 11.5 Apparent Power and Power Factor - -**11.38** For the power system in Fig. 11.67, find: (a) the average power, (b) the reactive power, (c) the power factor. Note that 440 V is an rms value. - -**11.39** An ac motor with impedance **Z***L* = 2 + *j*1.2 Ω is supplied by a 220-V, 60-Hz source. (a) Find pf, *P*, and *Q*. (b) Determine the capacitor required to be connected in parallel with the motor so that the power factor is corrected to unity. - -**11.40** Design a problem to help other students better understand apparent power and power factor. - -**11.41** Obtain the power factor for each of the circuits in Fig. 11.68. Specify each power factor as leading or lagging. - -# **Figure 11.68** - -For Prob. 11.41. - -# Section 11.6 Complex Power - -- **11.42** A 110-V rms, 60-Hz source is applied to a load impedance **Z**. The apparent power entering the load is 120 VA at a power factor of 0.707 lagging. - - (a) Calculate the complex power. - - (b) Find the rms current supplied to the load. - - (c) Determine **Z**. (d) Assuming that **Z** = *R* + *jωL*, find the values of *R* and *L*. - -**11.43** Design a problem to help other students understand complex power. - -**11.44** Find the complex power delivered by *vs* to the network in Fig. 11.69. Let *vs* = 100 cos 2000*t* V. - -# **Figure 11.69** For Prob. 11.44. - -**11.45** The voltage across a load and the current through it are given by - -$$ -v(t) = 20 + 60 \cos 100t -$$ - -$$ -i(t) = 1 - 0.5 \sin 100t -$$ - A - -Find: - -(a) the rms values of the voltage and of the current (b) the average power dissipated in the load - -**11.46** For the following voltage and current phasors, calculate the complex power, apparent power, real power, and reactive power. Specify whether the pf is leading or lagging. - -(a) -$$ -V = 220/30^{\circ} -$$ - V rms, $I = 0.5/60^{\circ}$ A rms - -(b) -$$ -V = 250 \div 10^{\circ} \text{ V rms} -$$ -, - -$$ -I = 6.2 \angle -25^{\circ} -$$ - A rms - -(c) -$$ -V = 120/0 -$$ -° V rms, $I = 2.4/15$ ° A rms - -- (d) **V** = 160⧸45*°* V rms, **I** = 8.5⧸90*°* A rms -- **11.47** For each of the following cases, find the complex power, the average power, and the reactive power: - -(a) -$$ -v(t) = 169.7 \sin(377t + 45^\circ) -$$ - V, - $i(t) = 5.657 \sin(377t)$ A - -(b) -$$ -v(t) = 339.4 \sin (377t + 90^\circ) -$$ - V, - -$$ -i(t) = 5.657 \sin (377t + 45^{\circ}) \text{ A} -$$ - -(c) V = -$$ -900/90^{\circ} -$$ - V rms, Z = $75/45^{\circ}$ $\Omega$ - -(d) -$$ -I = 100/60^{\circ} -$$ - A rms, $Z = 50/60^{\circ}$ $\Omega$ - -- **11.48** Determine the complex power for the following cases: - - (a) *P* = 269 W, *Q* = 150 VAR (capacitive) - - (b) *Q* = 2000 VAR, pf = 0.9 (leading) - -(c) -$$ -S = 600 -$$ - VA, $Q = 450$ VAR (inductive) - -(d) *V*rms = 220 V, *P* = 1 kW, - -∣**Z**∣ = 40 Ω (inductive) - -**11.49** Find the complex power for the following cases: - -- (a) *P* = 4 kW, pf = 0.86 (lagging) (b) *S* = 2 kVA, *P* = 1.6 kW (capacitive) (c) **V**rms = 208⧸20*°* V, **I**rms = 6.5⧸−50*°* A (d) **V**rms = 120⧸30*°* V, **Z** = 40 + *j*60 Ω -- **11.50** Obtain the overall impedance for the following cases: - - (a) *P* = 1000 W, pf = 0.8 (leading), *V*rms = 220 V - - (b) *P* = 1500 W, *Q* = 2000 VAR (inductive), *I*rms = 12 A - -(c) -$$ -S = 4500/60^{\circ} -$$ - VA, $V = 120/45^{\circ}$ V - -- **11.51** For the entire circuit in Fig. 11.70, calculate: - - (a) the power factor - - (b) the average power delivered by the source - - (c) the reactive power - - (d) the apparent power - - (e) the complex power - -# **Figure 11.70** - -For Prob. 11.51. - -- **11.52** In the circuit of Fig. 11.71, device *A* receives 2 kW at 0.8 pf lagging, device *B* receives 3 kVA at 0.4 pf leading, while device *C* is inductive and consumes 1 kW and receives 500 VAR. - - (a) Determine the power factor of the entire system. - - (b) Find **I** given that **V***s* = 120⧸45*°* V rms. - -**Figure 11.71** For Prob. 11.52. - -- **11.53** In the circuit of Fig. 11.72, load *A* receives 4 kVA at 0.8 pf leading. Load *B* receives 2.4 kVA at 0.6 pf lagging. Box *C* is an inductive load that consumes 1 kW and receives 500 VAR. - - (a) Determine **I**. - - (b) Calculate the power factor of the combination. - -**Figure 11.72** For Prob. 11.53. - -# Section 11.7 Conservation of AC Power - -**11.54** For the network in Fig. 11.73, find the complex power absorbed by each element. - -# **Figure 11.73** - -For Prob. 11.54. - -**Figure 11.74** - -For Prob. 11.55. - -**11.56** Obtain the complex power delivered by the source in the circuit of Fig. 11.75. - -**Figure 11.75** - -For Prob. 11.56. - -For Prob. 11.57. - -**11.58** Obtain the complex power delivered to the 10-kΩ resistor in Fig. 11.77 below. - -- **11.59** Calculate the reactive power in the inductor and capacitor in the circuit of Fig. 11.78. -- **Figure 11.78** 100 Ω 100 Ω j100 Ω 100 0° mA ‒j200 Ω 20 0°V + ‒ - -For Prob. 11.59. - -**11.60** For the circuit in Fig. 11.79, find **V***o* and the input power factor. - -**Figure 11.79** For Prob. 11.60. - -**11.61** Given the circuit in Fig. 11.80, find *Io* and the overall - -**Figure 11.80** For Prob. 11.61. - -**11.62** For the circuit in Fig. 11.81, find **V***s*. - -complex power supplied. - -**11.63** Find **I***o* in the circuit of Fig. 11.82. - -**11.64** Determine **I***s* in the circuit of Fig. 11.83, if the voltage source supplies 6 kW and 1.2 kVAR (leading). - -# **Figure 11.83** - -For Prob. 11.64. - -**11.65** In the op amp circuit of Fig. 11.84, *vs* = 4 cos 104 *t* V. Find the average power delivered to the 50-kΩ resistor. - -For Prob. 11.65. - -**11.66** Obtain the average power absorbed by the 10-Ω resistor in the op amp circuit in Fig. 11.85. - -**Figure 11.85** - -For Prob. 11.66. - -**11.67** For the op amp circuit in Fig. 11.86, calculate: - -- (a) the complex power delivered by the voltage source -- (b) the average power dissipated in the 10-Ω resistor - -**11.68** Compute the complex power supplied by the current source in the series *RLC* circuit in Fig. 11.87. - -# **Figure 11.87** - -For Prob. 11.68. - -# Section 11.8 Power Factor Correction - -**11.69** Refer to the circuit shown in Fig. 11.88. - -- (a) What is the power factor? -- (b) What is the average power dissipated? -- (c) What is the value of the capacitance that will give a unity power factor when connected to the load? - -# **Figure 11.88** - -For Prob. 11.69. - -- **11.70** Design a problem to help other students better understand power factor correction. - - **11.71** Three loads are connected in parallel to a 120⧸0*°* V rms source. Load 1 absorbs 60 kVAR at pf = 0.85 lagging, load 2 absorbs 90 kW and 50 kVAR leading, and load 3 absorbs 100 kW at pf = 1. (a) Find the equivalent impedance. (b) Calculate the power factor of the parallel combination. (c) Determine the current supplied by the source. - - **11.72** Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW at a 0.707 pf lagging. Determine: (a) the pf of the second load, (b) the parallel element required to correct the pf to 0.9 lagging for the two loads. - - **11.73** A 240-V rms 60-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find: - - (a) the apparent power - - (b) the current drawn from the supply - - (c) the kVAR rating and capacitance required to improve the power factor to 0.96 lagging - - (d) the current drawn from the supply under the new power-factor conditions - -For Prob. 11.67. - -- **11.74** A 120-V rms 60-Hz source supplies two loads connected in parallel, as shown in Fig. 11.89. - - (a) Find the power factor of the parallel combination. - - (b) Calculate the value of the capacitance connected in parallel that will raise the power factor to unity. - -**Figure 11.89** For Prob. 11.74. - -- **11.75** Consider the power system shown in Fig. 11.90. Calculate: - - (a) the total complex power - - (b) the power factor - - (c) the parallel capacitance necessary to establish a unity power factor - -For Prob. 11.77. - -**11.78** Find the wattmeter reading of the circuit shown in Fig. 11.93. - -**11.79** Determine the wattmeter reading of the circuit in - -# **Figure 11.94** - -For Prob. 11.79. - -For Prob. 11.80. - -- **11.80** The circuit of Fig. 11.95 portrays a wattmeter connected into an ac network. - - (a) Find the magnitude of the load current. - - (b) Calculate the wattmeter reading. - -For Prob. 11.78. - -Fig. 11.94. - -# Section 11.9 Applications - -**11.76** Obtain the wattmeter reading of the circuit in Fig. 11.91. - -For Prob. 11.76. - -- **11.81** Design a problem to help other students better understand how to correct power factor to values other than unity. -- **11.82** A 240-V rms 60-Hz source supplies a parallel combination of a 5-kW heater and a 30-kVA induction motor whose power factor is 0.82. Determine: - - (a) the system apparent power - - (b) the system reactive power - - (c) the kVA rating of a capacitor required to adjust the system power factor to 0.9 lagging - - (d) the value of the capacitor required -- **11.83** Oscilloscope measurements indicate that the peak voltage across a load and the peak current through it are, respectively, 210⧸ 60*°* V and 8⧸25*°* A. Determine: - - (a) the real power - - (b) the apparent power - - (c) the reactive power - - (d) the power factor - -**11.84** A consumer has an annual consumption of 1200 MWh with a maximum demand of 2.4 MVA. The maximum demand charge is \$30 per kVA per annum, and the energy charge per kWh is 4 cents. - -(a) Determine the annual cost of energy. - -# Comprehensive Problems - -- **11.86** A transmitter delivers maximum power to an antenna when the antenna is adjusted to represent a load of 75-Ω resistance in series with an inductance of 4 *μ*H. If the transmitter operates at 4.12 MHz, find its internal impedance. -- **11.87** In a TV transmitter, a series circuit has an impedance of 3 kΩ and a total current of 50 mA. If the voltage across the resistor is 80 V, what is the power factor of the circuit? -- **11.88** A certain electronic circuit is connected to a 110-V ac line. The root-mean-square value of the current drawn is 2 A, with a phase angle of 55°. - - (a) Find the true power drawn by the circuit. - - (b) Calculate the apparent power. - -**11.89** An industrial heater has a nameplate that reads: - -- 210 V 60 Hz 12 kVA 0.78 pf lagging Determine: - - (a) the apparent and the complex power - - (b) the impedance of the heater -- **11.90** A 2000-kW turbine-generator of 0.85 power factor operates at the rated load. An additional load of 300 kW at 0.8 power factor is added.What kVAR \* - -- (b) Calculate the charge per kWh with a flat-rate tariff if the revenue to the utility company is to remain the same as for the two-part tariff. -- **11.85** A regular household system of a single-phase threewire circuit allows the operation of both 120-V and 240-V, 60-Hz appliances. The household circuit is modeled as shown in Fig. 11.96. Calculate: - - (a) the currents **I**1, **I**2, and I*n* - - (b) the total complex power supplied - - (c) the overall power factor of the circuit - -# **Figure 11.96** - -For Prob. 11.85. - -of capacitors is required to operate the turbinegenerator but keep it from being overloaded? - -**11.91** The nameplate of an electric motor has the following information: - -> Line voltage: 220 V rms Line current: 15 A rms Line frequency: 60 Hz Power: 2700 W - - Determine the power factor (lagging) of the motor. Find the value of the capacitance *C* that must be connected across the motor to raise the pf to unity. - -- **11.92** As shown in Fig. 11.97, a 550-V feeder line supplies an industrial plant consisting of a motor drawing 90 kW at 0.8 pf (inductive), a capacitor with a rating of 20 kVAR, and lighting drawing 10 kW. - - (a) Calculate the total reactive power and apparent power absorbed by the plant. - - (b) Determine the overall pf. - - (c) Find the magnitude of the current in the feeder line. - -\* An asterisk indicates a challenging problem. - -- **11.93** A factory has the following four major loads: - - A motor rated at 5 hp, 0.8 pf lagging (1hp = 0.7457 kW). - - A heater rated at 1.2 kW, 1.0 pf. - - Ten 120-W lightbulbs. - - A synchronous motor rated at 1.6 kVAR, 0.6 pf leading. - - (a) Calculate the total real and reactive power. - - (b) Find the overall power factor. - -- (a) Calculate the cost of capacitors needed. -- (b) Find the savings in substation capacity released. -- (c) Are capacitors economical for releasing the amount of substation capacity? - -- (a) At what frequency is maximum power transferred to the speaker? -- (b) If *Vs* = 4.6 V rms, how much power is delivered to the speaker at that frequency? - -# **Figure 11.98** - -For Prob. 11.95. - -- **1.96** A power amplifier has an output impedance of 40 + *j*8 Ω. It produces a no-load output voltage of 146 V at 300 Hz. - - (a) Determine the impedance of the load that achieves maximum power transfer. - - (b) Calculate the load power under this matching condition. -- **1.97** A power transmission system is modeled as shown in Fig. 11.99. If **V***s* = 440⧸0*°* rms, find the average power absorbed by the load. - -**Figure 11.99** For Prob. 11.97. - -*This page intentionally left blank* - -# **chapter** - -# 12 - -# Three-Phase Circuits - -*He who cannot forgive others breaks the bridge over which he must pass himself.* - -—G. Herbert - -# Enhancing Your Skills and Your Career - -# **ABET EC 2000 criteria (3.e), "an ability to identify, formulate, and solve engineering problems."** - -Developing and enhancing your "ability to identify , formulate, and solve engineering problems" is a primary focus of te xtbook. Following our six-step problem-solving process is the best w ay to practice this skill. Our recommendation is that you use this process whene ver possible. You may be pleased to learn that this process w orks well for nonengineering courses. - -# **ABET EC 2000 criteria (f), "an understanding of professional and ethical responsibility."** - -"An understanding of professional and ethical responsibility" is required of every engineer. To some e xtent, this understanding is v ery personal for each of us. Let us identify some pointers to help you de velop this understanding. One of my f avorite examples is that an engineer has the responsibility to answer what I call the "unasked question." For instance, assume that you own a car that has a problem with the transmission. In the process of selling that car, the prospective buyer asks you if there is a problem in the right-front wheel bearing. You answer no. However, as an engineer, you are required to inform the buyer that there is a problem with the transmission without being asked. - -Your responsibility both professionally and ethically is to perform in a manner that does not harm those around you and to whom you are responsible. Clearly, developing this capability will tak e time and ma turity on your part. I recommend practicing this by looking for profes sional and ethical components in your day-to-day activities. - -Photo by Charles Alexander - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Understand balanced three-phase voltages. -- 2. Analyze balanced wye-wye circuits. -- 3. Understand and analyze balanced wye-delta circuits. -- 4. Analyze balanced delta-delta circuits. -- 5. Understand and analyze balanced delta-wye circuits. -- 6. Explain and analyze power in balanced three-phase circuits. -- 7. Analyze unbalanced three-phase circuits. - -# **12.1** Introduction - -So far in this text, we have dealt with single-phase circuits. A singlephase ac power system consists of a generator connected through a pair of wires (a transmission line) to a load. Figure 12.1(a) depicts a singlephase two-wire system, where *Vp* is the rms magnitude of the source voltage and *ϕ* is the phase. What is more common in practice is a singlephase three-wire system, shown in Fig. 12.1(b). It contains two identical sources (equal magnitude and the same phase) that are connected to two loads by two outer wires and the neutral. For example, the normal household system is a single-phase three-wire system because the terminal voltages have the same magnitude and the same phase. Such a system allows the connection of both 120- and 240-V appliances. - -Historical note: Thomas Edison invented a three-wire system, using three wires instead of four. - -Circuits or systems in which the ac sources operate at the same fre quency but different phases are known as *polyphase*. Figure 12.2 shows a two-phase three-wire system, and Fig. 12.3 sho ws a three-phase fourwire system. As distinct from a single-phase system, a two-phase system is produced by a generator consisting of tw o coils placed perpendicular to each other so that the voltage generated by one lags the other by 90°. By the same token, a three-phase system is produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120°. Because the three-phase system is by far the most pre valent and most economical polyphase system, discus sion in this chapter is mainly on three-phase systems. - -**Figure 12.2** Two-phase three-wire system. - -Three-phase systems are important for at least three reasons. First, nearly all electric po wer is generated and distrib uted in three-phase, - -# Historical - -**Nikola Tesla** (1856–1943) was a Croatian-American engineer whose inventions—among them the induction motor and the first polyphase ac power system—greatly influenced the settlement of the ac versus dc debate in favor of ac. He was also responsible for the adoption of 60 Hz as the standard for ac power systems in the United States. - -Born in Austria-Hungary (now Croatia), to a clergyman, Tesla had an incredible memory and a keen affinity for mathematics. He moved to the United States in 1884 and first worked for Thomas Edison. At that time, the country was in the "battle of the currents" with George Westinghouse (1846–1914) promoting ac and Thomas Edison rigidly leading the dc forces. Tesla left Edison and joined Westinghouse be cause of his interest in ac. Through Westinghouse, Tesla gained the reputation and acceptance of his polyphase ac generation, transmission, and distribution system. He held 700 patents in his lifetime. His other inventions include high-voltage apparatus (the tesla coil) and a wireless transmission system. The unit of magnetic flux density, the tesla, was named in honor of him. - -Library of Congress [LC-USZ62-61761] - -at the operating frequenc y of 60 Hz (or *ω* = 377 rad/s) in the United States or 50 Hz (or *ω* = 314 rad/s) in some other parts of the w orld. When one-phase or tw o-phase inputs are required, the y are taken from the three-phase system rather than generated independently. Even when more than three phases are needed—such as in the aluminum industry , where 48 phases are required for melting purposes—the y can be pro vided by manipulating the three phases supplied. Second, the instanta neous power in a three-phase system can be constant (not pulsating), as we will see in Section 12.7. This results in uniform power transmission and less vibration of three-phase machines. Third, for the same amount of power, the three-phase system is more economical than the singlephase. The amount of wire required for a three-phase system is less than that required for an equivalent single-phase system. - -We begin with a discussion of balanced three-phase v oltages. Then we analyze each of the four possible configurations of balanced threephase systems. We also discuss the analysis of unbalanced three-phase systems. We learn how to use *PSpice for Windows* to analyze a balanced or unbalanced three-phase system. Finally, we apply the concepts developed in this chapter to three-phase po wer measurement and residential electrical wiring. - -# **12.2** Balanced Three-Phase Voltages - -Three-phase voltages are often produced with a three-phase ac genera tor (or alternator) whose cross-sectional view is shown in Fig. 12.4. The generator basically consists of a rotating magnet (called the *rotor*) surrounded by a stationary winding (called the *stator*). Three separate windings or coils with terminals *a*-*a*′, *b*-*b*′, and *c*-*c*′ are physically placed 120° apart around the stator. Terminals *a* and *a*′, for example, stand for one of the ends of coils going into and the other end coming out of the - -**Figure 12.3** Three-phase four-wire system. - -A three-phase generator. - -**Figure 12.5** The generated voltages are 120° apart from each other. - -page. As the rotor rotates, its magnetic field "cuts" the flux from the three coils and induces voltages in the coils. Because the coils are placed 120° apart, the induced voltages in the coils are equal in magnitude but out of phase by 120° (Fig. 12.5). Since each coil can be regarded as a singlephase generator by itself, the three-phase generator can supply power to both single-phase and three-phase loads. - -A typical three-phase system consists of three voltage sources connected to loads by three or four wires (or transmission lines). (Threephase current sources are very scarce.) A three-phase system is equivalent to three single-phase circuits. The voltage sources can be either wyeconnected as shown in Fig. 12.6(a) or delta-connected as in Fig. 12.6(b). - -# **Figure 12.6** - -Three-phase voltage sources: (a) Y-connected source, (b) ∆-connected source. - -> Let us consider the wye-connected voltages in Fig. 12.6(a) for now. The voltages **V***an*, **V***bn*, and **V***cn* are respectively between lines *a*, *b*, and *c*, and the neutral line *n*. These voltages are called *phase voltages*. If the voltage sources have the same amplitude and frequency *ω* and are out of phase with each other by 120°, the voltages are said to be *balanced*. This implies that - -$$ -\mathbf{V}_{an} + \mathbf{V}_{bn} + \mathbf{V}_{cn} = 0 \tag{12.1} -$$ - -$$ -|\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| \tag{12.2} -$$ - -Thus, - -As a common tradition in power systems, voltage and current in this chapter are in rms values unless otherwise stated. - -Balanced phase voltages are equal in magnitude and are out of phase with each other by 120°. - -Because the three-phase voltages are 120° out of phase with each other, there are tw o possible combinations. One possibility is sho wn in Fig. 12.7(a) and expressed mathematically as - -$$ -\mathbf{V}_{an} = V_p / \mathbf{0}^{\circ} -$$ -\n -$$ -\mathbf{V}_{bn} = V_p / -120^{\circ} -$$ -\n -$$ -\mathbf{V}_{cn} = V_p / -240^{\circ} = V_p / +120^{\circ} -$$ -\n(12.3) - -where *Vp* is the effective or rms value of the phase voltages. This is known as the *abc sequence* or *positive sequence*. In this phase sequence, **V***an* leads **V***bn*, which in turn leads **V***cn*. This sequence is produced when the rotor in Fig. 12.4 rotates counterclockwise. The other possibility is shown in Fig. 12.7(b) and is given by - -$$ -\mathbf{V}_{an} = V_p \underline{/0^{\circ}} -$$ - -\n -$$ -\mathbf{V}_{cn} = V_p \underline{/ -120^{\circ}} -$$ - -\n -$$ -\mathbf{V}_{bn} = V_p \underline{/ -240^{\circ}} = V_p \underline{/ +120^{\circ}} -$$ - (12.4) - -This is called the *acb sequence* or *negative sequence*. For this phase sequence, **V***an* leads **V***cn*, which in turn leads **V***bn*. The *acb* sequence is produced when the rotor in Fig. 12.4 rotates in the clockwise direction. It is easy to show that the voltages in Eqs. (12.3) or (12.4) satisfy Eqs. (12.1) and (12.2). For example, from Eq. (12.3), - -$$ -\mathbf{V}_{an} + \mathbf{V}_{bn} + \mathbf{V}_{cn} = V_p \underline{\hspace{0.3cm}} \left( 0^{\circ} + V_p \underline{\hspace{0.3cm}} \right) + V_p \underline{\hspace{0.3cm}} \left( 120^{\circ} + V_p \underline{\hspace{0.3cm}} \right) -$$ -\n -$$ -= V_p (1.0 - 0.5 - j0.866 - 0.5 + j0.866) \quad (12.5) -$$ -\n -$$ -= 0 -$$ - -The phase sequence is the time order in which the voltages pass through their respective maximum values. - -The phase sequence is determined by the order in which the phasors pass through a fixed point in the phase diagram. - -In Fig. 12.7(a), as the phasors rotate in the counterclockwise direction with frequenc y *ω*, the y pass through the horizontal axis in a se quence *abcabca* . . . . Thus, the sequence is *abc* or *bca* or *cab*. Similarly, for the phasors in Fig. 12.7(b), as the y rotate in the counterclockwise direction, they pass the horizontal axis in a sequence *acbacba* . . . . This describes the *acb* sequence. The phase sequence is important in threephase power distribution. It determines the direction of the rotation of a motor connected to the power source, for example. - -Like the generator connections, a three-phase load can be either wye-connected or delta-connected, depending on the end application. Figure 12.8(a) sho ws a wye-connected load, and Fig. 12.8(b) sho ws a delta-connected load. The neutral line in Fig. 12.8(a) may or may not be there, depending on whether the system is four - or three-wire. (And, of course, a neutral connection is topologically impossible for a delta con nection.) A wye- or delta-connected load is said to be *unbalanced* if the phase impedances are not equal in magnitude or phase. - -**Figure 12.7** Phase sequences: (a) *abc* or positive sequence, (b) *acb* or negative sequence. - - The phase sequence may also be regarded as the order in which the phase voltages reach their peak (or maximum) values with respect to time. - -Reminder: As time increases, each phasor (or sinor) rotates at an angular velocity *ω*. - -# **Figure 12.8** - -Two possible three-phase load configurations: (a) a Y-connected load, (b) a ∆-connected load. - -A balanced load is one in which the phase impedances are equal in magnitude and in phase. - -For a *balanced* wye-connected load, - -$$ -\mathbf{Z}_1 = \mathbf{Z}_2 = \mathbf{Z}_3 = \mathbf{Z}_Y \tag{12.6} -$$ - -where **Z***Y* is the load impedance per phase. For a *balanced* delta-connected - -$$ -\mathbf{Z}_a = \mathbf{Z}_b = \mathbf{Z}_c = \mathbf{Z}_{\Delta} \tag{12.7} -$$ - -$$ -\mathbf{Z}_{\Delta} = 3\mathbf{Z}_{Y} \qquad \text{or} \qquad \mathbf{Z}_{Y} = \frac{1}{3}\mathbf{Z}_{\Delta} \qquad (12.8) -$$ - -Because both the three-phase source and the three-phase load can be either wye- or delta-connected, we have four possible connections: - -- Y-Y connection (i.e., Y-connected source with a Y-connected load). -- Y-∆ connection. - -load, - -Eq. (9.69) that - -- ∆-∆ connection. -- ∆-Y connection. - -In subsequent sections, we will consider each of these possible con figurations. - -It is appropriate to mention here that a balanced delta-connected load is more common than a balanced wye-connected load. This is due to the ease with which loads may be added or remo ved from each phase of a deltaconnected load. This is very difficult with a wye-connected load because the neutral may not be accessible. On the other hand, delta-connected sources are not common in practice because of the circulating current that will result in the delta-mesh if the three-phase voltages are slightly unbalanced. - -Example 12.1 Determine the phase sequence of the set of voltages - -$$ -v_{an} = 200 \cos(\omega t + 10^{\circ}) -$$ - -*vbn* = 200 cos(*ωt* − 230°), *vcn* = 200 cos(*ωt* − 110°) - -# **Solution:** - -The voltages can be expressed in phasor form as - -$$ -\mathbf{V}_{an} = 200 \underline{10^{\circ}} \text{ V}, \qquad \mathbf{V}_{bn} = 200 \underline{100^{\circ}} \text{ V}, \qquad \mathbf{V}_{cn} = 200 \underline{110^{\circ}} \text{ V} -$$ - -We notice that **V***an* leads **V***cn* by 120° and **V***cn* in turn leads **V***bn* by 120°. Hence, we have an *acb* sequence. - -Practice Problem 12.1 Given that **V***bn* = 220⧸ 30° V, find **V***an* and **V***cn*, assuming a positive (*abc*) sequence. - -**Answer:** 220⧸150° V, 220⧸−90° V. - -# **12.3** Balanced Wye-Wye Connection - -We begin with the Y-Y system, because any balanced three-phase sys tem can be reduced to an equivalent Y-Y system. Therefore, analysis of this system should be regarded as the key to solving all balanced threephase systems. - -A balanced Y-Y system is a three-phase system with a balanced Y-connected source and a balanced Y-connected load. - -Consider the balanced four-wire Y-Y system of Fig. 12.9, where a Y-connected load is connected to a Y-connected source. We assume a balanced load so that load impedances are equal. Although the impedance **Z***Y* is the total load impedance per phase, it may also be re garded as the sum of the source impedance **Z***s*, line impedance **Z***ℓ*, and load impedance **Z***L* for each phase, since these impedances are in series. As illustrated in Fig. 12.9, **Z***s* denotes the internal impedance of the phase winding of the generator; **Z***ℓ* is the impedance of the line join ing a phase of the source with a phase of the load; **Z***L* is the impedance of each phase of the load; and **Z***n* is the impedance of the neutral line. Thus, in general - -$$ -\mathbf{Z}_{Y} = \mathbf{Z}_{s} + \mathbf{Z}_{\ell} + \mathbf{Z}_{L} \tag{12.9} -$$ - -A balanced Y-Y system, showing the source, line, and load impedances. - -**Z***s* and **Z***ℓ* are often very small compared with **Z***L*, so one can assume that **Z***Y* = **Z***L* if no source or line impedance is given. In any event, by lump ing the impedances together, the Y-Y system in Fig. 12.9 can be simplified to that shown in Fig. 12.10. - -Assuming the positi ve sequence, the *phase* v oltages (or line-toneutral voltages) are - -$$ -\mathbf{V}_{an} = V_p / \underline{\mathbf{0}^{\circ}} -$$ - -$$ -\mathbf{V}_{bn} = V_p / \underline{-120^{\circ}}, \qquad \mathbf{V}_{cn} = V_p / \underline{+120^{\circ}} -$$ -(12.10) - -**Figure 12.10** Balanced Y-Y connection. - -The *line-to-line* voltages or simply *line* voltages **V***ab*, **V***bc*, and **V***ca* are related to the phase voltages. For example, - -$$ -\mathbf{V}_{ab} = \mathbf{V}_{an} + \mathbf{V}_{nb} = \mathbf{V}_{an} - \mathbf{V}_{bn} = V_p \left( \frac{0^\circ}{\rho} - V_p \right) - 120^\circ -$$ - -= $V_p \left( 1 + \frac{1}{2} + j \frac{\sqrt{3}}{2} \right) = \sqrt{3} V_p / 30^\circ$ (12.11a) - -Similarly, we can obtain - -$$ -\mathbf{V}_{bc} = \mathbf{V}_{bn} - \mathbf{V}_{cn} = \sqrt{3} V_p \sqrt{-90^\circ} -$$ - (12.11b) - -$$ -\mathbf{V}_{ca} = \mathbf{V}_{cn} - \mathbf{V}_{an} = \sqrt{3} V_p \underline{/-210^\circ} -$$ - (12.11c) - -Thus, the magnitude of the line voltages *VL* is √ 3 times the magnitude of the phase voltages *Vp*, or - -$$ -V_L = \sqrt{3} V_P \tag{12.12} -$$ - -where - -$$ -V_p = |\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| \tag{12.13} -$$ - -and - -$$ -V_L = |\mathbf{V}_{ab}| = |\mathbf{V}_{bc}| = |\mathbf{V}_{ca}| \tag{12.14} -$$ - -Also the line voltages lead their corresponding phase voltages by 30°. Figure 12.11(a) illustrates this. Figure 12.11(a) also shows how to determine **V***ab* from the phase voltages, while Fig. 12.11(b) shows the same for the three line voltages. Notice that **V***ab* leads **V***bc* by 120°, and **V***bc* leads **V***ca* by 120°, so that the line voltages sum up to zero as do the phase voltages. - -Applying KVL to each phase in Fig. 12.10, we obtain the line cur rents as - -$$ -\mathbf{I}_{a} = \frac{\mathbf{V}_{an}}{\mathbf{Z}_{Y}}, \qquad \mathbf{I}_{b} = \frac{\mathbf{V}_{bn}}{\mathbf{Z}_{Y}} = \frac{\mathbf{V}_{an} / -120^{\circ}}{\mathbf{Z}_{Y}} = \mathbf{I}_{a} / -120^{\circ} \tag{12.15} -$$ -\n -$$ -\mathbf{I}_{c} = \frac{\mathbf{V}_{cn}}{\mathbf{Z}_{Y}} = \frac{\mathbf{V}_{an} / -240^{\circ}}{\mathbf{Z}_{Y}} = \mathbf{I}_{a} / -240^{\circ} -$$ - -We can readily infer that the line currents add up to zero, - -$$ -\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0 \tag{12.16} -$$ - -so that - -$$ -\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = 0 \tag{12.17a} -$$ - -or - -$$ -\mathbf{V}_{nN} = \mathbf{Z}_n \mathbf{I}_n = 0 \tag{12.17b} -$$ - -that is, the voltage across the neutral wire is zero. The neutral line can thus be removed without affecting the system. In fact, in long distance power transmission, conductors in multiples of three are used with the earth itself acting as the neutral conductor. Power systems designed in this way are well grounded at all critical points to ensure safety. - -While the *line* current is the current in each line, the *phase* current is the current in each phase of the source or load. In the Y-Y system, the line current is the same as the phase current. We will use single subscripts - -**V**nb **V**ab = **V**an + **V**nb - -**V**cn - -**V**bn - -Phasor diagrams illustrating the relationship between line voltages and phase voltages. - -(b) - -**V**bc - -From **I***a*, we use the phase sequence to obtain other line currents. Thus, as long as the system is balanced, we need only analyze one phase. We may do this even if the neutral line is absent, as in the three-wire system. - -Calculate the line currents in the three-wire Y-Y system of Fig. 12.13. Example 12.2 - -**Figure 12.13** - -Three-wire Y-Y system; for Example 12.2. - -# **Solution:** - -The three-phase circuit in Fig. 12.13 is balanced; we may replace it with its single-phase equivalent circuit such as in Fig. 12.12. We obtain **I***a* from the single-phase analysis as - -$$ -\mathbf{I}_a = \frac{\mathbf{V}_{an}}{\mathbf{Z}_Y} -$$ - -where **Z***Y* = (5 − *j*2) + (10 + *j*8) = 15 + *j*6 = 16.155⧸21.8°. Hence, - -$$ -\mathbf{I}_a = \frac{110/0^{\circ}}{16.155/21.8^{\circ}} = 6.81/-21.8^{\circ} \text{ A} -$$ - -In as much as the source voltages in Fig. 12.13 are in positive sequence, the line currents are also in positive sequence: - -$$ -\mathbf{I}_b = \mathbf{I}_a \underline{/-120^\circ} = 6.81 \underline{/-141.8^\circ} \text{ A} -$$ -$$ -\mathbf{I}_c = \mathbf{I}_a \underline{/-240^\circ} = 6.81 \underline{/-261.8^\circ} \text{ A} = 6.81 \underline{/-98.2^\circ} \text{ A} -$$ - -# A Y-connected balanced three-phase generator with an impedance of 0.4 + *j*0.3 Ω per phase is connected to a Y-connected balanced load with an impedance of 24 + *j*19 Ω per phase. The line joining the generator and the load has an impedance of 0.6 + *j*0.7 Ω per phase. Assuming a positive sequence for the source voltages and that **V***an* = 120⧸30° V, find: (a) the line voltages, (b) the line currents. Practice Problem 12.2 - -**Answer:** (a) 207.8⧸60° V, 207.8⧸−60° V, 207.8⧸−180° V, (b) 3.75⧸−8.66° A, 3.75⧸−128.66° A, 3.75⧸ 111.34° A. - -# **12.4** Balanced Wye-Delta Connection - -A balanced Y-∆ system consists of a balanced Y-connected source feeding a balanced ∆-connected load. - -This is perhaps the most practical three-phase system, as the three-phase sources are usually Y-connected while the three-phase loads are usually ∆-connected. - -The balanced Y-delta system is sho wn in Fig. 12.14, where the source is Y-connected and the load is ∆-connected. There is, of course, no neutral connection from source to load for this case. Assuming the positive sequence, the phase voltages are again - -$$ -\mathbf{V}_{an} = V_p \underline{/0^{\circ}} \n\mathbf{V}_{bn} = V_p \underline{/ -120^{\circ}}, \qquad \mathbf{V}_{cn} = V_p \underline{/ +120^{\circ}}. -$$ -\n(12.19) - -As shown in Section 12.3, the line voltages are - -$$ -\mathbf{V}_{ab} = \sqrt{3} V_p \frac{\Delta 30^\circ}{\mathbf{V}_{ca}} = \mathbf{V}_{AB}, \qquad \mathbf{V}_{bc} = \sqrt{3} V_p \frac{\Delta 90^\circ}{\Delta 10^\circ} = \mathbf{V}_{BC} -$$ -\n -$$ -\mathbf{V}_{ca} = \sqrt{3} V_p \frac{\Delta 150^\circ}{\Delta 10^\circ} = \mathbf{V}_{CA} -$$ -\n(12.20) - -showing that the line voltages are equal to the voltages across the load impedances for this system configuration. From these voltages, we can obtain the phase currents as - -$$ -\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}}, \qquad \mathbf{I}_{BC} = \frac{\mathbf{V}_{BC}}{\mathbf{Z}_{\Delta}}, \qquad \mathbf{I}_{CA} = \frac{\mathbf{V}_{CA}}{\mathbf{Z}_{\Delta}} -$$ -(12.21) - -These currents have the same magnitude but are out of phase with each other by 120°. - -**Figure 12.14** Balanced Y-∆ connection. - -Another way to get these phase currents is to apply KVL. For example, applying KVL around loop *aABbna* gives - -$$ --\mathbf{V}_{an} + \mathbf{Z}_{\Delta} \mathbf{I}_{AB} + \mathbf{V}_{bn} = 0 -$$ - -or - -$$ -\mathbf{I}_{AB} = \frac{\mathbf{V}_{an} - \mathbf{V}_{bn}}{\mathbf{Z}_{\Delta}} = \frac{\mathbf{V}_{ab}}{\mathbf{Z}_{\Delta}} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} -$$ -(12.22) - -which is the same as Eq. (12.21). This is the more general way of finding the phase currents. - -The line currents are obtained from the phase currents by applying KCL at nodes *A*, *B*, and *C*. Thus, - -$$ -\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA}, \qquad \mathbf{I}_b = \mathbf{I}_{BC} - \mathbf{I}_{AB}, \qquad \mathbf{I}_c = \mathbf{I}_{CA} - \mathbf{I}_{BC} \quad (12.23) -$$ - -Since **I***CA* = **I***AB*⧸−240°, - -$$ -\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA} = \mathbf{I}_{AB} (1 - 1/240^\circ) -$$ - -= $\mathbf{I}_{AB} (1 + 0.5 - j0.866) = \mathbf{I}_{AB} \sqrt{3}/-30^\circ$ (12.24) - -showing that the magnitude *IL* of the line current is √ \_\_ 3 times the magnitude *Ip* of the phase current, or - -$$ -I^L = \sqrt{3}I_p \tag{12.25} -$$ - -where - -$$ -I_L = |\mathbf{I}_a| = |\mathbf{I}_b| = |\mathbf{I}_c| \tag{12.26} -$$ - -and - -$$ -I_p = |\mathbf{I}_{AB}| = |\mathbf{I}_{BC}| = |\mathbf{I}_{CA}| \tag{12.27} -$$ - -Also, the line currents lag the corresponding phase currents by 30°, assuming the positive sequence. Figure 12.15 is a phasor diagram illustrating the relationship between the phase and line currents. - -An alternative way of analyzing the Y-∆ circuit is to transform the ∆-connected load to an equi valent Y-connected load. Using the ∆-Y transformation formula in Eq. (12.8), - -$$ -Z_Y = \frac{Z_{\Delta}}{3} -$$ - (12.28) - -After this transformation, we now have a Y-Y system as in Fig. 12.10. The three-phase Y-∆ system in Fig. 12.14 can be replaced by the singlephase equivalent circuit in Fig. 12.16. This allows us to calculate only the line currents. The phase currents are obtained using Eq. (12.25) and utilizing the fact that each of the phase currents leads the corresponding line current by 30°. - -A balanced *abc*-sequence Y-connected source with **V***an* = 100⧸ 10° V is Example 12.3 connected to a ∆-connected balanced load (8 + *j*4) Ω per phase. Calculate the phase and line currents. - -Phasor diagram illustrating the relationship between phase and line currents. - -**Figure 12.16** A single-phase equivalent circuit of a balanced Y-∆ circuit. - -# **Solution:** - -This can be solved in two ways. - -■ **METHOD 1** The load impedance is - -$$ -\mathbf{Z}_{\Delta} = 8 + j4 = 8.944 / 26.57^{\circ} \,\Omega -$$ - -If the phase voltage **V***an* = 100⧸ 10°, then the line voltage is - -$$ -\mathbf{V}_{ab} = \mathbf{V}_{an} \sqrt{3} / 30^{\circ} = 100 \sqrt{3} / 10^{\circ} + 30^{\circ} = \mathbf{V}_{AB} -$$ - -or - -$$ -V_{AB} = 173.2 \angle 40^{\circ} -$$ - V - -The phase currents are - -$$ -\begin{aligned}\n\text{I}_{AB} &= \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} = \frac{173.2/40^{\circ}}{8.944/26.57^{\circ}} = 19.36/13.43^{\circ} \text{ A} \\ -\text{I}_{BC} &= \mathbf{I}_{AB} / -120^{\circ} = 19.36 / -106.57^{\circ} \text{ A} \\ -\text{I}_{CA} &= \mathbf{I}_{AB} / +120^{\circ} = 19.36 / 133.43^{\circ} \text{ A}\n\end{aligned} -$$ - -The line currents are - -$$ -\mathbf{I}_a = \mathbf{I}_{AB} \sqrt{3} \underline{/-30^\circ} = \sqrt{3} (19.36) \underline{/13.43^\circ - 30^\circ} -$$ - -= 33.53} \underline{/-16.57^\circ} A -$$ -\mathbf{I}_b = \mathbf{I}_a \underline{/-120^\circ} = 33.53 \underline{/-136.57^\circ} A -$$ -$$ -\mathbf{I}_c = \mathbf{I}_a \underline{/+120^\circ} = 33.53 \underline{/-103.43^\circ} A -$$ - -■ **METHOD 2** Alternatively, using single-phase analysis, - -$$ -I_a = \frac{V_{an}}{Z_{\Delta}/3} = \frac{100/10^{\circ}}{2.981/26.57^{\circ}} = 33.54/-16.57^{\circ} -$$ - A - -as above. Other line currents are obtained using the *abc* phase sequence. - -Practice Problem 12.3 One line voltage of a balanced Y-connected source is **V***AB* = 120⧸−20° V. If the source is connected to a ∆-connected load of 20 ⧸40° Ω, find the phase and line currents. Assume the *abc* sequence. - -> **Answer:** 6⧸−60° A, 6⧸−180° A, 6⧸60° A, 10.392⧸−90° A, 10.392⧸ 150° A, 10.392⧸30° A. - -# **12.5** Balanced Delta-Delta Connection - -A balanced ∆-∆ system is one in which both the balanced source and balanced load are ∆-connected. - -The source as well as the load may be delta-connected as sho wn in Fig. 12.17. Our goal is to obtain the phase and line currents as usual. - -Assuming a positive sequence, the phase voltages for a delta-connected source are - -$$ -\mathbf{V}_{ab} = V_p / \underline{\mathbf{0}^{\circ}} -$$ - -$$ -\mathbf{V}_{bc} = V_p / \underline{\mathbf{-120}^{\circ}}, \qquad \mathbf{V}_{ca} = V_p / \underline{\mathbf{+120}^{\circ}} -$$ -(12.29) - -The line voltages are the same as the phase voltages. From Fig. 12.17, assuming there is no line impedances, the phase voltages of the de ltaconnected source are equal to the voltages across the impedances; that is, - -$$ -\mathbf{V}_{ab} = \mathbf{V}_{AB}, \qquad \mathbf{V}_{bc} = \mathbf{V}_{BC}, \qquad \mathbf{V}_{ca} = \mathbf{V}_{CA} \tag{12.30} -$$ - -Hence, the phase currents are - -$$ -\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{Z_{\Delta}} = \frac{\mathbf{V}_{ab}}{Z_{\Delta}}, \qquad \mathbf{I}_{BC} = \frac{\mathbf{V}_{BC}}{Z_{\Delta}} = \frac{\mathbf{V}_{bc}}{Z_{\Delta}} -$$ -\n -$$ -\mathbf{I}_{CA} = \frac{\mathbf{V}_{CA}}{Z_{\Delta}} = \frac{\mathbf{V}_{ca}}{Z_{\Delta}} -$$ -\n(12.31) - -Because the load is delta-connected just as in the previous section, some of the formulas derived there apply here. The line currents are obtained from the phase currents by applying KCL at nodes *A*, *B*, and *C*, as we did in the previous section: - -$$ -\mathbf{I}_a = \mathbf{I}_{AB} - \mathbf{I}_{CA}, \qquad \mathbf{I}_b = \mathbf{I}_{BC} - \mathbf{I}_{AB}, \qquad \mathbf{I}_c = \mathbf{I}_{CA} - \mathbf{I}_{BC} \tag{12.32} -$$ - -Also, as shown in the last section, each line current lags the corre sponding phase current by 30°; the magnitude *IL* of the line current is √ \_\_ 3 times the magnitude *Ip* of the phase current, - -$$ -I_L = \sqrt{3}I_p \tag{12.33} -$$ - -An alternative way of analyzing the ∆-∆ circuit is to convert both the source and the load to their Y equivalents. We already kno w that **Z***Y* = **Z**∆∕3. To convert a ∆-connected source to a Y-connected source, see the next section. - -A balanced ∆-connected load ha ving an impedance 20 − *j*15 Ω is Example 12.4 connected to a ∆-connected, positi ve-sequence generator ha ving **V***ab* = 330⧸ 0° V. Calculate the phase currents of the load and the line currents. - -# **Solution:** - -The load impedance per phase is - -$$ -Z_{\Delta} = 20 - j15 = 25 \sqrt{-36.57^{\circ}} \,\Omega -$$ - -Since **V***AB* = **V***ab*, the phase currents are - -$$ -\mathbf{I}_{AB} = \frac{\mathbf{V}_{AB}}{\mathbf{Z}_{\Delta}} = \frac{330/0^{\circ}}{25/-36.87^{\circ}} = 13.2/36.87^{\circ} \text{ A} -$$ -$$ -\mathbf{I}_{BC} = \mathbf{I}_{AB}/-120^{\circ} = 13.2/-83.13^{\circ} \text{ A} -$$ -$$ -\mathbf{I}_{CA} = \mathbf{I}_{AB}/+120^{\circ} = 13.2/156.87^{\circ} \text{ A} -$$ - -For a delta load, the line current always lags the corresponding phase current by 30° and has a magnitude √ \_\_ 3 times that of the phase current. Hence, the line currents are - -$$ -\mathbf{I}_a = \mathbf{I}_{AB}\sqrt{3}/-30^\circ = (13.2/36.87^\circ)(\sqrt{3}/-30^\circ) -$$ - -= 22.86/6.87° A -$$ -\mathbf{I}_b = \mathbf{I}_a/-120^\circ = 22.86/-113.13^\circ -$$ - A -$$ -\mathbf{I}_c = \mathbf{I}_a/+120^\circ = 22.86/126.87^\circ -$$ - A - -A positive-sequence, balanced ∆-connected source supplies a balanced ∆-connected load. If the impedance per phase of the load is 18 + *j*12 Ω and **I***a* = 9.609⧸35° A, find **I***AB* and **V***AB*. Practice Problem 12.4 - -**Answer:** 5.548⧸65° A, 120⧸98.69° V. - -# **12.6** Balanced Delta-Wye Connection - -A balanced ∆-Y system consists of a balanced ∆-connected source feeding a balanced Y-connected load. - -Consider the ∆-Y circuit in Fig. 12.18. Again, assuming the *abc* sequence, the phase voltages of a delta-connected source are - -$$ -\mathbf{V}_{ab} = V_p \underline{\hat{\mathbf{O}}^{\circ}}, \qquad \mathbf{V}_{bc} = V_p \underline{\hat{\mathbf{O}}^{\circ}} = V_p \underline{\hat{\mathbf{O}}^{\circ}} -$$ -\n -$$ -\mathbf{V}_{ca} = V_p \underline{\hat{\mathbf{O}}^{\circ}} \tag{12.34} -$$ - -These are also the line voltages as well as the phase voltages. - -We can obtain the line currents in man y ways. One way is to apply KVL to loop *aANBba* in Fig. 12.18, writing - -$$ --\mathbf{V}_{ab} + \mathbf{Z}_{Y}\mathbf{I}_{a} - \mathbf{Z}_{Y}\mathbf{I}_{b} = 0 -$$ - -or - -Thus, - -$$ -\mathbf{Z}_{Y}(\mathbf{I}_{a}-\mathbf{I}_{b})=\mathbf{V}_{ab}=V_{p}\underline{\int_{0}^{\circ}} -$$ - -**I***a* − **I***b* = *Vp*⧸ 0° \_\_\_\_\_ **Z***Y* **(12.35)** - -**Figure 12.18** A balanced ∆-Y connection. - -But **I***b* lags **I***a* by 120°, since we assumed the *abc* sequence; that is, **I***b* = **I***a*⧸−120°. Hence, - -$$ -\mathbf{I}_a - \mathbf{I}_b = \mathbf{I}_a (1 - 1/ -120^\circ) -$$ - -= $\mathbf{I}_a \left( 1 + \frac{1}{2} + j\frac{\sqrt{3}}{2} \right) = \mathbf{I}_a \sqrt{3}/30^\circ$ (12.36) - -Substituting Eq. (12.36) into Eq. (12.35) gives - -$$ -I_a = \frac{\left(V_p / \sqrt{3}\right) / -30^{\circ}}{Z_Y} -$$ - (12.37) - -From this, we obtain the other line currents **I***b* and **I***c* using the positive phase sequence, i.e., **I***b* = **I***a*⧸−120°, **I***c* = **I***a*⧸+120°. The phase currents are equal to the line currents. - -Another w ay to obtain the line currents is to replace the deltaconnected source with its equi valent wye-connected source, as sho wn - -Transforming a ∆-connected source to an equivalent Y-connected source. - -per phase, according to Eq. (9.69). - -Once the source is transformed to wye, the circuit becomes a wyewye system. Therefore, we can use the equi valent single-phase circuit shown in Fig. 12.20, from which the line current for phase *a* is - -$$ -I_a = \frac{V_p / \sqrt{3} / -30^{\circ}}{Z_Y} -$$ - (12.39) - -**Z**Y **I**a **V**p ‒30° √3 + ‒ - -which is the same as Eq. (12.37). - -Alternatively, we may transform the wye-connected load to an equivalent delta-connected load. This results in a delta-delta system, which can be analyzed as in Section 12.5. Note that - -$$ -\mathbf{V}_{AN} = \mathbf{I}_a \mathbf{Z}_Y = \frac{V_p}{\sqrt{3}} \frac{1 - 30^\circ}{\sqrt{3}} \tag{12.40} -$$ -\n -$$ -\mathbf{V}_{BN} = \mathbf{V}_{AN} \frac{1 - 120^\circ}{\sqrt{3}} \qquad \mathbf{V}_{CN} = \mathbf{V}_{AN} \frac{1 + 120^\circ}{\sqrt{3}} \tag{12.40} -$$ - -As stated earlier , the delta-connected load is more desirable than the wye-connected load. It is easier to alter the loads in any one phase of the delta-connected loads, as the individual loads are connected directly across the lines. Ho wever, the delta-connected source is hardly used in practice because any slight imbalance in the phase voltages will result in unwanted circulating currents. - -Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four connections. Students are advised not to memorize the formulas but to understand ho w - -## **TABLE 12.1** - -| | Summary of phase and line voltages/currents for | | -|-------------------------------|-------------------------------------------------|--| -| balanced three-phase systems. | 1 | | - -| Connection | Phase voltages/currents | Line voltages/currents | -|-------------|-----------------------------|--------------------------------------------------| -| Y-Y | Van =
Vp
⧸ 0° | __
√
Vab =
3
Vp
⧸ 30° | -| | Vbn =
Vp
−120°
⧸ | Vbc =
Vab
−120°
⧸ | -| | Vcn =
Vp
+120°
⧸ | Vca =
Vab
+120°
⧸ | -| | Same as line currents | I
Van
Z
a =
∕
Y | -| | | I
I
−120°
b =
⧸
a | -| | | I
I
+120°
c =
⧸
a | -| Y-
∆ | Van =
Vp
⧸ 0° | __
√
Vab =
VAB =
3
Vp
30°
⧸ | -| | Vbn =
Vp
−120°
⧸ | Vbc =
VBC =
Vab
−120°
⧸ | -| | Vcn =
Vp
+120°
⧸ | Vca =
VCA =
Vab
+120°
⧸
__ | -| | IAB =
VAB
Z
∕
∆ | √
I
IAB
3
−30°
a =
⧸ | -| | IBC =
VBC
Z
∕
∆ | I
I
−120°
b =
⧸
a | -| | ICA =
VCA
Z
∕
∆ | I
I
+120°
c =
⧸
a | -| -
∆
∆ | Vab =
Vp
⧸ 0° | Same as phase voltages | -| | Vbc =
Vp
−120°
⧸ | | -| | Vca =
Vp
+120°
⧸ | __ | -| | IAB =
Vab
Z
∕
∆ | √
I
a =
IAB
3
−30°
⧸ | -| | IBC =
Vbc
Z
∕
∆ | I
b =
I
−120°
⧸
a | -| | ICA =
Vca
Z
∕
∆ | I
c =
I
+120°
⧸
a | -| ∆-Y | Vab =
Vp
⧸ 0° | Same as phase voltages | -| | Vbc =
Vp
−120°
⧸ | | -| | Vca =
Vp
+120°
⧸ | −30° ________
Vp
⧸ | -| | Same as line currents | I
a =
__
√
3
Z
Y | -| | | I
I
−120°
b =
⧸
a | -| | | I
I
+120°
c =
⧸
a | - -1 Positive or *abc* sequence is assumed. they are derived. The formulas can always be obtained by directly applying KCL and KVL to the appropriate three-phase circuits. - -A balanced Y-connected load with a phase impedance of 40 + *j*25 Ω is supplied by a balanced, positive sequence ∆-connected source with a line voltage of 210 V. Calculate the phase currents. Use **V***ab* as a reference. - -# **Solution:** - -The load impedance is - -$$ -\mathbf{Z}_Y = 40 + j25 = 47.17 / 32^{\circ} \,\Omega -$$ - -and the source voltage is - -$$ -\mathbf{V}_{ab} = 210 \underline{\big/ 0^{\circ}} \, \mathrm{V} -$$ - -When the ∆-connected source is transformed to a Y-connected source, - -$$ -\mathbf{V}_{an} = \frac{\mathbf{V}_{ab}}{\sqrt{3}} \underline{/-30^{\circ}} = 121.2 \underline{/-30^{\circ}} \text{ V} -$$ - -The line currents are - -$$ -\mathbf{I}_a = \frac{\mathbf{V}_{an}}{\mathbf{Z}_Y} = \frac{121.2 \div 30^\circ}{47.12 \div 32^\circ} = 2.57 \div 62^\circ \text{ A} -$$ -\n -$$ -\mathbf{I}_b = \mathbf{I}_a \div 120^\circ = 2.57 \div 178^\circ \text{ A} -$$ -\n -$$ -\mathbf{I}_c = \mathbf{I}_a \div 120^\circ = 2.57 \div 58^\circ \text{ A} -$$ - -which are the same as the phase currents. - -In a balanced ∆-Y circuit, **V***ab* = 440⧸ 15° and **Z***Y* = (12 + *j*15) Ω. Practice Problem 12.5 Calculate the line currents. - -**Answer:** 13.224⧸−66.34° A, 13.224⧸+173.66° A, 13.224⧸ 53.66° A. - -# **12.7** Power in a Balanced System - -Let us now consider the power in a balanced three-phase system. We begin by examining the instantaneous power absorbed by the load. This requires that the analysis be done in the time domain. For a Y-connected load, the phase voltages are - -$$ -v_{AN} = \sqrt{2} V_p \cos \omega t, \qquad v_{BN} = \sqrt{2} V_p \cos(\omega t - 120^\circ) -$$ - -$$ -v_{CN} = \sqrt{2} V_p \cos(\omega t + 120^\circ) -$$ - (12.41) - -where the factor √ 2is necessary because *Vp* has been defined as the rms value of the phase voltage. If **Z***Y* = *Z*⧸*θ*, the phase currents lag behind their corresponding phase voltages by *θ*. Thus, - -$$ -i_a = \sqrt{2} I_p \cos(\omega t - \theta), \qquad i_b = \sqrt{2} I_p \cos(\omega t - \theta - 120^\circ) \quad (12.42) -$$ -$$ -i_c = \sqrt{2} I_p \cos(\omega t - \theta + 120^\circ) -$$ - -Example 12.5 - -where *Ip* is the rms value of the phase current. The total instantaneous power in the load is the sum of the instantaneous powers in the three phases; that is, - -$$ -p = p_a + p_b + p_c = v_{AN}i_a + v_{BN}i_b + v_{CN}i_c -$$ - -= $2V_pI_p[\cos \omega t \cos(\omega t - \theta)$ -+ $\cos(\omega t - 120^\circ) \cos(\omega t - \theta - 120^\circ)$ -+ $\cos(\omega t + 120^\circ) \cos(\omega t - \theta + 120^\circ)$ (12.43) - -Applying the trigonometric identity - -$$ -\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)] \quad (12.44) -$$ - -gives - -$$ -p = V_p I_p [3 \cos \theta + \cos(2\omega t - \theta) + \cos(2\omega t - \theta - 240^\circ) + \cos(2\omega t - \theta + 240^\circ)] -$$ - -= $V_p I_p [3 \cos \theta + \cos \alpha + \cos \alpha \cos 240^\circ + \sin \alpha \sin 240^\circ + \cos \alpha \cos 240^\circ - \sin \alpha \sin 240^\circ]$ (12.45) -where $\alpha = 2\omega t - \theta$ -= $V_p I_p [3 \cos \theta + \cos \alpha + 2(-\frac{1}{2}) \cos \alpha] = 3V_p I_p \cos \theta$ - -Thus the total instantaneous power in a balanced three-phase system is constant—it does not change with time as the instantaneous power of each phase does. This result is true whether the load is Y- or ∆- connected. This is one important reason for using a three-phase system to generate and distribute power. We will look into another reason a little later. - -Since the total instantaneous po wer is independent of time, the average po wer per phase *Pp* for either the ∆-connected load or the Y-connected load is *p*∕3, or - -$$ -P_p = V_p I_p \cos \theta \tag{12.46} -$$ - -and the reactive power per phase is - -$$ -Q_p = V_p I_p \sin \theta \tag{12.47} -$$ - -The apparent power per phase is - -$$ -S_p = V_p I_p \tag{12.48} -$$ - -The complex power per phase is - -$$ -S_p = P_p + jQ_p = V_p I_p^* \tag{12.49} -$$ - -\_\_ - -where **V***p* and **I***p* are the phase voltage and phase current with magnitudes *Vp* and *Ip*, respectively. The total average power is the sum of the average powers in the phases: - -$$ -P = P_a + P_b + P_c = 3P_p = 3V_p I_p \cos \theta = \sqrt{3} V_L I_L \cos \theta \qquad (12.50) -$$ - -For a Y-connected load, *IL* = *Ip* but *VL* = √ 3 *Vp*, whereas for a ∆-connected load, *IL* = √ 3 *Ip* but *VL* = *Vp*. Thus, Eq. (12.50) applies for both Y-connected and ∆-connected loads. Similarly, the total reactive power is - -$$ -Q = 3V_p I_p \sin \theta = 3Q_p = \sqrt{3} V_L I_L \sin \theta \qquad (12.51) -$$ - -and the total complex power is - -$$ -\mathbf{S} = 3\mathbf{S}_p = 3\mathbf{V}_p \mathbf{I}_p^* = 3I_p^2 \mathbf{Z}_p = \frac{3V_p^2}{\mathbf{Z}_p^*} -$$ - (12.52) - -where **Z***p* = *Zp*⧸*θ* is the load impedance per phase. (**Z***p* could be **Z***Y* or **Z**∆.) Alternatively, we may write Eq. (12.52) as - -$$ -\mathbf{S} = P + jQ = \sqrt{3} V_L I_L \underline{\theta} \tag{12.53} -$$ - -Remember that *Vp*, *Ip*, *VL*, and *IL* are all rms values and that *θ* is the angle of the load impedance or the angle between the phase voltage and the phase current. - -A second major adv antage of three-phase systems for po wer distribution is that the three-phase system uses a lesser amount of wire than the single-phase system for the same line voltage *VL* and the same absorbed power *PL*. We will compare these cases and assume in both that the wires are of the same material (e.g., copper with resisti vity *ρ*), of the same length *ℓ*, and that the loads are resisti ve (i.e., unity power factor). For the tw o-wire single-phase system in Fig. 12.21(a), *IL* = *PL*∕*VL*, so the power loss in the two wires is - -$$ -P_{\text{loss}} = 2I_L^2 R = 2R \frac{P_L^2}{V_L^2} -$$ - (12.54) - -# **Figure 12.21** - -Comparing the power loss in (a) a single-phase system, and (b) a three-phase system. - -For the three-wire three-phase system in Fig. 12.21(b), *IL*′ = |**I***a*| = |**I***b*| = |**I***c*| = *PL*∕ √ \_\_ 3 *VL* from Eq. (12.50). The power loss in the three wires is - -$$ -P'_{\text{loss}} = 3(I'_L)^2 R' = 3R' \frac{P_L^2}{3V_L^2} = R' \frac{P_L^2}{V_L^2} -$$ - (12.55) - -Equations (12.54) and (12.55) show that for the same total power delivered *PL* and same line voltage *VL*, - -$$ -\frac{P_{\text{loss}}}{P'_{\text{loss}}} = \frac{2R}{R'} -$$ -\n(12.56) - -But from Chapter 2, *R* = *ρℓ*∕*πr* 2 and *R*′ = *ρℓ*∕*πr*′ 2 , where *r* and *r*′ are the radii of the wires. Thus, - -$$ -\frac{P_{\text{loss}}}{P'_{\text{loss}}} = \frac{2r'^2}{r^2} -$$ -\n(12.57) - -If the same power loss is tolerated in both systems, then *r* 2 = 2r′ 2 . The ratio of material required is determined by the number of wires and their volumes, so \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ Material for three-phase = 2(π*r* - -$$ -\frac{\text{Material for single-phase}}{\text{Material for three-phase}} = \frac{2(\pi r^2 \ell)}{3(\pi r^2 \ell)} = \frac{2r^2}{3r^2} -$$ -\n -$$ -= \frac{2}{3}(2) = 1.333 -$$ -\n(12.58) - -since *r* 2 = 2*r*′ 2 . Equation (12.58) shows that the single-phase system uses 33 percent more material than the three-phase system or that the threephase system uses only 75 percent of the material used in the equivalent single-phase system. In other words, considerably less material is needed to deliver the same power with a three-phase system than is required for a single-phase system. - -Refer to the circuit in Fig. 12.13 (in Example 12.2). Determine the total average power, reactive power, and complex power at the source and at the load. - -# **Solution:** - -It is sufficient to consider one phase, as the system is balanced. For phase *a*, - -$$ -V_p = 110\frac{0}{0} \text{ V} -$$ - and $I_p = 6.81\frac{1}{21.8^{\circ}} \text{ A}$ - -Thus, at the source, the complex power absorbed is - -$$ -\mathbf{S}_s = -3\mathbf{V}_p \mathbf{I}_p^* = -3(110/0^\circ)(6.81/21.8^\circ) -$$ - -= -2247/21.8° = -(2087 + j834.6) VA - -The real or average power absorbed is −2087 W and the reactive power is −834.6 VAR. - -At the load, the complex power absorbed is - -$$ -\mathbf{S}_{L} = 3|\mathbf{I}_{p}|^{2}\mathbf{Z}_{p} -$$ - -where $\mathbf{Z}_{p} = 10 + j8 = 12.81 \underline{ / 38.66^{\circ}}$ and $\mathbf{I}_{p} = \mathbf{I}_{a} = 6.81 \underline{ / -21.8^{\circ}}$ . Hence, -$$ -\mathbf{S}_{L} = 3(6.81)^{2} 12.81 \underline{ / 38.66^{\circ}} = 1782 \underline{ / 38.66^{\circ}} -$$ -$$ -= (1392 + j1113) \text{ VA} -$$ - -The real power absorbed is 1391.7 W and the reactive power absorbed is 1113.3 VAR. The difference between the two complex powers is ab sorbed by the line impedance (5 − *j*2) Ω. To show that this is the case, we find the complex power absorbed by the line as - -$$ -\mathbf{S}_{\ell} = 3|\mathbf{I}_{p}|^{2}\mathbf{Z}_{\ell} = 3(6.81)^{2}(5 - j2) = 695.6 - j278.3 \text{ VA} -$$ - -which is the difference between **S***s* and **S***L*; that is, **S***s* + **S***ℓ* + **S***L* = 0, as expected. - -Example 12.6 - -For the Y-Y circuit in Practice Prob. 12.2, calculate the complex power at the source and at the load. - -**Answer:** −(1054.2 + *j*843.3) VA, (1012 + *j*801.6) VA. - -A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the motor. - -# **Solution:** - -The apparent power is - -$$ -S = \sqrt{3}V_L I_L = \sqrt{3}(220)(18.2) = 6935.13 \text{ VA} -$$ - -Since the real power is - -$$ -P = S \cos \theta = 5600 \text{ W} -$$ - -the power factor is - -pf = -$$ -\cos \theta = \frac{P}{S} = \frac{5600}{6935.13} = 0.8075 -$$ - -Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85 lagging if it is connected to a balanced source with a line voltage of 550 V. - -**Answer:** 37.05 A. - -Two balanced loads are connected to a 240-kV rms 60-Hz line, as shown in Fig. 12.22(a). Load 1 dra ws 30 kW at a po wer factor of 0.6 lagging, while load 2 draws 45 kVAR at a power factor of 0.8 lagging. Assuming the *abc* sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load, (b) the line currents, and (c) the kVAR rating of the three capacitors ∆-connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor. - -# **Solution:** - -(a) For load 1, given that *P*1 = 30 kW and cos *θ*1 = 0.6, then sin *θ*1 = 0.8. Hence, - -$$ -S_1 = \frac{P_1}{\cos \theta_1} = \frac{30 \text{ kW}}{0.6} = 50 \text{ kVA} -$$ - -and *Q*1 = *S*1 sin *θ*1 = 50(0.8) = 40 kVAR. Thus, the complex power due to load 1 is - -$$ -S_1 = P_1 + jQ_1 = 30 + j40 \text{ kVA} -$$ - (12.8.1) - -Practice Problem 12.7 - -Example 12.8 - -Practice Problem 12.6 - -Example 12.7 - -# **Figure 12.22** - -For Example 12.8: (a) The original balanced loads, (b) the combined load with improved power factor. - -For load 2, if -$$ -Q_2 = 45 -$$ - kVAR and $\cos \theta_2 = 0.8$ , then $\sin \theta_2 = 0.6$ . We find - -$$ -S_2 = \frac{Q_2}{\sin \theta_2} = \frac{45 \text{ kVA}}{0.6} = 75 \text{ kVA} -$$ - -and *P*2 = *S*2 cos *θ*2 75(0.8) = 60 kW. Therefore the complex power due to load 2 is - -$$ -S_2 = P_2 + jQ_2 = 60 + j45 \text{ kVA} -$$ - (12.8.2) - -From Eqs. (12.8.1) and (12.8.2), the total complex power absorbed by the load is - -$$ -S = S_1 + S_2 = 90 + j85 \text{ kVA} = 123.8 \underline{43.36^{\circ}} \text{ kVA} \quad (12.8.3) -$$ - -which has a power factor of cos 43.36° = 0.727 lagging. The real power is then 90 kW, while the reactive power is 85 kVAR. - -It will help with the calculations to assume that the loads are wye connected and then to work with the phase voltages, i.e. the magnitude of *VAN* = (240∕√ \_\_ 3 ) kV. - -(b) Since -$$ -S = 3((240 \text{ kV}/\sqrt{3})I_L) = \sqrt{3}(240 \text{ kV})I_L -$$ - -the magnitude of the line current is - -$$ -I_L = \frac{S}{\sqrt{3}(240,000)} -$$ -(12.8.4) - -We apply this to each load keeping in mind that the magnitude of the phase voltages is equal to (240∕√ \_\_ 3 ) kV. For load 1, - -$$ -I_{L1} = \frac{50,000}{\sqrt{3} \, 240,000} = 120.28 \, \text{mA} -$$ - -Since the power factor is lagging, the line current lags the line voltage by *θ*1 = cos−1 0.6 = 53.13°. Thus, - -$$ -I_{a1} = 120.28 \sqrt{-53.13^{\circ}} -$$ - -For load 2, - -$$ -I_{L2} = \frac{75,000}{\sqrt{3} \, 240,000} = 180.42 \, \text{mA} -$$ - -and the line current lags the line voltage by *θ*2 = cos−1 0.8 = 36.87°. Hence, - -$$ -I_{a2} = 180.42 \underline{/-36.87^{\circ}} -$$ - -The total line current is - -$$ -\mathbf{I}_a = \mathbf{I}_{a1} + \mathbf{I}_{a2} = 120.28 \underline{/ -53.13^\circ} + 180.42 \underline{/ -36.87^\circ} -$$ - -= (72.168 - j96.224) + (144.336 - j108.252) -= 216.5 - j204.472 = 297.8 \underline{/ -43.36^\circ} mA - -Alternatively, we could obtain the current from the total complex power using Eq. (12.8.4) as - -$$ -I_L = \frac{123,800}{\sqrt{3} \, 240,000} = 297.82 \, \text{mA} -$$ - -and - -$$ -I_a = 297.82 \div 43.36^\circ \text{ mA} -$$ - -which is the same as before. The other line currents, **I***b*2 and **I***ca*, can be obtained according to the *abc* sequence (i.e., **I***b* = 297.82⧸−163.36° mA and **I***c* = 297.82⧸76.64° mA). - -(c) We can find the reactive power needed to bring the power factor to 0.9 lagging using Eq. (11.59), - -$$ -QC = P(\tan \theta_{\text{old}} - \tan \theta_{\text{new}}) -$$ - -where *P* = 90 kW, *θ*old = 43.36°, and *θ*new = cos−1 0.9 = 25.84°. Hence, - -$$ -Q_C = 90,000 \text{(tan } 43.36^\circ - \text{tan } 25.84^\circ) = 41.4 \text{ kVAR} -$$ - -This reactive power is for the three capacitors. For each capacitor, the rating *QC*′ = 13.8 kVAR. From Eq. (11.60), the required capacitance is - -$$ -C = \frac{Q'_C}{\omega V_{\text{rms}}^2} -$$ - -Since the capacitors are ∆-connected as shown in Fig. 12.22(b), *V*rms in the above formula is the line-to-line or line voltage, which is 240 kV. Thus, - -is the line-to-ine of the voltage, -$$ -C = \frac{13,800}{(2\pi60)(240,000)^2} = 635.5 \text{ pF} -$$ - -Assume that the two balanced loads in Fig. 12.22(a) are supplied by an 840-V rms 60-Hz line. Load 1 is Y-connected with 30 + *j*40 Ω per phase, while load 2 is a balanced three-phase motor drawing 48 kW at a power factor of 0.8 lagging. Assuming the *abc* sequence, calculate: (a) the complex power absorbed by the combined load, (b) the kVAR rating of each of the three capacitors ∆-connected in parallel with the load to raise the power factor to unity, and (c) the current drawn from the supply at unity power factor condition. - -**Answer:** (a) 56.47 + *j*47.29 kVA, (b) 15.76 kVAR, (c) 38.81 A. - -# **12.8** Unbalanced Three-Phase Systems - -This chapter would be incomplete without mentioning unbalanced threephase systems. An unbalanced system is caused by two possible situa tions: (1) The source voltages are not equal in magnitude and/or differ in phase by angles that are unequal, or (2) load impedances are unequal. Thus, - -An unbalanced system is due to unbalanced voltage sources or an unbalanced load. - -To simplify analysis, we will assume balanced source voltages, but an unbalanced load. - -Unbalanced three-phase systems are solved by direct application of mesh and nodal analysis. Figure 12.23 sho ws an e xample of an unbal anced three-phase system that consists of balanced source v oltages (not shown in the figure) and an unbalanced Y-connected load (shown in the figure). Since the load is unbalanced, **Z***A*, **Z***B*, and **Z***C* are not equal. The line currents are determined by Ohm's law as - -$$ -\mathbf{I}_a = \frac{\mathbf{V}_{AN}}{\mathbf{Z}_A}, \qquad \mathbf{I}_b = \frac{\mathbf{V}_{BN}}{\mathbf{Z}_B}, \qquad \mathbf{I}_c = \frac{\mathbf{V}_{CN}}{\mathbf{Z}_C} -$$ -(12.59) - -# **Figure 12.23** Unbalanced three-phase Y-connected load. - -A special technique for handling unbalanced three-phase systems is the method of symmetrical components, which is beyond the scope of this text. - -**I**a - -Practice Problem 12.8 - -This set of unbalanced line currents produces current in the neutral line, which is not zero as in a balanced system. Applying KCL at node *N* gives the neutral line current as - -$$ -\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) \tag{12.60} -$$ - -In a three-wire system where the neutral line is absent, we can still find the line currents **I***a*, **I***b*, and **I***c* using mesh analysis. At node *N*, KCL must be satisfied so that **I***a* + **I***b* + **I***c* = 0 in this case. The same could be done for an unbalanced ∆-Y, Y-∆, or ∆-∆ three-wire system. As mentioned earlier, in long distance po wer transmission, conductors in mul tiples of three (multiple three-wire systems) are used, with the earth itself acting as the neutral conductor. - -To calculate po wer in an unbalanced three-phase system requires that we find the power in each phase using Eqs. (12.46) to (12.49). The total power is not simply three times the power in one phase but the sum of the powers in the three phases. - -The unbalanced Y-load of Fig. 12.23 has balanced voltages of 100 V and the *acb* sequence. Calculate the line currents and the neutral current. Take **Z***A* = 15 Ω, **Z***B* = 10 + *j*5 Ω, **Z***C* = 6 − *j*8 Ω. - -# **Solution:** - -Using Eq. (12.59), the line currents are - -$$ -\mathbf{I}_a = \frac{100/0^{\circ}}{15} = 6.67/0^{\circ} \text{ A} -$$ -$$ -\mathbf{I}_b = \frac{100/120^{\circ}}{10 + j5} = \frac{100/120^{\circ}}{11.18/26.56^{\circ}} = 8.94/93.44^{\circ} \text{ A} -$$ -$$ -\mathbf{I}_c = \frac{100/-120^{\circ}}{6 - j8} = \frac{100/-120^{\circ}}{10/-53.13^{\circ}} = 10/-66.87^{\circ} \text{ A} -$$ - -Using Eq. (12.60), the current in the neutral line is - -$$ -\mathbf{I}_n = -(\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c) = -(6.67 - 0.54 + j8.92 + 3.93 - j9.2) -$$ - -= -10.06 + j0.28 = 10.06/178.4° A - -The unbalanced ∆-load of Fig. 12.24 is supplied by balanced line-to-line voltages of 440 V in the positive sequence. Find the line currents. Take **V***ab* as reference. - -**Answer:** -$$ -39.71 \underline{/-41.06^{\circ}} -$$ - A, $64.12 \underline{/-139.8^{\circ}}$ A, $70.13 \underline{/74.27^{\circ}}$ A. - -Example 12.9 - -Practice Problem 12.9 - -**Figure 12.24** Unbalanced ∆-load; for Practice Prob. 12.9. For the unbalanced circuit in Fig. 12.25, find: (a) the line currents, (b) the total complex power absorbed by the load, and (c) the total complex power absorbed by the source. - -For Example 12.10. - -# **Solution:** - -(a) We use mesh analysis to find the required currents. For mesh 1, - -$$ -120 \underline{/- 120^{\circ}} - 120 \underline{/0^{\circ}} + (10 + j5)I_1 - 10I_2 = 0 -$$ - -or - -$$ -(10+j5)\mathbf{I}_1 - 10\mathbf{I}_2 = 120\sqrt{3}/30^{\circ} -$$ - (12.10.1) - -For mesh 2, - -$$ -120\underline{\bigg/120^{\circ}} - 120\underline{\bigg/}-120^{\circ} + (10 - j10)\mathbf{I}_2 - 10\mathbf{I}_1 = 0 -$$ - -or - -$$ --10I1 + (10 - j10)I2 = 120\sqrt{3} / -90^{\circ} -$$ - (12.10.2) - -Equations (12.10.1) and (12.10.2) form a matrix equation: - -2.10.1) and (12.10.2) form a matrix equation: -\n -$$ -\begin{bmatrix}\n10 + j5 & -10 \\ --10 & 10 - j10\n\end{bmatrix}\n\begin{bmatrix}\n\mathbf{I}_1 \\ -\mathbf{I}_2\n\end{bmatrix} = \n\begin{bmatrix}\n120\sqrt{3}/30^\circ \\ -120\sqrt{3}/-90^\circ\n\end{bmatrix} -$$ - -The determinants are - -erminants are -\n -$$ -\Delta = \begin{vmatrix} 10 + j5 & -10 \\ -10 & 10 - j10 \end{vmatrix} = 50 - j50 = 70.71 \underline{/-45^{\circ}} -$$ -\n -$$ -\Delta_1 = \begin{vmatrix} 120\sqrt{3}/30^{\circ} & -10 \\ 120\sqrt{3}/-90^{\circ} & 10 - j10 \end{vmatrix} = 207.85(13.66 - j13.66) -$$ -\n -$$ -= 4015 \underline{/-45^{\circ}} -$$ -\n -$$ -\Delta_2 = \begin{vmatrix} 10 + j5 & 120\sqrt{3}/30^{\circ} \\ -10 & 120\sqrt{3}/-90^{\circ} \end{vmatrix} = 207.85(13.66 - j5) -$$ -\n -$$ -= 3023.4 \underline{/-20.1^{\circ}} -$$ - -The mesh currents are - -rents are -\n -$$ -\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{4015.23/-45^{\circ}}{70.71/-45^{\circ}} = 56.78 \text{ A} -$$ -\n -$$ -\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{3023.4/-20.1^{\circ}}{70.71/-45^{\circ}} = 42.75/24.9^{\circ} \text{ A} -$$ - -The line currents are - -$$ -\mathbf{I}_a = \mathbf{I}_1 = 56.78 \text{ A}, \qquad \mathbf{I}_c = -\mathbf{I}_2 = 42.75 \underline{\smash{\big)}\, - 155.1^\circ} \text{ A} -$$ -\n -$$ -\mathbf{I}_b = \mathbf{I}_2 - \mathbf{I}_1 = 38.78 + j18 - 56.78 = 25.46 \underline{\smash{\big)}\, 135^\circ} \text{ A} -$$ - -(b) We can now calculate the complex power absorbed by the load. For phase A, - -$$ -\mathbf{S}_A = \mathbf{I} \mathbf{I}_a \mathbf{I}^2 \mathbf{Z}_A = (56.78)^2 (j5) = j16,120 \text{ VA} -$$ - -For phase B, - -$$ -\mathbf{S}_B = |\mathbf{I}_b|^2 \mathbf{Z}_B = (25.46)^2 (10) = 6480 \text{ VA} -$$ - -For phase C, - -$$ -S_C = I I_c^2 Z_C = (42.75)^2(-j10) = -j18,276 \text{ VA} -$$ - -The total complex power absorbed by the load is - -$$ -S_L = S_A + S_B + S_C = 6480 - j2156 -$$ - VA - -(c) We check the result above by finding the power absorbed by the source. For the voltage source in phase *a,* - -$$ -S_a = -V_{an}I_a^* = -(120/0^\circ)(56.78) = -6813.6 -$$ - VA - -For the source in phase *b,* - -$$ -\mathbf{S}_b = -\mathbf{V}_{bn}\mathbf{I}_b^* = -(120/-120°)(25.46/-135°) -$$ - -= -3055.2/105° = 790 - j2951.1 VA - -For the source in phase *c*, - -$$ -\mathbf{S}_c = -\mathbf{V}_{bn}\mathbf{I}_c^* = -(120/120^\circ)(42.75/155.1^\circ) -$$ - -= -5130/275.1° = -456.03 + j5109.7 VA - -The total complex power absorbed by the three-phase source is - -$$ -S_s = S_a + S_b + S_c = -6480 + j2156 -$$ - VA - -showing that **S***s* + **S***L* = 0 and confirming the conservation principle of ac power. - -**Answer:** 128.01⧸ 80.1° A, 76.21⧸−60° A, 85⧸−135° A, 19.36 kW. - -# **12.9** PSpice for Three-Phase Circuits - -*PSpice* can be used to analyze three-phase balanced or unbalanced cir cuits in the same way it is used to analyze single-phase ac circuits. However, a delta-connected source presents two major problems to *PSpice*. First, a delta-connected source is a loop of voltage sources—which *PSpice* does not like. To avoid this problem, we insert a resistor of neg ligible resistance (say, 1 *μ*Ω per phase) into each phase of the deltaconnected source. Second, the delta-connected source does not provide a convenient node for the ground node, which is necessary to run *PSpice*. This problem can be eliminated by inserting balanced wye-connected large resistors (say, 1 MΩ per phase) in the delta-connected source so that the neutral node of the wye-connected resistors serves as the ground node 0. Example 12.12 will illustrate this. - -For the balanced Y-∆ circuit in Fig. 12.27, use *PSpice* to find the line current **I***aA*, the phase voltage **V***AB*, and the phase current **I***AC*. Assume that the source frequency is 60 Hz. - -# **Solution:** - -The schematic is shown in Fig. 12.28. The pseudocomponents IPRINT are inserted in the appropriate lines to obtain **I***aA* and **I***AC*, while VPRINT2 is inserted between nodes A and B to print differential voltage **V***AB*. We set the attributes of IPRINT and VPRINT2 each to *AC* = *yes*, *MAG* = *yes*, *PHASE* = *yes*, to print only the magnitude and phase of the currents and voltages. As a single-frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 60, and *Final Freq* = 60. Once the circuit is saved, it is simulated by selecting **Analysis/Simulate**. The output file includes the following: - -| FREQ | V(A,B) | VP(A,B) | -|-----------|--------------|--------------| -| 6.000E+01 | 1.699E+02 | 3.081E+01 | -| FREQ | IM(V_PRINT2) | IP(V_PRINT2) | -| 6.000E+01 | 2.350E+00 | -3.620E+01 | -| FREQ | IM(V_PRINT3) | IP(V_PRINT3) | -| 6.000E+01 | 1.357E+00 | -6.620E+01 | - -# Example 12.11 - -Schematic for the circuit in Fig. 12.27. - -From this, we obtain - -$$ -I_{aA} = 2.35 \underline{/-36.2^{\circ}} A -$$ - -$$ -V_{AB} = 169.9 \underline{/30.81^{\circ}} V, \quad I_{AC} = 1.357 \underline{/-66.2^{\circ}} A -$$ - -# Practice Problem 12.11 - -Refer to the balanced Y-Y circuit of Fig. 12.29. Use *PSpice* to find the line current **I***bB* and the phase voltage **V***AN*. Take *f* = 100 Hz. - -**Answer:** 100.9⧸ 60.87° V, 8.547⧸−91.27° A. - -Consider the unbalanced ∆-∆ circuit in Fig. 12.30. Use *PSpice* t o find the generator current **I***ab*, the line current **I***bB*, and the phase current **I***BC*. - -Example 12.12 - -# **Solution:** - -- 1. **Define.** The problem and solution process are clearly defined. -- 2. **Present.** We are to find the generator current flowing from *a* to *b,* the line current flowing from *b* to *B,* and the phase current flowing from *B* to *C*. -- 3. **Alternative.** Although there are different approaches to solving this problem, the use of *PSpice* is mandated. Therefore, we will not use another approach. -- 4. **Attempt.** As mentioned above, we avoid the loop of voltage sources by inserting a 1-*μ*Ω series resistor in the delta-connected source. To provide a ground node 0, we insert balanced wyeconnected resistors (1 MΩ per phase) in the delta-connected source, as shown in the schematic in Fig. 12.31. Three IPRINT pseudocomponents with their attributes are inserted to be able - -**Figure 12.31** Schematic for the circuit in Fig. 12.30. - -to get the required currents **I***ab*, **I***bB*, and **I***BC*. Since the operating frequency is not given and the inductances and capacitances should be specified instead of impedances, we assume *ω* = 1 rad/s so that *f* = 1∕2*π* = 0.159155 Hz. Thus, - -$$ -L = \frac{X_L}{\omega} \quad \text{and} \quad C = \frac{1}{\omega X_C} -$$ - - We select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 0.159155, and *Final Freq* = 0.159155. Once the schematic is saved, we select **Analysis/Simulate** to simulate the circuit. The output file includes: - -| FREQ | IM(V_PRINT1) | IP(V_PRINT1) | -|-----------|--------------|--------------| -| 1.592E-01 | 9.106E+00 | 1.685E+02 | -| FREQ | IM(V_PRINT2) | IP(V_PRINT2) | -| 1.592E-01 | 5.959E+00 | -1.772E+02 | -| FREQ | IM(V_PRINT3) | IP(V_PRINT3) | -| 1.592E-01 | 5.500E+00 | 1.725E+02 | - -which yields - -$$ -I_{ab} = 5.595 \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \big/ \ -$$ - -5. **Evaluate.** We can check our results by using mesh analysis. Let the loop *aABb* be loop 1, the loop *bBCc* be loop 2, and the loop *ACB* be loop 3, with the three loop currents all flowing in the clockwise direction. We then end up with the following loop equations: - -Loop 1 - -(54 + *j*10)*I*1 − (2 + *j*5)*I*2 − (50)*I*3 = 208⧸ 10° = 204.8 + *j*36.12 - -Loop 2 - -$$ --(2+j5)I1 + (4+j40)I2 - (j30)I3 = 208/ -110o -$$ - -= -71.14 - j195.46 - -Loop 3 - -$$ --(50)I_1 - (j30)I_2 + (50 - j10)I_3 = 0 -$$ - -Using MATLAB to solve this we get, - -``` ->>Z = [(54+10i),(-2-5i),-50;(-2-5i),(4+40i), --30i;-50,-30i,(50-10i)] -``` - -Z = - -``` -54.0000+10.0000i-2.0000-5.0000i-50.0000 --2.0000-5.0000i 4.0000 + 40.0000i 0-30.0000i --50.0000 0-30.0000i 50.0000-10.0000i -``` - -``` ->>V = [(204.8+36.12i);(-71.14-195.46i);0] -V = -1.0e+002* -2.0480+0.3612i --0.7114-1.9546i - 0 ->>I = inv(Z)*V -I = -8.9317+2.6983i -0.0096+4.5175i -5.4619+3.7964i - IbB = −I1 + I2 = −(8.932 + j2.698) + (0.0096 + j4.518) - = −8.922 + j1.82 = 9.106⧸168.47° A Answer checks - IBC = I2 − I3 = (0.0096 + j4.518) − (5.462 + j3.796) - = −5.452 + j0.722 = 5.5⧸172.46° A Answer checks -``` - -Now to solve for *Iab*. If we assume a small internal impedance for each source, we can obtain a reasonably good estimate for *Iab*. Adding in internal resistors of 0.01 Ω, and adding a fourth loop around the source circuit, we now get - -Loop 1 - -$$ -(54.01 + j10)I1 - (2 + j5)I2 - (50)I3 - 0.01I4 = 208/10o -$$ - -= 204.8 + j36.12 - -Loop 2 - -$$ --(2+j5)I1 + (4.01+j40)I2 - (j30)I3 - 0.01I4 -$$ - -= 208/ $-110^{\circ}$ = -71.14 - j195.46 - -Loop 3 - -$$ --(50)I1 - (j30)I2 + (50 - j10)I3 = 0 -$$ - -Loop 4 - -$$ --(0.01)I_1 - (0.01)I_2 + (0.03)I_4 = 0 -$$ - ->>Z = [(54.01+10i),(-2-5i),-50,-0.01;(-2-5i), (4.01+40i),-30i,-0.01;-50,-30i,(50-10i), 0;-0.01,-0.01,0,0.03] Z = 54.0100 + 10.0000i -2.0000-5.0000i, -50.0000 -0.0100 -2.0000-5.0000i 4.0100-40.0000i 0-30.0000i 0.0100 -50.0000 0-30.0000i 50.0000-10.0000i 0 -0.0100 -0.0100 0 0.0300 >>V = [(204.8 + 36.12i);(-71.14-195.46i);0;0] - -``` -V = -1.0e+002* -2.0480+0.3612i --0.7114-1.9546i - 0 - 0 ->>I = inv(Z)*V -I = -8.9309+2.6973i -0.0093+4.5159i -5.4623+3.7954i -2.9801+2.4044i - Iab = −I1 + I4 = −(8.931 + j2.697) + (2.98 + j2.404) -= −5.951 − j0.293 = 5.958⧸−177.18° A. Answer checks. -``` - -6. **Satisfactory?** We have a satisfactory solution and an adequate check for the solution. We can now present the results as a solution to the problem. - -Practice Problem 12.12 For the unbalanced circuit in Fig. 12.32, use *PSpice* to find the generator current **I***ca*, the line current **I***cC*, and the phase current **I***AB*. - -**Answer:** 24.68⧸−90° A, 37.25⧸ 83.79° A, 15.55⧸−75.01° A. - -# **12.10** Applications - -Both wye and delta source connections have important practical applications. The wye source connection is used for long distance transmission of electric power, where resistive losses ( *I* 2 *R*) should be minimal. This - -is due to the fact that the wye connection gives a line voltage that is √ \_\_ 3 greater than the delta connection; hence, for the same power, the line current is √ \_\_ 3smaller. In addition, delta connected are also undesirable due to the potential of having disastrous circulating currents. Sometimes, using transformers, we create the equivalent of delta connect source. This conversion from three-phase to single-phase is required in residential wiring, because household lighting and appliances use single-phase power. Three-phase power is used in industrial wiring where a large power is required. In some applications, it is immaterial whether the load is wye- or delta-connected. For example, both connections are satisfac tory with induction motors. In fact, some manufacturers connect a motor in delta for 220 V and in wye for 440 V so that one line of motors can be readily adapted to two different voltages. - -Here we consider two practical applications of those concepts co vered in this chapter: power measurement in three-phase circuits and residential wiring. - -# **12.10.1** Three-Phase Power Measurement - -Section 11.9 presented the wattmeter as the instrument for measuring the average (or real) power in single-phase circuits. A single wattmeter can also measure the average power in a three-phase system that is bal anced, so that *P*1 = *P*2 = *P*3; the total power is three times the reading of that one wattmeter. However, two or three single-phase wattmeters are necessary to measure power if the system is unbalanced. The *threewattmeter method* of power measurement, shown in Fig. 12.33, will work regardless of whether the load is balanced or unbalanced, wye- or delta-connected. The three-wattmeter method is well suited for power measurement in a three-phase system where the power factor is con stantly changing. The total average power is the algebraic sum of the three wattmeter readings, - -$$ -P_T = P_1 + P_2 + P_3 \tag{12.61} -$$ - -where *P*1, *P*2, and *P*3 correspond to the readings of wattmeters *W*1, *W*2, and *W*3, respectively. Notice that the common or reference point *o* in Fig. 12.33 is selected arbitrarily. If the load is wye-connected, point *o* can be connected to the neutral point *n*. For a delta-connected load, point *o* can be connected to any point. If point *o* is connected to point *b*, for example, the voltage coil in wattmeter *W*2 reads zero and *P*2 = 0, indicating that wattmeter *W*2 is not necessary. Thus, two wattmeters are sufficient to measure the total power. - -The *two-wattmeter method* is the most commonly used method for three-phase power measurement. The two wattmeters must be properly connected to an y two phases, as sho wn typically in Fig. 12.34. Notice that the current coil of each w attmeter measures the line current, while the respective voltage coil is connected between the line and the third line and measures the line v oltage. Also notice that the ± terminal of the voltage coil is connected to the line to which the corresponding current coil is connected. Although the individual wattmeters no longer read the power taken by any particular phase, the algebraic sum of the two wattmeter readings equals the total a verage power absorbed by the load, regardless of whether it is wye- or delta-connected, balanced or - -unbalanced. The total real power is equal to the algebraic sum of the two wattmeter readings, - -$$ -P_T = P_1 + P_2 \tag{12.62} -$$ - -We will show here that the method works for a balanced three-phase system. - -Consider the balanced, wye-connected load in Fig. 12.35. Our objective is to apply the two-wattmeter method to find the average power absorbed by the load. Assume the source is in the *abc* sequence and the load impedance **Z***Y* = Z*Y*⧸*θ*. Due to the load impedance, each voltage coil leads its current coil by *θ*, so that the po wer factor is cos *θ*. We recall that each line voltage leads the corresponding phase voltage by 30°. Thus, the total phase dif ference between the phase current **I***a* and line v oltage **V***ab* is *θ* + 30°, and the a verage po wer read by wattmeter *W*1 is - -$$ -P_1 = \text{Re}[\mathbf{V}_{ab}\mathbf{I}_a^*] = V_{ab}I_a \cos(\theta + 30^\circ) = V_L I_L \cos(\theta + 30^\circ) \qquad (12.63) -$$ - -**Figure 12.35** Two-wattmeter method applied to a balanced wye load. - -Similarly, we can show that the average power read by wattmeter 2 is - -$$ -P_2 = \text{Re}[\mathbf{V}_{cb}\mathbf{I}_c^*] = V_{cb}I_c \cos(\theta - 30^\circ) = V_L I_L \cos(\theta - 30^\circ) \qquad (12.64) -$$ - -We now use the trigonometric identities - -$$ -\cos(A + B) = \cos A \cos B - \sin A \sin B -$$ - -\n -$$ -\cos(A - B) = \cos A \cos B + \sin A \sin B -$$ - (12.65) - -to find the sum and the difference of the two wattmeter readings in Eqs. (12.63) and (12.64): - -$$ -P_1 + P_2 = V_L I_L [\cos(\theta + 30^\circ) + \cos(\theta - 30^\circ)] -$$ - -= $V_L I_L (\cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ$ -+ $\cos \theta \cos 30^\circ + \sin \theta \sin 30^\circ)$ -= $V_L I_L 2 \cos 30^\circ \cos \theta = \sqrt{3} V_L I_L \cos \theta$ (12.66) - -since 2 cos 30° = √ \_\_ 3 . Comparing Eq. (12.66) with Eq. (12.50) shows that the sum of the wattmeter readings gives the total average power, - -$$ -P_T = P_1 + P_2 \t\t(12.67) -$$ - -Similarly, - -$$ -P_1 - P_2 = V_L I_L [\cos(\theta + 30^\circ) - \cos(\theta - 30^\circ)] -$$ - -= $V_L I_L (\cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ$ - $- \cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ)$ (12.68) -= $-V_L I_L 2 \sin 30^\circ \sin \theta$ - $P_2 - P_1 = V_L I_L \sin \theta$ - -since 2 sin 30° = 1. Comparing Eq. (12.68) with Eq. (12.51) shows that the difference of the wattmeter readings is proportional to the total reactive power, or - -$$ -Q_T = \sqrt{3}(P_2 - P_1) -$$ - (12.69) - -From Eqs. (12.67) and (12.69), the total apparent power can be obtained as - -$$ -S_T = \sqrt{P_T^2 + Q_T^2} -$$ - (12.70) - -Dividing Eq. (12.69) by Eq. (12.67) gives the tangent of the power fac tor angle as - -$$ -\tan \theta = \frac{Q_T}{P_T} = \sqrt{3} \frac{P_2 - P_1}{P_2 + P_1} -$$ -\n(12.71) - -from which we can obtain the power factor as pf = cos *θ*. Thus, the two-wattmeter method not only provides the total real and reactive powers, it can also be used to compute the power factor. From Eqs. (12.67), (12.69), and (12.71), we conclude that: - -- 1. If *P*2 = *P*1, the load is resistive. -- 2. If *P*2 > *P*1, the load is inductive. -- 3. If *P*2 < *P*1, the load is capacitive. - -Although these results are derived from a balanced wye-connected load, they are equally valid for a balanced delta-connected load. However, the two-wattmeter method cannot be used for power measurement in a three-phase four-wire system unless the current through the neutral line is zero. We use the three-wattmeter method to measure the real power in a three-phase four-wire system. - -# Three wattmeters *W*1, *W*2, and *W*3 are connected, respectively, to phases *a, b,* and *c* to measure the total power absorbed by the unbalanced wye-connected load in Example 12.9 (see Fig. 12.23). (a) Predict the wattmeter readings. (b) Find the total power absorbed. - -# **Solution:** - -# Example 12.13 - -Part of the problem is already solved in Example 12.9. Assume that the wattmeters are properly connected as in Fig. 12.36. - -**Figure 12.36** For Example 12.13. - -(a) From Example 12.9, - -$$ -\mathbf{V}_{AN} = 100 \underline{\text{/}0^{\circ}}, \qquad \mathbf{V}_{BN} = 100 \underline{\text{/}120^{\circ}}, \qquad \mathbf{V}_{CN} = 100 \underline{\text{/} -120^{\circ}} \text{ V} -$$ - -while - -$$ -\mathbf{I}_a = 6.67 \underline{\bigcirc}^{\circ}, \qquad \mathbf{I}_b = 8.94 \underline{\bigcirc} 3.44^{\circ}, \qquad \mathbf{I}_c = 10 \underline{\bigcirc} -66.87^{\circ} \text{ A} -$$ - -We calculate the wattmeter readings as follows: - -$$ -P_1 = \text{Re}(\mathbf{V}_{AN}\mathbf{I}_{a}^{*}) = V_{AN}I_a \cos(\theta_{\mathbf{V}_{AN}} - \theta_{\mathbf{I}_a}) -$$ - -= 100 × 6.67 × cos(0° – 0°) = 667 W -$$ -P_2 = \text{Re}(\mathbf{V}_{BN}\mathbf{I}_{b}^{*}) = V_{BN}I_b \cos(\theta_{\mathbf{V}_{BN}} - \theta_{\mathbf{I}_b}) -$$ - -= 100 × 8.94 × cos(120° – 93.44°) = 800 W -$$ -P_3 = \text{Re}(\mathbf{V}_{CN}\mathbf{I}_{c}^{*}) = V_{CN}I_c \cos(\theta_{\mathbf{V}_{CN}} - \theta_{\mathbf{I}_c}) -$$ - -= 100 × 10 × cos(-120° + 66.87°) = 600 W - -(b) The total power absorbed is - -$$ -P_T = P_1 + P_2 + P_3 = 667 + 800 + 600 = 2067 -$$ - W - -We can find the power absorbed by the resistors in Fig. 12.36 and use that to check or confirm this result - -$$ -P_T = |I_a|^2(15) + |I_b|^2(10) + |I_c|^2(6) -$$ - -= 6.672(15) + 8.942(10) + 102(6) -= 667 + 800 + 600 = 2067 W - -which is exactly the same thing. - -| Practice Problem 12.13 | Repeat Example 12.13 for the network in Fig. 12.24 (see Practice
Prob. 12.9). Hint: Connect the reference point o in Fig. 12.33 to point B.
Answer: (a) 13.175 kW, 0 W, 29.91 kW, (b) 43.08 kW. | -|------------------------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------| -| Example 12.14 | The two-wattmeter method produces w attmeter readings P1 = 1560 W
and P2 = 2100 W when connected to a delta-connected load. If the line
voltage is 220 V, calculate: (a) the per-phase average power, (b) the per
phase reactive power, (c) the power factor, and (d) the phase impedance. | - -# **Solution:** - -We can apply the given results to the delta-connected load. (a) The total real or average power is - -$$ -P_T = P_1 + P_2 = 1560 + 2100 = 3660 -$$ - W - -The per-phase average power is then - -$$ -P_p = \frac{1}{3}P_T = 1220 \text{ W} -$$ - -(b) The total reactive power is - -$$ -Q_T = \sqrt{3}(P_2 - P_1) = \sqrt{3}(2100 - 1560) = 935.3 \text{ VAR} -$$ - -so that the per-phase reactive power is - -$$ -Q_p = \frac{1}{3}Q_T = 311.77 \text{ VAR} -$$ - -(c) The power angle is - -$$ -\theta = \tan^{-1} \frac{Q_T}{P_T} = \tan^{-1} \frac{935.3}{3660} = 14.33^{\circ} -$$ - -Hence, the power factor is - -$$ -\cos \theta = 0.9689(\text{lagging}) -$$ - -It is a lagging pf because *QT* is positive or *P*2 > *P*1. (d) The phase impedance is **Z***p* = *Zp*⧸*θ*. We know that *θ* is the same as the pf angle; that is, *θ* = 14.33°. - -$$ -Z_p = \frac{V_p}{I_p} -$$ - -We recall that for a delta-connected load, *Vp* = *VL* = 220 V. From Eq. (12.46), - -2.46), -\n -$$ -P_p = V_p I_p \cos \theta \implies I_p = \frac{1220}{220 \times 0.9689} = 5.723 \text{ A} -$$ - -Hence, - -$$ -Z_p = \frac{V_p}{I_p} = \frac{220}{5.723} = 38.44 \ \Omega -$$ - -and - -$$ -Z_p = 38.44 / 14.33^{\circ} \,\Omega -$$ - -Let the line voltage *VL* = 208 V and the wattmeter readings of the balanced system in Fig. 12.35 be *P*1 = −560 W and *P*2 = 800 W. Determine: - -(a) the total average power - -- (b) the total reactive power -- (c) the power factor - -(d) the phase impedance - -Is the impedance inductive or capacitive? - -**Answer:** (a) 240 W, (b) 2.356 kV AR, (c) 0.1014, (d) 18.25 ⧸ 84.18° Ω, inductive. - -Practice Problem 12.14 - -Example 12.15 - -The three-phase balanced load in Fig. 12.35 has impedance per phase of **Z***Y* = 8 + *j*6 Ω. If the load is connected to 208-V lines, predict the read ings of the wattmeters *W*1 and *W*2. Find *PT* and *QT*. - -# **Solution:** - -The impedance per phase is - -$$ -\mathbf{Z}_{Y} = 8 + j6 = 10/36.87^{\circ} \,\Omega -$$ - -so that the pf angle is 36.87°. Since the line voltage *VL* = 208 V, the line current is \_\_ - -$$ -I_L = \frac{V_p}{|\mathbf{Z}_Y|} = \frac{208/\sqrt{3}}{10} = 12 \text{ A} -$$ - -Then - -$$ -P_1 = V_L I_L \cos(\theta + 30^\circ) = 208 \times 12 \times \cos(36.87^\circ + 30^\circ) -$$ - -= 980.48 W -$$ -P_2 = V_L I_L \cos(\theta - 30^\circ) = 208 \times 12 \times \cos(36.87^\circ - 30^\circ) -$$ - -= 2478.1 W - -Thus, wattmeter 1 reads 980.48 W, while wattmeter 2 reads 2478.1 W. Since *P*2 > *P*1, the load is inductive. This is evident from the load **Z***Y* itself. Next, - -$$ -P_T = P_1 + P_2 = 3.459 \text{ kW} -$$ - -and - -$$ -Q_T = \sqrt{3}(P_2 - P_1) = \sqrt{3}(1497.6) -$$ - VAR = 2.594 kVAR - -If the load in Fig. 12.35 is delta-connected with impedance per phase of **Z***p* = 30 + *j*40 Ω and *VL* = 220 V, predict the readings of the wattmeters *W*1 and *W*2. Calculate *PT* and *QT*. Practice Problem 12.15 - -**Answer:** 200.6 W, 1.5418 kW, 1.7424 kW, 2.323 kVAR. - -# **12.10.2** Residential Wiring - -In the United States, most household lighting and appliances operate on 120-V, 60-Hz, single-phase alternating current. (The electricity may also be supplied at 110, 115, or 117 V, depending on the location.) The local power company supplies the house with a three-wire ac system. Typically, as in Fig. 12.37, the line voltage of, say, 12,000 V is stepped down to 120/240 V with a transformer (more details on transformers in the next chapter). The three wires coming from the transformer are typically colored red (hot), black (hot), and white (neutral). As shown in Fig. 12.38, the two 120-V voltages are opposite in phase and hence add up to zero. - -That is, -$$ -\mathbf{V}_W = 0/\underline{0}^\circ -$$ -, $\mathbf{V}_B = 120/\underline{0}^\circ$ , $\mathbf{V}_R = 120/\underline{180}^\circ = -\mathbf{V}_B$ . -\n -$$ -\mathbf{V}_{BR} = \mathbf{V}_B - \mathbf{V}_R = \mathbf{V}_B - (-\mathbf{V}_B) = 2\mathbf{V}_B = 240/\underline{0}^\circ \qquad (12.72) -$$ - -# **Figure 12.37** - -A 120/240 household power system. Source: A. Marcus and C. M. Thomson, *Electricity for Technicians,* 2nd edition, © 1975, p. 324. Pearson Education, Inc., Upper Saddle River, NJ. - -# **Figure 12.38** - -Single-phase three-wire residential wiring. - -Since most appliances are designed to operate with 120 V, the light ing and appliances are connected to the 120-V lines, as illustrated in Fig. 12.39 for a room. Notice in Fig. 12.37 that all appliances are connected in parallel. Hea vy appliances that consume lar ge currents, such as air conditioners, dishwashers, ovens, and laundry machines, are con nected to the 240-V power line. - -Because of the dangers of electricity , house wiring is carefully regulated by a code dra wn by local ordinances and by the National Electrical Code (NEC). To avoid trouble, insulation, grounding, fuses, and circuit breakers are used. Modern wiring codes require a third wire for a separate ground. The ground wire does not carry po wer like the neutral wire but enables appliances to have a separate ground connection. Figure 12.40 shows the connection of the receptacle to a 120-V rms line and to the ground. As shown in the figure, the neutral line is connected to the ground (the earth) at man y critical locations. Although the ground line seems redundant, grounding is important for many reasons. First, it is required by NEC. Second, grounding provides - -# **Figure 12.39** - -A typical wiring diagram of a room. Source: A. Marcus and C. M. Thomson, *Electricity for Technicians,* 2nd edition, © 1975, p. 325. Pearson Education, Inc., Upper Saddle River, NJ. - -a convenient path to ground for lightning that strik es the po wer line. Third, grounds minimize the risk of electric shock. What causes shock is the passage of current from one part of the body to another . The human body is lik e a big resistor *R*. If *V* is the potential dif ference between the body and the ground, the current through the body is determined by Ohm's law as - -$$ -I = \frac{V}{R} \tag{12.73} -$$ - -The value of *R* varies from person to person and depends on whether the body is wet or dry. How great or how deadly the shock is depends on the amount of current, the pathway of the current through the body, and the length of time the body is e xposed to the current. Currents less than 1 mA may not be harmful to the body , but currents greater than 10 mA can cause se vere shock. A modern safety de vice is the *ground-fault circuit interrupter* (GFCI), used in outdoor circuits and in bathrooms, where the risk of electric shock is greatest. It is essentially a circuit break er that opens when the sum of the currents *iR*, *iW*, and *iB* through the red, white, and the black lines is not equal to zero, or *iR* + *iW* + *iB* ≠ 0. - -The best w ay to a void electric shock is to follo w safety guide lines concerning electrical systems and appliances. Here are some of them: - -- Never assume that an electrical circuit is dead. Always check to be sure. -- Use safety de vices when necessary , and wear suitable clothing (insulated shoes, gloves, etc.). -- Never use tw o hands when testing high-v oltage circuits, since the current through one hand to the other hand has a direct path through your chest and heart. -- Do not touch an electrical appliance when you are wet. Remember that water conducts electricity. -- Be extremely careful when working with electronic appliances such as radio and TV because these appliances ha ve lar ge capacitors in them. The capacitors tak e time to dischar ge after the po wer is disconnected. -- Always ha ve another person present when w orking on a wiring system, just in case of an accident. - -# **12.11** Summary - -- 1. The phase sequence is the order in which the phase v oltages of a three-phase generator occur with respect to time. In an *abc* sequence of balanced source v oltages, **V***an* leads **V***bn* by 120°, which in turn leads **V***cn* by 120°. In an *acb* sequence of balanced v oltages, **V***an* leads **V***cn* by 120°, which in turn leads **V***bn* by 120°. -- 2. A balanced wye- or delta-connected load is one in which the threephase impedances are equal. -- 3. The easiest way to analyze a balanced three-phase circuit is to transform both the source and the load to a Y-Y system and then analyze the single-phase equivalent circuit. Table 12.1 presents a summary of the formulas for phase currents and voltages and line currents and voltages for the four possible configurations. -- 4. The line current *IL* is the current flowing from the generator to the load in each transmission line in a three-phase system. The line voltage *VL* is the v oltage between each pair of lines, e xcluding the neutral line if it e xists. The phase current *Ip* is the current flowing through each phase in a three-phase load. The phase voltage *Vp* is the voltage of each phase. For a wye-connected load, - -$$ -V_L = \sqrt{3} V_p \qquad \text{and} \qquad I_L = I_p -$$ - -For a delta-connected load, - -$$ -V_L = V_p \qquad \text{and} \qquad I_L = \sqrt{3}I_p -$$ - -- 5. The total instantaneous po wer in a balanced three-phase system is constant and equal to the average power. -- 6. The total comple x po wer absorbed by a balanced three-phase Y-connected or ∆-connected load is - -$$ -\mathbf{S} = P + jQ = \sqrt{3} V_L I_L \underline{\theta} -$$ - -where *θ* is the angle of the load impedances. - -- 7. An unbalanced three-phase system can be analyzed using nodal or mesh analysis. -- 8. *PSpice* is used to analyze three-phase circuits in the same w ay as it is used for analyzing single-phase circuits. -- 9. The total real power is measured in three-phase systems using either the three-wattmeter method or the two-wattmeter method. -- 10. Residential wiring uses a 120/240-V, single-phase, three-wire system. - -# Review Questions - -**12.1** What is the phase sequence of a three-phase motor for which **V***AN* = 220⧸−100° V and **V***BN* = 220⧸ 140° V? - -(a) *abc* (b) *acb* - -**12.2** If in an *acb* phase sequence, *Van* = 100⧸−20°, then **V***cn* is: - -(a) -$$ -100 \div 140^{\circ} -$$ - (b) $100 \div 100^{\circ}$ - -(c) -$$ -100\div 50^{\circ} -$$ - (d) $100\div 10^{\circ}$ - -**12.3** Which of these is not a required condition for a balanced system: - -$$ -(a) |\mathbf{V}_{an}| = |\mathbf{V}_{bn}| = |\mathbf{V}_{cn}| -$$ - -(b) -$$ -\mathbf{I}_a + \mathbf{I}_b + \mathbf{I}_c = 0 -$$ - -(c) -$$ -V_{an} + V_{bn} + V_{cn} = 0 -$$ - -- (d) Source voltages are 120° out of phase with each other. -- (e) Load impedances for the three phases are equal. - -**12.4** In a Y-connected load, the line current and phase current are equal. - -(a) True (b) False - -**12.5** In a ∆-connected load, the line current and phase current are equal. - -(a) True (b) False - -**12.6** In a Y-Y system, a line voltage of 220 V produces a phase voltage of: - -(a) 381 V (b) 311 V (c) 220 V (d) 156 V (e) 127 V - -**12.7** In a ∆-∆ system, a phase voltage of 100 V produces a line voltage of: - -(a) 58 V (b) 71 V (c) 100 V (d) 173 V (e) 141 V - -Problems1 - -# Section 12.2 Balanced Three-Phase Voltages - -**12.1** If **V***ab* = 400 V in a balanced Y-connected threephase generator, find the phase voltages, assuming the phase sequence is: - -(*a*) *abc* (*b*) *acb* - -- **12.2** What is the phase sequence of a balanced threephase circuit for which **V***an* = 120⧸30° V and **V***cn* = 120⧸−90° V? Find **V***bn*. -- **12.3** Given a balanced Y-connected three-phase generator with a line-to-line voltage of **V***ab* = 100⧸45° V and **V***bc* = 100⧸165° V, determine the phase sequence and the value of **V***ca*. -- **12.4** A three-phase system with *abc* sequence and *VL* = 440 V feeds a Y-connected load with *ZL* = 40⧸30° Ω. Find the line currents. -- **12.5** For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at the terminals are: - -*vAN* = 120 cos(*ωt* + 32°) V *vBN* = 120 cos(*ωt* – 88°) V *vCN* = 120 cos(*ωt* + 152°) V - - Write the time-domain expressions for the line-toline voltages *vAB*, *vBC*, and *vCA*. - -# Section 12.3 Balanced Wye-Wye Connection - -**12.6** Using Fig. 12.41, design a problem to help other students better understand balanced wye-wye connected circuits. - -**12.8** When a Y-connected load is supplied by voltages in *abc* phase sequence, the line voltages lag the corresponding phase voltages by 30°. - -(a) True (b) False - -**12.9** In a balanced three-phase circuit, the total instantaneous power is equal to the average power. - -(a) True (b) False - -**12.10** The total power supplied to a balanced ∆-load is found in the same way as for a balanced Y-load. - -(a) True (b) False - -*Answers: 12.1a, 12.2a, 12.3c, 12.4a, 12.5b, 12.6e, 12.7c, 12.8b, 12.9a, 12.10a.* - -# **Figure 12.41** - -For Prob. 12.6. - -- **12.7** Obtain the line currents in the three-phase circuit of Fig. 12.42 on the next page. -- **12.8** In a balanced three-phase Y-Y system, the source is an *acb* sequence of voltages and **V***cn* = 120⧸ 35° V rms. The line impedance per phase is (1 + *j*2) Ω, while the per-phase impedance of the load is (11 + *j*14) Ω. Calculate the line currents and the load voltages. -- **12.9** A balanced Y-Y four-wire system has phase voltages - -$$ -\mathbf{V}_{an} = 120 \underline{\text{O}^{\circ}}, \qquad \mathbf{V}_{bn} = 120 \underline{\text{O} - 120^{\circ}} -$$ -$$ -\mathbf{V}_{cn} = 120 \underline{\text{O} \cdot 120^{\circ}} \text{ V} -$$ - - The load impedance per phase is 19 + *j*13 Ω, and the line impedance per phase is 1 + *j*2 Ω. Solve for the line currents and neutral current. - -1 Remember that unless stated otherwise, all given voltages and currents are rms values. - -Problems **543** - -**12.10** For the circuit in Fig. 12.43, determine the current in the neutral line. - -For Prob. 12.10. - -# Section 12.4 Balanced Wye-Delta Connection - -**12.11** In the Y-∆ system shown in Fig. 12.44, the source is a positive sequence with **V***an* = 440⧸ 0° V and phase impedance **Z***p* = 2 – *j*3 Ω. Calculate the line voltage **V***L* and the line current **I***L*. - -**Figure 12.44** For Prob. 12.11. - -**Figure 12.46** For Prob. 12.13. - -**12.14** Obtain the line currents in the three-phase circuit of Fig. 12.47 on the next page. - -For Prob. 12.14. - -**12.15** The circuit in Fig. 12.48 is excited by a balanced three-phase source with a line voltage of 210 V. If **Z***l* = 1 + *j*1 Ω, **Z**∆ = 24 − *j*30 Ω, and **Z***Y* = 12 + *j*5 Ω, determine the magnitude of the line current of the combined loads. - -For Prob. 12.15. - -- **12.16** A balanced delta-connected load has a phase current **I***AC* = 5⧸−30° A. - - (a) Determine the three line currents assuming that the circuit operates in the positive phase sequence. - - (b) Calculate the load impedance if the line voltage is **V***AB* = 440⧸ 0° V. -- **12.17** A positive sequence wye-connected source where **V***an* = 120⧸ 90° V, is connected to a delta-connected load where **Z***L* = (60 + *j*45) Ω. Determine the line currents. -- **12.18** If **V***an* = 220⧸ 60° V in the network of Fig. 12.49, find the load phase currents **I***AB*, **I***BC*, and **I***CA*. - -**Figure 12.49** For Prob. 12.18. - -# Section 12.5 Balanced Delta-Delta Connection - -**12.19** For the ∆-∆ circuit of Fig. 12.50, calculate the phase and line currents. - -# **Figure 12.50** - -For Prob. 12.19. - -**12.20** Using Fig. 12.51, design a problem to help other students better understand balanced delta-delta connected circuits. - -For Prob. 12.20. - -**12.21** Three 440-V generators form a delta-connected source that is connected to a balanced deltaconnected load of *ZL* = (8.66 + *j*5) Ω per phase as shown in Fig. 12.52. Determine the value of *IBC* and *IaA*. What is the pf of the load? - -b - -+ - -+ ‒ - -- **12.22** Find the line currents *IaA*, *IbB*, and *IcC* in the three-phase network of Fig. 12.53. Take **Z***L* = (114 + *j*87) Ω and **Z***l* = (2 + *j*) Ω. -- **12.23** A balanced delta connected source is connected to a balanced delta connected load where *ZL* = (80 + *j*60) Ω and *Zl* = (2 + *j*) Ω. Given that the load voltages are **V***AB* = 100⧸ 0° V, **V***BC* = 100⧸ 120° V, and **V***CA* = 100⧸−120° V. Calculate the source voltages **V***ab*, **V***bc*, and **V***ca*. -- **12.24** A balanced delta-connected source has phase voltage **V***ab* = 880⧸ 30° V and a positive phase sequence. If this is connected to a balanced delta-connected load, find the line and phase currents. Take the load impedance per phase as 60⧸ 30° Ω and line impedance per phase as 1 + *j*1 Ω. -- **12.26** Using Fig. 12.55, design a problem to help other students better understand balanced delta connected sources delivering power to balanced wye connected loads. - -**Figure 12.55** For Prob. 12.26. - -10 ‒ j8 Ω - -10 ‒ j8 Ω - -# Section 12.6 Balanced Delta-Wye Connection - -12.25 In the circuit of Fig. 12.54, if -$$ -\mathbf{V}_{ab} = 440/10^{\circ} -$$ -, -\n $V_{bc} = 440/-110^{\circ}$ , $V_{ca} = 440/130^{\circ}$ V, find the line currents. - -3 + j2 Ω 10 ‒ j8 W - -**I**b - -**I**a - -a 3+ j2 W - -Vab - -- **12.27** A ∆-connected source supplies power to a Yconnected load in a three-phase balanced system. Given that the line impedance is 2 + *j*1 Ω per phase while the load impedance is 6 + *j*4 Ω per phase, find the magnitude of the line voltage at the load. Assume the source phase voltage **V***ab* = 208⧸ 0° V rms. -- **12.28** The line-to-line voltages in a Y-load have a magnitude of 880 V and are in the positive sequence at 60 Hz. If the loads are balanced with *Z*1 = *Z*2 = *Z*3 = 25⧸30°, find all line currents and phase voltages. - -# Section 12.7 Power in a Balanced System - -- **12.29** A balanced three-phase Y-∆ system has **V***an* = 240⧸ 0° V rms and **Z**∆ = 51 + *j*45 Ω. If the line impedance per phase is 0.4 + *j*1.2 Ω, find the total complex power delivered to the load. -- **12.30** In Fig. 12.56, the rms value of the line voltage is 208 V. Find the average power delivered to the load. - -**Figure 12.56** - -For Prob. 12.30. - -- **12.31** A balanced delta-connected load is supplied by a 60-Hz three-phase source with a line voltage of 480 V. Each load phase draws 24 kW at a lagging power factor of 0.8. Find: - - (a) the load impedance per phase - - (b) the line current - - (c) the value of capacitance needed to be connected in parallel with each load phase to minimize the current from the source - -- **12.32** Design a problem to help other students better understand power in a balanced three-phase system. -- **12.33** A three-phase source delivers 4.8 kVA to a wyeconnected load with a phase voltage of 208 V and a power factor of 0.9 lagging. Calculate the source line current and the source line voltage. -- **12.34** A balanced wye-connected load with a phase impedance of 10 – *j*16 Ω is connected to a balanced three-phase generator with a line voltage of 220 V. Determine the line current and the complex power absorbed by the load. -- **12.35** Three equal impedances, 60 + *j*30 Ω each, are delta-connected to a 230-V rms, three-phase circuit. Another three equal impedances, 40 + *j*10 Ω each, are wye-connected across the same circuit at the same points. Determine: - - (a) the line current - - (b) the total complex power supplied to the two loads - - (c) the power factor of the two loads combined -- **12.36** A 4200-V, three-phase transmission line has an impedance of 4 + *j* Ω per phase. If it supplies a load of 1 MVA at 0.75 power factor (lagging), find: - - (a) the complex power - - (b) the power loss in the line - - (c) the voltage at the sending end -- **12.37** The total power measured in a three-phase system feeding a balanced wye-connected load is 12 kW at a power factor of 0.6 leading. If the line voltage is 440 V, calculate the line current *IL* and the load impedance **Z***Y*. -- **12.38** Given the circuit in Fig. 12.57 below, find the total complex power absorbed by the load. - -**Figure 12.57** For Prob. 12.38. - -# For Prob. 12.39. - -**12.40** For the three-phase circuit in Fig. 12.59, find the average power absorbed by the delta-connected load with **Z**∆ = 21 + *j*24 Ω. - -# **Figure 12.59** For Prob. 12.40. - -- **12.41** A balanced delta-connected load draws 5 kW at a power factor of 0.8 lagging. If the three-phase system has an effective line voltage of 400 V, find the line current. -- **12.42** A balanced three-phase generator delivers 7.2 kW to a wye-connected load with impedance 30 – *j*40 Ω per phase. Find the line current *IL* and the line voltage *VL*. -- **12.43** Refer to Fig. 12.48. Obtain the complex power absorbed by the combined loads. -- **12.44** A three-phase line has an impedance of 1 + *j*3 Ω per phase. The line feeds a balanced delta-connected load, which absorbs a total complex power of 12 + *j*5 kVA. If the line voltage at the load end has a magnitude of 240 V, calculate the magnitude of the line voltage at the source end and the source power factor. -- **12.45** A balanced wye-connected load is connected to the generator by a balanced transmission line with an impedance of 0.5 + *j*2 Ω per phase. If the load is rated at 450 kW, 0.708 power factor lagging, 440-V line voltage, find the line voltage at the generator. -- **12.46** A three-phase load consists of three 100-Ω resistors that can be wye- or delta-connected. Determine which connection will absorb the most average - -power from a three-phase source with a line voltage of 110 V. Assume zero line impedance. - -**12.47** The following three parallel-connected three-phase loads are fed by a balanced three-phase source: - -> Load 1: 250 kVA, 0.8 pf lagging Load 2: 300 kVA, 0.95 pf leading Load 3: 450 kVA, unity pf - - If the line voltage is 13.8 kV, calculate the line current and the power factor of the source. Assume that the line impedance is zero. - -- **12.48** A balanced, positive-sequence wye-connected source has **V***an* = 240⧸ 0° V rms and supplies an unbalanced delta-connected load via a transmission line with impedance 2 + *j*3 Ω per phase. - - (a) Calculate the line currents if **Z***AB* = 40 + *j*15 Ω, **Z***BC* = 60 Ω, **Z***CA* = 18 – *j*12 Ω. - - (b) Find the complex power supplied by the source. -- **12.49** Each phase load consists of a 20-Ω resistor and a 10-Ω inductive reactance. With a line voltage of 480 V rms, calculate the average power taken by the load if: - -(a) the three-phase loads are delta-connected (b) the loads are wye-connected - -**12.50** A balanced three-phase source with **V***L* = 240 V rms is supplying 8 kVA at 0.6 power factor lagging to two wye-connected parallel loads. If one load draws 3 kW at unity power factor, calculate the impedance per phase of the second load. - -# Section 12.8 Unbalanced Three-Phase Systems - -**12.51** Consider the wye-delta system shown in Fig. 12.60. Let **Z**1 = 100 Ω, **Z**2 = *j*100 Ω, and **Z**3 = –*j*100 Ω. Determine the phase currents, **I***AB*, **I***BC*, and **I***CA*, and the line currents, **I***aA*, **I***bB* , and **I***cC*. - -**Figure 12.60** For Prob. 12.51. - -$$ -\mathbf{V}_{an} = 220/120^{\circ}, \qquad \mathbf{V}_{bn} = 220/0^{\circ} -$$ -$$ -\mathbf{V}_{cn} = 220/-120^{\circ} \text{ V} -$$ - -If the impedances are - -$$ -\mathbf{Z}_{AN} = 20/60^\circ, \qquad \mathbf{Z}_{BN} = 30/0^\circ -$$ -$$ -\mathbf{Z}_{cn} = 40/30^\circ \ \Omega -$$ - -find the current in the neutral line. - -**12.53** Using Fig. 12.61, design a problem that will help other students better understand unbalanced threephase systems. - -- For Prob. 12.53. -- **12.54** A balanced three-phase Y-source with *VP* = 880 V rms drives a Y-connected three-phase load with phase impedance **Z***A* = 80 Ω, **Z***B* = 60 + *j*90 Ω, and **Z***C* = *j*80 Ω. Calculate the line currents and total complex power delivered to the load. Assume that the neutrals are connected. -- **12.55** A three-phase supply, with the line-to-line voltage of 240 V rms, has the unbalanced load as shown in Fig. 12.62. Find the line currents and the total complex power delivered to the load. - -**Figure 12.62** For Prob. 12.55. - -**12.56** Using Fig. 12.63, design a problem to help other students to better understand unbalanced three-phase systems. - -# **Figure 12.63** - -For Prob. 12.56. - -**12.57** Determine the line currents for the three-phase circuit of Fig. 12.64. Let **V***a* = 220⧸ 0°, **V***b* = 220⧸−120°, **V***c* = 220⧸ 120° V. - -For Prob. 12.57. - -# Section 12.9 PSpice for Three-Phase Circuits - -- **12.58** Solve Prob. 12.10 using *PSpice or MultiSim*. -- **12.59** The source in Fig. 12.65 is balanced and exhibits a positive phase sequence. If *f* = 60 Hz, use *PSpice or MultiSim* to find **V***AN*,**V***BN*, and **V***CN*. - -For Prob. 12.59. - -**12.60** Use *PSpice or MultiSim* to determine **I***o* in the single-phase, three-wire circuit of Fig. 12.66. Let **Z**1 = 15 – *j*10 Ω, **Z**2 = 30 + *j*20 Ω, and **Z**3 = 12 + *j*5 Ω. - -**12.61** Given the circuit in Fig. 12.67, use *PSpice or MultiSim* to determine currents **I***aA* and voltage **V***BN*. - -**Figure 12.67** - -For Prob. 12.61. - -**12.62** Using Fig. 12.68, design a problem to help other students better understand how to use *PSpice or MultiSim* to analyze three-phase circuits. - -**12.63** Use *PSpice or MultiSim* to find currents **I***aA* and **I***AC* in the unbal anced three-phase system shown in Fig. 12.69. Let - -$$ -Zl = 2 + j, \t Z1 = 40 + j20 Ω,\nZ2 = 50 - j30 Ω, \t Z3 = 25 Ω -$$ - -# **Figure 12.69** - -For Prob. 12.63. - -- **12.64** For the circuit in Fig. 12.58, use *PSpice or MultiSim* to find the line currents and the phase currents. -- **12.65** A balanced three-phase circuit is shown in Fig. 12.70 on the next page. Use *PSpice or MultiSim* to find the line currents **I***aA*, **I***bB*, and **I***cC*. - -# Section 12.10 Applications - -- **12.66** A three-phase, four-wire system operating with a 480-V line voltage is shown in Fig. 12.71. The source voltages are balanced. The power absorbed by the resistive wye-connected load is measured by the three-wattmeter method. Calculate: - - (a) the voltage to neutral - - (b) the currents **I**1, **I**2, **I**3, and **I***n* - - (c) the readings of the wattmeters - - (d) the total power absorbed by the load -- **12.67** As shown in Fig. 12.72, a three-phase four-wire line with a phase voltage of 120 V rms and positive phase sequence supplies a balanced motor load at 260 kVA at 0.85 pf lagging. The motor load is connected to the three main lines marked *a, b,* and *c*. In addition, incandescent lamps (unity pf) are connected as follows: 24 kW from line *c* to the neutral, 15 kW from line b to the neutral, and 9 kW from line *c* to the neutral. - - (a) If three wattmeters are arranged to measure the power in each line, calculate the reading of each meter. - - (b) Find the magnitude of the current in the neutral line. - -\* An asterisk indicates a challenging problem. - -**Figure 12.68** For Prob. 12.62. - -**Figure 12.70** For Prob. 12.65. - -- **12.68** Meter readings for a three-phase wye-connected alternator supplying power to a motor indicate that the line voltages are 330 V, the line currents are 8.4 A, and the total line power is 4.5 kW. Find: - - (a) the load in VA - - (b) the load pf - - (c) the phase current - - (d) the phase voltage -- **12.69** A certain store contains three balanced three-phase loads. The three loads are: - -Load 1: 16 kVA at 0.85 pf lagging Load 2: 12 kVA at 0.6 pf lagging Load 3: 8 kW at unity pf - - The line voltage at the load is 208 V rms at 60 Hz, and the line impedance is 0.4 + *j*0.8 Ω. Determine the line current and the complex power delivered to the loads. - -# **Figure 12.72** - -For Prob. 12.67. - -- **12.70** The two-wattmeter method gives *P*1 = 1200 W and *P*2 = −400 W for a three-phase motor running on a 240-V line. Assume that the motor load is wye- connected and that it draws a line current of 6 A. Calculate the pf of the motor and its phase impedance. -- **12.71** In Fig. 12.73, two wattmeters are properly connected to the unbalanced load supplied by a balanced source such that **V***ab* = 208⧸ 0° V with positive phase sequence. - - (a) Determine the reading of each wattmeter. - - (b) Calculate the total apparent power absorbed by the load. - -**Figure 12.73** For Prob. 12.71. - -- **12.72** If wattmeters *W*1 and *W*2 are properly connected respectively between lines *a* and *b* and lines *b* and *c* to measure the power absorbed by the deltaconnected load in Fig. 12.44, predict their readings. -- **12.73** For the circuit displayed in Fig. 12.74, find the wattmeter readings. - -# **Figure 12.74** - -For Prob. 12.73. - -**12.74** Predict the wattmeter readings for the circuit in Fig. 12.75. - -# **Figure 12.75** For Prob. 12.74. - -- **12.75** A man has a body resistance of 600 Ω. How much current flows through his ungrounded body: - - (a) when he touches the terminals of a 12-V autobattery? - - (b) when he sticks his finger into a 120-V light socket? - -# Comprehensive Problems - -- **12.77** A three-phase generator supplied 10 kVA at a power factor of 0.85 lagging. If 7,500 W are delivered to the load and line losses are 160 W per phase, what are the losses in the generator? -- **12.78** A three-phase 440-V, 51-kW, 60-kVA inductive load operates at 60 Hz and is wye-connected. It is desired to correct the power factor to 0.95 lagging. What value of capacitor should be placed in parallel with each load impedance? -- **12.79** A balanced three-phase generator has an *abc* phase sequence with phase voltage **V***an* = 554.3⧸ 0° V. The generator feeds an induction motor which may be represented by a balanced Y-connected load with an impedance of 12 + *j*5 Ω per phase. Find the line currents and the load voltages. Assume a line impedance of 2 Ω per phase. -- **12.80** A balanced three-phase source furnishes power to the following three loads: - -Load 1: 6 kVA at 0.83 pf lagging Load 2: unknown Load 3: 8 kW at 0.7071 pf leading - - If the line current is 84.6 A rms, the line voltage at the load is 208 V rms, and the combined load has a 0.8 pf lagging, determine the unknown load. - -**12.81** A professional center is supplied by a balanced three-phase source. The center has four balanced three-phase loads as follows: - -> Load 1: 150 kVA at 0.8 pf leading Load 2: 100 kW at unity pf Load 3: 200 kVA at 0.6 pf lagging Load 4: 80 kW and 95 kVAR (inductive) - - If the line impedance is 0.02 + *j*0.05 Ω per phase and the line voltage at the loads is 480 V, find the magnitude of the line voltage at the source. - -- **12.82** A balanced three-phase system has a distribution wire with impedance 2 + *j*6 Ω per phase. The system supplies two three-phase loads that are connected in parallel. The first is a balanced wye-connected load that absorbs 400 kVA at a power factor of 0.8 lagging. The second load is a balanced delta-connected load with impedance of 10 + *j*8 Ω per phase. If the magnitude of the line voltage at the loads is 2400 V rms, calculate the magnitude of the line voltage at the source and the total complex power supplied to the two loads. -- **12.83** A commercially available three-phase inductive motor operates at a full load of 120 hp (1 hp = 746 W) at 95 percent efficiency at a lagging power - -factor of 0.707. The motor is connected in parallel to a 80-kW balanced three-phase heater at unity power factor. If the magnitude of the line voltage is 480 V rms, calculate the line current. - -**12.84** Figure 12.76 displays a three-phase delta-connected motor load which is connected to a line voltage of 440 V and draws 4 kVA at a power factor of 72 per cent lagging. In addition, a single 1.8 kVAR capacitor is connected between lines *a* and *b*, while a 800-W lighting load is connected between line *c* and neutral. Assuming the *abc* sequence and taking **V***an* = *Vp*⧸0°, find the magnitude and phase angle of currents **I***a*, **I***b*, **I***c*, and **I***n*. - -**Figure 12.76** - -For Prob. 12.84. - -**12.85** Design a three-phase heater with suitable symmetric loads using wye-connected pure resistance. Assume that the heater is supplied by a 240-V line voltage and is to give 27 kW of heat. - -**12.86** For the single-phase three-wire system in Fig. 12.77, find currents **I***aA*, **I***bB*, and **I***nN*. - -# **Figure 12.77** - -For Prob. 12.86. - -**12.87** Consider the single-phase three-wire system shown in Fig. 12.78. Find the current in the neutral wire and the complex power supplied by each source. Take **V***s* as a 220⧸ 0°-V, 60-Hz source. - -**Figure 12.78** For Prob. 12.87. - -# Magnetically Coupled Circuits - -*If you would increase your happiness and prolong your life, forget your neighbor's faults. . . . Forget the peculiarities of your friends, and only remember the good points which make you fond of them. . . . Obliterate everything disagreeable from yesterday; write upon today's clean sheet those things lovely and lovable.* - -—Anonymous - -# **chapter** - -13 - -# Enhancing Your Career - -# **Career in Electromagnetics** - -Electromagnetics (EM) is the branch of electrical engineering (or ph ysics) that deals with the analysis and application of electric and magnetic fields. In electromagnetics, electric circuit analysis is applied at low frequencies. - -The principles of EM are applied in various allied disciplines, such as electric machines, electromechanical ener gy conversion, radar meteorology, remote sensing, satellite communications, bioelectromagnetics, electromagnetic interference and compatibility, plasmas, and fiber optics. EM devices include electric motors and generators, transformers, electromagnets, magnetic levitation, antennas, radars, microwave ovens, microwave dishes, superconductors, and electrocardiograms. The design of these de vices requires a thorough knowledge of the laws and principles of EM. - -EM is regarded as one of the more dif ficult disciplines in electrical engineering. One reason is that EM phenomena are rather abstract. But if one enjoys working with mathematics and can visualize the invisible, one should consider being a specialist in EM, inasmuch as few electrical engineers specialize in this area. Electrical engineers who specialize in EM are needed in micro wave industries, radio/TV broadcasting stations, electromagnetic research laboratories, and se veral communications industries. - -Telemetry receiving station for space satellites. © DV169/Getty Images RF - -# Historical - -© Bettmann/Corbis - -**James Clerk Maxwell** (1831–1879), a graduate in mathematics from Cambridge University, in 1865 wrote a most remarkable paper in which he mathematically unified the laws of Faraday and Ampere. This relationship between the electric field and magnetic field served as the basis for what was later called electromagnetic fields and waves, a major field of study in electrical engineering. The Institute of Electrical and Electron ics Engineers (IEEE) uses a graphical representation of this principle in its logo, in which a straight arrow represents current and a curved arrow represents the electromagnetic field. This relationship is commonly known as *the right-hand rule* . Maxwell was a very active theoretician and scientist. He is best known for the "Maxwell equations." The max well, a unit of magnetic flux, was named after him. - -# Learning Objectives - -By using the information and exercises in this chapter you will be able to: - -- 1. Understand the physics behind mutually coupled circuits and how to analyze circuits containing mutually coupled inductors. -- 2. Understand how energy is stored in mutually coupled circuits. -- 3. Understand how linear transformers work and how to analyze circuits containing them. -- 4. Understand how ideal transformers work and how to analyze circuits containing them. -- 5. Understand how ideal auto transformers work and know how to analyze them when used in a variety of circuits. - -# **13.1** Introduction - -The circuits we have considered so far may be regarded as *conductively coupled*, because one loop affects the neighboring loop through current conduction. When tw o loops with or without contacts between them affect each other through the magnetic field generated by one of them, they are said to be *magnetically coupled*. - -The transformer is an electrical de vice designed on the basis of the concept of magnetic coupling. It uses magnetically coupled coils to transfer energy from one circuit to another. Transformers are key circuit elements. They are used in power systems for stepping up or stepping down ac voltages or currents. They are used in electronic circuits such as radio and television receivers for such purposes as impedance matching, isolating one part of a circuit from another, and again for stepping up or down ac voltages and currents. - -We will begin with the concept of mutual inductance and introduce the dot convention used for determining the v oltage polarities of inductively coupled components. Based on the notion of mutual inductance, - -we then introduce the circuit element known as the *transformer*. We will consider the linear transformer, the ideal transformer, the ideal autotransformer, and the three-phase transformer. Finally, among their important applications, we look at transformers as isolating and matching devices and their use in power distribution. - -# **13.2** Mutual Inductance - -When two inductors (or coils) are in a close proximity to each other, the magnetic flux caused by current in one coil links with the other coil, thereby inducing v oltage in the latter . This phenomenon is kno wn as *mutual inductance*. - -Let us first consider a single inductor, a coil with *N* turns. When current *i* flows through the coil, a magnetic flux *ϕ* is produced around it (Fig. 13.1). According to Faraday's law, the voltage *v* induced in the coil is proportional to the number of turns *N* and the time rate of change of the magnetic flux *ϕ*; that is, - -$$ -v = N \frac{d\phi}{dt} \tag{13.1} -$$ - -But the flux *ϕ* is produced by current *i* so that any change in *ϕ* is caused by a change in the current. Hence, Eq. (13.1) can be written as - -$$ -v = N \frac{d\phi}{di} \frac{di}{dt} -$$ - (13.2) - -or - -$$ -v = L \frac{di}{dt} -$$ - (13.3) - -which is the voltage-current relationship for the inductor. From Eqs. (13.2) and (13.3), the inductance *L* of the inductor is thus given by - -$$ -L = N \frac{d\phi}{di} -$$ - (13.4) - -This inductance is commonly called *self-inductance,* because it relates the voltage induced in a coil by a time-varying current in the same coil. - -Now consider two coils with self-inductances *L*1 and *L*2 that are in close proximity with each other (Fig. 13.2). Coil 1 has *N*1 turns, while coil 2 has *N*2 turns. F or the sak e of simplicity, assume that the second inductor carries no current. The magnetic flux *ϕ*1 emanating from coil 1 has two components: One component *ϕ*11 links only coil 1, and another component *ϕ*12 links both coils. Hence, - -$$ -\phi_1 = \phi_{11} + \phi_{12} \tag{13.5} -$$ - -Although the tw o coils are ph ysically separated, the y are said to be *magnetically coupled*. Since the entire flux *ϕ*1 links coil 1, the v oltage induced in coil 1 is - -$$ -v_1 = N_1 \frac{d\phi_1}{dt} \tag{13.6} -$$ - -Only flux *ϕ*12 links coil 2, so the voltage induced in coil 2 is - -$$ -v_2 = N_2 \frac{d\phi_{12}}{dt} -$$ - (13.7) - -**Figure 13.1** Magnetic flux produced by a single coil with *N* turns. - -Again, as the fluxes are caused by the current *i*1 flowing in coil 1, Eq. (13.6) can be written as - -$$ -v_1 = N_1 \frac{d\phi_1}{di_1} \frac{di_1}{dt} = L_1 \frac{di_1}{dt} -$$ - (13.8) - -where *L*1 =  *N*1 *dϕ*1∕*di*1 is the self-inductance of coil 1. Similarly, Eq. (13.7) can be written as - -$$ -v_2 = N_2 \frac{d\phi_{12}}{di_1} \frac{di_1}{dt} = M_{21} \frac{di_1}{dt} -$$ - (13.9) - -where - -$$ -M_{21} = N_2 \frac{d\phi_{12}}{di_1} -$$ - (13.10) - -*M*21 is known as the *mutual inductance* of coil 2 with respect to coil 1. Subscript 21 indicates that the inductance *M*21 relates the voltage induced in coil 2 to the current in coil 1. Thus, the open-circuit *mutual voltage* (or induced voltage) across coil 2 is - -$$ -v_2 = M_{21} \frac{di_1}{dt} -$$ - (13.11) - -Suppose we now let current *i*2 flow in coil 2, while coil 1 carries no current (Fig. 13.3). The magnetic flux *ϕ*2 emanating from coil 2 comprises flux *ϕ*22 that links only coil 2 and flux *ϕ*21 that links both coils. Hence, - -$$ -\phi_2 = \phi_{21} + \phi_{22} \tag{13.12} -$$ - -The entire flux *ϕ*2 links coil 2, so the voltage induced in coil 2 is - -$$ -v_2 = N_2 \frac{d\phi_2}{dt} = N_2 \frac{d\phi_2}{di_2} \frac{di_2}{dt} = L_2 \frac{di_2}{dt} -$$ - (13.13) - -where *L*2 = *N*2 *dϕ*2∕*di*2 is the self-inductance of coil 2. Since only flux *ϕ*21 links coil 1, the voltage induced in coil 1 is - -$$ -v_1 = N_1 \frac{d\phi_{21}}{dt} = N_1 \frac{d\phi_{21}}{di_2} \frac{di_2}{dt} = M_{12} \frac{di_2}{dt} -$$ - (13.14) - -where - -$$ -M_{12} = N_1 \frac{d\phi_{21}}{di_2} -$$ - (13.15) - -which is the *mutual inductance* of coil 1 with respect to coil 2. Thus, the open-circuit *mutual voltage* across coil 1 is - -$$ -v_1 = M_{12} \frac{di_2}{dt} -$$ - (13.16) - -We will see in the next section that *M*12 and *M*21 are equal; that is, - -$$ -M_{12} = M_{21} = M \tag{13.17} -$$ - -and we refer to *M* as the mutual inductance between the tw o coils. Like self-inductance *L*, mutual inductance *M* is measured in henrys (H). Keep in mind that mutual coupling only exists when the inductors or coils are in close proximity, and the circuits are dri ven by time-varying sources. We recall that inductors act like short circuits to dc. - -From the two cases in Figs. 13.2 and 13.3, we conclude that mutual inductance results if a v oltage is induced by a time-v arying current in another circuit. It is the property of an inductor to produce a v oltage in reaction to a time-varying current in another inductor near it. Thus, - -Mutual inductance *M*12 of coil 1 with respect to coil 2. - -Mutual inductance is the ability of one inductor to induce a voltage across a neighboring inductor, measured in henrys (H). - -Although mutual inductance *M* is al ways a positi ve quantity, the mutual voltage *M di*∕*dt* may be negative or positive, just like the self-induced voltage *L di*∕*dt*. However, unlike the self-induced *L di*∕*dt*, whose polarity is determined by the reference direction of the current and the reference polarity of the v oltage (according to the passi ve sign convention), the polarity of mutual v oltage *M di*∕*dt* is not easy to determine, because four terminals are involved. The choice of the correct polarity for *M di*∕*dt* is made by examining the orientation or particular way in which both coils are physically wound and applying Lenz's law in conjunction with the right-hand rule. Since it is inconvenient to show the construction details of coils on a circuit schematic, we apply the *dot convention* in circuit analysis. By this convention, a dot is placed in the circuit at one end of each of the two magnetically coupled coils to indicate the direction of the magnetic flux if current enters that dotted terminal of the coil. This is illustrated in Fig. 13.4. Given a circuit, the dots are already placed beside the coils so that we need not bother about how to place them. The dots are used along with the dot con vention to determine the polarity of the mutual voltage. The dot convention is stated as follows: - -If a current enters the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is positive at the dotted terminal of the second coil. - -# Alternatively, - -If a current leaves the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is negative at the dotted terminal of the second coil. - -Thus, the reference polarity of the mutual v oltage depends on the ref erence direction of the inducing current and the dots on the coupled coils. Application of the dot con vention is illustrated in the four pairs of mutually coupled coils in Fig. 13.5. Fo r the coupled coils in Fig. 13.5(a), the sign of the mutual voltage *v*2 is determined by the reference polarity for *v*2 and the direction of *i*1. Since *i*1 enters the dotted terminal of coil 1 and *v*2 is positive at the dotted terminal of coil 2, the mutual voltage is +*M di*1∕*dt*. For the coils in Fig. 13.5(b), the current *i*1 enters - -Illustration of the dot convention. - -**Figure 13.5** Examples illustrating how to apply the dot convention. - -the dotted terminal of coil 1 and *v*2 is ne gative at the dotted terminal of coil 2. Hence, the mutual v oltage is −*M di*1∕*dt*. The same reasoning applies to the coils in Figs. 13.5(c) and 13.5(d). - -Figure 13.6 shows the dot convention for coupled coils in series. For the coils in Fig. 13.6(a), the total inductance is - -$$ -L = L_1 + L_2 + 2M -$$ - (Series-aiding connection) (13.18) - -For the coils in Fig. 13.6(b), - -$$ -L = L_1 + L_2 - 2M -$$ - (Series-opposing connection) (13.19) - -Now that we know how to determine the polarity of the mutual voltage, we are prepared to analyze circuits involving mutual inductance. As the first example, consider the circuit in Fig. 13.7(a). Applying KVL to coil 1 gives - -$$ -v_1 = i_1 R_1 + L_1 \frac{di_1}{dt} + M \frac{di_2}{dt} -$$ - (13.20a) - -For coil 2, KVL gives - -$$ -v_2 = i_2 R_2 + L_2 \frac{di_2}{dt} + M \frac{di_1}{dt} -$$ - (13.20b) - -We can write Eq. (13.20) in the frequency domain as - -$$ -\mathbf{V}_1 = (R_1 + j\omega L_1)\mathbf{I}_1 + j\omega M \mathbf{I}_2 \tag{13.21a} -$$ - -$$ -\mathbf{V}_2 = j\omega M \mathbf{I}_1 + (R_2 + j\omega L_2) \mathbf{I}_2 \tag{13.21b} -$$ - -As a second e xample, consider the circuit in Fig. 13.7(b). We analyze this in the frequency domain. Applying KVL to coil 1, we get - -$$ -\mathbf{V} = (\mathbf{Z}_1 + j\omega L_1)\mathbf{I}_1 - j\omega M\mathbf{I}_2 -$$ - (13.22a) - -For coil 2, KVL yields - -$$ -0 = -j\omega M I_1 + (Z_L + j\omega L_2)I_2 \tag{13.22b} -$$ - -Equations (13.21) and (13.22) are solv ed in the usual manner to deter mine the currents. - -One of the most important things in making sure one solv es problems accurately is to be able to check each step during the solution pro cess and to mak e sure assumptions can be v erified. Too often, solving mutually coupled circuits requires the problem solv er to track tw o or more steps made at once re garding the sign and v alues of the mutually induced voltages. - -# **Figure 13.7** - -Time-domain analysis of a circuit containing coupled coils (a) and frequency-domain analysis of a circuit containing coupled coils (b). - -# **Figure 13.6** - -Dot convention for coils in series; the sign indicates the polarity of the mutual voltage: (a) seriesaiding connection, (b) seriesopposing connection. - -Model that makes analysis of mutually coupled easier to solve. - -Experience has sho wn that if we break the problem into steps of solving for the value and the sign into separate steps, the decisions made are easier to track. We suggest that model (Figure 13.8 (b)) be used when analyzing circuits containing a mutually c oupled circuit shown in Figure 13.8(a): - -Notice that we have not included the signs in the model. The reason for that is that we first determine the value of the induced voltages and then we determine the appropriate signs. Clearly, I1 induces a voltage within the second coil represented by the value *jω*I1 and I2 induces a voltage of *jω*I2 in the first coil. Once we have the values we next use both circuits to find the correct signs for the dependent sources as shown in Figure 13.8(c). - -Since I1 enters *L*1 at the dotted end, it induces a voltage in *L*2 that tries to force a current out of the dotted end of *L*2 which means that the source must have a plus on top and a minus on the bottom as sho wn in Figure 13.8(c). I2 leaves the dotted end of *L*2 which means that it induces a voltage in *L*1 which tries to force a current into the dotted end of *L*1 requiring a dependent source that has a plus on the bottom and a minus on top as shown in Figure 13.8(c). No w all we have to do is to analyze a circuit with two dependent sources. This process allows you to check each of your assumptions. - -At this introductory level we are not concerned with the determination of the mutual inductances of the coils and their dot placements. Lik e *R*, *L*, and *C*, calculation of *M* would involve applying the theory of elect romagnetics to the actual ph ysical properties of the coils. In this te xt, we assume that the mutual inductance and the placement of the dots are the "gi vens'' of the circuit problem, like the circuit components *R*, *L*, and *C*. - -For Example 13.1. - -$$ --12 + (-j4 + j5)\mathbf{I}_1 - j3\mathbf{I}_2 = 0 -$$ - -$$ -jI_1 - j3I_2 = 12 \tag{13.1.1} -$$ - -For loop 2, KVL gives - -$$ --j3\mathbf{I}_1 + (12 + j6)\mathbf{I}_2 = 0 -$$ - -or - -$$ -\mathbf{I}_1 = \frac{(12 + j6)\mathbf{I}_2}{j3} = (2 - j4)\mathbf{I}_2 -$$ - (13.1.2) - -Substituting this in Eq. (13.1.1), we get - -( *j*2 + 4 − *j*3)**I**2 = (4 − *j*)**I**2 = 12 - -or - -$$ -\mathbf{I}_2 = \frac{12}{4 - j} = 2.91 \underline{ / 14.04^{\circ}} A \tag{13.1.3} -$$ - -From Eqs. (13.1.2) and (13.1.3), - -$$ -\mathbf{I}_1 = (2 - j4)\mathbf{I}_2 = (4.472 \angle -63.43^\circ)(2.91 \angle 14.04^\circ) -$$ - -= 13.01 \angle -49.39^\circ A - -Practice Problem 13.1 Determine the voltage **V***o* in the circuit of Fig. 13.10. - -**Figure 13.10** For Practice Prob. 13.1. - -**Answer:** 12⧸ −45° V. - -Example 13.2 Calculate the mesh currents in the circuit of Fig. 13.11. - -**Figure 13.11** For Example 13.2. - -# **Solution:** - -The key to analyzing a magnetically coupled circuit is knowing the polarity of the mutual voltage. We need to apply the dot rule. In Fig. 13.11, suppose coil 1 is the one whose reactance is 6 Ω, and coil 2 is the one whose reactance is 8 Ω. To figure out the polarity of the mutual voltage in coil 1 due to current **I**2, we observe that **I**2 leaves the dotted terminal of coil 2. Since we are applying KVL in the clockwise direction, it implies that the mutual voltage is negative, that is, −*j*2**I**2. - -Alternatively, it might be best to figure out the mutual voltage by redrawing the rele vant portion of the circuit, as sho wn in Fig. 13.12, where it becomes clear that the mutual voltage is **V**1 = −2*j* **I**2. - -Thus, for mesh 1 in Fig. 13.11, KVL gives - -$$ --100 + \mathbf{I}_1(4 - j3 + j6) - j6\mathbf{I}_2 - j2\mathbf{I}_2 = 0 -$$ - -or - -$$ -100 = (4+j3)\mathbf{I}_1 - j8\mathbf{I}_2 \tag{13.2.1} -$$ - -Similarly, to figure out the mutual voltage in coil 2 due to current **I**1, consider the relevant portion of the circuit, as shown in Fig. 13.12. Applying the dot convention gives the mutual voltage as **V**2 = −2*j***I**1. Also, current **I**2 sees the two coupled coils in series in Fig. 13.11; since it leaves the dotted terminals in both coils, Eq. (13.18) applies. Therefore, for mesh 2 in Fig. 13.11, KVL gives - -$$ -0 = -2jI_1 - j6I_1 + (j6 + j8 + j2 \times 2 + 5)I_2 -$$ - -or - -$$ -0 = -j8I_1 + (5+j18)I_2 \tag{13.2.2} -$$ - -Putting Eqs. (13.2.1) and (13.2.2) in matrix form, we get - -$$ -\begin{bmatrix} 100 \\ 0 \end{bmatrix} = \begin{bmatrix} 4+j3 & -j8 \\ -j8 & 5+j18 \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} -$$ - -The determinants are - -$$ -\Delta = \begin{vmatrix} 4+j3 & -j8 \\ -j8 & 5+j18 \end{vmatrix} = 30 + j87 -$$ - -\n -$$ -\Delta_1 = \begin{vmatrix} 100 & -j8 \\ 0 & 5+j18 \end{vmatrix} = 100(5+j18) -$$ - -\n -$$ -\Delta_2 = \begin{vmatrix} 4+j3 & 100 \\ -j8 & 0 \end{vmatrix} = j800 -$$ - -Thus, we obtain the mesh currents as - -Thus, we obtain the mesh currents as -\n -$$ -\mathbf{I}_1 = \frac{\Delta_1}{\Delta} = \frac{100(5 + j18)}{30 + j87} = \frac{1,868.2/74.5^{\circ}}{92.03/71^{\circ}} = 20.3/3.5^{\circ} \text{ A} -$$ -\n -$$ -\mathbf{I}_2 = \frac{\Delta_2}{\Delta} = \frac{j800}{30 + j87} = \frac{800/90^{\circ}}{92.03/71^{\circ}} = 8.693/19^{\circ} \text{ A} -$$ - -Determine the phasor currents **I**1 and **I**2 in the circuit of Fig. 13.13. Practice Problem 13.2 - -**Figure 13.13** For Practice Prob. 13.2. - -**Answer:** I1 = 17.889⧸86.57° A, I2 = 26.83⧸86.57° A. - -**Figure 13.12** Model for Example 13.2 showing the polarity of the induced voltages. - -# **13.3** Energy in a Coupled Circuit - -In Chapter 6, we saw that the energy stored in an inductor is given by - -$$ -w = \frac{1}{2} L i^2 -$$ - (13.23) - -We now want to determine the energy stored in magnetically coupled coils. - -Consider the circuit in Fig. 13.14. We assume that currents *i*1 and *i*2 are zero initially, so that the ener gy stored in the coils is zero. If we let *i*1 increase from zero to *I*1 while maintaining *i*2 = 0, the power in coil 1 is - -$$ -p_1(t) = v_1 i_1 = i_1 L_1 \frac{di_1}{dt} -$$ - (13.24) - -and the energy stored in the circuit is - -$$ -w_1 = \int p_1 dt = L_1 \int_0^{I_1} i_1 dt_1 = \frac{1}{2} L_1 I_1^2 -$$ - (13.25) - -If we now maintain *i*1 = *I*1 and increase *i*2 from zero to *I*2, the mutual voltage induced in coil 1 is *M*12 *di*2∕*dt*, while the mutual v oltage induced in coil 2 is zero, since *i*1 does not change. The power in the coils is now - -$$ -p_2(t) = i_1 M_{12} \frac{di_2}{dt} + i_2 v_2 = I_1 M_{12} \frac{di_2}{dt} + i_2 L_2 \frac{di_2}{dt} -$$ - (13.26) - -and the energy stored in the circuit is - -$$ -w_2 = \int p_2 dt = M_{12} I_1 \int_0^{I_2} di_2 + L_2 \int_0^{I_2} i_2 di_2 -$$ -$$ -= M_{12} I_1 I_2 + \frac{1}{2} L_2 I_2^2 -$$ -(13.27) - -The total ener gy stored in the coils when both *i*1 and *i*2 have reached constant values is - -$$ -w = w_1 + w_2 = \frac{1}{2} L_1 I_1^2 + \frac{1}{2} L_2 I_2^2 + M_{12} I_1 I_2 -$$ - (13.28) - -If we reverse the order by which the currents reach their final values, that is, if we first increase *i*2 from zero to *I*2 and later increase *i*1 from zero to *I*1, the total energy stored in the coils is - -$$ -w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + M_{21}I_1I_2 -$$ -\n(13.29) - -Because the total energy stored should be the same regardless of how we reach the final conditions, comparing Eqs. (13.28) and (13.29) leads us to conclude that - -$$ -M_{12} = M_{21} = M \tag{13.30a} -$$ - -and - -$$ -w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 + MI_1I_2 -$$ - (13.30b) - -This equation was derived based on the assumption that the coil currents both entered the dotted terminals. If one current enters one dotted - -**Figure 13.14** The circuit for deriving energy stored in a coupled circuit. - -terminal while the other current leaves the other dotted terminal, the mutual voltage is negative, so that the mutual energy *MI*1*I*2 is also negative. In that case, - -$$ -w = \frac{1}{2}L_1I_1^2 + \frac{1}{2}L_2I_2^2 - MI_1I_2 -$$ - (13.31) - -Also, because *I*1 and *I*2 are arbitrary v alues, they may be replaced by *i*1 and *i*2, which gives the instantaneous ener gy stored in the circuit the general expression - -$$ -w = \frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 \pm Mi_1i_2 -$$ - (13.32) - -The positive sign is selected for the mutual term if both currents enter or leave the dotted terminals of the coils; the ne gative sign is selected otherwise. - -We will now establish an upper limit for the mutual inductance M. The energy stored in the circuit cannot be negative because the circuit is passive. This means that the quantity 1∕2*L*1 *i* 1 2 + 1∕2*L*2 *i* 2 2 − *Mi*1*i*2 must be greater than or equal to zero: - -$$ -\frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 - Mi_1i_2 \ge 0 -$$ -\n(13.33) - -To complete the square, we both add and subtract the term *i*1*i*2( √ *L*1*L*2 on the right-hand side of Eq. (13.33) and obtain - -$$ -\frac{1}{2}(i_1\sqrt{L_1} - i_2\sqrt{L_2})^2 + i_1i_2(\sqrt{L_1L_2} - M) \ge 0 -$$ - (13.34) - -The squared term is ne ver negative; at its least it is zero. Therefore, the sec ond term on the right-hand side of Eq. (13.34) must be greater than zero; that is, - -√ - -$$ -\overline{L_1 L_2} - M \ge 0 -$$ - -$$ -M \le \sqrt{L_1 L_2} -$$ - (13.35) - -Thus, the mutual inductance cannot be greater than the geometric mean of the self-inductances of the coils. The e xtent to which the mutual inductance *M* approaches the upper limit is specified by the *coefficient of coupling k*, given by - -$$ -k = \frac{M}{\sqrt{L_1 L_2}}\tag{13.36} -$$ - -$$ -M = k\sqrt{L_1 L_2} \tag{13.37} -$$ - -where 0 ≤ *k* ≤ 1 or equivalently 0 ≤ *M* ≤  √ \_\_\_\_ *L*1*L*2 . The coupling coefficient is the fraction of the total flux emanating from one coil that links the other coil. For example, in Fig. 13.2, - -$$ -k = \frac{\phi_{12}}{\phi_1} = \frac{\phi_{12}}{\phi_{11} + \phi_{12}} -$$ - (13.38) - -or - -or - -and in Fig. 13.3, - -# **Figure 13.15** - -Windings: (a) loosely coupled, (b) tightly coupled; cutaway view demonstrates both windings. - -$$ -k = \frac{\phi_{21}}{\phi_2} = \frac{\phi_{21}}{\phi_{21} + \phi_{22}} -$$ - (13.39) - -If the entire flux produced by one coil links another coil, then *k* = 1 and we have 100 percent coupling, or the coils are said to be *perfectly coupled*. For *k* < 0.5, coils are said to be *loosely coupled*; and for *k* > 0.5, they are said to be *tightly coupled*. Thus, - -The coupling coefficient k is a measure of the magnetic coupling between two coils; 0 ≤ k ≤ 1. - -We expect *k* to depend on the closeness of the two coils, their core, their orientation, and their windings. Figure 13.15 sho ws loosely coupled windings and tightly coupled windings. The air -core trans formers used in radio frequenc y circuits are loosely coupled, whereas iron-core transformers used in po wer systems are tightly coupled. The linear transformers discussed in Section 3.4 are mostly air -core; the ideal transformers discussed in Sections 13.5 and 13.6 are principally iron-core. - -**Figure 13.16** For Example 13.3. - -Consider the circuit in Fig. 13.16. Determine the coupling coef ficient. Calculate the ener gy stored in the coupled inductors at time *t* = 1 s if *v* = 60 cos(4*t* + 30°) V. - -# **Solution:** - -The coupling coefficient is - -$$ -k = \frac{M}{\sqrt{L_1 L_2}} = \frac{2.5}{\sqrt{20}} = 0.56 -$$ - -indicating that the inductors are tightly coupled. To find the energy stored, we need to calculate the current. To find the current, we need to obtain the frequency-domain equivalent of the circuit. - -60 cos(4*t* + 30°) -$$ -\Rightarrow -$$ - 60/30°, $\omega = 4$ rad/s -\n5 H $\Rightarrow$ $j\omega L_1 = j20 \Omega$ -\n2.5 H $\Rightarrow$ $j\omega M = j10 \Omega$ -\n4 H $\Rightarrow$ $j\omega L_2 = j16 \Omega$ -\n $\frac{1}{16}F \Rightarrow \frac{1}{j\omega C} = -j4 \Omega$ - -The frequency-domain equivalent is shown in Fig. 13.17. We now apply mesh analysis. For mesh 1, - -$$ -(10 + j20)\mathbf{I}_1 + j10\mathbf{I}_2 = 60/30^{\circ} -$$ - (13.3.1) - -For mesh 2, - -$$ -j10\mathbf{I}_1 + (j16 - j4)\mathbf{I}_2 = 0 -$$ - -$$ -\overline{a} -$$ - -$$ -I_1 = -1.2I_2 \tag{13.3.2} -$$ - -Substituting this into Eq. (13.3.1) yields - -$$ -I_2(-12 - j14) = 60/30^{\circ} \qquad \Rightarrow \qquad I_2 = 3.254/160.6^{\circ} \text{ A} -$$ - -and - -$$ -I_1 = -1.2I_2 = 3.905 \div 19.4^{\circ} -$$ - A - -In the time-domain, - -$$ -i_1 = 3.905 \cos(4t - 19.4^\circ), -$$ - $i_2 = 3.254 \cos(4t + 160.6^\circ)$ - -At time *t* = 1 s, 4*t* = 4 rad = 229.2°, and - -$$ -i_1 = 3.905 \cos(229.2^\circ - 19.4^\circ) = -3.389 \text{ A} -$$ - -$$ -i_2 = 3.254 \cos(229.2^\circ + 160.6^\circ) = 2.824 \text{ A} -$$ - -The total energy stored in the coupled inductors is - -$$ -w = \frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 + Mi_1i_2 -$$ - -= $\frac{1}{2}(5)(-3.389)^2 + \frac{1}{2}(4)(2.824)^2 + 2.5(-3.389)(2.824) = 20.73 \text{ J}$ - -**Figure 13.17** - -Frequency-domain equivalent of the circuit in Fig. 13.16. - -For the circuit in Fig. 13.18, determine the coupling coefficient and the energy stored in the coupled inductors at *t* = 1.5 s. Practice Problem 13.3 - -**Answer:** 0.7071, 246.2 J. - -# **13.4** Linear Transformers - -Here we introduce the transformer as a ne w circuit element. A transformer is a magnetic device that takes advantage of the phenomenon of mutual inductance. - -A transformer is generally a four-terminal device comprising two (or more) magnetically coupled coils. - -As shown in Fig. 13.19, the coil that is directly connected to the voltage source is called the *primary winding*. The coil connected to the load is called the *secondary winding*. The resistances *R*1 and *R*2 are included to account for the losses (po wer dissipation) in the coils. The transformer is said to be *linear* if the coils are w ound on a magnetically linear material—a material for which the magnetic permeability is constant. Such materials include air, plastic, Bakelite, and wood. In fact, most materials are magnetically linear. Linear transformers are sometimes called *air-core transformers*, although not all of them are necessarily air -core. They are used in radio and TV sets. Figure 13.20 portrays different types of transformers. - -# **Figure 13.20** - -Different types of transformers: (a) copper wound dry power transformer, (b) audio transformers. (a) © Electric Service Co., Cincinnati Ohio, (b) © Jensen Transformers, Inc., Chatsworth, CA - -A linear transformer may also be regarded as one whose flux is proportional to the currents in its windings. - -It should be noted that the result in Eq. (13.41) or (13.42) is not affected by the location of the dots on the transformer, because the same result is produced when *M* is replaced by −*M*. - -The little bit of experience gained in Sections 13.2 and 13.3 in analyzing magnetically coupled circuits is enough to con vince anyone that analyzing these circuits is not as easy as circuits in previous chapters. For this reason, it is sometimes convenient to replace a magnetically coupled circuit by an equi valent circuit with no magnetic coupling. We want to replace the linear transformer in Fig. 13.21 by an equi valent T or Π circuit, a circuit that would have no mutual inductance. - -The v oltage-current relationships for the primary and secondary coils give the matrix equation - -$$ -\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} j\omega L_1 & j\omega M \\ j\omega M & j\omega L_2 \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} -$$ - (13.43) - -By matrix inversion, this can be written as - -version, this can be written as -\n -$$ -\begin{bmatrix}\nI_1 \\ -I_2\n\end{bmatrix} = \begin{bmatrix}\n\frac{L_2}{j\omega(L_1L_2 - M^2)} & \frac{-M}{j\omega(L_1L_2 - M^2)} \\ -\frac{-M}{j\omega(L_1L_2 - M^2)} & \frac{L_1}{j\omega(L_1L_2 - M^2)}\n\end{bmatrix} \begin{bmatrix}\nV_1 \\ -V_2\n\end{bmatrix} -$$ -\n(13.44) - -Our goal is to match Eqs. (13.43) and (13.44) with the corresponding equations for the T and Π networks. - -For the T (or Y) network of Fig. 13.22, mesh analysis pro vides the terminal equations as - -$$ -\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} j\omega(L_a + L_c) & j\omega L_c \\ j\omega L_c & j\omega(L_b + L_c) \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} -$$ - (13.45) - -# **Figure 13.21** Determining the equivalent circuit of a linear transformer. - -**Figure 13.22** An equivalent T circuit. - -If the circuits in Figs. 13.21 and 13.22 are equivalents, Eqs. (13.43) and (13.45) must be identical. Equating terms in the impedance matrices of Eqs. (13.43) and (13.45) leads to - -$$ -L_a = L_1 - M, \qquad L_b = L_2 - M, \qquad L_c = M \qquad (13.46) -$$ - -For the Π (or Δ) network in Fig. 13.23, nodal analysis gi ves the terminal equations as - -$$ -\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{j\omega L_A} + \frac{1}{j\omega L_C} & -\frac{1}{j\omega L_C} \\ -\frac{1}{j\omega L_C} & \frac{1}{j\omega L_B} + \frac{1}{j\omega L_C} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} -$$ -(13.47) - -Equating terms in admittance matrices of Eqs. (13.44) and (13.47), we obtain - -$$ -L_{A} = \frac{L_{1}L_{2} - M^{2}}{L_{2} - M}, \qquad L_{B} = \frac{L_{1}L_{2} - M^{2}}{L_{1} - M} -$$ -$$ -L_{C} = \frac{L_{1}L_{2} - M^{2}}{M} -$$ -(13.48) - -Note that in Figs. 13.22 and 13.23, the inductors are not magnetically coupled. Also note that changing the locations of the dots in Fig. 13.21 can cause *M* to become − *M*. As Example 13.6 illustrates, a ne gative value of *M* is ph ysically unrealizable b ut the equi valent model is still mathematically valid. - -In the circuit of Fig. 13.24, calculate the input impedance and current **I**1. Take **Z**1 = 60 − *j*100 Ω, **Z**2 = 30 + *j*40 Ω, and **Z***L* = 80 + *j*60 Ω. - -# **Figure 13.24** - -# **Solution:** - -From Eq. (13.41), - -$$ -Zin = Z1 + j20 + \frac{(5)2}{j40 + Z2 + ZL} -$$ - -= 60 - j100 + j20 + $\frac{25}{110 + j140}$ -= 60 - j80 + 0.14/ $-51.84^{\circ}$ -= 60.09 - j80.11 = 100.14/ $-53.1^{\circ}$ Ω - -**Figure 13.23** An equivalent Π circuit. - -Example 13.4 - -For Example 13.4. - -Thus, - -$$ -\mathbf{I}_1 = \frac{\mathbf{V}}{\mathbf{Z}_{in}} = \frac{50/60^{\circ}}{100.14 / -53.1^{\circ}} = 0.5 / 113.1^{\circ} \text{ A} -$$ - -Find the input impedance of the circuit in Fig. 13.25 and the current from the voltage source. Practice Problem 13.4 - -**Answer:** 8.58⧸ 58.05° Ω, 4.662⧸− 58.05° A. - -2 H - -Determine the T-equivalent circuit of the linear transformer in Fig. 13.26(a). - -> (a) 10 H 4 H a b c d (b) a b c d 2 H **I**1 **I**2 8 H 2 H - -# **Solution:** - -Given that *L*1 = 10, *L*2 = 4, and *M* = 2, the T-equivalent network has the following parameters: - -$$ -L_a = L_1 - M = 10 - 2 = 8 -$$ -H -$$ -L_b = L_2 - M = 4 - 2 = 2 -$$ -H, -$$ -L_c = M = 2 -$$ -H - -The T-equivalent circuit is shown in Fig. 13.26(b). We have assumed that reference directions for currents and voltage polarities in the primary and secondary windings conform to those in Fig. 13.21. Otherwise, we may need to replace *M* with −*M*, as Example 13.6 illustrates. - -Example 13.5 - -Practice Problem 13.5 - -For the linear transformer in Fig. 13.26(a), find the Π equivalent network. - -**Answer:** *LA* = 18 H, *LB* = 4.5 H, *LC* = 18 H. - -| Example 13.6 | | | | -|--------------|--|--|--| -|--------------|--|--|--| - -Solve for **I**1, **I**2, and **V***o* in Fig. 13.27 (the same circuit as for Practice Prob. 13.1) using the T-equivalent circuit for the linear transformer. Example 13.6 - -# **Solution:** - -Notice that the circuit in Fig. 13.27 is the same as that in Fig. 13.10 except that the reference direction for current **I**2 has been reversed, just to make the reference directions for the currents for the magnetically coupled coils conform with those in Fig. 13.21. - -We need to replace the magnetically coupled coils with the T-equivalent circuit. The relevant portion of the circuit in Fig. 13.27 is shown in Fig. 13.28(a). Comparing Fig. 13.28(a) with Fig. 13.21 sho ws that there are tw o differences. First, due to the current reference direc tions and voltage polarities, we need to replace *M* by −*M* to make Fig. 13.28(a) conform with Fig. 13.21. Second, the circuit in Fig. 13.21 is in the time-domain, whereas the circuit in Fig. 13.28(a) is in the frequencydomain. The difference is the factor *jω*; that is, *L* in Fig. 13.21 has been replaced with *jωL* and *M* with *jωM*. Since *ω* is not specified, we can assume *ω* = 1 rad/s or any other value; it really does not matter. With these two differences in mind, - -$$ -L_a = L_1 - (-M) = 8 + 1 = 9 \text{ H} -$$ - -$$ -L_b = L_2 - (-M) = 5 + 1 = 6 \text{ H}, \qquad L_c = -M = -1 \text{ H} -$$ - -Thus, the T-equivalent circuit for the coupled coils is as shown in Fig. 13.28(b). - -Inserting the T-equivalent circuit in Fig. 13.28(b) to replace the two coils in Fig. 13.27 gives the equivalent circuit in Fig. 13.29, which can be solved using nodal or mesh analysis. Applying mesh analysis, we obtain - -$$ -j6 = I_1(4 + j9 - j1) + I_2(-j1) -$$ - (13.6.1) - -and - -$$ -0 = I1(-j1) + I2(10 + j6 - j1) -$$ - (13.6.2) - -From Eq. (13.6.2), - -$$ -\mathbf{I}_1 = \frac{(10 + j5)}{j} \mathbf{I}_2 = (5 - j10)\mathbf{I}_2 -$$ - (13.6.3) - -# **Figure 13.28** - -For Example 13.6: (a) circuit for coupled coils of Fig. 13.27, (b) T-equivalent circuit. - -For Example 13.6. - -Substituting Eq. (13.6.3) into Eq. (13.6.1) gives - -$$ -j6 = (4 + j8)(5 - j10)\mathbf{I}_2 - j\mathbf{I}_2 = (100 - j)\mathbf{I}_2 \simeq 100\mathbf{I}_2 -$$ - -Since 100 is very large compared with 1, the imaginary part of (100 − *j*) can be ignored so that 100 − *j* ≃ 100. Hence, - -$$ -I_2 = \frac{j6}{100} = j0.06 = 0.06 / 90^{\circ} A -$$ - -From Eq. (13.6.3), - -$$ -I_1 = (5 - j10)j0.06 = 0.6 + j0.3 -$$ - A - -and - -$$ -\mathbf{V}_o = -10\mathbf{I}_2 = -j0.6 = 0.6 \div 90^\circ \text{ V} -$$ - -This agrees with the answer to Practice Prob. 13.1. Of course, the direction of **I**2 in Fig. 13.10 is opposite to that in Fig. 13.27. This will not affect **V***o*, but the value of **I**2 in this example is the negative of that of **I**2 in Practice Prob. 13.1. The advantage of using the T-equivalent model for the magnetically coupled coils is that in Fig. 13.29 we do not need to bother with the dot on the coupled coils. - -Solve the problem in Example 13.1 (see Fig. 13.9) using the T-equivalent model for the magnetically coupled coils. - -Practice Problem 13.6 - -**Answer:** 13⧸ −49.4° A, 2.91⧸ 14.04° A. - -# **13.5** Ideal Transformers - -An ideal transformer is one with perfect coupling ( *k* =1). It consists of two (or more) coils with a lar ge number of turns w ound on a common core of high permeability. Because of this high permeability of the core, the flux links all the turns of both coils, thereby resulting in a perfect coupling. - -To see ho w an ideal transformer is the limiting case of tw o coupled inductors where the inductances approach infinity and the coupling is perfect, let us reexamine the circuit in Fig. 13.14. In the frequency domain, - -$$ -V_1 = j\omega L_1 I_1 + j\omega M I_2 -$$ -\n -$$ -V_2 = j\omega M I_1 + j\omega L_2 I_2 -$$ -\n(13.49b) - -\_\_\_\_\_ - -From Eq. (13.49a), **I**1 = (**V**1 − *jωM***I**2)∕*jωL*1 (we could have also used this equation to develop the current ratios instead of using the conservation of power which we will do shortly). Substituting this in Eq. (13.49b) gives - -$$ -\mathbf{V}_2 = j\omega L_2 \mathbf{I}_2 + \frac{M\mathbf{V}_1}{L_1} - \frac{j\omega M^2 \mathbf{I}_2}{L_1} -$$ - -But *M* = √ \_\_\_\_ *L*1*L*2 for perfect coupling (*k* = 1). Hence, - -$$ -\mathbf{V}_2 = j\omega L_2 \mathbf{I}_2 + \frac{\sqrt{L_1 L_2} \mathbf{V}_1}{L_1} - \frac{j\omega L_1 L_2 \mathbf{I}_2}{L_1} = \sqrt{\frac{L_2}{L_1}} \mathbf{V}_1 = n\mathbf{V}_1 -$$ - -where *n* =  √ *L*2∕*L*1 and is called the *turns ratio.* As *L*1, *L*2, *M* → ∞ such that *n* remains the same, the coupled coils become an ideal transformer. A transformer is said to be ideal if it has the following properties: - -- 1. Coils have very large reactances (*L*1, *L*2, *M* → ∞). -- 2. Coupling coefficient is equal to unity (*k* = 1). -- 3. Primary and secondary coils are lossless (*R*1 = 0 = *R*2). - -An ideal transformer is a unity-coupled, lossless transformer in which the primary and secondary coils have infinite self-inductances. - -Iron-core transformers are close approximations to ideal transformers. These are used in power systems and electronics. - -as shown in Fig. 13.31, the same magnetic flux *ϕ* goes through both windings. According to F araday's law, the v oltage across the primary winding is - -$$ -v_1 = N_1 \frac{d\phi}{dt} \tag{13.50a} -$$ - -while that across the secondary winding is - -$$ -v_2 = N_2 \frac{d\phi}{dt} \tag{13.50b} -$$ - -Dividing Eq. (13.50b) by Eq. (13.50a), we get - -$$ -\mathsf{w}\mathsf{c}\mathsf{g}\mathsf{c}\mathsf{t} -$$ - -$$ -\frac{v_2}{v_1} = \frac{N_2}{N_1} = n \tag{13.51} -$$ - -where *n* is, again, the *turns ratio* or *transformation ratio*. We can use the phasor voltages **V**1 and **V**2 rather than the instantaneous values *v*1 and *v*2. Thus, Eq. (13.51) may be written as - -$$ -\frac{\mathbf{V}_2}{\mathbf{V}_1} = \frac{N_2}{N_1} = \mathbf{0} -$$ - (13.52) - -**Figure 13.30** (a) Ideal transformer, (b) circuit symbol for an ideal transformer. - -**Figure 13.31** Relating primary and secondary quantities in an ideal transformer. - -For the reason of po wer conservation, the ener gy supplied to the pri mary must equal the ener gy absorbed by the secondary , since there are no losses in an ideal transformer. This implies that - -$$ -\sum_{i=1}^{n} v_i = v_1 i_1 = v_2 i_2 \quad \text{(power conserved)} \quad (13.53) -$$ - -In phasor form, Eq. (13.53) in conjunction with Eq. (13.52) becomes - -$$ -\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{\mathbf{V}_2}{\mathbf{V}_1} = \mathbf{\hat{u}} -$$ - (13.54) - -$$ -\frac{\mathbf{I}_2}{\mathbf{I}_1} = \frac{N_1}{N_2} = \frac{1}{n} -$$ - (13.55) - -When *n* = 1, we generally call the transformer an *isolation transformer*. The reason will become ob vious in Section 13.9.1. If *n* > 1, we ha ve a *step-up transformer*, as the voltage is increased from primary to secondary (**V**2 > **V**1). On the other hand, if *n* < 1, the transformer is a *step-down transformer*, since the v oltage is decreased from primary to secondary - -A step-down transformer is one whose secondary voltage is less than its primary voltage. - -A step-up transformer is one whose secondary voltage is greater than its primary voltage. - -The ratings of transformers are usually specified as *V*1∕*V*2. A transformer with rating 2400∕120 V should have 2400 V on the primary and 120 in the secondary (i.e., a step-down transformer). Keep in mind that the voltage ratings are in rms. - -Power companies often generate at some convenient voltage and use a step-up transformer to increase the v oltage so that the po wer can be transmitted at very high voltage and low current over transmission lines, resulting in significant cost savings. Near residential consumer premises, step-down transformers are used to bring the voltage down to 120 V. Section 13.9.3 will elaborate on this. - -It is important that we know how to get the proper polarity of the voltages and the direction of the currents for the transformer in Fig. 13.31. If the polarity of **V**1 or **V**2 or the direction of **I**1 or **I**2 is changed, n in Eqs. (13.51) to (13.55) may need to be replaced by − *n*. The two simple rules to follow are: - -- 1. If **V**1 and **V**2 are *both* positive or both negative at the dotted termi nals, use *+n* in Eq. (13.52). Otherwise, use −*n*. -- 2. If **I**1 and **I**2 *both* enter into or both lea ve the dotted terminals, use −*n* in Eq. (13.55). Otherwise, use *+n*. - -The rules are demonstrated with the four circuits in Fig. 13.32. - -# **Figure 13.32** Typical circuits illustrating proper voltage polarities and current directions in an ideal transformer. - -Using Eqs. (13.52) and (13.55), we can al ways express **V**1 in terms of **V**2 and **I**1 in terms of **I**2, or vice versa: - -$$ -V_1 = \frac{V_2}{n} -$$ - or $V_2 = nV_1$ (13.56) - -$$ -\mathbf{I}_1 = n\mathbf{I}_2 \qquad \text{or} \qquad \mathbf{I}_2 = \frac{\mathbf{I}_1}{n} \tag{13.57} -$$ - -The complex power in the primary winding is - -$$ -\hat{\mathbf{S}}_1 = \mathbf{V}_1 \mathbf{I}_1^* = \frac{\mathbf{V}_2}{n} (n \mathbf{I}_2)^* = \mathbf{V}_2 \mathbf{I}_2^* = \mathbf{S}_2 \tag{13.58} -$$ - -The input impedance as seen by the source in Fig. 13.31 is found from - -$$ -Z_{\text{in}} = \frac{V_1}{I_1} = \frac{1}{n^2} \frac{V_2}{I_2} -$$ - (13.59) - -$$ -V_2/I_2 = Z_L -$$ -, so that - -$$ -\mathbf{Z}_{\text{in}} = \frac{\mathbf{Z}_L}{n^2} -$$ - (13.60) - -The input impedance is also called the *reflected impedance,* inasmuch - -as it appears as if the load impedance is reflected to the primary side. This ability of the transformer to transform a gi ven impedance into another impedance provides us a means of *impedance matching* to ensure maximum power transfer. The idea of impedance matching is v ery useful in practice and will be discussed more in Section 13.9.2. - -In analyzing a circuit containing an ideal transformer , it is com mon practice to eliminate the transformer by reflecting impedances and sources from one side of the transformer to the other . In the circuit of Fig. 13.33, suppose we want to reflect the secondary side of the circuit to the primary side. We find the Thevenin equi valent of the circuit to the right of the terminals *a*-*b*. We obtain **V**Th as the open-circuit voltage at terminals *a*-*b*, as shown in Fig. 13.34(a). - -**Figure 13.33** Ideal transformer circuit whose equivalent circuits are to be found. - -Notice that an ideal transformer reflects an impedance as the square of the turns ratio. - -# **Figure 13.34** - -n2 - -by n. - -(a) Obtaining **V**Th for the circuit in Fig. 13.33, (b) obtaining **Z**Th for the circuit in Fig. 13.33. - -Because terminals *a*-*b* are open, **I**1 = 0 = **I**2 so that **V**2 = **V***s*2. Hence, - -$$ -V_{\text{Th}} = V_1 = \frac{V_2}{n} = \frac{(V_{s2})}{n} -$$ - (13.61) - -$$ -I_1 = nI_2 -$$ - and $V_1 = V_2/n$ , so that - -$$ -Z_{\text{Th}} = \frac{V_1}{I_1} = \frac{V_2/n}{nI_2} = \frac{Z_2}{n^2} \qquad V_2 = Z_2 I_2 \tag{13.62} -$$ - -**Figure 13.35** - -by reflecting the secondary circuit to the primary side. - -Equivalent circuit for Fig. 13.33 obtained - -The general rule for eliminating the transformer and reflecting the second- - -n2**Z**1 **Z**2 n**V**s1 **V**s2 c d **V**2 + ‒ ‒ + ‒ - -by reflecting the primary circuit to the secondary side. - -Equivalent circuit for Fig. 13.33 obtained - -According to Eq. (13.58), the po wer remains the same, whether calcu lated on the primary or the secondary side. But realize that this reflection approach only applies if there are no e xternal connections between the primary and secondary windings. When we have external connections between the primary and secondary windings, we simply use regular mesh and nodal analysis. Examples of circuits where there are external connections between the primary and secondary windings are in Figs. 13.39 and 13.40. Also note that if the locations of the dots in Fig. 13.33 are changed, we might have to replace *n* by −*n* in order to obey the dot rule, illustrated in Fig. 13.32. - -, multiply the primary voltage by n, and divide the primary current - -Example 13.7 - -An ideal transformer is rated at 2400 ∕120 V, 9.6 kVA, and has 50 turns on the secondary side. Calculate: (a) the turns ratio, (b) the number of turns on the primary side, and (c) the current ratings for the primary and secondary windings. - -# **Solution:** - -(a) This is a step-down transformer, since *V*1 = 2,400 V > *V*2 = 120 V. - -$$ -n = \frac{V_2}{V_1} = \frac{120}{2,400} = 0.05 -$$ - -⇒ 0.05 =  \_\_\_ 50 - -*N*1 - -(b) - -or - -$$ -N_1 = \frac{50}{0.05} = 1,000 \text{ turns} -$$ - -(c) -$$ -S = V_1 I_1 = V_2 I_2 = 9.6 -$$ - kVA. Hence, - -*n* = \_\_\_ *N*2 *N*1 - -$$ -I_1 = \frac{9,600}{V_1} = \frac{9,600}{2,400} = 4 \text{ A} -$$ -$$ -I_2 = \frac{9,600}{V_2} = \frac{9,600}{120} = 80 \text{ A} \qquad \text{or} \qquad I_2 = \frac{I_1}{n} = \frac{4}{0.05} = 80 \text{ A} -$$ - -Practice Problem 13.7 The primary current to an ideal transformer rated at 2200∕110 V is 25 A. Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current. - -**Answer:** (a) 1∕20, (b) 55 kVA, (c) 500 A. - -For the ideal transformer circuit of Fig. 13.37, find: (a) the source current **I**1, (b) the output voltage **V***o*, and (c) the complex power supplied by the source. Example 13.8 - -# **Solution:** - -(a) The 20-Ω impedance can be reflected to the primary side and we get - -$$ -Z_R = \frac{20}{n^2} = \frac{20}{4} = 5 \ \Omega -$$ - -Thus, - -$$ -\mathbf{Z}_{in} = 4 - j6 + \mathbf{Z}_R = 9 - j6 = 10.82 \underline{/ -33.69^\circ} \,\Omega -$$ -\n -$$ -\mathbf{I}_1 = \frac{120 \underline{/ 0^\circ}}{\mathbf{Z}_{in}} = \frac{120 \underline{/ 0^\circ}}{10.82 \underline{/ -33.69^\circ}} = 11.09 \underline{/ 33.69^\circ} \,\text{A} -$$ - -(b) Because both **I**1 and **I**2 leave the dotted terminals, - -$$ -\mathbf{I}_2 = -\frac{1}{n}\mathbf{I}_1 = -5.545/33.69^\circ \text{ A} -$$ - -$$ -\mathbf{V}_o = 20\mathbf{I}_2 = 110.9/213.69^\circ \text{ V} -$$ - -(c) The complex power supplied is - -$$ -S = V_s I_1^* = (120/0°)(11.09/-33.69°) = 1,330.8/-33.69° VA -$$ - -In the ideal transformer circuit of Fig. 13.38, find **V***o* and the complex power supplied by the source. Practice Problem 13.8 - -**Answer:** 429.4⧸ 116.57° V, 17.174⧸ −26.57° kVA. - -Calculate the power supplied to the 10-Ω resistor in the ideal transformer circuit of Fig. 13.39. - -Example 13.9 - -# **Solution:** - -Reflection to the secondary or primary side cannot be done with this circuit: There is direct connection between the primary and - -secondary sides due to the 30-Ω resistor. We apply mesh analysis. For mesh 1, - -$$ --120 + (20 + 30)I_1 - 30I_2 + V_1 = 0 -$$ - -or - -$$ -50I_1 - 30I_2 + V_1 = 120 \tag{13.9.1} -$$ - -For mesh 2, - -$$ --\mathbf{V}_2 + (10 + 30)\mathbf{I}_2 - 30\mathbf{I}_1 = 0 -$$ - -or - -$$ --30I_1 + 40I_2 - V_2 = 0 \tag{13.9.2} -$$ - -At the transformer terminals, - -$$ -V_2 = -\frac{1}{2} V_1 \tag{13.9.3} -$$ - -$$ -\mathbf{I}_2 = -2\mathbf{I}_1 \tag{13.9.4} -$$ - -(Note that *n* = 1∕2.) We now have four equations and four unknowns, but our goal is to get **I**2. So we substitute for **V**1 and **I**1 in terms of **V**2 and **I**2 in Eqs. (13.9.1) and (13.9.2). Equation (13.9.1) becomes - -$$ --55I_2 - 2V_2 = 120 \tag{13.9.5} -$$ - -and Eq. (13.9.2) becomes - -$$ -15I_2 + 40I_2 - V_2 = 0 \Rightarrow V_2 = 55I_2 \qquad (13.9.6) -$$ - -Substituting Eq. (13.9.6) in Eq. (13.9.5), - -$$ --165I_2 = 120 \qquad \Rightarrow \qquad I_2 = -\frac{120}{165} = -0.7272 \text{ A} -$$ - -The power absorbed by the 10-Ω resistor is - -$$ -P = (-0.7272)^{2}(10) = 5.3 -$$ - W - -# Find **V***o* Practice Problem 13.9 in the circuit of Fig. 13.40. - -For Practice Prob. 13.9. - -**Answer:** 96 V. - -# **13.6** Ideal Autotransformers - -Unlike the conventional two-winding transformer we have considered so far, an *autotransformer* has a single continuous winding with a connection point called a *tap* between the primary and secondary sides. The tap is often adjustable so as to provide the desired turns ratio for stepping up or stepping down the voltage. This way, a variable voltage is provided to the load connected to the autotransformer. - -An autotransformer is a transformer in which both the primary and the secondary are in a single winding. - -Figure 13.41 sho ws a typical autotransformer . As sho wn in Fig. 13.42, the autotransformer can operate in the step-down or step-up mode. The autotransformer is a type of po wer transformer. Its major advantage o ver the tw o-winding transformer is its ability to transfer larger apparent po wer. Example 13.10 will demonstrate this. Another advantage is that an autotransformer is smaller and lighter than an equivalent two-winding transformer. However, since both the primary and secondary windings are one winding, *electrical isolation* (no direct electrical connection) is lost. (We will see how the property of electri cal isolation in the conventional transformer is practically employed in Section 13.9.1.) The lack of electrical isolation between the primary and secondary windings is a major disadvantage of the autotransformer. - -Some of the formulas we deri ved for ideal transformers apply to ideal autotransformers as well. F or the step-do wn autotransformer cir cuit of Fig. 13.42(a), Eq. (13.52) gives - -$$ -\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1 + N_2}{N_2} = 1 + \frac{N_1}{N_2} -$$ - (13.63) - -As an ideal autotransformer, there are no losses, so the comple x power remains the same in the primary and secondary windings: - -$$ -S_1 = V_1 I_1^* = S_2 = V_2 I_2^* -$$ - (13.64) - -Equation (13.64) can also be expressed as - -$$ -V_1I_1=V_2I_2 -$$ - -or - -$$ -\frac{V_2}{V_1} = \frac{I_1}{I_2} -$$ - (13.65) - -Thus, the current relationship is - -$$ -\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_2}{N_1 + N_2} -$$ - (13.66) - -For the step-up autotransformer circuit of Fig. 13.42(b), - -$$ -\frac{\mathbf{V}_1}{N_1} = \frac{\mathbf{V}_2}{N_1 + N_2} -$$ - -**Figure 13.41** A typical autotransformer. © Todd Systems, Inc. - -# **Figure 13.42** - -(a) Step-down autotransformer, (b) step-up autotransformer. - -(b) - -$$ -\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1}{N_1 + N_2} \tag{13.67} -$$ - -$$ -\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_1 + N_2}{N_1} = 1 + \frac{N_2}{N_1} -$$ -\n(13.68) - -A major dif ference between con ventional transformers and auto transformers is that the primary and secondary sides of the autotrans former are not only coupled magnetically b ut also coupled conductively. The autotransformer can be used in place of a con ventional transformer when electrical isolation is not required. - -Example 13.10 - -Compare the po wer ratings of the tw o-winding transformer in Fig. 13.43(a) and the autotransformer in Fig. 13.43(b). - -# **Solution:** - -Although the primary and secondary windings of the autotransformer are together as a continuous winding, they are separated in Fig. 13.43(b) for clarity. We note that the current and voltage of each winding of the autotransformer in Fig. 13.43(b) are the same as those for the two-winding transformer in Fig. 13.43(a). This is the basis of comparing their power ratings. - -For the two-winding transformer, the power rating is - -*S*1 = 0.2(240) = 48 VA or *S*2 = 4(12) = 48 VA - -For the autotransformer, the power rating is - -*S*1 = 4.2(240) = 1,008 VA or *S*2 = 4(252) = 1,008 VA - -which is 21 times the power rating of the two-winding transformer. - -Refer to Fig. 13.43. If the two-winding transformer is a 60-VA, 120 V∕10 V transformer, what is the power rating of the autotransformer? Practice Problem 13.10 - -**Answer:** 780 VA. - -Refer to the autotransformer circuit in Fig. 13.44. Calculate: (a) **I**1, **I**2, and **I***o* if **Z***L* = 8 + *j*6 Ω, and (b) the complex power supplied to the load. Example 13.11 - -# **Solution:** - -(a) This is a step-up autotransformer with *N*1  =  80, *N*2  =  120, **V**1 = 120⧸ 30°, so Eq. (13.67) can be used to find **V**2 by - -$$ -\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1}{N_1 + N_2} = \frac{80}{200} -$$ - -or - -$$ -\mathbf{V}_2 = \frac{200}{80} \mathbf{V}_1 = \frac{200}{80} (120/30^\circ) = 300/30^\circ \text{ V} -$$ -$$ -\mathbf{I}_2 = \frac{\mathbf{V}_2}{\mathbf{Z}_L} = \frac{300/30^\circ}{8 + j6} = \frac{300/30^\circ}{10/36.87^\circ} = 30/-6.87^\circ \text{ A} -$$ - -But - -$$ -\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_1 + N_2}{N_1} = \frac{200}{80} -$$ - -or - -$$ -\mathbf{I}_1 = \frac{200}{80} \mathbf{I}_2 = \frac{200}{80} (30 \angle -6.87^\circ) = 75 \angle -6.87^\circ \text{ A} -$$ - -At the tap, KCL gives - -$$ -\mathbf{I}_1 + \mathbf{I}_o = \mathbf{I}_2 -$$ - -or - -$$ -I_o = I_2 - I_1 = 30 \underline{\textstyle{\frac{1}{6.87^\circ}} - 75 \underline{\textstyle{\frac{1}{6.87^\circ}}}} = 45 \underline{\textstyle{\frac{173.13^\circ}{173.13^\circ}}} -$$ - -(b) The complex power supplied to the load is - -$$ -\mathbf{S}_2 = \mathbf{V}_2 \mathbf{I}_2^* = |\mathbf{I}_2|^2 \mathbf{Z}_L = (30)^2 (10/36.87^\circ) = 9/36.87^\circ \text{ kVA} -$$ - -# Practice Problem 13.11 - -For Practice Prob. 13.11. - -In the autotransformer circuit of Fig. 13.45, find currents **I**1, **I**2, and **I***o*. Take **V**1 = 8 kV, **V**2 = 2 kV. - -**Answer:** 2 A, 8 A, 6 A. - -# **13.7** Three-Phase Transformers - -To meet the demand for three-phase po wer transmission, transformer connections compatible with three-phase operations are needed. We can achieve the transformer connections in tw o ways: by connecting three single-phase transformers, thereby forming a so-called *transformer bank*, or by using a special three-phase transformer . For the same kVA rating, a three-phase transformer is always smaller and cheaper than three single-phase transformers. When single-phase transformers are used, one must ensure that the y have the same turns ratio *n* to achie ve a balanced three-phase system. There are four standard w ays of con necting three single-phase transformers or a three-phase transformer for three-phase operations: Y-Y, Δ-Δ, Y-Δ, and Δ-Y. - -For any of the four connections, the total apparent po wer *ST*, real power *PT*, and reactive power *QT* are obtained as - -$$ -S_T = \sqrt{3} V_L I_L \tag{13.69a} -$$ - -$$ -P_T = S_T \cos \theta = \sqrt{3} V_L I_L \cos \theta \tag{13.69b} -$$ - -$$ -Q_T = S_T \sin \theta = \sqrt{3} V_L I_L \sin \theta \qquad (13.69c) -$$ - -where *VL* and *IL* are, respecti vely, equal to the line v oltage *VLp* and the line current *ILp* for the primary side, or the line v oltage *VLs* and the line current *ILs* for the secondary side. Notice From Eq. (13.69) that for each of the four connections, *VLs ILs* = *VLp ILp*, since power must be conserved in an ideal transformer. - -For the Y-Y connection (Fig. 13.46), the line v oltage *VLp* at the primary side, the line v oltage *VLs* on the secondary side, the line current *ILp* on the primary side, and the line current *ILs* on the secondary side are related to the transformer per phase turns ratio *n* according to Eqs. (13.52) and (13.55) as - -$$ -V_{Ls} = nV_{Lp} \tag{13.70a} -$$ - -$$ -I_{Ls} = \frac{I_{Lp}}{n} \tag{13.70b} -$$ - -For the Δ-Δ connection (Fig. 13.47), Eq. (13.70) also applies for the line voltages and line currents. This connection is unique in the sense - -**Figure 13.46** Y-Y three-phase transformer connection. - -**Figure 13.47** Δ-Δ three-phase transformer connection. - -that if one of the transformers is removed for repair or maintenance, the other two form an *open delta,* which can provide three-phase voltages at a reduced level of the original three-phase transformer. \_\_ - -For the Y-Δ connection (Fig. 13.48), there is a factor of √ 3 arising from the line-phase values in addition to the transformer per phase turns ratio *n*. Thus, - -\_\_ - -*n* √ \_\_ 3 - -$$ -V_{Ls} = \frac{nV_{Lp}}{\sqrt{3}} -$$ -(13.71a) - -$$ -I_{Ls} = \frac{\sqrt{3}I_{Lp}}{n} \tag{13.71b} -$$ - -Similarly, for the Δ-Y connection (Fig. 13.49), - -$$ -V_{Ls} = n\sqrt{3} V_{Lp} -$$ - (13.72a) -$$ -I_{Ls} = \frac{I_{Lp}}{\sqrt{3}} -$$ - (13.72b) - -Y-Δ three-phase transformer connection. - -**Figure 13.49** Δ-Y three-phase transformer connection. - -Example 13.12 - -The 42-kVA balanced load depicted in Fig. 13.50 is supplied by a threephase transformer. (a) Determine the type of transformer connections. (b) Find the line voltage and current on the primary side. (c) Determine the kVA rating of each transformer used in the transformer bank. Assume that the transformers are ideal. - -# **Solution:** - -(a) A careful observation of Fig. 13.50 shows that the primary side is Y-connected, while the secondary side is Δ-connected. Thus, the threephase transformer is Y-Δ, similar to the one shown in Fig. 13.48. (b) Given a load with total apparent power *ST* = 42 kVA, the turns ratio *n*  =  5, and the secondary line voltage *VLs*  =  240 V, we can find the secondary line current using Eq. (13.69a), by - -$$ -I_{Ls} = \frac{S_T}{\sqrt{3} V_{Ls}} = \frac{42,000}{\sqrt{3}(240)} = 101 \text{ A} -$$ - -From Eq. (13.71), - -$$ -I_{Lp} = \frac{n}{\sqrt{3}} I_{Ls} = \frac{5 \times 101}{\sqrt{3}} = 292 \text{ A} -$$ -$$ -V_{Lp} = \frac{\sqrt{3}}{n} V_{Ls} = \frac{\sqrt{3} \times 240}{5} = 83.14 \text{ V} -$$ - -(c) Because the load is balanced, each transformer equally shares the total load and since there are no losses (assuming ideal transformers), the kVA rating of each transformer is *S* = *ST*∕3 = 14 kVA. Alternatively, the transformer rating can be determined by the product of the phase current and phase voltage of the primary or secondary side. For the pri mary side, for example, we have a delta connection, so that the phase voltage is the same as the line voltage of 240 V, while the phase current is *ILp*∕√ \_\_ 3  = 58.34 A. Hence, *S* = 240 × 58.34 = 14 kVA. - -A three-phase Δ-Δ transformer is used to step down a line voltage of 625 kV, to supply a plant operating at a line voltage of 12.5 kV. The plant draws 40 MW with a lagging power factor of 85 percent. Find: (a) the current drawn by the plant, (b) the turns ratio, (c) the current on the primary side of the transformer, and (d) the load carried by each transformer. Practice Problem 13.12 - -**Answer:** (a) 2.174 kA, (b) 0.02, (c) 43.47 A, (d) 15.69 MVA. - -# **13.8** PSpice Analysis of Magnetically Coupled Circuits - -*PSpice* analyzes magnetically coupled circuits just lik e inductor cir cuits except that the dot convention must be followed. In *PSpice* Schematic, the dot (not sho wn) is al ways next to pin 1, which is the lefthand terminal of the inductor when the inductor with part name L is placed (horizontally) without rotation on a schematic. Thus, the dot or pin 1 will be at the bottom after one 90 ° counterclockwise rotation, since rotation is al ways about pin 1. Once the magnetically coupled inductors are arranged with the dot convention in mind and their value attributes are set in henries, we use the coupling symbol K\_LINEAR to define the coupling. For each pair of coupled inductors, take the following steps: - -- 1. Select **Draw/Get New Part** and type K\_LINEAR. -- 2. Hit or click **OK** and place the K\_LINEAR symbol on the schematic, as shown in Fig. 13.51. (Notice that K\_LINEAR is not a component and therefore has no pins.) -- 3. **DCLICKL** on COUPLING and set the v alue of the coupling coefficient *k*. -- 4. **DCLICKL** on the boxe d K (the coupling symbol) and enter the reference designator names for the coupled inductors as v alues of Li, *i* = 1, 2, …, 6. For example, if inductors L20 and L23 are coupled, we set L1 = L20 and L2 = L23. L1 and at least one other Li must be assigned values; other Li's may be left blank. - -In step 4, up to six coupled inductors with equal coupling can be specified. - -For the air -core transformer, the partname is XFRM\_LINEAR. It can be inserted in a circuit by selecting **Draw/Get P art Name** and then typing in the part name or by selecting the part name from the analog.slb library. As shown typically in Fig. 13.52(a), the main attributes of the linear transformer are the coupling coef ficient *k* and the inductance v alues L1 and L2 in henries. If the mutual induc tance *M* is specified, its value must be used along with L1 and L2 to calculate *k*. Keep in mind that the v alue of *k* should lie between 0 and 1. - -For the ideal transformer , the part name is XFRM\_NONLINEAR and is located in the breako ut.slb library. Select it by clicking **Draw/ Get Part Name** and then typing in the part name. Its attrib utes are the coupling coefficient and the numbers of turns associated with L1 and L2, as illustrated typically in Fig. 13.52(b). The value of the coef ficient of mutual coupling *k* = 1. - -*PSpice* has some additional transformer configurations that we will not discuss here. - -# **Figure 13.51** - -K\_Linear for defining coupling. - -COUPLING = 0.5 L1\_VALUE = 1 mH L2\_VALUE = 25 mH (a) - -**Figure 13.52** (a) Linear transformer XFRM\_LINEAR, (b) ideal transformer XFRM\_NONLINEAR. - -Use *PSpice* to find *i*1, *i*2, and *i*3 in the circuit displayed in Fig. 13.53. - -For Example 13.13. - -# **Solution:** - -The coupling coefficients of the three coupled inductors are determined as follows: - -$$ -k_{12} = \frac{M_{12}}{\sqrt{L_1 L_2}} = \frac{1}{\sqrt{3 \times 3}} = 0.3333 -$$ -$$ -k_{13} = \frac{M_{13}}{\sqrt{L_1 L_3}} = \frac{1.5}{\sqrt{3 \times 4}} = 0.433 -$$ -$$ -k_{23} = \frac{M_{23}}{\sqrt{L_2 L_3}} = \frac{2}{\sqrt{3 \times 4}} = 0.5774 -$$ - -The operating frequency *f* is obtained from Fig. 13.53 as *ω*= 12 *π*=  2 *πf* → *f* = 6 Hz. - -The schematic of the circuit is portrayed in Fig. 13.54. Notice ho w the dot convention is adhered to. For L2, the dot (not shown) is on pin 1 (the left-hand terminal) and is therefore placed without rotation. For L1, in order for the dot to be on the right-hand side of the inductor, the inductor must be rotated through 180 °. For L3, the inductor must be rotated through 90° so that the dot will be at the bottom. Note that the 2-H in ductor (*L*4) is not coupled. To handle the three coupled inductors, we use three K\_LINEAR parts provided in the analog library and set the following attributes (by double-clicking on the symbol K in the box): - -**Figure 13.54** Schematic of the circuit of Fig. 13.53. - -designators of the inductors on the schematic. - -Three IPRINT pseudocomponents are inserted in the appropriate branches to obtain the required currents *i*1, *i*2, and *i*3. As an AC singlefrequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* =  1, *Start Freq* =  6, and *Final Freq* =  6. After saving the schematic, we select **Analysis/Simulate** to simulate it. The output file includes: - -| FREQ | IM(V_PRINT2) | IP(V_PRINT2) | -|-----------|--------------|--------------| -| 6.000E+00 | 2.114E-01 | -7.575E+01 | -| FREQ | IM(V_PRINT1) | IP(V_PRINT1) | -| 6.000E+00 | 4.654E-01 | -7.025E+01 | -| FREQ | IM(V_PRINT3) | IP(V_PRINT3) | -| 6.000E+00 | 1.095E-01 | 1.715E+01 | - -From this we obtain - -$$ -\mathbf{I}_1 = 0.4654 \underline{\smash{\big)}\xspace - 70.25^{\circ}} -$$ -\n -$$ -\mathbf{I}_2 = 0.2114 \underline{\smash{\big)}\xspace - 75.75^{\circ}}, \qquad \mathbf{I}_3 = 0.1095 \underline{\smash{\big)}\xspace 17.15^{\circ}} -$$ - -Thus, - -$$ -i_1 = 0.4654 \cos (12 \pi t - 70.25^\circ) \text{ A} -$$ - -$$ -i_2 = 0.2114 \cos(12 \pi t - 75.75^\circ) \text{ A} -$$ - -$$ -i_3 = 0.1095 \cos(12 \pi t + 17.15^\circ) \text{ A} -$$ - -Find *io* in the circuit of Fig. 13.55, using *PSpice*. - -For Practice Prob. 13.13. - -**Answer:** 2.012 cos(4*t* + 68.52°) A. - -Find **V**1 and **V**2 in the ideal transformer circuit of Fig. 13.56, using *PSpice*. - -# Practice Problem 13.13 - -# Example 13.14 - -# **Solution:** - -- 1. **Define.** The problem is clearly defined and we can proceed to the next step. -- 2. **Present.** We have an ideal transformer and we are to find the input and the output voltages for that transformer. In addition, we are to use *PSpice* to solve for the voltages. -- 3. **Alternative.** We are required to use *PSpice*. We can use mesh analysis to perform a check. -- 4. **Attempt.** As usual, we assume *ω* = 1 and find the corresponding values of capacitance and inductance of the elements: - -$$ -j10 = j\omega L \qquad \Rightarrow \qquad L = 10 \text{ H} -$$ -$$ --j40 = \frac{1}{j\omega C} \qquad \Rightarrow \qquad C = 25 \text{ mF} -$$ - -Figure 13.57 shows the schematic. For the ideal transformer, we set the coupling factor to 0.99999 and the numbers of turns to 400,000 and 100,000. The two VPRINT2 pseudocomponents are connected across the transformer terminals to obtain **V**1 and **V**2. As a single-frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* =  0.1592, and *Final Freq* =  0.1592. After saving the schematic, we select **Analysis/Simulate** to simulate it. The output file includes: - -| FREQ | | VM(\$N_0003,\$N_0006) VP(\$N_0003,\$N_0006) | -|------|---------------------|---------------------------------------------| -| | 1.592E-01 9.112E+01 | 3.792E+01 | -| FREQ | | VM(\$N_0006,\$N_0005) VP(\$N_0006,\$N_0005) | -| | 1.592E-01 2.278E+01 | -1.421E+02 | - -This can be written as - -$$ -V_1 = 91.12 \times 37.92^{\circ} -$$ - V and $V_2 = 22.78 \times 142.1^{\circ}$ V - -5. **Evaluate.** We can check the answer by using mesh analysis as follows: - -Loop 1 -$$ --120\sqrt{30^{\circ}} + (80 - j40)I_1 + V_1 + 20(I_1 - I_2) = 0 -$$ - -Loop 2 -$$ -20(-I_1 + I_2) - V_2 + (6 + j10)I_2 = 0 -$$ - -**Figure 13.57** The schematic for the circuit in Fig. 13.56. - -Reminder: For an ideal transformer, the inductances of both the primary and secondary windings are infinitely large. - -But -$$ -V_2 = -V_1/4 -$$ - and $I_2 = -4I_1$ . This leads to -\n $-120/30^\circ + (80 - j40)I_1 + V_1 + 20(I_1 + 4I_1) = 0$ -\n $(180 - j40)I_1 + V_1 = 120/30^\circ$ -\n $20(-I_1 - 4I_1) + V_1/4 + (6 + j10)(-4I_1) = 0$ -\n $(-124 - j40)I_1 + 0.25V_1 = 0$ or $I_1 = V_1/(496 + j160)$ - -Substituting this into the first equation yields - -$$ -(180 - j40)V_1/(496 + j160) + V_1 = 120/30^{\circ} -$$ -$$ -(184.39/-12.53^{\circ}/521.2/17.88^{\circ})V_1 + V_1 -$$ -$$ -= (0.3538/-30.41^{\circ} + 1)V_1 = (0.3051 + 1 - j0.17909)V_1 = 120/30^{\circ} -$$ -$$ -V_1 = 120/30^{\circ}/1.3173/-7.81^{\circ} = 91.1/37.81^{\circ} V \qquad \text{and} -$$ -$$ -V_2 = 22.78/-142.19^{\circ} V -$$ - -Both answers check. - -6. **Satisfactory?** We have satisfactorily answered the problem and checked the solution. We can now present the entire solution to the problem. - -Obtain **V**1 and **V**2 in the circuit of Fig. 13.58 using *PSpice*. - -For Practice Prob. 13.14. - -**Answer:** V1 = 153⧸ 2.18° V, V2 = 230.2⧸ 2.09° V. - -# **13.9** Applications - -Transformers are the largest, the heaviest, and often the costliest of cir cuit components. Nevertheless, they are indispensable passive devices in electric circuits. They are among the most efficient machines, 95 percent efficiency being common and 99 percent being achie vable. They have numerous applications. For example, transformers are used: - -- To step up or step down voltage and current, making them useful for power transmission and distribution. -- To isolate one portion of a circuit from another (i.e., to transfer power without any electrical connection). -- As an impedance-matching device for maximum power transfer. -- In frequenc y-selective circuits whose operation depends on the response of inductances. - -Practice Problem 13.14 - -For more information on the many kinds of transformers, a good text is W. M. Flanagan, Handbook of Transformer Design and Applications, 2nd ed. (New York: McGraw-Hill, 1993). - -# **Figure 13.59** - -A transformer used to isolate an ac supply from a rectifier. - -Because of these di verse uses, there are man y special designs for transformers (only some of which are discussed in this chapter): v oltage transformers, current transformers, power transformers, distribution transformers, impedance-matching transformers, audio transformers, single-phase transformers, three-phase transformers, rectifier transformers, inverter transformers, and more. In this section, we consider three important applications: transformer as an isolation de vice, transformer as a matching device, and power distribution system. - -# **13.9.1** Transformer as an Isolation Device - -Electrical isolation is said to exist between two devices when there is no physical connection between them. In a transformer, energy is transferred by magnetic coupling, without electrical connection between the primary circuit and secondary circuit. We no w consider three simple practical examples of how we take advantage of this property. - -First, consider the circuit in Fig. 13.59. A rectifier is an electronic circuit that converts an ac supply to a dc supply. A transformer is often used to couple the ac supply to the rectifier. The transformer serves two purposes. First, it steps up or steps do wn the voltage. Second, it pro vides electrical isolation between the ac po wer supply and the rectifier, thereby reducing the risk of shock hazard in handling the electronic device. - -As a second e xample, a transformer is often used to couple tw o stages of an amplifier, to prevent any dc voltage in one stage from affecting the dc bias of the next stage. Biasing is the application of a dc voltage to a transistor amplifier or any other electronic device in order to produce a desired mode of operation. Each amplifier stage is biased separately to operate in a particular mode; the desired mode of operation will be compromised without a transformer providing dc isolation. As shown in Fig. 13.60, only the ac signal is coupled through the transformer from one stage to the next. We recall that magnetic coupling does not exist with a dc voltage source. Transformers are used in radio and TV receivers to couple stages of high-frequency amplifiers. When the sole purpose of a transformer is to pro vide isolation, its turns ratio *n* is made unity . Thus, an isolation transformer has *n* = 1. - -As a third example, consider measuring the voltage across 13.2-kV lines. It is obviously not safe to connect a voltmeter directly to such highvoltage lines. A transformer can be used both to electrically isolate the line power from the v oltmeter and to step do wn the v oltage to a safe level, as shown in Fig. 13.61. Once the voltmeter is used to measure the - -**Figure 13.60** A transformer providing dc isolation between two amplifier stages. - -# **Figure 13.61** - -A transformer providing isolation between the power lines and the voltmeter. - -secondary voltage, the turns ratio is used to determine the line v oltage on the primary side. - -Determine the voltage across the load in Fig. 13.62. - -# **Solution:** - -We can apply the superposition principle to find the load voltage. Let *vL* = *vL*1 + *vL*2, where *vL*1 is due to the dc source and *vL*2 is due to the ac source. We consider the dc and ac sources separately, as shown in Fig. 13.63. The load voltage due to the dc source is zero, because a timevarying voltage is necessary in the primary circuit to induce a voltage in the secondary circuit. Thus, *vL*1 = 0. For the ac source and a value of *Rs* so small it can be neglected, - -$$ -\frac{\mathbf{V}_2}{\mathbf{V}_1} = \frac{\mathbf{V}_2}{120} = \frac{1}{3} \quad \text{or} \quad \mathbf{V}_2 = \frac{120}{3} = 40 \text{ V} -$$ - -Hence, **V***L*2 = 40 V ac or *vL*2 = 40 cos *ωt*; that is, only the ac voltage is passed to the load by the transformer. This example shows how the transformer provides dc isolation. - -For Example 13.15: (a) dc source, (b) ac source. - -Refer to Fig. 13.61. Calculate the turns ratio required to step down the 14.4-kV line voltage to a safe level of 120 V. - -Practice Problem 13.15 - -**Answer:** 120. - -# **13.9.2** Transformer as a Matching Device - -We recall that for maximum po wer transfer, the load resistance *RL* must be matched with the source resistance *Rs*. In most cases, the two resistances are not matched; both are fixed and cannot be altered. However, an iron-core transformer can be used to match the load resistance to the source resistance. This is called *impedance matching*. For example, to connect a loudspeaker to an audio power amplifier requires a transformer, because the speak er's resistance is only a few ohms while the internal resistance of the amplifier is several thousand ohms. - -Consider the circuit shown in Fig. 13.64. We recall from Eq. (13.60) that the ideal transformer reflects its load back to the primary with a - -# **Figure 13.64** - -Transformer used as a matching device. - -**Figure 13.62** For Example 13.15. - -12 V dc - -+ ‒ - -scaling factor of *n*2 . To match this reflected load *RL*∕*n*2 with the source resistance *Rs*, we set them equal, - -$$ -R_s = \frac{R_L}{n^2} \tag{13.73} -$$ - -Equation (13.73) can be satisfied by proper selection of the turns ratio *n*. From Eq. (13.73), we notice that a step-down transformer (*n* < 1) is needed as the matching de vice when *Rs* > *RL*, and a step-up ( *n* > 1) is required when *Rs* < *RL*. - -The ideal transformer in Fig. 13.65 is used to match the amplifier circuit to the loudspeak er to achie ve maximum po wer transfer. The Thevenin (or output) impedance of the amplifier is 192 Ω, and the internal impedance of the speaker is 12 Ω. Determine the required turns ratio. - -# **Solution:** - -Speaker - -We replace the amplifier circuit with the Thevenin equivalent and reflect the impedance **Z***L* = 12 Ω of the speaker to the primary side of the ideal transformer. Figure 13.66 shows the result. For maximum power transfer, - -$$ -Z_{\text{Th}} = \frac{Z_L}{n^2} -$$ - or $n^2 = \frac{Z_L}{Z_{\text{Th}}} = \frac{12}{192} = \frac{1}{16}$ - -Thus, the turns ratio is *n* = 1∕4 = 0.25. - -Using *P* = *I* 2 *R*, we can sho w that indeed the po wer delivered to the speaker is much lar ger than without the ideal transformer. Without the ideal transformer, the amplifier is directly connected to the speaker. The power delivered to the speaker is - -$$ -P_L = \left(\frac{\mathbf{V}_{\text{Th}}}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L}\right)^2 \mathbf{Z}_L = 288 \text{ V}_{\text{Th}}^2 \mu \text{W} -$$ - -With the transformer in place, the primary and secondary currents are - -$$ -I_p = \frac{\mathbf{V}_{\text{Th}}}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L/n^2}, \qquad I_s = \frac{I_p}{n} -$$ - -Hence, - -$$ -P_L = I_s^2 \mathbf{Z}_L = \left(\frac{\mathbf{V}_{\text{Th}}/n}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L/n^2}\right)^2 \mathbf{Z}_L -$$ -$$ -= \left(\frac{n\mathbf{V}_{\text{Th}}}{n^2 \mathbf{Z}_{\text{Th}} + \mathbf{Z}_L}\right)^2 \mathbf{Z}_L = 1,302 \mathbf{V}_{\text{Th}}^2 \mu \mathbf{W} -$$ - -confirming what was said earlier. - -Calculate the turns ratio of an ideal transformer required to match a 8-Ω load to a source with internal impedance of 800 Ω. Find the load voltage when the source voltage is 300 V. - -**Answer:** 0.1, 15 V. - -**Figure 13.66** Equivalent circuit of the circuit in Fig. 13.65; for Example 13.16. - -Using an ideal transformer to match the speaker to the amplifier; for - -**Figure 13.65** - -Example 13.16. - -# Practice Problem 13.16 - -# **13.9.3** Power Distribution - -A po wer system basically consists of three components: generation, transmission, and distrib ution. The local electric compan y operates a plant that generates se veral hundreds of me gavolt-amperes (MV A), typically at about 18 kV . As Fig. 13.67 illustrates, three-phase step-up transformers are used to feed the generated po wer to the transmission line. Why do we need the transformer? Suppose we need to transmit 100,000 VA over a distance of 50 km. Since *S* = *VI*, using a line voltage of 1,000 V implies that the transmission line must carry 100 A and this requires a transmission line of a large diameter. If, on the other hand, we use a line voltage of 10,000 V, the current is only 10 A. The smaller current reduces the required conductor size, producing considerable savings as well as minimizing transmission line *I* 2 *R* losses. To minimize losses requires a step-up transformer. Without the transformer, the majority of the power generated would be lost on the transmission line. The ability of the transformer to step up or step down voltage and distribute power economically is one of the major reasons for generating ac rather than dc. Thus, for a gi ven power, the lar ger the v oltage, the better . Today, 1 MV is the lar gest voltage in use; the le vel may increase as a result of research and experiments. - -# **Figure 13.67** - -A typical power distribution system. - -Source: A. Marcus and C. M. Thomson, *Electricity for Technicians,* 2nd edition, © 1975, p. 337. Pearson Education, Inc., Upper Saddle River, NJ. - -Beyond the generation plant, the power is transmitted for hundreds of miles through an electric netw ork called the *power grid*. The threephase power in the po wer grid is con veyed by transmission lines hung overhead from steel towers which come in a v ariety of sizes and shapes. The (aluminum-conductor, steel-reinforced) lines typically ha ve overall diameters up to about 40 mm and can carry current of up to 1,380 A. - -At the substations, distrib ution transformers are used to step do wn the voltage. The step-do wn process is usually carried out in stages. Power may be distributed throughout a locality by means of either overhead or under ground cables. The substations distrib ute the po wer to residential, commercial, and industrial customers. At the receiving end, a residential customer is e ventually supplied with 120 ∕240 V, while industrial or commercial customers are fed with higher voltages such as One may ask, How would increasing the voltage not increase the current, thereby increasing I 2 R losses? Keep in mind that I = Vℓ∕R, where Vℓ is the potential difference between the sending and receiving ends of the line. The voltage that is stepped up is the sending end voltage V, not Vℓ. If the receiving end is VR, then Vℓ = V − VR. Since V and VR are close to each other, Vℓ is small even when V is stepped up. 460∕208 V. Residential customers are usually supplied by distrib ution transformers often mounted on the poles of the electric utility company. When direct current is needed, the alternating current is converted to dc electronically. - -# Example 13.17 - -A distribution transformer is used to supply a household as in Fig. 13.68. The load consists of eight 100-W bulbs, a 350-W TV, and a 15-kW kitchen range. If the secondary side of the transformer has 72 turns, calculate: (a) the number of turns of the primary winding, and (b) the current *Ip* in the primary winding. - -**Figure 13.68** For Example 13.17. - -# **Solution:** - -(a) The dot locations on the winding are not important, since we are only interested in the magnitudes of the variables involved. Since - -$$ -\frac{N_p}{N_s} = \frac{V_p}{V_s} -$$ - -we get - -$$ -N_p = N_s \frac{V_p}{V_s} = 72 \frac{2,400}{240} = 720 \text{ turns} -$$ - -(b) The total power absorbed by the load is - -$$ -S = 8 \times 100 + 350 + 15{,}000 = 16.15 -$$ - kW - -But *S* = *VpIp* = *VsIs*, so that - -$$ -I_p = \frac{S}{V_p} = \frac{16,150}{2,400} = 6.729 \text{ A} -$$ - -# Practice Problem 13.17 - -In Example 13.17, if the eight 100-W bulbs are replaced by twelve 60-W bulbs and the kitchen range is replaced by a 4.5-kW air- conditioner, find: (a) the total power supplied, (b) the current *Ip* in the primary winding. - -**Answer:** (a) 5.57 kW, (b) 2.321 A. - -# **13.10** Summary - -1. Two coils are said to be mutually coupled if the magnetic flux *ϕ* emanating from one passes through the other . The mutual induc tance between the two coils is given by - -$$ -M = k\sqrt{L_1 L_2} -$$ - -where *k* is the coupling coefficient, 0 < *k* < 1. - -2. If *v*1 and *i*1 are the voltage and current in coil 1, while *v*2 and *i*2 are the voltage and current in coil 2, then - -$$ -v_1 = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt} -$$ - and $v_2 = L_2 \frac{di_2}{dt} + M \frac{di_1}{dt}$ - - Thus, the voltage induced in a coupled coil consists of self-induced voltage and mutual voltage. - -- 3. The polarity of the mutually-induced v oltage is e xpressed in the schematic by the dot convention. -- 4. The energy stored in two coupled coils is - -$$ -\frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 \pm Mi_1i_2 -$$ - -- 5. A transformer is a four -terminal de vice containing tw o or more magnetically coupled coils. It is used in changing the current, v oltage, or impedance level in a circuit. -- 6. A linear (or loosely coupled) transformer has its coils wound on a magnetically linear material. It can be replaced by an equi valent T or Π network for the purposes of analysis. -- 7. An ideal (or iron-core) transformer is a lossless (*R*1 = *R*2 = 0) transformer with unity coupling coef ficient (*k* = 1) and infinite inductances (*L*1, *L*2, *M* → ∞). -- 8. For an ideal transformer, - -$$ -\mathbf{V}_2 = n\mathbf{V}_1, \qquad \mathbf{I}_2 = \frac{\mathbf{I}_1}{n}, \qquad \mathbf{S}_1 = \mathbf{S}_2, \qquad \mathbf{Z}_R = \frac{\mathbf{Z}_L}{n^2} -$$ - - where *n* = *N*2∕*N*1 is the turns ratio. *N*1 is the number of turns of the primary winding and *N*2 is the number of turns of the second ary winding. The transformer steps up the primary voltage when *n* > 1, steps it do wn when *n* < 1, or serv es as a matching de vice when *n* = 1. - -- 9. An autotransformer is a transformer with a single winding common to both the primary and the secondary circuits. -- 10. *PSpice* is a useful tool for analyzing magnetically coupled circuits. -- 11. Transformers are necessary in all stages of po wer distribution systems. Three-phase voltages may be stepped up or do wn by threephase transformers. -- 12. Important uses of transformers in electronics applications are as electrical isolation devices and impedance-matching devices. - -# Review Questions - -**13.1** Refer to the two magnetically coupled coils of Fig. 13.69(a). The polarity of the mutual voltage is: - -(a) Positive (b) Negative - -# **Figure 13.69** - -For Review Questions 13.1 and 13.2. - -**13.2** For the two magnetically coupled coils of Fig. 13.69(b), the polarity of the mutual voltage is: - -(a) Positive (b) Negative - -**13.3** The coefficient of coupling for two coils having *L*1 = 2 H, *L*2 = 8 H, *M* = 3 H is: - -| (a) 0.1875 | (b) 0.75 | -|------------|-----------| -| (c) 1.333 | (d) 5.333 | - -- **13.4** A transformer is used in stepping down or stepping up: - - (a) dc voltages (b) ac voltages (c) both dc and ac voltages -- **13.5** The ideal transformer in Fig. 13.70(a) has *N*2∕*N*1 = 10. The ratio *V*2∕*V*1 is: - -(a) 10 (b) 0.1 (c) −0.1 (d) −10 - -**Figure 13.70** For Review Questions 13.5 and 13.6. - -- **13.6** For the ideal transformer in Fig. 13.70(b), *N*2∕*N*1 = 10. The ratio *i*2∕*I*1 is: - - (a) 10 (b) 0.1 (c) −0.1 (d) −10 -- **13.7** A three-winding transformer is connected as portrayed in Fig. 13.71(a). The value of the output voltage *Vo* is: - -(a) 10 (b) 6 (c) −6 (d) −10 - -# **Figure 13.71** - -For Review Questions 13.7 and 13.8. - -- **13.8** If the three-winding transformer is connected as in Fig. 13.71(b), the value of the output voltage *Vo* is: (a) 10 (b) 6 (c) −6 (d) −10 -- **13.9** In order to match a source with internal impedance of 500 Ω to a 15-Ω load, what is needed is: - - (a) step-up linear transformer - - (b) step-down linear transformer - - (c) step-up ideal transformer - - (d) step-down ideal transformer - - (e) autotransformer -- **13.10** Which of these transformers can be used as an isolation device? - - (a) linear transformer (b) ideal transformer (c) autotransformer (d) all of the above - -*Answers: 13.1b, 13.2a, 13.3b, 13.4b, 13.5d, 13.6b, 13.7c, 13.8a, 13.9d, 13.10b.* - -# Problems1 - -# Section 13.2 Mutual Inductance - -**13.1** For the three coupled coils in Fig. 13.72, calculate the total inductance. - -**Figure 13.72** - -For Prob. 13.1. - -**13.2** Using Fig. 13.73, design a problem to help other students better understand mutual inductance. - -For Prob. 13.2. - -- **13.3** Two coils connected in series-aiding fashion have a total inductance of 500 mH. When connected in a series-opposing configuration, the coils have a total inductance of 300 mH. If the inductance of one coil (*L*1) is three times the other, find *L*1, *L*2, and *M*. What is the coupling coefficient? -- **13.4** (a) For the coupled coils in Fig. 13.74(a), show that - -$$ -L_{\text{eq}} = L_1 + L_2 + 2M -$$ - -(b) For the coupled coils in Fig. 13.74(b), show that - -eq: Cous in Fig. 13.74( -$$ -L_{eq} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M} -$$ - -**13.5** Two coils are mutually coupled, with *L*1 = 50 mH, *L*2 = 120 mH, and *k* = 0.5. Calculate the maximum possible equivalent inductance if: - -> (a) the two coils are connected in series (b) the coils are connected in parallel - -**13.6** Given the circuit shown in Fig. 13.75, determine the value of V1 and I2. - -# **Figure 13.76** - -For Prob. 13.7. - -**Figure 13.77** - -For Prob. 13.8. - -**13.9** Find **V***x* in the network shown in Fig. 13.78. - -1Remember, unless otherwise specified, assume all values of currents and voltages are rms. - -For Prob. 13.10. - -**13.11** Use mesh analysis to find *ix* in Fig. 13.80, where *is* = 4 cos(600*t*) A and *vs* = 110 cos(600*t* + 30°) - -**Figure 13.80** For Prob. 13.11. - -**13.12** Determine the equivalent *L*eq in the circuit of Fig. 13.81. - -For Prob. 13.13. - -in Fig. 13.83 at terminals *a*-*b*. - -# **Figure 13.83** - -For Prob. 13.14. - -**13.15** Find the Norton equivalent for the circuit in Fig. 13.84 at terminals *a*-*b*. - -# **Figure 13.84** - -For Prob. 13.15. - -**13.16** Obtain the Norton equivalent at terminals *a*-*b* of the circuit in Fig. 13.85. - -# **Figure 13.85** For Prob. 13.16. - -**13.17** In the circuit of Fig. 13.86, Z*L* is a 15-mH inductor having an impedance of *j*40 Ω. Determine *Z*in when *k* = 0.6. - -**Figure 13.86** For Prob. 13.17. - -**13.18** Find the Thevenin equivalent to the left of the load Z in the circuit of Fig. 13.87. - -# **Figure 13.87** - -For Prob. 13.18. - -**13.19** Determine an equivalent T-section that can be used to replace the transformer in Fig. 13.88. - -**Figure 13.88** - -# For Prob. 13.19. - -# Section 13.3 Energy in a Coupled Circuit - -**13.20** Determine currents **I**1, **I**2, and **I**3 in the circuit of Fig. 13.89. Find the energy stored in the coupled coils at *t* = 2 ms. Take *ω* = 1,000 rad/s. - -# **Figure 13.89** For Prob. 13.20. - -**13.21** Using Fig. 13.90, design a problem to help other students better understand energy in a coupled circuit. - -**Figure 13.90** For Prob. 13.21. - -# **Figure 13.91** For Prob. 13.22. - -**13.23** Let *is* = 5 cos (100*t*) A. Calculate the voltage across the capacitor, *vc*. Also calculate the value of the energy stored in the coupled coils at *t* = 2.5*π* ms. - -# **Figure 13.92** - -For Prob. 13.23. - -**13.24** In the circuit of Fig. 13.93, - -- (a) find the coupling coefficient, (b) calculate *vo*, -- (c) determine the energy stored in the coupled inductors at *t* = 2 s. - -\* An asterisk indicates a challenging problem. - -For Prob. 13.25. - -**13.26** Find **I***o* in the circuit of Fig. 13.95. Switch the dot on the winding on the right and calculate **I***o* again. - -For Prob. 13.26. - -# **Figure 13.96** - -For Prob. 13.27. - -For Prob. 13.28. - -# Section 13.4 Linear Transformers - -**13.29** In the circuit of Fig. 13.98, find the value of the coupling coefficient *k* that will make the 10-Ω resistor dissipate 1.28 kW. For this value of *k*, find the energy stored in the coupled coils at *t* = 1.5 s. - -# **Figure 13.98** - -For Prob. 13.29. - -- **13.30** (a) Find the input impedance of the circuit in Fig. 13.99 using the concept of reflected impedance. - - (b) Obtain the input impedance by replacing the linear transformer by its T equivalent. - -# **Figure 13.100** For Prob. 13.31. - -**\*13.32** Two linear transformers are cascaded as shown in Fig. 13.101. Show that - -$$ -\omega^{2}R(L_{a}^{2} + L_{a}L_{b} - M_{a}^{2}) -$$ -$$ -Z_{in} = \frac{+j\omega^{3}(L_{a}^{2}L_{b} + L_{a}L_{b}^{2} - L_{a}M_{b}^{2} - L_{b}M_{a}^{2})}{\omega^{2}(L_{a}L_{b} + L_{b}^{2} - M_{b}^{2}) - j\omega R(L_{a} + L_{b})} -$$ - -For Prob. 13.32. - -**13.33** Determine the input impedance of the air-core transformer circuit of Fig. 13.102. - -**Figure 13.102** For Prob. 13.33. - -**13.34** Using Fig. 13.103, design a problem to help other students better understand how to find the input impedance of circuits with transformers. - -**Figure 13.103** For Prob. 13.34. - -# Section 13.5 Ideal Transformers - -**13.36** As done in Fig. 13.32, obtain the relationships between terminal voltages and currents for each of the ideal transformers in Fig. 13.105. - -For Prob. 13.36. - -- **13.37** A 240∕2,400-V rms step-up ideal transformer delivers 50 kW to a resistive load. Calculate: - - (a) the turns ratio - - (b) the primary current - - (c) the secondary current - -**13.38** Design a problem to help other students better understand ideal transformers. - -**13.39** A 1,200∕240-V rms transformer has impedance 60⧸−30° Ω on the high-voltage side. If the transformer is connected to a 0.8⧸10°-Ω load on the low-voltage side, determine the primary and secondary currents when the transformer is - -**13.40** The primary of an ideal transformer with a turns ratio of 5 is connected to a voltage source with Thevenin parameters *v*Th = 10 cos 2000*t* V and *R*Th = 100 Ω. Determine the average power delivered to a 200-Ω load connected across the secondary winding. - -For Prob. 13.41. - -**13.42** For the circuit in Fig. 13.107, determine the power absorbed by the 2-Ω resistor. Assume the 120 V is an rms value. - -**Figure 13.107** For Prob. 13.42. - -For Prob. 13.43. - -**13.45** For the circuit shown in Fig. 13.110, find the value of the average power absorbed by the 8-Ω resistor. - -# **Figure 13.110** - -For Prob. 13.45. - -**13.46** (a) Find **I**1 and **I**2 in the circuit of Fig. 13.111 below. (b) Switch the dot on one of the windings. Find **I**1 and **I**2 again. - -For Prob. 13.47. - -For Prob. 13.46. - -# **Figure 13.113** For Prob. 13.48. - -**13.49** Find current *ix* in the ideal transformer circuit shown in Fig. 13.114. - -For Prob. 13.49. - -**13.50** Calculate the input impedance for the network in Fig. 13.115. - -**13.53** Refer to the network in Fig. 13.118. - -- (a) Find *n* for maximum power supplied to the 200-Ω load. -- (b) Determine the power in the 200-Ω load if *n* = 10. - -**Figure 13.118** For Prob. 13.53. - -- **13.51** Use the concept of reflected impedance to find -- the input impedance and current **I**1 in - -Fig. 13.116. - -**13.54** A transformer is used to match an amplifier with - -an 8-Ω load as shown in Fig. 13.119. The Thevenin equivalent of the amplifier is: *V*Th = 10 V, ZTh = 128 Ω. - -- (a) Find the required turns ratio for maximum energy power transfer. -- (b) Determine the primary and secondary currents. -- (c) Calculate the primary and secondary voltages. - -**Figure 13.119** - -For Prob. 13.54. - -**13.56** Find the power absorbed by the 100-Ω resistor in the ideal transformer circuit of Fig. 13.121. - -**Figure 13.121** For Prob. 13.56. - -**13.57** For the ideal transformer circuit of Fig. 13.122 below, find: - -- (b) **V**1, **V**2, and **V***o*, -- (c) the complex power supplied by the source. - -**13.58** Determine the average power absorbed by each resistor in the circuit of Fig. 13.123. - -# **Figure 13.123** - -For Prob. 13.58. - -# **Figure 13.124** - -For Prob. 13.59. - -**13.60** Refer to the circuit in Fig. 13.125 on the following page. - -(a) Find currents **I**1, **I**2, and **I**3. - -**Figure 13.125** For Prob. 13.60. - -**\*13.61** For the circuit in Fig. 13.126, find **I**1, **I**2, and **V***o*. - -**13.62** For the network in Fig. 13.127, find: (a) the complex power supplied by the source, (b) the average power delivered to the 18-Ω resistor. - -# **Figure 13.128** - -For Prob. 13.63. - -**13.64** For the circuit in Fig. 13.129, find the turns ratio so that the maximum power is delivered to the 30-kΩ resistor. - -**\*13.65** Calculate the average power dissipated by the 20-Ω resistor in Fig. 13.130. - -# Section 13.6 Ideal Autotransformers - -**13.66** Design a problem to help other students better understand how the ideal autotransformer works. - -**13.67** An autotransformer with a 40 percent tap is supplied by an 880-V, 60-Hz source and is used for stepdown operation. A 5-kVA load operating at unity power factor is connected to the secondary terminals. Find: - -(a) the secondary voltage, - -- (b) the secondary current, -- (c) the primary current. -- **13.68** In the ideal autotransformer of Fig. 13.131, calculate **I**1, **I**2, and **I***o*. Find the average power delivered to the load. - -For Prob. 13.68. - -**\*13.69** In the circuit of Fig. 13.131, *N*1 = 190 turns and *N*2 = 10 turns. Determine the Thevenin equivalent circuit looking into terminals *a* and *b*. What would be the value of **Z***L* that would absorb maximum power from the circuit? - -For Prob. 13.69. - -**13.70** In the ideal transformer circuit shown in Fig. 13.133, determine the average power delivered to the load. - -# **Figure 13.133** - -For Prob. 13.70. - -**13.71** When individuals travel, their electrical appliances need to have converters to match the voltages required by their appliances to the local voltage available to power their appliances. Today these converters use power electronics to convert voltages. In the past these converters were autotransformers. The autotransformer shown in Fig. 13.134 is used to convert 115 to 220 V. What is the value of the turns? If the maximum current available from the 115 V source is 15 A, what will be the maximum current available for the 220-V appliance? - -# **Figure 13.134** - -For Prob. 13.71. - -# Section 13.7 Three-Phase Transformers - -**13.72** In order to meet an emergency, three single-phase transformers with 12,470∕7,200 V rms are connected in Δ-Y to form a three-phase transformer which is fed by a 12,470-V transmission line. If the transformer supplies 60 MVA to a load, find: - -(a) the turns ratio for each transformer, - -- (b) the currents in the primary and secondary windings of the transformer, -- (c) the incoming and outgoing transmission line currents. - -**13.73** Figure 13.135 on the next page shows a three-phase transformer that supplies a Y-connected load. - -- (a) Identify the transformer connection. -- (b) Calculate currents **I**2 and **I***c*. -- (c) Find the average power absorbed by the load. - -**Figure 13.135** For Prob. 13.73. - -- **13.74** Consider the three-phase transformer shown in Fig. 13.136. The primary is fed by a three-phase source with line voltage of 2.4 kV rms, while the secondary supplies a three-phase 120-kW balanced load at pf of 0.8. Determine: - - (a) the type of transformer connections, - -(b) the values of *ILS* and *IPS*, - -(c) the values of *ILP* and *IPP*, - -- (d) the kVA rating of each phase of the transformer. -- **13.75** A balanced three-phase transformer bank with the Δ-Y connection depicted in Fig. 13.137 is used to step down line voltages from 4,500 V rms to 900 V rms. If the transformer feeds a 120-kVA load, find: - - (a) the turns ratio for the transformer, - - (b) the line currents at the primary and secondary sides. - -**Figure 13.137** For Prob. 13.75. - -**13.76** Using Fig. 13.138, design a problem to help other students better understand a Y-Δ, three-phase transformer and how they work. - -**Figure 13.138** For Prob. 13.76. - -- **13.77** The three-phase system of a town distributes power with a line voltage of 13.2 kV. A pole transformer connected to single wire and ground steps down the high-voltage wire to 120 V rms and serves a house as shown in Fig. 13.139. - - (a) Calculate the turns ratio of the pole transformer to get 120 V. - - (b) Determine how much current a 100-W lamp connected to the 120-V hot line draws from the high-voltage line. - -**Figure 13.139** For Prob. 13.77. - -# Section 13.8 PSpice Analysis of Magnetically Coupled Circuits - -**13.78** Use *PSpice* or *MultiSim* to determine the mesh currents in the circuit of Fig. 13.140. Take *ω* = 1 rad/s. Use *k* = 0.5 when solving this problem. - -For Prob. 13.78. - -**13.79** Use *PSpice* or *MultiSim* to find **I**1, **I**2, and **I**3 in the circuit of Fig. 13.141. - -# **Figure 13.141** For Prob. 13.79. - -- **13.80** Rework Prob. 13.22 using *PSpice* or *Multisim*. -- **13.81** Use *PSpice* or *MultiSim* to find **I**1, **I**2, and **I**3 in the circuit of Fig. 13.142. - -# **Figure 13.142** - -For Prob. 13.81. - -**13.82** Use *PSpice* or *MultiSim* to find **V**1, **V**2, and **I***o* in the circuit of Fig. 13.143. - -# **Figure 13.143** - -For Prob. 13.82. - -**13.83** Find **I***x* and **V***x* in the circuit of Fig. 13.144 using *PSpice* or *MultiSim*. - -**13.84** Determine **I**1, **I**2, and **I**3 in the ideal transformer circuit of Fig. 13.145 using *PSpice* or *MultiSim*. - -# Section 13.9 Applications - -- **13.85** A stereo amplifier circuit with an output impedance of 7.2 kΩ is to be matched to a speaker with an input impedance of 8 Ω by a transformer whose primary side has 3,000 turns. Calculate the number of turns required on the secondary side. -- **13.86** A transformer having 2,400 turns on the primary and 48 turns on the secondary is used as an impedancematching device. What is the reflected value of a 3-Ω load connected to the secondary? -- **13.87** A radio receiver has an input resistance of 300 Ω. When it is connected directly to an antenna system with a characteristic impedance of 75 Ω, an - -impedance mismatch occurs. By inserting an impedance-matching transformer ahead of the receiver, maximum power can be realized. Calculate the required turns ratio. - -- **13.88** A step-down power transformer with a turns ratio of *n* = 0.1 supplies 12.6 V rms to a resistive load. If the primary current is 2.5 A rms, how much power is delivered to the load? -- **13.89** A 240∕120-V rms power transformer is rated at 10 kVA. Determine the turns ratio, the primary current, and the secondary current. -- **13.90** A 4-kVA, 2,400∕240-V rms transformer has 250 turns on the primary side. Calculate: - - (a) the turns ratio, - - (b) the number of turns on the secondary side, - - (c) the primary and secondary currents. -- **13.91** A 25,000∕240-V rms distribution transformer has a primary current rating of 75 A. - - (a) Find the transformer kVA rating. - - (b) Calculate the secondary current. -- **13.92** A 4,800-V rms transmission line feeds a distribution transformer with 1,200 turns on the primary and 28 turns on the secondary. When a 10-Ω load is connected across the secondary, find: - - (a) the secondary voltage, - - (b) the primary and secondary currents, - - (c) the power supplied to the load. - -# Comprehensive Problems - -- **13.93** A four-winding transformer (Fig. 13.146) is often used in equipment (e.g., PCs, VCRs) that may be operated from either 110 or 220 V. This makes the equipment suitable for both domestic and foreign use. Show which connections are necessary to provide: - - (a) an output of 14 V with an input of 110 V, (b) an output of 50 V with an input of 220 V. - -# **Figure 13.146** - -For Prob. 13.93. - -**\*13.94** A 440∕110-V ideal transformer can be connected to become a 550∕440-V ideal autotransformer. There - -are four possible connections, two of which are wrong. Find the output voltage of: - -- (a) a wrong connection, -- (b) the right connection. -- **13.95** Ten bulbs in parallel are supplied by a 7,200∕120-V transformer as shown in Fig. 13.147, where the bulbs are modeled by the 144-Ω resistors. Find: - - (a) the turns ratio *n*, - - (b) the current through the primary winding. - -For Prob. 13.95. - -**\*13.96** Some modern power transmission systems now have major high-voltage DC transmission segments. There are a lot of good reasons for doing this but we will not go into them here. To go from the AC to DC, power electronics are used. We start with three-phase AC and then rectify it (using a full-wave rectifier). It was found that using a delta to wye and delta combination connected secondary would give us a much smaller ripple after the full-wave rectifier. How is this accomplished? Remember that these are real devices and are wound on common cores. - -*Hint:* Use Figs. 13.47 and 13.49, and the fact that each coil of the wye connected secondary and each coil of the delta connected secondary are wound around the same core of each coil of the delta connected primary so the voltage of each of the corresponding coils are in phase. When the output leads of both secondaries are connected through fullwave rectifiers with the same load, you will see that the ripple is now greatly reduced. Please consult the instructor for more help if necessary. - -# **chapter** - -# Frequency 14 Response - -*Dost thou love Life? Then do not squander Time; for that is the stuff Life is made.* - -—Benjamin Franklin - -# Enhancing Your Career - -# **Career in Control Systems** - -Control systems are another area of electrical engineering where circuit analysis is used. A control system is designed to re gulate the beha vior of one or more variables in some desired manner. Control systems play major roles in our everyday life. Household appliances such as heating and air -conditioning systems, switch-controlled thermostats, w ashers and dryers, cruise controllers in automobiles, elevators, traffic lights, manufacturing plants, navigation systems—all utilize control systems. In the aerospace field, precision guidance of space probes, the wide range of operational modes of the space shuttle, and the ability to maneuv er space vehicles remotely from earth all require knowledge of control systems. In the manuf acturing sector, repetitive production line operations are increasingly performed by robots, which are programmable control systems designed to operate for many hours without fatigue. - -Control engineering inte grates circuit theory and communication theory. It is not limited to an y specific engineering discipline but may involve en vironmental, chemical, aeronautical, mechanical, ci vil, and electrical engineering. For example, a typical task for a control system engineer might be to design a speed regulator for a disk drive head. - -A thorough understanding of control systems techniques is essen tial to the electrical engineer and is of great v alue for designing control systems to perform the desired task. - -A welding robot. © Vol. 1 PhotoDisc/Getty Images RF - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Understand what transfer functions are and how to determine them. -- 2. Understand the decibel scale, why we use it, and how to use it. -- 3. Understand Bode plots, and know why we use them and how to determine them. -- 4. Understand series and parallel resonance, why they are important, and how to find them. -- 5. Understand passive filters. -- 6. Understand active filters. -- 7. Discuss magnitude and frequency scaling and why they are important. - -# **14.1** Introduction - -In our sinusoidal circuit analysis, we ha ve learned ho w to find voltages and currents in a circuit with a constant frequency source. If we let the amplitude of the sinusoidal source remain constant and v ary the frequency, we obtain the circuit' s *frequency r esponse.* The frequency response may be re garded as a complete description of the sinusoidal steady-state behavior of a circuit as a function of frequency. - -The frequency response of a circuit is the variation in its behavior with change in signal frequency. - -The sinusoidal steady-state frequenc y responses of circuits are of significance in many applications, especially in communications and control systems. A specific application is in electric filters that block out or eliminate signals with unw anted frequencies and pass signals of the desired frequencies. Filters are used in radio, TV, and telephone systems to separate one broadcast frequency from another. - -We begin this chapter by considering the frequency response of simple circuits using their transfer functions. We then consider Bode plots, which are the industry-standard w ay of presenting frequenc y response. We also consider series and parallel resonant circuits and encounter important concepts such as resonance, quality f actor, cutoff frequency, and bandwidth. We discuss different kinds of filters and network scaling. In the last section, we consider one practical application of resonant circuits and two applications of filters. - -# **14.2** Transfer Function - -The transfer function **H**(*ω*) (also called the *network function*) is a useful analytical tool for finding the frequency response of a circuit. In fact, the frequency response of a circuit is the plot of the circuit' s transfer function **H**(*ω*) versus *ω*, with *ω* varying from *ω* = 0 to *ω* = ∞. - -A transfer function is the frequenc y-dependent ratio of a forced function to a forcing function (or of an output to an input). The idea of a transfer function was implicit when we used the concepts of impedance - -The frequency response of a circuit may also be considered as the variation of the gain and phase with frequency. - -and admittance to relate v oltage and current. In general, a linear net work can be represented by the block diagram shown in Fig. 14.1. - -The transfer function **H**(*ω*) of a circuit is the frequency-dependent ratio of a phasor output **Y**(*ω*) (an element voltage or current) to a phasor input **X**(*ω*) (source voltage or current). - -Thus, - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{Y}(\omega)}{\mathbf{X}(\omega)}\tag{14.1} -$$ - -assuming zero initial conditions. Since the input and output can be ei ther voltage or current at any place in the circuit, there are four possible transfer functions: - -$$ -\mathbf{H}(\omega) = \text{Voltage gain} = \frac{\mathbf{V}_o(\omega)}{\mathbf{V}_i(\omega)}\tag{14.2a} -$$ - -$$ -\mathbf{H}(\omega) = \text{Current gain} = \frac{\mathbf{I}_o(\omega)}{\mathbf{I}_i(\omega)}\tag{14.2b} -$$ - -$$ -\mathbf{H}(\omega) = \text{Transfer impedance} = \frac{\mathbf{V}_o(\omega)}{\mathbf{I}_i(\omega)} \tag{14.2c} -$$ - -$$ -\mathbf{H}(\omega) = \text{Transfer admittance} = \frac{\mathbf{I}_o(\omega)}{\mathbf{V}_i(\omega)}\tag{14.2d} -$$ - -where subscripts *i* and *o* denote input and output values. Being a complex quantity, **H**(*ω*) has a magnitude *H*(*ω*) and a phase *ϕ*; that is, **H**(*ω*) = *H*(*ω*)⧸*ϕ*. - -To obtain the transfer function using Eq. (14.2), we first obtain the frequenc y-domain equi valent of the circuit by replacing resistors, inductors, and capacitors with their impedances *R*, *jωL*, and 1∕*jωC*. We then use an y circuit technique(s) to obtain the appropriate quantity in Eq. (14.2). We can obtain the frequency response of the circuit by plot ting the magnitude and phase of the transfer function as the frequenc y varies. A computer is a real time-saver for plotting the transfer function. - -The transfer function **H**(*ω*) can be expressed in terms of its numerator polynomial **N**(*ω*) and denominator polynomial **D**(*ω*) as - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{N}(\omega)}{\mathbf{D}(\omega)} -$$ - (14.3) - -where **N**(*ω*) and **D**(*ω*) are not necessarily the same e xpressions for the input and output functions, respecti vely. The representation of **H**(*ω*) in Eq. (14.3) assumes that common numerator and denominator f actors in **H**(*ω*) have canceled, reducing the ratio to lo west terms. The roots of **N**(*ω*) = 0 are called the *zeros* of **H**(*ω*) and are usually represented as *jω* = *z*1, *z*2, …. Similarly, the roots of **D**(*ω*) = 0 are the *poles* of **H**(*ω*) and are represented as *jω* = *p*1, *p*2,…. - -A zero, as a root of the numerator polynomial, is a value that results in a zero value of the function. A pole, as a root of the denominator polynomial, is a value for which the function is infinite. - -To avoid complex algebra, it is expedient to replace *jω* temporarily with *s* when working with **H**(*ω*) and replace *s* with *jω* at the end. - -# **Figure 14.1** - -A block diagram representation of a linear network. - -In this context, **X**(*ω*) and **Y**(*ω*) denote the input and output phasors of a network; they should not be confused with the same symbolism used for reactance and admittance. The multiple usage of symbols is conventionally permissible due to lack of enough letters in the English language to express all circuit variables distinctly. - -Some authors use **H**( j*ω*) for transfer instead of **H**(*ω*), since *ω* and j are an inseparable pair. - -A zero may also be regarded as the value of s = j*ω* that makes H(s) zero, and a pole as the value of s = j*ω* that makes H(s) infinite. - -# **Solution:** - -The frequency-domain equivalent of the circuit is in Fig. 14.2(b). By voltage division, the transfer function is given by - -# **Figure 14.2** For Example 14.1: (a) time-domain *RC* circuit, - -Comparing this with Eq. (9.18e), we obtain the magnitude and phase of **H**(*ω*) as - -$$ -H = \frac{1}{\sqrt{1 + (\omega/\omega_0)^2}}, \qquad \phi = -\tan^{-1}\frac{\omega}{\omega_0} -$$ - -where *ω*0 = 1∕*RC*. To plot *H* and *ϕ* for 0 < *ω*< ∞, we obtain their values at some critical points and then sketch. - - At *ω* = 0, *H* = 1 and *ϕ* = 0. At *ω* = ∞, *H* = 0 and *ϕ* = −90°. Also, at *ω* = *ω*0, *H* = 1∕ √ \_\_ 2 and *ϕ* = −45°. With these and a few more points as shown in Table 14.1, we find that the frequency response is as shown in Fig. 14.3. Additional features of the frequency response in Fig. 14.3 will be explained in Section 14.6.1 on low-pass filters. - -# **TABLE 14.1** For Example 14.1. *ω*∕*ω***0** *H ϕ ω*∕*ω***0** *H ϕ* 0 1 0 10 0.1 −84° 1 0.71 −45° 20 0.05 −87° 2 0.45 −63° 100 0.01 −89° - -Practice Problem 14.1 Obtain the transfer function **V***o*∕**V***s* of the *RL* circuit in Fig. 14.4, assuming *vs* = *Vm* cos *ωt*. Sketch its frequency response. - -3 0.32 −72° ∞ 0 −90° - -**Answer:** *jωL*∕(*R* + *jωL*); see Fig. 14.5 for the response. - -H 1 - -0.707 - -Frequency response of the *RC* circuit: (a) amplitude response, (b) phase response. - -**Figure 14.4** *RL* circuit for Practice Prob. 14.1. - -(b) frequency-domain *RC* circuit. - -For the circuit in Fig. 14.6, calculate the g ain **I***o*(*ω*)∕**I***i*(*ω*) and its poles Example 14.2 and zeros. - -# **Solution:** - -By current division, - -$$ -\mathbf{I}_o(\omega) = \frac{4 + j2\omega}{4 + j2\omega + 1/j0.5\omega} \mathbf{I}_i(\omega) -$$ - -or - -$$ -\frac{\mathbf{I}_o(\omega)}{\mathbf{I}_i(\omega)} = \frac{j0.5\omega(4+j2\omega)}{1+j2\omega + (j\omega)^2} = \frac{s(s+2)}{s^2 + 2s + 1}, \qquad s = j\omega -$$ - -The zeros are at - -*s*(*s* + 2) =0 ⇒ *z*1 = 0, *z*2 = −2 - -The poles are at - -$$ -s^2 + 2s + 1 = (s + 1)^2 = 0 -$$ - -Thus, there is a repeated pole (or double pole) at *p* = −1. - -Find the transfer function **V***o*(*ω*)∕**I***i*(*ω*) for the circuit in Fig. 14.7. Obtain Practice Problem 14.2 its zeros and poles. - -its zeros and poles. -\n**Answer:** -$$ -\frac{10(s + 2)(s + 5)}{s^2 + 10s + 10} -$$ -, $s = j\omega$ ; zeros: -2, -5; poles: -1.127, -8.873. - -# **14.3** The Decibel Scale - -It is not always easy to get a quick plot of the magnitude and phase of the transfer function as we did above. A more systematic way of obtaining the frequency response is to use Bode plots. Before we begin to construct Bode plots, we should take care of two important issues: the use of logarithms and decibels in expressing gain. - -For Example 14.2. - -**Figure 14.7** For Practice Prob. 14.2. - -© Ingram Publishing RF - -# Historical - -**Alexander Graham Bell** (1847–1922) inventor of the telephone, was a Scottish-American scientist. - -Bell was born in Edinburgh, Scotland, a son of Alexander Melville Bell, a well-known speech teacher. Alexander the younger also became a speech teacher after graduating from the University of Edinburgh and the University of London. In 1866 he became interested in transmitting speech electrically. After his older brother died of tuberculosis, his father decided to move to Canada. Alexander was asked to come to Boston to work at the School for the Deaf. There he met Thomas A. Watson, who became his assistant in his electromagnetic transmitter experiment. On March 10, 1876, Alexander sent the famous first telephone message: "Watson, come here I want you." The bel, the logarithmic unit intro duced in this chapter, is named in his honor. - -Since Bode plots are based on log arithms, it is important that we keep the following properties of logarithms in mind: - -1. log *P*1*P*2 = log *P*1 + log *P*2 2. log *P*1∕*P*2 = log *P*1 − log *P*2 3. log *Pn* = *n* log *P* 4. log 1 = 0 - -Historical note: The bel is named after Alexander Graham Bell, the inventor of the telephone. - -In communications systems, g ain is measured in *bels*. Historically, the bel is used to measure the ratio of two levels of power or power gain *G*; that is, - -$$ -G = \text{Number of bels} = \log_{10} \frac{P_2}{P_1} \tag{14.4} -$$ - -The *decibel* (dB) provides us with a unit of less magnitude. It is 1∕10th of a bel and is given by - -$$ -G_{\text{dB}} = 10 \log_{10} \frac{P_2}{P_1} \tag{14.5} -$$ - -When *P*1 = *P*2, there is no change in po wer and the g ain is 0 dB. If *P*2 = 2*P*1, the gain is - -$$ -G_{\rm dB} = 10 \log_{10} 2 \simeq 3 \text{ dB} \tag{14.6} -$$ - -and when *P*2 = 0.5*P*1, the gain is - -$$ -G_{\rm dB} = 10 \log_{10} 0.5 \simeq -3 \text{ dB} \tag{14.7} -$$ - -Equations (14.6) and (14.7) sho w another reason wh y log arithms are greatly used: The logarithm of the reciprocal of a quantity is simply negative the logarithm of that quantity. - -Alternatively, the g ain *G* can be e xpressed in terms of v oltage or current ratio. To do so, consider the network shown in Fig. 14.8. If *P*1 is the input power, *P*2 is the output (load) power, *R*1 is the input resistance, - -Voltage-current relationships for a fourterminal network. - -and *R*2 is the load resistance, then *P*1 = 0.5*V*2 1∕*R*1 and *P*2 = 0.5*V*2 2 ∕*R*2, and Eq. (14.5) becomes - -$$ -G_{\text{dB}} = 10 \log_{10} \frac{P_2}{P_1} = 10 \log_{10} \frac{V_2^2 / R_2}{V_1^2 / R_1} -$$ - -= 10 \log\_{10} \left(\frac{V\_2}{V\_1}\right)^2 + 10 \log\_{10} \frac{R\_1}{R\_2} (14.8) - -$$ -G_{\text{dB}} = 20 \log_{10} \frac{V_2}{V_1} - 10 \log_{10} \frac{R_2}{R_1} -$$ - (14.9) - -For the case when *R*2 = *R*1, a condition that is often assumed when comparing voltage levels, Eq. (14.9) becomes - -$$ -G_{\text{dB}} = 20 \log_{10} \frac{V_2}{V_1} -$$ - (14.10) - -Instead, if *P*1 = *I* 1 2 *R*1 and *P*2 = *I*2 2 *R*2, for *R*1 = *R*2, we obtain - -$$ -G_{\text{dB}} = 20 \log_{10} \frac{I_2}{I_1} \tag{14.11} -$$ - -Three things are important to note from Eqs. (14.5), (14.10), and (14.11): - -- 1. That 10 log 10 is used for po wer, while 20 log 10 is used for v oltage or current, because of the square relationship between them (*P* = *V*2 ∕*R* = *I* 2 *R*). -- 2. That the dB value is a logarithmic measurement of the *ratio* of one variable to another *of the same type*. Therefore, it applies in expressing the transfer function *H* in Eqs. (14.2a) and (14.2b), which are dimensionless quantities, but not in expressing *H* in Eqs. (14.2c) and (14.2d). -- 3. It is important to note that we only use voltage and current magnitudes in Eqs. (14.10) and (14.11). Negative signs and angles will be handled independently as we will see in Section 14.4. - -With this in mind, we now apply the concepts of logarithms and decibels to construct Bode plots. - -# **14.4** Bode Plots - -Obtaining the frequenc y response from the transfer function as we did in Section 14.2 is an uphill task. The frequency range required in fre quency response is often so wide that it is incon venient to use a linear scale for the frequenc y axis. Also, there is a more systematic w ay of locating the important features of the magnitude and phase plots of the transfer function. For these reasons, it has become standard practice to plot the transfer function on a pair of semilogarithmic plots: The magnitude in decibels is plotted against the logarithm of the frequency; on a separate plot, the phase in degrees is plotted against the logarithm of the frequency. Such semilogarithmic plots of the transfer function—kno wn as *Bode plots*—have become the industry standard. - -Bode plots are semilog plots of the magnitude (in decibels) and phase (in degrees) of a transfer function versus frequency. - -Historical note: Named after Hendrik W. Bode (1905–1982), an engineer with the Bell Telephone Laboratories, for his pioneering work in the 1930s and 1940s. - -Bode plots contain the same information as the nonlogarithmic plots discussed in the previous section, but they are much easier to construct, as we shall see shortly. - -The transfer function can be written as - -$$ -\mathbf{H} = H/\phi = He^{j\phi} \tag{14.12} -$$ - -Taking the natural logarithm of both sides, - -$$ -\ln H = \ln H + \ln e^{j\phi} = \ln H + j\phi \tag{14.13} -$$ - -Thus, the real part of ln**H** is a function of the magnitude while the imaginary part is the phase. In a Bode magnitude plot, the gain - -$$ -H_{\rm dB} = 20 \log_{10} H \tag{14.14} -$$ - -is plotted in decibels (dB) v ersus frequency. Table 14.2 provides a few values of *H* with the corresponding v alues in decibels. In a Bode phase plot, *ϕ* is plotted in degrees versus frequency. Both magnitude and phase plots are made on semilog graph paper. - -A transfer function in the form of Eq. (14.3) may be written in terms of factors that have real and imaginary parts. One such representation might be - -A transfer function in the form of Eq. (14.3) may be written in terms of -\nors that have real and imaginary parts. One such representation might be -\n -$$ -\mathbf{H}(\omega) = \frac{K(j\omega)^{\pm 1}(1 + j\omega/z_1)[1 + j2\zeta_1\omega/\omega_k + (j\omega/\omega_k)^2] \cdots}{(1 + j\omega/p_1)[1 + j2\zeta_2\omega/\omega_n + (j\omega/\omega_n)^2] \cdots} -$$ -\n(14.15) - -which is obtained by di viding out the poles and zeros in **H**(*ω*). The representation of **H**(*ω*) as in Eq. (14.15) is called the *standard form.* **H**(*ω*) may include up to seven types of different factors that can appear in various combinations in a transfer function. These are: - -- 1. A gain *K* -- 2. A pole (*jω*) −1 or zero (*jω*) at the origin -- 3. A simple pole 1∕(1 + *jω*∕*p*1) or zero (1 + *jω*∕*z*1) -- 4. A quadratic pole 1 ∕[1 + *j*2*ζ*2*ω*∕*ωn* + ( *jω*∕*ωn*) 2 ] or zero [1 + *j*2*ζ*1*ω*∕*ωk* + ( *jω*∕*ωk*) 2 ] - -In constructing a Bode plot, we plot each factor separately and then add them graphically. The factors can be considered one at a time and then combined additively because of the logarithms involved. It is this mathematical convenience of the logarithm that makes Bode plots a powerful engineering tool. 0 *ϕ* - -We will now make straight-line plots of the factors listed above. We shall find that these straight-line plots known as Bode plots approximate the actual plots to a reasonable degree of accuracy. 0.1 1 10 100 *ω* - -**Constant term:** For the g ain *K*, the magnitude is 20 log 10 *K* and the phase is 0°; both are constant with frequenc y. Thus, the magnitude and phase plots of the gain are shown in Fig. 14.9. If *K* is negative, the magnitude remains 20 log10 ∣*K*∣ but the phase is ±180°. - -**Pole/zero at the origin:** For the zero (*jω*) at the origin, the magnitude is 20 log10 *ω* and the phase is 90°. These are plotted in Fig. 14.10, where we notice that the slope of the magnitude plot is 20 dB/decade, while the phase is constant with frequency. - -The Bode plots for the pole (*jω*) −1 are similar except that the slope of the magnitude plot is −20 dB/decade while the phase is −90°. In general, - -# **TABLE 14.2** - -Specific gain and their decibel values.\* - -| Magnitude H | 20 log10
H (dB) | -|--------------|--------------------| -| 0.001 | −60 | -| 0.01 | −40 | -| 0.1 | −20 | -| 0.5
__ | −6 | -| 1∕ √
2 | −3 | -| 1 | 0 | -| __
√
2 | 3 | -| 2 | 6 | -| 10 | 20 | -| 20 | 26 | -| 100 | 40 | -| 1000 | 60 | -| | | - -\* Some of these values are approximate. - -The origin is where *ω* = 1 or log *ω* = 0 and the gain is zero. - -**Figure 14.9** - -Bode plots for gain *K*: (a) magnitude plot, (b) phase plot. - -for (*jω*) *N*, where *N* is an integer, the magnitude plot will have a slope of 20*N* dB/decade, while the phase is 90*N* degrees. - -**Simple pole/zero:** For the simple zero (1 + *jω*∕*z*1), the magnitude is 20 log10 ∣1 + *jω*∕*z*1∣ and the phase is tan−1 *ω*∕*z*1. We notice that - -$$ -H_{\text{dB}} = 20 \log_{10} \left| 1 + \frac{j\omega}{z_1} \right| \qquad \Rightarrow \qquad 20 \log_{10} 1 = 0 \qquad \textbf{(14.16)} -$$ -\n -$$ -\text{as} \quad \omega \to 0 -$$ -\n -$$ -H_{\text{dB}} = 20 \log_{10} \left| 1 + \frac{j\omega}{z_1} \right| \qquad \Rightarrow \qquad 20 \log_{10} \frac{\omega}{z_1} \qquad \textbf{(14.17)} -$$ -\n -$$ -\text{as} \quad \omega \to \infty -$$ - -showing that we can approximate the magnitude as zero (a straight line with zero slope) for small v alues of *ω* and by a straight line with slope 20 dB/decade for large values of *ω*. The frequency *ω* = *z*1 where the two asymptotic lines meet is called the *corner frequency* or *break frequency*. Thus, the approximate magnitude plot is sho wn in Fig. 14.11(a), where the actual plot is also shown. Notice that the approximate plot is close to the actual plot except at the break frequency, where *ω* = *z*1 and the deviation is 20 log10 ∣(1 + *j*1)∣ = 20 log10 √ \_\_ 2 ≃ 3 dB. - -The phase tan−1 (*ω*∕*z*1) can be expressed as - -$$ -\phi = \tan^{-1}\left(\frac{\omega}{z_1}\right) = \begin{cases} 0, & \omega = 0 \\ 45^\circ, & \omega = z_1 \\ 90^\circ, & \omega \to \infty \end{cases} -$$ - (14.18) - -As a straight-line approximation, we let *ϕ*≃ 0 for *ω*≤ *z*1∕10, *ϕ* ≃ 45° for *ω* = *z*1, and *ϕ*≃ 90° for *ω* ≥ 10*z*1. As shown in Fig. 14.11(b) along with the actual plot, the straight-line plot has a slope of 45° per decade. - -The Bode plots for the pole 1 ∕(1 + *jω*∕*p*1) are similar to those in Fig. 14.11 except that the corner frequency is at *ω* = *p*1, the magnitude has a slope of −20 dB/decade, and the phase has a slope of −45° per decade. - -**Quadratic pole/zer o:** The magnitude of the quadratic pole 1 ∕[1 + *j*2*ζ*2*ω*∕*ωn* + (*jω*∕*ωn*) 2 ] is −20 log10∣1 + *j*2*ζ*2*ω*∕*ωn* + ( *jω*∕*ωn*) 2 ∣ and the phase is −tan−1 (2*ζ*2*ω*∕*ωn*)∕(1 − *ω*2 ∕*ωn* 2 ). But - -A decade is an interval between two frequencies with a ratio of 10; e.g., between *ω*0 and 10*ω*0, or between 10 and 100 Hz. Thus, 20 dB/decade means that the magnitude changes 20 dB whenever the frequency changes tenfold or one decade. - -The special case of dc (*ω* = 0) does not appear on Bode plots because log 0 = −∞, implying that zero frequency is infinitely far to the left of the origin of Bode plots. - -# **Figure 14.10** - -Bode plot for a zero ( *jω*) at the origin: (a) magnitude plot, (b) phase plot. - -**(14.19)** - -Bode plots of zero (1 + *jω*∕*z*1): (a) magnitude plot, (b) phase plot. - -and - -$$ -H_{\text{dB}} = -20 \log_{10} \left| 1 + \frac{j2\zeta_2 \omega}{\omega_n} + \left( \frac{j\omega}{\omega_n} \right)^2 \right| \qquad \Rightarrow \qquad -40 \log_{10} \frac{\omega}{\omega_n} -$$ -as $\omega \to \infty$ (14.20) - -Thus, the amplitude plot consists of tw o straight asymptotic lines: one with zero slope for *ω*< *ωn* and the other with slope −40 dB/decade for *ω* > *ωn*, with *ωn* as the corner frequenc y. Figure 14.12(a) sho ws the approximate and actual amplitude plots. Note that the actual plot depends on the damping factor *ζ*2 as well as the corner frequency *ωn*. The significant peaking in the neighborhood of the corner frequency should be added to the straight-line approximation if a high le vel of accurac y is desired. However, we will use the straight-line approximation for the sake of simplicity. - -**Figure 14.12** Bode plots of quadratic pole [1 + *j*2*ζω*∕*ωn* − *ω*2 ∕ *ωn* 2 ] −1 : (a) magnitude plot, (b) phase plot. - -There is another procedure for obtaining Bode plots that is faster and perhaps more efficient than the one we have just discussed. It consists in realizing that zeros cause an increase in slope, while poles cause a decrease. By starting with the low-frequency asymptote of the Bode plot, moving along the frequency axis, and increasing or decreasing the slope at each corner frequency, one can sketch the Bode plot immediately from the transfer function without the effort of making individual plots and adding them. This procedure can be used once you become proficient in the one discussed here. - - Digital computers have rendered the procedure discussed here almost obsolete. Several software packages such as PSpice, MATLAB, Mathcad, and Micro-Cap can be used to generate frequency response plots. We will discuss PSpice later in the chapter. - -# The phase can be expressed as - -$$ -\phi = -\tan^{-1} \frac{2\zeta_2 \omega / \omega_n}{1 - \omega^2 / \omega_n^2} = \begin{cases} 0, & \omega = 0 \\ -90^\circ, & \omega = \omega_n \\ -180^\circ, & \omega \to \infty \end{cases} -$$ - (14.21) - -The phase plot is a straight line with a slope of −90° per decade starting at *ωn*∕10 and ending at 10*ωn*, as shown in Fig. 14.12(b). We see again that the difference between the actual plot and the straight-line plot is due to the damping f actor. Notice that the straight-line approximations for both magnitude and phase plots for the quadratic pole are the same as those for a double pole, that is, (1 + *jω*∕*ωn*) −2 . We should e xpect this because the double pole (1 + *jω*∕*ωn*) −2 equals the quadratic pole 1∕[1 + *j*2*ζ*2*ω*∕*ωn* + (*jω*∕*ωn*) 2 ] when *ζ*2 = 1. Thus, the quadratic pole can be treated as a double pole as far as straight-line approximation is concerned. - -For the quadratic zero [1 + *j*2*ζ*1*ω*∕*ωk* + (*jω*∕*ωk*) 2 ], the plots in Fig. 14.12 are in verted because the magnitude plot has a slope of 40 dB/decade while the phase plot has a slope of 90° per decade. - -Table 14.3 presents a summary of Bode plots for the se ven factors. Of course, not every transfer function has all seven factors. To sketch the Bode plots for a function **H**(*ω*) in the form of Eq. (14.15), for e xample, we first record the corner frequencies on the semilog graph paper, sketch the factors one at a time as discussed above, and then combine additively - -the graphs of the f actors. The combined graph is often dra wn from left to right, changing slopes appropriately each time a corner frequenc y is encountered. The following examples illustrate this procedure. - -Example 14.3 Construct the Bode plots for the transfer function - -ots for the transfer function -\n -$$ -\mathbf{H}(\omega) = \frac{200j\omega}{(j\omega + 2)(j\omega + 10)} -$$ - -# **Solution:** - -We first put **H**(*ω*) in the standard form by dividing out the poles and zeros. Thus, - -We first put -$$ -\mathbf{H}(\omega) -$$ - in the standard form by dividing out the poles and -zeros. Thus, -$$ -\mathbf{H}(\omega) = \frac{10j\omega}{(1 + j\omega/2)(1 + j\omega/10)} -$$ -$$ -= \frac{10 |j\omega|}{|1 + j\omega/2||1 + j\omega/10|} \frac{(90^\circ - \tan^{-1} \omega/2 - \tan^{-1} \omega/10)}{1 - j\omega/10} -$$ - -Hence, the magnitude and phase are - -$$ -H_{\text{dB}} = 20 \log_{10} 10 + 20 \log_{10} |j\omega| - 20 \log_{10} \left| 1 + \frac{j\omega}{2} \right| -$$ -$$ -- 20 \log_{10} \left| 1 + \frac{j\omega}{10} \right| -$$ -$$ -\phi = 90^{\circ} - \tan^{-1} \frac{\omega}{2} - \tan^{-1} \frac{\omega}{10} -$$ - -We notice that there are two corner frequencies at *ω* = 2,10. For both the magnitude and phase plots, we sketch each term as shown by the dotted lines in Fig. 14.13. We add them up graphically to obtain the overall plots shown by the solid curves. - -**Figure 14.13** For Example 14.3: (a) magnitude plot, (b) phase plot. - -$$ -\mathbf{H}(\omega) = \frac{5(j\omega + 2)}{j\omega(j\omega + 10)} -$$ - -**Answer:** See Fig. 14.14. - -$$ -\mathbf{H}(\omega) = \frac{j\omega + 10}{j\omega(j\omega + 5)^2} -$$ - -# **Solution:** - -Putting **H**(*ω*) in the standard form, we get - -$$ -\text{H}(\omega) = \frac{0.4(1 + j\omega/10)}{j\omega(1 + j\omega/5)^2} -$$ - -From this, we obtain the magnitude and phase as - -$$ -H_{\text{dB}} = 20 \log_{10} 0.4 + 20 \log_{10} \left| 1 + \frac{j\omega}{10} \right| - 20 \log_{10} |j\omega| -$$ -$$ -- 40 \log_{10} \left| 1 + \frac{j\omega}{5} \right| -$$ -$$ -\phi = 0^{\circ} + \tan^{-1} \frac{\omega}{10} - 90^{\circ} - 2 \tan^{-1} \frac{\omega}{5} -$$ - -There are two corner frequencies at *ω* = 5, 10 rad/s. For the pole with corner frequency at *ω* = 5, the slope of the magnitude plot is −40 dB/decade and that of the phase plot is −90° per decade due to the power of 2. The - -Obtain the Bode plots for Example 14.4 - -magnitude and the phase plots for the individual terms (in dotted lines) and the entire **H**( *jω*) (in solid lines) are in Fig. 14.15. - -# **Figure 14.15** - -Bode plots for Example 14.4: (a) magnitude plot, (b) phase plot. - -# **Figure 14.16** - -For Practice Prob. 14.4: (a) magnitude plot, (b) phase plot. - -Example 14.5 Draw the Bode plots for - -$$ -H(s) = \frac{s+1}{s^2 + 12s + 100} -$$ - -# **Solution:** - -- 1. **Define.** The problem is clearly stated and we follo w the technique outlined in the chapter. -- 2. **Present.** We are to develop the approximate Bode plot for the given function, **H**(*s*). -- 3. **Alternative.** The two most effective choices would be the approximation technique outlined in the chapter , which we will use here, and *MATLAB*, which can actually give us the exact Bode plots. - -# 4. **Attempt.** We express **H**(*s*) as - -express **H**(*s*) as -$$ -\mathbf{H}(\omega) = \frac{1/100(1 + j\omega)}{1 + j\omega 1.2/10 + (j\omega/10)^2} -$$ - - For the quadratic pole, *ωn* = 10 rad/s, which serves as the corner frequency. The magnitude and phase are - -$$ -H_{\text{dB}} = -20 \log_{10} 100 + 20 \log_{10} |1 + j\omega| -$$ -$$ -- 20 \log_{10} \left| 1 + \frac{j\omega 1.2}{10} - \frac{\omega^2}{100} \right| -$$ -$$ -\phi = 0^\circ + \tan^{-1} \omega - \tan^{-1} \left[ \frac{\omega 1.2/10}{1 - \omega^2/100} \right] -$$ - - Figure 14.17 shows the Bode plots. Notice that the quadratic pole is treated as a repeated pole at *ωk*, that is, (1 + *jω*∕*ωk*) 2 , which is an approximation. - -**Figure 14.17** Bode plots for Example 14.5: (a) magnitude plot, (b) phase plot. - -5. **Evaluate.** Although we could use *MATLAB* to validate the solution, we will use a more straightforward approach. First, we must realize that the denominator assumes that *ζ* = 0 for the approximation, so we will use the following equation to check our answer: - -$$ -\mathbf{H}(s) \simeq \frac{s+1}{s^2 + 10^2} -$$ - - We also note that we need to actually solve for *H*dB and the corresponding phase angle *ϕ*. First, let *ω* = 0. - -$$ -H_{\text{dB}} = 20 \log_{10}(1/100) = -40 -$$ - and $\phi = 0^{\circ}$ - -Now try *ω* = 1. - -$$ -H_{\rm dB} = 20 \log_{10}(1.4142/99) = -36.9 \text{ dB} -$$ - -which is the expected 3 dB up from the corner frequency. - -$$ -\phi = 45^{\circ} -$$ - from $\mathbf{H}(j) = \frac{j+1}{-1+100}$ - -Now try *ω* = 100. - -*H*dB = 20 log10 (100) − 20 log10 (9900) = 39.91 dB - -*ϕ* is 90° from the numerator minus 180°, which gives −90°. We now have checked three different points and got close agreement, and, because this is an approximation, we can feel confident that we have worked the problem successfully. - - You can reasonably ask why did we not check at *ω* = 10? If we just use the approximate value we used above, we end up with an infinite value, which is to be expected from *ζ* = 0 (see Fig. 14.12a). If we used the actual value of **H**( *j*10) we will still end up being far from the approximate values, since *ζ* = 0.6 and Fig. 14.12a shows a significant deviation from the approximation. We could have reworked the problem with *ζ* = 0.707, which would have gotten us closer to the approximation. However, we really have enough points without doing this. - -6. **Satisfactory?** We are satisfied the problem has been worked successfully and we can present the results as a solution to the problem. - -Practice Problem 14.5 Construct the Bode plots for - -**Answer:** See Fig. 14.18. - -$$ -H(s) = \frac{10}{s(s^2 + 80s + 400)} -$$ - -For Practice Prob. 14.5: (a) magnitude plot, (b) phase plot. - -Example 14.6 Given the Bode plot in Fig. 14.19, obtain the transfer function **H**(*ω*). - -# **Solution:** - -To obtain **H**(*ω*) from the Bode plot, we keep in mind that a zero al ways causes an upward turn at a corner frequency, while a pole causes - -40 dB - -H - -0 - -**Figure 14.19** For Example 14.6. - -a downward turn. We notice from Fig. 14.19 that there is a zero *jω* at the origin, which should have intersected the frequency axis at *ω* = 1. This is indicated by the straight line with slope +20 dB/decade. The fact that this straight line is shifted by 40 dB indicates that there is a 40-dB gain; that is, - -$$ -40 = 20 \log_{10} K \qquad \Rightarrow \qquad \log_{10} K = 2 -$$ - -or - -$$ -K = 10^2 = 100 -$$ - -In addition to the zero *jω* at the origin, we notice that there are three factors with corner frequencies at *ω* = 1, 5, and 20 rad/s. Thus, we have: - -- 1. A pole at *p* = 1 with slope −20 dB/decade to cause a downward turn and counteract the zero at the origin. The pole at *p* = 1 is determined as 1∕(1 + *jω*∕1). -- 2. Another pole at *p* = 5 with slope −20 dB/decade causing a do wnward turn. The pole is 1∕(1 + *jω*∕5). -- 3. A third pole at *p* = 20 with slope −20 dB/decade causing a further downward turn. The pole is 1∕(1 + *jω*∕20). - -Putting all these together gives the corresponding transfer function as - -H turn. The pole is -$$ -1/(1 + j\omega/20) -$$ -. -\nthese together gives the corresponding transfer in -\n -$$ -\mathbf{H}(\omega) = \frac{100 j\omega}{(1 + j\omega/1)(1 + j\omega/5)(1 + j\omega/20)} -$$ -\n -$$ -= \frac{j\omega 10^4}{(j\omega + 1)(j\omega + 5)(j\omega + 20)} -$$ - -or - -$$ -(j\omega + 1)(j\omega + 3)(j\omega + 20) -$$ - -$$ -\mathbf{H}(s) = \frac{10^4 s}{(s+1)(s+5)(s+20)}, \qquad s = j\omega -$$ - -Obtain the transfer function H( *ω*) corresponding to the Bode plot in Practice Problem 14.6 Fig. 14.20. - -**Answer: H**(*ω*) = 2,000,000(*s* + 5) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*s* + 10)(*s* + 100)2 . - -To see how to use MATLAB to produce Bode plots, refer to Section 14.11. - -0.1 1 5 10 20 100 - -‒40 dB/decade - -+20 dB/decade - -# **14.5** Series Resonance - -The most prominent feature of the frequenc y response of a circuit may be the sharp peak (or *resonant peak* ) exhibited in its amplitude char acteristic. The concept of resonance applies in se veral areas of science and engineering. Resonance occurs in an y system that has a comple x conjugate pair of poles; it is the cause of oscillations of stored ener gy from one form to another . It is the phenomenon that allo ws frequency - -*ω* - -‒20 dB/decade - -discrimination in communications netw orks. Resonance occurs in an y circuit that has at least one inductor and one capacitor. - -Resonance is a condition in an RLC circuit in which the capacitive and inductive reactances are equal in magnitude, thereby resulting in a purely resistive impedance. - -Resonant circuits (series or parallel) are useful for constructing filters, as their transfer functions can be highly frequency selective. They are used in many applications such as selecting the desired stations in radio and TV receivers. - -Consider the series *RLC* circuit shown in Fig. 14.21 in the frequency domain. The input impedance is - -$$ -\mathbf{Z} = \mathbf{H}(\omega) = \frac{\mathbf{V}_s}{\mathbf{I}} = R + j\omega L + \frac{1}{j\omega C} -$$ - (14.22) - -or - -$$ -\mathbf{Z} = R + j \left( \omega L - \frac{1}{\omega C} \right) \tag{14.23} -$$ - -Resonance results when the imaginary part of the transfer function is zero, or - -$$ -\operatorname{Im}(\mathbf{Z}) = \omega L - \frac{1}{\omega C} = 0 \tag{14.24} -$$ - -The value of *ω* that satisfies this condition is called the *resonant frequency ω*0. Thus, the resonance condition is - -$$ -\omega_0 L = \frac{1}{\omega_0 C} \tag{14.25} -$$ - -or - -$$ -\omega_0 = \frac{1}{\sqrt{LC}} \text{rad/s} \tag{14.26} -$$ - -Since *ω*0 = 2 *π f*0, - -$$ -f_0 = \frac{1}{2\pi\sqrt{LC}} \text{Hz} -$$ - (14.27) - -Note that at resonance: - -- 1. The impedance is purely resistive, thus, **Z** = *R*. In other words, the *LC* series combination acts like a short circuit, and the entire voltage is across *R*. -- 2. The voltage **V***s* and the current **I** are in phase, so that the po wer factor is unity. -- 3. The magnitude of the transfer function **H**(*ω*) = **Z**(*ω*) is minimum. -- 4. The inductor v oltage and capacitor v oltage can be much more than the source voltage. - -The frequency response of the circuit's current magnitude - -response of the circuit's current magnitude -\n -$$ -I = |\mathbf{I}| = \frac{V_m}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}} -$$ -\n(14.28) - -$$ -|\mathbf{V}_L| = \frac{V_m}{R} \omega_0 L = Q V_m -$$ -$$ -|\mathbf{V}_C| = \frac{V_m}{R} \frac{1}{\omega_0 C} = Q V_m -$$ - -where Q is the quality factor, defined in Eq. (14.38). - -The series resonant circuit. - -is shown in Fig. 14.22; the plot only sho ws the symmetry illustrated in this graph when the frequenc y axis is a log arithm. The average power dissipated by the *RLC* circuit is - -$$ -P(\omega) = \frac{1}{2} \hat{I}^2 R \tag{14.29} -$$ - -The highest po wer dissipated occurs at resonance, when *I* = *Vm*∕*R*, so that - -$$ -P(\omega_0) = \frac{1}{2} \frac{V_m^2}{R} -$$ - (14.30) - -At certain frequencies *ω* = *ω*1, *ω*2, the dissipated power is half the maximum value; that is, - -$$ -P(\omega_1) = P(\omega_2) = \frac{(V_m/\sqrt{2})^2}{2R} = \frac{V_m^2}{4R} -$$ - (14.31) - -Hence, *ω*1 and *ω*2 are called the *half-power frequencies.* - -The half-power frequencies are obtained by setting *Z* equal to √ \_\_ 2 *R*, and writing - -equences are obtained by setting Z equal to V2R, -\n -$$ -\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} = \sqrt{2}R -$$ -\n(14.32) - -Solving for *ω*, we obtain - -$$ -\omega_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}} -$$ -\n -$$ -\omega_2 = \frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}} -$$ -\n(14.33) - -We can relate the half-po wer frequencies with the resonant frequenc y. From Eqs. (14.26) and (14.33), - -$$ -\omega_0 = \sqrt{\omega_1 \omega_2} \tag{14.34} -$$ - -showing that the resonant frequenc y is the geometric mean of the halfpower frequencies. Notice that *ω*1 and *ω*2 are in general not symmetrical around the resonant frequency *ω*0, because the frequency response is not generally symmetrical. However, as will be explained shortly, symmetry of the half-power frequencies around the resonant frequenc y is often a reasonable approximation. - -Although the height of the curv e in Fig. 14.22 is determined by *R*, the width of the curv e depends on other f actors. The width of the re sponse curve depends on the *bandwidth B*, which is defined as the difference between the two half-power frequencies, - -$$ -B = \omega_2 - \omega_1 \tag{14.35} -$$ - -This definition of bandwidth is just one of several that are commonly used. Strictly speaking, *B* in Eq. (14.35) is a half-po wer bandwidth, because it is the width of the frequenc y band between the half-po wer frequencies. - -The "sharpness" of the resonance in a resonant circuit is measured quantitatively by the *quality factor Q*. At resonance, the reactive energy - -# **Figure 14.22** - -The current amplitude versus frequency for the series resonant circuit of Fig. 14.21. - -Although the same symbol Q is used for the reactive power, the two are not equal and should not be confused. Q here is dimensionless, whereas reactive power Q is in VAR. This may help distinguish between the two. - -in the circuit oscillates between the inductor and the capacitor. The quality factor relates the maximum or peak ener gy stored to the energy dissipated in the circuit per cycle of oscillation: - -is the maximum or peak energy stored to the energy dis- -rcuit per cycle of oscillation: - -\n -$$ -Q = 2\pi \frac{\text{Peak energy stored in the circuit}}{\text{Energy dissipated by the circuit}} \qquad (14.36) -$$ -\nin one period at resonance - -It is also regarded as a measure of the energy storage property of a circuit in relation to its energy dissipation property. In the series *RLC* circuit, the peak energy stored is \_\_1 2 *LI*2 , while the energy dissipated in one period is \_\_1 2 (*I* 2 *R*)(1∕*f*0). Hence, - -$$ -Q = 2\pi \frac{\frac{1}{2}LI^2}{\frac{1}{2}I^2R(1/f_0)} = \frac{2\pi f_0L}{R} -$$ - (14.37) - -B3 Q3 (greatest selectivity) Q2 (medium selectivity) Q1 (least selectivity) B2 B1 *ω* - -# Amplitude - -The higher the circuit *Q*, the smaller the bandwidth. - -The quality factor is a measure of the selectivity (or "sharpness" of resonance) of the circuit. - -or - -$$ -Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR} -$$ - (14.38) - -Notice that the quality factor is dimensionless. The relationship between the bandwidth *B* and the quality f actor *Q* is obtained by substituting Eq. (14.33) into Eq. (14.35) and utilizing Eq. (14.38). - -$$ -B = \frac{R}{L} = \frac{\omega_0}{Q} \tag{14.39} -$$ - -or *B* = *ω*0 2 *CR*. Thus, - -> The quality factor of a resonant circuit is the ratio of its resonant frequency to its bandwidth. - -Keep in mind that Eqs. (14.33), (14.38), and (14.39) only apply to a series *RLC* circuit. - -As illustrated in Fig. 14.23, the higher the v alue of *Q*, the more selective the circuit is but the smaller the bandwidth. The *selectivity* of an *RLC* circuit is the ability of the circuit to respond to a certain frequenc y and discriminate against all other frequencies. If the band of frequencies to be selected or rejected is narrow, the quality f actor of the resonant circuit must be high. If the band of frequencies is wide, the quality factor must be low. - -A resonant circuit is designed to operate at or near its resonant fre quency. It is said to be a *high-Q circuit* when its quality factor is equal to or greater than 10. F or high *-Q* circuits (*Q* ≥ 10), the half- power frequencies are, for all practical purposes, symmetrical around the resonant frequency and can be approximated as - -$$ -\omega_1 \simeq \omega_0 - \frac{B}{2}, \qquad \omega_2 \simeq \omega_0 + \frac{B}{2} \qquad (14.40) -$$ - -High-*Q* circuits are used often in communications networks. - -We see that a resonant circuit is characterized by five related parameters: the two half-power frequencies *ω*1 and *ω*2, the resonant frequency *ω*0, the bandwidth *B*, and the quality factor *Q*. - -In the circuit of Fig. 14.24, *R* = 2 Ω, *L* = 1 mH, and *C* = 0.4 *μ*F. (a) Find Example 14.7 the resonant frequency and the half-power frequencies. (b) Calculate the quality factor and bandwidth. (c) Determine the amplitude of the current at *ω*0, *ω*1, and *ω*2. - -# **Solution:** - -(a) The resonant frequency is - -ant frequency is -\n -$$ -\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-3} \times 0.4 \times 10^{-6}}} = 50 \text{ krad/s} -$$ - -■ **METHOD 1** The lower half-power frequency is - -$$ -\omega_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}} -$$ - -= $-\frac{2}{2 \times 10^{-3}} + \sqrt{(10^3)^2 + (50 \times 10^3)^2}$ -= $-1 + \sqrt{1 + 2500}$ krad/s = 49 krad/s - -Similarly, the upper half-power frequency is - -$$ -\omega_2 = 1 + \sqrt{1 + 2500} \text{ krad/s} = 51 \text{ krad/s} -$$ - -(b) The bandwidth is - -$$ -B = \omega_2 - \omega_1 = 2 \text{ krad/s} -$$ - -or - -$$ -B = \frac{R}{L} = \frac{2}{10^{-3}} = 2 \text{ krad/s} -$$ - -The quality factor is - -$$ -Q = \frac{\omega_0}{B} = \frac{50}{2} = 25 -$$ - -■ **METHOD 2** Alternatively, we could find - -Alternatively, we could find -\n -$$ -Q = \frac{\omega_0 L}{R} = \frac{50 \times 10^3 \times 10^{-3}}{2} = 25 -$$ - -From *Q*, we find - -$$ -B = \frac{\omega_0}{Q} = \frac{50 \times 10^3}{25} = 2 \text{ krad/s} -$$ - -Since *Q* > 10, this is a high-*Q* circuit and we can obtain the half-power frequencies as - -$$ -\omega_1 = \omega_0 - \frac{B}{2} = 50 - 1 = 49 \text{ krad/s} -$$ -$$ -\omega_2 = \omega_0 + \frac{B}{2} = 50 + 1 = 51 \text{ krad/s} -$$ - -as obtained earlier. - -(c) At *ω* = *ω*0, - -$$ -I = \frac{V_m}{R} = \frac{20}{2} = 10 \text{ A} -$$ - -At *ω* = *ω*1, *ω*2, - -$$ -I = \frac{V_m}{\sqrt{2} R} = \frac{10}{\sqrt{2}} = 7.071 \text{ A} -$$ - -Practice Problem 14.7 A series-connected circuit has *R* = 4 Ω and *L* = 25 mH. (a) Calculate the value of *C* that will produce a quality factor of 50. (b) Find *ω*1, *ω*2, and *B*. (c) Determine the average power dissipated at *ω* = *ω*0, *ω*1, *ω*2. Take *Vm* = 100 V. - -> **Answer:** (a) 0.625 *μ*F, (b) 7920 rad/s, 8080 rad/s, 160 rad/s, (c) 1.25 kW , 0.625 kW, 0.625 kW. - -# **14.6** Parallel Resonance - -The parallel *RLC* circuit in Fig. 14.25 is the dual of the series *RLC* circuit. So we will avoid needless repetition. The admittance is - -$$ -Y = H(\omega) = \frac{I}{V} = \frac{1}{R} + j\omega C + \frac{1}{j\omega L} -$$ - (14.41) - -or - -or - -$$ -\mathbf{Y} = \frac{1}{R} + j \left( \omega C - \frac{1}{\omega L} \right) \tag{14.42} -$$ - -Resonance occurs when the imaginary part of **Y** is zero, - -$$ -\omega C - \frac{1}{\omega L} = 0 \tag{14.43} -$$ - -$$ -\omega_0 = \frac{1}{\sqrt{LC}} \text{ rad/s} -$$ - (14.44) - -which is the same as Eq. (14.26) for the series resonant circuit. The voltage ∣**V**∣ is sketched in Fig. 14.26 as a function of frequenc y. Notice that at resonance, the parallel *LC* combination acts like an open circuit, so that the entire current flows through *R*. Also, the inductor and capacitor current can be much more than the source current at resonance. - -We exploit the duality between Figs. 14.21 and 14.25 by comparing Eq. (14.42) with Eq. (14.23). By replacing *R*, *L*, and *C* in the expressions - -**Figure 14.25** - -The parallel resonant circuit. - -The current amplitude versus frequency for the series resonant circuit of Fig. 14.25. - -We can see this from the fact that - -$$ -|\mathbf{I}_L| = \frac{I_m R}{\omega_0 L} = Q I_m -$$ - -$$ -|\mathbf{I}_C| = \omega_0 C I_m R = Q I_m -$$ - -where Q is the quality factor, defined in Eq. (14.47). - -for the series circuit with 1 ∕*R*, *C*, and *L* respectively, we obtain for the parallel circuit - -$$ -\omega_1 = -\frac{1}{2RC} + \sqrt{\left(\frac{1}{2RC}\right)^2 + \frac{1}{LC}} -$$ -\n -$$ -\omega_2 = \frac{1}{2RC} + \sqrt{\left(\frac{1}{2RC}\right)^2 + \frac{1}{LC}} -$$ -\n(14.45) - -$$ -B = \omega_2 - \omega_1 = \frac{1}{RC} -$$ - (14.46) - -$$ -Q = \frac{\omega_0}{B} = \omega_0 RC = \frac{R}{\omega_0 L} -$$ - (14.47) - -It should be noted that Eqs. (14.45) to (14.47) apply only to a parallel *RLC* circuit. Using Eqs. (14.45) and (14.47), we can e xpress the halfpower frequencies in terms of the quality factor. The result is - -$$ -\omega_1 = \omega_0 \sqrt{1 + \left(\frac{1}{2Q}\right)^2} - \frac{\omega_0}{2Q}, \qquad \omega_2 = \omega_0 \sqrt{1 + \left(\frac{1}{2Q}\right)^2} + \frac{\omega_0}{2Q} -$$ -\n(14.48) - -Again, for high-*Q* circuits (*Q* ≥ 10) - -$$ -\omega_1 \simeq \omega_0 - \frac{B}{2}, \qquad \omega_2 \simeq \omega_0 + \frac{B}{2} -$$ - (14.49) - -Table 14.4 presents a summary of the characteristics of the series and parallel resonant circuits. Besides the series and parallel *RLC* considered here, other resonant circuits exist. Example 14.9 treats a typical example. - -## **TABLE 14.4** - -Summary of the characteristics of resonant RLC circuits. - -| Characteristic | Series circuit | Parallel circuit | -|--------------------------------|--------------------------------------------------------|--------------------------------------------------------| -| Resonant frequency, ω0 | ____ 1
___
√
LC | ____ 1
___
√
LC | -| Quality factor, Q | ω0L ____
or _____ 1
ω0 RC
R | ____ R
or ω0RC
ω0 L | -| Bandwidth, B | ω0
___
Q
__________ | ω0
___
Q
__________ | -| Half-power frequencies, ω1, ω2 | ω0
___1
2
± ___
1 + (
ω0 √
2Q)
2Q | ω0
___1
2
± ___
1 + (
ω0 √
2Q)
2Q | -| For Q ≥ 10, ω1, ω2 | B
± __
ω0
2 | B
± __
ω0
2 | - -For Example 14.8. - -Example 14.8 In the parallel *RLC* circuit of Fig. 14.27, let *R* = 8 kΩ, *L* = 0.2 mH, and *C* = 8 *μ*F. (a) Calculate *ω*0, *Q*, and *B*. (b) Find *ω*1 and *ω*2. (c) Determine the power dissipated at *ω*0, *ω*1, and *ω*2. - -# **Solution:** - -(a) - -$$ -\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.2 \times 10^{-3} \times 8 \times 10^{-6}}} = \frac{10^5}{4} = 25 \text{ krad/s} -$$ -$$ -Q = \frac{R}{\omega_0 L} = \frac{8 \times 10^3}{25 \times 10^3 \times 0.2 \times 10^{-3}} = 1,600 -$$ -$$ -B = \frac{\omega_0}{Q} = 15.625 \text{ rad/s} -$$ - -(b) Due to the high value of *Q*, we can regard this as a high- *Q* circuit, Hence, - -$$ -\omega_1 = \omega_0 - \frac{B}{2} = 25,000 - 7.812 = 24,992 \text{ rad/s} -$$ -$$ -\omega_2 = \omega_0 + \frac{B}{2} = 25,000 + 7.812 = 25,008 \text{ rad/s} -$$ - -(c) At *ω* = *ω*0, **Y** = 1∕*R* or **Z** = *R* = 8 kΩ. Then - -$$ -I_o = \frac{V}{Z} = \frac{10/-90^{\circ}}{8,000} = 1.25/-90^{\circ} \text{ mA} -$$ - -Since the entire current flows through *R* at resonance, the average power dissipated at *ω* = *ω*0 is - -$$ -P = \frac{1}{2} |\mathbf{I}_o|^2 R = \frac{1}{2} (1.25 \times 10^{-3})^2 (8 \times 10^3) = 6.25 \text{ mW} -$$ - -or - -$$ -P = \frac{V_m^2}{2R} = \frac{100}{2 \times 8 \times 10^3} = 6.25 -$$ - mW - -At *ω* = *ω*1, *ω*2, - -$$ -P = \frac{V_m^2}{4R} = 3.125 \text{ mW} -$$ - -Practice Problem 14.8 A parallel resonant circuit has *R* = 100 kΩ, *L* = 50 mH, and *C* = 2 nF. Calculate *ω*0, *ω*1, *ω*2, *Q*, and *B*. - -**Answer:** 100 krad/s, 97.5 krad/s,102.5 krad/s, 20, 5 krad/s. - -# **Solution:** - -The input admittance is - -$$ -\mathbf{Y} = j\omega 0.1 + \frac{1}{10} + \frac{1}{2 + j\omega 2} = 0.1 + j\omega 0.1 + \frac{2 - j\omega 2}{4 + 4\omega^2} -$$ - -At resonance, Im(**Y**) = 0 and - -$$ -\omega_0 0.1 - \frac{2\omega_0}{4 + 4\omega_0^2} = 0 \qquad \Rightarrow \qquad \omega_0 = 2 \text{ rad/s} -$$ - -Calculate the resonant frequency of the circuit in Fig. 14.29. Practice Problem 14.9 - -**Answer:** 173.21 rad/s. - -# **Figure 14.29 14.7** Passive Filters For Practice Prob. 14.9 - -The concept of filters has been an integral part of the evolution of electrical engineering from the beginning. Several technological achievements would not have been possible without electrical filters. Because of this prominent role of filters, much effort has been expended on the theory, design, and construction of filters and many articles and books have been written on them. Our discussion in this chapter should be considered introductory. - -A filter is a circuit that is designed to pass signals with desired frequencies and reject or attenuate others. - -As a frequency-selective device, a filter can be used to limit the frequency spectrum of a signal to some specified band of frequencies. Filters are the circuits used in radio and TV receivers to allow us to select one desired signal out of a multitude of broadcast signals in the environment. - -A filter is a *passive filter* if it consists of only passive elements *R*, *L*, and *C*. It is said to be an *active filter* if it consists of acti ve elements (such as transistors and op amps) in addition to passi ve elements *R*, *L*, and *C*. We consider passive filters in this section and active filters in the next section. *LC* filters have been used in practical applications for more than eight decades. *LC* filter technology feeds related areas such as equalizers, impedance-matching networks, transformers, shaping networks, power dividers, attenuators, and directional couplers, and is continuously providing practicing engineers with opportunities to inno vate and experiment. Besides the *LC* filters we study in these sections, there are other kinds of filters—such as digital filters, electromechanical filters, and microwave filters—which are beyond the level of this text. - -0 (b) *ω*c *ω* 1 0 (a) *ω*c *ω* 1 0 (c) *ω*1 *ω*2 *ω* 1 0 Passband Passband Passband Stopband Stopband Stopband Passband Passband Stopband Stopband *ω*1 *ω*2 *ω* 1 │H(*ω*)│ │H(*ω*)│ │H(*ω*)│ │H(*ω*)│ - -# **Figure 14.30** - -Ideal frequency response of four types of filters: (a) low-pass filter, (b) high-pass filter, (c) band-pass filter, (d) band-stop filter. - -(d) - -**Figure 14.31** A low-pass filter. - -**Figure 14.32** Ideal and actual frequency response of a low-pass filter. - -As shown in Fig. 14.30, there are four types of filters whether passive or active: - -- 1. A *low-pass filter* passes low frequencies and stops high frequencies, as shown ideally in Fig. 14.30(a). -- 2. A *high-pass filter* passes high frequencies and rejects low frequencies, as shown ideally in Fig. 14.30(b). -- 3. A *band-pass filter* passes frequencies within a frequenc y band and blocks or attenuates frequencies outside the band, as sho wn ideally in Fig. 14.30(c). -- 4. A *band-stop filter* passes frequencies outside a frequenc y band and blocks or attenuates frequencies within the band, as sho wn ideally in Fig. 14.30(d). - -Table 14.5 presents a summary of the characteristics of these filters. Be aware that the characteristics in Table 14.5 are only valid for first- or second-order filters—but one should not have the impression that only these kinds of filter exist. We now consider typical circuits for realizing the filters shown in Table 14.5. - -# **TABLE 14.5** - -Summary of the characteristics of ideal filters. - -| Type of Filter | H(0) | H(∞) | H(ωc) or H(ω0) | -|----------------|------|------|-----------------------| -| Low-pass | 1 | 0 | __
1∕ √
2
__ | -| High-pass | 0 | 1 | 1∕ √
2 | -| Band-pass | 0 | 0 | 1 | -| Band-stop | 1 | 1 | 0 | - -*ωc* is the cutoff frequency for low-pass and high-pass filters; *ω*0 is the center frequency for band-pass and band-stop filters. - -# **14.7.1** Low-Pass Filter - -A typical low-pass filter is formed when the output of an *RC* circuit is taken off the capacitor as shown in Fig. 14.31. The transfer function (see also Example 14.1) is - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{1/j\omega C}{R + 1/j\omega C} -$$ -$$ -\mathbf{H}(\omega) = \frac{1}{1 + j\omega RC} -$$ -(14.50) - -Note that **H**(0) = 1, **H**(∞) = 0. Figure 14.32 shows the plot of ∣*H*(*ω*)∣, along with the ideal characteristic. The half-power frequency, which is equivalent to the corner frequency on the Bode plots but in the context of filters is usually known as the *cutoff frequency ωc*, is obtained by setting the magnitude of **H**(*ω*) equal to 1∕ √ \_\_ 2 , thus, - -$$ -H(\omega_c) = \frac{1}{\sqrt{1 + \omega_c^2 R^2 C^2}} = \frac{1}{\sqrt{2}} -$$ - -or - -$$ -\omega_c = \frac{1}{RC} \tag{14.51} -$$ - -The cutoff frequency is also called the *rolloff frequency*. - -A low-pass filter is designed to pass only frequencies from dc up to the cutoff frequency *ω*c. - -A low-pass filter can also be formed when the output of an *RL* circuit is taken off the resistor. Of course, there are many other circuits for low-pass filters. - -# **14.7.2** High-Pass Filter - -A high-pass filter is formed when the output of an *RC* circuit is taken off the resistor as shown in Fig. 14.33. The transfer function is - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{R}{R + 1/j\omega C} -$$ -$$ -\mathbf{H}(\omega) = \frac{j\omega RC}{1 + j\omega RC} -$$ -(14.52) - -Note that **H**(0) = 0, **H**(∞) = 1. Figure 14.34 shows the plot of ∣*H*(*ω*)∣. Again, the corner or cutoff frequency is - -$$ -\omega_c = \frac{1}{RC} \tag{14.53} -$$ - -A high-pass filter is designed to pass all frequencies above its cutoff frequency *ω*c. - -A high-pass filter can also be formed when the output of an *RL* circuit is taken off the inductor. - -# **14.7.3** Band-Pass Filter - -The *RLC* series resonant circuit provides a band-pass filter when the output is taken off the resistor as shown in Fig. 14.35. The transfer function is - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{R}{R + j(\omega L - 1/\omega C)} -$$ -(14.54) - -We observe that **H**(0) = 0, **H**(∞) = 0. Figure 14.36 shows the plot of ∣*H*(*ω*)∣. The band-pass filter passes a band of frequencies (*ω*1 < *ω*< *ω*2) centered on *ω*0, the center frequency, which is given by - -$$ -\omega_0 = \frac{1}{\sqrt{LC}}\tag{14.55} -$$ - -A band-pass filter is designed to pass all frequencies within a band of frequencies, *ω*1 < *ω*< *ω*2. - -Because the band-pass filter in Fig. 14.35 is a series resonant circuit, the half-power frequencies, the bandwidth, and the quality factor are determined as in Section 14.5. A band-pass filter can also be formed by cascading the low-pass filter (where *ω*2 = *ωc*) in Fig. 14.31 with the The cutoff frequency is the frequency at which the transfer function **H** drops in magnitude to 70.71% of its maximum value. It is also regarded as the frequency at which the power dissipated in a circuit is half of its maximum value. - -# **Figure 14.34** - -Ideal and actual frequency response of a high-pass filter. - -**Figure 14.35** - -# **Figure 14.36** Ideal and actual frequency response of a band-pass filter. - -high-pass filter (where *ω*1 = *ωc*) in Fig. 14.33. However, the result would not be the same as just adding the output of the low-pass filter to the input of the high-pass filter, because one circuit loads the other and alters the desired transfer function. - -# **14.7.4** Band-Stop Filter - -A filter that prevents a band of frequencies between two designated values (*ω*1 and *ω*2) from passing is variably known as a *band-stop, bandreject*, or *notch* filter. A band-stop filter is formed when the output *RLC* series resonant circuit is taken off the *LC* series combination as shown in Fig. 14.37. The transfer function is - -nster function is -\n -$$ -\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{j(\omega L - 1/\omega C)}{R + j(\omega L - 1/\omega C)} -$$ -\n(14.56) - -Notice that **H**(0) = 1, **H**(∞) = 1. Figure 14.38 shows the plot of ∣*H*(*ω*)∣. Again, the center frequency is given by - -$$ -\omega_0 = \frac{1}{\sqrt{LC}}\tag{14.57} -$$ - -while the half-power frequencies, the bandwidth, and the quality factor are calculated using the formulas in Section 14.5 for a series reso nant circuit. Here, *ω*0 is called the *frequency of rejection*, while the corresponding bandwidth (*B* = *ω*2 − *ω*1) is known as the *bandwidth of rejection*. Thus, - -A band-stop filter is designed to stop or eliminate all frequencies within a band of frequencies, *ω*1 < *ω*< *ω*2. - -Notice that adding the transfer functions of the band-pass and the band-stop gives unity at any frequency for the same values of *R, L,* and *C*. Of course, this is not true in general b ut true for the circuits treated here. This is due to the fact that the characteristic of one is the inverse of the other. - -In concluding this section, we should note that: - -- 1. From Eqs. (14.50), (14.52), (14.54), and (14.56), the maximum gain of a passive filter is unity. To generate a gain greater than unity, one should use an active filter as the next section shows. -- 2. There are other ways to get the types of filters treated in this section. -- 3. The filters treated here are the simple types. Many other filters have sharper and complex frequency responses. - -Example 14.10 Determine what type of filter is shown in Fig. 14.39. Calculate the corner or cutoff frequency. Take *R* = 2 kΩ, *L* = 2 H, and *C* = 2 *μ*F. - -# **Solution:** - -The transfer function is - -ction is -\n -$$ -\mathbf{H}(s) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \frac{R||1/sC}{sL + R||1/sC}, \qquad s = j\omega -$$ -\n(14.10.1) - -**Figure 14.37** A band-stop filter. - -band-stop filter. - -Ideal and actual frequency response of a - -But - -$$ -R\left\|\frac{1}{sC} = \frac{R/sC}{R+1/sC} = \frac{R}{1+sRC} -$$ - -Substituting this into Eq. (14.10.1) gives - -uting this into Eq. (14.10.1) gives -\n -$$ -\mathbf{H}(s) = \frac{R/(1+sRC)}{sL + R/(1+sRC)} = \frac{R}{s^2RLC + sL + R}, \qquad s = j\omega -$$ - -or - -$$ -R/(1 + sRC) \t s2RLC + sL + R' -$$ - -$$ -H(\omega) = \frac{R}{-\omega^2 RLC + j\omega L + R} -$$ - (14.10.2) - -Because **H**(0) = 1 and **H**(∞) = 0, we conclude from Table 14.5 that the circuit in Fig. 14.39 is a second-order low-pass filter. The magnitude of **H** is *H* =  \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ *R √* - -a second-order low-pass filter. The magnitude of -$$ -H = \frac{R}{\sqrt{(R - \omega^2 R LC)^2 + \omega^2 L^2}} -$$ -(14.10.3) - -The corner frequency is the same as the half-power frequency, that is, where **H** is reduced by a factor of 1 ∕ *√ \_\_* 2 . Because the dc value of *H*(*ω*) is 1, at the corner frequency, Eq. (14.10.3) becomes after squaring - -where **H** is reduced by a factor of 1 / -$$ -\sqrt{2} -$$ -. Because the d -is 1, at the corner frequency, Eq. (14.10.3) becomes after -$$ -H^2 = \frac{1}{2} = \frac{R^2}{(R - \omega_c^2 R LC)^2 + \omega_c^2 L^2} -$$ -or - -$$ -2 = (1 - \omega_c^2 LC)^2 + \left(\frac{\omega_c L}{R}\right)^2 -$$ - -Substituting the values of *R*, *L*, and *C*, we obtain - -$$ -2 = (1 - \omega_c^2 4 \times 10^{-6})^2 + (\omega_c 10^{-3})^2 -$$ - -Assuming that *ωc* is in krad/s, - -$$ -2 = (1 - 4\omega_c^2)^2 + \omega_c^2 \qquad \text{or} \qquad 16\omega_c^4 - 7\omega_c^2 - 1 = 0 -$$ - -Solving the quadratic equation in *ωc* 2 , we get *ωc* 2  = 0.5509 and −0.1134. Because *ωc* is real, - -$$ -\omega_c = 0.742 \text{ krad/s} = 742 \text{ rad/s} -$$ - -For the circuit in Fig. 14.40, obtain the transfer function **V***o*(*ω*)∕**V***i*(*ω*). Practice Problem 14.10 Identify the type of filter the circuit represents and determine the corner frequency. Take *R*1 = 100 Ω = *R*2, *L* = 2 mH. - -Answer: -$$ -\frac{R_2}{R_1 + R_2} \left( \frac{j\omega}{j\omega + \omega_c} \right) -$$ -, high-pass filter -$$ -\omega_c = \frac{R_1 R_2}{(R_1 + R_2)L} = 25 \text{ krad/s}. -$$ - -For Practice Prob. 14.10. - -Example 14.11 If the band-stop filter in Fig. 14.37 is to reject a 200-Hz sinusoid while passing other frequencies, calculate the v alues of *L* and *C*. Take *R* = 150 Ω and the bandwidth as 100 Hz. - -# **Solution:** - -We use the formulas for a series resonant circuit in Section 14.5. - -$$ -B = 2\pi(100) = 200\pi \text{ rad/s} -$$ - -But - -*B* = \_\_ *R L* ⇒ *L* = \_\_ *R B* = \_\_\_\_\_ 150 200*π* = 0.2387 H - -Rejection of the 200-Hz sinusoid means that *f*0 is 200 Hz, so that *ω*0 in Fig. 14.38 is - -$$ -\omega_0 = 2\pi f_0 = 2\pi (200) = 400\pi -$$ - -Given that *ω*0 = 1∕ √ \_\_\_ *LC* , - -$$ -= 1/\sqrt{LC}, -$$ - -\n -$$ -C = \frac{1}{\omega_0^2 L} = \frac{1}{(400\pi)^2 (0.2387)} = 2.653 \pi F -$$ - -Practice Problem 14.11 Design a band-pass filter of the form in Fig. 14.35 with a lower cutoff frequency of 20.1 kHz and an upper cutoff frequency of 20.3 kHz. Take *R* = 30 kΩ. Calculate *L*, *C*, and *Q*. - -**Answer:** 23.87 H, 2.6 pF, 101. - -# **14.8** Active Filters - -There are three major limitations to the passive filters considered in the previous section. First, they cannot generate gain greater than 1; passive elements cannot add energy to the network. Second, they may require bulky and expensive inductors. Third, they perform poorly at frequencies below the audio frequency range (300 Hz < *f* < 3,000Hz). Nevertheless, passive filters are useful at high frequencies. - -Active filters consist of combinations of resistors, capacitors, and op amps. They offer some adv antages over passive *RLC* filters. First, they are often smaller and less expensive, because they do not require inductors. This makes feasible the integrated circuit realizations of filters. Second, they can provide amplifier gain in addition to pro viding the same frequenc y response as *RLC* filters. Third, active filters can be combined with buffer amplifiers (voltage followers) to isolate each stage of the filter from source and load impedance effects. This isolation allows designing the stages independently and then cascading them to realize the desired transfer function. (Bode plots, being log arithmic, may be added when transfer functions are cascaded.) Ho wever, active filters are less reliable and less stable. The practical limit of most active filters is about 100 kHz—most active filters operate well below that frequency. - -Filters are often classified according to their order (or number of poles) or their specific design type. - -# **14.8.1** First-Order Low-Pass Filter - -One type of first-order filter is shown in Fig. 14.41. The components selected for *Zi* and *Zf* determine whether the filter is low-pass or high-pass, but one of the components must be reactive. - -Figure 14.42 shows a typical active low-pass filter. For this filter, the transfer function is - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{\mathbf{Z}_f}{\mathbf{Z}_i} -$$ - (14.58) - -where **Z***i* = *Ri* and - -$$ -\mathbf{Z}_f = R_f \left\| \frac{1}{j\omega C_f} = \frac{R_f/j\omega C_f}{R_f + 1/j\omega C_f} = \frac{R_f}{1 + j\omega C_f R_f} \right. \tag{14.59} -$$ - -Therefore, - -$$ -\mathbf{H}(\omega) = -\frac{R_f}{R_i} \frac{1}{1 + j\omega C_f R_f} -$$ -(14.60) - -We notice that Eq. (14.60) is similar to Eq. (14.50), except that there is a low frequency ( *ω* → 0) gain or dc gain of −*Rf*∕*Ri*. Also, the corner frequency is - -$$ -\omega_c = \frac{1}{R_f C_f} \tag{14.61} -$$ - -which does not depend on *Ri*. This means that several inputs with dif ferent *Ri* could be summed if required, and the corner frequency would remain the same for each input. - -# **14.8.2** First-Order High-Pass Filter - -Figure 14.43 shows a typical high-pass filter. As before, - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{\mathbf{Z}_f}{\mathbf{Z}_i} -$$ - (14.62) - -where **Z***i* = *Ri* + 1∕*jωCi* and **Z***f* = *Rf* so that - -$$ -\mathbf{H}(\omega) = -\frac{R_f}{R_i + 1/j\omega C_i} = -\frac{j\omega C_i R_f}{1 + j\omega C_i R_i} -$$ -(14.63) - -This is similar to Eq. (14.52), except that at very high frequencies (*ω* → ∞), the gain tends to −*Rf*∕*Ri*. The corner frequency is - -$$ -\omega_c = \frac{1}{R_i C_i} \tag{14.64} -$$ - -# **14.8.3** Band-Pass Filter - -The circuit in Fig. 14.42 may be combined with that in Fig. 14.43 to form a band-pass filter that will have a gain *K* over the required range of frequencies. By cascading a unity-gain low-pass filter, a unity-gain - -Active first-order high-pass filter. - -This way of creating a band-pass filter, not necessarily the best, is perhaps the easiest to understand. - -**Figure 14.41** A general first-order active filter. - -**Figure 14.42** Active first-order low-pass filter. - -high-pass filter, and an inverter with gain −*Rf*∕*Ri*, as shown in the block diagram of Fig. 14.44(a), we can construct a band-pass filter whose frequency response is that in Fig. 14.44(b). The actual construction of the band-pass filter is shown in Fig. 14.45. - -# **Figure 14.44** - -Active band-pass filter: (a) block diagram, (b) frequency response. - -The analysis of the band-pass filter is relatively simple. Its transfer function is obtained by multiplying Eqs. (14.60) and (14.63) with the gain of the inverter; that is, - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = \left(-\frac{1}{1 + j\omega C_1 R}\right) \left(-\frac{j\omega C_2 R}{1 + j\omega C_2 R}\right) \left(-\frac{R_f}{R_i}\right) -$$ -$$ -= -\frac{R_f}{R_i} \frac{1}{1 + j\omega C_1 R} \frac{j\omega C_2 R}{1 + j\omega C_2 R} -$$ -(14.65) - -The low-pass section sets the upper corner frequency as - -$$ -\omega_2 = \frac{1}{RC_1} \tag{14.66} -$$ - -while the high-pass section sets the lower corner frequency as - -$$ -\omega_1 = \frac{1}{RC_2} \tag{14.67} -$$ - -With these values of *ω*1 and *ω*2, the center frequency, bandwidth, and quality factor are found as follows: - -$$ -\omega_0 = \sqrt{\omega_1 \omega_2} \tag{14.68} -$$ - -$$ -B = \omega_2 - \omega_1 \tag{14.69} -$$ - -$$ -Q = \frac{\omega_0}{B} \tag{14.70} -$$ - -To find the passband gain *K*, we write Eq. (14.65) in the standard form of Eq. (14.15), - -To find the passband gain *K*, we write Eq. (14.65) in the standard -\nn of Eq. (14.15), -\n -$$ -\mathbf{H}(\omega) = -\frac{R_f}{R_i} \frac{j\omega/\omega_1}{(1 + j\omega/\omega_1)(1 + j\omega/\omega_2)} = -\frac{Rf}{R_i} \frac{j\omega\omega_2}{(\omega_1 + j\omega)(\omega_2 + j\omega)} -$$ -\n(14.71) - -At the center frequency *ω*0 = √ \_\_\_\_\_ *ω*1*ω*2 , the magnitude of the transfer function is - -center frequency -$$ -\omega_0 = \sqrt{\omega_1 \omega_2} -$$ -, the magnitude of the transfer -\nn is -\n -$$ -|\mathbf{H}(\omega_0)| = \left| \frac{R_f}{R_i} \frac{j \omega_0 \omega_2}{(\omega_1 + j \omega_0)(\omega_2 + j \omega_0)} \right| = \frac{R_f}{R_i} \frac{\omega_2}{\omega_1 + \omega_2} -$$ -(14.72) - -Thus, the passband gain is - -$$ -K = \frac{R_f}{R_i} \frac{\omega_2}{\omega_1 + \omega_2} \tag{14.73} -$$ - -# **14.8.4** Band-Reject (or Notch) Filter - -A band-reject filter may be constructed by parallel combination of a lowpass filter and a high-pass filter and a summing amplifier, as shown in the block diagram of Fig. 14.46(a). The circuit is designed such that the lower cutoff frequency *ω*1 is set by the low-pass filter while the upper cutoff frequency *ω*2 is set by the high-pass filter. The gap between *ω*1 and *ω*2 is the bandwidth of the filter. As shown in Fig. 14.46(b), the filter passes frequencies below *ω*1 and above *ω*2. The block diagram in Fig. 14.46(a) is actually constructed as shown in Fig. 14.47. The transfer function is - -$$ -\mathbf{H}(\omega) = \frac{\mathbf{V}_o}{\mathbf{V}_i} = -\frac{R_f}{R_i} \left( -\frac{1}{1 + j\omega C_1 R} - \frac{j\omega C_2 R}{1 + j\omega C_2 R} \right) \tag{14.74} -$$ - -**Figure 14.46** Active band-reject filter: (a) block diagram, (b) frequency response. - -Active band-reject filter. - -The formulas for calculating the values of *ω*1, *ω*2, the center frequency, bandwidth, and quality factor are the same as in Eqs. (14.66) to (14.70). - -To determine the passband gain *K* of the filter, we can write Eq. (14.74) in terms of the upper and lower corner frequencies as - -$$ -\mathbf{H}(\omega) = \frac{R_f}{R_i} \left( \frac{1}{1 + j\omega/\omega_2} + \frac{j\omega/\omega_1}{1 + j\omega/\omega_1} \right) -$$ - -= $\frac{R_f}{R_i} \frac{(1 + j2\omega/\omega_1 + (j\omega)^2/\omega_1\omega_1)}{(1 + j\omega/\omega_2)(1 + j\omega/\omega_1)}$ (14.75) - -Comparing this with the standard form in Eq. (14.15) indicates that in the two passbands (*ω* → 0 and *ω* → ∞) the gain is - -$$ -K = \frac{R_f}{R_i} \tag{14.76} -$$ - -We can also find the gain at the center frequency by finding the magnitude of the transfer function at *ω*0 = √ \_\_\_\_\_ *ω*1*ω*2 , writing - -and the gain at the center frequency by finding the magni- -\nnsfer function at -$$ -\omega_0 = \sqrt{\omega_1 \omega_2} -$$ -, writing -\n -$$ -H(\omega_0) = \left| \frac{R_f (1 + j2\omega_0/\omega_1 + (j\omega_0)^2/\omega_1 \omega_1)}{R_i (1 + j\omega_0/\omega_2)(1 + j\omega_0/\omega_1)} \right| -$$ -\n -$$ -= \frac{R_f}{R_i} \frac{2\omega_1}{\omega_1 + \omega_2} -$$ -\n(14.77) - -Again, the filters treated in this section are only typical. There are many other active filters that are more complex. - -Example 14.12 Design a low-pass active filter with a dc gain of 4 and a corner frequency of 500 Hz. - -# **Solution:** - -From Eq. (14.61), we find - -$$ -\omega_c = 2\pi f_c = 2\pi (500) = \frac{1}{R_f C_f} \tag{14.12.1} -$$ - -The dc gain is - -$$ -H(0) = -\frac{R_f}{R_i} = -4\tag{14.12.2} -$$ - -We have two equations and three unknowns. If we select *Cf* = 0.2*μ*F, then - -tions and three unknowns. If we se -$$ -R_f = \frac{1}{2\pi (500) 0.2 \times 10^{-6}} = 1.59 \text{ k}\Omega -$$ - -and - -$$ -R_i = \frac{R_f}{4} = 397.5 \ \Omega -$$ - -We use a 1.6-kΩ resistor for *Rf* and a 400-Ω resistor for *Ri* . Figure 14.42 shows the filter. - -Design a high-pass filter with a high-frequency gain of 5 and a corner Practice Problem 14.12 frequency of 2 kHz. Use a 50-nF capacitor in your design. - -**Answer:** *Ri* = 1,600 Ω and *Rf* = 8 kΩ. - -Design a band-pass filter in the form of Fig. 14.45 to pass frequencies Example 14.13 between 250 and 3,000 Hz and with *K* = 10. Select *R* = 20 kΩ. - -# **Solution:** - -- 1. **Define.** The problem is clearly stated and the circuit to be used in the design is specified. -- 2. **Present.** We are ask ed to use the op amp circuit specified in Fig. 14.45 to design a band-pass filter. We are given the value of *R* to use (20 kΩ). In addition, the frequency range of the signals to be passed is 250 Hz to 3 kHz. -- 3. **Alternative.** We will use the equations developed in Section 14.8.3 to obtain a solution. We will then use the resulting transfer function to validate the answer. -- 4. **Attempt.** Because *ω*1 = 1∕*RC*2, we obtain - -te the answer. -\n**apt.** Because -$$ -\omega_1 = 1/RC_2 -$$ -, we obtain -\n -$$ -C_2 = \frac{1}{R\omega_1} = \frac{1}{2\pi f_1 R} = \frac{1}{2\pi \times 250 \times 20 \times 10^3} = 31.83 \text{ nF} -$$ - -Similarly, since *ω*2 = 1∕*RC*1, - -$$ -K\omega_1 = 2\pi f_1 K - 2\pi \times 250 \times 20 \times 10^8 -$$ - -arly, since $\omega_2 = 1/RC_1$ , -$$ -C_1 = \frac{1}{R\omega_2} = \frac{1}{2\pi f_2 R} = \frac{1}{2\pi \times 3,000 \times 20 \times 10^3} = 2.65 \text{ nF} -$$ - -From Eq. (14.73), - -$$ -\frac{R_f}{R_i} = K \frac{\omega_1 + \omega_2}{\omega_2} = K \frac{f_1 + f_2}{f_2} = \frac{10(3,250)}{3,000} = 10.83 -$$ - -If we select *Ri* = **10 k**Ω, then *Rf* = 10.83*Ri* ≃ **108.3 k**Ω. 5. **Evaluate.** The output of the first op amp is given by - -ct -$$ -R_i = 10 \text{ k}\Omega -$$ -, then $R_f = 10.83R_i \approx 108.3 \text{ k}\Omega$ . -The output of the first op amp is given by -$$ -\frac{V_i - 0}{20 \text{ k}\Omega} + \frac{V_1 - 0}{20 \text{ k}\Omega} + \frac{s2.65 \times 10^{-9} (V_1 - 0)}{1} -$$ -$$ -= 0 \rightarrow V_1 = -\frac{V_i}{1 + 5.3 \times 10^{-5} s} -$$ - -The output of the second op amp is given by - -$$ -t \text{ of the second op amp is given by} -$$ -\n -$$ -\frac{V_1 - 0}{20 \text{ k}\Omega + \frac{1}{s31.83 \text{ nF}}} + \frac{V_2 - 0}{20 \text{ k}\Omega} = 0 \rightarrow -$$ -\n -$$ -V_2 = -\frac{6.366 \times 10^{-4} s V_1}{1 + 6.366 \times 10^{-4} s} -$$ -\n -$$ -= \frac{6.366 \times 10^{-4} s V_i}{(1 + 6.366 \times 10^{-4} s)(1 + 5.3 \times 10^{-5} s)} -$$ - -The output of the third op amp is given by - -$$ -\frac{V_2 - 0}{10 \text{ k}\Omega} + \frac{V_o - 0}{108.3 \text{ k}\Omega} = 0 \to V_o = 10.83 V_2 \to j2\pi \times 25^\circ -$$ -$$ -V_o = -\frac{6.894 \times 10^{-3} sV_i}{(1 + 6.366 \times 10^{-4} s)(1 + 5.3 \times 10^{-5} s)} -$$ - -Let *j*2*π* × 25° and solve for the magnitude of *Vo*∕*Vi*. - -$$ -\frac{V_o}{V_i} = \frac{-j10.829}{(1+j1)(1)} -$$ - - ∣*Vo*∕*Vi* ∣ = **(0.7071)10.829**, which is the lower corner frequency point. Let *s* = *j*2*π* × 3000 = *j*18.849 kΩ. We then get - -Let -$$ -s = j2\pi \times 3000 = j18.849 \text{ k}\Omega -$$ -. We then get -\n -$$ -\frac{V_o}{V_i} = \frac{-j129.94}{(1+j12)(1+j1)} -$$ -\n -$$ -= \frac{129.94/-90^{\circ}}{(12.042/85.24^{\circ})(1.4142/45^{\circ})} = (0.7071)10.791/-18.61^{\circ} -$$ - -Clearly this is the upper corner frequency and the answer checks. - -6. **Satisfactory?** We have satisfactorily designed the circuit and can present the results as a solution to the problem. - -| Practice Problem 14.13 | Design a notch filter based on Fig. 14.47 for ω0 = 20 krad/s, K = 5, and | -|------------------------|--------------------------------------------------------------------------| -| | Q = 10. Use R = Ri = 10 kΩ. | - -**Answer:** *C*1 = 4.762 nF, *C*2 = 5.263 nF, and *Rf* = 50 kΩ. - -# **14.9** Scaling - -In designing and analyzing filters and resonant circuits or in circuit analysis in general, it is sometimes convenient to work with element values of 1 Ω, 1 H, or 1 F, and then transform the values to realistic values by - -*scaling*. We have taken advantage of this idea by not using realistic element values in most of our examples and problems; mastering circuit analysis is made easy by using convenient component values. We have thus eased calculations, knowing that we could use scaling to then make the values realistic. - -There are two ways of scaling a circuit: *magnitude* or *impedance scaling*, and *frequency scaling* . Both are useful in scaling responses and circuit elements to values within the practical ranges. While magnitude scaling leaves the frequency response of a circuit unaltered, frequency scaling shifts the frequency response up or down the frequency spectrum. - -# **14.9.1** Magnitude Scaling - -Magnitude scaling is the process of increasing all impedances in a network by a factor, the frequency response remaining unchanged. - -Recall that impedances of indi vidual elements *R*, *L*, and *C* are given by - -$$ -\mathbf{Z}_R = R, \qquad \mathbf{Z}_L = j\omega L, \qquad \mathbf{Z}_C = \frac{1}{j\omega C} \tag{14.78} -$$ - -In magnitude scaling, we multiply the impedance of each circuit element by a factor *Km* and let the frequency remain constant. This gives the new impedances as - -$$ -\mathbf{Z}'_R = K_m \mathbf{Z}_R = K_m R, \qquad \mathbf{Z}'_L = K_m \mathbf{Z}_L = j\omega K_m L -$$ -$$ -\mathbf{Z}'_C = K_m \mathbf{Z}_C = \frac{1}{j\omega C/K_m} -$$ -(14.79) - -Comparing Eq. (14.79) with Eq. (14.78), we notice the following changes in the element values: *R* → *KmR*, *L* → *Km L*, and *C* → *C*∕*Km*. Thus, in magnitude scaling, the new values of the elements and frequency are - -$$ -R' = K_m R, \qquad L' = K_m L -$$ - -$$ -C' = \frac{C}{K_m}, \qquad \omega' = \omega -$$ - (14.80) - -The primed variables are the new values and the unprimed variables are the old values. Consider the series or parallel *RLC* circuit. We now have - -$$ -\omega_0' = \frac{1}{\sqrt{LC'}} = \frac{1}{\sqrt{K_m LC/K_m}} = \frac{1}{\sqrt{LC}} = \omega_0 -$$ -(14.81) - -showing that the resonant frequency, as expected, has not changed. Similarly, the quality factor and the bandwidth are not affected by magnitude scaling. Also, magnitude scaling does not affect transfer functions in the forms of Eqs. (14.2a) and (14.2b), which are dimen sionless quantities. - -# **14.9.2** Frequency Scaling - -Frequency scaling is equivalent to relabeling the frequency axis of a frequency response plot. It is needed when translating frequencies such as a resonant frequency, a corner frequency, a bandwidth, etc., to a realistic level. It can be used to bring capacitance and inductance values into a range that is convenient to work with. - -Frequency scaling is the process of shifting the frequency response of a network up or down the frequency axis while leaving the impedance the same. - -We achieve frequency scaling by multiplying the frequency by a factor *Kf* while keeping the impedance the same. - -From Eq. (14.78), we see that the impedances of *L* and *C* are frequency-dependent. If we apply frequenc y scaling to **Z***L*(*ω*) and **Z***C*(*ω*) in Eq. (14.78), we obtain - -$$ -\mathbf{Z}_L = j(\omega K_f)L' = j\omega L \qquad \Rightarrow \qquad L' = \frac{L}{K_f} \tag{14.82a} -$$ - -$$ -Z_C = \frac{1}{j(\omega K_f)C'} = \frac{1}{j\omega C} \qquad \Rightarrow \qquad C' = \frac{C}{K_f} \tag{14.82b} -$$ - -since the impedance of the inductor and capacitor must remain the same after frequency scaling. We notice the following changes in the element values: *L* → *L*∕*Kf* and *C* → *C*∕*Kf*. The value of *R* is not affected, since its impedance does not depend on frequency. Thus, in frequency scaling, the new values of the elements and frequency are - -$$ -R' = R, \qquad L' = \frac{L}{K_f} -$$ - -$$ -C' = \frac{C}{K_f}, \qquad \omega' = K_f \omega -$$ - (14.83) - -Again, if we consider the series or parallel *RLC* circuit, for the resonant frequency - -$$ -\omega'_{0} = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(L/K_{f})(C/K_{f})}} = \frac{K_{f}}{\sqrt{LC}} = K_{f}\omega_{0} -$$ -(14.84) - -and for the bandwidth - -$$ -B' = K_f B \tag{14.85} -$$ - -but the quality factor remains the same (*Q*′ = *Q*). - -# **14.9.3** Magnitude and Frequency Scaling - -If a circuit is scaled in magnitude and frequency at the same time, then - -$$ -R' = K_m R, \qquad L' = \frac{K_m}{K_f} L -$$ - -$$ -C' = \frac{1}{K_m K_f} C, \qquad \omega' = K_f \omega -$$ - (14.86) - -These are more general formulas than those in Eqs. (14.80) and (14.83). We set *Km* = 1 in Eq. (14.86) when there is no magnitude scaling or *Kf* = 1 when there is no frequency scaling. - -A fourth-order Butterworth low-pass filter is shown in Fig. 14.48(a). The Example 14.14 filter is designed such that the cutoff frequency *ωc* = 1 rad/s. Scale the circuit for a cutoff frequency of 50 kHz using 10-kΩ resistors. - -# **Figure 14.48** - -For Example 14.14: (a) Normalized Butterworth low-pass filter, (b) scaled version of the same low-pass filter. - -# **Solution:** - -If the cutoff frequency is to shift from *ωc* = 1 rad/s to *ω*′ *c* = 2*π*(50) krad/s, then the frequency scale factor is - -$$ -K_f = \frac{\omega_c'}{\omega_c} = \frac{100\pi \times 10^3}{1} = \pi \times 10^5 -$$ - -Also, if each 1-Ω resistor is to be replaced by a 10-k Ω resistor, then the magnitude scale factor must be - -$$ -K_m = \frac{R'}{R} = \frac{10 \times 10^3}{1} = 10^4 -$$ - -Using Eq. (14.86), - -$$ -L'_1 = \frac{K_m}{K_f} L_1 = \frac{10^4}{\pi \times 10^5} (1.848) = 58.82 \text{ mH} -$$ - -\n -$$ -L'_2 = \frac{K_m}{K_f} L_2 = \frac{10^4}{\pi \times 10^5} (0.765) = 24.35 \text{ mH} -$$ - -\n -$$ -C'_1 = \frac{C_1}{K_m K_f} = \frac{0.765}{\pi \times 10^9} = 243.5 \text{ pF} -$$ - -\n -$$ -C'_2 = \frac{C_2}{K_m K_f} = \frac{1.848}{\pi \times 10^9} = 588.2 \text{ pF} -$$ - -The scaled circuit is shown in Fig. 14.48(b). This circuit uses practical values and will provide the same transfer function as the prototype in Fig. 14.48(a), but shifted in frequency. - -A third-order Butterworth filter normalized to *ωc* = 1 rad/s is shown Practice Problem 14.14 in Fig. 14.49. Scale the circuit to a cutoff frequency of 10 kHz. Use 15-nF capacitors. - -**Answer:** *R*′ 1 = *R*′ 2 = 1.061 kΩ, *C*′ 1 = *C*′ 2 = 15 nF, *L*′ = 33.77 mH. - -# **14.10** Frequency Response Using PSpice - -*PSpice* is a useful tool in the hands of the modern circuit designer for obtaining the frequency response of circuits. The frequency response is obtained using the AC Sweep as discussed in Section D.5 (Appendix D). This requires that we specify in the AC Sweep dialog box *Total Pts, Start Freq, End Freq*, and the sweep type. *Total Pts* is the number of points in the frequency sweep, and *Start Freq* and *End Freq* are, respectively, the starting and final frequencies, in hertz. In order to know what frequencies to select for *Start Freq* and *End Freq*, one must have an idea of the frequency range of interest by making a rough sketch of the frequency response. In a complex circuit where this may not be possible, one may use a trial-and-error approach. - -There are three types of sweeps: - -- *Linear:* The frequency is varied linearly from *Start Freq* to *End Freq* with *Total Pts* equally spaced points (or responses). -- *Octave:* The frequency is swept logarithmically by octaves from *Start Freq* to *End Freq* with *Total Pts* per octave. An octave is a factor of 2 (e.g., 2 to 4, 4 to 8, 8 to 16). -- *Decade:* The frequency is varied logarithmically by decades from *Start Freq* to *End Freq* with *Total Pts* per decade. A decade is a factor of 10 (e.g., from 2 to 20 Hz, 20 to 200 Hz, 200 Hz to 2 kHz). - -It is best to use a linear sweep when displaying a narrow frequency range of interest, as a linear sweep displays the frequency range well in a nar row range. Conversely, it is best to use a logarithmic (octave or decade) sweep for displaying a wide frequency range of interest—if a linear sweep is used for a wide range, all the data will be crowded at the highor low-frequency end and insufficient data at the other end. - -With the abo ve specifications, *PSpice* performs a steady-state sinusoidal analysis of the circuit as the frequency of all the independent sources is varied (or swept) from *Start Freq* to *End Freq*. - -The *PSpice* A/D program produces a graphical output. The output data type may be specified in the *Trace Command Box* by adding one of the following suffixes to V or I: - -- M Amplitude of the sinusoid. -- P Phase of the sinusoid. -- dB Amplitude of the sinusoid in decibels, that is, 20 log 10 (amplitude). - -Example 14.15 Determine the frequency response of the circuit shown in Fig. 14.50. - -# **Solution:** - -We let the input voltage *vs* be a sinusoid of amplitude 1 V and phase 0°. Figure 14.51 is the schematic for the circuit. The capacitor is rotated 270° counterclockwise to ensure that pin 1 (the positive terminal) is on top. The voltage marker is inserted to the output voltage across the capacitor. To perform a linear sweep for 1 < *f* < 1,000 Hz with 50 points, we select **Analysis/Setup/AC Sweep, DCLICK** *Linear*, type 50 in the *Total Pts* box, type 1 in the *Start Freq* box, and type 1000 in the *End Freq* box. After saving the file, we select **Analysis/Simulate** to simulate the circuit. If there are no errors, the *PSpice* A/D window will - -**Figure 14.50** For Example 14.15. - -The schematic for the circuit in Fig. 14.50. - -display the plot of V(C1:1), which is the same as *Vo* or *H*(*ω*) = *Vo*∕1, as shown in Fig. 14.52(a). This is the magnitude plot, since V(C1:1) is the same as VM(C1:1). To obtain the phase plot, select **Trace/Add** in the *PSpice* A/D menu and type VP(C1:1) in the **Trace Command** box. Figure 14.52(b) shows the result. By hand, the transfer function is - -$$ -H(\omega) = \frac{V_o}{V_s} = \frac{1,000}{9,000 + j\omega 8} -$$ - -or - -$$ -H(\omega) = \frac{1}{9 + j16\pi \times 10^{-3}} -$$ - -showing that the circuit is a low-pass filter as demonstrated in Fig. 14.52. Notice that the plots in Fig. 14.52 are similar to those in Fig. 14.3 (note that the horizontal axis in Fig. 14.52 is logrithic while the horizontal axis in Fig. 14.3 is linear.) - -**Figure 14.53** For Practice Prob. 14.15. - -For Practice Problem 14.15: (a) magnitude plot, (b) phase plot of the frequency response. - -For Example 14.16. - -Example 14.16 Use *PSpice* to generate the gain and phase Bode plots of *V* in the circuit of Fig. 14.55. - -# **Solution:** - -The circuit treated in Example 14.15 is first-order while the one in this example is second-order. Since we are interested in Bode plots, we use decade frequency sweep for 300 < *f* < 3,000 Hz with 50 points per de cade. We select this range because we know that the resonant frequency of the circuit is within the range. Recall that - -$$ -\omega_0 = \frac{1}{\sqrt{LC}} = 5 \text{ krad/s} \qquad \text{or} \qquad f_0 = \frac{\omega}{2\pi} = 795.8 \text{ Hz} -$$ - -After drawing the circuit as in Fig. 14.55, we select **Analysis/Setup/AC Sweep, DCLICK** *Decade*, enter 50 in the *Total Pts* box, 300 as the *Start Freq*, and 3,000 in the *End Freq* box. Upon saving the file, we simulate it by selecting **Analysis/Simulate**. This will automatically bring up the *PSpice* A/D window and display V(C1:1) if there are no errors. Since we are interested in the Bode plot, we select **Trace/Add** in the *PSpice* A/D menu and type dB(V(C1:1)) in the **Trace Command** box. The result is the Bode magnitude plot in Fig. 14.56(a). For the phase plot, we select - -For Example 14.16: (a) Bode plot, (b) phase plot of the response. - -**Trace/Add** in the *PSpice* A/D menu and type VP(C1:1) in the **Trace Command** box. The result is the Bode phase plot of Fig. 14.56(b). Notice that the plots confirm the resonant frequency of 795.8 Hz. - -Consider the network in Fig. 14.57. Use *PSpice* to obtain the Bode plots Practice Problem 14.16 for *Vo* over a frequency from 1 to 100 kHz using 20 points per decade. - -For Practice Prob. 14.16. - -**Answer:** See Fig. 14.58. - -# **14.11** Computation Using MATLAB - -*MATLAB* is a software package that is widely used for engineering computation and simulation. A review of *MATLAB* is provided in Appen dix E for the beginner. This section shows how to use the software to numerically perform most of the operations presented in this chapter and Chapter 15. The key to describing a system in *MATLAB* is to specify the numerator (num) and denominator (den) of the transfer function of the system. Once this is done, we can use several *MATLAB* commands to obtain the system's Bode plots (frequency response) and the system's response to a given input. - -The command **bode** produces the Bode plots (both magnitude and phase) of a given transfer function *H*(*s*). The format of the com mand is **bode** (num, den), where num is the numerator of *H*(*s*) and den is its denominator. The frequency range and number of points are automatically selected. For example, consider the transfer function in Example 14.3. It is better to first write the numerator and denominator in polynomial forms. - -Thus, - -$$ -H(s) = \frac{200 j\omega}{(j\omega + 2)(j\omega + 10)} = \frac{200s}{s^2 + 12s + 20}, \qquad s = j\omega -$$ - -Using the following commands, the Bode plots are generated as shown in Fig. 14.59. If necessary, the command **logspace** can be included to generate a logarithmically spaced frequency and the command **semilogx** can be used to produce a semilog scale. - ->> num = [200 0]; % specify the numerator of H(s) >> den = [1 12 20]; % specify the denominator of H(s) >> bode(num, den); % determine and draw Bode plots - -The step response *y*(*t*) of a system is the output when the input *x*(*t*) is the unit step function. The command **step** plots the step response of a system given the numerator and denominator of the transfer function of that sys tem. The time range and number of points are automatically selected. F or example, consider a second-order system with the transfer function - -$$ -H(s) = \frac{12}{s^2 + 3s + 12} -$$ - -We obtain the step response of the system shown in Fig. 14.60 by using the following commands. - ->> n = 12; >> d = [1 3 12]; >> step(n,d); - -We can verify the plot in Fig. 14.60 by obtaining *y*(*t*) = *x*(*t*) \* *u*(*t*) or *Y*(*s*) = *X*(*s*)*H*(*s*). - -The command **lsim** is a more general command than **step**. It calculates the time response of a system to any arbitrary input signal. The format of the command is *y* = **lsim** (num, den, *x*, *t*), where *x*(*t*) is the input signal, *t* is the time vector, and *y*(*t*) is the output generated. For example, assume a system is described by the transfer function - -for, and -$$ -y(t) -$$ - is the output gen- -cribed by the transfer function -$$ -H(s) = \frac{s+4}{s^3 + 2s^2 + 5s + 10} -$$ - -To find the response *y*(*t*) of the system to input *x*(*t*) = 10*e*−*t u*(*t*), we use the following *MATLAB* commands. Both the response *y*(*t*) and the input *x*(*t*) are plotted in Fig. 14.61. - -``` ->> t = 0:0.02:5; % time vector 0 < t < 5 with increment - 0.02 ->> x = 10*exp(-t); ->> num = [1 4]; ->> den = [1 2 5 10]; ->> y = lsim(num,den,x,t); ->> plot(t,x,t,y) -``` - -# **Figure 14.61** - -The response of the system described by *H*(*s*) = (*s* + 4)∕(*s* 2 + 2*s* 2 + 5*s* + 10) to an exponential input. - -# **14.12** Applications - -Resonant circuits and filters are widely used, particularly in electronics, power systems, and communications systems. For example, a Notch filter with a cutoff frequency at 60 Hz may be used to eliminate the 60-Hz power line noise in various communications electronics. Filtering of signals in communications systems is necessary in order to select the desired signal from a host of others in the same range (as in the case of radio receivers discussed next) and also to minimize the effects of noise and interference on the desired signal. In this section, we consider one practical application of resonant circuits and two applications of filters. The focus of each application is not to understand the details of how each device works but to see how the circuits considered in this chapter are applied in the practical devices. - -# **14.12.1** Radio Receiver - -Series and parallel resonant circuits are commonly used in radio and TV receivers to tune in stations and to separate the audio signal from the radio-frequency carrier wave. As an example, consider the block diagram of an AM radio receiver shown in Fig. 14.62. Incoming amplitudemodulated radio waves (thousands of them at different frequencies from different broadcasting stations) are received by the antenna. A resonant circuit (or a band-pass filter) is needed to select just one of the incoming waves. The selected signal is very weak and is amplified in stages in order to generate an audible audio-frequency wave. Thus, we have the radiofrequency (RF) amplifier to amplify the selected broadcast signal, the intermediate frequency (IF) amplifier to amplify an internally generated signal based on the RF signal, and the audio amplifier to amplify the audio signal just before it reaches the loudspeaker. It is much easier to amplify the signal at three stages than to build an amplifier to provide the same amplification for the entire band. - -# **Figure 14.62** - -A simplified block diagram of a superheterodyne AM radio receiver. - -The type of AM receiver shown in Fig. 14.62 is known as the *superheterodyne receiver*. In the early de velopment of radio, each amplification stage had to be tuned to the frequenc y of the incoming signal. This way, each stage must have several tuned circuits to cover the entire AM band (540 to 1600 kHz). To avoid the problem of having several resonant circuits, modern recei vers use a *frequency mixer* or *heterodyne* circuit, which always produces the same IF signal (445 kHz) but retains the audio frequencies carried on the incoming signal. To produce the constant IF frequency, the rotors of two separate variable capacitors are mechanically coupled with one another so that they can be rotated simultaneously with a single control; this is called *ganged tuning*. A *local oscillator* ganged with the RF amplifier produces an RF signal that is combined with the incoming wave by the frequenc y mixer to produce a n output signal that contains the sum and the difference frequencies of the two signals. For example, if the resonant circuit is tuned to recei ve an 800-kHz incoming signal, the local oscillator must produce a 1,255-kHz signal, so that the sum (1,255 + 800 = 2,055 kHz) and the difference (1,255 – 800 = 455 kHz) of frequencies are a vailable at the output of the mix er. However, only - -the difference, 455 kHz, is used in practice. This is the only frequency to which all the IF amplifier stages are tuned, regardless of the station dialed. The original audio signal (containing the "intelligence") is e xtracted in the detector stage. The detector basically removes the IF signal, leaving the audio signal. The audio signal is amplified to drive the loudspeaker, which acts as a transducer converting the electrical signal to sound. - -Our major concern here is the tuning circuit for the AM radio re ceiver. The operation of the FM radio recei ver is different from that of the AM receiver discussed here, and in a much dif ferent range of fre quencies, but the tuning is similar. - -The resonant or tuner circuit of an AM radio is portrayed in Fig. 14.63. Example 14.17 Given that *L* = 1*μ*H, what must be the range of *C* to have the resonant frequency adjustable from one end of the AM band to another? - -# **Solution:** - -The frequency range for AM broadcasting is 540 to 1,600 kHz. We consider the low and high ends of the band. Since the resonant circuit in Fig. 14.63 is a parallel type, we apply the ideas in Section 14.6. From Eq. (14.44), - -$$ -\omega_0 = 2\pi f_0 = \frac{1}{\sqrt{LC}} -$$ - -or - -$$ -C = \frac{1}{4\pi^2 f_0^2 L} -$$ - -For the high end of the AM band, *f*0 = 1,600 kHz, and the corresponding *C* is - -$$ -4\pi J_0 L -$$ - -d of the AM band, $f_0 = 1,600$ kHz, and the -$$ -C_1 = \frac{1}{4\pi^2 \times 1,600^2 \times 10^6 \times 10^{-6}} = 9.9 -$$ - nF - -For the low end of the AM band, *f*0 = 540 kHz, and the corresponding *C* is - -of the AM band, -$$ -f_0 = 540 -$$ - kHz, and the co- -$$ -C_2 = \frac{1}{4\pi^2 \times 540^2 \times 10^6 \times 10^{-6}} = 86.9 -$$ - nF - -Thus, *C* must be an adjustable (gang) capacitor varying from 9.9 to 86.9 nF. - -For an FM radio receiver, the incoming wave is in the frequency range Practice Problem 14.17 from 88 to 108 MHz. The tuner circuit is a parallel *RLC* circuit with a 4-*μ*H coil. Calculate the range of the variable capacitor necessary to cover the entire band. - -**Answer:** From 0.543 to 0.818 pF. - -**Figure 14.63** The tuner circuit for Example 14.17. - -# **14.12.2** Touch-Tone Telephone - -A typical application of filtering is the touch-tone telephone set shown in Fig. 14.64. The keypad has 12 buttons arranged in four rows and three columns. The arrangement provides 12 distinct signals by using seven tones divided into two groups: the low-frequency group (697 to 941 Hz) and the high-frequency group (1,209 to 1,477 Hz). Pressing a button generates a sum of two sinusoids corresponding to its unique pair of frequencies. For example, pressing the number 6 button generates sinusoidal tones with frequencies 770 and 1,477 Hz. - -**Figure 14.64** Frequency assignments for touch-tone dialing. - -When a caller dials a telephone number, a set of signals is transmitted to the telephone of fice, where the touch-tone signals are de coded by detecting the frequencies they contain. Figure 14.65 shows the block diagram for the detection scheme. The signals are f irst amplified and separated into their respective groups by the low-pass (LP) and high-pass (HP) f ilters. The limiters (L) are used to con vert the separated tones into square w aves. The individual tones are identified using seven band-pass (BP) filters, each filter passing one tone and rejecting other tones. Each f ilter is follow ed by a detec tor (D), which is energized when its input voltage exceeds a certain level. The outputs of the detectors pro vide the required dc signals needed by the switching system to connect the caller to the party being called. - -Block diagram of detection scheme. - -Using the standard 600-Ω resistor used in telephone circuits and a series Example 14.18 *RLC* circuit, design the band-pass filter *BP*2 in Fig. 14.65. - -# **Solution:** - -The band-pass filter is the series *RLC* circuit in Fig. 14.35. Inasmuch as *BP*2 passes frequencies 697 to 852 Hz and is centered at *f*0 = 770 Hz, its bandwidth is - -$$ -B = 2\pi(f_2 - f_1) = 2\pi(852 - 697) = 973.89 \text{ rad/s} -$$ - -From Eq. (14.39), - -$$ -L = \frac{R}{B} = \frac{600}{973.89} = 0.616 \text{ H} -$$ - -From Eq. (14.27) or (14.55), - -(14.27) or (14.55), -\n -$$ -C = \frac{1}{\omega_0^2 L} = \frac{1}{4\pi^2 f_0^2 L} = \frac{1}{4\pi^2 \times 770^2 \times 0.616} = 69.36 \text{ nF} -$$ - -Repeat Example 14.18 for band-pass filter *BP*6. Practice Problem 14.18 - -**Answer:** 356 mH, 39.83 nF. - -# **14.12.3** Crossover Network - -Another typical application of filters is the *crossover network* that couples an audio amplifier to woofer and tweeter speakers, as shown in Fig. 14.66(a). The network basically consists of one high-pass *RC* filter - -**Figure 14.66** - -(a) A crossover network for two loudspeakers, (b) equivalent model. - -**Figure 14.67** Frequency responses of the crossover network in Fig. 14.66. - -and one low-pass *RL* filter. It routes frequencies higher than a prescribed crossover frequency *fc* to the tweeter (high-frequency l oudspeaker) and frequencies below *fc* into the woofer (low-frequency loudspeaker). These loudspeakers have been designed to accommodate certain frequency responses. A woofer is a low- frequency loudspeaker designed to repro duce the lower part of the frequency range, up to about 3 kHz. A tweeter can reproduce audio frequencies from about 3 kHz to about 20 kHz. The two speaker types can be combined to reproduce the entire audio range of interest and provide the optimum in frequency response. - -By replacing the amplifier with a voltage source, the approximate equivalent circuit of the crosso ver network is sho wn in Fig. 14.66(b), where the loudspeak ers are modeled by resistors. As a high-pass filter, the transfer function *V*1∕*Vs* is given by - -$$ -H_1(\omega) = \frac{V_1}{V_s} = \frac{j\omega R_1 C}{1 + j\omega R_1 C} -$$ - (14.87) - -Similarly, the transfer function of the low-pass filter is given by - -$$ -H_2(\omega) = \frac{V_2}{V_s} = \frac{R_2}{R_2 + j\omega L} -$$ - (14.88) - -The values of *R*1, *R*2, *L*, and *C* may be selected such that the two filters have the same cutoff frequency, known as the *crossover frequency*, as shown in Fig. 14.67. - -The principle behind the crossover network is also used in the resonant circuit for a TV receiver, where it is necessary to separate the video and audio bands of RF carrier frequencies. The lower-frequency band (picture information in the range from about 30 Hz to about 4 MHz) is channeled into the receiver's video amplifier, while the high-frequency band (sound information around 4.5 MHz) is channeled to the receiver's sound amplifier. - -Example 14.19 In the crossover network of Fig. 14.66, suppose each speak er acts as a 6-Ω resistance. Find *C* and *L* if the crossover frequency is 2.5 kHz. - -# **Solution:** - -For the high-pass filter, - -$$ -\omega_c = 2\pi f_c = \frac{1}{R_1 C} -$$ - -or - -or - -$$ -C = \frac{1}{2\pi f_c R_1} = \frac{1}{2\pi \times 2.5 \times 10^3 \times 6} = 10.61 \,\mu\text{F} -$$ - -For the low-pass filter, - -$$ -\omega_c = 2\pi f_c = \frac{R_2}{L} -$$ - -$$ -L = \frac{R_2}{2\pi f_c} = \frac{6}{2\pi \times 2.5 \times 10^3} = 382 \,\mu\text{H} -$$ - -If each speaker in Fig. 14.66 has an 8-Ω resistance and *C* = 10*μ*F, find *L* Practice Problem 14.19 and the crossover frequency. - -**Answer:** 0.64 mH, 1.989 kHz. - -# **14.13** Summary - -- 1. The transfer function **H**(*ω*) is the ratio of the output response **Y**(*ω*) to the input excitation **X**(*ω*); that is, **H**(*ω*) = **Y**(*ω*)∕**X**(*ω*). -- 2. The frequency response is the variation of the transfer function with frequency. -- 3. Zeros of a transfer function **H**(*s*) are the values of *s* = *jω* that make *H*(*s*) = 0, while poles are the values of *s* that make *H*(*s*) → ∞. -- 4. The decibel is the unit of log arithmic gain. For a voltage or current gain *G*, its decibel equivalent is *G*dB = 20 log10 *G*. -- 5. Bode plots are semilog plots of the magnitude and phase of the transfer function as it v aries with frequenc y. The straight-line approximations of *H* (in dB) and *ϕ* (in degrees) are constructed using the corner frequencies defined by the poles and zeros of **H**(*ω*). -- 6. The resonant frequency is that frequency at which the imaginary part of a transfer function vanishes. For series and parallel *RLC* circuits. - -$$ -\omega_0=\frac{1}{\sqrt{LC}} -$$ - -7. The half-power frequencies (*ω*1, *ω*2) are those frequencies at which the power dissipated is one-half of that dissipated at the resonant frequency. The geometric mean between the half-power frequencies is the resonant frequency, or - -$$ -\omega_0 = \sqrt{\omega_1 \omega_2} -$$ - -8. The bandwidth is the frequenc y band between half-po wer frequencies: - -$$ -B=\omega_2-\omega_1 -$$ - -9. The quality f actor is a measure of the sharpness of the resonance peak. It is the ratio of the resonant (angular) frequency to the bandwidth, - -$$ -Q = \frac{\omega_0}{B} -$$ - -- 10. A filter is a circuit designed to pass a band of frequencies and reject others. Passive filters are constructed with resistors, capacitors, and inductors. Active filters are constructed with resistors, capacitors, and an active device, usually an op amp. -- 11. Four common types of filters are low-pass, high-pass, band-pass, and band-stop. A low-pass filter passes only signals whose frequencies are below the cutoff frequency *ωc*. A high-pass filter passes only signals whose frequencies are abo ve the cutof f frequency *ωc*. A band-pass filter passes only signals whose frequencies are within a prescribed - -range (*ω*1 < *ω*< *ω*2). A band-stop filter passes only signals whose frequencies are outside a prescribed range (*ω*1 > *ω*> *ω*2). - -12. Scaling is the process whereby unrealistic element v alues are magnitude-scaled by a factor *Km* and/or frequency-scaled by a factor *Kf* to produce realistic values. - -$$ -R' = K_m R, \qquad L' = \frac{K_m}{K_f} L, \qquad C' = \frac{1}{K_m K_f} C -$$ - -- 13. *PSpice* can be used to obtain the frequency response of a circuit if a frequency range for the response and the desired number of points within the range are specified in the AC Sweep. -- 14. The radio receiver—one practical application of resonant circuits employs a band-pass resonant circuit to tune in one frequenc y among all the broadcast signals picked up by the antenna. -- 15. The touch-tone telephone and the crosso ver network are tw o typical applications of filters. The touch-tone telephone system employs filters to separate tones of different frequencies to activate electronic switches. The crossover network separates signals in dif ferent frequency ranges so that they can be delivered to different devices such as tweeters and woofers in a loudspeaker system. - -# Review Questions - -**14.1** A zero of the transfer function - -$$ -H(s) = \frac{10(s + 1)}{(s + 2)(s + 3)} -$$ - -is at - -(a) 10 (b) −1 (c) −2 (d) −3 - -- **14.2** On the Bode magnitude plot, the slope of 1∕(5 + *jω*) 2 for large values of *ω* is -- (a) 20 dB/decade (b) 40 dB/decade - -(c) −40 dB/decade (d) −20 dB/decade - - **14.3** On the Bode phase plot for 0.5 < *ω*< 50, the slope of [1 + *j*10*ω*− *ω*2 ∕25]2 is - -| (a) 45°/decade | (b) 90°/decade | -|----------------|----------------| -| | | - -- (c) 135°/decade (d) 180°/decade - - **14.4** How much inductance is needed to resonate at 5 kHz with a capacitance of 12 nF? - -| (a) 2,652 H | (b) 11.844 H | -|-------------|--------------| -| (c) 3.333 H | (d) 84.43 mH | - - **14.5** The difference between the half-power frequencies is called the: - -| (a) quality factor | (b) resonant frequency | -|--------------------|------------------------| -|--------------------|------------------------| - -- (c) bandwidth (d) cutoff frequency - - **14.6** In a series *RLC* circuit, which of these quality factors has the steepest magnitude response curve near resonance? - -| (a) Q = 20 | (b) Q = 12 | -|------------|------------| -| (c) Q = 8 | (d) Q = 4 | - - **14.7** In a parallel *RLC* circuit, the bandwidth *B* is directly proportional to *R*. - -(a) True (b) False - - **14.8** When the elements of an *RLC* circuit are both magnitude-scaled and frequency-scaled, which quality is unaffected? - -| (a) resistor | (b) resonant frequency | -|---------------|------------------------| -| (c) bandwidth | (d) quality factor | - - **14.9** What kind of filter can be used to select a signal of one particular radio station? - -(a) low-pass (b) high-pass - -- (c) band-pass (d) band-stop -- **14.10** A voltage source supplies a signal of constant amplitude, from 0 to 40 kHz, to an RC low-pass filter. A load resistor, connected in parallel across the capacitor, experiences the maximum voltage at: - -| (a) dc | (b) 10 kHz | -|------------|------------| -| (c) 20 kHz | (d) 40 kHz | - -*Answers: 14.1b, 14.2c, 14.3d, 14.4d, 14.5c, 14.6a, 14.7b, 14.8d, 14.9c, 14.10a.* - -# Problems - -# Section 14.2 Transfer Function - -**14.1** Find the transfer function *Io*∕*Ii* of the *RL* circuit in Fig. 14.68. Express it using *ω*0 = *R*∕*L*. - -**Figure 14.68** For Prob. 14.1. - -**14.2** Using Fig. 14.69, design a problem to help other students better understand how to determine transfer functions. - -# **Figure 14.69** - -- For Prob. 14.2. -- **14.3** For the circuit shown in Fig. 14.70, find **H**(*s*) = **V***o*(*s*)/**I***i* (*s*). - -# **Figure 14.70** - -For Prob. 14.3. - -**14.4** Find the transfer function **H**(*s*) = **V***o*∕**V***i* of the circuit shown in Fig. 14.71. - -**14.5** For the circuit shown in Fig. 14.72, find **H**(*s*) = **V***o*∕**I***s*. - -# **Figure 14.72** For Prob. 14.5. - -**14.6** For the circuit shown in Fig. 14.73, find **H**(*s*) = **V***o*(*s*)∕**V***s*(*s*). - -**Figure 14.73** For Prob. 14.6. - -# Section 14.3 The Decibel Scale - -**14.7** Calculate ∣H(*ω*)∣ if *H*dB equals - -(a) 0.1 dB (b) −5 dB (c) 215 dB - -**14.8** Design a problem to help other students calculate the magnitude in dB and phase in degrees of a variety of transfer functions at a single value of *ω*. - -# Section 14.4 Bode Plots - -**14.9** A ladder network has a voltage gain of - -etwork has a voltage gain of - -\n -$$ -\mathbf{H}(\omega) = \frac{10}{(1 + j\omega)(10 + j\omega)} -$$ - -Sketch the Bode plots for the gain. - -- **14.10** Design a problem to help other students better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of *jω*. -- **14.11** Sketch the Bode plots for - -$$ -\mathbf{H}(\omega) = \frac{0.2(10 + j\omega)}{j\omega(2 + j\omega)} -$$ - - **14.12** A transfer function is given by - -$$ -T(s) = \frac{100(s+10)}{s(s+10)} -$$ - -Sketch the magnitude and phase Bode plots. - - **14.13** Construct the Bode plots for - -$$ -G(s) = \frac{0.1(s+1)}{s^2(s+10)}, \qquad s = j\omega -$$ - - **14.14** Draw the Bode plots for - -the Bode plots for -\n -$$ -\mathbf{H}(\omega) = \frac{250(j\omega + 1)}{j\omega(-\omega^2 + 10j\omega + 25)} -$$ - - **14.15** Construct the Bode magnitude and phase plots for - -$$ -H(s) = \frac{2(s+1)}{(s+2)(s+10)}, \qquad s = j\omega -$$ - - **14.16** Sketch Bode magnitude and phase plots for - -h Bode magnitude and phase pl -$$ -H(s) = \frac{1.6}{s(s^2 + s + 16)}, \quad s = j\omega -$$ - - **14.17** Sketch the Bode plots for - -Let the Bode plots for - -\n -$$ -G(s) = \frac{s}{(s+2)^2(s+1)}, \qquad s = j\omega -$$ - -**14.18** A linear network has this transfer function - -ar network has this transfer function -\n -$$ -H(s) = \frac{7s^2 + s + 4}{s^3 + 8s^2 + 14s + 5}, \qquad s = j\omega -$$ - - Use *MATLAB* or equivalent to plot the magnitude and phase (in degrees) of the transfer function. Take 0.1 < *ω*< 10 rad/s. - -**14.19** Sketch the asymptotic Bode plots of the magnitude and phase for \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*s* + 10)(*s* + 20)(*s* + 40) , *s* = *jω* - -$$ -H(s) = \frac{80s}{(s+10)(s+20)(s+40)}, \qquad s = -$$ - -**14.20** Design a more complex problem than given in Prob. 14.10, to help other students better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of *jω*. Include at least a second order repeated root. - -**14.21** Sketch the magnitude Bode plot for - -ch the magnitude Bode plot for -\n -$$ -H(s) = \frac{10s(s + 20)}{(s + 1)(s^2 + 60s + 400)}, \qquad s = j\omega -$$ - -**14.22** Find the transfer function **H**(*ω*) with the Bode magnitude plot shown in Fig. 14.74. - -**Figure 14.74** For Prob. 14.22. - -**14.23** The Bode magnitude plot of **H**(*ω*) is shown in Fig. 14.75. Find **H**(*ω*). - -# **Figure 14.75** - -- For Prob. 14.23. - - **14.24** The magnitude plot in Fig. 14.76 represents the transfer function of a preamplifier. Find *H*(*s*). - -For Prob. 14.24. - -## Problems **665** - -# Section 14.5 Series Resonance - -- **14.25** A series *RLC* network has *R* = 2 kΩ, *L* = 40 mH, and *C* = 1*μ*F. Calculate the impedance at resonance and at one-fourth, one-half, twice, and four times the resonant frequency. - -**14.26** Design a problem to help other students better understand *ω*0, *Q*, and *B* at resonance in series *RLC* circuits. - -- **14.27** Design a series *RLC* resonant circuit with *ω*0 = 40 rad/s and *B* = 10 rad/s. -- **14.28** Design a series *RLC* circuit with *B* = 20 rad/s and *ω*0 = 1,000 rad/s. Find the circuit's *Q*. Let *R* = 10 Ω. -- **14.29** Let *vs* = 20 cos(*at*) V in the circuit of Fig. 14.77. Find *ω*0, *Q*, and *B*, as seen by the capacitor. - -**Figure 14.77** For Prob. 14.29. - -- **14.30** A circuit consisting of a coil with inductance 10 mH and resistance 20 Ω is connected in series with a capacitor and a generator with an rms voltage of 120 V. Find: - - (a) the value of the capacitance that will cause the circuit to be in resonance at 15 kHz - - (b) the current through the coil at resonance - - (c) the *Q* of the circuit - -# Section 14.6 Parallel Resonance - -- **14.31** Design a parallel resonant *RLC* circuit with *ω*0 = 100 krad/s and a bandwidth of 10 krad/s. Additionally what is the value of *Q*? -- **14.32** Design a problem to help other students better understand the quality factor, the resonant frequency, and bandwidth of a parallel *RLC* circuit. -- **14.33** A parallel resonant circuit with a bandwidth of 40 krad/s and the half-power frequencies are *ω*1 = 4.98 Mrad/s and *ω*2 = 5.02 Mrad/s, calculate the quality factor and resonant frequency. -- **14.34** A parallel *RLC* circuit has *R* = 100 kΩ, *L* = 100 mH, and a *C* = 10 *μ*F. Determine the value of *Q*, the resonant frequency, and the bandwidth. If - -*R* = 200 kΩ, how does that affect the values of *Q*, resonant frequency, and the bandwidth? - -- **14.35** A parallel *RLC* circuit has *R* = 10 kΩ, *L* = 100 mH, and a resonant frequency of 200 krad/s. Calculate the value of *C*, the value of the quality factor, and the bandwidth. -- **14.36** It is expected that a parallel *RLC* resonant circuit has a midband admittance of 25 × 10−3 S, quality factor of 120, and a resonant frequency of 200 krad/s. Calculate the values of *R*, *L*, and *C*. Find the bandwidth and the half-power frequencies. -- **14.37** Rework Prob. 14.25 if the elements are connected in parallel. -- **14.38** Find the resonant frequency of the circuit in Fig. 14.78. - -**Figure 14.78** For Prob. 14.38. - -**14.39** For the "tank" circuit in Fig. 14.79, find the resonant frequency. - -# **Figure 14.79** - -- **14.40** A parallel resonance circuit has a resistance of 2 kΩ and half-power frequencies of 86 kHz and 90 kHz. Determine: - - (a) the capacitance - - (b) the inductance - - (c) the resonant frequency - - (d) the bandwidth - - (e) the quality factor -- **14.41** Using Fig. 14.80, design a problem to help other students better understand the quality factor, the resonant frequency, and bandwidth of *RLC* circuits. - -**Figure 14.80** - -For Prob. 14.41. - -**14.42** For the circuits in Fig. 14.81, find the resonant frequency *ω*0, the quality factor *Q*, and the bandwidth *B*. - -For Prob. 14.42. - -**14.43** Calculate the resonant frequency of each of the circuits in Fig. 14.82. - -**Figure 14.82** For Prob. 14.43. - -**\*14.44** For the circuit in Fig. 14.83, find: - -(a) the resonant frequency *ω*0 - -(b) **Z**in(*ω*0) - -**Figure 14.83** For Prob. 14.44. - -> **14.45** For the circuit shown in Fig. 14.84, find *ω*0, *B*, and *Q*, as seen by the voltage across the inductor. - -**Figure 14.84** For Prob. 14.45. - -- **14.46** For the network illustrated in Fig. 14.85, find - - (a) the transfer function **H**(*ω*) = **V***o*(*ω*)∕**I**(*ω*), - -(b) the magnitude of **H** at *ω*0 = 1 rad/s. - -**Figure 14.85** - -For Probs. 14.46, 14.78, and 14.92. - -# Section 14.7 Passive Filters - -- **14.47** Show that a series *LR* circuit is a low-pass filter if the output is taken across the resistor. Calculate the corner frequency *fc* if *L* = 2 mH and *R* = 10 kΩ. -- **14.48** Find the transfer function **V***o*∕**V***s* of the circuit in Fig. 14.86. Show that the circuit is a low-pass filter. - -For Prob. 14.48. - -**14.49** Design a problem to help other students better understand low-pass filters described by transfer functions. - -**14.50** Determine what type of filter is in Fig. 14.87. Calculate the corner frequency *fc*. - -**Figure 14.87** For Prob. 14.50. - -\* An asterisk indicates a challenging problem. - -Problems **667** - -- **14.55** Determine the range of frequencies that will be passed by a series *RLC* band-pass filter with *R* = 10 Ω, *L* = 25 mH, and *C* = 0.4*μ*F. Find the quality factor. -- **14.56** (a) Show that for a band-pass filter, - -$$ -\mathbf{H}(s) = \frac{sB}{s^2 + sB + \omega_0^2}, \qquad s = j\omega -$$ - - where *B* = bandwidth of the filter and *ω*0 is the center frequency. - -(b) Similarly, show that for a band-stop filter, - -$$ -\mathbf{H}(s) = \frac{s^2 + \omega_0^2}{s^2 + s + \omega_0^2}, \qquad s = j\omega -$$ - -**14.57** Determine the center frequency and bandwidth of the band-pass filters in Fig. 14.88. - -# **Figure 14.88** - -For Prob. 14.57. - -- **14.58** The circuit parameters for a series *RLC* bandstop filter are *R* = 250 Ω, *L* = 1 mH, *C* = 40 pF. Calculate: - - (a) the center frequency - - (b) the half-power frequencies - - (c) the quality factor -- **14.59** Find the bandwidth and center frequency of the band-stop filter of Fig. 14.89. - -# **Figure 14.89** For Prob. 14.59. - -# Section 14.8 Active Filters - -- **14.60** Obtain the transfer function of a high-pass filter with a passband gain of 100 and a cutoff frequency of 40 rad/s. -- **14.61** Find the transfer function for each of the active filters in Fig. 14.90. - -# **Figure 14.90** - -**14.62** The filter in Fig. 14.90(b) has a 3-dB cutoff frequency at 1 kHz. If its input is connected to a 120-mV variable frequency signal, find the output voltage at: - -(a) 200 Hz (b) 2 kHz (c) 10 kHz - -**14.63** Design an active first-order high-pass filter with - -$$ -\mathbf{H}(s) = -\frac{100s}{s+10}, \qquad s = j\omega -$$ - -Use a 1-*μ*F capacitor. - -**14.64** Obtain the transfer function of the active filter in Fig. 14.91 on the next page. What kind of filter is it? - -For Prob. 14.64. - -**14.65** A high-pass filter is shown in Fig. 14.92. Show that the transfer function is - -**Figure 14.92** For Prob. 14.65. - -- **14.66** A "general" first-order filter is shown in Fig. 14.93. - - (a) Show that the transfer function is - -(a) Show that the transfer function is -\n -$$ -\mathbf{H}(s) = \frac{R_4}{R_3 + R_4} \times \frac{s + (1/R_1C)[R_1/R_2 - R_3/R_4]}{s + 1/R_2C}, -$$ -\n -$$ -s = j\omega -$$ - -- (b) What condition must be satisfied for the circuit to operate as a high-pass filter? -- (c) What condition must be satisfied for the circuit to operate as a low-pass filter? - -**14.67** Design an active low-pass filter with dc gain of 0.25 and a corner frequency of 500 Hz. - - **14.68** Design a problem to help other students better understand the design of active high-pass filters when specifying a high-frequency gain and a corner frequency. - -**14.69** Design the filter in Fig. 14.94 to meet the following requirements: - -- (a) It must attenuate a signal at 2 kHz by 3 dB compared with its value at 10 MHz. -- (b) It must provide a steady-state output of *vo*(*t*) = 10 sin(2*π* × 108 *t* + 180°) V for an input *vs*(*t*) = 4 sin(2*π* × 108 *t*) V. - -# **Figure 14.94** - -For Prob. 14.69. - -- **\*14.70** A second-order active filter known as a Butterworth filter is shown in Fig. 14.95. - - (a) Find the transfer function **V***o*∕**V***i*. - - (b) Show that it is a low-pass filter. - -# **Figure 14.95** - -For Prob. 14.70. - -# Section 14.9 Scaling - -**14.71** Use magnitude and frequency scaling on the circuit of Fig. 14.79 to obtain an equivalent circuit in which the inductor and capacitor have magnitude 1 H and 1 F respectively. - -**14.72** Design a problem to help other students better understand magnitude and frequency scaling. - -**14.73** Calculate the values of *R*, *L*, and *C* that will result in *R* = 12 kΩ, *L* = 40*μ*H, and *C* = 300 nF respectively when magnitude-scaled by 800 and frequency-scaled by 1000. - -Problems **669** - -- **14.74** A circuit has *R*1 = 3 Ω, *R*2 = 10 Ω, *L* = 2H, and *C* = 1∕10 F. After the circuit is magnitude-scaled by 100 and frequency-scaled by 106 , find the new values of the circuit elements. -- **14.75** In an *RLC* circuit, *R* = 20 Ω, *L* = 4 H, and *C* = 1 F. The circuit is magnitude-scaled by 10 and frequency-scaled by 105 . Calculate the new values of the elements. -- **14.76** Given a parallel *RLC* circuit with *R* = 5 kΩ, *L* = 10 mH, and *C* = 20*μ*F, if the circuit is magnitude-scaled by *Km* = 500 and frequencyscaled by *Kf* = 105 , find the resulting values of *R*, *L*, and *C*. -- **14.77** A series *RLC* circuit has *R* = 10 Ω, *ω*0 = 40 rad/s, and *B* = 5 rad/s. Find *L* and *C* when the circuit is scaled: - - (a) in magnitude by a factor of 600, - - (b) in frequency by a factor of 1,000, - - (c) in magnitude by a factor of 400 and in frequency by a factor of 105 . -- **14.78** Redesign the circuit in Fig. 14.85 so that all resistive elements are scaled by a factor of 1,000 and all frequency-sensitive elements are frequency-scaled by a factor of 104 . -- **\*14.79** Refer to the network in Fig. 14.96. - - (a) Find **Z**in(*s*). - - (b) Scale the elements by *Km* = 10 and *Kf* = 100. Find **Z**in(*s*) and *ω*0. - -# **Figure 14.96** - -For Prob. 14.79. - -- **14.80** (a) For the circuit in Fig. 14.97, draw the new circuit after it has been scaled by *Km* = 200 and *Kf* = 104 . - - (b) Obtain the Thevenin equivalent impedance at terminals *a-b* of the scaled circuit at *ω* = 104 rad/s. - -**Figure 14.97** For Prob. 14.80. - - **14.81** The circuit shown in Fig. 14.98 has the impedance - -The circuit shown in Fig. 14.98 has the impedance -$$ -Z(s) = \frac{1,000(s + 1)}{(s + 1 + j50)(s + 1 - j50)}, \qquad s = j\omega -$$ - -Find: - -- (a) the values of *R*, *L*, *C*, and *G* -- (b) the element values that will raise the resonant frequency by a factor of 103 by frequency scaling - -# **Figure 14.98** - -For Prob. 14.81. - -**14.82** Scale the low-pass active filter in Fig. 14.99 so that its corner frequency increases from 1 rad/s to 200 rad/s. Use a 1-*μ*F capacitor. - -# **Figure 14.99** - -For Prob. 14.82. - -**14.83** The op amp circuit in Fig. 14.100 is to be magnitude-scaled by 100 and frequency-scaled by 105 . Find the resulting element values. - -# **Figure 14.100** - -For Prob. 14.83. - -# Section 14.10 Frequency Response Using PSpice - -**14.84** Using *PSpice or MultiSim,* obtain the frequency response of the circuit in Fig. 14.101 on the next page. - -# **Figure 14.101** - -**14.85** Use *PSpice or MultiSim* to obtain the magnitude and phase plots of **V***o*∕**I***s* of the circuit in Fig. 14.102. - -# **Figure 14.102** - -For Prob. 14.85. - -**14.86** Using Fig. 14.103, design a problem to help other students better understand how to use *PSpice* to obtain the frequency response (magnitude and phase of I) in electrical circuits. - -**Figure 14.103** - -- For Prob. 14.86. - - **14.87** In the interval 0.1 < *f* < 100 Hz, plot the response of the network in Fig. 14.104. Classify this filter and obtain *ω*0. - -**Figure 14.104** For Prob. 14.87. - -**14.88** Use *PSpice or MultiSim* to generate the magnitude and phase Bode plots of **V***o* in the circuit of Fig. 14.105. - -For Prob. 14.88. - - **14.89** Obtain the magnitude plot of the response **V***o* in the network of Fig. 14.106 for the frequency interval 100 < *f* < 1,000 Hz. - -# **Figure 14.106** - -For Prob. 14.89. - -- **14.90** Obtain the frequency response of the circuit in Fig. 14.40 (see Practice Problem 14.10). Take *R*1 = *R*2 = 100 Ω, *L* = 2 mH. Use 1 < *f* < 100,000 Hz. -- **14.91** For the "tank" circuit of Fig. 14.79, obtain the frequency response (voltage across the capacitor) using *PSpice or MultiSim*. Determine the resonant frequency of the circuit. -- **14.92** Using *PSpice or MultiSim*, plot the magnitude of the frequency response of the circuit in Fig. 14.85. - -# Section 14.12 Applications - -**14.93** For the phase shifter circuit shown in Fig. 14.107, find H = *Vo*∕*Vs*. - -# **Figure 14.107** - -For Prob. 14.93. - -**14.94** For an emergency situation, an engineer needs to make an *RC* high-pass filter. He has one 10-pF capacitor, one 30-pF capacitor, one 1.8-kΩ resistor, and one 3.3-kΩ resistor available. Find the greatest cutoff frequency possible using these elements. - -**14.95** A series-tuned antenna circuit consists of a variable capacitor (40 pF to 360 pF) and a 240-*μ*H antenna coil that has a dc resistance of 12 Ω. - -- (a) Find the frequency range of radio signals to which the radio is tunable. -- (b) Determine the value of *Q* at each end of the frequency range. - -**14.96** The crossover circuit in Fig. 14.108 is a low-pass filter that is connected to a woofer. Find the transfer function **H**(*ω*) = **V***o*(*ω*)∕**V***i*(*ω*). - -For Prob. 14.96. - -# Comprehensive Problems - -- **14.98** A certain electronic test circuit produced a resonant curve with half-power points at 432 Hz and 454 Hz. If *Q* = 20, what is the resonant frequency of the circuit? -- **14.99** In an electronic device, a series circuit is employed that has a resistance of 100 Ω, a capacitive reactance of 5 kΩ, and an inductive reactance of 300 Ω when used at 2 MHz. Find the resonant frequency and bandwidth of the circuit. -- **14.100** In a certain application, a simple *RC* low-pass filter is designed to reduce high frequency noise. If the desired corner frequency is 20 kHz and *C* = 0.5*μ*F, find the value of *R*. -- **14.101** In an amplifier circuit, a simple *RC* high-pass filter is needed to block the dc component while passing the time-varying component. If the desired rolloff frequency is 15 Hz and *C* = 10*μ*F, find the value of *R*. -- **14.102** Practical *RC* filter design should allow for source and load resistances as shown in Fig. 14.110. Let *R* = 4 kΩ and *C* = 40-nF. Obtain the cutoff frequency when: - -**Figure 14.110** For Prob. 14.102. - -**14.103** The *RC* circuit in Fig. 14.111 is used for a lead compensator in a system design. Obtain the transfer function of the circuit. - -# **Figure 14.111** For Prob. 14.103. - - **14.104** A low-quality-factor, double-tuned band-pass filter is shown in Fig. 14.112. Use *PSpice or MultiSim* to generate the magnitude plot of **V***o*(*ω*). - -**Figure 14.112** For Prob. 14.104. - -# **PART THREE** - -# Advanced Circuit Analysis - -# OUTLINE - -- 15 Introduction to the Laplace Transform -- 16 Applications of the Laplace Transform -- 17 The Fourier Series -- 18 Fourier Transform -- 19 Two-Port Networks - -# **chapter** - -# 15 - -# Introduction to the Laplace Transform - -*The important thing about a problem is not its solution, but the strength we gain in finding the solution.* - -—Anonymous - -# Enhancing Your Skills and Your Career - -# **ABET EC 2000 criteria (3.h), "the broad education necessary to understand the impact of engineering solutions in a global and societal context."** - -As a student, you must mak e sure you acquire "the broad education necessary to understand the impact of engineering solutions in a global and societal context." To some extent, if you are already enrolled in an ABET-accredited engineering program, then some of the courses you are required to take must meet this criteria. My recommendation is that even if you are in such a program, you look at all the elective courses you take to make sure that you e xpand your awareness of global issues and societal concerns. The engineers of the future must fully understand that they and their activities affect all of us in one way or another. - -# **ABET EC 2000 criteria (3.i), "need for, and an ability to engage in life-long learning."** - -You must be fully aware of and recognize the "need for, and an ability to engage in life-long learning. " It almost seems absurd that this need and ability must be stated. Yet, you w ould be surprised at ho w m any engineers do not really understand this concept. The only way to be really able to keep up with the explosion in technology we are facing now and will be f acing in the future is through constant learning. This learning must include nontechnical issues as well as the latest technology in your field. - -The best w ay to k eep up with the state of the art in your field is through your colleagues and association with indi viduals you meet through your technical organization or organizations (especially IEEE). Reading state-of-the-art technical articles is the next best way to stay current. - -Photo by Charles Alexander - -**Pierre Simon Laplace** (1749–1827), a French astronomer and mathematician, first presented the transform that bears his name and its applications to differential equations in 1779. - -Born of humble origins in Beaumont-en-Auge, Normandy, France, Laplace became a professor of mathematics at the age of 20. His math ematical abilities inspired the f amous mathematician Simeon Poisson, who called Laplace the Isaac Newton of France. He made important contributions in potential theory, probability theory, astronomy, and celestial mechanics. He w as widely kno wn for his w ork, *Traite de Mecanique Celeste (Celestial Mechanics)*, which supplemented the work of Newton on astronomy . The Laplace transform, the subject of this chapter , is named after him. - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Understand the Laplace transform, its importance in circuit analysis, and how to determine the Laplace transform of functions common to circuit analysis. -- 2. Understand the properties of the Laplace transform. -- 3. Understand the inverse Laplace transform and how to determine its given functions in the s-domain. -- 4. Understand the convolution integral and how to use it in the time domain and its equivalence in the s-domain. - -# **15.1** Introduction - -Our goal in this and the follo wing chapters is to develop techniques for analyzing circuits with a wide variety of inputs and responses. Such circuits are modeled by *differential equations* whose solutions describe the total response behavior of the circuits. Mathematical methods have been devised to systematically determine the solutions of dif ferential equa tions. We now introduce the po werful method of *Laplace transformation*, which involves turning differential equations into *algebraic equations*, thus greatly facilitating the solution process. - -The idea of transformation should be f amiliar by now. When using phasors for the analysis of circuits, we transform the circuit from the time domain to the frequency or phasor domain. Once we obtain the phasor result, we transform it back to the time domain. The Laplace transform method follows the same process: We use the Laplace transformation to transform the circuit from the time domain to the frequenc y domain, obtain the solution, and apply the inverse Laplace transform to the result to transform it back to the time domain. - -The Laplace transform is significant for a number of reasons. First, it can be applied to a wider variety of inputs than phasor analysis. Second, it provides an easy way to solve circuit problems involving initial conditions, because it allo ws us to w ork with algebraic equations instead of differential equations. Third, the Laplace transform is capable of providing us, in one single operation, the total response of the circuit comprising both the natural and forced responses. - -We begin with the definition of the Laplace transform which gives rise to its most essential properties. By e xamining these properties, we shall see ho w and wh y the method w orks. This also helps us to better appreciate the idea of mathematical transformations. We also consider some properties of the Laplace transform that are very helpful in circuit analysis. We then consider the inverse Laplace transform, transfer functions, and convolution. In this chapter , we will focus on the mechanics of the Laplace transformation. In Chapter 16 we will e xamine how the Laplace transform is applied in circuit analysis, netw ork stability, and network synthesis. - -# **15.2** Definition of the Laplace Transform - -Given a function *f*(*t*), its Laplace transform, denoted by *F*(*s*) or [ *f*(*t*)], is defined by - -$$ -\mathcal{L}[f(t)] = F(s) = \int_{0^{-}}^{\infty} f(t)e^{-st} dt -$$ - (15.1) - -where *s* is a complex variable given by - -$$ -s = \sigma + j\omega \tag{15.2} -$$ - -Because the argument *st* of the exponent *e* in Eq. (15.1) must be dimensionless, it follows that *s* has the dimensions of frequenc y and units of inverse seconds (s −1 ) or "frequenc y." In Eq. (15.1), the lo wer limit is specified as 0− to indicate a time just before *t* = 0. We use 0− as the lower limit to include the origin and capture an y discontinuity of *f*(*t*) at *t* = 0; this will accommodate functions—such as singularity functions—that may be discontinuous at *t* = 0. - -It should be noted that the integral in Eq. (15.1) is a definite integral with respect to time. Hence, the result of inte gration is indepen dent of time and only involves the variable "*s*." - -Equation (15.1) illustrates the general concept of transformation. The function *f*(*t*) is transformed into the function *F*(*s*). Whereas the former function involves *t* as its argument, the latter involves *s*. We say the transformation is from *t*-domain to *s*-domain. Gi ven the interpre tation of *s* as frequenc y, we arri ve at the follo wing description of the Laplace transform: - -The Laplace transform is an integral transformation of a function f(t) from the time domain into the complex frequency domain, giving F(s). - -When the Laplace transform is applied to circuit analysis, the differential equations represent the circuit in the time domain. The terms in the differential equations take the place of *f*(*t*). Their Laplace transform, which corresponds to *F*(*s*), constitutes algebraic equations representing the circuit in the frequency domain. - -For an ordinary function f(t), the lower limit can be replaced by 0. - -We assume in Eq. (15.1) that *f* (*t*) is ignored for *t* < 0. To ensure that this is the case, a function is often multiplied by the unit step. Thus, *f* (*t*) is written as *f* (*t*)*u*(*t*) or *f* (*t*), *t* ≥ 0. - -The Laplace transform in Eq. (15.1) is kno wn as the *one-sided* (or *unilateral*) Laplace transform. The *two-sided* (or *bilateral*) Laplace transform is given by - -$$ -F(s) = \int_{-\infty}^{\infty} f(t)e^{-st} dt -$$ - (15.3) - -The one-sided Laplace transform in Eq. (15.1), being adequate for our purposes, is the only type of Laplace transform that we will treat in this book. - -A function *f* (*t*) may not ha ve a Laplace transform. F or *f* (*t*) to have a Laplace transform, the inte gral in Eq. (15.1) must con verge to a finite value. Because |*ejωt* | = 1 for any value of *t*, the integral converges when - -$$ -\int_0^\infty e^{-\sigma t} |f(t)| \, dt < \infty \tag{15.4} -$$ - -for some real v alue *σ* = *σc*. Thus, the re gion of con vergence for the Laplace transform is Re(*s*) = *σ* > *σc*, as shown in Fig. 15.1. In thisregion, |*F*(*s*)| < ∞ and *F*(*s*) exists. *F*(*s*) is undefined outside the region of convergence. Fortunately, all functions of interest in circuit analysis satisfy the convergence criterion in Eq. (15.4) and have Laplace transforms. Therefore, it is not necessary to specify *σc* in what follows. - -A companion to the direct Laplace transform in Eq. (15.1) is the *inverse* Laplace transform given by - -$$ -\mathcal{L}^{-1}[F(s)] = f(t) = \frac{1}{2\pi j} \int_{\sigma_1 - j\infty}^{\sigma_1 + j\infty} F(s)e^{st} ds -$$ - (15.5) - -where the integration is performed along a straight line (*σ*1 + *jω*, −∞ < *ω* < ∞) in the region of convergence, *σ*1 > *σc*. See Fig. 15.1. The direct application of Eq. (15.5) in volves some kno wledge about comple x analysis beyond the scope of this book. F or this reason, we will not use Eq. (15.5) to find the inverse Laplace transform. We will rather use a look-up table, to be de veloped in Section 15.3. The functions *f*(*t*) and *F*(*s*) are regarded as a Laplace transform pair where - -$$ -f(t) \qquad \Leftrightarrow \qquad F(s) \tag{15.6} -$$ - -meaning that there is one-to-one correspondence between *f*(*t*) and *F*(*s*). The following examples derive the Laplace transforms of some important functions. - -Example 15.1 - -Determine the Laplace transform of each of the following functions: (a) *u*(*t*), (b) *e*−*at u*(*t*), *a* ≥ 0, and (c) *δ*(*t*). - -# **Solution:** - -(a) For the unit step function *u*(*t*), shown in Fig. 15.2(a), the Laplace transform is - -$$ -\mathcal{L}[u(t)] = \int_{0^{-}}^{\infty} 1e^{-st} dt = -\frac{1}{s} e^{-st} \Big|_{0}^{\infty} -$$ - -= $-\frac{1}{s}(0) + \frac{1}{s}(1) = \frac{1}{s}$ (15.1.1) - -*ω*t + sin*2* - -*ω*t = 1 - -**Figure 15.1** Region of convergence for the Laplace transform. - -|ej*ω*t - -| = cos*2* - -(b) F or the e xponential function, sho wn in Fig. 15.2(b), the Laplace transform is - -$$ -\mathcal{L}[e^{-at} u(t)] = \int_{0^{-}}^{\infty} e^{-at} e^{-st} dt -$$ - -= $-\frac{1}{s+a} e^{-(s+a)t} \Big|_{0}^{\infty} = \frac{1}{s+a}$ (15.1.2) - -(c) For the unit impulse function, shown in Fig. 15.2(c), - -$$ -\mathcal{L}[\delta(t)] = \int_{0^{-}}^{\infty} \delta(t)e^{-st} dt = e^{-0} = 1 -$$ - (15.1.3) - -since the impulse function *δ*(*t*) is zero e verywhere except at *t* = 0. The sifting property in Eq. (7.33) has been applied in Eq. (15.1.3). - -# **Figure 15.2** - -For Example 15.1: (a) unit step function, (b) exponential function, (c) unit impulse function. - -Find the Laplace transforms of these functions: *r*(*t*) = *tu*(*t*), that is, the Practice Problem 15.1 ramp function; *Ae*−*at u*(*t*); and *Be*−*jωt u*(*t*). - -**Answer:** 1∕*s 2* , *A*∕(*s* + *a*), *B*∕(*s* + *jω*). - -Determine the Laplace transform of *f*(*t*) = sin *ωt u*(*t*). Example 15.2 - -# **Solution:** - -Using Eq. (B.27) in addition to Eq. (15.1), we obtain the Laplace trans form of the sine function as - -$$ -F(s) = \mathcal{L} \left[ \sin \omega t \right] = \int_0^\infty (\sin \omega t) e^{-st} dt = \int_0^\infty \left( \frac{e^{j\omega t} - e^{-j\omega t}}{2j} \right) e^{-st} dt -$$ -$$ -= \frac{1}{2j} \int_0^\infty (e^{-(s-j\omega)t} - e^{-(s+j\omega)t}) dt -$$ -$$ -= \frac{1}{2j} \left( \frac{1}{s - j\omega} - \frac{1}{s + j\omega} \right) = \frac{\omega}{s^2 + \omega^2} -$$ - -Find the Laplace transform of *f*(*t*) = 15 cos(3 *t*) using the e xponential Practice Problem 15.2 representation for the cosine function. - -**Answer:** 15*s*∕(*s* 2 + 9). - -# **15.3** Properties of the Laplace Transform - -The properties of the Laplace transform help us to obtain transform pairs without directly using Eq. (15.1) as we did in Examples 15.1 and 15.2. As we derive each of these properties, we should keep in mind the definition of the Laplace transform in Eq. (15.1). - -# **Linearity** - -If *F*1(*s*) and *F*2(*s*) are, respectively, the Laplace transforms of *f*1(*t*) and *f*2(*t*), then - -$$ -\mathcal{L}[a_1 f_1(t) + a_2 f_2(t)] = a_1 F_1(s) + a_2 F_2(s) -$$ - (15.7) - -where *a*1 and *a*2 are constants. Equation 15.7 e xpresses the linearity property of the Laplace transform. The proof of Eq. (15.7) follows readily from the definition of the Laplace transform in Eq. (15.1). - -For example, by the linearity property in Eq. (15.7), we may write - -$$ -\mathcal{L}[\cos \omega t \, u(t)] = \mathcal{L} \left[ \frac{1}{2} (e^{j\omega t} + e^{-j\omega t}) \right] = \frac{1}{2} \mathcal{L} [e^{j\omega t}] + \frac{1}{2} \mathcal{L} [e^{-j\omega t}] \tag{15.8} -$$ - -But from Example 15.1(b), [*e*−*at*] = 1∕(*s* + *a*). Hence, - -$$ -\mathcal{L}[\cos \omega t \, u(t)] = \frac{1}{2} \left( \frac{1}{s - j\omega} + \frac{1}{s + j\omega} \right) = \frac{s}{s^2 + \omega^2} -$$ - (15.9) - -# **Scaling** - -If *F*(*s*) is the Laplace transform of *f*(*t*), then - -$$ -\mathcal{L}[f(at)] = \int_{0-}^{\infty} f(at)e^{-st} dt -$$ - (15.10) - -where *a* is a constant and *a* > 0. If we let *x* = *at*, *dx* = *a dt*, then - -$$ -\mathcal{L}[f(at)] = \int_{0-}^{\infty} f(x) e^{-x(s/a)} \frac{dx}{a} = \frac{1}{a} \int_{0-}^{\infty} f(x) e^{-x(s/a)} dx \tag{15.11} -$$ - -Comparing this inte gral with the definition of the Laplace transform in Eq. (15.1) shows that *s* in Eq. (15.1) must be replaced by *s*∕*a* while the dummy v ariable *t* is replaced by *x*. Hence, we obtain the scaling property as - -$$ -\mathcal{L}[f(at)] = \frac{1}{a} F(\frac{s}{a}) -$$ -\n(15.12) - -For example, we know from Example 15.2 that - -$$ -\mathcal{L}[\sin \omega t \, u(t)] = \frac{\omega}{s^2 + \omega^2} \tag{15.13} -$$ - -Using the scaling property in Eq. (15.12), - -$$ -\mathcal{L}[\sin 2\omega t \, u(t)] = \frac{1}{2} \frac{\omega}{\left(s/2\right)^2 + \omega^2} = \frac{2\omega}{s^2 + 4\omega^2} \tag{15.14} -$$ - -which may also be obtained from Eq. (15.13) by replacing *ω* with 2*ω*. - -# **Time Shift** - -If *F*(*s*) is the Laplace transform of *f*(*t*), then - -$$ -\mathcal{L}[f(t-a)u(t-a)] = \int_{0-}^{\infty} f(t-a)u(t-a)e^{-st} dt -$$ - (15.15) - -But *u*(*t* − *a*) = 0 for *t* < *a* and *u*(*t* − *a*) = 1 for *t* > *a*. Hence, - -$$ -\mathcal{L}[f(t-a)u(t-a)] = \int_{a}^{\infty} f(t-a)e^{-st} dt -$$ - (15.16) - -If we let *x* = *t* − *a*, then *dx* = *dt* and *t* = *x* + *a*. As *t* → *a*, *x* → 0 and as *t* → ∞, *x* → ∞. Thus, - -$$ -\mathcal{L}[f(t-a)u(t-a)] = \int_0^\infty f(x)e^{-s(x+a)} dx -$$ -$$ -= e^{-as} \int_0^\infty f(x)e^{-sx} dx = e^{-as} F(s) -$$ - -or - -$$ -\mathcal{L}[f(t-a)u(t-a)] = e^{-as} F(s) -$$ - (15.17) - -In other w ords, if a function is delayed in time by *a*, the result in the *s*-domain is found by multiplying the Laplace transform of the function (without the delay) by *e*−*as*. This is called the *time-delay* or *time-shift property* of the Laplace transform. - -As an example, we know from Eq. (15.9) that - -$$ -\mathcal{L}[\cos \omega t \, u(t)] = \frac{s}{s^2 + \omega^2} -$$ - -Using the time-shift property in Eq. (15.17), - -$$ -\mathcal{L}[\cos \omega(t-a)u(t-a)] = e^{-as} \frac{s}{s^2 + \omega^2} -$$ - (15.18) - -# **Frequency Shift** - -If *F*(*s*) is the Laplace transform of *f*(*t*), then - -$$ -\mathcal{L}[e^{-at}f(t)u(t)] = \int_0^\infty e^{-at}f(t)e^{-st} dt -$$ -$$ -= \int_0^\infty f(t)e^{-(s+a)t} dt = F(s+a) -$$ - -or - -$$ -\mathcal{L}[e^{-at}f(t)u(t)] = F(s+a) -$$ - (15.19) - -That is, the Laplace transform of *e*−*at f*(*t*) can be obtained from the Laplace transform of *f*(*t*) by replacing every *s* with *s* + *a*. This is known as *frequency shift* or *frequency translation.* - -As an example, we know that - -cos *ωt u*(*t*) ⇔ \_\_\_\_\_\_\_ *s s* 2 + *ω*2 - -and - -**(15.20)** - -$$ -\sin \omega t \, u(t) \qquad \Leftrightarrow \qquad \frac{\omega}{s^2 + \omega^2} -$$ - -Using the shift property in Eq. (15.19), we obtain the Laplace transform of the damped sine and damped cosine functions as - -and damped cosine functions as -\n -$$ -\mathcal{L}[e^{-at} \cos \omega t \, u(t)] = \frac{s+a}{(s+a)^2 + \omega^2} -$$ -\n(15.21a) - -$$ -(s + a)^{2} + \omega^{2} -$$ - -\n -$$ -\mathcal{L}[e^{-at} \sin \omega t \, u(t)] = \frac{\omega}{(s + a)^{2} + \omega^{2}} -$$ -\n(15.21b) - -# **Time Differentiation** - -Given that *F*(*s*) is the Laplace transform of *f*(*t*), the Laplace transform of its derivative is - -$$ -\mathcal{L}\left[\frac{df}{dt}u(t)\right] = \int_{0^{-}}^{\infty} \frac{df}{dt} e^{-st} dt -$$ - (15.22) - -To inte grate this by parts, we let *u* = *e*−*st*, *du* = −*se*−*st dt*, and *dv* = (*df*∕*dt*) *dt* = *df*(*t*), *v* = *f*(*t*). Then - -$$ -\mathcal{L}\left[\frac{df}{dt}u(t)\right] = f(t)e^{-st}\Big|_{0^{-}}^{\infty} - \int_{0^{-}}^{\infty}f(t)[-se^{-st}] dt -$$ - -= 0 - f(0-) + s $\int_{0^{-}}^{\infty}f(t)e^{-st} dt = sF(s) - f(0^{-})$ - -or - -$$ -\mathcal{L}[f'(t)] = sF(s) - f(0^{-}) -$$ -\n(15.23) - -The Laplace transform of the second deri vative of *f*(*t*) is a repeated application of Eq. (15.23) as - -$$ -\mathcal{L}\left[\frac{d^2f}{dt^2}\right] = s\mathcal{L}[f'(t)] - f'(0^-) = s[sF(s) - f(0^-)] - f'(0^-) -$$ - -= $s^2F(s) - sf(0^-) - f'(0^-)$ - -or - -$$ -\mathcal{L}[f''(t)] = s^2 F(s) - sf(0^-) - f'(0^-) -$$ - (15.24) - -Continuing in this manner , we can obtain the Laplace transform of the *n*th derivative of *f*(*t*) as - -$$ -\mathcal{L}\left[\frac{d^n f}{dt^n}\right] = s^n F(s) - s^{n-1} f(0^-) - s^{n-2} f'(0^-) - \dots - s^0 f^{(n-1)}(0^-) -$$ -\n(15.25) - -As an example, we can use Eq. (15.23) to obtain the Laplace transform of the sine from that of the cosine. If we let *f*(*t*) = cos *ωt u*(*t*), then *f*(0) = 1 and *f* ′(*t*) = −*ω* sin *ωt u*(*t*). Using Eq. (15.23) and the scaling property, - -$$ -\mathcal{L}[\sin \omega t \, u(t)] = -\frac{1}{\omega} \mathcal{L}[f'(t)] = -\frac{1}{\omega} [sF(s) - f(0^{-})] -$$ -$$ -= -\frac{1}{\omega} \left(s \frac{s}{s^{2} + \omega^{2}} - 1\right) = \frac{\omega}{s^{2} + \omega^{2}} -$$ -(15.26) - -as expected. - -# **Time Integration** - -If *F*(*s*) is the Laplace transform of *f*(*t*), the Laplace transform of its integral is - -$$ -\mathcal{L}\bigg[\int_0^t f(x)dx\bigg] = \int_0^\infty \bigg[\int_0^t f(x)dx\bigg] e^{-st} dt \qquad (15.27) -$$ - -To integrate this by parts, we let - -$$ -u = \int_0^t f(x)dx, \qquad du = f(t)dt -$$ - -and - -$$ -dv = e^{-st} dt, \qquad v = -\frac{1}{s}e^{-st} -$$ - -Then - -$$ -C\left[\int_0^t f(x)dx\right] = \left[\int_0^t f(x)dx\right] \left(-\frac{1}{s}e^{-st}\right)\Big|_0^{\infty} -$$ -$$ --\int_0^{\infty} \left(-\frac{1}{s}\right)e^{-st}f(t)dt -$$ - -For the first term on the right-hand side of the equation, evaluating the term at *t* = ∞ yields zero due to *e*−*s*∞ and evaluating it at *t* = 0 gives \_\_1 *s* ∫ 0 0 *f*(*x*) *dx* = 0. Thus, the first term is zero, and *t* ∞ - -[∫ 0 *f*(*x*)*dx* ] = \_\_1 *s* ∫ 0− *f*(*t*)*e*−*st dt* = \_\_1 *s F*(*s*) - -or simply, - -$$ -\mathcal{L}\left[\int_0^t f(x)dx\right] = \frac{1}{s}F(s) -$$ - (15.28) - -As an example, if we let *f*(*t*) = *u*(*t*), from Example 15.1(a), *F*(*s*) = 1∕*s*. Using Eq. (15.28), - -$$ -\mathcal{L}\bigg[\int_0^t f(x)dx\bigg] = \mathcal{L}[t] = \frac{1}{s}\bigg(\frac{1}{s}\bigg) -$$ - -Thus, the Laplace transform of the ramp function is - -$$ -\mathcal{L}[t] = \frac{1}{s^2} \tag{15.29} -$$ - -Applying Eq. (15.28), this gives - -$$ -\left[\int_0^t x dx\right] = \mathcal{L}\left[\frac{t^2}{2}\right] = \frac{1}{s} \frac{1}{s^2} -$$ - -$$ -\mathcal{L}[t^2] = \frac{2}{s^3} \tag{15.30} -$$ - -or - -Repeated applications of Eq. (15.28) lead to - -$$ -\mathcal{L}[t^n] = \frac{n!}{s^{n+1}} \tag{15.31} -$$ - -Similarly, using integration by parts, we can show that - -$$ -\mathcal{L}\bigg[\int_{-\infty}^{t} f(x)dx\bigg] = \frac{1}{s}F(s) + \frac{1}{s}f^{-1}(0^{-}) -$$ -\n(15.32) - -where - -$$ -f^{-1}(0^{-}) = \int_{-\infty}^{0^{-}} f(t)dt -$$ - -# **Frequency Differentiation** - -If *F*(*s*) is the Laplace transform of *f*(*t*), then - -$$ -F(s) = \int_{0^-}^{\infty} f(t)e^{-st} dt -$$ - -Taking the derivative with respect to *s*, - -$$ -\frac{dF(s)}{ds} = \int_{0^{-}}^{\infty} f(t)(-te^{-st}) dt = \int_{0^{-}}^{\infty} (-tf(t))e^{-st} dt = \mathcal{L}[-tf(t)] -$$ - -and the frequency differentiation property becomes - -$$ -\mathcal{L}[tf(t)] = -\frac{dF(s)}{ds} -$$ - (15.33) - -Repeated applications of this equation lead to - -$$ -\mathcal{L}[t^n f(t)] = (-1)^n \frac{d^n F(s)}{ds^n} -$$ - (15.34) - -For example, we know from Example 15.1(b) that [*e* −*at*] = 1∕(*s* + *a*). Using the property in Eq. (15.33), - -$$ -\mathcal{L}[te^{-at}u(t)] = -\frac{d}{ds}\left(\frac{1}{s+a}\right) = \frac{1}{(s+a)^2} -$$ -(15.35) - -Note that if *a* = 0, we obtain [*t*] = 1∕*s* 2 as in Eq. (15.29), and repeated applications of Eq. (15.33) will yield Eq. (15.31). - -# **Time Periodicity** - -If function *f*(*t*) is a periodic function such as sho wn in Fig. 15.3, it can be represented as the sum of time-shifted functions sho wn in Fig. 15.4. Thus, - -$$ -f(t) = f_1(t) + f_2(t) + f_3(t) + \cdots -$$ - -= $f_1(t) + f_1(t - T)u(t - T)$ -+ $f_1(t - 2T)u(t - 2T) + \cdots$ (15.36) - -where *f*1(*t*) is the same as the function *f*(*t*) g ated o ver the interv al 0 < *t* < *T*, that is, - -$$ -f_1(t) = f(t)[u(t) - u(t - T)] -$$ - (15.37a) - -A periodic function. - -$$ -f_1(t) = \begin{cases} f(t), & 0 < t < T \\ 0, & \text{otherwise} \end{cases} \tag{15.37b} -$$ - -We no w transform each term in Eq. (15.36) and apply the time-shift property in Eq. (15.17). We obtain - -$$ -F(s) = F_1(s) + F_1(s)e^{-Ts} + F_1(s)e^{-2Ts} + F_1(s)e^{-3Ts} + \cdots -$$ - -= $F_1(s)[1 + e^{-Ts} + e^{-2Ts} + e^{-3Ts} + \cdots]$ (15.38) - -But - -$$ -1 + x + x2 + x3 + \dots = \frac{1}{1 - x} -$$ - (15.39) - -if |*x*| < 1. Hence, - -$$ -F(s) = \frac{F_1(s)}{1 - e^{-Ts}} \tag{15.40} -$$ - -where *F*1(*s*) is the Laplace transform of *f*1(*t*); in other words, *F*1(*s*) is the transform *f*(*t*) defined over its first period only. Equation (15.40) shows that the Laplace transform of a periodic function is the transform of the first period of the function divided by 1 − *e*−*Ts*. - -# **Initial and Final Values** - -The initial-value and final-value properties allo w us to find the initial value *f*(0) and the final value *f*(∞) of *f*(*t*) directly from its Laplace transform *F*(*s*). To obtain these properties, we be gin with the dif ferentiation property in Eq. (15.23), namely, - -$$ -sF(s) - f(0) = \mathcal{L}\left[\frac{df}{dt}\right] = \int_{0^-}^{\infty} \frac{df}{dt} e^{-st} dt -$$ - (15.41) - -If we let *s* → ∞, the integrand in Eq. (15.41) vanishes due to the damping exponential factor, and Eq. (15.41) becomes - -$$ -\lim_{s \to \infty} [sF(s) - f(0)] = 0 -$$ - -Because *f* (0) is independent of *s*, we can write - -$$ -f(0) = \lim_{s \to \infty} sF(s) \tag{15.42} -$$ - -This is known as the *initial-value theorem*. For example, we know from Eq. (15.21a) that - -5.21a) that -\n -$$ -f(t) = e^{-2t} \cos 10t \, u(t) \qquad \Leftrightarrow \qquad F(s) = \frac{s+2}{(s+2)^2 + 10^2} \quad \textbf{(15.43)} -$$ - -Using the initial-value theorem, - -$$ -f(0) = \lim_{s \to \infty} sF(s) = \lim_{s \to \infty} \frac{s^2 + 2s}{s^2 + 4s + 104} -$$ -$$ -= \lim_{s \to \infty} \frac{1 + 2/s}{1 + 4/s + 104/s^2} = 1 -$$ - -which confirms what we would expect from the given *f*(*t*). - -In Eq. (15.41), we let *s* → 0; then - -$$ -\lim_{s \to 0} [sF(s) - f(0^{-})] = \int_{0^{-}}^{\infty} \frac{df}{dt} e^{0t} dt = \int_{0^{-}}^{\infty} df = f(\infty) - f(0^{-}) -$$ - -or - -$$ -f(\infty) = \lim_{s \to 0} sF(s) \tag{15.44} -$$ - -This is referred to as the *final-value theorem*. In order for the final-value theorem to hold, all poles of *F*(*s*) must be located in the left half of the *s* plane (see Fig. 15.1 or 15.9); that is, the poles must have negative real parts. The only e xception to this requirement is the case in which *F*(*s*) has a simple pole at *s* = 0, because the effect of 1∕*s* will be nullified by *sF*(*s*) in Eq. (15.44). For example, from Eq. (15.21b), - -$$ -f(t) = e^{-2t} \sin 5t \, u(t) \qquad \Leftrightarrow \qquad F(s) = \frac{5}{(s+2)^2 + 5^2} \quad \textbf{(15.45)} -$$ - -Applying the final-value theorem, - -$$ -f(\infty) = \lim_{s \to 0} s F(s) = \lim_{s \to 0} \frac{5s}{s^2 + 4s + 29} = 0 -$$ - -as expected from the given *f*(*t*). As another example, - -$$ -f(t) = \sin t \, u(t) \qquad \Leftrightarrow \qquad f(s) = \frac{1}{s^2 + 1} -$$ - (15.46) - -so that - -$$ -f(\infty) = \lim_{s \to 0} sF(s) = \lim_{s \to 0} \frac{s}{s^2 + 1} = 0 -$$ - -This is incorrect, because *f*(*t*) = sin *t* oscillates between +1 and −1 and does not ha ve a limit as *t* → ∞. Thus, the final-value theorem cannot be used to find the final value of *f*(*t*) = sin *t*, because *F*(*s*) has poles at *s* = ±*j*, which are not in the left half of the *s* plane. In gen eral, the final-value theorem does not apply in finding the final values of sinusoidal functions—these functions oscillate forever and do not have final values. - -The initial-value and final-value theorems depict the relationship between the origin and infinity in the time domain and the *s*-domain. They serve as useful checks on Laplace transforms. - -Table 15.1 provides a list of the properties of the Laplace trans form. The last property (on con volution) will be pro ved in Sec tion 15.5. There are other properties, b ut these are enough for pres ent purposes. Table 15.2 summarizes the Laplace transforms of some common functions. We have omitted the f actor *u*(*t*) except where it is necessary. - -We should mention that many software packages, such as Mathcad, *MATLAB*, Maple, and Mathematica, offer symbolic math. For example, Mathcad has symbolic math for the Laplace, F ourier, and Z transforms as well as the inverse function. - -| TABLE 15.1 | | | TABLE 15.2 | | -|--------------------------------------|------------------------|-------------------------------------------------------------------------------|------------------------------------------|-------------------------------------------------------| -| Properties of the Laplace transform. | | Laplace transform pairs.* | | | -| Property | f(t) | F(s) | f(t) | F(s) | -| Linearity | a1f1(t) + a2f2(t) | a1F1(s) + a2F2(s) | δ(t) | 1 | -| Scaling | f(at) | __1
__s
a F(
a ) | u(t) | __1
s | -| Time shift | f(t − a)u(t − a) | e−as F(s) | e−at | _____ 1
s + a | -| Frequency shift | e−at f(t) | F(s + a) | t | __1 | -| Time differentiation | df
__
dt | sF(s) − f(0−) | t n | 2
s
____ n!
n+1 | -| | d2
f ___
dt2 | 2
F(s) − sf(0−) −
f ′(0−)
s | te−at | s
_______ 1
2
(s + a) | -| | d3
f ___
dt3 | 3
2
f(0−) −
sf ′(0−) −
f ″(0−)
s
F(s) − s | n
e−at
t | ________ n!
n+1
(s + a) | -| | n
d
f ___
dtn | n
n−1 f(0−) −
n−2 f ′(0−)
s
F(s) − s
s
(n−1)(0−)
− ⋯ − f | sin ωt | _______ ω
2
+ ω2
s | -| Time integration | t
f(x)dx
∫
0 | __1
F(s)
s | cos ωt | _______ s
2
+ ω2
s | -| Frequency
differentiation | tf(t) | − __d
ds F(s) | sin(ωt + θ) | s sin θ + ω cos θ
______________
2
+ ω2
s | -| Frequency
integration | f(t) ___
t | ∞
F(s)ds
∫
s | cos(ωt + θ) | s cos θ − ω sin θ
______________
2
+ ω2
s | -| Time periodicity | f(t) = f(t + nT) | F1(s) _______
1 − e−sT | e−at sin ωt | ___________ ω
2
+ ω2
(s + a) | -| Initial value | f(0) | s→∞ sF(s)
lim | e−at cos ωt | ___________ s + a
2
+ ω2
(s + a) | -| Final value | f(∞) | lim
sF(s)
s→0 | | | -| Convolution | f1(t) * f2(t) | F1(s)F2(s) | *Defined for t ≥ 0; f(t) = 0, for t < 0. | | - -Obtain the Laplace transform of *f*(*t*) = *δ*(*t*) + 2*u*(*t*) − 3*e* Example 15.3 −2*t u*(*t*). - -# **Solution:** - -By the linearity property, - -$$ -F(s) = \mathcal{L}[\delta(t)] + 2\mathcal{L}[u(t)] - 3\mathcal{L}[e^{-2t} u(t)] -$$ -$$ -= 1 + 2\frac{1}{s} - 3\frac{1}{s+2} = \frac{s^2 + s + 4}{s(s+2)} -$$ - -Find the Laplace transform of *f*(*t*) = (cos (2*t*) + *e* Practice Problem 15.3 −4*t* )*u*(*t*). - -Find the Laplace transform -\n**Answer:** -$$ -\frac{2s^2 + 4s + 4}{(s + 4)(s^2 + 4)} -$$ - -Example 15.4 Determine the Laplace transform of *f*(*t*) = *t* 2 sin 2*t u*(*t*). - -# **Solution:** - -We know that - -$$ -[\sin 2t] = \frac{2}{s^2 + 2^2} -$$ - -Using frequency differentiation in Eq. (15.34), - -$$ -F(s) = \mathcal{L}[t^2 \sin 2t] = (-1)^2 \frac{d^2}{ds^2} \left(\frac{2}{s^2 + 4}\right) -$$ -$$ -= \frac{d}{ds} \left(\frac{-4s}{(s^2 + 4)^2}\right) = \frac{12s^2 - 16}{(s^2 + 4)^3} -$$ - -Practice Problem 15.4 Find the Laplace transform of *f*(*t*) = *t* 2 cos 3*t u*(*t*). - -**Answer:** -$$ -\frac{2s(s^2 - 27)}{(s^2 + 9)^3} -$$ - -**Answer:** - -\_\_\_ 10 *s* - -(2 − *e*−4*s* − *e*−8*s* - -We can express the gate function in Fig. 15.5 as - -$$ -g(t) = 10[u(t-2) - u(t-3)] -$$ - -Given that we kno w the Laplace transform of *u*(*t*), we apply the timeshift property and obtain - -$$ -G(s) = 10\left(\frac{e^{-2s}}{s} - \frac{e^{-3s}}{s}\right) = \frac{10}{s}(e^{-2s} - e^{-3s}) -$$ - -The gate function; for Example 15.5. - -Practice Problem 15.5 Find the Laplace transform of the function *h*(*t*) in Fig. 15.6. - -). - -| $h(t)$ | | | | -|--------|---|---|---| -| 20 | | | | -| 10 | | | | -| 0 | 4 | 8 | t | - -\n**Figure 15.6** - -For Practice Prob. 15.5. - -$$ -\mathbf{36}^{\dagger} -$$ - -Calculate the Laplace transform of the periodic function in Fig. 15.7. Example 15.6 - -# **Solution:** - -The period of the function is *T* = 2. To apply Eq. (15.40), we first obtain the transform of the first period of the function. - -$$ -f_1(t) = 2t[u(t) - u(t-1)] = 2tu(t) - 2tu(t-1) -$$ - -= 2tu(t) - 2(t - 1 + 1)u(t - 1) -= 2tu(t) - 2(t - 1)u(t-1) - 2u(t-1) - -Using the time-shift property, - -$$ -F_1(s) = \frac{2}{s^2} - 2\frac{e^{-s}}{s^2} - \frac{2}{s}e^{-s} = \frac{2}{s^2}(1 - e^{-s} - se^{-s}) -$$ - -Thus, the transform of the periodic function in Fig. 15.7 is - -$$ -F(s) = \frac{F_1(s)}{1 - e^{-Ts}} = \frac{2}{s^2(1 - e^{-2s})}(1 - e^{-s} - se^{-s}) -$$ - -Determine the Laplace transform of the periodic function in Fig. 15.8. Practice Problem 15.6 - -For Practice Prob. 15.6. - -Find the initial and final values of the function whose Laplace trans- Example 15.7 form is - -inal values of the function who -$$ -H(s) = \frac{20}{(s+3)(s^2+8s+25)} -$$ - -# **Solution:** - -Applying the initial-value theorem, - -**lution:** -\nplying the initial-value theorem, -\n -$$ -h(0) = \lim_{s \to \infty} sH(s) = \lim_{s \to \infty} \frac{20s}{(s+3)(s^2+8s+25)} -$$ -\n -$$ -= \lim_{s \to \infty} \frac{20/s^2}{(1+3/s)(1+8/s+25/s^2)} = \frac{0}{(1+0)(1+0+0)} = 0 -$$ - -To be sure that the final-value theorem is applicable, we check where the poles of *H*(*s*) are located. The poles of *H*(*s*) are *s* = −3, −4 ± *j*3, which all have negative real parts: They are all located on the left half of the *s* plane (Fig. 15.9). Hence, the final-value theorem applies and \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ 20*s* - -$$ -h(\infty) = \lim_{s \to 0} sH(s) = \lim_{s \to 0} \frac{20s}{(s+3)(s^2+8s+25)} -$$ -$$ -= \frac{0}{(0+3)(0+0+25)} = 0 -$$ - -**Figure 15.9** For Example 15.7: Poles of *H*(*s*). - -**Figure 15.7** For Example 15.6. Both the initial and final values could be determined from *h*(*t*) if we knew it. See Example 15.11, where *h*(*t*) is given. - -Practice Problem 15.7 Obtain the initial and the final values of - -he final values of -\n -$$ -G(s) = \frac{3s^3 + 2s + 6}{s(s+1)^2(s+1.5)} -$$ - -**Answer:** 3, 4. - -# **15.4** The Inverse Laplace Transform - -Given *F*(*s*), how do we transform it back to the time domain and obtain the corresponding *f*(*t*)? By matching entries in Table 15.2, we a void using Eq. (15.5) to find *f*(*t*). - -Suppose *F*(*s*) has the general form of - -$$ -F(s) = \frac{N(s)}{D(s)} -$$ -(15.47) - -where *N*(*s*) is the numerator polynomial and *D*(*s*) is the denominator polynomial. The roots of *N*(*s*) = 0 are called the *zeros* of *F*(*s*), while the roots of *D*(*s*) = 0 are the *poles* of *F*(*s*). Although Eq. (15.47) is similar in form to Eq. (14.3), here *F*(*s*) is the Laplace transform of a function, which is not necessarily a transfer function. We use *partial fraction expansion* to break *F*(*s*) down into simple terms whose inverse transform we obtain from Table 15.2. Thus, finding the inverse Laplace transform of *F*(*s*) involves two steps. - -Steps to Find the Inverse Laplace Transform: - -- 1. Decompose *F*(*s*) into simple terms using partial fraction expansion. -- 2. Find the inverse of each term by matching entries in Table 15.2. - -Let us consider the three possible forms *F*(*s*) may take and how to apply the two steps to each form. - -# **15.4.1** Simple Poles - -Recall from Chapter 14 that a simple pole is a first-order pole. If *F*(*s*) has only simple poles, then *D*(*s*) becomes a product of factors, so that - -then -$$ -D(s) -$$ - becomes a product of factors, so that -\n -$$ -F(s) = \frac{N(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)} -$$ -\n(15.48) - -where *s* = −*p*1, −*p*2, … , −*pn* are the simple poles, and *pi* ≠ *pj* for all *i* ≠ *j* (i.e., the poles are distinct). Assuming that the degree of *N*(*s*) is - -Software packages such as MATLAB, Mathcad, and Maple are capable of finding partial fraction expansions quite easily. - -Otherwise, we must first apply long division so that F(s) = N(s)∕D(s) = Q(s) + R(s)∕D(s), where the degree of R(s), the remainder of the long division, is less than the degree of D(s). - -less than the degree of *D*(*s*), we use partial fraction expansion to decompose *F*(*s*) in Eq. (15.48) as - -$$ -F(s) = \frac{k_1}{s + p_1} + \frac{k_2}{s + p_2} + \dots + \frac{k_n}{s + p_n} -$$ - (15.49) - -The expansion coefficients *k*1, *k*2, … , *kn* are kno wn as the *residues* of *F*(*s*). There are man y ways of finding the expansion coefficients. One way is using the *residue method*. If we multiply both sides of Eq. (15.49) by (*s* + *p*1), we obtain - -$$ -(s+p_1)F(s) = k_1 + \frac{(s+p_1)k_2}{s+p_2} + \dots + \frac{(s+p_1)k_n}{s+p_n} -$$ - (15.50) - -Because *pi* ≠ *pj*, setting *s* = −*p*1 in Eq. (15.50) leaves only *k*1 on the righthand side of Eq. (15.50). Hence, - -$$ -(s+p_1)F(s)\big|_{s=-p_1} = k_1\tag{15.51} -$$ - -Thus, in general, - -$$ -k_i = (s + p_i)F(s) \Big|_{s = -p_i} -$$ - (15.52) - -This is known as *Heaviside's theorem*. Once the values of *ki* are known, we proceed to find the inverse of *F*(*s*) using Eq. (15.49). Since the inverse transform of each term in Eq. (15.49) is −1 [*k*∕(*s* + *a*)] = *ke*−*at u*(*t*), then, from Table 15.2, - -$$ -f(t) = (k_1 e^{-p_1 t} + k_2 e^{-p_2 t} + \dots + k_n e^{-p_n t}) u(t) -$$ - (15.53) - -**15.4.2** Repeated Poles - -Suppose *F*(*s*) has *n* repeated poles at *s* = −*p*. Then we may represent *F*(*s*) as - -$$ -F(s) = \frac{k_n}{(s+p)^n} + \frac{k_{n-1}}{(s+p)^{n-1}} + \dots + \frac{k_2}{(s+p)^2} + \frac{k_1}{s+p} + F_1(s) -$$ -\n(15.54) - -where *F*1(*s*) is the remaining part of *F*(*s*) that does not ha ve a pole at *s* = −*p*. We determine the expansion coefficient *kn* as - -$$ -k_n = (s+p)^n F(s) \Big|_{s=-p} \tag{15.55} -$$ - -as we did above. To determine *kn*−1, we multiply each term in Eq. (15.54) by (*s* + *p*) *n* and differentiate to get rid of *kn*, then evaluate the result at *s* = −*p* to get rid of the other coefficients except *kn*−1. Thus, we obtain - -$$ -k_{n-1} = \frac{d}{ds} [(s+p)^n F(s)] \Big|_{s=-p} -$$ - (15.56) - -Repeating this gives - -$$ -k_{n-2} = \frac{1}{2!} \frac{d^2}{ds^2} [(s+p)^n F(s)] \Big|_{s=-p} -$$ - (15.57) - -Historical note: Named after Oliver Heaviside (1850–1925), an English engineer, the pioneer of operational calculus. - -The *m*th term becomes - -$$ -k_{n-m} = \frac{1}{m!} \frac{d^m}{ds^m} [(s+p)^n F(s)] \Big|_{s=-p} -$$ - (15.58) - -where *m* = 1, 2, … , *n* − 1. One can expect the differentiation to be difficult to handle as *m* increases. Once we obtain the values of *k*1, *k*2, … , *kn* by partial fraction expansion, we apply the inverse transform - -$$ -\mathcal{L}^{-1}\left[\frac{1}{(s+a)^n}\right] = \frac{t^{n-1}e^{-at}}{(n-1)!}u(t) -$$ -\n(15.59) - -to each term on the right-hand side of Eq. (15.54) and obtain - -$$ -f(t) = \left(k_1 e^{-pt} + k_2 t e^{-pt} + \frac{k_3}{2!} t^2 e^{-pt} + \dots + \frac{k_n}{(n-1)!} t^{n-1} e^{-pt} \right) u(t) + f_1(t) -$$ -\n(15.60) - -# **15.4.3** Complex Poles - -A pair of comple x poles is simple if it is not repeated; it is a double or multiple pole if repeated. Simple comple x poles may be handled the same way as simple real poles, but because complex algebra is involved the result is always cumbersome. An easier approach is a method known as *completing the square*. The idea is to express each complex pole pair (or quadratic term) in *D*(*s*) as a complete square such as (*s* + *α*) 2 + *β*2 and then use Table 15.2 to find the inverse of the term. - -Because *N*(*s*) and *D*(*s*) always have real coefficients and we know that the complex roots of polynomials with real coef ficients must occur in conjugate pairs, *F*(*s*) may have the general form - -$$ -F(s) = \frac{A_1 s + A_2}{s^2 + as + b} + F_1(s) -$$ -\n(15.61) - -where *F*1(*s*) is the remaining part of *F*(*s*) that does not ha ve this pair of complex poles. If we complete the square by letting - -$$ -s^{2} + as + b = s^{2} + 2as + a^{2} + \beta^{2} = (s + a)^{2} + \beta^{2} -$$ - (15.62) - -and we also let - -$$ -A_1s + A_2 = A_1(s + \alpha) + B_1\beta \tag{15.63} -$$ - -then Eq. (15.61) becomes - -$$ -F(s) = \frac{A_1(s+\alpha)}{(s+\alpha)^2 + \beta^2} + \frac{B_1\beta}{(s+\alpha)^2 + \beta^2} + F_1(s) -$$ -(15.64) - -From Table 15.2, the inverse transform is - -$$ -f(t) = (A_1 e^{-\alpha t} \cos \beta t + B_1 e^{-\alpha t} \sin \beta t) u(t) + f_1(t) -$$ - (15.65) - -The sine and cosine terms can be combined using Eq. (9.11). - -Whether the pole is simple, repeated, or comple x, a general approach that can al ways be used in finding the expansion coefficients is the *method of algebra*, illustrated in Examples 15.9 to 15.11. To apply the method, we first set *F*(*s*) = *N*(*s*)∕*D*(*s*) equal to an expansion containing unknown constants. We multiply the result through by a common denominator. Then we determine the unkno wn constants by equating coefficients (i.e., by algebraically solving a set of simultaneous equations for these coefficients at like powers of *s*). - -Another general approach is to substitute specific, convenient values of *s* to obtain as man y simultaneous equations as the number of unknown coefficients, and then solve for the unkno wn coefficients. We must make sure that each selected v alue of *s* is not one of the poles of *F*(*s*). Example 15.11 illustrates this idea. - -Find the inverse Laplace transform of Example 15.8 - -$$ -F(s) = \frac{3}{s} - \frac{5}{s+1} + \frac{6}{s^2+4} -$$ - -# **Solution:** - -The inverse transform is given by - -$$ -f(t) = \mathcal{L}^{-1}[F(s)] = \mathcal{L}^{-1}\left(\frac{3}{s}\right) - \mathcal{L}^{-1}\left(\frac{5}{s+1}\right) + \mathcal{L}^{-1}\left(\frac{6}{s^2+4}\right) -$$ -$$ -= (3 - 5e^{-t} + 3\sin 2t)u(t), \qquad t \ge 0 -$$ - -where Table 15.2 has been consulted for the inverse of each term. - -$$ -F(s) = 5 + \frac{6}{s+4} - \frac{7s}{s^2 + 25} -$$ - -**Answer:** 5*δ*(*t*) + (6*e*−4*t* − 7 cos(5*t*))*u*(*t*). - -$$ -F(s) = \frac{s^2 + 12}{s(s+2)(s+3)} -$$ - -# **Solution:** - -Unlike in the pre vious example where the partial fractions ha ve been provided, we first need to determine the partial fractions. Given that there are three poles, we let - -les, we let -\n -$$ -\frac{s^2 + 12}{s(s+2)(s+3)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+3} -$$ -\n(15.9.1) - -where *A*, *B*, and *C* are the constants to be determined. We can find the constants using two approaches. - -Determine the inverse Laplace transform of Practice Problem 15.8 - -Find *f*(*t*) given that Example 15.9 - -■ **METHOD 1 Residue method:** - -**PROOF Residue method:** -\n -$$ -A = sF(s) \Big|_{s=0} = \frac{s^2 + 12}{(s+2)(s+3)} \Big|_{s=0} = \frac{12}{(2)(3)} = 2 -$$ -\n -$$ -B = (s+2)F(s) \Big|_{s=-2} = \frac{s^2 + 12}{s(s+3)} \Big|_{s=-2} = \frac{4+12}{(-2)(1)} = -8 -$$ -\n -$$ -C = (s+3)F(s) \Big|_{s=-3} = \frac{s^2 + 12}{s(s+2)} \Big|_{s=-3} = \frac{9+12}{(-3)(-1)} = 7 -$$ - -■ **METHOD 2 Algebraic method:** Multiplying both sides of Eq. (15.9.1) by *s*(*s* + 2)(*s* + 3) gives - -$$ -s^2 + 12 = A(s+2)(s+3) + Bs(s+3) + Cs(s+2) -$$ - -or - -$$ -s^2 + 12 = A(s^2 + 5s + 6) + B(s^2 + 3s) + C(s^2 + 2s) -$$ - -Equating the coefficients of like powers of *s* gives - -Constant: -$$ -12 = 6A \Rightarrow A = 2 -$$ - -\ns: $0 = 5A + 3B + 2C \Rightarrow 3B + 2C = -10$ - -\ns: $1 = A + B + C \Rightarrow B + C = -1$ - -Thus, *A* = 2, *B* = −8, *C* = 7, and Eq. (15.9.1) becomes - -$$ -F(s) = \frac{2}{s} - \frac{8}{s+2} + \frac{7}{s+3} -$$ - -By finding the inverse transform of each term, we obtain - -$$ -f(t) = (2 - 8e^{-2t} + 7e^{-3t})u(t) -$$ - -Practice Problem 15.9 Find *f*(*t*) if - -$$ -F(s) = \frac{48(s+2)}{(s+1)(s+3)(s+4)} -$$ - -**Answer:** -$$ -f(t) = (8e^{-t} + 24e^{-3t} - 32e^{-4t})u(t) -$$ -. - -Example 15.10 Calculate *v*(*t*) given that - -$$ -V(s) = \frac{10s^2 + 4}{s(s+1)(s+2)^2} -$$ - -# **Solution:** - -While the pre vious e xample is on simple roots, this e xample is on repeated roots. Let - -$$ -V(s) = \frac{10s^2 + 4}{s(s+1)(s+2)^2} -$$ - -= $\frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+2)^2} + \frac{D}{s+2}$ (15.10.1) - -# ■ **METHOD 1 Residue method:** - -**THEOREM 1** Residue method: -\n -$$ -A = sV(s) \Big|_{s=0} = \frac{10s^2 + 4}{(s+1)(s+2)^2} \Big|_{s=0} = \frac{4}{(1)(2)^2} = 1 -$$ -\n -$$ -B = (s+1)V(s) \Big|_{s=-1} = \frac{10s^2 + 4}{s(s+2)^2} \Big|_{s=-1} = \frac{14}{(-1)(1)^2} = -14 -$$ -\n -$$ -C = (s+2)^2 V(s) \Big|_{s=-2} = \frac{10s^2 + 4}{s(s+1)} \Big|_{s=-2} = \frac{44}{(-2)(-1)} = 22 -$$ -\n -$$ -D = \frac{d}{ds} [(s+2)^2 V(s)] \Big|_{s=-2} = \frac{d}{ds} \left( \frac{10s^2 + 4}{s^2 + s} \right) \Big|_{s=-2} = \frac{(s^2 + s)(20s) - (10s^2 + 4)(2s + 1)}{(s^2 + s)^2} \Big|_{s=-2} = \frac{52}{4} = 13 -$$ - -■ **METHOD 2 Algebraic method:** Multiplying Eq. (15.10.1) by *s*(*s* + 1)(*s* + 2)2 , we obtain - -$$ -10s2 + 4 = A(s + 1)(s + 2)2 + Bs(s + 2)2 + Cs(s + 1) + Ds(s + 1)(s + 2) -$$ - -or - -$$ -10s2 + 4 = A(s3 + 5s2 + 8s + 4) + B(s3 + 4s2 + 4s) -$$ - -+ C(s2 + s) + D(s3 + 3s2 + 2s) - -Equating coefficients, - -Constant: -$$ -4 = 4A -$$ - $\Rightarrow$ $A = 1$ - -\ns: $0 = 8A + 4B + C + 2D$ $\Rightarrow$ $4B + C + 2D = -8$ - -\ns2: $10 = 5A + 4B + C + 3D$ $\Rightarrow$ $4B + C + 3D = 5$ - -\ns3: $0 = A + B + D$ $\Rightarrow$ $B + D = -1$ - -Solving these simultaneous equations gi ves *A* = 1, *B* = −14, *C* = 22, *D* = 13, so that - -$$ -V(s) = \frac{1}{s} - \frac{14}{s+1} + \frac{13}{s+2} + \frac{22}{(s+2)^2} -$$ - -Taking the inverse transform of each term, we get - -$$ -v(t) = (1 - 14e^{-t} + 13e^{-2t} + 22te^{-2t})u(t) -$$ - -Obtain *g*(*t*) if Practice Problem 15.10 - -$$ -G(s) = \frac{s^3 + 2s + 6}{s(s+1)^2(s+3)} -$$ - -**Answer:** (2 − 3.25*e*−*t* − 1.5*te*−*t* + 2.25*e*−3*t* )*u*(*t*). - -Find the in verse transform of the frequenc y-domain function in Example 15.11 Example 15.7: *H*(*s*) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ 20 (*s* + 3)(*s* - -$$ -H(s) = \frac{20}{(s+3)(s^2+8s+25)} -$$ - -# **Solution:** - -In this example, *H*(*s*) has a pair of complex poles at *s* 2 + 8*s* + 25 = 0 or *s* = −4 ± *j*3. We let - -is example, -$$ -H(s) -$$ - has a pair of complex poles at $s^2 + 8s + 25 = 0$ or --4 ± j3. We let -$$ -H(s) = \frac{20}{(s+3)(s^2+8s+25)} = \frac{A}{s+3} + \frac{Bs+C}{(s^2+8s+25)} -$$ -(15.11.1) - -We now determine the expansion coefficients in two ways. - -■ **METHOD 1 Combination of methods:** We can obtain *A* using the method of residue, - -$$ -A = (s + 3)H(s)\Big|_{s=-3} = \frac{20}{s^2 + 8s + 25}\Big|_{s=-3} = \frac{20}{10} = 2 -$$ - -Although *B* and *C* can be obtained using the method of residue, we will not do so, to a void comple x algebra. Rather , we can substitute two specific values of *s* [say *s* = 0, 1, which are not poles of *F*(*s*)] into Eq. (15.11.1). This will give us two simultaneous equations from which to find *B* and *C*. If we let *s* = 0 in Eq. (15.11.1), we obtain - -$$ -\frac{20}{75} = \frac{A}{3} + \frac{C}{25} -$$ - -or - -$$ -20 = 25A + 3C \tag{15.11.2} -$$ - -Because *A* = 2, Eq. (15.11.2) gi ves *C* = −10. Substituting *s* = 1 into Eq. (15.11.1) gives - -$$ -\frac{20}{(4)(34)} = \frac{A}{4} + \frac{B+C}{34} -$$ - -or - -$$ -20 = 34A + 4B + 4C \tag{15.11.3} -$$ - -But *A* = 2, *C* = −10, so that Eq. (15.11.3) gives *B* = −2. - -■ **METHOD 2 Algebraic method:** Multiplying both sides of Eq. (15.11.1) by (*s* + 3)(*s* 2 + 8*s* + 25) yields - -$$ -20 = A(s2 + 8s + 25) + (Bs + C)(s + 3) -$$ - -= A(s2 + 8s + 25) + B(s2 + 3s) + C(s + 3) (15.11.4) - -Equating coefficients gives - -*s* 2 : 0 = *A* + *B* ⇒ *A* = −*B s*: 0 = 8*A* + 3*B* + *C* = 5*A* + *C* ⇒ *C* = −5*A* Constant: 20 = 25*A* + 3*C* = 25*A* − 15*A* ⇒ *A* = 2 - -That is, *B* = −2, *C* = −10. Thus, - -$$ -B = -2, C = -10. \text{ Thus,} -$$ - -\n -$$ -H(s) = \frac{2}{s+3} - \frac{2s+10}{(s^2+8s+25)} = \frac{2}{s+3} - \frac{2(s+4)+2}{(s+4)^2+9} -$$ - -\n -$$ -= \frac{2}{s+3} - \frac{2(s+4)}{(s+4)^2+9} - \frac{2}{3} \frac{3}{(s+4)^2+9} -$$ - -Taking the inverse of each term, we obtain - -$$ -h(t) = \left(2e^{-3t} - 2e^{-4t}\cos 3t - \frac{2}{3}e^{-4t}\sin 3t\right)u(t) \tag{15.11.5} -$$ - -It is alright to lea ve the result this w ay. However, we can combine the cosine and sine terms as - -$$ -h(t) = (2e^{-3t} - Re^{-4t}\cos(3t - \theta))u(t) -$$ - (15.11.6) - -To obtain Eq. (15.11.6) from Eq. (15.11.5), we apply Eq. (9.11). Next, we determine the coefficient *R* and the phase angle *θ*: - -$$ -R = \sqrt{2^2 + \left(\frac{2}{3}\right)^2} = 2.108, \qquad \theta = \tan^{-1} \frac{\frac{2}{3}}{2} = 18.43^{\circ} -$$ - -Thus, - -$$ -h(t) = (2e^{-3t} - 2.108e^{-4t}\cos(3t - 18.43^\circ))u(t) -$$ - -$$ -G(s) = \frac{20}{(s+1)(s^2+4s+13)} -$$ - -**Answer:** 2*e*−*t* − 2*e*−2*t* cos 3*t* − 0.6667*e*−2*t* sin 3*t*, *t* ≥ 0. - -# **15.5** The Convolution Integral - -The term *convolution* means "folding." Convolution is an invaluable tool to the engineer because it provides a means of viewing and characterizing physical systems. For example, it is used in finding the response *y*(*t*) of a system to an e xcitation *x*(*t*), knowing the system impulse response *h*(*t*). This is achieved through the *convolution integral*, defined as - -$$ -y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t - \lambda) d\lambda -$$ - (15.66) - -or simply - -$$ -y(t) = x(t) * h(t) -$$ - (15.67) - -where *λ* is a dummy v ariable and the asterisk denotes con volution. Equation (15.66) or (15.67) states that the output is equal to the input convolved with the unit impulse response. The convolution process is commutative: - -$$ -y(t) = x(t) * h(t) = h(t) * x(t) -$$ -\n(15.68a) - -or - -$$ -y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t-\lambda) d\lambda = \int_{-\infty}^{\infty} h(\lambda)x(t-\lambda) d\lambda -$$ - (15.68b) - -This implies that the order in which the tw o functions are convolved is immaterial. We will see shortly how to take advantage of this commutative property when performing graphical computation of the convolution integral. - -Find *g*(*t*) given that Practice Problem 15.11 - -The convolution of two signals consists of time-reversing one of the signals, shifting it, and multiplying it point by point with the second signal, and integrating the product. - -The convolution integral in Eq. (15.66) is the general one; it applies to any linear system. However, the convolution integral can be simplified if we assume that a system has two properties. First, if *x*(*t*) = 0 for *t* < 0, then - -$$ -y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t-\lambda) \, d\lambda = \int_{0}^{\infty} x(\lambda)h(t-\lambda) \, d\lambda \tag{15.69} -$$ - -Second, if the system's impulse response is *causal* (i.e., *h*(*t*) = 0 for *t* < 0), then *h*(*t* − *λ*) = 0 for *t* − *λ* < 0 or *λ* > *t*, so that Eq. (15.69) becomes - -$$ -y(t) = h(t) * x(t) = \int_0^t x(\lambda)h(t - \lambda)d\lambda -$$ - (15.70) - -Here are some properties of the convolution integral. - -1. *x*(*t*) \* *h*(*t*) = *h*(*t*) \* *x*(*t*) (Commutative) 2. *f* (*t*) \* [*x*(*t*) + *y*(*t*)] = *f* (*t*) \* *x*(*t*) + *f* (*t*) \* *y*(*t*) (Distributive) 3. *f* (*t*) \* [*x*(*t*) \* *y*(*t*)] = [ *f*(*t*) \* *x*(*t*)] \* *y*(*t*) (Associative) 4. *f* (*t*) \* *δ*(*t*) = ∫ −∞ ∞ *f* (*λ*)*δ*(*t* − *λ*) *dλ* = *f* (*t*) 5. *f* (*t*) \* *δ*(*t* − *to*) = *f* (*t* − *to*) 6. *f* (*t*) \* *δ*′(*t*) = ∫ −∞ ∞ *f* (*λ*)*δ*′(*t* − *λ*) *dλ* = *f* ′(*t*) 7. *f* (*t*) \* *u*(*t*) = ∫ −∞ ∞ *f* (*λ*)*u*(*t* − *λ*) *dλ* = ∫ −∞ *t f* (*λ*) *dλ* - -Before learning ho w to e valuate the con volution inte gral in Eq. (15.70), let us establish the link between the Laplace transform and the convolution integral. Given two functions *f*1(*t*) and *f*2(*t*) with Laplace transforms *F*1(*s*) and *F*2(*s*), respectively, their convolution is - -$$ -f(t) = f_1(t) * f_2(t) = \int_0^t f_1(\lambda) f_2(t - \lambda) d\lambda -$$ - (15.71) - -Taking the Laplace transform gives - -$$ -F(s) = \mathcal{L}[f_1(t) * f_2(t)] = F_1(s)F_2(s) -$$ -\n(15.72) - -To prove that Eq. (15.72) is true, we begin with the fact that *F*1(*s*) is defined as - -$$ -F_1(s) = \int_{0^-}^{\infty} f_1(\lambda) e^{-s\lambda} d\lambda \qquad (15.73) -$$ - -Multiplying this with *F*2(*s*) gives - -$$ -F_1(s)F_2(s) = \int_{0^-}^{\infty} f_1(\lambda)[F_2(s)e^{-s\lambda}] d\lambda -$$ - (15.74) - -We recall from the time shift property in Eq. (15.17) that the term in brackets can be written as - -$$ -F_2(s)e^{-s\lambda} = \mathcal{L}[f_2(t-\lambda)u(t-\lambda)] -$$ - -= -$$ -\int_0^\infty f_2(t-\lambda)u(t-\lambda)e^{-st} dt -$$ - (15.75) - -Substituting Eq. (15.75) into Eq. (15.74) gives - -$$ -F_1(s)F_2(s) = \int_0^\infty f_1(\lambda) \left[ \int_0^\infty f_2(t-\lambda)u(t-\lambda)e^{-st} dt \right] d\lambda \qquad (15.76) -$$ - -Interchanging the order of integration results in - -$$ -F_1(s)F_2(s) = \int_0^\infty \left[ \int_0^t f_1(\lambda)f_2(t-\lambda) d\lambda \right] e^{-st} dt \qquad (15.77) -$$ - -The integral in brackets extends only from 0 to *t* because the delayed unit step *u*(*t* − *λ*) = 1 for *λ* < *t* and *u*(*t* − *λ*) = 0 for *λ* > *t*. We notice that the integral is the convolution of *f*1(*t*) and *f*2(*t*) as in Eq. (15.71). Hence, - -$$ -F_1(s)F_2(s) = \mathcal{L}[f_1(t) * f_2(t)] -$$ - (15.78) - -as desired. This indicates that convolution in the time domain is equivalent to multiplication in the *s*-domain. For example, if *x*(*t*) = 4*e*−*t* and *h*(*t*) = 5*e*−2*t* , applying the property in Eq. (15.78), we get - -$$ -h(t) * x(t) = \mathcal{L}^{-1}[H(s)X(s)] = \mathcal{L}^{-1}\left[\left(\frac{5}{s+2}\right)\left(\frac{4}{s+1}\right)\right] -$$ -$$ -= \mathcal{L}^{-1}\left[\frac{20}{s+1} + \frac{-20}{s+2}\right] -$$ -(15.79) -$$ -= 20(e^{-t} - e^{-2t}), \qquad t \ge 0 -$$ - -Although we can find the convolution of tw o signals using Eq. (15.78), as we have just done, if the product *F*1(*s*)*F*2(*s*) is very complicated, finding the inverse may be tough. Also, there are situations in which *f*1(*t*) and *f*2(*t*) are a vailable in the form of e xperimental data and there are no explicit Laplace transforms. In these cases, one must do the convolution in the time domain. - -The process of convolving two signals in the time domain is better appreciated from a graphical point of view. The graphical procedure for evaluating the convolution integral in Eq. (15.70) usually in volves four steps. - -Steps to Evaluate the Convolution Integral: - -- 1. Folding: Take the mirror image of *h*(*λ*) about the ordinate axis to obtain *h*(−*λ*). -- 2. Displacement: Shift or delay *h*(−*λ*) by *t* to obtain *h*(*t* − *λ*). -- 3. Multiplication: Find the product of *h*(*t* − *λ*) and *x*(*λ*). -- 4. Integration: F or a gi ven time *t*, calculate the area under the product *h*(*t* − *λ*)*x*(*λ*) for 0 < *λ* < *t* to get *y*(*t*) at *t*. - -The folding operation in step 1 is the reason for the term *convolution*. The function *h*(*t* − *λ*) scans or slides over *x*(*λ*). In view of this superposition procedure, the convolution integral is also known as the *superposition integral*. - -To apply the four steps, it is necessary to be able to sk etch *x*(*λ*) and *h*(*t* − *λ*). To get *x*(*λ*) from the original function *x*(*t*) involves merely replacing every *t* with *λ*. Sketching *h*(*t* − *λ*) is the k ey to the con volution process. It involves reflecting *h*(*λ*) about the vertical axis and shifting it by *t*. Analytically, we obtain *h*(*t* − *λ*) by replacing every *t* in *h*(*t*) by *t* − *λ*. Given that con volution is commutati ve, it may be more con venient to apply steps 1 and 2 to *x*(*t*) instead of *h*(*t*). The best way to illustrate the procedure is with some examples. - -Example 15.12 Find the convolution of the two signals in Fig. 15.10. - -# **Solution:** - -We follow the four steps to get *y*(*t*) = *x*1(*t*) \* *x*2(*t*). First, we fold *x*1(*t*) as shown in Fig. 15.11(a) and shift it by *t* as sho wn in Fig. 15.11(b). F or different values of *t*, we now multiply the two functions and integrate to determine the area of the overlapping region. - -For 0 < *t* < 1, there is no overlap of the two functions, as shown in Fig. 15.12(a). Hence, - -$$ -y(t) = x_1(t) * x_2(t) = 0, \qquad 0 < t < 1 \tag{15.12.1} -$$ - -For 1 < *t* < 2, the tw o signals o verlap between 1 and *t*, as sho wn in Fig. 15.12(b). - -$$ -y(t) = \int_1^t (2)(1) \, d\lambda = 2\lambda \bigg|_1^t = 2(t - 1), \qquad 1 < t < 2 \tag{15.12.2} -$$ - -For 2 < *t* < 3, the two signals completely overlap between (*t* − 1) and *t*, as shown in Fig. 15.12(c). It is easy to see that the area under the curve is 2. Or - -$$ -y(t) = \int_{t-1}^{t} (2)(1) \, d\lambda = 2\lambda \Big|_{t-1}^{t} = 2, \qquad 2 < t < 3 \tag{15.12.3} -$$ - -For 3 < *t* < 4, the two signals overlap between (*t* − 1) and 3, as sho wn in Fig. 15.12(d). - -$$ -y(t) = \int_{t-1}^{3} (2)(1) \, d\lambda = 2\lambda \Big|_{t-1}^{3} -$$ -\n -$$ -= 2(3 - t + 1) = 8 - 2t, \qquad 3 < t < 4 \tag{15.12.4} -$$ - -For *t* > 4, the two signals do not overlap [Fig. 15.12(e)], and - -$$ -y(t) = 0, \qquad t > 4 \tag{15.12.5} -$$ - -Combining Eqs. (15.12.1) to (15.12.5), we obtain - -$$ -y(t) = \begin{cases} 0, & 0 \le t \le 1 \\ 2t - 2, & 1 \le t \le 2 \\ 2, & 2 \le t \le 3 \\ 8 - 2t, & 3 \le t \le 4 \\ 0, & t \ge 4 \end{cases} -$$ -(15.12.6) - -0 1 t t - -0 1 2 3 - -1 - -# **Figure 15.11** - -x1(t) x2(t) - -**Figure 15.10** For Example 15.12. - -2 - -which is sketched in Fig. 15.13. Notice that *y*(*t*) in this equation is continuous. This fact can be used to check the results as we move from one range of *t* to another. The result in Eq. (15.12.6) can be obtained without using the graphical procedure—by directly using Eq. (15.70) and the properties of step functions. This will be illustrated in Example 15.14. - -Convolution of signals *x*1(*t*) and *x*2(*t*) in Fig. 15.10. - -Graphically con volve the two functions in Fig. 15.14. To sho w ho w Practice Problem 15.12 powerful working in the *s*-domain is, verify your answer by performing the equivalent operation in the *s*-domain. - -**Answer:** The result of the convolution *y*(*t*) is shown in Fig. 15.15, where - -For Example 15.13. - -Example 15.13 Graphically convolve *g*(*t*) and *u*(*t*) shown in Fig. 15.16. - -# **Solution:** - -Let *y*(*t*) = *g*(*t*) \* *u*(*t*). We can find *y*(*t*) in two ways. - -■ **METHOD 1** Suppose we fold *g*(*t*), as in Fig. 15.17(a), and shift it by *t*, as in Fig. 15.17(b). Since *g*(*t*) = *t*, 0 < *t* < 1 originally, we expect that *g*(*t* − *λ*) = *t* − *λ*, 0 < *t* − *λ* < 1 or *t* − 1 < *λ* < *t*. There is no overlap of the two functions when *t* < 0 so that *y*(0) = 0 for this case. - -**Figure 15.17** Convolution of *g*(*t*) and *u*(*t*) in Fig. 15.16 with *g*(*t*) folded. - -For 0 < *t* < 1, *g*(*t* − *λ*) and *u*(*λ*) o verlap from 0 to *t*, as e vident in Fig. 15.17(b). Therefore, - -$$ -y(t) = \int_0^t (1)(t - \lambda) d\lambda = \left(t\lambda - \frac{1}{2}\lambda^2\right)\Big|_0^t -$$ - -= $t^2 - \frac{t^2}{2} = \frac{t^2}{2}$ , $0 \le t \le 1$ (15.13.1) - -For *t* > 1, the two functions overlap completely between (*t* − 1) and *t* [see Fig. 15.17(c)]. Hence, - -$$ -y(t) = \int_{t-1}^{t} (1)(t - \lambda) d\lambda -$$ - -= $\left(t\lambda - \frac{1}{2}\lambda^2\right)\Big|_{t-1}^{t} = \frac{1}{2}, \qquad t \ge 1$ (15.13.2) - -Thus, from Eqs. (15.13.1) and (15.13.2), - -$$ -y(t) = \begin{cases} \frac{1}{2}t^2, & 0 \le t \le 1\\ \frac{1}{2}, & t \ge 1 \end{cases} -$$ - -■ **METHOD 2** Instead of folding *g,* suppose we fold the unit step function *u*(*t*), as in Fig. 15.18(a), and then shift it by *t*, as in Fig. 15.18(b). Because *u*(*t*) = 1 for *t* > 0, *u*(*t* − *λ*) = 1 for *t* − *λ* > 0 or *λ* < *t*, the two functions overlap from 0 to *t*, so that - -$$ -y(t) = \int_0^t (1)\lambda \, d\lambda = \frac{1}{2}\lambda^2 \bigg|_0^t = \frac{t^2}{2}, \qquad 0 \le t \le 1 \quad (15.13.3) -$$ - -# **Figure 15.18** - -Convolution of *g*(*t*) and *u*(*t*) in Fig. 15.16 with *u*(*t*) folded. - -For *t* > 1, the tw o functions o verlap between 0 and 1, as sho wn in Fig. 15.18(c). Hence, - -$$ -y(t) = \int_0^1 (1)\lambda \, d\lambda = \frac{1}{2}\lambda^2 \bigg|_0^1 = \frac{1}{2}, \qquad t \ge 1 \tag{15.13.4} -$$ - -And, from Eqs. (15.13.3) and (15.13.4), - -$$ -y(t) = \begin{cases} \frac{1}{2}t^2, & 0 \le t \le 1\\ \frac{1}{2}, & t \ge 1 \end{cases} -$$ - -Although the two methods give the same result, as expected, notice that it is more convenient to fold the unit step function *u*(*t*) than fold *g*(*t*) in this example. Figure 15.19 shows *y*(*t*). - -Result of Example 15.13. - -Given *g*(*t*) and *f*(*t*) in Fig. 15.20, graphically find *y*(*t*) = *g*(*t*) \* *f*(*t*). Practice Problem 15.13 - -Answer: -$$ -y(t) = \begin{cases} 3(1 - e^{-t}), & 0 \le t \le 1 \\ 3(e - 1)e^{-t}, & t \ge 1 \\ 0, & \text{elsewhere.} \end{cases} -$$ - -For the *RL* circuit in Fig. 15.21(a), use the convolution integral to find the Example 15.14 response *io*(*t*) due to the excitation shown in Fig. 15.21(b). - -# **Solution:** - -1. **Define.** The problem is clearly stated and the method of solution is also specified. - -**Figure 15.21** For Example 15.14. - -For the circuit in Fig. 15.21(a): (a) its *s*-domain equivalent, (b) its impulse response. - -- 2. **Present.** We are to use the con volution inte gral to solv e for the response *io*(*t*) due to *is*(*t*) shown in Fig. 15.21(b). -- 3. **Alternative.** We have learned to do con volution by using the con volution integral and how to do it graphically. In addition, we could always work in the *s*-domain to solve for the current. We will solve for the current using the convolution integral and then check it using the graphical approach. -- 4. **Attempt.** As we stated, this problem can be solv ed in tw o ways: directly using the con volution inte gral or using the graphical technique. To use either approach, we first need the unit impulse response *h*(*t*) of the circuit. In the *s*-domain, applying the current division principle to the circuit in Fig. 15.22(a) gives - -$$ -I_o = \frac{1}{s+1} I_s -$$ - -Hence, - -$$ -H(s) = \frac{I_o}{I_s} = \frac{1}{s+1} -$$ - (15.14.1) - -and the inverse Laplace transform of this gives - -$$ -h(t) = e^{-t} u(t) -$$ - (15.14.2) - -Figure 15.22(b) shows the impulse response *h*(*t*) of the circuit. - -To use the convolution integral directly, recall that the response is given in the *s*-domain as - -$$ -I_o(s) = H(s)I_s(s) -$$ - -With the given *is*(*t*) in Fig. 15.21(b), - -$$ -i_s(t) = u(t) - u(t-2) -$$ - -so that - -$$ -i_o(t) = h(t) * i_s(t) = \int_0^t i_s(\lambda)h(t - \lambda) d\lambda -$$ - -= -$$ -\int_0^t [u(\lambda) - u(\lambda - 2)]e^{-(t - \lambda)} d\lambda -$$ - (15.14.3) - -Since *u*(*λ* − 2) = 0 for 0 < *λ* < 2, the inte grand involving *u*(*λ*) is nonzero for all *λ* > 0, whereas the inte grand involving *u*(*λ* − 2) is nonzero only for λ > 2. The best way to handle the integral is to do the two parts separately. For 0 < *t* < 2, - -$$ -i'_{o}(t) = \int_{0}^{t} (1)e^{-(t-\lambda)} d\lambda = e^{-t} \int_{0}^{t} (1)e^{\lambda} d\lambda -$$ - -= $e^{-t}(e^{t} - 1) = 1 - e^{-t}, \qquad 0 < t < 2$ (15.14.4) - -For *t* > 2, - -$$ -i''_o(t) = \int_2^t (1)e^{-(t-\lambda)} d\lambda = e^{-t} \int_2^t e^{\lambda} d\lambda -$$ - -= $e^{-t}(e^t - e^2) = 1 - e^2 e^{-t}, \qquad t > 2$ (15.14.5) - -Substituting Eqs. (15.14.4) and (15.14.5) into Eq. (15.14.3) gives - -$$ -i_o(t) = i'_o(t) - i''_o(t) -$$ - -= $(1 - e^{-t})[u(t - 2) - u(t)] - (1 - e^2 e^{-t})u(t - 2)$ -= $\begin{cases} 1 - e^{-t}A, & 0 < t < 2 \\ (e^2 - 1)e^{-t}A, & t > 2 \end{cases}$ (15.14.6) - -5. **Evaluate.** To use the graphical technique, we may fold *is*(*t*) in Fig. 15.21(b) and shift by *t*, as shown in Fig. 15.23(a). For 0 < *t* < 2, the overlap between *is*(*t* − *λ*) and *h*(*λ*) is from 0 to *t*, so that - -$$ -i_o(t) = \int_0^t (1)e^{-\lambda} d\lambda = -e^{-\lambda} \Big|_0^t = (1 - e^{-t}) \mathbf{A}, \qquad 0 \le t \le 2 \quad (15.14.7) -$$ - -For *t* > 2, the tw o functions o verlap between ( *t* − 2) and *t*, as in Fig. 15.23(b). Hence, - -$$ -i_o(t) = \int_{t-2}^{t} (1)e^{-\lambda} d\lambda = -e^{-\lambda} \Big|_{t-2}^{t} = -e^{-t} + e^{-(t-2)} -$$ - -= $(e^2 - 1)e^{-t} A$ , $t \ge 0$ (15.14.8) - -From Eqs. (15.14.7) and (15.14.8), the response is - -15.14.7) and (15.14.8), the response is -\n -$$ -i_o(t) = \begin{cases} 1 - e^{-t} A, & 0 \le t \le 2 \\ (e^2 - 1)e^{-t} A, & t \ge 2 \end{cases} -$$ -\n(15.14.9) - -which is the same as in Eq. (15.14.6). Thus, the response *io*(*t*) along the excitation *is*(*t*) is as shown in Fig. 15.24. - -6. **Satisfactory?** We have satisf actorily solv ed the problem and can present the results as a solution to the problem. - -Use convolution to find *vo*(*t*) in the circuit of Fig. 15.25(a) when the exci- Practice Problem 15.14 tation is the signal shown in Fig. 15.25(b). To show how powerful working in the *s*-domain is, verify your answer by performing the equivalent operation in the *s*-domain. - -**Answer:** -$$ -20(e^{-t} - e^{-2t})u(t) -$$ - V. - -# **15.6** Application to Integrodifferential Equations - -The Laplace transform is useful in solving linear inte grodifferential equations. Using the differentiation and integration properties of Laplace transforms, each term in the integrodifferential equation is transformed. - -**Figure 15.23** For Example 15.14. - -# **Figure 15.24** - -For Example 15.14; excitation and response. - -Initial conditions are automatically taken into account. We solve the resulting algebraic equation in the *s*-domain. We then con vert the solution back to the time domain by using the in verse transform. The following examples illustrate the process. - -Example 15.15 Use the Laplace transform to solve the differential equation - -$$ -\frac{d^2v(t)}{dt^2} + 6\frac{dv(t)}{dt} + 8v(t) = 2u(t) -$$ - -subject to *v*(0) = 1, *v*′(0) = −2. - -# **Solution:** - -We tak e the Laplace transform of each term in the gi ven dif ferential equation and obtain - -$$ -[s^2V(s) - sv(0) - v'(0)] + 6[sV(s) - v(0)] + 8V(s) = \frac{2}{s} -$$ - -Substituting *v*(0) = 1, *v*′(0) = −2, - -$$ -s^2V(s) - s + 2 + 6sV(s) - 6 + 8V(s) = \frac{2}{s} -$$ - -or - -$$ -(s2 + 6s + 8)V(s) = s + 4 + \frac{2}{s} = \frac{s2 + 4s + 2}{s} -$$ - -Hence, - -$$ -V(s) = \frac{s^2 + 4s + 2}{s(s+2)(s+4)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+4} -$$ - -where - -$$ -A = sV(s) \Big|_{s=0} = \frac{s^2 + 4s + 2}{(s+2)(s+4)} \Big|_{s=0} = \frac{2}{(2)(4)} = \frac{1}{4} -$$ -$$ -B = (s+2)V(s) \Big|_{s=-2} = \frac{s^2 + 4s + 2}{s(s+4)} \Big|_{s=-2} = \frac{-2}{(-2)(2)} = \frac{1}{2} -$$ -$$ -C = (s+4)V(s) \Big|_{s=-4} = \frac{s^2 + 4s + 2}{s(s+2)} \Big|_{s=-4} = \frac{2}{(-4)(-2)} = \frac{1}{4} -$$ - -Hence, - -$$ -V(s) = \frac{\frac{1}{4}}{s} + \frac{\frac{1}{2}}{s+2} + \frac{\frac{1}{4}}{s+4} -$$ - -By the inverse Laplace transform, - -$$ -v(t) = \frac{1}{4}(1 + 2e^{-2t} + e^{-4t})u(t) -$$ - -Solve the follo wing differential equation using the Laplace transform Practice Problem 15.15 method. - -$$ -\frac{d^2v(t)}{dt^2} + 4\frac{dv(t)}{dt} + 4v(t) = 7e^{-t} -$$ - -if *v*(0) = *v*′(0) = 2. - -**Answer:** -$$ -(7e^{-t} - 5e^{-2t} - te^{-2t})u(t) -$$ -. - -Solve for the response *y*(*t*) in the following integrodifferential equation. Example 15.16 - -$$ -\frac{dy}{dt} + 5y(t) + 6 \int_0^t y(\tau)d\tau = u(t), \qquad y(0) = 2 -$$ - -# **Solution:** - -Taking the Laplace transform of each term, we get - -$$ -[sY(s) - y(0)] + 5Y(s) + \frac{6}{s}Y(s) = \frac{1}{s} -$$ - -Substituting *y*(0) = 2 and multiplying through by *s*, - -$$ -Y(s)(s^2 + 5s + 6) = 1 + 2s -$$ - -or - -$$ -Y(s) = \frac{2s+1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3} -$$ - -where - -$$ -A = (s + 2)Y(s) \Big|_{s=-2} = \frac{2s+1}{s+3} \Big|_{s=-2} = \frac{-3}{1} = -3 -$$ -$$ -B = (s+3)Y(s) \Big|_{s=-3} = \frac{2s+1}{s+2} \Big|_{s=-3} = \frac{-5}{-1} = 5 -$$ - -*s* + 2 - -−1 - -Thus, - -$$ -Y(s) = \frac{-3}{s+2} + \frac{5}{s+3} -$$ - -Its inverse transform is - -$$ -y(t) = (-3e^{-2t} + 5e^{-3t})u(t) -$$ - -Use the Laplace transform to solve the integrodifferential equation Practice Problem 15.16 - -$$ -\frac{dy}{dt} + 3y(t) + 2\int_0^t y(\tau)d\tau = 2e^{-3t}, \qquad y(0) = 0 -$$ - -**Answer:** (−*e*−*t* + 4*e*−2*t* − 3*e*−3*t* )*u*(*t*). - -# **15.7** Summary - -1. The Laplace transform allows a signal represented by a function in the time domain to be analyzed in the *s*-domain (or comple x frequency domain). It is defined as - -$$ -\mathcal{L}[f(t)] = F(s) = \int_0^\infty f(t)e^{-st} dt -$$ - -- 2. Properties of the Laplace transform are listed in Table 15.1, while the Laplace transforms of basic common functions are listed in Table 15.2. -- 3. The inverse Laplace transform can be found using partial fraction expansions and using the Laplace transform pairs in Table 15.2 as a look-up table. Real poles lead to exponential functions and complex poles to damped sinusoids. -- 4. The convolution of tw o signals consists of time-re versing one of the signals, shifting it, multiplying it point by point with the second signal, and integrating the product. The convolution integral relates the convolution of two signals in the time domain to the in verse of the product of their Laplace transforms: - -$$ -\mathcal{L}^{-1}[F_1(s)F_2(s)] = f_1(t) * f_2(t) = \int_0^t f_1(\lambda)f_2(t - \lambda) d\lambda -$$ - -5. In the time domain, the output *y*(*t*) of the network is the convolution of the impulse response with the input *x*(*t*), - -$$ -y(t) = h(t) * x(t) -$$ - -Convolution may be re garded as the flip-shift-multiply-time-area method. - -6. The Laplace transform can be used to solve a linear integrodifferential equation. - -# Review Questions - -**15.1** Every function *f*(*t*) has a Laplace transform. - -(a) True (b) False - -**15.2** The variable *s* in the Laplace transform *H*(*s*) is called - -| (a) complex frequency | (b) transfer function | -|-----------------------|-----------------------| -| (c) zero | (d) pole | - -**15.3** The Laplace transform of *u*(*t* − 2) is: - -(a) -$$ -\frac{1}{s+2} -$$ - (b) $\frac{1}{s-2}$ -(c) $\frac{e^{2s}}{s}$ (d) $\frac{e^{-2s}}{s}$ - -**15.4** The zero of the function - -of the function -$$ -F(s) = \frac{s+1}{(s+2)(s+3)(s+4)} -$$ - -| is at | | -|--------|--------| -| (a) −4 | (b) −3 | -| (c) −2 | (d) −1 | - -**15.5** The poles of the function - -$$ -F(s) = \frac{s+1}{(s+2)(s+3)(s+4)} -$$ - -are at - -| (a) −4 | (b) −3 | -|--------|--------| -| (c) −2 | (d) −1 | - -**15.6** If *F*(*s*) = 1∕(*s* + 2), then *f*(*t*) is - -(a) *e*2*t u*(*t*) (b) *e*−2*t u*(*t*) (c) *u*(*t* − 2) (d) *u*(*t* + 2) - -Problems **707** - -**15.7** Given that *F*(*s*) = *e*−2*s* ∕(*s* + 1), then *f*(*t*) is (a) *e*−2(*t*−1)*u*(*t* − 1) (b) *e*−(*t*−2)*u*(*t* − 2) (c) *e*−*t u*(*t* − 2) (d) *e*−*t u*(*t* + 1) (e) *e*−(*t*−2)*u*(*t*) - -**15.8** The initial value of *f*(*t*) with transform - -$$ -F(s) = \frac{s+1}{(s+2)(s+3)} -$$ - -is: - -| (a) nonexistent | (b) ∞ | (c) 0 | -|-----------------|--------------|-------| -| (d) 1 | (e) __1
6 | | - -**15.9** The inverse Laplace transform of - -$$ -\frac{s+2}{\left(s+2\right)^2+1} -$$ - -# Problems - -# Sections 15.2 and 15.3 Definition and Properties of the Laplace Transform - -- **15.1** Find the Laplace transform of 5 sin(*at*) cos(*bt*). (*Hint*: Using the exponential representation for both functions may make this problem easier.) -- **15.2** Determine the Laplace transform of 3.5 cos (5*t* − 45°). -- **15.3** Obtain the Laplace transform of each of the following functions: - -| (a) e−2t | (b) e−2t | -|------------------------|-------------| -| cos 3tu(t) | sin 4 tu(t) | -| (c) e−3t | (d) e−4t | -| cosh 2tu(t) | sinh tu(t) | -| (e) te−t
sin 2tu(t) | | - -- **15.4** Design a problem to help other students better understand how to find the Laplace transform of different time varying functions. -- **15.5** Find the Laplace transform of each of the following functions: - - (a) *t* 2 cos(2*t* + 30°)*u*(*t*) - -(b) -$$ -3t^4e^{-2t}u(t) -$$ - -(c) -$$ -2tu(t) - 4\frac{d}{dt}\delta(t) -$$ - -(d) -$$ -2e^{-(t-1)}u(t) -$$ - -$$ -(e) 5u(t/2) -$$ - -(f) -$$ -6e^{-t/3} u(t) -$$ - -(g) *dn* \_\_\_ *dtn δ*(*t*) - -is: -\n(a) -$$ -e^{-t} \cos 2t -$$ - -\n(b) $e^{-t} \sin 2t$ -\n(c) $e^{-2t} \cos t$ -\n(d) $e^{-2t} \sin 2t$ -\n(e) none of the above - -**15.10** The result of *u*(*t*) \* *u*(*t*) is: - -| (a) u2
(t) | (b) tu(t) | -|--------------------|-----------| -| 2
(c) t
u(t) | (d) δ(t) | - -*Answers: 15.1b, 15.2a, 15.3d, 15.4d, 15.5a,b,c, 15.6b, 15.7b, 15.8c, 15.9c, 15.10b.* - -- **15.6** Find *G*(*s*) given that *g*(*t*) = 2*r*(*t*) 2*r*(*t* − 2). -- **15.7** Find the Laplace transform of the following signals: - -$$ -(a) f(t) = (2t + 4)u(t) -$$ - -(b) -$$ -g(t) = (4 + 3e^{-2t})u(t) -$$ - -- (c) *h*(*t*) = (6 sin(3*t*) + 8 cos(3*t*))*u*(*t*) -- (d) *x*(*t*) = (*e*−2*t* cosh(4*t*))*u*(*t*) -- **15.8** Find the Laplace transform *F*(*s*), given that *f*(*t*) is: - - (a) 2*tu*(*t* − 4) (b) 5 cos(*t*) *δ*(*t* − 2) (c) *e*−*t u*(*t* − *t*) (d) sin(2*t*)*u*(*t* − *τ*) -- **15.9** Determine the Laplace transforms of these functions: - -(a) -$$ -f(t) = (t - 4)u(t - 2) -$$ - -\n(b) $g(t) = 2e^{-4t}u(t - 1)$ -\n(c) $h(t) = 5 \cos(2t - 1)u(t)$ - -- (d) *p*(*t*) = 6[*u*(*t* − 2) − *u*(*t* − 4)] -- **15.10** In two different ways, find the Laplace transform of - -$$ -g(t) = \frac{d}{dt}(te^{-t}\cos t) -$$ - -**15.11** Find *F*(*s*) if: - -(a) -$$ -f(t) = 6e^{-t} \cosh 2t -$$ - (b) $f(t) = 3te^{-2t} \sinh 4t$ -(c) $f(t) = 8e^{-3t} \cosh tu(t - 2)$ - -- **15.12** If *g*(*t*) = 4*e*−2*t* cos 4*t*, find *G*(*s*). -- **15.13** Find the Laplace transform of the following functions: (a) *t* cos *tu*(*t*) (b) *e*−*t t* sin *tu*(*t*) - -(a) -$$ -t \cos u(t) -$$ - -(c) $\frac{\sin \beta t}{t} u(t)$ - -*t* - -**15.14** Find the Laplace transform of the signal in Fig. 15.26. - -# **Figure 15.26** - -For Prob. 15.14. - -**15.15** Determine the Laplace transform of the function in Fig. 15.27. - -For Prob. 15.15. - -**15.16** Obtain the Laplace transform of *f*(*t*) in Fig. 15.28. - -**Figure 15.29** - -For Prob. 15.17. - -**15.18** Obtain the Laplace transforms of the functions in Fig. 15.30. - -**Figure 15.30** - -For Prob. 15.18. - -**15.19** Calculate the Laplace transform of the infinite train of unit impulses in Fig. 15.31. - -**Figure 15.31** For Prob. 15.19. - -**15.17** Using Fig. 15.29, design a problem to help other students better understand the Laplace transform of a simple, non-periodic waveshape. - -**Figure 15.32** For Prob. 15.20. - -**15.21** Obtain the Laplace transform of the periodic waveform in Fig. 15.33. - -# **Figure 15.33** - -For Prob. 15.21. - -**15.22** Find the Laplace transforms of the functions in Fig. 15.34. - -**15.23** Determine the Laplace transforms of the periodic functions in Fig. 15.35. - -# **Figure 15.35** - -For Prob. 15.23. - -**15.24** Design a problem to help other students better understand how to find the initial and final value of a transfer function. - -**15.25** Let - -$$ -F(s) = \frac{18(s+1)}{(s+2)(s+3)} -$$ - -- (a) Use the initial and final value theorems to find *f*(0) and *f* (∞). -- (b) Verify your answer in part (a) by finding *f*(*t*), using partial fractions. - -**15.26** Determine the initial and final values of *f*(*t*), if they exist, given that: - -(a) -$$ -F(s) = \frac{5s^2 + 3}{s^3 + 4s^2 + 6} -$$ - -\n(b) $F(s) = \frac{s^2 - 2s + 1}{4(s - 2)(s^2 + 2s + 4)}$ - -# Section 15.4 The Inverse Laplace Transform - -**15.27** Determine the inverse Laplace transform of each of the following functions: - -Problems 709 -\nDetermine the initial and final values of -$$ -f(t) -$$ -, if they exist, given that: -\n(a) $F(s) = \frac{s^2 + 3}{s^3 + 4s^2 + 6}$ -\n(b) $F(s) = \frac{s^2 - 2s + 1}{4(s - 2)(s^2 + 2s + 4)}$ -\nOn 15.4 The Inverse Laplace Transform -\n7 Determine the inverse Laplace transform of each of the following functions: -\n(a) $F(s) = \frac{1}{s} + \frac{2}{s + 1}$ -\n(b) $G(s) = \frac{3s + 1}{s + 4}$ -\n(c) $H(s) = \frac{12}{(s + 1)(s + 3)}$ -\n(d) $J(s) = \frac{12}{(s + 2)^2(s + 4)}$ -\nDesign a problem to help other students better understand how to find the inverse Laplace transform. -\nFind the inverse Laplace transform of: -\n $F(s) = \frac{s^2 + 2}{s^3 + 2s^2 + 2s}$ -\nFind the inverse Laplace transform of: -\n(a) $F_1(s) = \frac{6s^2 + 8s + 3}{s(s^2 + 2s + 5)}$ -\n(b) $F_2(s) = \frac{s^2 + 5s + 6}{(s + 1)^2(s + 4)}$ -\n(c) $F_3(s) = \frac{10}{(s + 1)(s^2 + 4s + 8)}$ -\nFind $f(t)$ for each $F(s)$ : -\n(a) $\frac{10s}{(s + 1)(s + 2)^3}$ -\n(b) $\frac{2s^2 + 4s + 1}{(s + 1)(s + 2)^3}$ -\n(c) $\frac{s + 1}{(s + 2)(s^2 + 2s + 5)}$ -\nDetermine the inverse Laplace transform of each of the following functions: -\n(a) $\frac{8(s + 1)(s + 2)}{s(s + 2)(s + 4)}$ -\n(b) $\frac{s^2 - 2s + 4}{s(s + 1)(s + 2)^2}$ -\n(c) $\frac{s^2 + 1}{(s + 1)(s + 2)^2}$ - -- **15.28** Design a problem to help other students better understand how to find the inverse Laplace transform. -- **15.29** Find the inverse Laplace transform of: - -$$ -F(s) = \frac{s^2 + 2}{s^3 + 2s^2 + 2s} -$$ - -**15.30** Find the inverse Laplace transform of: - -Find the inverse Laplace transform -\n(a) -$$ -F_1(s) = \frac{6s^2 + 8s + 3}{s(s^2 + 2s + 5)} -$$ - -\n(b) $F_2(s) = \frac{s^2 + 5s + 6}{(s+1)^2(s+4)}$ -\n(c) $F_3(s) = \frac{10}{(s+1)(s^2 + 4s + 8)}$ - -**15.31** Find *f*(*t*) for each *F*(*s*): - -Find -$$ -f(t) -$$ - for each $F(s)$ : -\n(a) $\frac{10s}{(s + 1)(s + 2)(s + 3)}$ -\n(b) $\frac{2s^2 + 4s + 1}{(s + 1)(s + 2)^3}$ -\n(c) $\frac{s + 1}{(s + 2)(s^2 + 2s + 5)}$ - -**15.32** Determine the inverse Laplace transform of each of the following functions: - -the following functions: -\n(a) -$$ -\frac{8(s + 1)(s + 3)}{s(s + 2)(s + 4)} -$$ -\n(b) -$$ -\frac{s^2 - 2s + 4}{(s + 1)(s + 2)^2} -$$ -\n(c) -$$ -\frac{s^2 + 1}{(s + 3)(s^2 + 4s + 5)} -$$ - -**15.33** Calculate the inverse Laplace transform of: - -(a) -$$ -\frac{6(s-1)}{s^4 - 1} -$$ - (b) $\frac{se^{-\pi s}}{s^2 + 1}$ -(c) $\frac{8}{s(s+1)^3}$ - -**15.34** Find the time functions that have the following Laplace transforms: - -(a) -$$ -F(s) = 10 + \frac{s^2 + 1}{s^2 + 4} -$$ - -\n(b) $G(s) = \frac{e^{-s} + 4e^{-2s}}{s^2 + 6s + 8}$ -\n(c) $H(s) = \frac{(s + 1)e^{-2s}}{s(s + 3)(s + 4)}$ - -**15.35** Obtain *f*(*t*) for the following transforms: - -Obtain -$$ -f(t) -$$ - for the following -\n(a) $F(s) = \frac{(s+3)e^{-6s}}{(s+1)(s+2)}$ -\n(b) $F(s) = \frac{4 - e^{-2s}}{s^2 + 5s + 4}$ -\n(c) $F(s) = \frac{se^{-s}}{(s+3)(s^2 + 4)}$ - -**15.36** Obtain the inverse Laplace transforms of the following functions: - -following functions: -\n(a) -$$ -X(s) = \frac{3}{s^2(s+2)(s+3)} -$$ - -\n(b) $Y(s) = \frac{2}{s(s+1)^2}$ -\n(c) $Z(s) = \frac{5}{s(s+1)(s^2+6s+10)}$ - -Laplace transform of: -\n**15.37** Find the inverse Laplace transform of: -\n**(b)** -$$ -\frac{se^{-xs}}{s^2 + 1} -$$ - -\n**(c)** $F(s) = \frac{s^2 + 4s + 5}{(s + 2)(s^2 + 2s + 2)}$ -\n**(d)** $D(s) = \frac{10s}{(s^2 + 1)(s^2 + 4)}$ -\n**15.38** Find $f(t)$ given that: -\n**(a)** $F(s) = \frac{10s}{(s^2 + 1)(s^2 + 4)}$ -\n**15.38** Find $f(t)$ given that: -\n**(a)** $F(s) = \frac{s^2 + 4s}{s^2 + 10s + 26}$ -\n**(b)** $F(s) = \frac{5s^2 + 7s + 29}{s(s^2 + 4s + 29)}$ -\nIlowing transforms: -\n**(a)** $F(s) = \frac{5s^2 + 7s + 29}{(s^2 + 2s + 17)(s^2 + 4s + 20)}$ -\n**(b)** $F(s) = \frac{2s^3 + 4s^2 + 1}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$ -\n**(c)** $F(s) = \frac{2s^3 + 4s^2 + 1}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$ -\n**(d)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 + 4s + 20)}$ -\n**(e)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$ -\n**15.40** Show that -\n**(f)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$ -\n**(g)** Show that -\n**(h)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 + 6s + 3)}$ -\n**(i)** $F(s) = \frac{s^2 + 4}{(s^2 + 2s + 17)(s^2 +$ - -**15.38** Find *f*(*t*) given that: - -Find -$$ -f(t) -$$ - given that: -\n(a) $F(s) = \frac{s^2 + 4s}{s^2 + 10s + 26}$ -\n(b) $F(s) = \frac{5s^2 + 7s + 29}{s(s^2 + 4s + 29)}$ - -**15.39** Determine *f*(*t*) if: \* - -$$ -s(s^{2} + 4s + 29) -$$ - -Determine $f(t)$ if: -(a) $F(s) = \frac{2s^{3} + 4s^{2} + 1}{(s^{2} + 2s + 17)(s^{2} + 4s + 20)}$ -(b) $F(s) = \frac{s^{2} + 4}{(s^{2} + 9)(s^{2} + 6s + 3)}$ - -**15.40** Show that - -at -\n -$$ -\mathcal{L}^{-1} \left[ \frac{4s^2 + 7s + 13}{(s+2)(s^2 + 2s + 5)} \right] = -$$ -\n -$$ -\left[ \sqrt{2}e^{-t} \cos(2t + 45^\circ) + 3e^{-2t} \right] u(t) -$$ - -# Section 15.5 The Convolution Integral - -**15.41** Let *x*(*t*) and *y*(*t*) be as shown in Fig. 15.36. Find *z*(*t*) = *x*(*t*) \* *y*(*t*). \* - -**15.42** Design a problem to help other students better understand how to convolve two functions together. - -For Prob. 15.41. - -\* An asterisk indicates a challenging problem. - -**Figure 15.37** - -- For Prob. 15.43. -- **15.44** Obtain the convolution of the pairs of signals in Fig. 15.38. - -**15.45** Given *h*(*t*) = 4*e*−2*t u*(*t*) and *x*(*t*) = *δ*(*t*) − 2*e*−2*t u*(*t*), find *y*(*t*) = *x*(*t*) \* *h*(*t*). - -**15.46** Given the following functions - -*x*(*t*) = 2*δ*(*t*), *y*(*t*) = 4*u*(*t*), *z*(*t*) = *e*−2*t u*(*t*), evaluate the following convolution operations. - -\n- (a) - $$ - x(t) \ast y(t) - $$ -\n- (b) $x(t) \ast z(t)$ -\n- (c) $y(t) \ast z(t)$ -\n- (d) $y(t) \ast [y(t) + z(t)]$ -\n - -**15.47** A system has the transfer function - -as the transfer function -$$ -H(s) = \frac{6s}{(s+1)(s+2)} -$$ - -- (a) Find the impulse response of the system. -- (b) Determine the output *y*(*t*), given that the input is *x*(*t*) = *u*(*t*). -- **15.48** Find *f*(*t*) using convolution given that: - -Find -$$ -f(t) -$$ - using convolution -\n(a) $F(s) = \frac{4}{(s^2 + 2s + 5)^2}$ -\n(b) $F(s) = \frac{2s}{(s + 1)(s^2 + 4)}$ - -**15.49** Use the convolution integral to find: \* - -(a) -$$ -t * e^{at}u(t) -$$ - -(b) $\cos(t) * \cos(t)u(t)$ - -# Section 15.6 Application to Integrodifferential Equations - -**15.50** Use the Laplace transform to solve the differential equation - -$$ -\frac{d^2v(t)}{dt^2} + 2\frac{dv(t)}{dt} + 10v(t) = 3\cos 2t -$$ - -subject to *v*(0) = 1, *dv*(0)∕*dt* = −2. - -**15.51** Given that *v*(0) = 5 and *dv*(0)∕*dt* = 10, solve - -$$ -\frac{d^2v}{dt^2} + 5\frac{dv}{dt} + 6v = 25e^{-t}u(t) -$$ - -**15.52** Use the Laplace transform to find *i*(*t*) for *t* > 0 if - -$$ -\frac{d^2i}{dt^2} + 3\frac{di}{dt} + 2i + \delta(t) = 0, -$$ - -$$ -i(0) = 0, \qquad i'(0) = 3 -$$ - -**15.53** Use Laplace transforms to solve for *x*(*t*) in \* - -$$ -x(t) = \cos t + \int_0^t e^{\lambda - t} x(\lambda) d\lambda -$$ - -**15.54** Design a problem to help other students better understand solving second order differential equations with a time varying input. - -**15.55** Solve for *y*(*t*) in the following differential equation if the initial conditions are zero. - -$$ -\frac{d^3y}{dt^3} + 6\frac{d^2y}{dt^2} + 8\frac{dy}{dt} = e^{-t}\cos 2t -$$ - -**15.56** Solve for *v*(*t*) in the integrodifferential equation - -$$ -12\frac{dv}{dt} + 36 \int_0^t v \, d\tau = 0 -$$ - -given that *v*(0) = 2. - -**15.57** Design a problem to help other students better understand solving integrodifferential equations with a periodic input, using Laplace transforms. - -**15.58** Given that - -$$ -\frac{dv}{dt} + 2v + 5 \int_0^t v(\lambda) d\lambda = 4u(t) -$$ - -with *v*(0) = −1, determine *v*(*t*) for *t* > 0. - -**15.59** Solve the integrodifferential equation - -$$ -\frac{dy}{dt} + 4y + 3 \int_0^t y \, d\tau = 18e^{-2t} u(t), \qquad y(0) = -3 -$$ - -**15.60** Solve the following integrodifferential equation - -$$ -2\frac{dx}{dt} + 5x + 3\int_0^t x\,dt + 4 = \sin 4t, \qquad x(0) = 1 -$$ - -- **15.61** Solve the following differential equations subject to the specified initial conditions. - - (a) *d*2 *v*/*dt*2 + 4*v* = 12, *v*(0) = 0, *dv*(0)/*dt* = 2 (b) *d*2 *i*/*dt*2 + 5*di*/*dt* + 4*i* = 8, *i*(0) = −1, *di*(0)/*dt* = 0 (c) *d*2 *v*/*dt*2 + 2*dv*/*dt* + *v* = 3, *v*(0) = 5, *dv*(0)/*dt* = 1 (d) *d*2 *i*/*dt*2 + 2*di*/*dt* + 5*i* = 10, *i*(0) = 4, *di*(0)/*dt* = −2 - -# **chapter** - -# 16 - -# Applications of the Laplace Transform - -*Communication skills are the most important skills any engineer can have. A very critical element in this tool set is the ability to ask a ques tion and understand the answer, a very simple thing and yet it may make the difference between success and failure!* - -—James A. Watson - -# Enhancing Your Skills and Your Career - -# **Asking Questions** - -In more than 30 years of teaching, I ha ve struggled with determining ho w best to help students learn. Regardless of how much time students spend in studying for a course, the most helpful activity for students is learning how to ask questions in class and then asking those questions. The student, by asking questions, becomes actively involved in the learning process and no longer is merely a passive receptor of information. I think this acti ve involvement contributes so much to the learning process that it is probably the single most important aspect to the de velopment of a modern engineer . In fact, asking questions is the basis of science. As Charles P. Steinmetz rightly said, "No man really becomes a fool until he stops asking questions." - -It seems very straightforward and quite easy to ask questions. Have we not been doing that all our li ves? The truth is to ask questions in an appropriate manner and to maximize the learning process tak es some thought and preparation. - -I am sure that there are se veral models one could ef fectively use. Let me share what has w orked for me. The most important thing to k eep in mind is that you do not ha ve to form a perfect question. Because the question-and-answer format allows the question to be developed in an iterative manner, the original question can easily be refined as you go. I frequently tell students that they are most welcome to read their questions in class. - -Here are three things you should keep in mind when asking questions. First, prepare your question. If you are lik e many students who are either shy or ha ve not learned to ask questions in class, you may wish to start with a question you ha ve written down outside of class. Second, w ait for an appropriate time to ask the question. Simply use your judgment on that. Third, be prepared to clarify your question by paraphrasing it or saying it in a different way in case you are asked to repeat the question. - -Photo by Charles Alexander - -# Learning Objectives - -*By using the information and exercises in this chapter you will be able to:* - -- 1. Understand and use effectively circuit element models in the *s*-domain. -- 2. Understand how to perform circuit analysis in the *s*-domain and how to transform the results back into the time domain. -- 3. Understand what a transfer function is and how it is used. -- 4. Understand state variables and how to apply and use them in circuit analysis. - -One last comment: Not all professors like students to ask questions in class even though they may say the y do. You need to find out which professors like classroom questions. Good luck in enhancing one of your most important skills as an engineer. - -# **16.1** Introduction - -Now that we have introduced the Laplace transform, let us see what we can do with it. Keep in mind that with the Laplace transform we actually have one of the most powerful mathematical tools for analysis, synthesis, and design. Being able to look at circuits and systems in the *s*-domain can help us to understand ho w our circuits and systems really function. In this chapter we will tak e an in-depth look at ho w easy it is to w ork with circuits in the *s*-domain. In addition, we will briefly look at physical systems. We are sure you have studied some mechanical systems and may have used the same differential equations to describe them as we use to describe our electric circuits. Actually that is a wonderful thing about the physical universe in which we li ve; the same dif ferential equations can be used to describe any linear circuit, system, or process. The key is the term *linear*. - -A system is a mathematical model of a physical process relating the input to the output. - -It is entirely appropriate to consider circuits as systems. Historically, circuits have been discussed as a separate topic from systems, so we will actually talk about circuits and systems in this chapter realizing that circuits are nothing more than a class of electrical systems. - -The most important thing to remember is that everything we discussed in the last chapter and in this chapter applies to any linear system. In the last chapter, we saw how we can use Laplace transforms to solv e linear differential equations and integral equations. In this chapter, we introduce the concept of modeling circuits in the *s*-domain. We can use that principle to help us solv e just about an y kind of linear circuit. W e will take a quick look at how state variables can be used to analyze systems with multiple inputs and multiple outputs. Finally, we examine how the Laplace transform is used in netw ork stability analysis and in netw ork synthesis. - -# **16.2** Circuit Element Models - -Having mastered how to obtain the Laplace transform and its inverse, we are now prepared to employ the Laplace transform to analyze circuits. This usually involves three steps. - -# Steps in Applying the Laplace Transform: - -- 1. Transform the circuit from the time domain to the *s*-domain. -- 2. Solve the circuit using nodal analysis, mesh analysis, source transformation, superposition, or any circuit analysis technique with which we are familiar. -- 3. Take the inverse transform of the solution and thus obtain the solution in the time domain. - -Only the first step is new and will be discussed here. As we did in phasor analysis, we transform a circuit in the time domain to the frequenc y or *s*-domain by Laplace transforming each term in the circuit. - -For a resistor, the voltage-current relationship in the time domain is - -$$ -v(t) = Ri(t) \tag{16.1} -$$ - -Taking the Laplace transform, we get - -$$ -V(s) = RI(s) \tag{16.2} -$$ - -For an inductor, - -$$ -v(t) = L \frac{di(t)}{dt} -$$ - (16.3) - -Taking the Laplace transform of both sides gives - -$$ -V(s) = L[sI(s) - i(0^{-})] = sLI(s) - Li(0^{-}) -$$ -\n(16.4) - -or - -$$ -I(s) = \frac{1}{sL} V(s) + \frac{i(0^{-})}{s} -$$ - (16.5) - -The *s*-domain equivalents are shown in Fig. 16.1, where the initial condition is modeled as a voltage or current source. - -For a capacitor, - -$$ -i(t) = C \frac{dv(t)}{dt} -$$ - (16.6) - -which transforms into the *s*-domain as - -$$ -I(s) = C[sV(s) - v(0^{-})] = sCV(s) - Cv(0^{-}) -$$ -\n(16.7) - -$$ -\sum_{i=1}^{n} x_i -$$ - -$$ -V(s) = \frac{1}{sC} I(s) + \frac{v(0^{-})}{s} -$$ - (16.8) - -i(t) + ‒ *v*(t) i(0) L (a) I(s) + ‒ V(s) sL (b) Li(0‒) (c) V(s) I(s) + ‒ sL i(0‒) s + ‒ **Figure 16.1** - -Representation of an inductor: (a) timedomain, (b,c) *s*-domain equivalents. - -As one can infer from step 2, all the circuit analysis techniques applied for dc circuits are applicable to the s-domain. - -The elegance of using the Laplace transform in circuit analysis lies in the automatic inclusion of the initial conditions in the transformation process, thus providing a complete (transient and steady-state) solution. - -# **Figure 16.3** - -Time-domain and *s*-domain representations of passive elements under zero initial conditions. - -The *s*-domain equi valents are sho wn in Fig. 16.2. With the *s*-domain equivalents, the Laplace transform can be used readily to solv e first- and second-order circuits such as those we considered in Chapters 7 and 8. We should observe from Eqs. (16.3) to (16.8) that the initial conditions are part of the transformation. This is one advantage of using the Laplace transform in circuit analysis. Another advantage is that a complete re sponse—transient and steady state—of a netw ork is obtained. We will illustrate this with Examples 16.2 and 16.3. Also, observe the duality of Eqs. (16.5) and (16.8), confirming what we already know from Chapter 8 (see Table 8.1), namely, that *L* and *C, I*(*s*) and *V*(*s*), and *v*(0) and *i*(0) are dual pairs. - -If we assume zero initial conditions for the inductor and the capacitor, the above equations reduce to: - -Resistor: -$$ -V(s) = RI(s) -$$ - -Inductor: $V(s) = sLI(s)$ (16.9) -Capacitor: $V(s) = \frac{1}{sC}I(s)$ - -The *s*-domain equivalents are shown in Fig. 16.3. - -We define the impedance in the *s*-domain as the ratio of the voltage transform to the current transform under zero initial conditions; that is, - -$$ -Z(s) = \frac{V(s)}{I(s)} -$$ -(16.10) - -Thus, the impedances of the three circuit elements are - -Resistor: -$$ -Z(s) = R -$$ - -Inductor: $Z(s) = sL$ (16.11) -Capacitor: $Z(s) = \frac{1}{sC}$ - -# Table 16.1 summarizes these. The admittance in the *s*-domain is the reciprocal of the impedance, or - -$$ -Y(s) = \frac{1}{Z(s)} = \frac{I(s)}{V(s)} -$$ -(16.12) - -The use of the Laplace transform in circuit analysis f acilitates the use of various signal sources such as impulse, step, ramp, e xponential, and sinusoidal. - -The models for dependent sources and op amps are easy to develop drawing from the simple fact that if the Laplace transform of *f*(*t*) is *F*(*s*), - -# **TABLE 16.1** - -Impedance of an element in the s-domain.\* - -| Element | Z(s) = V(s)∕I(s) | -|-----------|------------------| -| Resistor | R | -| Inductor | sL | -| Capacitor | 1∕sC | -| | | - -\* Assuming zero initial conditions - -then the Laplace transform of *af*(*t*) is *aF*(*s*)—the linearity property. The dependent source model is a little easier in that we deal with a single value. The dependent source can have only two controlling values, a constant times either a voltage or a current. Thus, - -$$ -\mathcal{L}[av(t)] = aV(s) \tag{16.13} -$$ - -$$ -\mathcal{L}[ai(t)] = aI(s) \tag{16.14} -$$ - -The ideal op amp can be treated just lik e a resistor. Nothing within an op amp, either real or ideal, does an ything more than multiply a voltage by a constant. Thus, we only need to write the equations as we always do using the constraint that the input voltage to the op amp has to be zero and the input current has to be zero. - -Find *vo*(*t*) in the circuit of Fig. 16.4, assuming zero initial conditions. Example 16.1 - -# **Solution:** - -We first transform the circuit from the time domain to the *s*-domain. - -*u*(*t*) ⇒ \_\_1 *s* 1 H ⇒ *sL* = *s* \_\_1 3 F⇒ \_\_\_1 *sC* = \_\_3 *s* - -The resulting *s*-domain circuit is in Fig. 16.5. We now apply mesh analysis. For mesh 1, - -$$ -\frac{1}{s} = \left(1 + \frac{3}{s}\right)I_1 - \frac{3}{s}I_2\tag{16.1.1} -$$ - -For mesh 2, - -$$ -0 = -\frac{3}{s}I_1 + \left(s + 5 + \frac{3}{s}\right)I_2 -$$ - -or - -$$ -I_1 = \frac{1}{3}(s^2 + 5s + 3)I_2 -$$ - (16.1.2) - -Substituting this into Eq. (16.1.1), - -$$ -\frac{1}{s} = \left(1 + \frac{3}{s}\right)\frac{1}{3}(s^2 + 5s + 3)I_2 - \frac{3}{s}I_2 -$$ - -Multiplying through by 3*s* gives - -ng through by 3*s* gives -\n -$$ -3 = (s^3 + 8s^2 + 18s)I_2 \implies I_2 = \frac{3}{s^3 + 8s^2 + 18s} -$$ -\n -$$ -V_o(s) = sI_2 = \frac{3}{s^2 + 8s + 18} = \frac{3}{\sqrt{2}} \frac{\sqrt{2}}{(s + 4)^2 + (\sqrt{2})^2} -$$ - -Taking the inverse transform yields - -$$ -v_o(t) = \frac{3}{\sqrt{2}} e^{-4t} \sin \sqrt{2t} \text{ V}, \qquad t \ge 0 -$$ - -**Figure 16.4** For Example 16.1. - -# **Figure 16.5** - -Mesh analysis of the frequency-domain equivalent of the same circuit. - -+ ‒ 4 Ω *v*o(t) 1 H F 2.5u(t) V 1 4 - -+ - -u(t) V (t) 2*δ*(t) A + - -‒ - -0.1 F - -**Answer:** 10(1 − *e*−2*t* − 2*te*−2*t* )*u*(*t*) V. - -**Figure 16.6** - -For Practice Prob. 16.1. - -Example 16.2 Find *vo*(*t*) in the circuit of Fig. 16.7. Assume *vo*(0) = 5 V. - -10 Ω - -10 Ω *v*o 10e‒t - -‒ - -**Figure 16.7** For Example 16.2. - -# **Solution:** - -We transform the circuit to the *s*-domain as shown in Fig. 16.8. The initial condition is included in the form of the current source *Cvo*(0) = 0.1(5) = 0.5 A. [See Fig. 16.2(c).] We apply nodal analysis. At the top node, - -16.2(c).] We apply nodal analysis. At the to -\n -$$ -\frac{10/(s+1) - V_o}{10} + 2 + 0.5 = \frac{V_o}{10} + \frac{V_o}{10/s} -$$ - -or - -$$ -\frac{1}{s+1} + 2.5 = \frac{2V_o}{10} + \frac{sV_o}{10} = \frac{1}{10}V_o(s+2) -$$ - -Multiplying through by 10, - -$$ -\frac{10}{s+1} + 25 = V_o(s+2) -$$ - -or - -$$ -V_o = \frac{25s + 35}{(s + 1)(s + 2)} = \frac{A}{s + 1} + \frac{B}{s + 2} -$$ - -where - -$$ -A = (s+1)V_o(s) \Big|_{s=-1} = \frac{25s+35}{(s+2)} \Big|_{s=-1} = \frac{10}{1} = 10 -$$ - -$$ -B = (s+2)V_o(s) \Big|_{s=-2} = \frac{25s+35}{(s+1)} \Big|_{s=-2} = \frac{-15}{-1} = 15 -$$ - -Thus, - -$$ -V_o(s) = \frac{10}{s+1} + \frac{15}{s+2} -$$ - -Taking the inverse Laplace transform, we obtain - -$$ -v_o(t) = (10e^{-t} + 15e^{-2t})u(t) \text{ V} -$$ - -Find *vo*(*t*) in the circuit shown in Fig. 16.9. Note that, since the voltage input is multiplied by *u*(*t*), the voltage source is a short for all *t* < 0 and *iL*(0) = 0. - -**Answer:** -$$ -(12e^{-2t} - 2e^{-t/3})u(t) -$$ - V. - -**Figure 16.9** For Practice Prob. 16.2. - -**Figure 16.10** For Example 16.3. - -In the circuit of Fig. 16.10(a), the switch moves from position *a* to posi- Example 16.3 tion *b* at *t* = 0. Find *i*(*t*) for *t* > 0. - -# **Solution:** - -The initial current through the inductor is *i*(0) = *Io*. For *t* > 0, Fig. 16.10(b) shows the circuit transformed to the *s*-domain. The initial condition is incorporated in the form of a voltage source as *Li*(0) = *LIo*. Using mesh analysis, - -$$ -I(s)(R + sL) - L I_o - \frac{V_o}{s} = 0 -$$ -\n(16.3.1) - -or - -$$ -I(s) = \frac{L I_o}{R + sL} + \frac{V_o}{s(R + sL)} = \frac{I_o}{s + R/L} + \frac{V_o/L}{s(s + R/L)} -$$ -(16.3.2) - -Applying partial fraction expansion on the second term on the right-hand side of Eq. (16.3.2) yields - -$$ -I(s) = \frac{I_o}{s + R/L} + \frac{V_o/R}{s} - \frac{V_o/R}{(s + R/L)} -$$ -(16.3.3) - -The inverse Laplace transform of this gives - -$$ -i(t) = \left(I_o - \frac{V_o}{R}\right)e^{-t/\tau} + \frac{V_o}{R}, \qquad t \ge 0 -$$ - (16.3.4) - -where *τ* = *R*∕*L*. The term in parentheses is the transient response, while the second term is the steady-state response. In other words, the final value is *i*(∞) = *Vo*∕*R*, which we could have predicted by applying the final-value theorem on Eq. (16.3.2) or (16.3.3); that is, - -$$ -\lim_{s \to 0} sI(s) = \lim_{s \to 0} \left( \frac{sI_o}{s + R/L} + \frac{V_o/L}{s + R/L} \right) = \frac{V_o}{R} -$$ - (16.3.5) - -Equation (16.3.4) may also be written as - -$$ -i(t) = I_0 e^{-t/\tau} + \frac{V_o}{R} (1 - e^{-t/\tau}), \qquad t \ge 0 -$$ - (16.3.6) - -The first term is the natural response, while the second term is the forced response. If the initial condition *Io* = 0, Eq. (16.3.6) becomes - -$$ -i(t) = \frac{V_o}{R}(1 - e^{-t/\tau}), \qquad t \ge 0 -$$ - (16.3.7) - -which is the step response, since it is due to the step input *Vo* with no initial energy. - -**Figure 16.11** For Practice Prob. 16.3. - -Practice Problem 16.3 The switch in Fig. 16.11 has been in position *b* for a long time. It is moved to position *a* at *t* = 0. Determine *v*(*t*) for *t* > 0. - -> **Answer:** *v*(*t*) = (*Vo* − *IoR*)*e*−*t*∕*τ* + *IoR*, *t* > 0, where *τ* = *RC*. - -# **16.3** Circuit Analysis - -Circuit analysis is again relatively easy to do when we are in the *s*-domain. We merely need to transform a complicated set of mathematical relationships in the time domain into the *s*-domain where we convert operators (derivatives and integrals) into simple multipliers of *s* and 1∕*s*. This now allows us to use algebra to set up and solv e our circuit equations. The exciting thing about this is that *all* of the circuit theorems and relation ships we developed for dc circuits are perfectly valid in the *s*-domain. - -Remember, equivalent circuits, with capacitors and inductors, only exist in the s-domain; they cannot be transformed back into the time domain. - -Example 16.4 Consider the circuit in Fig. 16.12(a). Find the value of the voltage across the capacitor assuming that the value of *vs*(*t*) = 10*u*(*t*) V and assume that at *t* = 0, −1 A flows through the inductor and +5 V is across the capacitor. - -# **Solution:** - -Figure 16.12(b) represents the entire circuit in the *s*-domain with the initial conditions incorporated. We now have a straightforward nodal analysis problem. Because the value of *V*1 is also the value of the capacitor voltage in the time domain and is the only unknown node voltage, we only need to write one equation. - -$$ -\frac{V_1 - 10/s}{10/3} + \frac{V_1 - 0}{5s} + \frac{i(0)}{s} + \frac{V_1 - [v(0)/s]}{1/(0.1s)} = 0 -$$ - (16.4.1) - -or - -$$ -0.1\left(s+3+\frac{2}{s}\right)V_1 = \frac{3}{s} + \frac{1}{s} + 0.5\tag{16.4.2} -$$ - -where *v*(0) = 5 V and *i*(0) = −1 A. Simplifying we get - -$$ -(s^2 + 3s + 2) V_1 = 40 + 5s -$$ - -or - -$$ -V_1 = \frac{40 + 5s}{(s+1)(s+2)} = \frac{35}{s+1} - \frac{30}{s+2} -$$ - (16.4.3) - -Taking the inverse Laplace transform yields - -$$ -v_1(t) = (35e^{-t} - 30e^{-2t})u(t) \text{ V} -$$ - (16.4.4) - -For the circuit shown in Fig. 16.12 with the same initial conditions, find Practice Problem 16.4 the current through the inductor for all time *t* > 0. - -**Answer:** -$$ -i(t) = (3 - 7e^{-t} + 3e^{-2t})u(t) -$$ - A. - -For the circuit sho wn in Fig. 16.12, and the initial conditions used Example 16.5 in Example 16.4, use superposition to find the value of the capacitor voltage. - -# **Solution:** - -Inasmuch as the circuit in the *s*-domain actually has three independent sources, we can look at the solution one source at a time. Figure 16.13 presents the circuits in the *s*-domain considering one source at a time. We now have three nodal analysis problems. First, let us solve for the capacitor voltage in the circuit shown in Fig. 16.13(a). - -$$ -\frac{V_1 - 10/s}{10/3} + \frac{V_1 - 0}{5s} + 0 + \frac{V_1 - 0}{1/(0.1s)} = 0 -$$ - -or - -$$ -0.1\left(s+3+\frac{2}{s}\right)V_1 = \frac{3}{s} -$$ - -Simplifying we get - -2 - -(*s* - - + 3*s* + 2)*V*1 = 30 *V*1 = \_\_\_\_\_\_\_\_\_\_\_\_ 30 (*s* + 1)(*s* + 2) = \_\_\_\_\_ 30 *s* + 1 − \_\_\_\_\_ 30 *s* + 2 - -or - -$$ -v_1(t) = (30e^{-t} - 30e^{-2t})u(t) \text{ V} \tag{16.5.1} -$$ - -For Fig. 16.13(b) we get, - -$$ -\frac{V_2 - 0}{10/3} + \frac{V_2 - 0}{5s} - \frac{1}{s} + \frac{V_2 - 0}{1/(0.1s)} = 0 -$$ - -or - -$$ -0.1\left(s+3+\frac{2}{s}\right)V_2 = \frac{1}{s} -$$ - -This leads to - -$$ -V_2 = \frac{10}{(s+1)(s+2)} = \frac{10}{s+1} - \frac{10}{s+2} -$$ - -Taking the inverse Laplace transform, we get - -$$ -v_2(t) = (10e^{-t} - 10e^{-2t})u(t) \text{ V} -$$ - (16.5.2) - -For Fig. 16.13(c), - -$$ -\frac{V_3 - 0}{10/3} + \frac{V_3 - 0}{5s} - 0 + \frac{V_3 - 5/s}{1/(0.1s)} = 0 -$$ - -$$ -0.1\left(s+3+\frac{2}{s}\right)V_3 = 0.5 -$$ -$$ -V_3 = \frac{5s}{(s+1)(s+2)} = \frac{-5}{s+1} + \frac{10}{s+2} -$$ - -This leads to - -$$ -v_3(t) = (-5e^{-t} + 10e^{-2t})u(t) \text{ V} -$$ - (16.5.3) - -Now all we need to do is to add Eqs. (16.5.1), (16.5.2), and (16.5.3): - -$$ -v(t) = v_1(t) + v_2(t) + v_3(t) -$$ - -= { (30 + 10 - 5)e-t + (-30 + 10 - 10)e-2t}u(t) V - -or - -$$ -v(t) = (35e^{-t} - 30e^{-2t})u(t) \text{ V} -$$ - -which agrees with our answer in Example 16.4. - -Practice Problem 16.5 For the circuit shown in Fig. 16.12, and the same initial conditions in Example 16.4, find the current through the inductor for all time *t* > 0 using superposition. - -**Answer:** -$$ -i(t) = (3 - 7e^{-t} + 3e^{-2t})u(t) -$$ - A. - -**Figure 16.14** For Example 16.6. - -Example 16.6 Assume that there is no initial ener gy stored in the circuit of Fig. 16.14 at *t* = 0 and that *is* = 10*u*(*t*) A. (a) Find *Vo*(*s*) using Thevenin's theorem. (b) Apply the initial- and final-value theorems to find *vo*(0+) and *vo*(∞). (c) Determine *vo*(*t*). - -# **Solution:** - -Because there is no initial energy stored in the circuit, we assume that the initial inductor current and initial capacitor voltage are zero at *t* = 0. - -(a) To find the Thevenin equivalent circuit, we remove the 5-Ω resistor and then find *V*oc (*V*Th) and *I*sc. To find *V*Th, we use the Laplacetransformed circuit in Fig. 16.15(a). Since *Ix* = 0, the dependent voltage source contributes nothing, so - -$$ -V_{\text{oc}} = V_{\text{Th}} = 5\left(\frac{10}{s}\right) = \frac{50}{s} -$$ - -To find *Z*Th, we consider the circuit in Fig. 16.15(b), where we first find *I*sc. We can use nodal analysis to solve for *V*1 which then leads to *I*sc(*I*sc = *Ix* = *V*1∕2*s*). - -s). -$$ --\frac{10}{s} + \frac{(V_1 - 2I_x) - 0}{5} + \frac{V_1 - 0}{2s} = 0 -$$ - -along with - -$$ -I_x = \frac{V_1}{2s} -$$ - -leads to - -$$ -V_1 = \frac{100}{2s + 3} -$$ - -Hence, - -$$ -I_{\rm sc} = \frac{V_1}{2s} = \frac{100/(2s+3)}{2s} = \frac{50}{s(2s+3)} -$$ - -and - -$$ -Z_{\text{Th}} = \frac{V_{\text{oc}}}{I_{\text{sc}}} = \frac{50/s}{50/[s(2s+3)]} = 2s + 3 -$$ - -The given circuit is replaced by its Thevenin equivalent at terminals *a*-*b* as shown in Fig. 16.16. From Fig. 16.16, - -$$ -V_o = \frac{5}{5 + Z_{\text{Th}}} V_{\text{Th}} = \frac{5}{5 + 2s + 3} \left(\frac{50}{s}\right) = \frac{250}{s(2s + 8)} = \frac{125}{s(s + 4)} -$$ - -(b) Using the initial-value theorem we find - -$$ -v_o(0) = \lim_{s \to \infty} sV_o(s) = \lim_{s \to \infty} \frac{125}{s+4} = \lim_{s \to \infty} \frac{125/s}{1+4/s} = \frac{0}{1} = 0 -$$ - -Using the final-value theorem we find - -$$ -v_o(\infty) = \lim_{s \to 0} sV_o(s) = \lim_{s \to 0} \frac{125}{s+4} = \frac{125}{4} = 31.25 \text{ V} -$$ - -(c) By partial fraction, - -$$ -V_o = \frac{125}{s(s+4)} = \frac{A}{s} + \frac{B}{s+4} -$$ - -\n -$$ -A = sV_o(s) \Big|_{s=0} = \frac{125}{s+4} \Big|_{s=0} = 31.25 -$$ - -\n -$$ -B = (s+4)V_o(s) \Big|_{s=-4} = \frac{125}{s} \Big|_{s=-4} = -31.25 -$$ - -\n -$$ -V_o = \frac{31.25}{s} - \frac{31.25}{s+4} -$$ - -Taking the inverse Laplace transform gives - -$$ -v_o(t) = 31.25(1 - e^{-4t})u(t) -$$ - V - -Notice that the values of *vo*(0) and *vo*(∞) obtained in part (b) are confirmed. - -The initial energy in the circuit of Fig. 16.17 is zero at *t* = 0. Assume that Practice Problem 16.6 *vs* = 360*u*(*t*) V. (a) Find *Vo*(*s*) using the Thevenin theorem. (b) Apply the initial- and final-value theorems to find *vo*(0) and *vo*(∞). (c) Obtain *vo*(*t*). - -**Answer:** (a) -$$ -V_o(s) = \frac{288(s+0.25)}{s(s+0.3)} -$$ -, (b) 288 V, 240 V, -(c) $(240 + 48e^{-0.3t})u(t)$ V. - -(b) - -# **Figure 16.15** For Example 16.6: (a) finding *V*Th, (b) determining *Z*Th. - -**Figure 16.16** The Thevenin equivalent of the circuit in Fig. 16.14 in the *s*-domain. - -**Figure 16.17** For Practice Prob. 16.6. - -# **16.4** Transfer Functions - -The *transfer function* is a key concept in signal processing because it indicates how a signal is processed as it passes through a network. It is a fitting tool for finding the network response, determining (or designing for) network stability, and network synthesis. The transfer function of a network describes how the output behaves with respect to the input. It specifies the transfer from the input to the output in the *s*-domain, assuming no initial energy. - -The transfer function H(s) is the ratio of the output response Y(s) to the input excitation X(s), assuming all initial conditions are zero. - -Thus, - -$$ -H(s) = \frac{Y(s)}{X(s)} -$$ -\n(16.15) - -The transfer function depends on what we define as input and output. Because the input and output can be either current or voltage at any place in the circuit, there are four possible transfer functions: - -$$ -H(s) = \text{Voltage gain} = \frac{V_o(s)}{V_i(s)}\tag{16.16a} -$$ - -$$ -H(s) = \text{Current gain} = \frac{I_o(s)}{I_i(s)}\tag{16.16b} -$$ - -$$ -H(s) = \text{Impedance} = \frac{V(s)}{I(s)}\tag{16.16c} -$$ - -$$ -H(s) = \text{Admittance} = \frac{I(s)}{V(s)}\tag{16.16d} -$$ - -Thus, a circuit can ha ve man y transfer functions. Note that *H*(*s*) is dimensionless in Eqs. (16.16a) and (16.16b). - -Each of the transfer functions in Eq. (16.16) can be found in two ways. One w ay is to assume any convenient input *X*(*s*), use any circuit analysis technique (such as current or v oltage division, nodal or mesh analysis) to find the output *Y*(*s*), and then obtain the ratio of the two. The other approach is to apply the *ladder method*, which involves walking our way through the circuit. By this approach, we assume that the output is 1 V or 1 A as appropriate and use the basic laws of Ohm and Kirchhoff (KCL only) to obtain the input. The transfer function becomes unity divided by the input. This approach may be more convenient to use when the circuit has many loops or nodes so that applying nodal or mesh analysis becomes cumbersome. In the first method, we assume an input and find the output; in the second method, we assume the output and find the input. In both methods, we calculate *H*(*s*) as the ratio of output to input transforms. The two methods rely on the linearity property, since we only deal with linear circuits in this book. Example 16.8 illustrates these methods. - -Some authors would not consider Eqs. (16.16c) and (16.16d) transfer functions. - -For electrical networks, the transfer function is also known as the network - -function. - -Equation (16.15) assumes that both *X*(*s*) and *Y*(*s*) are known. Sometimes, we know the input *X*(*s*) and the transfer function *H*(*s*). We find the output *Y*(*s*) as - -$$ -Y(s) = H(s)X(s) \tag{16.17} -$$ - -and take the inverse transform to get *y*(*t*). A special case is when the input is the unit impulse function, *x*(*t*) = *δ*(*t*), so that *X*(*s*) = 1. For this case, - -$$ -Y(s) = H(s) -$$ - or $y(t) = h(t)$ (16.18) - -where - -$$ -h(t) = \mathcal{L}^{-1}[H(s)] -$$ - (16.19) - -The term *h*(*t*) represents the *unit impulse response*—it is the time-domain response of the netw ork to a unit impulse. Thus, Eq. (16.19) pro vides a new interpretation for the transfer function: *H*(*s*) is the Laplace transform of the unit impulse response of the netw ork. Once we kno w the impulse response *h*(*t*) of a network, we can obtain the response of the netw ork to *any* input signal using Eq. (16.17) in the *s*-domain or using the convolution integral (section 15.5) in the time domain. - -The output of a linear system is *y*(*t*) = 10*e* Example 16.7 −*t* cos 4*t u*(*t*) when the input is *x*(*t*) = *e*−*t u*(*t*). Find the transfer function of the system and its impulse response. - -# **Solution:** - -If *x*(*t*) = *e*−*t u*(*t*) and *y*(*t*) = 10*e*−*t* cos 4*t u*(*t*), then - -$$ -X(s) = \frac{1}{s+1} -$$ - and $Y(s) = \frac{10(s+1)}{(s+1)^2 + 4^2}$ - -Hence, - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{10(s+1)^2}{(s+1)^2 + 16} = \frac{10(s^2 + 2s + 1)}{s^2 + 2s + 17} -$$ - -To find *h*(*t*), we write *H*(*s*) as - -$$ -H(s) = 10 - 40 \frac{4}{(s+1)^2 + 4^2} -$$ - -From Table 15.2, we obtain - -$$ -h(t) = 10\delta(t) - 40e^{-t}\sin 4t \,u(t) -$$ - -The transfer function of a linear system is Practice Problem 16.7 - -$$ -H(s) = \frac{2s}{s+6} -$$ - -Find the output *y*(*t*) due to the input 45*e* −3*t u*(*t*) and its impulse response. - -**Answer:** −90*e*−3*t* + 180*e*−6*t* , *t* ≥ 0, 2*δ*(*t*) − 12*e*−6*t u*(*t*). The unit impulse response is the output response of a circuit when the input is a unit impulse. - -I 1 - -Example 16.8 Determine the transfer function *H*(*s*) = *Vo*(*s*)∕*Io*(*s*) of the circuit in Fig. 16.18. - -# **Solution:** - -■ **METHOD 1** By current division, - -$$ -I_2 = \frac{(s+4)I_o}{s+4+2+1/2s} -$$ - -But - -+ ‒ Vo - -2 Ω - -4 Ω - -s - -$$ -V_o = 2I_2 = \frac{2(s+4)I_o}{s+6+1/2s} -$$ - -Hence, - -$$ -H(s) = \frac{V_o(s)}{I_o(s)} = \frac{4s(s+4)}{2s^2 + 12s + 1} -$$ - -■ **METHOD 2** We can apply the ladder method. We let *Vo* = 1 V. By Ohm's law, *I*2 = *Vo*∕2 = 1∕2 A. The voltage across the (2 + 1∕2*s*) impedance is - -$$ -V_1 = I_2 \left( 2 + \frac{1}{2s} \right) = 1 + \frac{1}{4s} = \frac{4s + 1}{4s} -$$ - -This is the same as the voltage across the (*s* + 4) impedance. Hence, - -$$ -I_1 = \frac{V_1}{s+4} = \frac{4s+1}{4s(s+4)} -$$ - -Applying KCL at the top node yields - -$$ -I_o = I_1 + I_2 = \frac{4s + 1}{4s(s + 4)} + \frac{1}{2} = \frac{2s^2 + 12s + 1}{4s(s + 4)} -$$ - -Hence, - -$$ -H(s) = \frac{V_o}{I_o} = \frac{1}{I_o} = \frac{4s(s+4)}{2s^2 + 12s + 1} -$$ - -as before. - -Practice Problem 16.8 Find the transfer function *H*(*s*) = *I*1(*s*)∕*Io*(*s*) in the circuit of Fig. 16.18. - -Answer: -$$ -\frac{4s+1}{2s^2+12s+1} -$$ -. - -V(s) - -+ ‒ - -**Figure 16.18** For Example 16.8. - -# **Solution:** - -(a) Using voltage division, - -$$ -V_o = \frac{1}{s+1} V_{ab} -$$ - (16.9.1) - -But - -$$ -V_{ab} = \frac{1|| (s+1)}{1+1|| (s+1)} V_i = \frac{(s+1)/(s+2)}{1+(s+1)/(s+2)} V_i -$$ - -or - -$$ -V_{ab} = \frac{s+1}{2s+3} V_i -$$ - (16.9.2) - -Substituting Eq. (16.9.2) into Eq. (16.9.1) results in - -$$ -V_o = \frac{V_i}{2s + 3} -$$ - -Thus, the transfer function is - -$$ -H(s) = \frac{V_o}{V_i} = \frac{1}{2s + 3} -$$ - -(b) We may write *H*(*s*) as - -$$ -H(s) = \frac{1}{2} \frac{1}{s + \frac{3}{2}} -$$ - -Its inverse Laplace transform is the required impulse response: - -$$ -h(t) = \frac{1}{2}e^{-3t/2}u(t) -$$ - -(c) When *vi*(*t*) = *u*(*t*), *Vi*(*s*) = 1∕*s*, and - -$$ -V_o(s) = H(s)V_i(s) = \frac{1}{2s(s + \frac{3}{2})} = \frac{A}{s} + \frac{B}{s + \frac{3}{2}} -$$ - -where - -$$ -A = sV_o(s)|_{s=0} = \frac{1}{2(s + \frac{3}{2})}|_{s=0} = \frac{1}{3} -$$ -$$ -B = \left(s + \frac{3}{2}\right) V_o(s)|_{s=-3/2} = \frac{1}{2s}|_{s=-3/2} = -\frac{1}{3} -$$ - -Hence, for *vi*(*t*) = *u*(*t*), - -$$ -V_o(s) = \frac{1}{3} \left( \frac{1}{s} - \frac{1}{s + \frac{3}{2}} \right) -$$ - -and its inverse Laplace transform is - -$$ -v_o(t) = \frac{1}{3}(1 - e^{-3t/2})u(t) \text{ V} -$$ - -For Example 16.9. - -(d) When -$$ -v_i(t) = 8 \cos 2t -$$ -, then $V_i(s) = \frac{8s}{s^2 + 4}$ , and -\n -$$ -V_o(s) = H(s)V_i(s) = \frac{4s}{(s + \frac{3}{2})(s^2 + 4)} -$$ -\n -$$ -= \frac{A}{s + \frac{3}{2}} + \frac{Bs + C}{s^2 + 4} -$$ -\n(16.9.3) - -where - -$$ -A = \left(s + \frac{3}{2}\right) V_o(s) \Big|_{s = -3/2} = \frac{4s}{s^2 + 4} \Big|_{s = -3/2} = -\frac{24}{25} -$$ - -To get *B* and *C*, we multiply Eq. (16.9.3) by (*s* + 3∕2)(*s* 2 + 4). We get - -$$ -4s = A(s^{2} + 4) + B(s^{2} + \frac{3}{2}s) + C(s + \frac{3}{2}) -$$ - -Equating coefficients, - -Constant: -$$ -0 = 4A + \frac{3}{2}C -$$ - $\Rightarrow$ $C = -\frac{8}{3}A$ - -\ns: $4 = \frac{3}{2}B + C$ - -\ns2: $0 = A + B$ $\Rightarrow$ $B = -A$ - -Solving these gives *A* = −24∕25, *B* = 24∕25, *C* = 64∕25. Hence, for *vi*(*t*) = 8 cos 2*t* V, - -$$ -V_o(s) = \frac{-\frac{24}{25}}{s + \frac{3}{2}} + \frac{24}{25} \frac{s}{s^2 + 4} + \frac{32}{25} \frac{2}{s^2 + 4} -$$ - -and its inverse is - -$$ -v_o(t) = \frac{24}{25} \left( -e^{-3t/2} + \cos 2t + \frac{4}{3} \sin 2t \right) u(t) \text{ V} -$$ - -**Figure 16.20** - -**Figure 16.21** - -A linear system with *m* inputs and *p* outputs. - -Practice Problem 16.9 Rework Example 16.9 for the circuit shown in Fig. 16.20. - -**Answer:** (a) -$$ -2/(s + 4) -$$ -, (b) $2e^{-4t}u(t)$ , (c) $\frac{1}{2}(1 - e^{-4t})u(t)$ V, -(d) $3.2(-e^{-4t} + \cos 2t + \frac{1}{2} \sin 2t)u(t)$ V. - -# For Practice Prob. 16.9. **16.5** State Variables - -Thus far in this book we have considered techniques for analyzing systems with only one input and only one output. Man y engineering systems have many inputs and many outputs, as shown in Fig. 16.21. The state variable method is a v ery important tool in analyzing systems and understanding such highly complex systems. Thus, the state variable model is more gen eral than the single-input, single-output model, such as a transfer function. Although the topic cannot be adequately co vered in one chapter, let alone one section of a chapter, we will cover it briefly at this point. - -In the state variable model, we specify a collection of variables that describe the internal behavior of the system. These variables are known as the *state variables* of the system. They are the variables that determine the future behavior of a system when the present state of the system and the input signals are kno wn. In other w ords, the y are those v ariables which, if known, allow all other system parameters to be determined by using only algebraic equations. - -A state variable is a physical property that characterizes the state of a system, regardless of how the system got to that state. - -Common examples of state variables are the pressure, volume, and temperature. In an electric circuit, the state v ariables are the inductor current and capacitor voltage since they collectively describe the energy state of the system. - -The standard way to represent the state equations is to arrange them as a set of first-order differential equations: - -$$ -\dot{x} = Ax + Bz \tag{16.20} -$$ - -where - -$$ -\dot{\mathbf{x}}(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \\ \vdots \\ x_n(t) \end{bmatrix} = \text{state vector representing } n \text{ state vectors} -$$ - -and the dot represents the first derivative with respect to time, i.e., - -$$ -\dot{\mathbf{x}}(t) = \begin{bmatrix} \dot{x}_1(t) \\ \dot{x}_2(t) \\ \vdots \\ \dot{x}_n(t) \end{bmatrix} -$$ - -and - -$$ -\mathbf{z}(t) = \begin{bmatrix} z_1(t) \\ z_2(t) \\ \vdots \\ z_m(t) \end{bmatrix} = \text{input vector representing } m \text{ inputs} -$$ - -**A** and **B** are respectively *n* × *n* and *n* × *m* matrices. In addition to the state equation in Eq. (16.20), we need the output equation. The complete state model or state space is - -$$ -\dot{x} = Ax + Bz -$$ - (16.21a) -y = Cx + Dz (16.21b) - -where - -$$ -\mathbf{y}(t) = \begin{bmatrix} y_1(t) \\ y_2(t) \\ \vdots \\ y_p(t) \end{bmatrix} = \text{the output vector representing } p \text{ outputs} -$$ - -and **C** and **D** are, respectively, *p* × *n* and *p* × *m* matrices. For the special case of single-input single-output, *n* = *m* = *p* = 1. - -Assuming zero initial conditions, the transfer function of the system is found by taking the Laplace transform of Eq. (16.21a); we obtain - -$$ -s\mathbf{X}(s) = \mathbf{A}\mathbf{X}(s) + \mathbf{B}\mathbf{Z}(s) \qquad \rightarrow \qquad (s\mathbf{I} - \mathbf{A})\mathbf{X}(s) = \mathbf{B}\mathbf{Z}(s) -$$ - -or - -$$ -\mathbf{X}(s) = (s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} \mathbf{Z}(s) -$$ - (16.22) - -where **I** is the identity matrix. Taking the Laplace transform of Eq. (16.21b) yields - -$$ -\mathbf{Y}(s) = \mathbf{C}\mathbf{X}(s) + \mathbf{D}\mathbf{Z}(s) \tag{16.23} -$$ - -Substituting Eq. (16.22) into Eq. (16.23) and di viding by **Z**(*s*) gives the transfer function as - -$$ -H(s) = \frac{Y(s)}{Z(s)} = C(sI - A)^{-1}B + D -$$ - (16.24) - -where - -**A** = system matrix **B** = input coupling matrix **C** = output matrix **D** = feedforward matrix - -In most cases, **D** = **0**, so the degree of the numerator of *H*(*s*) in Eq. (16.24) is less than that of the denominator. Thus, - -$$ -H(s) = C(sI - A)^{-1}B -$$ - (16.25) - -Because of the matrix computation in volved, *MATLAB* can be used to find the transfer function. - -To apply state variable analysis to a circuit, we follow the following three steps. - -Steps to Apply the State Variable Method to Circuit Analysis: - -- 1. Select the inductor current *i* and capacitor voltage *v* as the state variables, making sure they are consistent with the passive sign convention. -- 2. Apply KCL and KVL to the circuit and obtain circuit variables (voltages and currents) in terms of the state v ariables. This should lead to a set of first-order differential equations necessary and sufficient to determine all state variables. -- 3. Obtain the output equation and put the final result in state-space representation. - -Steps 1 and 3 are usually straightforward; the major task is in step 2. We will illustrate this with examples. - -# **Solution:** - -We select the inductor current *i* and capacitor voltage *v* as the state variables. - -$$ -v_L = L \frac{di}{dt} \tag{16.10.1} -$$ - -$$ -i_C = C \frac{dv}{dt} \tag{16.10.2} -$$ - -Applying KCL at node 1 gives - -$$ -i = i_x + i_C -$$ - $\rightarrow$ $C\frac{dv}{dt} = i - \frac{v}{R}$ - -or - -$$ -\dot{v} = -\frac{v}{RC} + \frac{i}{C} -$$ - (16.10.3) - -since the same voltage *v* is across both *R* and *C*. Applying KVL around the outer loop yields - -$$ -v_s = v_L + v \rightarrow L\frac{di}{dt} = -v + v_s -$$ - -$$ -\dot{i} = -\frac{v}{L} + \frac{v_s}{L} -$$ - (16.10.4) - -Equations (16.10.3) and (16.10.4) constitute the state equations. If we regard *ix* as the output, - -$$ -i_x = \frac{v}{R} \tag{16.10.5} -$$ - -Putting Eqs. (16.10.3), (16.10.4), and (16.10.5) in the standard form leads to - -$$ -\begin{bmatrix} \dot{v} \\ \dot{i} \end{bmatrix} = \begin{bmatrix} \frac{-1}{RC} & \frac{1}{C} \\ \frac{-1}{L} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} 0 \\ \frac{1}{L} \end{bmatrix} v_s -$$ - (16.10.6a) -$$ -i_x = \begin{bmatrix} \frac{1}{R} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} -$$ - (16.10.6b) - -If *R* = 1, *C* = \_1 4 , and *L* = \_1 2 , we obtain from Eq. (16.10.6) matrices - -$$ -\mathbf{A} = \begin{bmatrix} \frac{-1}{RC} & \frac{1}{C} \\ \frac{-1}{L} & 0 \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -2 & 0 \end{bmatrix}, \qquad \mathbf{B} = \begin{bmatrix} 0 \\ \frac{1}{L} \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \end{bmatrix}, -$$ -$$ -\mathbf{C} = \begin{bmatrix} \frac{1}{R} & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \end{bmatrix} -$$ -$$ -s\mathbf{I} - \mathbf{A} = \begin{bmatrix} s & 0 \\ 0 & s \end{bmatrix} - \begin{bmatrix} -4 & 4 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} s+4 & -4 \\ 2 & s \end{bmatrix} -$$ - -Taking the inverse of this gives - -inverse of this gives -\n -$$ -(s\mathbf{I} - \mathbf{A})^{-1} = \frac{\text{adjoint of } \mathbf{A}}{\text{determinant of } \mathbf{A}} = \frac{\begin{bmatrix} s & 4\\ -2 & s+4 \end{bmatrix}}{s^2 + 4s + 8} -$$ - -Thus, the transfer function is given by - -Thus, the transfer function is given by - -\n -$$ -\mathbf{H}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} = \frac{\begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} s & 4 \\ -2 & s+4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \end{bmatrix}}{s^2 + 4s + 8} = \frac{8}{s^2 + 4s + 8} -$$ -\n -$$ -= \frac{8}{s^2 + 4s + 8} -$$ - -which is the same thing we would get by directly Laplace transforming the circuit and obtaining **H**(*s*) = *Ix*(*s*)∕*Vs*(*s*). The real advantage of the state variable approach comes with multiple inputs and multiple outputs. In this case, we have one input *vs* and one output *ix*. In the next example, we will have two inputs and two outputs. - -Example 16.11 Consider the circuit in Fig. 16.24, which may be regarded as a two- input, two-output system. Determine the state v ariable model and find the transfer function of the system. - -**Figure 16.24** For Example 16.11. - -# **Solution:** - -In this case, we have two inputs *vs* and *vi* and two outputs *vo* and *io*. Again, we select the inductor current *i* and capacitor voltage *v* as the state variables. Applying KVL around the left-hand loop gives - -$$ --v_s + i_1 + \frac{1}{6}i = 0 \quad \to \quad i = 6v_s - 6i_1 \quad (16.11.1) -$$ - -We need to eliminate *i*1. Applying KVL around the loop containing *vs*,1-Ω resistor, 2-Ω resistor, and \_1 3 -F capacitor yields - -$$ -v_s = i_1 + v_o + v \tag{16.11.2} -$$ - -But at node 1, KCL gives - -$$ -i_1 = i + \frac{v_o}{2} \rightarrow v_o = 2(i_1 - i) -$$ - (16.11.3) - -For Practice Prob. 16.10. - -Substituting this in Eq. (16.11.2), - -$$ -v_s = 3i_1 + v - 2i \rightarrow i_1 = \frac{2i - v + v_s}{3} -$$ - (16.11.4) - -Substituting this in Eq. (16.11.1) gives - -$$ -\dot{i} = 2v - 4i + 4v_s \tag{16.11.5} -$$ - -which is one state equation. To obtain the second one, we apply KCL at node 2. - -$$ -\frac{v_o}{2} = \frac{1}{3} \dot{v} + i_o \rightarrow \dot{v} = \frac{3}{2} v_o - 3i_o \tag{16.11.6} -$$ - -We need to eliminate *vo* and *io*. From the right-hand loop, it is evident that - -$$ -i_o = \frac{v - v_i}{3} \tag{16.11.7} -$$ - -Substituting Eq. (16.11.4) into Eq. (16.11.3) gives - -*i* - -$$ -v_o = 2\left(\frac{2i - v + v_s}{3} - i\right) = -\frac{2}{3}(v + i - v_s) \tag{16.11.8} -$$ - -Substituting Eqs. (16.11.7) and (16.11.8) into Eq. (16.11.6) yields the second state equation as - -$$ -\dot{v} = -2v - i + v_s + v_i \tag{16.11.9} -$$ - -The two output equations are already obtained in Eqs. (16.11.7) and (16.11.8). Putting Eqs. (16.11.5) and (16.11.7) to (16.11.9) together in the standard form leads to the state model for the circuit, namely, - -$$ -\begin{bmatrix} \dot{v} \\ \dot{i} \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ 2 & -4 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} 1 & 1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} v_s \\ v_i \end{bmatrix} -$$ -(16.11.10a) -$$ -\begin{bmatrix} v_o \\ i_o \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} & -\frac{2}{3} \\ \frac{1}{3} & 0 \end{bmatrix} \begin{bmatrix} v \\ i \end{bmatrix} + \begin{bmatrix} \frac{2}{3} & 0 \\ 0 & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} v_s \\ v_i \end{bmatrix} -$$ -(16.11.10b) - -For the electric circuit in Fig. 16.25, determine the state model. Take *vo* Practice Problem 16.11 and *io* as the output variables. - -**Answer:** - -**Figure 16.25** For Practice Prob. 16.11. - -Example 16.12 Assume we have a system where the output is *y*(*t*) and the input is *z*(*t*). Let the following differential equation describe the relationship between the input and the output. - -$$ -\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = 5z(t) -$$ - (16.12.1) - -Obtain the state model and the transfer function of the system. - -*x .* - -# **Solution:** - -First, we select the state variables. Let *x*1 = *y*(*t*), therefore - -$$ -x_1 = \dot{y}(t) \tag{16.12.2} -$$ - -Now let - -$$ -x_2 = \dot{x}_1 = \dot{y}(t) \tag{16.12.3} -$$ - -Note that at this time we are looking at a second-order system that would normally have two first-order terms in the solution. - -Now we have *x .* 2 = *y ..*(*t*), where we can find the value *x .* 2 from Eq. (16.12.1), i.e., - -$$ -\dot{x}_2 = \ddot{y}(t) = -2y(t) - 3\dot{y}(t) + 5z(t) = -2x_1 - 3x_2 + 5z(t) -$$ - (16.12.4) - -From Eqs. (16.12.2) to (16.12.4), we can now write the following matrix equations: - -$$ -\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 5 \end{bmatrix} z(t) -$$ - (16.12.5) - -$$ -\mathbf{y}(t) = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \tag{16.12.6} -$$ - -We now obtain the transfer function. - -$$ -s\mathbf{I} - \mathbf{A} = s \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} = \begin{bmatrix} s & -1 \\ 2 & s + 3 \end{bmatrix} -$$ - -The inverse is - -$$ -(s\mathbf{I} - \mathbf{A})^{-1} = \frac{\begin{bmatrix} s+3 & 1\\ -2 & s \end{bmatrix}}{s(s+3)+2} -$$ - -The transfer function is - -ne transfer function is -\n -$$ -\mathbf{H}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1} \mathbf{B} = \frac{(1 \quad 0) \begin{bmatrix} s+3 & 1 \\ -2 & s \end{bmatrix} \begin{pmatrix} 0 \\ 5 \end{pmatrix}}{s(s+3)+2} = \frac{(1 \quad 0) \begin{pmatrix} 5 \\ 5s \end{pmatrix}}{(s+1)(s+2)} -$$ - -To check this, we directly apply the Laplace transfer to each term in Eq. (16.12.1). Given that initial conditions are zero, we get - -$$ -[s2 + 3s + 2]Y(s) = 5Z(s) \rightarrow H(s) = \frac{Y(s)}{Z(s)} = \frac{5}{s2 + 3s + 2} -$$ - -which is in agreement with what we got previously. - -$$ -\frac{d^3y}{dt^3} + 18\frac{d^2y}{dt^2} + 20\frac{dy}{dt} + 5y = z(t) -$$ - -**Answer:** - -$$ -\mathbf{A} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -20 & -18 \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \quad \mathbf{C} = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}. -$$ - -# **16.6** Applications - -So far we have considered three applications of Laplace's transform: circuit analysis in general, obtaining transfer functions, and solving linear integrodifferential equations. The Laplace transform also finds application in other areas in circuit analysis, signal processing, and control systems. Here we will consider tw o more important applications: netw ork stability and network synthesis. - -# **16.6.1** Network Stability - -A circuit is *stable* if its impulse response *h*(*t*) is bounded (i.e., *h*(*t*) converges to a finite value) as *t* → ∞; it is *unstable* if *h*(*t*) grows without bound as *t* → ∞. In mathematical terms, a circuit is stable when - -$$ -\lim_{t \to \infty} |h(t)| = \text{finite} \tag{16.26} -$$ - -Because the transfer function *H*(*s*) is the Laplace transform of the impulse response *h*(*t*), *H*(*s*) must meet certain requirements for Eq. (16.26) to hold. Recall that *H*(*s*) may be written as - -$$ -H(s) = \frac{N(s)}{D(s)} -$$ -(16.27) - -where the roots of *N*(*s*) = 0 are called the *zeros* of *H*(*s*) because the y make *H*(*s*) = 0, while the roots of *D*(*s*) = 0 are called the *poles* of *H*(*s*) since they cause *H*(*s*) → ∞. The zeros and poles of *H*(*s*) are often located in the *s* plane as sho wn in Fig. 16.26(a). Recall from Eqs. (15.47) and (15.48) that *H*(*s*) may also be written in terms of its poles as \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*s* + *p*1)(*s* + *p*2) ⋯ (*s* + *pn*) - -$$ -H(s) = \frac{N(s)}{D(s)} = \frac{N(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)} -$$ -(16.28) - -*H*(*s*) must meet tw o requirements for the circuit to be stable. First, the degree of *N*(*s*) must be less than the degree of *D*(*s*); otherwise, long division would produce - -$$ -H(s) = k_n s^n + k_{n-1} s^{n-1} + \dots + k_1 s + k_0 + \frac{R(s)}{D(s)} -$$ -(16.29) - -where the de gree of *R*(*s*), the remainder of the long di vision, is less than the degree of *D*(*s*). The inverse of *H*(*s*) in Eq. (16.29) does not meet the condition in Eq. (16.26). Second, all the poles of *H*(*s*) in - -# **Figure 16.26** The complex *s* plane: (a) poles and zeros plotted, (b) left-half plane. - -Eq. (16.27) (i.e., all the roots of *D*(*s*) = 0) must have negative real parts; in other words, all the poles must lie in the left half of the *s* plane, as shown typically in Fig. 16.26(b). The reason for this will be apparent if we take the inverse Laplace transform of *H*(*s*) in Eq. (16.27). Because Eq. (16.27) is similar to Eq. (15.48), its partial fraction expansion is similar to the one in Eq. (15.49) so that the inverse of *H*(*s*) is similar to that in Eq. (15.53). Hence, - -$$ -h(t) = (k_1 e^{-p_1 t} + k_2 e^{-p_2 t} + \dots + k_n e^{-p_n t}) u(t) -$$ -\n(16.30) - -We see from this equation that each pole *pi* must be positive (i.e., pole *s* = −*pi* in the left-half plane) for *e*−*pi t* to decrease with increasing *t*. Thus, - -A circuit is stable when all the poles of its transfer function H(s) lie in the left half of the s plane. - -An unstable circuit never reaches steady state because the transient response does not decay to zero. Consequently , steady-state analysis is only applicable to stable circuits. - -A circuit made up exclusively of passive elements (*R*, *L*, and *C*) and independent sources cannot be unstable, because that w ould imply that some branch currents or v oltages would grow indefinitely with sources set to zero. Passive elements cannot generate such indefinite growth. Passive circuits either are stable or have poles with zero real parts. To show that this is the case, consider the series *RLC* circuit in Fig. 16.27. The transfer function is given by - -$$ -H(s) = \frac{V_o}{V_s} = \frac{1/sC}{R + sL + 1/sC} -$$ - -$$ -H(s) = \frac{1/L}{s^2 + sR/L + 1/LC} -$$ - (16.31) - -Notice that *D*(*s*) = *s* 2 + *sR*∕*L* + 1∕*LC* = 0 is the same as the characteristic equation obtained for the series *RLC* circuit in Eq. (8.8). The circuit has poles at \_\_\_\_\_\_\_ - -$$ -p_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} -$$ - (16.32) - -where - -or - -$$ -\alpha = \frac{R}{2L}, \qquad \omega_0 = \frac{1}{LC} -$$ - -For *R*, *L*, *C* > 0, the tw o poles always lie in the left half of the *s* plane, implying that the circuit is al ways stable. However, when *R* = 0, *α* = 0 and the circuit becomes unstable. Although ideally this is possible, it does not really happen, because *R* is never zero. - -On the other hand, active circuits or passive circuits with controlled sources can supply energy, and they can be unstable. In fact, an oscillator is a typical example of a circuit designed to be unstable. An oscillator is designed such that its transfer function is of the form - -imple of a circuit designed to be unstable. An oscula- -ch that its transfer function is of the form -$$ -H(s) = \frac{N(s)}{s^2 + \omega_0^2} = \frac{N(s)}{(s + j\omega_0)(s - j\omega_0)} -$$ -(16.33) - -so that its output is sinusoidal. - -**Figure 16.27** A typical *RLC* circuit. - -Determine the values of *k* for which the circuit in Fig. 16.28 is stable. Example 16.13 - -# **Solution:** - -Applying mesh analysis to the first-order circuit in Fig. 16.28 gives - -$$ -V_i = \left(R + \frac{1}{sC}\right)I_1 - \frac{I_2}{sC} -$$ - (16.13.1) - -and - -$$ -0 = -kI_1 + \left(R + \frac{1}{sC}\right)I_2 - \frac{I_1}{sC} -$$ - -or - -$$ -0 = -\left(k + \frac{1}{sC}\right)I_1 + \left(R + \frac{1}{sC}\right)I_2\tag{16.13.2} -$$ - -We can write Eqs. (16.13.1) and (16.13.2) in matrix form as - -$$ -\begin{bmatrix} V_i \\ 0 \end{bmatrix} = \begin{bmatrix} \left(R + \frac{1}{sC}\right) & -\frac{1}{sC} \\ -\left(k + \frac{1}{sC}\right) & \left(R + \frac{1}{sC}\right) \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} -$$ - -The determinant is - -inant is -\n -$$ -\Delta = \left(R + \frac{1}{sC}\right)^2 - \frac{k}{sC} - \frac{1}{s^2C^2} = \frac{sR^2C + 2R - k}{sC} -$$ -\n(16.13.3) - -The characteristic equation (∆ = 0) gives the single pole as - -$$ -p = \frac{k - 2R}{R^2 C} -$$ - -which is negative when *k* < 2*R*. Thus, we conclude the circuit is stable when *k* < 2*R* and unstable for *k* > 2*R*. - -**Answer:** *β* > −1∕*R*. - -For Practice Prob. 16.13. - -An active filter has the transfer function Example 16.14 - -$$ -H(s) = \frac{k}{s^2 + s(4 - k) + 1} -$$ - -For what values of *k* is the filter stable? - -+ - -1 - -I1 I2 kI1 - -R R - -‒ + - -# **Solution:** - -As a second-order circuit, *H*(*s*) may be written as - -$$ -H(s) = \frac{N(s)}{s^2 + bs + c} -$$ - -where *b* = 4 − *k*, *c* = 1, and *N*(*s*) = *k*. This has poles at *p*2 + *bp* + *c* = 0; that is, - -$$ -p_{1,2} = \frac{-b \pm \sqrt{b^2 - 4c}}{2} -$$ - -For the circuit to be stable, the poles must be located in the left half of the *s* plane. This implies that *b* > 0. - -Applying this to the given *H*(*s*) means that for the circuit to be stable, 4 − *k* > 0 or *k* < 4. - -Practice Problem 16.14 A second-order active circuit has the transfer function - -circuit has the transfer function -$$ -H(s) = \frac{1}{s^2 + s(25 + \alpha) + 25} -$$ - -Find the range of the values of *α* for which the circuit is stable. What is the value of *α* that will cause oscillation? - -**Answer:** *α* > −25, *α* = −25. - -# **16.6.2** Network Synthesis - -Network synthesis may be regarded as the process of obtaining an appropriate network to represent a given transfer function. Network synthesis is easier in the *s*-domain than in the time domain. - -In network analysis, we find the transfer function of a given network. In netw ork synthesis, we re verse the approach: Gi ven a transfer function, we are required to find a suitable network. - -Network synthesis is finding a network that represents a given transfer function. - -Keep in mind that in synthesis, there may be man y dif ferent answers—or possibly no answers—because there are many circuits that can be used to represent the same transfer function; in network analysis, there is only one answer. - -Network synthesis is an exciting field of prime engineering importance. Being able to look at a transfer function and come up with the type of circuit it represents is a great asset to a circuit de signer. Although network synthesis constitutes a whole course by itself and requires some e xperience, the follo wing e xamples are meant to stimulate your appetite. - -Given the transfer function Example 16.15 - -$$ -H(s) = \frac{V_o(s)}{V_i(s)} = \frac{10}{s^2 + 3s + 10} -$$ - -realize the function using the circuit in Fig. 16.30(a). (a) Select *R* = 5 Ω, and find *L* and *C*. (b) Select *R* = 1 Ω, and find *L* and *C*. - -# **Solution:** - -1. **Define.** The problem is clearly and completely defined. This problem is what we call a synthesis problem: Given a transfer function, synthesize a circuit that produces the given transfer function. However, to keep the problem more manageable, we give a circuit that produces the desired transfer function. - -Had one of the variables, *R* in this case, not been given a value, then the problem would have had an infinite number of answers. An open-ended problem of this kind would require some additional assumptions that would have narrowed the set of solutions. - -- 2. **Present.** A transfer function of the voltage out versus the voltage in is equal to 10∕(*s* 2 + 3*s* + 10). A circuit, Fig. 16.30, is also given that should be able to produce the required transfer function. Two different values of *R*, 5 and 1 Ω, are to be used to calculate the values of *L* and *C* that produce the given transfer function. -- 3. **Alternative.** All solution paths involve determining the transfer function of Fig. 16.30 and then matching the various terms of the transfer function. Two approaches would be to use mesh analysis or nodal analysis. Because we are looking for a ratio of voltages, nodal analysis makes the most sense. -- 4. **Attempt.** Using nodal analysis leads to - -$$ -\frac{V_o(s) - V_i(s)}{sL} + \frac{V_o(s) - 0}{1/(sC)} + \frac{V_o(s) - 0}{R} = 0 -$$ - -Now multiply through by *sLR*: - -$$ -RV_o(s) - RV_i(s) + s^2 R LCV_o(s) + sLV_o(s) = 0 -$$ - -Collecting terms we get - -$$ -(s^2RLC + sL + R)V_o(s) = RV_i(s) -$$ - -or - -$$ -(s2RLC + sL + R)Vo(s) = RVi(s) -$$ -$$ -\frac{Vo(s)}{Vi(s)} = \frac{1/(LC)}{s2 + [1/(RC)]s + 1/(LC)} -$$ - - Matching the two transfer functions produces two equations with three unknowns. - -$$ -LC = 0.1 \qquad \text{or} \qquad L = \frac{0.1}{C} -$$ - -and - -$$ -RC = \frac{1}{3} \qquad \text{or} \qquad C = \frac{1}{3R} -$$ - -We have a constraint equation, *R* = 5 Ω for (a) and = 1 Ω for (b). - -(a) *C* = 1∕(3 × 5) = **66.67 mF** and *L* = **1.5 H** (b) *C* = 1∕(3 × 1) = **333.3 mF** and *L* = **300 mH** - -**Figure 16.30** For Example 16.15. - -5. **Evaluate.** There are different ways of checking the answer. Solving for the transfer function by using mesh analysis seems the most straightforward and the approach we could use here. However, it should be pointed out that this is mathematically more complex and will take longer than the original nodal analysis approach. Other approaches also exist. We can assume an input for *vi*(*t*), *vi*(*t*) = *u*(*t*) V and, using either nodal analysis or mesh analysis, see if we get the same answer we would get with just using the transfer function. That is the approach we will try using mesh analysis. - -Let *vi*(*t*) = *u*(*t*) *V* or *Vi*(*s*) = 1∕*s*. This will produce - -$$ -V_o(s) = 10/(s^3 + 3s^2 + 10s) -$$ - -Based on Fig. 16.30, mesh analysis leads to (a) For loop 1, - -$$ --(1/s) + 1.5sI_1 + [1/(0.06667s)] (I_1 - I_2) = 0 -$$ - -or - -$$ -(1.5s^2 + 15)I_1 - 15I_2 = 1 -$$ - -For loop 2, - -$$ -(15/s)(I_2 - I_1) + 5I_2 = 0 -$$ - -or - -$$ --15I_1 + (5s + 15)I_2 = 0 \qquad \text{or} \qquad I_1 = (0.3333s + 1)I_2 -$$ - -Substituting into the first equation we get - -$$ -(0.5s3 + 1.5s2 + 5s + 15)I2 - 15I2 = 1 -$$ - -or - -$$ -I_2 = 2/(s^3 + 3s^2 + 10s) -$$ - -but - -$$ -V_o(s) = 5I_2 = 10/(s^3 + 3s^2 + 10s) -$$ - -and the answer checks. - -(b) For loop 1, - -$$ --(1/s) + 0.3sI_1 + [1/(0.3333s)] (I_1 - I_2) = 0 -$$ - -or - -$$ -(0.3s^2 + 3)I_1 - 3I_2 = 1 -$$ - -For loop 2, - -$$ -(3/s)(I_2 - I_1) + I_2 = 0 -$$ - -or - -$$ --3I_1 + (s+3)I_2 = 0 -$$ - or $I_1 = (0.3333s + 1)I_2$ - -Substituting into the first equation we get - -$$ -(0.09999s3 + 0.3s2 + s + 3)I2 - 3I2 = 1 -$$ - -or - -$$ -I_2 = 10/(s^3 + 3s^2 + 10s) -$$ - -but *Vo*(*s*) = 1 × *I*2 = 10∕(*s* 3 + 3*s* 2 + 10*s*) and the answer checks. - -6. **Satisfactory?** We have clearly identified values of *L* and *C* for each of the conditions. In addition, we have carefully checked the answers to see if they are correct. The problem has been adequately solved. The results can now be presented as a solution to the problem. - -Realize the function Practice Problem 16.15 - -$$ -G(s) = \frac{V_o(s)}{V_i(s)} = \frac{4s}{s^2 + 4s + 20} -$$ - -using the circuit in Fig. 16.31. Select *R* = 2 Ω, and determine *L* and *C*. - -**Answer:** 500 mH, 100 mF. - -Synthesize the function Example 16.16 - -ion -$$ -T(s) = \frac{V_o(s)}{V_s(s)} = \frac{10^6}{s^2 + 100s + 10^6} -$$ - -using the topology in Fig. 16.32. - -For Example 16.16. - -# **Solution:** - -We apply nodal analysis to nodes 1 and 2. At node 1, - -$$ -(V_s - V_1)Y_1 = (V_1 - V_o)Y_2 + (V_1 - V_2)Y_3 \tag{16.16.1} -$$ - -At node 2, - -$$ -(V_1 - V_2)Y_3 = (V_2 - 0)Y_4 -$$ - (16.16.2) - -But *V*2 = *Vo*, so Eq. (16.16.1) becomes - -$$ -Y_1 V_s = (Y_1 + Y_2 + Y_3) V_1 - (Y_2 + Y_3) V_o \tag{16.16.3} -$$ - -and Eq. (16.16.2) becomes - -$$ -V_1 Y_3 = (Y_3 + Y_4) V_o -$$ - -or - -$$ -V_1 = \frac{1}{Y_3} (Y_3 + Y_4) V_o -$$ - (16.16.4) - -Substituting Eq. (16.16.4) into Eq. (16.16.3) gives - -$$ -Y_1 V_s = (Y_1 + Y_2 + Y_3) \frac{1}{Y_3} (Y_3 + Y_4) V_o - (Y_2 + Y_3) V_o -$$ - -or - -$$ -Y_1Y_3V_s = [Y_1Y_3 + Y_4(Y_1 + Y_2 + Y_3)]V_o -$$ - -Thus, - -$$ -V_s = [Y_1Y_3 + Y_4(Y_1 + Y_2 + Y_3)]V_o -$$ - -$$ -\frac{V_o}{V_s} = \frac{Y_1Y_3}{Y_1Y_3 + Y_4(Y_1 + Y_2 + Y_3)} -$$ -(16.16.5) - -To synthesize the given transfer function *T*(*s*), compare it with the one in Eq. (16.16.5). Notice two things: (1) *Y*1*Y*3 must not involve *s* because the numerator of *T*(*s*) is constant; (2) the given transfer function is second-order, which implies that we must have two capacitors. Therefore, we must make *Y*1 and *Y*3 resistive, while *Y*2 and *Y*4 are capacitive. So we select - -$$ -Y_1 = \frac{1}{R_1} -$$ -, $Y_2 = sC_1$ , $Y_3 = \frac{1}{R_2}$ , $Y_4 = sC_2$ (16.16.6) - -Substituting Eq. (16.16.6) into Eq. (16.16.5) gives - -q. (16.16.6) into Eq. (16.16.5) gives -\n -$$ -\frac{V_o}{V_s} = \frac{1/(R_1R_2)}{1/(R_1R_2) + sC_2(1/R_1 + 1/R_2 + sC_1)} -$$ -\n -$$ -= \frac{1/(R_1R_2C_1C_2)}{s^2 + s(R_1 + R_2)/(R_1R_2C_1) + 1/(R_1R_2C_1C_2)} -$$ - -Comparing this with the given transfer function *T*(*s*), we notice that - -$$ -\frac{1}{R_1 R_2 C_1 C_2} = 10^6, \qquad \frac{R_1 + R_2}{R_1 R_2 C_1} = 100 -$$ - -If we select *R*1 = *R*2 = 10 kΩ, then - -$$ -R_1 = R_2 = 10 \text{ k}\Omega, \text{ then} -$$ - -\n -$$ -C_1 = \frac{R_1 + R_2}{100R_1R_2} = \frac{20 \times 10^3}{100 \times 100 \times 10^6} = 2 \text{ }\mu\text{F} -$$ - -\n -$$ -C_2 = \frac{10^{-6}}{R_1R_2C_1} = \frac{10^{-6}}{100 \times 10^6 \times 2 \times 10^{-6}} = 5 \text{ nF} -$$ - -Thus, the given transfer function is realized using the circuit shown in Fig. 16.33. - -For Example 16.16. - -Synthesize the function Practice Problem 16.16 - -$$ -\frac{V_o(s)}{V_{\text{in}}} = \frac{-2s}{s^2 + 6s + 10} -$$ - -using the op amp circuit shown in Fig. 16.34. Select - -$$ -Y_1 = \frac{1}{R_1} -$$ -, $Y_2 = sC_1$ , $Y_3 = sC_2$ , $Y_4 = \frac{1}{R_2}$ - -Let *R*1 = 1 kΩ, and determine *C*1, *C*2, and *R*2. - -For Practice Prob. 16.16. - -**Answer:** 100 *µ*F, 500 *µ*F, 2 kΩ. - -# **16.7** Summary - -- 1. The Laplace transform can be used to analyze a circuit. We convert each element from the time domain to the *s*-domain, solve the problem using an y circuit analysis technique, and con vert the result to the time domain using the inverse transform. -- 2. In the *s*-domain, the circuit elements are replaced with the initial condition at *t* = 0 as follo ws. (Please note, v oltage models are - -given below, but the corresponding current models w ork equally well.): - -Resistor: -$$ -v_R = Ri \rightarrow V_R = RI -$$ - -\nInductor: $v_L = L\frac{di}{dt} \rightarrow V_L = sLI - Li(0^-)$ -\nCapacitor: $v_C = \int i \, dt \rightarrow V_C = \frac{1}{sC} - \frac{v(0^-)}{s}$ - -- 3. Using the Laplace transform to analyze a circuit results in a com plete (both transient and steady state) response because the initial conditions are incorporated in the transformation process. -- 4. The transfer function *H*(*s*) of a network is the Laplace transform of the impulse response *h*(*t*). -- 5. In the *s*-domain, the transfer function *H*(*s*) relates the output response *Y*(*s*) and an input excitation *X*(*s*); that is, *H*(*s*) = *Y*(*s*)∕*X*(*s*). -- 6. The state v ariable model is a useful tool for analyzing comple x systems with several inputs and outputs. State variable analysis is a powerful technique that is most popularly used in circuit theory and control. The state of a system is the smallest set of quanti ties (known as state v ariables) that we must kno w to determine its future response to an y given input. The state equation in state variable form is - -$$ -\dot{\mathbf{x}} = \mathbf{A}x + \mathbf{B}z -$$ - -while the output equation is - -$$ -\mathbf{y} = \mathbf{C}x + \mathbf{D}z -$$ - -- 7. For an electric circuit, we first select capacitor voltages and inductor current as state v ariables. We then apply KCL and KVL to obtain the state equations. -- 8. Two other areas of applications of the Laplace transform co vered in this chapter are circuit stability and synthesis. A circuit is stable when all the poles of its transfer function lie in the left half of the *s* plane. Network synthesis is the process of obtaining an appropriate network to represent a given transfer function for which analysis in the *s*-domain is well suited. - -# Review Questions - -**16.1** The voltage through a resistor with current *i*(*t*) in the *s*-domain is *sRI*(*s*). - -(a) True (b) False - -**16.2** The current through an *RL* series circuit with input voltage *v*(*t*) is given in the *s*-domain as: - -(a) -$$ -V(s) \left[ R + \frac{1}{sL} \right] -$$ - (b) $V(s)(R + sL)$ -(c) $\frac{V(s)}{R + 1/sL}$ (d) $\frac{V(s)}{R + sL}$ - -**16.3** The impedance of a 10-F capacitor is: - -(a) 10∕*s* (b) *s*∕10 (c) 1∕10*s* (d) 10*s* - -**16.4** We can usually obtain the Thevenin equivalent in the time domain. - -(a) True (b) False - -**16.5** A transfer function is defined only when all initial conditions are zero. - -(a) True (b) False - -Problems **745** - -**16.6** If the input to a linear system is *δ*(*t*) and the output is *e*−2*t u*(*t*), the transfer function of the system is: - -(a) -$$ -\frac{1}{s+2} -$$ - (b) $\frac{1}{s-2}$ (c) $\frac{s}{s+2}$ (d) $\frac{s}{s-2}$ - -(e) None of the above - -**16.7** If the transfer function of a system is - -sfer function of a system is -\n -$$ -H(s) = \frac{s^2 + s + 2}{s^3 + 4s^2 + 5s + 1} -$$ - -it follows that the input is *X*(*s*) = *s* 3 + 4*s* 2 + 5*s* + 1, while the output is *Y*(*s*) = *s* 2 + *s* + 2. - -$$ -(a) True \t\t (b) False -$$ - -**16.8** A network has its transfer function as - -has its transfer function a -$$ -H(s) = \frac{s+1}{(s-2)(s+3)} -$$ - -The network is stable. - -$$ -(a) True \t\t (b) False -$$ - -# Problems - -- Sections 16.2 and 16.3 Circuit Element Models and Circuit Analysis -- **16.1** The current in an *RLC* circuit is described by - -$$ -\frac{d^2i}{dt^2} + 10\frac{di}{dt} + 25i = 0 -$$ - -If -$$ -i(0) = 7 -$$ - A and $di(0)/dt = 0$ , find $i(t)$ for $t > 0$ . - -**16.2** The differential equation that describes the voltage in an *RLC* network is - -$$ -3\frac{d^2i}{dt^2} + 15\frac{di}{dt} + 12i = 0 -$$ - -Given that *v*(0) = 0, *dv*(0)/*dt* = 6 mA∕s, obtain *i*(*t*). - -**16.3** The natural response of an *RLC* circuit is described by the differential equation - -$$ -\frac{d^2v}{dt^2} + 2\frac{dv}{dt} + v = 0 -$$ - - for which the initial conditions are *v*(0) = 350 V and *dv*(0)/*dt* = 0. Solve for *v*(*t*). - -- **16.4** If *R* = 20 Ω, *L* = 0.6 H, what value of *C* will make an *RLC* series circuit: - - (a) overdamped? - - (b) critically damped? - - (c) underdamped? - -**16.9** Which of the following equations is called the state equation? - -(a) -$$ -\dot{\mathbf{x}} = \mathbf{A}\mathbf{x} + \mathbf{B}\mathbf{z} -$$ - -\n(b) $\mathbf{y} = \mathbf{C}\mathbf{x} + \mathbf{D}\mathbf{z}$ -\n(c) $\mathbf{H}(s) = \mathbf{Y}(s)/\mathbf{Z}(s)$ -\n(d) $\mathbf{H}(s) = \mathbf{C}(s\mathbf{I} - \mathbf{A})^{-1}$ - -**16.10** A single-input, single-output system is described by the state model as: - -**B** - -$$ -\dot{x}_1 = 2x_1 - x_2 + 3z -$$ - -\n -$$ -\dot{x}_2 = -4x_2 - z -$$ - -\n -$$ -y = 3x_1 - 2x_2 + z -$$ - -Which of the following matrices is incorrect? - -(a) -$$ -A = \begin{bmatrix} 2 & -1 \\ 0 & -4 \end{bmatrix} -$$ - (b) $B = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$ -(c) $C = \begin{bmatrix} 3 & -2 \end{bmatrix}$ (d) $D = 0$ - -*Answers: 16.1b, 16.2d, 16.3c, 16.4b, 16.5b, 16.6a, 16.7b, 16.8b, 16.9a, 16.10d.* - -**16.5** The responses of a series *RLC* circuit are - -$$ -v_c(t) = [30 - 10e^{-20t} + 30e^{-10t}]u(t) -$$ - -$$ -i_L(t) = [40e^{-20t} - 60e^{-10t}]u(t) -$$ -mA - - where *vC*(*t*) and *iL*(*t*) are the capacitor voltage and inductor current, respectively. Determine the values of *R*, *L*, and *C*. - -**16.6** Design a parallel *RLC* circuit that has the characteristic equation - -$$ -s^2 + 100s + 10^6 = 0. -$$ - -**16.7** The step response of an *RLC* circuit is given by - -$$ -\frac{d^2i}{dt^2} + 2\frac{di}{dt} + 5i = 30 -$$ - - Given that *i*(0) = 18 A and *di*(0)/*dt* = 36 A/s, solve for *i*(*t*). - -**16.8** A branch voltage in an *RLC* circuit is described by - -$$ -\frac{d^2v}{dt^2} + 4\frac{dv}{dt} + 8v = 120 -$$ - - If the initial conditions are *v*(0) = 0 = *dv*(0)/*dt*, find *v*(*t*). - -**16.9** A series *RLC* circuit is described by - -$$ -L\frac{d^2i(t)}{dt} + R\frac{di(t)}{dt} + \frac{i(t)}{C} = 15 -$$ - - Find the response when *L* = 0.5*H*, *R* = 4Ω, and *C* = 0.2 *F*. Let *i*(0−) = 7.5 A and [*di*(0−)/*dt*] = 0. **16.10** The step responses of a series *RLC* circuit are - -$$ -V_c = 40 - 10e^{-2000t} - 10e^{-4000t} \text{ V}, t > 0 -$$ -$$ -i_L(t) = 3e^{-2000t} + 6e^{-4000t} mA, t > 0 -$$ - -- (a) Find *C*. - - (b) Determine what type of damping is exhibited by the circuit. -- **16.11** The step response of a parallel RLC circuit is - -$$ -v = 10 + 20e^{-300t} (\cos 400t - 2 \sin 400t) \text{V}, t \ge 0 -$$ - -when the inductor is 50 mH. Find *R* and *C*. - -**16.12** Determine *i*(*t*) in the circuit of Fig. 16.35 by means of the Laplace transform. - -# **Figure 16.35** - -For Prob. 16.12. - -**16.13** Using Fig. 16.36, design a problem to help other students better understand circuit analysis using Laplace transforms. - -- **16.14** Find *i*(*t*) for *t* > 0 for the circuit in Fig. 16.37. Assume *is*(*t*) = [6(*t*) + 3*δ*(*t*)]mA. - -**16.15** For the circuit in Fig. 16.38, calculate the value of *R* needed to have a critically damped response. - -# **Figure 16.38** - -**16.16** The capacitor in the circuit of Fig. 16.39 is initially uncharged. Find *v*0(*t*) for *t* > 0. - -# **Figure 16.39** - -**16.17** If *is*(*t*) = 7.5*e* −2*t u*(*t*) A in the circuit shown in Fig. 16.40, find the value of *io*(*t*). - -**Figure 16.40** - -For Prob. 16.17. - -**16.18** Find *v*(*t*), *t* > 0 in the circuit of Fig. 16.41. Let *vs* = 12 V. - -# **Figure 16.41** - -For Prob. 16.18. - -**16.19** The switch in Fig. 16.42 moves from position A to position B at *t* = 0 (please note that the switch must connect to point B before it breaks the connection at A, a make before break switch). Find *v*(*t*) for *t* > 0. - -For Prob. 16.19. - -**16.20** Find *i*(*t*) for *t* > 0 in the circuit of Fig. 16.43. - -**16.21** In the circuit of Fig. 16.44, the switch moves (make before break switch) from position *A* to *B* at *t* = 0. Find *v*(*t*) for all *t* ≥ 0. - -**16.22** Find the voltage across the capacitor as a function of time for *t* > 0 for the circuit in Fig. 16.45. Assume steady-state conditions exist at *t* = 0−. - -**Figure 16.45** For Prob. 16.22. - -**16.23** Obtain *v*(*t*) for *t* > 0 in the circuit of Fig. 16.46. - -**Figure 16.46** For Prob. 16.23. - -**16.24** The switch in the circuit of Fig. 16.47 has been closed for a long time but is opened at *t* = 0. Determine *i*(*t*) for *t* > 0. - -**Figure 16.47** For Prob. 16.24. - -**16.25** Calculate *v*(*t*) for *t* > 0 in the circuit of Fig. 16.48. - -**Figure 16.48** For Prob. 16.25. - -**16.26** The switch in Fig. 16.49 moves from position A to position B at *t* = 0 (please note that the switch must connect to point B before it breaks the connection at A, a make before break switch). Determine *i*(*t*) for *t* > 0. Also assume that the initial voltage on the capacitor is zero. - -**Figure 16.49** For Prob. 16.26. - -**16.27** Find *v*(*t*) for *t* > 0 in the circuit in Fig. 16.50. - -**16.28** For the circuit in Fig. 16.51, find *v*(*t*) for *t* > 0. - -**Figure 16.55** For Prob. 16.32. - -**16.33** Using Fig. 16.56, design a problem to help other students understand how to use Thevenin's theorem (in the *s*-domain) to aid in circuit analysis. - -**Figure 16.56** For Prob. 16.33. - -**16.30** Find *vo*(*t*), for all *t* > 0, in the circuit of Fig. 16.53. - -**16.31** Obtain *v*(*t*) and *i*(*t*) for *t* > 0 in the circuit in Fig. 16.54. - -**Figure 16.54** For Prob. 16.31. - -**16.34** Solve for the mesh currents in the circuit of Fig. 16.57. You may leave your results in the *s*-domain. - -**Figure 16.57** For Prob. 16.34. - -**16.35** Find *vo*(*t*) in the circuit of Fig. 16.58. - -**Figure 16.58** For Prob. 16.35. - -**16.36** Refer to the circuit in Fig. 16.59. Calculate *i*(*t*) for *t* > 0. - -# **Figure 16.59** - -For Prob. 16.36. - -**16.37** Determine *v* for *t* > 0 in the circuit in Fig. 16.60. - -For Prob. 16.37. - -**16.38** The switch in the circuit of Fig. 16.61 is moved from position *a* to *b* (a make before break switch) at *t* = 0. Determine *i*(*t*) for *t* > 0. - -# **Figure 16.61** - -For Prob. 16.38. - -**16.39** For the network in Fig. 16.62, find *i*(*t*) for *t* > 0. - -**Figure 16.62** For Prob. 16.39. - -**16.40** In the circuit of Fig. 16.63, find *v*(*t*) and *i*(*t*) for *t* > 0. Assume *v*(0) = 0 V and *i*(0) = 1.25 A. - -# **Figure 16.63** For Prob. 16.40. - -**16.41** Find the output voltage *vo*(*t*) in the circuit of Fig. 16.64. - -For Prob. 16.41. - -**16.42** Given the circuit in Fig. 16.65, find *i*(*t*) and *v*(*t*) for *t* > 0. - -# **Figure 16.65** For Prob. 16.42. - -**16.43** Determine *i*(*t*) for *t* > 0 in the circuit of Fig. 16.66. - -**Figure 16.67** - -**16.45** Find *v*(*t*) for *t* > 0 in the circuit in Fig. 16.68. - -For Prob. 16.45. - -**16.46** Determine *io*(*t*) in the circuit in Fig. 16.69. - -# **Figure 16.69** - -For Prob. 16.46. - -**16.47** Determine *io*(*t*) in the network shown in Fig. 16.70. - -# **Figure 16.70** - -For Prob. 16.47. - -**Figure 16.71** For Prob. 16.48. - -**16.49** Find *i*0(*t*) for *t* > 0 in the circuit in Fig. 16.72. - -# **Figure 16.72** - -For Prob. 16.49. - -**16.50** For the circuit in Fig. 16.73, find *v*(*t*) for *t* > 0. Assume that *i*(0) = 2 A. - -# **Figure 16.73** - -For Prob. 16.50. - -# **Figure 16.74** - -For Prob. 16.51. - -**16.52** Given the circuit shown in Fig. 16.75, determine the values for *i*(*t*) and *v*(*t*) for all *t* > 0. - -# **Figure 16.75** - -For Prob. 16.52. - -**16.53** In the circuit of Fig. 16.76, the switch has been in position 1 for a long time but moved to position 2 at *t* = 0. Find: - -> (a) *v*(0+), *dv*(0+)/*dt* (b) *v*(*t*) for *t* ≥ 0. - -For Prob. 16.44. - -Problems **751** - -**Figure 16.76** - -For Prob. 16.53. - -**16.54** The switch in Fig. 16.77 has been in position 1 for *t* < 0. At *t* = 0, it is moved from position 1 to the top of the capacitor at *t* = 0. Please note that the switch is a make before break switch; it stays in contact with position 1 until it makes contact with the top of the capacitor and then breaks the contact at position 1. Determine *v*(*t*). - -**16.55** Obtain *i*1 and *i*2 for *t* > 0 in the circuit of Fig. 16.78. - -**Figure 16.78** For Prob. 16.55. - -**16.56** Calculate *io*(*t*) for *t* > 0 in the network of Fig. 16.79. - -**16.57** (a) Find the Laplace transform of the voltage shown in Fig. 16.80(a). (b) Using that value of *vs*(*t*) in the circuit shown in Fig. 16.80(b), find the value of *vo*(*t*). - -**Figure 16.80** For Prob. 16.57. - -**16.58** Using Fig. 16.81, design a problem to help other students better understand circuit analysis in the *s*-domain with circuits that have dependent sources. - -**Figure 16.81** For Prob. 16.58. - -**16.59** Find *vo*(*t*) in the circuit of Fig. 16.82 if *vx*(0) = 10 V and *i*(0) = 5 A. - -# **Figure 16.82** - -For Prob. 16.59. - -For Prob. 16.60. - -**16.60** Find the response *v*(*t*) for *t* > 0 in the circuit in Fig. 16.83. Let *R* = 8 Ω, *L* = 2 H, and *C* = 125 mF. - -**16.61** Find the voltage *vo*(*t*) in the circuit of Fig. 16.84 by means of the Laplace transform. \* - -# **Figure 16.84** - -For Prob. 16.61. - -**16.62** Using Fig. 16.85, design a problem to help other - -students better understand solving for node voltages by working in the *s*-domain. - -# **Figure 16.85** - -For Prob. 16.62. - -**16.63** Consider the parallel *RLC* circuit of Fig. 16.86. Find *v*(*t*) and *i*(*t*) given that *v*(0) = 7.5 V and *i*(0) = −3 A. - -# **Figure 16.86** - -For Prob. 16.63. - -**16.64** The switch in Fig. 16.87 moves from position 1 to position 2 at *t* = 0. Find *v*(*t*), for all *t* > 0. - -# **Figure 16.87** - -For Prob. 16.64. - -**16.65** For the *RLC* circuit shown in Fig. 16.88, find the complete response if *v*(0) = 100 V when the switch is closed. - -# **Figure 16.88** - -For Prob. 16.65. - -\* An asterisk indicates a challenging problem. For Prob. 16.69. - -**16.66** For the op amp circuit in Fig. 16.89, find *v*0(*t*) for *t* > 0. Take *vs* = 12 *e*−5*t u*(*t*) V. - -# **Figure 16.89** - -For Prob. 16.66. - -**16.67** Given the op amp circuit in Fig. 16.90, if *v*1(0+) = 2 V and *v*2(0+) = 0 V, find *v*0 for *t* > 0. Let *R* = 100 kΩ and *C* = 1 *μ*F. - -# **Figure 16.90** - -For Prob. 16.67. - -**16.68** Obtain *V*0/*Vs* in the op amp circuit in Fig. 16.91. - -# **Figure 16.91** - -For Prob. 16.68. - -**16.69** Find *I*1(*s*) and *I*2(*s*) in the circuit of Fig. 16.92. - -# **Figure 16.93** - -For Prob. 16.70. - -**16.71** For the ideal transformer circuit in Fig. 16.94, determine *io*(*t*). - -# **Figure 16.94** - -For Prob. 16.71. - -# Section 16.4 Transfer Functions - -**16.72** The transfer function of a system is - -$$ -H(s) = \frac{s^2}{3s+1} -$$ - - Find the output when the system has an input of 14*e*−*t*∕3 *u*(*t*). - -- **16.73** When the input to a system is a unit step function, the response is 120 cos 2*tu*(*t*). Obtain the transfer function of the system. -- **16.74** Design a problem to help other students better -- understand how to find outputs when given a transfer function and an input. -- **16.75** When a unit step is applied to a system at *t* = 0, its response is - -*y*(*t*) = [6 + 0.75 *e*−3*t* − *e*−2*t* (3 cos 4*t* + 4.5 sin 4*t*)]*u*(*t*) - -What is the transfer function of the system? - -**16.76** For the circuit in Fig. 16.95, find *H*(*s*) = *Vo*(*s*)∕*Vs*(*s*). Assume zero initial conditions. - -**Figure 16.95** For Prob. 16.76. - -**16.77** Obtain the transfer function *H*(*s*) = *Vo*∕*Vs* for the circuit of Fig. 16.96. - -# **Figure 16.96** - -For Prob. 16.77. - -**16.78** The transfer function of a certain circuit is - -$$ -H(s) = \frac{10}{s+1} - \frac{6}{s+2} + \frac{12}{s+4} -$$ - -Find the impulse response of the circuit. - -**16.79** For the circuit in Fig. 16.97, find: - -# **Figure 16.97** - -For Prob. 16.79. - -**16.80** Refer to the network in Fig. 16.98. Find the following transfer functions: - -(a) -$$ -H_1(s) = V_o(s)/V_s(s) -$$ - -\n(b) $H_2(s) = V_o(s)/I_s(s)$ -\n(c) $H_3(s) = I_o(s)/I_s(s)$ -\n(d) $H_4(s) = I_o(s)/V_s(s)$ - -$$ -v_{s} \xrightarrow{I_{s}} 1 \Omega \xrightarrow{1 H} 1 \Omega \xrightarrow{I_{0}} -$$ -\n -$$ -1 \Gamma \xrightarrow{1 \Gamma} 1 \Gamma \xrightarrow{1 \Omega} \xleftarrow{1} v_{0} -$$ - -# **Figure 16.98** - -For Prob. 16.80. - -**16.81** For the op-amp circuit in Fig. 16.99, find the transfer function, *T*(*s*) = *I*(*s*)/*Vs*(*s*). Assume all initial conditions are zero. - -**Figure 16.99** For Prob. 16.81. - -**16.82** Calculate the gain *H*(*s*) = *Vo*∕*Vs* in the op amp circuit of Fig. 16.100. - -- **16.83** Refer to the *RL* circuit in Fig. 16.101. Find: - - (a) the impulse response *h*(*t*) of the circuit. - - (b) the unit step response of the circuit. - -**Figure 16.101** - -- For Prob. 16.83. -- **16.84** A parallel *RL* circuit has *R* = 4 Ω and *L* = 1 H. The input to the circuit is *is*(*t*) = 1.4*e*−*t u*(*t*) A. Find the inductor current *iL*(*t*) for all *t* > 0 and assume that *iL*(0) = −1.4 A. -- **16.85** A circuit has a transfer function - -s a transfer function -$$ -H(s) = \frac{3(s + 4)}{(s + 1)(s + 2)^2} -$$ - -Find the impulse response. - -# Section 16.5 State Variables - -- **16.86** Develop the state equations for Prob. 16.12. -- **16.87** Develop the state equations for the problem you designed in Prob. 16.13. -- **16.88** Develop the state equations for the circuit shown in Fig. 16.102. - -**Figure 16.102** For Prob. 16.88. - -**16.89** Develop the state equations for the circuit shown in Fig. 16.103. - -**16.90** Develop the state equations for the circuit shown in Fig. 16.104. - -For Prob. 16.90. - -**16.91** Develop the state equations for the following differential equation. - -$$ -\frac{d^2y(t)}{dt^2} + \frac{6\ dy(t)}{dt} + 7y(t) = z(t) -$$ - -**16.92** Develop the state equations for the following differential equation. \* - -$$ -\frac{d^2y(t)}{dt^2} + \frac{7\,dy(t)}{dt} + 9y(t) = \frac{dz(t)}{dt} + z(t) -$$ - -**16.93** Develop the state equations for the following differential equation. \* - -$$ -\frac{d^3y(t)}{dt^3} + \frac{6 \ d^2y(t)}{dt^2} + \frac{11 \ dy(t)}{dt} + 6y(t) = z(t) -$$ - -**16.94** Given the following state equation, solve for *y*(*t*): \* - -$$ -\dot{\mathbf{x}} = \begin{bmatrix} -4 & 4 \\ -2 & 0 \end{bmatrix} x + \begin{bmatrix} 0 \\ 2 \end{bmatrix} u(t) -$$ -$$ -\mathbf{y}(t) = \begin{bmatrix} 1 & 0 \end{bmatrix} x -$$ - -**16.95** Given the following state equation, solve for *y*1(*t*) and *y*2(*t*). \* - -$$ -\dot{\mathbf{x}} = \begin{bmatrix} -2 & -1 \\ 2 & -4 \end{bmatrix} x + \begin{bmatrix} 1 & 1 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} u(t) \\ 2u(t) \end{bmatrix} -$$ -$$ -\mathbf{y} = \begin{bmatrix} -2 & -2 \\ 1 & 0 \end{bmatrix} x + \begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} u(t) \\ 2u(t) \end{bmatrix} -$$ - -# Section 16.6 Applications - -**16.96** Show that the parallel *RLC* circuit shown in Fig. 16.105 is stable. - -# **Figure 16.105** - -For Prob. 16.96. - -**16.97** A system is formed by cascading two systems as shown in Fig. 16.106. Given that the impulse responses of the systems are - -$$ -h_1(t) = 21e^{-t}u(t), \qquad h_2(t) = e^{-4t}u(t) -$$ - -- (a) Obtain the impulse response of the overall system. -- (b) Check if the overall system is stable. - -**Figure 16.106** - -For Prob. 16.97. - -**16.98** Determine whether the op amp circuit in Fig. 16.107 is stable. - -**Figure 16.107** - -For Prob. 16.98. - -**16.99** It is desired to realize the transfer function - -$$ -\frac{V_2(s)}{V_1(s)} = \frac{2s}{s^2 + 2s + 6} -$$ - - using the circuit in Fig. 16.108. Choose *R* = 1 kΩ and find *L* and *C*. - -**Figure 16.108** For Prob. 16.99. - -**16.100** Design an op amp circuit, using Fig. 16.109, that will realize the following transfer function: - -$$ -\frac{V_o(s)}{V_i(s)} = -\frac{s + 1000}{2(s + 4000)} -$$ - -Choose *C*1 = 10 *μ*F; determine *R*1, *R*2, and *C*2. - -# **Figure 16.109** - -For Prob. 16.100. - -**16.101** Realize the transfer function - -$$ -\frac{V_o(s)}{V_s(s)} = -\frac{s}{s+10} -$$ - - using the circuit in Fig. 16.110. Let *Y*1 = *sC*1, *Y*2 = 1∕*R*1, *Y*3 = *sC*2. Choose *R*1 = 1 kΩ and determine *C*1 and *C*2. - -# **Figure 16.110** - -For Prob. 16.101. - -**16.102** Synthesize the transfer function - -ize the transfer function -\n -$$ -\frac{V_o(s)}{V_{in}(s)} = \frac{10^6}{s^2 + 100s + 10^6} -$$ - - using the topology of Fig. 16.111. Let *Y*1 = 1∕*R*1, *Y*2 = 1∕*R*2, *Y*3 = *sC*1, *Y*4 = *sC*2. Choose *R*1 = 1 kΩ and determine *C*1, *C*2, and *R*2. - -# Comprehensive Problems - -**16.103** Obtain the transfer function of the op amp circuit in Fig. 16.112 in the form of - -$$ -\frac{V_o(s)}{V_i(s)} = \frac{as}{s^2 + bs + c} -$$ - - where *a*, *b*, and *c* are constants. Determine the constants. - -**16.104** A certain network has an input admittance *Y*(*s*). The admittance has a pole at *s* = −3, a zero at *s* = −1, and *Y*(∞) = 0.25 S. - -- (a) Find *Y*(*s*). -- (b) An 8-V battery is connected to the network via a switch. If the switch is closed at *t* = 0, find the current *i*(*t*) through *Y*(*s*) using the Laplace transform. - -**16.105** A gyrator is a device for simulating an inductor in a network. A basic gyrator circuit is shown in - -**Figure 16.113** For Prob. 16.105. - -# **chapter** - -17 - -# The Fourier Series - -*Research is to see what everybody else has seen, and think what nobody has thought.* - -—Albert Szent Györgyi - -# Enhancing Your Skills and Your Career - -# **ABET EC 2000 criteria (3.j), "a knowledge of contemporary issues."** - -Engineers must have knowledge of contemporary issues. To have a truly meaningful career in the twenty-first century, you must have knowledge of contemporary issues, especially those that may directly af fect your job and/or work. One of the easiest ways to achieve this is to read a lot newspapers, magazines, and contemporary books. As a student enrolled in an ABET-accredited program, some of the courses you take will be directed toward meeting this criteria. - -# **ABET EC 2000 criteria (3.k), "an ability to use the techniques, skills, and modern engineering tools necessary for engineering practice."** - -The successful engineer must ha ve the "ability to use the techniques, skills, and modern engineering tools necessary for engineering practice." Clearly, a major focus of this te xtbook is to do just that. Learning to use skillfully the tools that facilitate your working in a modern "knowledge capturing integrated design environment" (KCIDE) is fundamental to your performance as an engineer . The ability to w ork in a modern KCIDE environment requires a thorough understanding of the tools associated with that environment. - -The successful engineer , therefore, must k eep abreast of the ne w design, analysis, and simulation tools. That engineer must also use those tools until he or she is comfortable with using them. The engineer also must make sure software results are consistent with real-world actualities. It is probably in this area that most engineers have the greatest difficulty. Thus, successful use of these tools requires constant learning and relearning the fundamentals of the area in which the engineer is working. - -Photo by Charles Alexander - -# Historical - -**Jean Baptiste Joseph Fourier** (1768–1830), a French mathematician, first presented the series and transform that bear his name. Fourier's results were not en thusiastically received by the scientific world. He could not even get his work published as a paper. - -Born in Auxerre, France, Fourier was orphaned at age 8. He attended a local military college run by Benedictine monks, where hedemonstrated great proficiency in mathematics. Like most of his contemporaries, Fourier was swept into the politics of the French Revolution. He played an important role in Napoleon's expeditions to Egypt in the later 1790s. Due to his political involvement, he narrowly escaped death twice. - -# Learning Objectives - -By using the information and exercises in this chapter you will be able to: - -- 1. Understand the trigonometric Fourier series and know how to determine the Fourier series with a variety of periodic functions. -- 2. Effectively use the Fourier series to analyze the response of circuits to a variety of periodic sources. -- 3. Know how the symmetrical characteristics of some wave shapes can make determining the Fourier series of classes of periodic functions easier to determine. -- 4. Understand how to determine average power and rms values associated with periodic functions. -- 5. Understand the use of the discrete Fourier transform and the fast Fourier transform. - -# **17.1** Introduction - -We have spent a considerable amount of time on the analysis of circuits with sinusoidal sources. This chapter is concerned with a means of ana lyzing circuits with periodic, nonsinusoidal e xcitations. The notion of periodic functions w as introduced in Chapter 9; it w as mentioned there that the sinusoid is the most simple and useful periodic function. This chapter introduces the F ourier series, a technique for e xpressing a periodic function in terms of sinusoids. Once the source function is expressed in terms of sinusoids, we can apply the phasor method to analyze circuits. - -The F ourier series is named after Jean Baptiste Joseph F ourier (1768–1830). In 1822, F ourier's genius came up with the insight that any practical periodic function can be represented as a sum of sinusoids. Such a representation, along with the superposition theorem, allo ws us to find the response of circuits to arbitrary periodic inputs using phasor techniques. - -We begin with the trigonometric F ourier series. Later we consider the exponential Fourier series. We then apply F ourier series in circuit analysis. Finally, practical applications of Fourier series in spectrum analyzers and filters are demonstrated. - -# **17.2** Trigonometric Fourier Series - -While studying heat flow, Fourier discovered that a nonsinusoidal periodic function can be expressed as an infinite sum of sinusoidal functions. Recall that a periodic function is one that repeats every *T* seconds. In other words, a periodic function *f* (*t*) satisfies - -$$ -f(t) = f(t + nT) -$$ - (17.1) - -where *n* is an integer and *T* is the period of the function. - -According to the *Fourier theorem*, any practical periodic function of angular frequenc y *ω*0 can be e xpressed as an infinite sum of sine or cosine functions that are inte gral multiples of *ω*0. Thus, *f*(*t*) can be expressed as - -$$ -f(t) = a_0 + a_1 \cos \omega_0 t + b_1 \sin \omega_0 t + a_2 \cos 2\omega_0 t + b_2 \sin 2\omega_0 t + a_3 \cos 3\omega_0 t + b_3 \sin 3\omega_0 t + \cdots -$$ - (17.2) - -or - -$$ -f(t) = a_0 + \sum_{n=1}^{\infty} (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t) -$$ - (17.3) - -where *ω*0 = 2*π*∕*T* is called the *fundamental angular frequency* in radians per second. The sinusoid sin *nω*0*t* or cos *nω*0*t* is called the *n*th harmonic of *f*(*t*); it is an odd harmonic if *n* is odd and an even harmonic if *n* is even. Equation 17.3 is called the *trigonometric Fourier series* of *f*(*t*). The constants *an* and *bn* are the *Fourier coefficients*. The coefficient *a*0 is the dc component or the average value of *f*(*t*). (Recall that sinusoids have zero average values.) The coefficients *an* and *bn* (for *n* ≠ 0) are the amplitudes of the sinusoids in the ac component. Thus, - -The Fourier series of a periodic function f (t) is a representation that resolves f (t) into a dc component and an ac component comprising an infinite series of harmonic sinusoids. - -A function that can be represented by a Fourier series as in Eq. (17.3) must meet certain requirements, because the infinite series in Eq. (17.3) may or may not converge. These conditions on *f*(*t*) to yield a convergent Fourier series are as follows: - -1. *f*(*t*) is single-valued everywhere. - -- 2. *f*(*t*) has a finite number of finite discontinuities in any one period. -- 3. *f*(*t*) has a finite number of maxima and minima in any one period. -- 4. The integral ∫ *t*0 *t*0+*T* ∣ *f*(*t*)∣ *dt* < ∞ for any *t*0. - -The harmonic angular frequency *ω*n is an integer multiple of the fundamental angular frequency *ω*0, i.e., *ω*n = n*ω*0. - -Historical note: Although Fourier published his theorem in 1822, it was P. G. L. Dirichlet (1805–1859) who later supplied an acceptable proof of the theorem. - -A software package like Mathcad or Maple can be used to evaluate the Fourier coefficients. - -These conditions are called *Dirichlet conditions*. Although they are not necessary conditions, the y are sufficient conditions for a Fourier series to exist. - -A major task in Fourier series is the determination of the Fourier coefficients *a*0, *an*, and *bn*. The process of determining the coefficients is called *Fourier analysis*. The following trigonometric inte grals are very helpful in Fourier analysis. For any integers *m* and *n*, - -$$ -\int_0^T \sin n\omega_0 t \, dt = 0 \tag{17.4a} -$$ - -$$ -\int_0^T \cos n\omega_0 t \, dt = 0 \tag{17.4b} -$$ - -$$ -\int_0^T \sin n\omega_0 t \cos m\omega_0 t \, dt = 0 \tag{17.4c} -$$ - -$$ -\int_0^T \sin n\omega_0 t \sin m\omega_0 t \, dt = 0, \qquad (m \neq n) \tag{17.4d} -$$ - -$$ -\int_0^T \cos n\omega_0 t \cos m\omega_0 t \, dt = 0, \qquad (m \neq n) \tag{17.4e} -$$ - -$$ -\int_0^T \sin^2 n\omega_0 t \, dt = \frac{T}{2} \tag{17.4f} -$$ - -$$ -\int_0^T \cos^2 n\omega_0 t \, dt = \frac{T}{2} \tag{17.4g} -$$ - -Let us use these identities to evaluate the Fourier coefficients. - -We begin by finding *a*0. We integrate both sides of Eq. (17.3) o ver one period and obtain - -$$ -\int_0^T f(t) dt = \int_0^T \left[ a_0 + \sum_{n=1}^\infty (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t) \right] dt -$$ - -= -$$ -\int_0^T a_0 dt + \sum_{n=1}^\infty \left[ \int_0^T a_n \cos n\omega_0 t dt + \int_0^T b_n \sin n\omega_0 t dt \right] dt -$$ - (17.5) - -Invoking the identities of Eqs. (17.4a) and (17.4b), the tw o inte grals involving the ac terms vanish. Hence, - -$$ -\int_0^T f(t) \, dt = \int_0^T a_0 \, dt = a_0 \, T -$$ - -or - -$$ -a_0 = \frac{1}{T} \int_0^T f(t) \, dt \tag{17.6} -$$ - -showing that *a*0 is the average value of *f*(*t*). - -To evaluate *an*, we multiply both sides of Eq. (17.3) by cos *mω*0*t* and integrate over one period: - -$$ -\int_0^T f(t) \cos m\omega_0 t \, dt -$$ -\n -$$ -= \int_0^T \left[ a_0 + \sum_{n=1}^\infty (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t) \right] \cos m\omega_0 t \, dt -$$ -\n -$$ -= \int_0^T a_0 \cos m\omega_0 t \, dt + \sum_{n=1}^\infty \left[ \int_0^T a_n \cos n\omega_0 t \cos m\omega_0 t \, dt \right] -$$ -\n -$$ -+ \int_0^T b_n \sin n\omega_0 t \cos m\omega_0 t \, dt \right] dt \tag{17.7} -$$ - -The inte gral containing *a*0 is zero in vie w of Eq. (17.4b), while the integral containing *bn* vanishes according to Eq. (17.4c). The inte gral containing *an* will be zero e xcept when *m* = *n*, in which case it is *T*∕2, according to Eqs. (17.4e) and (17.4g). Thus, - -$$ -\int_0^T f(t) \cos m\omega_0 t \, dt = a_n \frac{T}{2}, \qquad \text{for } m = n -$$ - -or - -$$ -a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t \, dt \qquad (17.8) -$$ - -In a similar vein, we obtain *bn* by multiplying both sides of Eq. (17.3) by sin *mω*0*t* and integrating over the period. The result is - -$$ -b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t \, dt \qquad (17.9) -$$ - -Be aware that because *f*(*t*) is periodic, it may be more convenient to carry the integrations above from −*T*∕2 to *T*∕2 or generally from *t*0 to *t*0 + *T* instead of 0 to *T*. The result will be the same. - -An alternative form of Eq. (17.3) is the *amplitude-phase* form - -$$ -f(t) = a_0 + \sum_{n=1}^{\infty} A_n \cos(n\omega_0 t + \phi_n) -$$ - (17.10) - -We can use Eqs. (9.11) and (9.12) to relate Eq. (17.3) to Eq. (17.10), or we can apply the trigonometric identity - -$$ -\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \tag{17.11} -$$ - -to the ac terms in Eq. (17.10) so that - -$$ -a_0 + \sum_{n=1}^{\infty} A_n \cos(n\omega_0 t + \phi_n) = a_0 + \sum_{n=1}^{\infty} (A_n \cos \phi_n) \cos n\omega_0 t -$$ - - -$$ --(A_n \sin \phi_n) \sin n\omega_0 t -$$ - (17.12) - -Equating the coef ficients of the series expansions in Eqs. (17.3) and (17.12) shows that - -$$ -a_n = A_n \cos \phi_n, \qquad b_n = -A_n \sin \phi_n \tag{17.13a} -$$ - -or - -$$ -A_n = \sqrt{a_n^2 + b_n^2}, \qquad \phi_n = -\tan^{-1}\frac{b_n}{a_n} -$$ - (17.13b) - -To avoid any confusion in determining *n*, it may be better to relate the terms in complex form as - -$$ -A_n / \underline{\phi_n} = a_n - jb_n \tag{17.14} -$$ - -The convenience of this relationship will become evident in Section 17.6. The plot of the amplitude *An* of the harmonics v ersus *nω*0 is called the *amplitude spectrum* of *f*(*t*); the plot of the phase *n* v ersus *nω*0 is the *phase spectrum* of *f*(*t*). Both the amplitude and phase spectra form the *frequency spectrum* of *f*(*t*). - -The frequency spectrum of a signal consists of the plots of the amplitudes and phases of the harmonics versus frequency. - -Thus, the F ourier analysis is also a mathematical tool for finding the spectrum of a periodic signal. Section 17.6 will elaborate more on the spectrum of a signal. - -To evaluate the Fourier coefficients *a*0, *an*, and *bn*, we often need to apply the following integrals: - -$$ -\int \cos at \, dt = \frac{1}{a} \sin at \tag{17.15a} -$$ - -$$ -\int \sin at \, dt = -\frac{1}{a} \cos at \tag{17.15b} -$$ - -$$ -\int t \cos at \, dt = \frac{1}{a^2} \cos at + \frac{1}{a} t \sin at \tag{17.15c} -$$ - -$$ -\int t \sin at \, dt = \frac{1}{a^2} \sin at - \frac{1}{a} \, t \cos at \tag{17.15d} -$$ - -It is also useful to kno w the v alues of the cosine, sine, and e xponential functions for inte gral multiples of *π*. These are given in Table 17.1, where *n* is an integer. - -For Example 17.1; a square wave. - -Determine the Fourier series of the waveform shown in Fig. 17.1. Obtain the amplitude and phase spectra. - -# **Solution:** - -The Fourier series is given by Eq. (17.3), namely, - -$$ -f(t) = a_0 + \sum_{n=1}^{\infty} (a_n \cos n\omega_0 t + b_n \sin n\omega_0 t) -$$ - (17.1.1) - -The frequency spectrum is also known as the line spectrum in view of the discrete frequency components. - -# **TABLE 17.1** - -Values of cosine, sine, and exponential functions for integral multiples of *π*. - -| Function | Value | -|-----------------|---------------------------------------------------------------| -| cos 2nπ | 1 | -| sin 2nπ | 0 | -| cos nπ | (−1)n | -| sin nπ | 0 | -| cos ___ nπ
2 | (−1)n∕2
,
n = even
0,
n = odd
{ | -| sin ___ nπ
2 | (−1)(n−1)∕2
,
n = odd
0,
{
n = even | -| e j2nπ | 1 | -| e jnπ | (−1)n | -| e jnπ∕2 | (−1)n∕2
,
n = even
n = odd
{
j(−1)(n−1)∕2
, | - -Our goal is to obtain the Fourier coefficients *a*0, *an*, and *bn* using Eqs. (17.6), (17.8), and (17.9). First, we describe the waveform as - -$$ -f(t) = \begin{cases} 1, & 0 < t < 1 \\ 0, & 1 < t < 2 \end{cases} \tag{17.1.2} -$$ - -and *f*(*t*) = *f*(*t* + *T*). Because *T* = 2, *ω*0 = 2*π*∕*T* = *π*. Thus, - -$$ -a_0 = \frac{1}{T} \int_0^T f(t) dt = \frac{1}{2} \left[ \int_0^1 1 dt + \int_1^2 0 dt \right] = \frac{1}{2} t \Big|_0^1 = \frac{1}{2} -$$ - (17.1.3) - -Using Eq. (17.8) along with Eq. (17.15a), - -$$ -a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t \, dt -$$ - -= $\frac{2}{2} \left[ \int_0^1 1 \cos n\pi t \, dt + \int_1^2 0 \cos n\pi t \, dt \right]$ -= $\frac{1}{n\pi} \sin n\pi t \Big|_0^1 = \frac{1}{n\pi} [\sin n\pi - \sin(0)] = 0$ (17.1.4) - -From Eq. (17.9) with the aid of Eq. (17.15b), - -$$ -b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t \, dt -$$ - -= $\frac{2}{2} \left[ \int_0^1 1 \sin n\pi t \, dt + \int_1^2 0 \sin n\pi t \, dt \right]$ -= $-\frac{1}{n\pi} \cos n\pi t \Big|_0^1$ (17.1.5) -= $-\frac{1}{n\pi} (\cos n\pi - 1)$ , $\cos n\pi = (-1)^n$ -= $\frac{1}{n\pi} [1 - (-1)^n] = \begin{cases} \frac{2}{n\pi}, & n = \text{odd} \\ 0, & n = \text{even} \end{cases}$ - -Substituting the Fourier coefficients in Eqs. (17.1.3) to (17.1.5) into Eq. (17.1.1) gives the Fourier series as - -$$ -f(t) = \frac{1}{2} + \frac{2}{\pi} \sin \pi t + \frac{2}{3\pi} \sin 3 \pi t + \frac{2}{5\pi} \sin 5 \pi t + \dots -$$ - (17.1.6) - -Given that *f*(*t*) contains only the dc component and the sine terms with the fundamental component and odd harmonics, it may be written as - -$$ -f(t) = \frac{1}{2} + \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin n\pi t, \qquad n = 2k - 1 -$$ - (17.1.7) - -By summing the terms one by one as demonstrated in Fig. 17.2, we notice how superposition of the terms can evolve into the original square. As more and more F ourier components are added, the sum gets closer and closer to the square wave. However, it is not possible in practice to sum the series in Eq. (17.1.6) or (17.1.7) to infinity. Only a partial sum (*n* = 1, 2, 3, … , *N*, where *N* is finite) is possible. If we plot the partial sum (or truncated series) o ver one period for a lar ge *N* as in - -Sum of first three ac components - -# **Figure 17.2** - -Evolution of a square wave from its Fourier components. - - Summing the Fourier terms by hand calculation may be tedious. A computer is helpful to compute the terms and plot the sum like those shown in Fig. 17.2. - -# **Figure 17.4** - -For Example 17.1: (a) amplitude and (b) phase spectrum of the function shown in Fig. 17.1. - -Truncating the Fourier series at *N* = 11; Gibbs phenomenon. - -Fig. 17.3, we notice that the partial sum oscillates abo ve and below the actual value of *f*(*t*). At the neighborhood of the points of discontinuity (*x* = 0, 1, 2, …), there is o vershoot and damped oscillation. In f act, an overshoot of about 9 percent of the peak value is always present, regardless of the number of terms used to approximate *f*(*t*). This is called the *Gibbs phenomenon*. - -Finally, let us obtain the amplitude and phase spectra for the signal in Fig. 17.1. Since *an* = 0, - -$$ -A_n = \sqrt{a_n^2 + b_n^2} = |b_n| = \begin{cases} \frac{2}{n\pi}, & n = \text{odd} \\ 0, & n = \text{even} \end{cases} -$$ - (17.1.8) - -and - -$$ -\phi_n = -\tan^{-1} \frac{b_n}{a_n} = \begin{cases} -90^\circ, & n = \text{odd} \\ 0, & n = \text{even} \end{cases} -$$ - (17.1.9) - -The plots of *An* and *n* for different values of *nω*0 = *nπ* provide the amplitude and phase spectra in Fig. 17.4. Notice that the amplitudes of the harmonics decay very fast with frequency. - -**Figure 17.5** For Practice Prob. 17.1. - -Find the Fourier series of the square wave in Fig. 17.5. Plot the ampli tude and phase spectra. - -# **Figure 17.6** - -the amplitude and phase spectra. - -# **Solution:** - -The function is described as - -$$ -f(t) = \begin{cases} t, & 0 < t < 1 \\ 0, & 1 < t < 2 \end{cases} -$$ - -Because *T* = 2, *ω*0 = 2*π*∕*T* = *π*. Then - -$$ -a_0 = \frac{1}{T} \int_0^T f(t) \, dt = \frac{1}{2} \left[ \int_0^1 t \, dt + \int_1^2 0 \, dt \right] = \frac{1}{2} \frac{t^2}{2} \Big|_0^1 = \frac{1}{4} \quad (17.2.1) -$$ - -To evaluate *an* and *bn*, we need the integrals in Eq. (17.15): - -$$ -a_n = \frac{2}{T} \int_0^T f(t) \cos n\omega_0 t \, dt -$$ - -= $\frac{2}{2} \left[ \int_0^1 t \cos n\pi t \, dt + \int_1^2 0 \cos n\pi t \, dt \right]$ -= $\left[ \frac{1}{n^2 \pi^2} \cos n\pi t + \frac{t}{n\pi} \sin n\pi t \right]_0^1$ -= $\frac{1}{n^2 \pi^2} (\cos n\pi - 1) + 0 = \frac{(-1)^n - 1}{n^2 \pi^2}$ (17.2.2) - -since cos *nπ* = (−1)*n* ; and - -$$ -b_n = \frac{2}{T} \int_0^T f(t) \sin n\omega_0 t \, dt -$$ - -= $\frac{2}{2} \left[ \int_0^1 t \sin n\pi t \, dt + \int_1^2 0 \sin n\pi t \, dt \right]$ -= $\left[ \frac{1}{n^2 \pi^2} \sin n\pi t - \frac{t}{n\pi} \cos n\pi t \right]_0^1$ -= $0 - \frac{\cos n\pi}{n\pi} = \frac{(-1)^{n+1}}{n\pi}$ (17.2.3) - -Substituting the Fourier coefficients just found into Eq. (17.3) yields - -$$ -f(t) = \frac{1}{4} + \sum_{n=1}^{\infty} \left[ \frac{[(-1)^n - 1]}{(n\pi)^2} \cos n\pi t + \frac{(-1)^{n+1}}{n\pi} \sin n\pi t \right] -$$ - -To obtain the amplitude and phase spectra, we notice that, for e ven harmonics, *an* = 0, *bn* = −1∕*nπ*, so that - -$$ -A_n / \underline{\phi_n} = a_n - jb_n = 0 + j\frac{1}{n\pi} \tag{17.2.4} -$$ - -Hence, - -$$ -A_n = |b_n| = \frac{1}{n\pi}, \qquad n = 2, 4, ... -$$ - -\n -$$ -\phi_n = 90^\circ, \qquad n = 2, 4, ... -$$ -\n(17.2.5) - -For odd harmonics, *an* = −2∕(*n*2 *π*2 ),*bn* = 1∕(*nπ*) so that - -$$ -A_n / \underline{\phi_n} = a_n - jb_n = -\frac{2}{n^2 \pi^2} - j \frac{1}{n \pi} -$$ - (17.2.6) - -That is, - -$$ -A_n = \sqrt{a_n^2 + b_n^2} = \sqrt{\frac{4}{n^4 \pi^4} + \frac{1}{n^2 \pi^2}} -$$ - -= $\frac{1}{n^2 \pi^2} \sqrt{4 + n^2 \pi^2}$ , $n = 1, 3, ...$ (17.2.7) - -From Eq. (17.2.6), we observe that lies in the third quadrant, so that - -$$ -\phi_n = 180^\circ + \tan^{-1} \frac{n\pi}{2}, \qquad n = 1, 3, ... \tag{17.2.8} -$$ - -From Eqs. (17.2.5), (17.2.7), and (17.2.8), we plot *An* and *n* for different values of *nω*0 = *nπ* to obtain the amplitude spectrum and phase spectrum as shown in Fig. 17.8. - -**Figure 17.8** For Example 17.2: (a) amplitude spectrum, (b) phase spectrum. - -# Practice Problem 17.2 - -**Figure 17.9** For Practice Prob. 17.2. - -Determine the Fourier series of the sawtooth waveform in Fig. 17.9. - -**Answer:** -$$ -f(t) = 4.5 - \frac{9}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \sin 2 \pi nt -$$ -. - -# **17.3** Symmetry Considerations - -We noticed that the Fourier series of Example 17.1 consisted only of the sine terms. One may w onder if a method e xists whereby one can kno w in advance that some F ourier coefficients would be zero and a void the unnecessary work involved in the tedious process of calculating them. Such a method does e xist; it is based on recognizing the e xistence of symmetry. Here we discuss three types of symmetry: (1) even symmetry, (2) odd symmetry, (3) half-wave symmetry. - -# **17.3.1** Even Symmetry - -A function *f*(*t*) is *even* if its plot is symmetrical about the vertical axis; that is, - -$$ -f(t) = f(-t) \tag{17.16} -$$ - -**(17.18)** - -Examples of e ven functions are *t* 2 , *t* 4 , and cos *t*. Figure 17.10 sho ws more e xamples of periodic e ven functions. Note that each of these examples satisfies Eq. (17.16). A main property of an e ven function *fe*(*t*) is that: - -$$ -\int_{-T/2}^{T/2} f_e(t) \, dt = 2 \int_0^{T/2} f_e(t) \, dt \tag{17.17} -$$ - -because integrating from −*T*∕2 to 0 is the same as inte grating from 0 to *T*∕2. Utilizing this property, the Fourier coefficients for an even function become - -$$ -a_0 = \frac{2}{T} \int_0^{T/2} f(t)dt -$$ - -\n -$$ -a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t dt -$$ - -\n -$$ -b_n = 0 -$$ - -Because *bn* = 0, Eq. (17.3) becomes a *Fourier cosine series.* This makes sense because the cosine function is itself e ven. It also mak es intuitive sense that an e ven function contains no sine terms gi ven that the sine function is odd. - -To confirm Eq. (17.18) quantitatively, we apply the property of an even function in Eq. (17.17) in evaluating the Fourier coefficients in Eqs. (17.6), (17.8), and (17.9). It is convenient in each case to integrate over the interval −*T*∕2 < *t* < *T*∕2, which is symmetrical about the origin. Thus, - -$$ -a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(t) dt = \frac{1}{T} \left[ \int_{-T/2}^{0} f(t) dt + \int_{0}^{T/2} f(t) dt \right] -$$ - (17.19) - -We change variables for the inte gral over the interval −*T*∕2 < *t* < 0 by letting *t* = −*x*, so that *dt* = −*dx*, *f*(*t*) = *f*(−*t*) = *f*(*x*), since *f*(*t*) is an even function, and when *t* = −*T*∕2, *x* = *T*∕2. Then, - -$$ -a_0 = \frac{1}{T} \left[ \int_{T/2}^0 f(x)(-dx) + \int_0^{T/2} f(t) dt \right] -$$ - -= -$$ -\frac{1}{T} \left[ \int_0^{T/2} f(x) dx + \int_0^{T/2} f(t) dt \right] -$$ - (17.20) - -showing that the two integrals are identical. Hence, - -$$ -a_0 = \frac{2}{T} \int_0^{T/2} f(t) dt -$$ - (17.21) - -as expected. Similarly, from Eq. (17.8), - -$$ -a_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \cos n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \cos n\omega_0 t \, dt \right] -$$ - (17.22) - -Typical examples of even periodic functions. - -We make the same change of v ariables that led to Eq. (17.20) and note that both *f*(*t*) and cos *nω*0*t* are even functions, implying that *f*(−*t*) = *f*(*t*) and cos(−*nω*0*t*) = cos *nω*0*t*. Equation (17.22) becomes - -$$ -a_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \cos(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right] -$$ - -= -$$ -\frac{2}{T} \left[ \int_{T/2}^0 f(x) \cos(n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right] -$$ - -= -$$ -\frac{2}{T} \left[ \int_0^{T/2} f(x) \cos(n\omega_0 x) dx + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right] -$$ -(17.23a) - -or - -t - -$$ -a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \tag{17.23b} -$$ - -as expected. For *bn*, we apply Eq. (17.9), - -$$ -b_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \sin n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \sin n\omega_0 t \, dt \right] \tag{17.24} -$$ - -We make the same change of variables but keep in mind that *f*(−*t*) = *f*(*t*) but sin(−*nω*0*t*) = −sin *nω*0*t*. Equation (17.24) yields - -$$ -b_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \sin(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right] -$$ - -= $\frac{2}{T} \left[ \int_{T/2}^0 f(x) \sin n\omega_0 x dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$ -= $\frac{2}{T} \left[ - \int_0^{T/2} f(x) \sin (n\omega_0 x) dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right]$ -= 0 (17.25) - -confirming Eq. (17.18). - -# **17.3.2** Odd Symmetry - -A function *f*(*t*) is said to be *odd* if its plot is antisymmetrical about the vertical axis: - -$$ -f(-t) = -f(t) \tag{17.26} -$$ - -Examples of odd functions are *t*, *t* 3 , and sin *t*. Figure 17.11 sho ws more examples of periodic odd functions. All these e xamples satisfy Eq. (17.26). An odd function *fo*(*t*) has this major characteristic: - -∫ - -$$ -f_{-T/2}^{T/2} f_o(t) dt = 0 -$$ -\n(17.27) - -f(t) - -(c) - -**Figure 17.11** Typical examples of odd periodic functions. - -because integration from −*T*∕2 to 0 is the negative of that from 0 to *T*∕2. With this property, the Fourier coefficients for an odd function become - -$$ -a_0 = 0, \t a_n = 0 -$$ - -$$ -b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt -$$ - (17.28) - -which give us a *Fourier sine series*. Again, this makes sense because the sine function is itself an odd function. Also, note that there is no dc term for the Fourier series expansion of an odd function. - -The quantitati ve proof of Eq. (17.28) follo ws the same proce dure tak en to pro ve Eq. (17.18) e xcept that *f*(*t*) is no w odd, so that *f*(*t*) = −*f*(*t*). With this fundamental b ut simple difference, it is easy to see that *a*0 = 0 in Eq. (17.20), *an* = 0 in Eq. (17.23a), and *bn* in Eq. (17.24) becomes - -$$ -b_n = \frac{2}{T} \left[ \int_{T/2}^0 f(-x) \sin(-n\omega_0 x)(-dx) + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right] -$$ - -$$ -= \frac{2}{T} \left[ -\int_{T/2}^0 f(x) \sin n\omega_0 x \, dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right] -$$ - -$$ -= \frac{2}{T} \left[ \int_0^{T/2} f(x) \sin(n\omega_0 x) \, dx + \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \right] -$$ - -$$ -b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt \qquad (17.29) -$$ - -as expected. - -It is interesting to note that an y periodic function *f*(*t*) with neither even nor odd symmetry may be decomposed into e ven and odd parts. Using the properties of even and odd functions from Eqs. (17.16) and (17.26), we can write - -$$ -f(t) = \underbrace{\frac{1}{2} [f(t) + f(-t)]}_{\text{even}} + \underbrace{\frac{1}{2} [f(t) - f(-t)]}_{\text{odd}} = f_e(t) + f_o(t) -$$ -(17.30) - -Notice that *fe*(*t*) = \_\_1 2 [ *f*(*t*) + *f*(−*t*)] satisfies the property of an even function in Eq. (17.16), while *fo*(*t*) = \_\_1 2 [ *f*(*t*) − *f*(−*t*)] satisfies the property of an odd function in Eq. (17.26). The fact that *fe*(*t*) contains only the dc term and the cosine terms, while *fo*(*t*) has only the sine terms, can be e xploited in grouping the F ourier series expansion of *f*(*t*) as - -$$ -f(t) = a_0 + \sum_{n=1}^{\infty} a_n \cos n\omega_0 t + \sum_{n=1}^{\infty} b_n \sin n\omega_0 t = f_e(t) + f_o(t) -$$ - (17.31) -even -odd - -It follows readily from Eq. (17.31) that when *f*(*t*) is e ven, *bn* = 0, and when *f*(*t*) is odd, *a*0 = 0 = *an*. - -Also, note the following properties of odd and even functions: - -- 1. The product of two even functions is also an even function. -- 2. The product of two odd functions is an even function. -- 3. The product of an e ven function and an odd function is an odd function. -- 4. The sum (or dif ference) of two even functions is also an even function. -- 5. The sum (or difference) of two odd functions is an odd function. -- 6. The sum (or difference) of an even function and an odd function is neither even nor odd. - -Each of these properties can be proved using Eqs. (17.16) and (17.26). - -# **17.3.3** Half-Wave Symmetry - -A function is half-wave (odd) symmetric if - -$$ -f\left(t - \frac{T}{2}\right) = -f(t) \tag{17.32} -$$ - -which means that each half-c ycle is the mirror image of the ne xt halfcycle. Notice that functions cos *nω*0*t* and sin *nω*0*t* satisfy Eq. (17.32) for odd values of *n* and therefore possess half-wave symmetry when *n* is odd. Figure 17.12 sho ws other examples of half-wave symmetric func tions. The functions in Figs. 17.11(a) and 17.11(b) are also half-w ave symmetric. Notice that for each function, one half-c ycle is the inverted version of the adjacent half-cycle. The Fourier coefficients become - -**Figure 17.12** Typical examples of half-wave odd symmetric functions. - -showing that the Fourier series of a half-wave symmetric function con tains only odd harmonics. - -To deri ve Eq. (17.33), we apply the property of half-w ave sym metric functions in Eq. (17.32) in e valuating the Fourier coefficients in Eqs. (17.6), (17.8), and (17.9). Thus, - -$$ -a_0 = \frac{1}{T} \int_{-T/2}^{T/2} f(t) dt = \frac{1}{T} \left[ \int_{-T/2}^{0} f(t) dt + \int_{0}^{T/2} f(t) dt \right] -$$ - (17.34) - -We change v ariables for the inte gral o ver the interv al −*T*∕2 < *t* < 0 by letting *x* = *t* + *T*∕2, so that *dx* = *dt*; when *t* = −*T*∕2, *x* = 0; and when *t* = 0, *x* = *T*∕2. Also, we k eep Eq. (17.32) in mind; that is, *f*(*x* − *T*∕2) = −*f* (*x*). Then, - -$$ -a_0 = \frac{1}{T} \left[ \int_0^{T/2} f\left(x - \frac{T}{2}\right) dx + \int_0^{T/2} f(t) dt \right] -$$ - -= $\frac{1}{T} \left[ - \int_0^{T/2} f(x) dx + \int_0^{T/2} f(t) dt \right] = 0$ (17.35) - -confirming the expression for *a*0 in Eq. (17.33). Similarly, - -$$ -a_n = \frac{2}{T} \left[ \int_{-T/2}^{0} f(t) \cos n\omega_0 t \, dt + \int_{0}^{T/2} f(t) \cos n\omega_0 t \, dt \right] \quad (17.36) -$$ - -We make the same change of variables that led to Eq. (17.35) so that Eq. (17.36) becomes - -$$ -a_n = \frac{2}{T} \left[ \int_0^{T/2} f\left(x - \frac{T}{2}\right) \cos n\omega_0 \left(x - \frac{T}{2}\right) dx + \int_0^{T/2} f(t) \cos n\omega_0 t \, dt \right] -$$ - (17.37) - -Because *f*(*x* − *T*∕2) = −*f*(*x*) and - -$$ -\cos n\omega_0 \left( x - \frac{T}{2} \right) = \cos(n\omega_0 t - n\pi) -$$ - -= $\cos n\omega_0 t \cos n\pi + \sin n\omega_0 t \sin n\pi$ (17.38) -= $(-1)^n \cos n\omega_0 t$ - -substituting these in Eq. (17.37) leads to - -$$ -a_n = \frac{2}{T} \left[ 1 - (-1)^n \right] \int_0^{T/2} f(t) \cos n\omega_0 t \, dt -$$ - -= -$$ -\begin{cases} \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t \, dt, & \text{for } n \text{ odd} \\ 0, & \text{for } n \text{ even} \end{cases} -$$ -(17.39) - -confirming Eq. (17.33). By following a similar procedure, we can derive *bn* as in Eq. (17.33). - -Table 17.2 summarizes the ef fects of these symmetries on the Fourier coef ficients. Table 17.3 pro vides the F ourier series of some common periodic functions. - -# **TABLE 17.2** - -# Effects of symmetry on Fourier coefficients. - -| Symmetry | a0 | an | bn | Remarks | -|-----------|--------|-----------|-----------|---------------------------------------------------------------| -| Even | a0 ≠ 0 | an ≠ 0 | bn = 0 | Integrate over T∕2 and multiply by 2 to get the coefficients. | -| Odd | a0 = 0 | an = 0 | bn ≠ 0 | Integrate over T∕2 and multiply by 2 to get the coefficients. | -| Half-wave | a0 = 0 | a2n = 0 | b2n = 0 | Integrate over T∕2 and multiply by 2 to get the coefficients. | -| | | a2n+1 ≠ 0 | b2n+1 ≠ 0 | | - -# **TABLE 17.3** - -Find the Fourier series expansion of *f*(*t*) given in Fig. 17.13. - -**Figure 17.13** For Example 17.3. - -# **Solution:** - -The function *f*(*t*) is an odd function. Hence *a*0 = 0 = *an*. The period is *T* = 4, and *ω*0 = 2*π*∕*T* = *π*∕2, so that - -$$ -b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt -$$ - -= $\frac{4}{4} \left[ \int_0^1 1 \sin \frac{n\pi}{2} t \, dt + \int_1^2 0 \sin \frac{n\pi}{2} t \, dt \right]$ -= $-\frac{2}{n\pi} \cos \frac{n\pi t}{2} \Big|_0^1 = \frac{2}{n\pi} \left( 1 - \cos \frac{n\pi}{2} \right)$ - -Hence, - -$$ -f(t) = \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \left( 1 - \cos \frac{n\pi}{2} \right) \sin \frac{n\pi}{2} t -$$ - -which is a Fourier sine series. - -Find the Fourier series of the function *f*(*t*) in Fig. 17.14. - -t f(t) ‒2*𝜋* ‒*𝜋 𝜋* 0 2*𝜋* 3*𝜋* 20 ‒20 - -**Figure 17.14** For Practice Prob. 17.3. - -**Answer:** -$$ -f(t) = -\frac{80}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin nt -$$ -, $n = 2k - 1$ . - -Practice Problem 17.3 - -Example 17.4 - -Determine the Fourier series for the half-w ave rectified cosine function shown in Fig. 17.15. - -A half-wave rectified cosine function; for Example 17.4. - -# **Solution:** - -This is an even function so that *bn* = 0. Also, *T* = 4, *ω*0 = 2*π*∕*T* = *π*∕2. Over a period, - -$$ -f(t) = \begin{cases} 0, & -2 < t < -1 \\ \cos \frac{\pi}{2} t, & -1 < t < 1 \end{cases} -$$ - -$$ -a_0 = \frac{2}{T} \int_0^{T/2} f(t) dt = \frac{2}{4} \left[ \int_0^1 \cos \frac{\pi}{2} t dt + \int_1^2 0 dt \right] -$$ - -$$ -= \frac{1}{2} \frac{2}{\pi} \sin \frac{\pi}{2} t \Big|_0^1 = \frac{1}{\pi} -$$ - -$$ -a_n = \frac{4}{T} \int_0^{T/2} f(t) \cos n\omega_0 t dt = \frac{4}{4} \left[ \int_0^1 \cos \frac{\pi}{2} t \cos \frac{n\pi t}{2} dt + 0 \right] -$$ - -Put $\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]$ . Then - -But cos *A* cos *B* = \_\_1 2 [cos(*A* + *B*) + cos(*A* − *B*)]. Then *an* = \_\_1 2 ∫ 0 1 [ cos \_\_ *π* 2 (*n* + 1)*t* + cos \_\_ *π* 2 (*n* − 1)*t* ] *dt* - -For *n* = 1, - -$$ -a_1 = \frac{1}{2} \int_0^1 \left[ \cos \pi t + 1 \right] dt = \frac{1}{2} \left[ \frac{\sin \pi t}{\pi} + t \right] \Big|_0^1 = \frac{1}{2} -$$ - -For *n* > 1, - -$$ -a_n = \frac{1}{\pi(n+1)} \sin \frac{\pi}{2} (n+1) + \frac{1}{\pi(n-1)} \sin \frac{\pi}{2} (n-1) -$$ - -For *n* = odd (*n* = 1, 3, 5, …), (*n* + 1) and (*n* − 1) are both even, so - -$$ -\sin \frac{\pi}{2}(n+1) = 0 = \sin \frac{\pi}{2}(n-1), \quad n = \text{odd} -$$ - -For *n* = even (*n* = 2, 4, 6, …), (*n* + 1) and (*n* − 1) are both odd. Also, - -$$ -\sin\frac{\pi}{2}(n+1) = -\sin\frac{\pi}{2}(n-1) = \cos\frac{n\pi}{2} = (-1)^{n/2}, \qquad n = \text{even} -$$ - -Hence, - -$$ -a_n = \frac{(-1)^{n/2}}{\pi(n+1)} + \frac{-(-1)^{n/2}}{\pi(n-1)} = \frac{-2(-1)^{n/2}}{\pi(n^2 - 1)}, \qquad n = \text{even} -$$ - -Thus, - -$$ -f(t) = \frac{1}{\pi} + \frac{1}{2}\cos\frac{\pi}{2}t - \frac{2}{\pi}\sum_{n=\text{even}}^{\infty}\frac{(-1)^{n/2}}{(n^2 - 1)}\cos\frac{n\pi}{2}t -$$ - -To avoid using *n* = 2, 4, 6, … and also to ease computation, we can replace *n* by 2*k*, where *k* = 1, 2, 3, … and obtain - -$$ -f(t) = \frac{1}{\pi} + \frac{1}{2}\cos\frac{\pi}{2}t - \frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^k}{(4k^2 - 1)}\cos k\pi t -$$ - -which is a Fourier cosine series. - -Find the Fourier series expansion of the function in Fig. 17.16. - -**Answer:** -$$ -f(t) = 16 - \frac{128}{\pi^2} \sum_{k=1}^{\infty} \frac{1}{n^2} \cos nt, n = 2k - 1. -$$ - -**Figure 17.16** For Practice Prob. 17.4. - -Calculate the Fourier series for the function in Fig. 17.17. - -# **Solution:** - -The function in Fig. 17.17 is half-wave odd symmetric, so that *a*0 = 0 = *an*. It is described over half the period as - -$$ -f(t) = t, \qquad -1 < t < 1 -$$ - -*T* = 4, *ω*0 = 2*π*∕*T* = *π*∕2. Hence, - -$$ -b_n = \frac{4}{T} \int_0^{T/2} f(t) \sin n\omega_0 t \, dt -$$ - -Instead of integrating *f*(*t*) from 0 to 2, it is more convenient to integrate from −1 to 1. Applying Eq. (17.15d), - -$$ -b_n = \frac{4}{4} \int_{-1}^{1} t \sin \frac{n\pi t}{2} dt = \left[ \frac{\sin n\pi t/2}{n^2 \pi^2/4} - \frac{t \cos n\pi t/2}{n\pi/2} \right] \Big|_{-1}^{1} -$$ - -= $\frac{4}{n^2 \pi^2} \left[ \sin \frac{n\pi}{2} - \sin \left( -\frac{n\pi}{2} \right) \right] - \frac{2}{n\pi} \left[ \cos \frac{n\pi}{2} - \cos \left( -\frac{n\pi}{2} \right) \right]$ -= $\frac{8}{n^2 \pi^2} \sin \frac{n\pi}{2}$ - -since sin(−*x*) = −sin *x* is an odd function, while cos( −*x*) = cos *x* is an even function. Using the identities for sin *nπ*∕2 in Table 17.1, - -$$ -b_n = \frac{8}{n^2 \pi^2} (-1)^{(n-1)/2}, \quad n = \text{odd} = 1, 3, 5, ... -$$ - -**Figure 17.17** For Example 17.5. - -Thus, - -$$ -f(t) = \sum_{n=1,3,5}^{\infty} b_n \sin \frac{n\pi}{2} t. -$$ - -# Practice Problem 17.5 - -Determine the Fourier series of the function in Fig. 17.12(a). Take *A* = 8 and *T* = 2*π*. - -**Answer:** -$$ -f(t) = \frac{16}{\pi} \sum_{k=1}^{\infty} \left( \frac{-2}{n^2 \pi} \cos nt + \frac{1}{n} \sin nt \right), n = 2k - 1. -$$ - -# **17.4** Circuit Applications - -We find that in practice, many circuits are driven by nonsinusoidal periodic functions. To find the steady-state response of a circuit to a nonsinusoidal periodic excitation requires the application of a Fourier series, ac phasor analysis, and the superposition principle. The procedure usually involves four steps. - -# Steps for Applying Fourier Series: - -- 1. Express the excitation as a Fourier series. -- 2. Transform the circuit from the time domain to the frequenc y domain. -- 3. Find the response of the dc and ac components in the Fourier series. -- 4. Add the individual dc and ac responses using the superposition principle. - -The first step is to determine the Fourier series e xpansion of the excitation. For the periodic v oltage source sho wn in Fig. 17.18(a), for example, the Fourier series is expressed as - -$$ -v(t) = V_0 + \sum_{n=1}^{\infty} V_n \cos(n\omega_0 t + \theta_n) -$$ - (17.40) - -(The same could be done for a periodic current source.) Equation (17.40) shows that *v*(*t*) consists of tw o parts: the dc component *V*0 and the ac component **V***n* = *Vn*⧸*θn* with several harmonics. This Fourier series representation may be re garded as a set of series-connected sinusoidal sources, with each source ha ving its own amplitude and frequency, as shown in Fig. 17.18(b). - -The third step is finding the response to each term in the Fourier series. The response to the dc component can be determined in the - -(a) Linear network excited by a periodic voltage source, (b) Fourier series representation (time-domain). - -frequency domain by setting *n* = 0 or *ω* = 0 as in Fig. 17.19(a), or in the time domain by replacing all inductors with short circuits and all capacitors with open circuits. The response to the ac component is obtained by applying the phasor techniques co vered in Chapter 9, as shown in Fig. 17.19(b). The network is represented by its impedance **Z**(*nω*0) or admittance **Y**(*nω*0). **Z**(*nω*0) is the input impedance at the source when *ω* is everywhere replaced by *nω*0, and **Y**(*nω*0) is the reciprocal of **Z**(*nω*0). - -Finally, following the principle of superposition, we add all the individual responses. For the case shown in Fig. 17.19, - -$$ -i(t) = i_0(t) + i_1(t) + i_2(t) + \cdots -$$ - -= $\mathbf{I}_0 + \sum_{n=1}^{\infty} |\mathbf{I}_n| \cos(n\omega_0 t + \psi_n)$ (17.41) - -where each component **I***n* with frequenc y *nω*0 has been transformed to the time domain to get *in*(*t*), and *ψn* is the argument of **I***n*. - -# **Figure 17.19** - -Steady-state responses: (a) dc component, (b) ac component (frequency domain). - -Let the function *f*(*t*) in Example 17.1 be the v oltage source *vs*(*t*) in the circuit of Fig. 17.20. Find the response *vo*(*t*) of the circuit. - -# **Solution:** - -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, - -> **I***o* = \_\_\_\_\_\_\_\_ 4 4 + *jωn*2 **I** = \_\_\_\_\_\_\_\_ **V** 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 - -**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*2 + 2*ab* + *b*2 , 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 -$$ -+ \sum_{n=1}^\infty \sum_{m=1}^\infty A_n A_m \frac{1}{T} \int_0^T \cos(n\omega_0 t + \phi_n) \cos(m\omega_0 t + \phi_m) dt -$$ -\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 2 = *a*0 2 +\_\_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* 0 2 is the power in the dc component, while \_\_1 2 ( *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. - -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. - -# **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* = √ \_\_\_\_\_\_ *a n* 2 + *b n* 2 \_\_\_\_\_\_\_\_ 2 ⧸ −tan−1 *bn*∕*an* **(17.61)** - -if only *an* > 0. Note that the phase *θn* 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 2 ∕*R*; if *f*(*t*) is the current through *R*, the power is *F*rms 2 *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*∣ ⧸*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. - -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. - -# **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 | -| C User Defined | C Restricted [analog] | -| OHz
to $10Hz$ | 10Hz
to TKHz | -| Scale | Processing Options | -| C Linear | $\nabla$ Fourier | -| $C$ Log | Performance Analysis | - -**Figure 17.35** *X* axis settings dialog box. - -| Transient | | -|---------------------------------|--------| -| Transient Analysis | | -| Print Step: | 0.01 | -| Final Time: | 12s | -| No-Print Delay: | | -| Step Ceiling: | 10ms | -| Detailed Bias Pt. | | -| Skip initial transient solution | | -| Fourier Analysis | | -| Ⅳ Enable Fourier | | -| Center Frequency: | 0.5 | -| Number of harmonics: | | -| Output Vars.: V(1) | | -| | | -| OK | 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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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*−0.5*t* − 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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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 | -| | | | | | | - -# **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
–106
GHz | -| X-rays | 108
–109
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 *ωc* 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 *ωc* ≪ *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 - -$$ -y(t) = \frac{1}{2} - \frac{1}{\pi} \sin 2\pi t -$$ - -**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:** -$$ -y(t) = -\frac{1}{3\pi} \sin 3\omega_0 t - \frac{1}{4\pi} \sin 4\omega_0 t - \frac{1}{5\pi} \sin 5\omega_0 t -$$ -. - -# **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 *θn* 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. - -# Review Questions - -- **17.1** Which of the following cannot be a Fourier series? - - (a) *t* − *t* 2 \_\_ 2 +*t* 3 \_\_ 3 − *t* 4 \_\_ 4 +*t* 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
(a) t + t | 2
(b) t
cos t | (c) et2 | -|------------------------|---------------------|---------| -| 2
4
(d) t
+ 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
(d) t
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 - -- **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.* - -# 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* 2 − 1 (c) cos *nπt* sin *nπt* (d) sin2 *πt* (e) *e*−*t* - -**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 *n* and *b*ʹ *n*, and *ω*ʹ *o*, 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*−*t* , 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) - - - -# **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. - -# 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 - -# 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 - -# 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. - -# **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). - -# **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 jω*0*t* , (c) cos *ω*0 *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 jω*0*t* 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 jω*0*t* , - -$$ -\mathcal{F}[e^{j\omega_0 t}] = 2\pi\delta(\omega - \omega_0) -$$ - (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*0 − *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*−*jω* − *e*−*j*2*ω*)∕*jω*, (b) 12*ej*2*ω*, (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) \_\_\_\_\_\_\_\_\_\_\_\_ *jω*. - -**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, - -$$ -F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt = \int_{0}^{\infty} e^{-at} e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(a+j\omega)t} dt -$$ -$$ -= \frac{-1}{a+j\omega} e^{-(a+j\omega)t} \Big|_{0}^{\infty} = \frac{1}{a+j\omega} -$$ - -*f*(*t*) = *e*−*atu*(*t*) = {*e*−*at*, - -For Example 18.3. - -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* (*ejω*0*t* − *e*−*jω*0*t* ). 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*0 *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*−(*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 jω*0*t* + *e*−*jω*0*t* ). 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*−*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*∣ , 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** - -# Properties of the Fourier transform. - -| Property | f(t) | F(ω) | -|-----------------|---------------------------|-------------------------------------| -| Linearity | a1
f1(t) + a2
f2(t) | a1F1(ω) + a2F2(ω) | -| Scaling | f(at) | ___1
ω
__
F(
a )
∣a∣ | -| Time shift | f(t − a) | e−jωaF(ω) | -| Frequency shift | e jω0t
f(t) | F(ω − ω0) | - -| TABLE 18.1 | (continued) | | -|---------------------------|-------------------------|-------------------------------------| -| Property | f(t) | F(ω) | -| Modulation | cos(ω0t)f(t) | __1
[F(ω + ω0) + F(ω − ω0)]
2 | -| Time differentiation | df
__
dt | jωF(ω) | -| | d n
f ___
dtn | n
( jω)
F(ω) | -| Time integration | t
f(t) dt
∫
−∞ | F(ω) _____
jω    + πF(0)δ(ω) | -| Frequency differentiation | n
t
f(t) | d n ____
n
( j)
dωn F(ω) | -| Reversal | f(−t) | F(−ω)
or
F*(ω) | -| Duality | F(t) | 2πf(−ω) | -| Convolution in t | f1(t) *
f2(t) | F1(ω)F2(ω) | -| Convolution in ω | f1(t)f2(t) | ___1
F1(ω) *
F2(ω)
2π | - -## **TABLE 18.2** - -# Fourier transform pairs. - -| f(t) | F(ω) | -|----------------------|-------------------------------------------------------| -| δ(t) | 1 | -| 1 | 2π δ(ω) | -| u(t) | π δ(ω) + ___1
jω | -| u(t + τ) − u(t − τ) | 2 ______ sin ωτ
ω | -| ∣t∣ | −2
___
ω2 | -| sgn(t) | ___2
jω | -| e−at u(t) | ______ 1
a + jω | -| eat u(−t) | ______ 1
a − jω | -| n
e−at u(t)
t | __________ n!
n+1
(a + jω) | -| e−a∣t∣ | _______ 2a
a2
+ 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
2
2
(a + jω)
+ ω 0 | -| e−at cos ω0
tu(t) | a + jω
____________
2
2
(a + jω)
+ 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*−*a*∣*t*∣ , 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* −*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 - -$$ -G(\omega) = \frac{\omega^2 + 21}{\omega^2 + 9} = 1 + \frac{12}{\omega^2 + 9} -$$ - -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*−4*t* )*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. - -# **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*2 (*t*) = *i* 2 (*t*) = *f* 2 (*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 f (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*2 + *y*2 = ∣*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*1 < *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- Ω resistor is *v*(*t*) = 5*e* Example 18.9 −3*t 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- Ω resistor with *i*(*t*) = 10*e*−2∣*t*∣ 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*−2*t u*(*t*). - -# **Solution:** - -Given that *f*(*t*) = *v*(*t*) = *e*−2*t 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 - -$$ -W = \frac{1}{\pi} \int_0^{10} |F(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^{10} \frac{d\omega}{4 + \omega^2} = \frac{1}{\pi} \left( \frac{1}{2} \tan^1 \frac{\omega}{2} \Big|_0^{10} \right) -$$ -$$ -= \frac{1}{2\pi} \tan^{-1} 5 = \frac{1}{2\pi} \left( \frac{78.69^\circ}{180^\circ} \pi \right) = 0.218 \text{ J} -$$ - -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 −*t 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). - -# **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) - -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 - -$$ -F(\omega) = \mathcal{F}[f(t)] = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt -$$ - -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 *ej*2*t* 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*−*jω* \_\_\_\_\_\_ 2 + *jω* is (a) *e*−2*t* (b) *e*−2*t u*(*t* − 1) (c) *e*−2(*t*−1) (d) *e*−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*2 *dω* results in: - -(a) 0 (b) 2 (c) 2.5 (d) ∞ - -**18.7** The integral ∫ −∞ ∞ 10*δ*(*ω*− 1) \_\_\_\_\_\_\_\_\_\_ 4 + *ω*2 *dω* gives: (a) 0 (b) 2 (c) 2.5 (d) ∞ **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
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
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*−3*t* sin(10*t*)*u*(*t*) (b) *f*2(*t*) = *e*−4*t* 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*−*jω* − 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) $G(\omega) = 2u(\omega + 1) - 2u(\omega - 1)$ - -\* 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) $i(t) = tf(t)$ - -# 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*−4*t 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*−*t 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*−2*t 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*−*t 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. - -# 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*−(*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*−2*t 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. - -# 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*−2*t* − 6*e*−4*t* V. Find the output when the input is *vi*(*t*) = 4*e*−*t u*(*t*) V. - -**Figure 18.54** For Prob. 18.9. - -# **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. \* - -# Comprehensive Problems - -**18.68** The voltage signal at the input of a filter is *v*(*t*) = 50*e*−2∣*t*∣ 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 - -# 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 - -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. - -# **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. - -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 **z**21 = Open-circuit transfer impedance from port 2 to port 1 **(19.4)** -- **z**22 = Open-circuit output impedance - -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**21 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. - -# **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) - -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**1 \_\_\_\_\_\_\_ − **V***o* 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, - -$$ -\frac{\mathbf{V}_o - 0}{4} + 2\mathbf{I}_1 + \mathbf{I}_2 = 0 -$$ - -$$ --I_2 = 0.25V_o - 1.5V_o = -1.25V_o -$$ - -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**12 ≠ **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. - -# **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**22 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 **g**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, - -$$ -\mathbf{I}_1 = \frac{\mathbf{V}_1}{s+1} -$$ - -$$ -\mathbf{g}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} = \frac{1}{s+1} -$$ - -By voltage division, - -**V**2 = \_\_\_\_\_ 1 *s* + 1 **V**1 - -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 = − \_\_\_\_\_ 1 *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**]=[ \_\_\_\_\_\_\_\_\_\_ *s* + 2 *s* 2 + 3*s* + 1 −\_\_\_\_\_\_\_\_\_\_ 1 *s* 2 + 3*s* + 1 \_\_\_\_\_\_\_\_\_\_ 1 *s* 2 + 3*s* + 1 *s*(*s* + 2) \_\_\_\_\_\_\_\_\_\_ *s* 2 + 3*s* + 1 ]. - -# **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**1 **V**2 - -(a) - -1/s s - -**I**1 - -+ - -+ ‒ **I I**1 2 = 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 **I**1 ] =[ **A B C D**] [ **V**2 −**I**2 ] = [**T**] [ **V**2 −**I**2 ] **(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 **b** = 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**21 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. - -## **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), - -$$ -\mathbf{V}_1 = \mathbf{z}_{11}\mathbf{I}_1 + \mathbf{z}_{12}\mathbf{I}_2 \tag{19.37a} -$$ - -$$ -V_2 = z_{21}I_1 + z_{22}I_2 \tag{19.37b} -$$ - -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
___
∆y | y12
− ___
∆y | ∆h
___
h22 | h12
___
h22 | ___1
g11 | g12
− ___
g11 | A
__
C | ∆T
___
C | d
__
c | __1
c | -| | z21 | z22 | y21
− ___
∆y | y11
___
∆y | h21
− ___
h22 | ___1
h22 | g21
___
g11 | ∆g
___
g11 | __1
C | D
__
C | ∆t
__
c | __a
c | -| y | z22
___
∆z | z12
− ___
∆z | y11 | y12 | ___1
h11 | h12
− ___
h11 | ∆g
___
g22 | g12
___
g22 | D
__
B | ∆T
− ___
B | __a
b | − __1
b | -| | z21
− ___
∆z | z11
___
∆z | y21 | y22 | h21
___
h11 | ∆h
___
h11 | g21
− ___
g22 | ___1
g22 | − __1
B | A
__
B | ∆t
− __
b | d
__
b | -| h | ∆z
___
z22 | z12
___
z22 | ___1
y11 | y12
− ___
y11 | h11 | h12 | g22
___
∆g | g12
− ___
∆g | B
__
D | ∆T
___
D | b
__
a | __1
a | -| | z21
− ___
z22 | ___1
z22 | y21
___
y11 | ∆y
___
y11 | h21 | h22 | g21
− ___
∆g | g11
___
∆g | − __1
D | C
__
D | ∆t
__
a | __c
a | -| g | ___1
z11 | z12
− ___
z11 | ∆y
___
y22 | y12
___
y22 | h22
___
∆h | h12
− ___
∆h | g11 | g12 | C
__
A | ∆T
− ___
A | __c
d | − __1
d | -| | z21
___
z11 | ∆z
___
z11 | y21
− ___
y22 | ___1
y22 | h21
− ___
∆h | h11
___
∆h | g21 | g22 | __1
A | B
__
A | ∆t
__
d | b
− __
d | -| T | z11
___
z21 | ∆z
___
z21 | y22
− ___
y21 | − ___1
y21 | ∆h
− ___
h21 | h11
− ___
h21 | ___1
g21 | g22
___
g21 | A | B | __d
∆t | __b
∆t | -| | ___1
z21 | z22
___
z21 | − ∆y
___
y21 | y11
− ___
y21 | h22
− ___
h21 | − ___1
h21 | g11
___
g21 | ∆g
___
g21 | C | D | __c
∆t | __a
∆t | -| t | z22
___
z12 | ∆z
___
z12 | y11
− ___
y12 | − ___1
y12 | ___1
h12 | h11
___
h12 | − ∆g
___
g12 | g22
− ___
g12 | ___ D
∆T | ___B
∆T | a | b | -| | ___1
z12 | z11
___
z12 | − ∆y
___
y12 | y22
− ___
y12 | h22
___
h12 | ∆h
___
h12 | g11
− ___
g12 | − ___1
g12 | ___ C
∆T | ___ A
∆T | c | d | - -**∆***z* = **z**11**z**22 − **z**12**z**21, **∆***h* = **h**11**h**22 − **h**12**h**21, **∆***T* = **AD** − **BC** - -**∆***y* = **y**11**y**22 − **y**12**y**21, **∆***g* = **g**11**g**22 − **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 2 ] Ω, [**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 - -$$ -I_1 = 0V_1 + 0V_2 \tag{19.11.1} -$$ - -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:** [ **z**] = [ *R*1 −*R*2 0 0] . Because [ **z**] −1 does not e xist, [ **y**] does not exist. - -**Figure 19.38** For Practice Prob. 19.11. - -# **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**1*a* = **I**1*b*, **I**2 = **I**2*a* = **I**2*b* **(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**1*a* + **I**1*b*, **I**2 = **I**2*a* + **I**2*b* **(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. - -Notice that - -$$ -\Delta_{T_a}=\Delta_{T_b}=\Delta_T=1 -$$ - -showing that the network is reciprocal. - -# **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, - -$$ -\mathbf{z}_{12} = \frac{\mathbf{V}_1}{1} = 19.70 \underline{\text{/} 175.7^{\circ}} \,\Omega, \qquad \mathbf{z}_{22} = \frac{\mathbf{V}_2}{1} = 19.56 \underline{\text{/} 175.7^{\circ}} \,\Omega -$$ - -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 s* = 0.8 k Ω and *R L* = 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. - -# **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**1 in Fig. 19.63(a) or by confirming the required *y*21. - -C2 - -ZB - -L1 L3 - -(a) - -L1 L3 - -C2 **V**2 1 Ω - -+ - -‒ - -(b) - -y22 = - -1 ZA - -**Figure 19.63** For Example 19.18. - -**V**1 + - -‒ - -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. - -# **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. - -# 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**2 **V**2 = 30**I**1 + 20**I**2 - -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.* - -# 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*2 = 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* = 106 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. - - - -# **Figure 19.133** - -For Prob. 19.97. - -For Prob. 19.98. **19.98** A two-stage amplifier in Fig. 19.134 contains two 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** - -# 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. - -# 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 - -$$ -\begin{aligned}\nx_1 &= \frac{\Delta_1}{\Delta} \\ -x_2 &= \frac{\Delta_2}{\Delta} \\ -&\vdots \\ -x_n &= \frac{\Delta_n}{\Delta}\n\end{aligned} -$$ -\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 ∆ *k* 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. - -$$ -= a_{11}a_{22}a_{33} + a_{21}a_{32}a_{13} + a_{31}a_{12}a_{23} - a_{13}a_{22}a_{31} - a_{23}a_{32}a_{11} -$$ - --a33a12a21 (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*1 − *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 | | -| [
−5 | −4 | ]
9 | [
]
x3 | | [
]
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*1 − *x*2 − 2*x*3 = 1 −*x*1 + 6*x*2 − 3*x*3 = 0 −2*x*1 − 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**−1 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**−1 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)*i+j* 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. - -$$ -x_1 + x_2 + x_3 = 5 -$$ - --x1 + 2x2 = 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. - -$$ -y_1 - y_3 = 1 -$$ - -2y1 + 3y2 - y3 = 1 -$$ -y_1 - y_2 - y_3 = 3 -$$ - -**Answer:** *y*1 *=* 6*, y*2 *= −*2, *y*3 *=* 5. - -# 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) - -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*ej*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 j = 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*−*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 j*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 j*306.9° , (b) 13⧸ 67.38°, 13*ej*67.38° , (c) 9.487⧸ 251.6°, 9.487*ej*251.6° , (d) 7.071⧸ 171.9°, 7.071*e j*171.9° . - -Convert the following complex numbers into rectangular form: Example B.2 (a) 12⧸ −60°, (b) −50⧸ 285°, (c) 8*e j*10° , (d) 20*e*−*jπ*∕3 . - -# **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* −*j*30° , (d) 50*e jπ*∕2 . - -**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⧸*θ*1 and *z*2 = *x*2 + *jy*2 = *r*2 ⧸*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 − *j*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*2 ∕*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) √ Example B.5 \_\_ *A* , (b) *A*4 . - -# **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*1∕3 - -**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 - -# 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*2 + *bx* + *c* = 0 are - -$$ -x_1, x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} -$$ - -# **C.2** Trigonometric Identities - -| 1
csc x = ____
sin x
cot x = _____ 1
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
− y) | -| 2 cos x cos y = cos(x + y) + cos(x
− 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 -$$ -\cos^2 x = \frac{1}{2} (1 + \cos 2x) -$$ - -\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. - -# 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*−30*t* [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*−4*t 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 -- **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
out), P60W = 9.6 W, P100W = 16 W. The best way to | | | -| | wire the bulbs is to connect the 100-W bulb in series
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
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)
(b) 24.61 V (with), 26.39 V (without)
(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
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
20 Ω, −84 V | | -| 4.41
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
10 Ω, 166.67 V, 16.667 A | R____
1R3
5.15 | -| 4.59
22.5 Ω, 40 V, 1.7778 A | (a) −(R1 + R3 +
), (b) −92 kΩ
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
V0
= 24 − 5I0 | 5.21
−3 V | -| 4.67
25 kΩ, 49 mW | Rf
___
5.23
−
R1 | -| 4.69 | 5.25 | -| ∞ (theoretically) | 9.375 V | -| 4.71 | 5.27 | -| 8 kΩ, 1.152 W | 2.7 V | -| 4.73
20.77 W | R___2
5.29
R1 | -| 4.75 | 5.31 | -| 250 Ω, 12 mW | 4.545 mA | -| 4.77
(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*5 − *R*4∕*R*6 1 − *R*2*R*4∕*R*3*R*5 **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*1 *v*2 *v*3 *v*4 *v*5 *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*1 − (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*−3*t* A, 30*t*(1 − 3*t*)*e*−6*t* 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*−*t* mA, 2*e*−*t* mA (b) −20*e*−*t 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 × 10−5 (*e*−200*t* − 1) + 4 A, 1.25 × 10−5 (*e*−200*t* − 1) − 2 A (d) 6.25 × 10−5 (*e*−200*t* − 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*−*t* V for 0 < *t* < 1 sec, 5.518*e*−2(*t*−1) V for 1 sec < *t* < ∞ -- **7.9** 10*e*−*t*/12 V - -- **7.11** 1.2*e*−3*t* 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*−2*t* ] V for all *t* > 0. -- **7.19** 5*e*−5*t u*(*t*) A -- **7.21** 1.618 Ω -- **7.23** 10*e*−4*t* V, *t* > 0, 2.5*e*−4*t* 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 −9 , (b) 7 -- **7.33** 4.5*u*(*t* − 2) A -- **7.35** (a) −*e* −2*t u*(*t*) V, (b) 2*e*1.5*t u*(*t*) A -- **7.37** (a) 4 s, (b) 10 V, (c) (10 − 8*e*−*t*∕4 ) *u*(*t*) V -- **7.39** (a) 4 V, *t* < 0, 20 −16*e* −*t*∕8 , *t* > 0, (b) 4 V, *t* < 0, 12 − 8*e*−*t*∕6 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*−14.286*t* ] *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*−*t*∕2 *u*(*t*) A, (b) 6 A, 6*e*−2*t*∕3 *u*(*t*) A -- **7.55** 96 V, 96*e*−4*t u*(*t*) V -- **7.57** 2.4*e*−2*t u*(*t*) A, 600*e*−5*t u*(*t*) mA -- **7.59** 120*e*−4*t u*(*t*) volts -- **7.61** 20*e*−8*t u*(*t*) V, (10 − 5*e*−8*t* )*u*(*t*) A -- **7.63** 2*e*−8*t u*(*t*) A, −8*e*−8*t u*(*t*) V -- **7.65** { 2(1 − *e*−2*t* )A 1.729*e*−2(*t*−1)A 0 < *t* < 1 *t* > 1 -- **7.67** 10*e*–*t*/6*u*(*t*) V -- **7.69** 48(*e*−*t*∕3000−1) *u*(*t*) V -- **7.71** [−5 + 5*e*–*t* ]*u*(*t*) V -- **7.73** −9*e*−5*t u*(*t*) V -- **7.75** [20 10*e*–*t* ]*u*(*t*) V, 100*μ*A -- **7.77** See Fig. D.15. - -**7.79** [1.75 – 0.75*e*–2*t* ]*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*−5*t* ] A **8.11** [(10 + 10*t*)*e*−*t* ] V -- **8.13** 120 Ω -- **8.15** 750 Ω, 200 *μ*F, 25 H -- **8.17** [21.55*e*−2.679*t* − 1.55*e*−37.32*t* ] V -- **8.19** 24 sin(0.5*t*) V -- **8.21** 18*e*−*t* − 2*e*−9*t* 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*−2*t* ] volts -- **8.29** (a) 3 − 3 cos 2*t* + sin 2*t* V, (b) 2 − 4*e*−*t* + *e*−4*t* A, - -- (c) 3 + (2 + 3*t*)*e*−*t* V, (d) 2 + 2 cos 2*te*−*t* A - - **8.31** 80 V, 40 V - - **8.33** [30 + 0.3078*e*−4.95*t* − 15.308*e*−0.05*t* ] V -- **8.35** This is a design problem with multiple answers. -- **8.37** 5*e*−4*t* A -- **8.39** (−60 + [−0.2102*e*−47.83*t* + 60.21*e*−0.167*t* ]) V -- **8.41** [8.7 sin(4.583*t*)*e*–2*t* ]*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*−*t*∕2 ] A, [7.559 sin(1.3229*t*)*e*−*t*∕2 ] V - - **8.47** (400*te*−10*t* ) V - - **8.49** {9 + [(3 + 6*t*)*e*–2*t* ]} *u*(*t*) A - - **8.51** [ − *i* \_\_\_\_0 *oC* sin(*ot*) ] V where *o* = 1∕ √ \_\_\_ LC - - **8.53** (*d*2 *i*∕*dt*2 ) + 1.25(*di*∕*dt*) + 400*i* = 200 - - **8.55** 2*e*–*t*/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*t* V -- **8.61** 2.4 2.667*e*−2*t* + 0.2667*e*−5*t* A, 9.6 – 16*e*−2*t* + 6.4*e*−5*t* 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*−*t 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** *d*2 \_\_\_*v dt*2 + \_\_ *R L* \_\_\_ *dv dt* + \_\_\_*R LC iD* + \_\_1 *C* \_\_\_ *diD dt* = \_\_\_ *vs 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*2 -- **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(106 *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 × 103 *t* + 4.06°) mA -- **10.7** 124.08⧸−154° V - -- **10.9** 6.154 cos(103 *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, − *j* \_\_ *R* √ \_\_ \_\_*L C* , (b) 0, 1, *j* \_\_ *R* √ \_\_ \_\_*L C* **10.23** (1 − *ω*2 *LC*)*Vs* \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_1− *ω*2 *LC* + *jωRC*(2 − *ω*2 *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,
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
________________________
(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,
(b) 40 kΩ | | | -| 10.93 Proof | | | | -| 10.95 Proof | | | | -| | Chapter 11 | | | -| | (Assume all values of currents and voltages are rms unless
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,
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,
(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,
average power = 339.4 W,
reactive power = 339.4 VAR
(b) S = (678.8 – j678.8) VA,
average power = 678.8 W,
reactive power = −678.8 VAR
(c) S = (7.637 + j7.637) kVA, average power =
7.637 W, reactive power = 7.637 VAR
(d) S = (250 + j433) kVA, average power =
250 kW, reactive power = 433 kVAR | -| | 11.49 (a) 4 + j2.373 kVA,
(b) 1.6 – j1.2 kVA,
(c) 0.4624 + j1.2705 kVA,
(d) 110.77 + j166.16 VA | -| | 11.51 (a) 0.9956 (lagging),
(b) 304 W,
(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* 2 + 4*s* + 1) - -14.5 -$$ -\frac{(Ls + R)}{(LCs^2 + RCs + 1)}. -$$ - -**14.7** (a) 1.0116, (b) 0.5623, (c) 5.623 × 1010 - -. - -**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 × 10−4 , *Km* = 5 × 10−3 -- **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** *s* 3 *LRLC*1*C*2 \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (*sRiC*1 + 1)(*s* 2 *LC*2 + *sRLC*2 + 1) + *s* 2 *LC*1(*sRLC*2 + 1) -- **14.99** 8.165 MHz, 4.188 × 106 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*−*t u*(*t*), (b) 3*δ*(*t*) − 11*e*−4*t u*(*t*), (c) (2*e*−*t* − 2*e*−3*t* )*u*(*t*), (d) (3*e*−4*t* − 3*e*−2*t* + 6*te*−*2t* )*u*(*t*) -- **15.29** [1 + 2*e*−*t* cos (*t* + 90°)] *u*(*t*) -- **15.31** (a) (−5*e*−*t* + 20*e*−2*t* − 15*e*−3*t* )*u*(*t*) (b) (−*e*−*t* +(1 + 3*t* − *t* 2 \_\_ 2 )*e*−2*t* ) *u*(*t*), (c) (−0.2*e*−2*t* + 0.2*e*−*t* cos(2*t*) + 0.4*e*−*t* sin(2*t*))*u*(*t*) -- **15.33** (a) (3*e*−*t* + 3 sin(*t*) − 3 cos(*t*))*u*(*t*), (b) cos(*t* −*π*)*u*(*t* − *π*), (c) 8 [1 − *e*−*t* − *te*−*t* − 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*−*t* − 7.5*e*–3*t* ]*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*−*t* + 36*e*−2*t* − 31.5*e*−3*t* ]*u*(*t*) -- **15.61** (a) [3 + 3.162 cos (2*t* − 161.12°)]*u*(*t*) volts, (b) [2 − 4*e*−*t* + *e*−4*t* ]*u*(*t*) amps, (c) [3 + 2*e*−*t* + 3*te*−*t* ]*u*(*t*) volts, (d) [2 + 2*e*−*t* cos(2*t*)]*u*(*t*) amps - -# Chapter 16 - -- **16.1** [(7 + 35*t*)*e*−5*t* ] *u*(*t*) A -- **16.3** [(20 + 20*t*)*e*−*t* ]*u*(*t*) V -- **16.5** 750 Ω, 25 H, 200 *μ*F -- **16.7** [6 + 12*e*−*t* cos(2*t*) + 2 sin(2*t*))]*u*(*t*) A -- **16.9** [3 + 5.924*e*−1.5505*t* − 1.4235*e*−6.45*t* ]*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*−2*t* −\_\_\_2 √ \_\_ 7 *e*−0.5*t* sin ( √ \_\_ \_\_\_7 2 *t* )) *u*(*t*) A -- **16.19** [−2.333*e*−*t*∕2 + 2.333*e*−2*t* ]*u*(*t*) volts -- **16.21** [10.776 *e*−2.679*t* − 0.774*e*−37.32*t* ]*u*(*t*) volts -- **16.23** 24 cos(0.5*t* + 90°)*u*(*t*) volts -- **16.25** [45*e*−*t* − 5*e*−9*t* ]*u*(*t*) volts -- **16.27** [30 − 15.309*e*−0.05051*t* + 0.3078*e*−4.949*t* ]*u*(*t*) volts -- **16.29** 17.5 cos(8*t* + 90°)*u*(*t*) amps -- **16.31** [−16 + 66.67*e*−0.8*t* cos(0.6*t* − 53.13°)]*u*(*t*) volts, 13.333*e*−0.8*t* [cos(0.6*t* + 90°)]*u*(*t*) amps -- **16.33** This is a design problem with multiple answers. -- **16.35** [9.091*e*−*t* + 19.653*e*−0.0625*t* cos(0.7044*t* − 117.55°] *u*(*t*) V. -- **16.37** [−60 + 60.21*e*−0.1672*t* − 0.21*e*−47.84*t* ]*u*(*t*) volts -- **16.39** [4.364*e*−2*t* cos(4.583*t* − 90°)]*u*(*t*) amps -- **16.41** [100*te*−10*t* ]*u*(*t*) volts -- **16.43** [9 + 9*e*−2*t* + 6*te*−2*t* ]*u*(*t*) amps -- **16.45** [*io*∕(*ωC*)] cos(*ωt* + 90°)*u*(*t*) volts -- **16.47** [60 − 40*e*−0.6*t* cos(0.2*t*) − sin(0.2*t*))]*u*(*t*) A -- **16.49** [1.0714*e*−2*t* − 2.572*e*−0.5*t* cos(1.25*t*) + 4.791*e*−0.5*t* sin(1.25*t*)]*u*(*t*) A - -- **16.51** [−12 + 41.17*e*−15.125*t* cos(4.608*t* − 73.06°)]*u*(*t*) amps -- **16.53** [11.547*e*−*t* cos(1.7321*t* + 30°)]*u*(*t*) volts -- **16.55** [5−4*e*−*t* − 1*e*−6*t* ]*u*(*t*) amps, [2*e*−*t* − 2*e*−6*t* ]*u*(*t*) amps -- **16.57** (a) (3∕*s*)[1 − *e*−*s* ], (b) [(2 − 2*e*−1.5*t* )*u*(*t*) − (2 − 2*e*−1.5(*t*−1))*u*(*t* − 1)] V -- **16.59** [5*e*−*t* − 10*e*−*t*∕2 cos (*t*∕2)]*u*(*t*) V -- **16.61** [2.333 − 2.38*e*−1.2306*t* + 2.033*e*−0.6347*t* cos(1.4265*t* + 88.68°)]*u*(*t*) V -- **16.63** [7.5*e*−4*t* cos (2*t*) + 345*e*−4*t* sin (2*t*)]*u*(*t*) V, [6 − 9*e*−4*t* cos (2*t*) − 17.062*e*−4*t* sin (2*t*)]*u*(*t*) A -- **16.65** {110.1*e*−3*t* + 192*te*−3*t* − 10.1 cos(4*t*) + 34.58 sin(4*t*)}*u*(*t*) V -- **16.67** [*e*10*t* − *e*−10*t* ] *u*(*t*) volts; this is an unstable circuit! -- **16.69** 240(*s* + 1)∕[*s*(*s* + 3)(3*s* 2 + 8*s* + 1)], −120(*s* −1)∕ [*s*(*s* + 3)(3*s* 2 + 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*2 ) **16.83** (a) \_\_ *R L e*−*Rt*∕*L u*(*t*), (b) (1 − *e*−*Rt*∕*L* )*u*(*t*) **16.85** [9*e*−*t* − 9*e*−2*t* − 6*te*−2*t* ]*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*−*t* − *e*−4*t* )*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*2 *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*3 -- **17.11** ∑*n*=−∞ ∞ 75 4*n*2 *π*2 [2 − 2 cos(*nπ*∕2) − 2*j* sin(*nπ*∕2) +*jnπ* cos(*nπ*∕2) + *nπ* (sin(*nπ*∕2))]*ejn π t*∕2 -- **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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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
NO | FREQUENCY
(HZ) | FOURIER
COMPONENT | NORMALIZED
COMPONENT | PHASE
(DEG) | NORMALIZED
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** 6*j \_\_ ω*− 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*2*t u*(*t*), (b) 6*e*−2*t* , (c) (−10*et u*(*t*) + 10*e*2*t* )*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 × 103 \_\_\_\_\_\_\_\_ 106 + *jω* ( \_\_\_1 *jω* + \_\_\_1 *ω*2 − \_\_\_1 *ω*2 *e*−*jω* ) - -**18.41** 2*jω*(4.5 + *j*2*ω*) \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (2 + *jω*)(4 − 2*ω*2 + *jω*) - -**18.43** 1000(*e*−1*t* − *e*−1.25*t* )*u*(*t*) V - -**18.45** 5(*e*−*t* − *e*−*2*t )*u*(*t*) A - -- **18.47** 16(*e*−*t* − *e*−*2*t )*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*−*t* − 20*e*−*2*t + 4*e*−4t )*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] S -- **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 ]S -- **19.29** (a) 44 V, 16 V, (b) same -- **19.31** [3.8 Ω −3.6 0.4 0.2 S] **19.33** [(3.077 + *j*1.2821) Ω −0.3846 + j0.2564 0.3846 − *j*0.2564 (76.9 + 282.1) mS] -- **19.35** [ 2Ω −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** [ 4 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 × 10−4] - -*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 - -# 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. - -# 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 \ No newline at end of file diff --git a/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 b/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 deleted file mode 100644 index 539a54fd2df040d8324ba8ee55e081ddb0fc147f..0000000000000000000000000000000000000000 --- a/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 +++ /dev/null @@ -1,3 +0,0 @@ -version https://git-lfs.github.com/spec/v1 -oid sha256:08b183286acd5cffae4f53364f258ab4ac2b5414a58374e89a94675fe152b991 -size 19937331 diff --git a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/catalog_entry.json b/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/catalog_entry.json deleted file mode 100644 index e0bf6237c092a9e9c138c3fdcb1a1965fd3e8e27..0000000000000000000000000000000000000000 --- a/engineering/fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku/catalog_entry.json +++ /dev/null @@ -1,3664 +0,0 @@ -{ - "doc_id": "fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku", - "file_name": "fundamentals-of-electric-circuits-6th-edition-by-charles-k.-alexander-and-matthew-n.o.-sadiku.pdf", - "title": "Fundamentals of Electric Circuits", - "author": "Charles K. 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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 -- Layout Coordinates JSON: 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.layout.json -- 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 \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/001_Cover.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/001_Cover.md deleted file mode 100644 index 05a91a819e69cebcd8c96f92529962e825d247d6..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/001_Cover.md +++ /dev/null @@ -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 - -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 - -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 - -# **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 - -- [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 - -- [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 -- [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 *n* 247 - - [3.3-4 Discrete-Time Sinusoid cos](#page-270-0)(*n*+θ ) 251 - - [3.3-5 Discrete-Time Complex Exponential](#page-271-0) *ejn* 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 - -### 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)− and at 0+ 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*3 422 - - [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 -- [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 *n*) 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 - -[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 - -### 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 - -### 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 - -[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 - -# **[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* - - - -# **[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*2 + 1 = 0, leading to *x*2 = −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 *i* (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*2 + 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*2 + 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*3 + 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*3 + 6*x* − 20 = 0. But when Cardano tried to solve the equation *x*3 −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*3 − 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]. - -† This equation is known as the *depressed cubic* equation. A general cubic equation - -*y*3 +*py*2 +*qy*+*r* = 0 - -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. - -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*θ , 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*θ = 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*θ 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*∗ are depicted in Fig. B.2. Observe that *z*∗ 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*θ 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 π 2 (more generally, π 2 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*θ . This picture is also helpful in several other applications. For example, to determine the limit of *e*(α+*j*ω)*t* as *t* → ∞, we note that - -*ej*ω*t* - -$$ -e^{(\alpha+j\omega)t} = e^{\alpha t} e^{j\omega t} -$$ -\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*θ . - -### 8 CHAPTER B BACKGROUND - -Now the magnitude of *ej*ω*t* is unity regardless of the value of ω or *t* because *ej*ω*t* = *rej*θ with *r* = 1. Therefore, *e*α*t* determines the behavior of *e*(α+*j*ω)*t* 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*θ 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*θ [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−1(−3/−2). This result is not the same as tan−1(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−1(−3/−2) as tan−1(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−1(1/−2) = −26.6◦, is off by 180◦. The correct answer is - -† 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, - -$$ --2 + j1 = \sqrt{5}e^{j153.4^{\circ}} -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/002_Half title.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/002_Half title.md deleted file mode 100644 index b0ce7b15eaa4666e9faa80d113e48ae1e05ce8cb..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/002_Half title.md +++ /dev/null @@ -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−1(−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 π . 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*ej*153.4◦ = 2.2361*ej*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*3π , **(e)** 2*ej*4π , and **(f)** 2*e*−*j*4π . - -**(a)** 2*ej*π/3 = 2(cos π/3+*j*sin π/3) = 1+*j* √3 (see Fig. B.5a) **(b)** 4*e*−*j*3π/4 = 4(cos 3π/4−*j*sin 3π/4) = −2 √2−*j*2 √2 (see Fig. B.5b) **(c)** 2*ej*π/2 = 2(cos π/2+*j*sin π/2) = 2(0+*j*1) = *j*2 (see Fig. B.5c) **(d)** 3*e*−*j*3π = 3(cos 3π −*j*sin 3π ) = 3(−1+*j*0) = −3 (see Fig. B.5d) **(e)** 2*ej*4π = 2(cos 4π +*j*sin 4π ) = 2(1+*j*0) = 2 (see Fig. B.5e) **(f)** 2*e*−*j*4π = 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/3 = 1+*j* √3 = 1+*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*1 and *z*2 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*1 and *z*2 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/*n* (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/*n* given in Eq. (B.11) is the *principal value* of *z*1/*n*, 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*2 and *z*1/*z*2 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*2 = −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*2 = 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*1 = 2*ej*π/4 and *z*2 = 8*ej*π/3, find the following: **(a)** 2*z*1 −*z*2, **(b)** 1/*z*1, **(c)** *z*1/*z*2 2, and **(d)** √3 *z*2. - -**(a)** Since subtraction cannot be performed directly in polar form, we convert *z*1 and *z*2 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 $X_i(\omega) = \frac{-5\omega}{9 + 16\omega^2}$ - -**(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 - -$$ -log(z1z2) = log z1 + log z2 -$$ -$$ -log(z1/z2) = log z1 - log z2 -$$ -$$ -a(z1+z2) = az1 × az2 -$$ -$$ -zc = ecln z -$$ -$$ -az = ezln a -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/003_B.2 SINUSOIDS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/003_B.2 SINUSOIDS.md deleted file mode 100644 index 15c6fcd6656fa032cc357c7b40cd14f202889ab4..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/003_B.2 SINUSOIDS.md +++ /dev/null @@ -1,186 +0,0 @@ -## **[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*0 seconds. As a result, there are *f*0 repetitions per second. This is the *frequency* of the sinusoid, and the repetition interval *T*0 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*0 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 ω0 (*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*0 and frequency ω0 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 ω0 as the frequency of the signal cos(ω0*t*+θ ), it should be clearly understood that ω0 is the *radian frequency*; the *hertzian frequency* of this sinusoid is *f*0 = 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 - -$$ -C\cos{(\omega_0 t - \pi/2)} = C\sin{\omega_0 t} -$$ - -#### 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*θ . 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)** *x*(*t*) = cos ω0*t* − √3 sin ω0*t* - -**(b)** *x*(*t*) = −3 cos ω0*t* +4 sin ω0*t* - -**(a)** In this case, *a* = 1 and *b* = −√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, − √3 sin ω0*t* is represented by a phasor of length √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* √3 = 2*ej*π/3. Hence, *C* = 2 and θ = 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 $\theta = \tan^{-1}\left(\frac{-4}{-3}\right) = -126.9^{\circ}$ - -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*ϕ = cos ϕ + *j*sin ϕ and *e*−*j*ϕ = 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) diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/004_B.3 SKETCHING SIGNALS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/004_B.3 SKETCHING SIGNALS.md deleted file mode 100644 index f9aadc1f2d8ca9baf64670d27f59110cf4fd3bb8..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/004_B.3 SKETCHING SIGNALS.md +++ /dev/null @@ -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*t* . - -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*1 and *t*2 separated by one time constant so that - -$$ -t_2 - t_1 = \frac{1}{a} -$$ - -Now the ratio of *e*−*at*2 to *e*−*at*1 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 manner, we see that *x*(*t*) = (1/*e*)3 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*t* and cos(6*t* −60◦) separately and then multiply them: - -- **(a) Sketching 4***e***−2***t* **.** 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*t* , as illustrated in Fig. B.11a. -- **(b) Sketching cos***(***6***t* **− 60◦***)***.** The procedure for sketching cos(6*t* − 60◦) is discussed in Sec. B.2 (Fig. B.6c). Here, the period of the sinusoid is *T*0 = 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***t* **cos***(***6***t* **− 60◦***)***.** 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*t* cos(6*t*−60◦) touches 4*e*−2*t* at the instants at which the sinusoid cos(6*t* −60◦) is at its positive peaks. Clearly, 4*e*−2*t* is an envelope for positive amplitudes of 4*e*−2*t* cos(6*t* − 60◦). Similar argument shows that 4*e*−2*t* cos(6*t* − 60◦) touches −4*e*−2*t* at its negative peaks. Therefore, −4*e*−2*t* is an envelope for negative amplitudes of 4*e*−2*t* cos(6*t* − 60◦). Thus, to sketch 4*e*−2*t* cos(6*t* − 60◦), we first draw the envelopes 4*e*−2*t* and −4*e*−2*t* (the mirror image of 4*e*−2*t* 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 ). - -† 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/ √*e*. Thus, at *t* = 0.25, *x*(*t*) = 1/ √*e*, and at *t* = 0.75, *x*(*t*) = 1/*e* √*e*, and so on. - -**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. diff --git "a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/005_B.4 CRAMER\342\200\231S RULE.md" "b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/005_B.4 CRAMER\342\200\231S RULE.md" deleted file mode 100644 index 5378ab3cf86f31ca7f6dd5b5b5f1a8c22dc4d189..0000000000000000000000000000000000000000 --- "a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/005_B.4 CRAMER\342\200\231S RULE.md" +++ /dev/null @@ -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 - -| ⎡
2 | 1 | ⎤
1 | ⎡
x1 | ⎤ | ⎡
⎤
3 | -|--------|---|---------|---------|-----|-------------| -| 1
⎣ | 3 | −1
⎦ | x2
⎣ | ⎦ = | 7
⎣
⎦ | -| 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: - -$$ -x_1 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 3 & 1 & 1 \\ 7 & 3 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \frac{8}{4} = 2 -$$ - -$$ -x_2 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 2 & 3 & 1 \\ 1 & 7 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \frac{4}{4} = 1 -$$ - -$$ -x_3 = \frac{1}{|\mathbf{A}|} \begin{vmatrix} 2 & 1 & 3 \\ 1 & 3 & 7 \\ 1 & 1 & 1 \end{vmatrix} = \frac{-8}{4} = -2 -$$ - -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*1 = 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 -``` diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/006_B.5 PARTIAL FRACTION EXPANSION.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/006_B.5 PARTIAL FRACTION EXPANSION.md deleted file mode 100644 index fad507f67d722698113e763b3031b6ce064f9a6d..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/006_B.5 PARTIAL FRACTION EXPANSION.md +++ /dev/null @@ -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. - -$$ -x^{2} + 4x + 3 \quad \begin{array}{c} 2x + 1 \\ 2x^{3} + 9x^{2} + 11x + 2 \\ 2x^{3} + 8x^{2} + 6x \\ x^{2} + 5x + 2 \\ x^{2} + 4x + 3 \\ x - 1 \end{array} -$$ - -† Some sources classify *F*(*x*) as strictly proper if *m* &lt; *n*, proper if *m* ≤ *n*, and improper if *m* &gt; *n*. - -### 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)2 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 -$$ -18k_1 + 9k_2 + 6k_3 + 2k_4 = 6 -$$ - -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*−λ1 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*1 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*1 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*2 and *k*3 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*1 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*1 and *c*2 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+c1)x2 + (8+c1+c2)x + (26+c2) - -Equating terms of similar powers yields *c*1 = 2, *c*2 = −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*1 and *c*2 in Eq. (B.26) can also be determined by using shortcuts. After computing *k*1 = 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*2 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*1 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*1 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*1 +*c*2 = 3, we see that *c*2 = 2 and therefore, - -$$ -F(x) = \frac{1}{x} + \frac{x+2}{x^2 + 2x + 5} -$$ - -## **[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)*r* . 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*0 is obtained by concealing the factor (*x*−λ)*r* 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*1 is obtained by concealing the factor (*x*−λ)*r* 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)*r F*(*x*) is obtained from *F*(*x*) by omitting the factor (*x* − λ)*r* from its denominator. Therefore, the coefficient *aj* is obtained by concealing the factor (*x* − λ)*r* in *F*(*x*), taking the *j*th derivative of the remaining expression, and then letting *x* = λ (while dividing by *j*!). - -### **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)3 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)3 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*0 are relatively simple to determine by the Heaviside expansion method. These values were found to be *k*1 = 1 and *a*0 = 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+a2)x3 + (a1+4a2+3)x2 + (5+3a1+5a2)x + (4+2a1+2a2+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*1 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*1 and *x*0 terms, we obtain - -$$ -23 = 5 + 3a_1 + 5a_2 -$$ - -$$ -13 = 4 + 2a_1 + 2a_2 + 1 -$$ - -These equations are satisfied by the values *a*1 = 1 and *a*2 = 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*0 = 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*1 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, - -$$ -\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2} -$$ - -#### 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*0 (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)** *F(x)* **with** *m* **=** *n* - -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*2 = 3. Therefore, - -$$ -F(x) = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} = 3 + \frac{k_1}{x - 2} + \frac{k_2}{x + 3} -$$ - -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*)*r* rather than *k*/(*x* −λ*i*)*r* . 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*1 = 1, *a*2 = 1, *a*3 = −2, and *a*4 = 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*)*r* . - -## **[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*)*m*×*n*, 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 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. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/007_B.6 VECTORS AND MATRICES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/007_B.6 VECTORS AND MATRICES.md deleted file mode 100644 index 72b8f855f0fb0f3894fdab8fa8a55b71a3b172d2..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/007_B.6 VECTORS AND MATRICES.md +++ /dev/null @@ -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 - -| ⎡
2 | 0 | 0 | ⎤ | -|--------|---|---|---| -| 0
⎣ | 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: - -| | ⎡
1 | 0 | 0 | ··· | ⎤
0 | -|--------|-------------|---|---|-----|-------------| -| | 0
⎢ | 1 | 0 | ··· | 0
⎥ | -| I
= | ⎢
0
⎢ | 0 | 1 | ··· | ⎥
0
⎥ | -| | ⎢
⎢
⎣ | | | ··· | ⎥
⎥
⎦ | -| | 0 | 0 | 0 | ··· | 1 | - -The order of the unit matrix is sometimes indicated by a subscript. Thus, **I***n* 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 **A** and is denoted by **A***T* . It is evident that **A***T* is an *n*×*m* matrix. For example, if - -$$ -\mathbf{A} = \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ 1 & 3 \end{bmatrix}, \quad \text{then} \quad \mathbf{A}^T = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 2 & 3 \end{bmatrix} -$$ - -### 38 CHAPTER B BACKGROUND - -Using the abbreviated notation, if **A** = (*aij*)*m*×*n*, then **A***T* = (*aji*)*n*×*m*. Intuitively, further notice that (**A***T* )*T* = **A**. - -### **[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 | !⎡
2 | 2 | 3 | ⎤ | -|---|---|---|---------|--------|---|---| -| | | | | 1
⎣ | 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)*i*+*j* 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**−1 , the inverse of a square matrix **A**, with the property - -> **A**−1 **A** = **I** (unit matrix) - -Then, premultiplying both sides of Eq. (B.36) by **A**−1 , 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**−1 **A** = **I** = **AA**−1 , we further note that the matrices **A** and **A**−1 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**−1 if - -**A** = ⎡ ⎣ 211 123 321 ⎤ ⎦ - -† These two conditions imply that the number of equations is equal to the number of unknowns and that all the equations are independent. - -‡ To prove **AA**−1 = **I**, notice first that we define **A**−1**A** = **I**. Thus, **IA** = **AI** = **A**(**A**−1**A**) = (**AA**−1)**A**. Subtracting (**AA**−1)**A**, we see that **IA**−(**AA**−1)**A** = 0 or (**I**−**AA**−1)**A** = 0. This requires **AA**−1 = **I**. - -Here, - -$$ -|\mathbf{D}_{11}| = -4, \t |\mathbf{D}_{12}| = 8, \t |\mathbf{D}_{13}| = -4 |\mathbf{D}_{21}| = 1, \t |\mathbf{D}_{22}| = -1, \t |\mathbf{D}_{23}| = -1 |\mathbf{D}_{31}| = 1, \t |\mathbf{D}_{32}| = -5, \t |\mathbf{D}_{33}| = 3 \text{and } |\mathbf{A}| = -4. \text{ Therefore, } \mathbf{A}^{-1} = -\frac{1}{4} \begin{bmatrix} -4 & 1 & 1 \\ 8 & -1 & -5 \\ -4 & -1 & 3 \end{bmatrix} -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/008_B.7 MATLAB - ELEMENTARY OPERATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/008_B.7 MATLAB - ELEMENTARY OPERATIONS.md deleted file mode 100644 index 8f6c887a2b98f0be0f1413521471341883f16d33..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/008_B.7 MATLAB - ELEMENTARY OPERATIONS.md +++ /dev/null @@ -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 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 i = j = √−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 |*z*| = |−3 − *j*4| = (−3)2 +(−4)2 = 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 |*z*| = √*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 -``` - -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*3 = −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*100 = −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. 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); -``` - -† Some other programming languages, such as C, begin indexing at 0. Careful attention is warranted. - -‡ 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*100 = −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*100 = −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*100 = −1. - -### 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*t* . 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 - -&gt;> h = f.\*g; - -† While grossly inefficient, element-by-element multiplication can be accomplished by extracting the main diagonal from the outer product of two *N*-length vectors. - -**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 -``` - -† 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*t*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**α 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]. - -† The repmat command provides a more flexible method to replicate or tile objects. Equivalently, T = repmat(t,1,11). - -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. - -† The benefits of vectorization are less pronounced in recent versions of MATLAB. - -## **[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 *e* ≈ 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 -$$ -\sum_{k=0}^{n} k^{2}r^{k} = \frac{r[(1+r)(1-r^{n}) - 2n(1-r)r^{n} - n^{2}(1-r)^{2}r^{n}]}{(1-r)^{3}} \qquad r \neq 1 -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/009_B.8 APPENDIX - USEFUL MATHEMATICAL FORMULAS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/009_B.8 APPENDIX - USEFUL MATHEMATICAL FORMULAS.md deleted file mode 100644 index bd693fd36b25a6a0f1f3abcf78f5d94404d709b6..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/009_B.8 APPENDIX - USEFUL MATHEMATICAL FORMULAS.md +++ /dev/null @@ -1,342 +0,0 @@ -### **[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 -$$ -\sin^2 x = \frac{1}{2} (1 - \cos 2x) -$$ - - -$$ -\cos^3 x = \frac{1}{4} (3 \cos x + \cos 3x) -$$ - -\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 -$$ -\frac{d}{dx}(\tan^{-1}ax) = \frac{a}{1 + a^2x^2} -$$ - -### **[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 -$$ -\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \ln(x -$$ - -### **[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 - -$$ -y = x - \frac{p}{3} -$$ diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/010_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/010_REFERENCES.md deleted file mode 100644 index 7e31c59887c28268c1069319fdcdd0b8ac26432d..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/010_REFERENCES.md +++ /dev/null @@ -1,79 +0,0 @@ -### **[REFERENCES](#page-7-0)** - -- 1. Asimov, Isaac. *Asimov on Numbers.* Bell Publishing, New York, 1982. -- 2. Calinger, R., ed. *Classics of Mathematics.* Moore Publishing, Oak Park, IL, 1982. -- 3. Hogben, Lancelot. *Mathematics in the Making.* Doubleday, New York, 1960. - -- 4. Cajori, Florian. *A History of Mathematics,* 4th ed. Chelsea, New York, 1985. -- 5. Encyclopaedia Britannica. *Micropaedia* IV, 15th ed., vol. 11, p. 1043. Chicago, 1982. -- 6. Singh, Jagjit. *Great Ideas of Modern Mathematics.* Dover, New York, 1959. -- 7. Dunham, William. *Journey Through Genius.* Wiley, New York, 1990. - -## **[PROBLEMS](#page-7-0)** - -- **B.1-1** Given a complex number *w* = *x* + *jy*, the complex conjugate of *w* is defined in rectangular coordinates as *w*∗ =*x*−*jy*. Use this fact to derive complex conjugation in polar form. -- **B.1-2** Express the following numbers in polar form: - - (a) *w*a = 1+*j* - - (b) *w*b = 1+*ej* - - (c) *w*c = −4+*j*3 - - (d) *w*d = (1+*j*)(−4+*j*3) - - (e) *w*e = *ej*π/4 +2*e*−*j*π/4 - -(f) -$$ -w_f = \frac{1+j}{2j} -$$ - -(g) -$$ -w_g = (1+j)/(-4+j3) -$$ - -(h) -$$ -w_h = \frac{1-j}{\sin(j)} -$$ - -- **B.1-3** Express the following numbers in Cartesian (rectangular) form: - - (a) *w*a = *j*+*ej* - - (b) *w*b = 3*ej*π/4 - - (c) *w*c = 1/*ej* - - (d) *w*d = (1+*j*)(−4+*j*3) - -(e) -$$ -w_e = e^{j\pi/4} + 2e^{-j\pi/4} -$$ - -- (f) *w*f = *ej* +1 -- (g) *w*g = 1/2*j* -- (h) *w*h = *j j j* (*j* raised to the *j* raised to the *j*) -- **B.1-4** Showing all work and simplifying your answer, determine the real part of the following numbers: - - (a) *w*a = 1 *j* (*j*−5*e*2−3*j* ) - - (b) *w*b = (1+*j*)ln(1+*j*) -- **B.1-5** Showing all work and simplifying your answer, determine the imaginary part of the following numbers: - - (a) *w*a = −*jej*π/4 - -(b) -$$ -w_b = 1 - 2je^{2-4j} -$$ - -- (c) *w*c = tan(*j*) -- **B.1-6** For complex constant *w*, prove: - - (a) Re(*w*) = (*w* +*w*∗)/2 - - (b) Im(*w*) = (*w* −*w*∗)/2*j* - -- **B.1-7** Given *w* = *x* −*jy*, determine: (a) Re(*ew*) - - (b) Im(*ew*) -- **B.1-8** For arbitrary complex constants *w*1 and *w*2, prove or disprove the following: - - (a) Re(*jw*1) = −Im(*w*1) - - (b) Im(*jw*1) = Re(*w*1) - - (c) Re(*w*1)+Re(*w*2) = Re(*w*1 +*w*2) - - (d) Im(*w*1)+Im(*w*2) = Im(*w*1 +*w*2) - - (e) Re(*w*1)Re(*w*2) = Re(*w*1*w*2) - - (f) Im(*w*1)/Im(*w*2) = Im(*w*1/*w*2) -- **B.1-9** Given *w*1 = 3+*j*4 and *w*2 = 2*ej*π/4. - - (a) Express *w*1 in standard polar form. - - (b) Express *w*2 in standard rectangular form. - - (c) Determine |*w*1| 2 and |*w*2| 2. \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/011_PROBLEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/011_PROBLEMS.md deleted file mode 100644 index fc397d1fb7fc3879f5b5a10f11ea484eae67d254..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/011_PROBLEMS.md +++ /dev/null @@ -1,293 +0,0 @@ - - (d) Express *w*1 + *w*2 in standard rectangular form. - - (e) Express *w*1 −*w*2 in standard polar form. - - (f) Express *w*1*w*2 in standard rectangular form. - - (g) Express *w*1/*w*2 in standard polar form. -- **B.1-10** Repeat Prob. B.1-9 using *w*1 = (3 + *j*4)2 and *w*2 = 2.5*je*−*j*40π . -- **B.1-11** Repeat Prob. B.1-9 using *w*1 = *j* + *e*π/4 and *w*2 = cos(*j*). -- **B.1-12** Using the complex plane: - - (a) Evaluate and locate the distinct solutions to (*w*) 4 = −1. - - (b) Evaluate and locate the distinct solutions to (*w* −(1+*j*2))5 = (32/ √2)(1+*j*). - - (c) Sketch the solution to |*w* −2*j*| = 3. - - (d) Graph *w*(*t*) = (1+*t*)*ejt* for (−10 ≤ *t* ≤ 10). -- **B.1-13** The distinct solutions to (*w* −*w*1) *n* = *w*2 lie on a circle in the complex plane, as shown in Fig. PB.1-13. One solution is located on the real axis at √3 + 1 = 2.732, and one solution is located on the imaginary axis at √3−1=0.732. Determine *w*1, *w*2, and *n*. - -- **B.1-14** Find the distinct solutions to each of the following. Use MATLAB to graph each solution set in the complex plane. - - (a) *w*3 = − 8 27 - - (b) (*w* +1)8 = 1 - - (c) *w*2 +*j* = 0 - - (d) 16(*w* −1)4 +81 = 0 - - (e) (*w* +2*j*)3 = −8 - - (f) (*j*−*w*)1.5 = 2+*j*2 - - (g) (*w* −1)2.5 = *j*4 √ 2 -- **B.1-15** If *j* = √−1, what is √*j*? -- **B.1-16** Find all the values of ln(−*e*), expressing your answer in Cartesian form. -- **B.1-17** Determine all values of log10(−1), expressing your answer in Cartesian form. Notice that the logarithm has base 10, not *e*. -- **B.1-18** Express the following in standard rectangular coordinates: - - (a) *w*a = ln(1/(1+*j*)) - - (b) *w*b = cos(1+*j*) - - (c) *w*c = (1−*j*)*j* -- **B.1-19** By constraining *w* to be purely imaginary, show that the equation cos(*w*) = 2 can be represented as a standard quadratic equation. Solve this equation for *w*. -- **B.1-20** Certain integrals, although expressed in relatively simple form, are quite difficult to solve. For example, \$ *e*−*x*2 *dx* cannot be evaluated in terms of elementary functions; most calculators that perform integration cannot handle this indefinite integral. Fortunately, you are smarter than most calculators. - -- (a) Express *e*−*x*2 using a Taylor series expansion. -- (b) Using your series expansion for *e*−*x*2 , determine \$ *e*−*x*2 *dx*. -- (c) Using a suitably truncated series, evaluate the definite integral \$ 1 0 *e*−*x*2 *dx*. -- **B.1-21** Repeat Prob. B.1-20 for \$ *e*−*x*3 *dx*. -- **B.1-22** Repeat Prob. B.1-20 for \$ cos *x*2 *dx*. -- **B.1-23** For each function, determine a suitable series expansion. - - (a) *f*a(*x*) = (2−*x*2)−1 - - (b) *f*b(*x*) = (0.5)*x* -- **B.1-24** Consider the function *f*(*x*) = 1+*x*+*x*2+*x*3. - - (a) Express *f*(*x*) using a Taylor series with expansion point of *a* = 1. Explicitly write out every term. [*Hint:* See Sec. B.8-4.] - - (b) Describe a good reason why you might want to express a function that is already a simple polynomial using such a series. -- **B.1-25** Determine the Maclaurin series expansion of each of the following. [*Hint:* See Sec. B.8-4.] (a) *f*a(*x*) = 2*x* - - (b) *f*b(*x*) = 1 3 *x* -- **B.2-1** Determine the fundamental period *T*0, frequency *f*0, and radian frequency ω0 for the following sinusoids: - - (a) cos(5π*t* +3) - - (b) 7 sin 2*t*−π 3 -- **B.2-2** Determine an expression for a sinusoid that oscillates 15 times per second, that has a value of -1 at *t* = 0, and whose peak amplitude is 3. Use MATLAB to plot the signal over 0 ≤ *t* ≤ 1. -- **B.2-3** Let *x*1(*t*) = 2 cos(3*t* + 1) and *x*2(*t*) = −3 cos (3*t* −2). - - (a) Determine *a*1 and *b*1 so that *x*1(*t*) = *a*1 cos(3*t*)+*b*1 sin(3*t*). - - (b) Determine *a*2 and *b*2 so that *x*2(*t*) = *a*2 cos(3*t*)+*b*2 sin(3*t*). - - (c) Determine *C* and θ so that *x*1(*t*) + *x*2(*t*) = *C*cos(3*t* +θ ). -- **B.2-4** In addition to the traditional sine and cosine functions, there are the *hyperbolic* sine and cosine functions, which are defined by sinh(*w*) = (*ew* −*e*−*w*)/2 and cosh(*w*) = (*ew* +*e*−*w*)/2. In general, the argument is a complex constant *w* = *x* +*jy*. - -- (a) Show that cosh(*w*) = cosh(*x*) cos(*y*) + *j*sinh(*x*)sin(*y*). -- (b) Determine a similar expression for sinh(*w*) in rectangular form that only uses functions of real arguments, such as sin(*x*), cosh(*y*), and so on. -- **B.2-5** Use Euler's identity to solve or prove the following: - - (a) Find real, positive constants *c* and φ for all real *t* such that 2.5 cos(3*t*) − 1.5 sin(3*t* + π/3) = *c* cos(3*t* + φ). Sketch the resulting sinusoid. - - (b) Prove that cos(θ ± φ) = cos(θ ) cos(φ) ∓ sin(θ )sin(φ). - - (c) Given real constants *a*, *b*, and α, complex constant *w*, and the fact that - -$$ -\int_a^b e^{wx} dx = \frac{1}{w} (e^{wb} - e^{wa}) -$$ - -evaluate the integral - -$$ -\int_a^b e^{wx} \sin(\alpha x) dx -$$ - -- **B.2-6** A particularly boring stretch of interstate highway has a posted speed limit of 70 mph. A highway engineer wants to install "rumble bars" (raised ridges on the side of the road) so that cars traveling the speed limit will produce quarter-second bursts of 1 kHz sound every second, a strategy that is particularly effective at startling sleepy drivers awake. Provide design specifications for the engineer. -- **B.3-1** By hand, accurately sketch the following signals over (0 ≤ *t* ≤ 1): - - (a) *x*a(*t*) = *e*−*t* - - (b) *x*b(*t*) = sin(2π5*t*) - -(c) -$$ -x_c(t) = e^{-t} \sin(2\pi 5t) -$$ - -- **B.3-2** In 1950, the human population was approximately 2.5 billion people. Assuming a doubling time of 40 years, formulate an exponential model for human population in the form *p*(*t*) = *aebt*, where *t* is measured in years. Sketch *p*(*t*) over the interval 1950 ≤ *t* ≤ 2100. According to this model, in what year can we expect the population to reach the estimated 15 billion carrying capacity of the earth? -- **B.3-3** Determine an expression for an exponentially decaying sinusoid that oscillates three - -times per second and whose amplitude envelope decreases by 50% every 2 seconds. Use MATLAB to plot the signal over −2 ≤ *t* ≤ 2. - -**B.3-4** By hand, sketch the following against independent variable *t*: - -(a) -$$ -x_a(t) = \text{Re}\left(2e^{(-1+j2\pi)t}\right) -$$ - -(b) -$$ -x_b(t) = \text{Im}\left(3 - e^{(1-j2\pi)t}\right) -$$ - -(c) $x_c(t) = 3 - \text{Im}\left(e^{(1-j2\pi)t}\right)$ - -**B.4-1** Consider the following system of equations: - -$$ -\left[\begin{array}{cc} -1 & 2 \\ 3 & -4 \end{array}\right] \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{c} 3 \\ -1 \end{array}\right] -$$ - -Expressing all answers in rational form (ratio of integers), use Cramer's rule to determine *x*1 and *x*2. Perform all calculations by hand, including matrix determinants. - -**B.4-2** Consider the following system of equations: - -| ⎡
1 | 2 | ⎤
0 | ⎡
⎤
x1 | | ⎡
⎤
7 | -|--------|---|--------|--------------|-----|-------------| -| 0
⎣ | 3 | 4
⎦ | x2
⎣ | ⎦ = | 8
⎣
⎦ | -| 5 | 0 | 6 | x3 | | 9 | - -Expressing all answers in rational form (ratio of integers), use Cramer's rule to determine *x*1, *x*2, and *x*3. Perform all calculations by hand, including matrix determinants. - -**B.4-3** Consider the following system of equations. - -$$ -x_1 + x_2 + x_3 = 1 -$$ - -$$ -x_1 + 2x_2 + 3x_3 = 3 -$$ - -$$ -x_1 - x_2 = -3 -$$ - -Use Cramer's rule to determine *x*1, *x*2, and *x*3. Matrix determinants can be computed by using MATLAB's det command. - -**B.5-1** Determine the constants *a*0, *a*1, and *a*2 of the partial fraction expansion - -$$ -F(s) = \frac{s}{(s+1)^3} -$$ - -= $\frac{a_0}{(s+1)^3} + \frac{a_1}{(s+1)^2} + \frac{a_2}{(s+1)}$ - -- **B.5-2** Compute by hand the partial fraction expansions of the following rational functions: - - (a) *H*a(*s*) = *s*2+5*s*+6 *s*3+*s*2+*s*+1 , which has denominator poles at *s* = ±*j* and *s* = −1 - -(b) -$$ -H_b(s) = \frac{1}{H_1(s)} = \frac{s^3 + s^2 + s + 1}{s^2 + 5s + 6} -$$ - -(c) -$$ -H_c(s) = \frac{1}{(s+1)^2(s^2+1)} -$$ - -(d) $H_d(s) = \frac{s^2+5s+6}{3s^2+2s+1}$ - -- **B.5-3** Compute by hand the partial fraction expansions of the following rational functions: (a) *F*a(*x*) = (*x*−1)(*x*−2) (*x*−3)2 (b) *F*b(*x*) = (*x*−1)2 (3*x*−1)(2*x*−1) (c) *F*c(*x*) = (*x*−1)2 (3*x*−1)2(2*x*−1) (d) *F*d(*x*) = *x*2−5*x*+6 2*x*2+8*x*+6 (e) *F*e(*x*) = 2*x*2−3*x*−11 *x*2−*x*−2 (f) *F*f(*x*) = 3+2*x*2 −3+2*x*+*x*2 - - (g) *F*g(*x*) = *x*3+2*x*2+3*x*+4 *x*2+1 (h) *F*h(*x*) = 1+2*x*+3*x*2 *x*2+5*x*+6 (i) *F*i(*x*) = 3*x*3−*x*2+14*x*+4 *x*2+4 (j) *F*j(*x*) = 2*x*−1−1+2*x x*−5+6*x*−1 (k) *F*k(*x*) = 3 −5*x*2−9*x*+23 -- *x*2+*x*−2 **B.6-1** A system of equations in terms of unknowns *x*1 and *x*2 and arbitrary constants *a*, *b*, *c*, *d*, *e*, and *f* is given by - -$$ -ax_1 + bx_2 = c -$$ - -$$ -dx_1 + ex_2 = f -$$ - -- (a) Represent this system of equations in matrix form. -- (b) Identify specific constants *a*, *b*, *c*, *d*, *e*, and *f* such that *x*1 = 3 and *x*2 = −2. Are the constants you selected unique? -- (c) Identify nonzero constants *a*, *b*, *c*, *d*, *e*, and *f* such that no solutions *x*1 and *x*2 exist. -- (d) Identify nonzero constants *a*, *b*, *c*, *d*, *e*, and *f* such that an infinite number of solutions *x*1 and *x*2 exist. -- **B.6-2** Using a matrix approach, solve the following system of equations: - -$$ -x_1 + x_2 + x_3 + x_4 = 4 -$$ - -\n -$$ -x_1 + x_2 + x_3 - x_4 = 2 -$$ - -\n -$$ -x_1 + x_2 - x_3 - x_4 = 0 -$$ - -\n -$$ -x_1 - x_2 - x_3 - x_4 = -2 -$$ - -**B.6-3** Using a matrix approach, solve the following system of equations: - -$$ -x_1 + x_2 + x_3 + x_4 = 1 -$$ -$$ -x_1 - 2x_2 + 3x_3 = 2 -$$ -$$ -x_1 - x_3 + 7x_4 = 3 -$$ -$$ --2x_2 + 3x_3 - 4x_4 = 4 -$$ - -**B.6-4** A signal *f*(*t*) = *a*cos(3*t*) + *b*sin(3*t*) reaches a peak amplitude of 5 at *t* = 1.8799 and has a zero crossing at *t* = 0.3091. Use a matrix-based approach to determine the constants *a* and *b*. - -### **B.6-5** Define - -$$ -\mathbf{x} = \begin{bmatrix} 1 & 3 \\ -2 & 4 \end{bmatrix}, \quad \mathbf{y} = \begin{bmatrix} -5 \\ 2 \end{bmatrix}, -$$ - -and -$$ -\mathbf{z} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} -$$ - -By hand, calculate the following: - -- (a) **f**a = **y***T* **y** (b) **f**b = **yy***T* (c) **f**c = **xy** (d) **f**d = **x***T* **y** (e) **f**e = **y***T* **x** (f) **f**f = **xz** (g) **f**g = **zxz** (h) **f**h = **x***T* −**z** -- **B.7-1** Use MATLAB to produce the plots requested in Prob. B.3-4. -- **B.7-2** Use MATLAB to plot the function *x*(*t*) = *t*sin(2π*t*) over 0 ≤ *t* ≤ 10 using 501 equally spaced points. What is the maximum value of *x*(*t*) over this range of *t*? -- **B.7-3** Use MATLAB to plot *x*(*t*) = cos(*t*)sin(20*t*) over a suitable range of *t*. -- **B.7-4** Use MATLAB to plot *x*(*t*) = %10 *k*=1 cos(2π*kt*) over a suitable range of *t*. The MATLAB command sum may prove useful. -- **B.7-5** When a bell is struck with a mallet, it produces a ringing sound. Write an equation that approximates the sound produced by a small, light bell. Carefully identify your assumptions. How does your equation change if the bell is large and heavy? You can assess the quality of your models by using the MATLAB sound command to listen to your "bell." - -- **B.7-6** You are working on a digital quadrature amplitude modulation (QAM) communication receiver. The QAM receiver requires a pair of quadrature signals: cos*n* and sin*n*. These can be simultaneously generated by following a simple procedure: (1) choose a point *w* on the unit circle, (2) multiply *w* by itself and store the result, (3) multiply *w* by the last result and store, and (4) repeat step 3. - - (a) Show that this method can generate the desired pair of quadrature sinusoids. - - (b) Determine a suitable value of *w* so that good-quality, periodic, 2π × 100,000 rad/s signals can be generated. How much time is available for the processing unit to compute each sample? - - (c) Simulate this procedure by using MATLAB and report your results. - - (d) Identify as many assumptions and limitations to this technique as possible. For example, can your system operate correctly for an indefinite period of time? -- **B.7-7** Using MATLAB's residue command, - - (a) Verify the results of Prob. B.5-2a. - - (b) Verify the results of Prob. B.5-2b. - - (c) Verify the results of Prob. B.5-2c. - - (d) Verify the results of Prob. B.5-2d. -- **B.7-8** Using MATLAB's residue command, - - (a) Verify the results of Prob. B.5-3a. - - (b) Verify the results of Prob. B.5-3b. - - (c) Verify the results of Prob. B.5-3c. - - (d) Verify the results of Prob. B.5-3d. - - (e) Verify the results of Prob. B.5-3e. - - (f) Verify the results of Prob. B.5-3f. - - (g) Verify the results of Prob. B.5-3g. - - (h) Verify the results of Prob. B.5-3h. - - (i) Verify the results of Prob. B.5-3i. - - (j) Verify the results of Prob. B.5-3j. - - (k) Verify the results of Prob. B.5-3k. -- **B.7-9** Determine the original length-3 vectors a and b need to produce the MATLAB output: - - >> [r,p,k] = residue(b,a) r = 0 + 2.0000i 0 - 2.0000i - -$$ -p = 3 -$$ - --3 -$$ -k = 0 + 1.0000i -$$ - -**B.7-10** Let *N* = [*n*7,*n*6,*n*5,...,*n*2,*n*1] represent the seven digits of your phone number. Construct a rational function according to - -$$ -H_N(s) = \frac{n_7s^2 + n_6s + n_5 + n_4s^{-1}}{n_3s^2 + n_2s + n_1} -$$ - -Use MATLAB's residue command to compute the partial fraction expansion of *HN*(*s*). - -- **B.7-11** When plotted in the complex plane for −π ≤ ω ≤ π, the function *f*(ω) = cos(ω) + *j*0.1 sin(2ω) results in a so-called Lissajous figure that resembles a two-bladed propeller. - - (a) In MATLAB, create two row vectors fr and fi corresponding to the real and imaginary portions of *f*(ω), respectively, over a suitable number *N* samples of ω. Plot the real portion against the imaginary portion and verify the figure resembles a propeller. - - (b) Let complex constant *w* = *x* + *jy* be represented in vector form - -$$ -\mathbf{w} = \left[ \begin{array}{c} x \\ y \end{array} \right] -$$ - -Consider the 2×2 rotational matrix **R**: - -$$ -\mathbf{R} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} -$$ - -Show that **Rw** rotates vector **w** by θ radians. - -- (c) Create a rotational matrix R corresponding to 10◦ and multiply it by the 2×*N* matrix f = [fr;fi];. Plot the result to verify that the "propeller" has indeed rotated counterclockwise. -- (d) Given the matrix R determined in part (c), what is the effect of performing RRf? How about RRRf? Generalize the result. -- (e) Investigate the behavior of multiplying *f*(ω) by the function *ej*θ . - - diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/012_1 SIGNALS AND SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/012_1 SIGNALS AND SYSTEMS.md deleted file mode 100644 index 8986cf0ae38c5d4a7dcb36d7d32fb6b5b393e088..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/012_1 SIGNALS AND SYSTEMS.md +++ /dev/null @@ -1,11 +0,0 @@ -# **[SIGNALS AND](#page-7-0) SYSTEMS** - -In this chapter we shall discuss basic aspects of signals and systems. We shall also introduce fundamental concepts and qualitative explanations of the hows and whys of systems theory, thus building a solid foundation for understanding the quantitative analysis in the remainder of the book. For simplicity, the focus of this chapter is on continuous-time signals and systems. Chapter 3 presents the same ideas for discrete-time signals and systems. - -### SIGNALS - -A *signal* is a set of data or information. Examples include a telephone or a television signal, monthly sales of a corporation, or daily closing prices of a stock market (e.g., the Dow Jones averages). In all these examples, the signals are functions of the independent variable *time*. This is not always the case, however. When an electrical charge is distributed over a body, for instance, the signal is the charge density, a function of *space* rather than time. In this book we deal almost exclusively with signals that are functions of time. The discussion, however, applies equally well to other independent variables. - -## SYSTEMS - -Signals may be processed further by *systems,* which may modify them or extract additional information from them. For example, an anti-aircraft gun operator may want to know the future location of a hostile moving target that is being tracked by his radar. Knowing the radar signal, he knows the past location and velocity of the target. By properly processing the radar signal (the input), he can approximately estimate the future location of the target. Thus, a system is an entity that *processes* a set of signals (*inputs*) to yield another set of signals (*outputs*). A system may be made up of physical components, as in electrical, mechanical, or hydraulic systems (hardware realization), or it may be an algorithm that computes an output from an input signal (software realization). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/013_1.1 SIZE OF A SIGNAL.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/013_1.1 SIZE OF A SIGNAL.md deleted file mode 100644 index aad20395d0d9d7136c18e38c8a668ee77e4c5811..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/013_1.1 SIZE OF A SIGNAL.md +++ /dev/null @@ -1,167 +0,0 @@ -## **[1.1 SIZE OF A](#page-7-0) SIGNAL** - -The size of any entity is a number that indicates the largeness or strength of that entity. Generally speaking, the signal amplitude varies with time. How can a signal that exists over a certain time interval with varying amplitude be measured by one number that will indicate the signal size or signal strength? Such a measure must consider not only the signal amplitude, but also its duration. For instance, if we are to devise a single number *V* as a measure of the size of a human being, we must consider not only his or her width (girth), but also the height. If we make a simplifying assumption that the shape of a person is a cylinder of variable radius *r* (which varies with the height *h*), then one possible measure of the size of a person of height *H* is the person's volume *V*, given by - -$$ -V = \pi \int_0^H r^2(h) \, dh -$$ - -### **[1.1-1 Signal Energy](#page-7-0)** - -Arguing in this manner, we may consider the area under a signal *x*(*t*) as a possible measure of its size, because it takes account not only of the amplitude but also of the duration. However, this will be a defective measure because even for a large signal *x*(*t*), its positive and negative areas could cancel each other, indicating a signal of small size. This difficulty can be corrected by defining the signal size as the area under |*x*(*t*)| 2, which is always positive. We call this measure the *signal energy Ex*, defined as - -$$ -E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt -$$ - (1.1) - -This definition simplifies for a real-valued signal *x*(*t*) to *Ex* = \$ ∞ −∞ *x*2(*t*)*dt*. There are also other possible measures of signal size, such as the area under |*x*(*t*)|. The energy measure, however, is not only more tractable mathematically but is also more meaningful (as shown later) in the sense that it is indicative of the energy that can be extracted from the signal. - -### **[1.1-2 Signal Power](#page-7-0)** - -Signal energy must be finite for it to be a meaningful measure of signal size. A necessary condition for the energy to be finite is that the signal amplitude → 0 as |*t*|→∞ (Fig. 1.1a). Otherwise the integral in Eq. (1.1) will not converge. - -When the amplitude of *x*(*t*) does not → 0 as |*t*|→∞ (Fig. 1.1b), the signal energy is infinite. A more meaningful measure of the signal size in such a case would be the time average of the energy, if it exists. This measure is called the *power* of the signal. For a signal *x*(*t*), we define its power *Px* as - -$$ -P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |x(t)|^2 dt -$$ - (1.2) - -This definition simplifies for a real-valued signal *x*(*t*) to *Px* = lim*T*→∞ 1 *T* \$ *T*/2 −*T*/2 *x*2(*t*)*dt*. Observe that the signal power *Px* is the time average (mean) of the signal magnitude squared, that is, the *mean-square* value of |*x*(*t*)|. Indeed, the square root of *Px* is the familiar *rms* (root-mean-square) value of *x*(*t*). - -Generally, the mean of an entity averaged over a large time interval approaching infinity exists if the entity either is periodic or has a statistical regularity. If such a condition is not satisfied, the average may not exist. For instance, a ramp signal *x*(*t*) = *t* increases indefinitely as |*t*|→∞, and neither the energy nor the power exists for this signal. However, the unit step function, which is not periodic nor has statistical regularity, does have a finite power. - -**Figure 1.1** Examples of signals: **(a)** a signal with finite energy and **(b)** a signal with finite power. - -When *x*(*t*) is periodic, |*x*(*t*)| 2 is also periodic. Hence, the power of *x*(*t*) can be computed from Eq. (1.2) by averaging |*x*(*t*)| 2 over one period. - -**Comments.** The signal energy as defined in Eq. (1.1) does not indicate the actual energy (in the conventional sense) of the signal because the signal energy depends not only on the signal, but also on the load. It can, however, be interpreted as the energy dissipated in a normalized load of a 1 ohm resistor if a voltage *x*(*t*) were to be applied across the 1 ohm resistor [or if a current *x*(*t*) were to be passed through the 1 ohm resistor]. The measure of "energy" is therefore indicative of the energy capability of the signal, not the actual energy. For this reason the concepts of conservation of energy should not be applied to this "signal energy." Parallel observation applies to "signal power" defined in Eq. (1.2). These measures are but convenient indicators of the signal size, which prove useful in many applications. For instance, if we approximate a signal *x*(*t*) by another signal *g*(*t*), the error in the approximation is *e*(*t*) = *x*(*t*) − *g*(*t*). The energy (or power) of *e*(*t*) is a convenient indicator of the goodness of the approximation. It provides us with a quantitative measure of determining the closeness of the approximation. In communication systems, during transmission over a channel, message signals are corrupted by unwanted signals (noise). The quality of the received signal is judged by the relative sizes of the desired signal and the unwanted signal (noise). In this case the ratio of the message signal and noise signal powers (signal-to-noise power ratio) is a good indication of the received signal quality. - -**Units of Energy and Power.** Equation (1.1) is not correct dimensionally. This is because here we are using the term *energy* not in its conventional sense, but to indicate the signal size. The same observation applies to Eq. (1.2) for power. The units of energy and power, as defined here, depend on the nature of the signal *x*(*t*). If *x*(*t*) is a voltage signal, its energy *Ex* has units of volts squared-seconds (V2 s), and its power *Px* has units of volts squared. If *x*(*t*) is a current signal, these units will be amperes squared-seconds (A2 s) and amperes squared, respectively. - -**Figure 1.2** Signals for Ex. 1.1 - -In Fig. 1.2a, the signal amplitude → 0 as |*t*|→∞. Therefore the suitable measure for this signal is its energy *Ex* given by - -$$ -E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-1}^{0} (2)^2 dt + \int_{0}^{\infty} 4e^{-t} dt = 4 + 4 = 8 -$$ - -In Fig. 1.2b, the signal magnitude does not → 0 as |*t*|→∞. However, it is periodic, and therefore its power exists. We can use Eq. (1.2) to determine its power. We can simplify the procedure for periodic signals by observing that a periodic signal repeats regularly each period (2 seconds in this case). Therefore, averaging |*x*(*t*)| 2 over an infinitely large interval is identical to averaging this quantity over one period (2 seconds in this case). Thus - -$$ -P_x = \frac{1}{2} \int_{-1}^{1} |x(t)|^2 dt = \frac{1}{2} \int_{-1}^{1} t^2 dt = \frac{1}{3} -$$ - -Recall that the signal power is the square of its rms value. Therefore, the rms value of this signal is 1/ √3. - -### **EXAMPLE 1.2 Determining Power and RMS Value** - -Determine the power and the rms value of - -- **(a)** *x*(*t*) = *C* cos(ω0*t* +θ ) -- **(b)** *x*(*t*) = *C*1 cos(ω1*t* +θ1)+*C*2 cos(ω2*t* +θ2) ω1 = ω2 -- **(c)** *x*(*t*) = *Dej*ω0*t* - -**(a)** This is a periodic signal with period *T*0 = 2π/ω0. The suitable measure of this signal is its power. Because it is a periodic signal, we may compute its power by averaging its energy over one period *T*0 = 2π/ω0. However, for the sake of demonstration, we shall use Eq. (1.2) to solve this problem by averaging over an infinitely large time interval. - -$$ -P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C^2 \cos^2 (\omega_0 t + \theta) dt = \lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} [1 + \cos (2\omega_0 t + 2\theta)] dt -$$ - -= -$$ -\lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} dt + \lim_{T \to \infty} \frac{C^2}{2T} \int_{-T/2}^{T/2} \cos (2\omega_0 t + 2\theta) dt -$$ - -The first term on the right-hand side is equal to *C*2/2. The second term, however, is zero because the integral appearing in this term represents the area under a sinusoid over a very large time interval *T* with *T* → ∞. This area is at most equal to the area of half the cycle because of cancellations of the positive and negative areas of a sinusoid. The second term is this area multiplied by *C*2/2*T* with *T* → ∞. Clearly this term is zero, and - -$$ -P_x = \frac{C^2}{2} -$$ - -This shows that a sinusoid of amplitude *C* has a power *C*2/2 regardless of the value of its frequency ω0 (ω0 = 0) and phase θ. The rms value is *C*/ √2. If the signal frequency is zero (dc or a constant signal of amplitude *C*), the reader can show that the power is *C*2. - -**(b)** In Ch. 6, we shall show that a sum of two sinusoids may or may not be periodic, depending on whether the ratio ω1/ω2 is a rational number. Therefore, the period of this signal is not known. Hence, its power will be determined by averaging its energy over *T* seconds with *T* → ∞. Thus, - -$$ -P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} [C_1 \cos(\omega_1 t + \theta_1) + C_2 \cos(\omega_2 t + \theta_2)]^2 dt -$$ - -= -$$ -\lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C_1^2 \cos^2(\omega_1 t + \theta_1) dt + \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} C_2^2 \cos^2(\omega_2 t + \theta_2) dt -$$ - -+ -$$ -\lim_{T \to \infty} \frac{2C_1 C_2}{T} \int_{-T/2}^{T/2} \cos(\omega_1 t + \theta_1) \cos(\omega_2 t + \theta_2) dt -$$ - -The first and second integrals on the right-hand side are the powers of the two sinusoids, which are *C*1 2/2 and *C*2 2/2, as found in part (a). The third term, the product of two sinusoids, can be expressed as a sum of two sinusoids cos[(ω1+ω2)*t*+(θ1+θ2)] and cos[(ω1−ω2)*t*+(θ1−θ2)], respectively. Now, arguing as in part (a), we see that the third term is zero. Hence, we have† - -$$ -P_x = \frac{C_1^2}{2} + \frac{C_2}{2} -$$ - -2 - -and the rms value is (*C*1 2 +*C*2 2)/2. - -We can readily extend this result to a sum of any number of sinusoids with distinct frequencies. Thus, if - -$$ -x(t) = \sum_{n=1}^{\infty} C_n \cos{(\omega_n t + \theta_n)} -$$ - -assuming that none of the two sinusoids have identical frequencies and ω*n* = 0, then - -$$ -P_x = \frac{1}{2} \sum_{n=1}^{\infty} C_n^2 -$$ - -If *x*(*t*) also has a dc term, as - -$$ -x(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(\omega_n t + \theta_n) -$$ - -then - -$$ -P_x = C_0^2 + \frac{1}{2} \sum_{n=1}^{\infty} C_n^2 -$$ - (1.3) - -**(c)** In this case the signal is complex, and we use Eq. (1.2) to compute the power. - -$$ -P_x = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |De^{j\omega_0 t}|^2 dt -$$ - -Recall that |*ej*ω0*t* | = 1 so that |*Dej*ω0*t* | 2 = |*D*| 2, and - -$$ -P_x = |D|^2 \tag{1.4} -$$ - -The rms value is |*D*|. - -**Comment.** In part (b) of Ex. 1.2, we have shown that the power of the sum of two sinusoids is equal to the sum of the powers of the sinusoids. It may appear that the power of *x*1(*t*) + *x*2(*t*) - -† This is true only if ω1 = ω2. If ω1 = ω2, the integrand of the third term contains a constant cos(θ1 − θ2), and the third term → 2*C*1*C*2 cos(θ1 −θ2) as *T* → ∞. - -is *Px*1 + *Px*2 . Unfortunately, this conclusion is not true in general. It is true only under a certain condition (orthogonality), discussed later (Sec. 6.5-3). - -### **DR ILL 1.1 Computing Energy, Power, and RMS Value** - -Show that the energies of the signals in Figs. 1.3a, 1.3b, 1.3c, and 1.3d are 4, 1, 4/3, and 4/3, respectively. Observe that doubling a signal quadruples the energy, and time-shifting a signal has no effect on the energy. Show also that the power of the signal in Fig. 1.3e is 0.4323. What is the rms value of signal in Fig. 1.3e? - -## **DR ILL 1.2 Computing Power over a Period** - -Redo Ex. 1.1a to find the power of a sinusoid *C* cos(ω0*t* + θ ) by averaging the signal energy over one period *T*0 = 2π/ω0 (rather than averaging over the infinitely large interval). Show also that the power of a dc signal *x*(*t*) = *C*0 is *C*2 0, and its rms value is *C*0. - -## **DR ILL 1.3 Power of a Sum of Two Equal-Frequency Sinusoids** - -Show that if ω1 = ω2, the power of *x*(*t*) = *C*1 cos(ω1*t* +θ1)+*C*2 cos(ω2*t* +θ2) is [*C*1 2 +*C*2 2 + 2*C*1*C*2 cos(θ1 −θ2)]/2, which is not equal to the Ex. 1.2b result of (*C*1 2 +*C*2 2)/2. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/014_1.2 SOME USEFUL SIGNAL OPERATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/014_1.2 SOME USEFUL SIGNAL OPERATIONS.md deleted file mode 100644 index 02419fecae61b07b6e74faca3552f666a209bb34..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/014_1.2 SOME USEFUL SIGNAL OPERATIONS.md +++ /dev/null @@ -1,132 +0,0 @@ -## **1.2 SOME USEFUL SIGNAL [OPERATIONS](#page-7-0)** - -We discuss here three useful signal operations: shifting, scaling, and inversion. Since the independent variable in our signal description is time, these operations are discussed as *time shifting, time scaling,* and *time reversal* (inversion). However, this discussion is valid for functions having independent variables other than time (e.g., frequency or distance). - -### **[1.2-1 Time Shifting](#page-7-0)** - -Consider a signal *x*(*t*) (Fig. 1.4a) and the same signal delayed by *T* seconds (Fig. 1.4b), which we shall denote by φ(*t*). Whatever happens in *x*(*t*) (Fig. 1.4a) at some instant *t* also happens in φ(*t*) (Fig. 1.4b) *T* seconds later at the instant *t* +*T*. Therefore - -$$ -\phi(t+T) = x(t) \qquad \text{and} \qquad \phi(t) = x(t-T) -$$ - -Therefore, to time-shift a signal by *T*, we replace *t* with *t* − *T*. Thus *x*(*t* − *T*) represents *x*(*t*) time-shifted by *T* seconds. If *T* is positive, the shift is to the right (delay), as in Fig. 1.4b. If *T* is negative, the shift is to the left (advance), as in Fig. 1.4c. Clearly, *x*(*t* − 2) is *x*(*t*) delayed (right-shifted) by 2 seconds, and *x*(*t* + 2) is *x*(*t*) advanced (left-shifted) by 2 seconds. - -**Figure 1.4** Time-shifting a signal. - -### **EXAMPLE 1.3 Time Shifting** - -An exponential function *x*(*t*) = *e*−2*t* shown in Fig. 1.5a is delayed by 1 second. Sketch and mathematically describe the delayed function. Repeat the problem with *x*(*t*) advanced by 1 second. - -**Figure 1.5 (a)** Signal *x*(*t*). **(b)** Signal *x*(*t*) delayed by 1 second. **(c)** Signal *x*(*t*) advanced by 1 second. - -The function *x*(*t*) can be described mathematically as - -$$ -x(t) = \begin{cases} e^{-2t} & t \ge 0\\ 0 & t < 0 \end{cases} \tag{1.5} -$$ - -Let *xd*(*t*) represent the function *x*(*t*) delayed (right-shifted) by 1 second, as illustrated in Fig. 1.5b. This function is *x*(*t* − 1); its mathematical description can be obtained from *x*(*t*) by replacing *t* with *t* −1 in Eq. (1.5). Thus, - -$$ -x_d(t) = x(t-1) = \begin{cases} e^{-2(t-1)} & t-1 \ge 0 \text{ or } t \ge 1\\ 0 & t-1 < 0 \text{ or } t < 1 \end{cases} -$$ - -Let *xa*(*t*) represent the function *x*(*t*) advanced (left-shifted) by 1 second, as depicted in Fig. 1.5c. This function is *x*(*t* + 1); its mathematical description can be obtained from *x*(*t*) by replacing *t* with *t* +1 in Eq. (1.5). Thus, - -$$ -x_a(t) = x(t+1) = \begin{cases} e^{-2(t+1)} & t+1 \ge 0 \text{ or } t \ge -1\\ 0 & t+1 < 0 \text{ or } t < -1 \end{cases} -$$ - -### **DR ILL 1.4 Working with Time Delay and Time Advance** - -Write a mathematical description of the signal *x*3(*t*) in Fig. 1.3c. Next, delay this signal by 2 seconds. Sketch the delayed signal. Show that this delayed signal *xd*(*t*) can be described mathematically as *xd*(*t*) = 2(*t* − 2) for 2 ≤ *t* ≤ 3, and equal to 0 otherwise. Now repeat the procedure with the signal advanced (left-shifted) by 1 second. Show that this advanced signal *xa*(*t*) can be described as *xa*(*t*) = 2(*t* +1) for −1 ≤ *t* ≤ 0, and 0 otherwise. - -### **[1.2-2 Time Scaling](#page-7-0)** - -The compression or expansion of a signal in time is known as *time scaling*. Consider the signal *x*(*t*) of Fig. 1.6a. The signal φ(*t*) in Fig. 1.6b is *x*(*t*) compressed in time by a factor of 2. Therefore, whatever happens in *x*(*t*) at some instant *t* also happens to φ(*t*) at the instant *t*/2 so that - -$$ -\phi\left(\frac{t}{2}\right) = x(t) -$$ - and $\phi(t) = x(2t)$ - -Observe that because *x*(*t*) = 0 at *t* = *T*1 and *T*2, we must have φ(*t*) = 0 at *t* = *T*1/2 and *T*2/2, as shown in Fig. 1.6b. If *x*(*t*) were recorded on a tape and played back at twice the normal recording speed, we would obtain *x*(2*t*). In general, if *x*(*t*) is compressed in time by a factor *a* (*a* > 1), the resulting signal φ(*t*) is given by - -$$ -\phi(t) = x(at) -$$ - -Using a similar argument, we can show that *x*(*t*) expanded (slowed down) in time by a factor *a* (*a* > 1) is given by - -$$ -\phi(t) = x \left(\frac{t}{a}\right) -$$ - -Figure 1.6c shows *x*(*t*/2), which is *x*(*t*) expanded in time by a factor of 2. Observe that in a time-scaling operation, the origin *t* = 0 is the anchor point, which remains unchanged under the scaling operation because at *t* = 0, *x*(*t*) = *x*(*at*) = *x*(0). - -In summary, to time-scale a signal by a factor *a*, we replace *t* with *at*. If *a* > 1, the scaling results in compression, and if *a* < 1, the scaling results in expansion. - -### **EXAMPLE 1.4 Continuous Time-Scaling Operation** - -Figure 1.7a shows a signal *x*(*t*). Sketch and describe mathematically this signal time-compressed by factor 3. Repeat the problem for the same signal time-expanded by factor 2. - -The signal *x*(*t*) can be described as - -$$ -x(t) = \begin{cases} 2 & -1.5 \le t < 0 \\ 2e^{-t/2} & 0 \le t < 3 \\ 0 & \text{otherwise} \end{cases} -$$ - (1.6) - -Figure 1.7b shows *xc*(*t*), which is *x*(*t*) time-compressed by factor 3; consequently, it can be described mathematically as *x*(3*t*), which is obtained by replacing *t* with 3*t* in the right-hand side of Eq. (1.6). Thus, - -$$ -x_c(t) = x(3t) = \begin{cases} 2 & -1.5 \le 3t < 0 \text{ or } -0.5 \le t < 0 \\ 2e^{-3t/2} & 0 \le 3t < 3 \text{ or } 0 \le t < 1 \\ 0 & \text{otherwise} \end{cases} -$$ - -Observe that the instants *t* = −1.5 and 3 in *x*(*t*) correspond to the instants *t* = −0.5, and 1 in the compressed signal *x*(3*t*). - -Figure 1.7c shows *xe*(*t*), which is *x*(*t*) time-expanded by factor 2; consequently, it can be described mathematically as *x*(*t*/2), which is obtained by replacing *t* with *t*/2 in *x*(*t*). Thus, - -$$ -x_e(t) = x\left(\frac{t}{2}\right) = \begin{cases} 2 & -1.5 \le \frac{t}{2} < 0 \text{ or } -3 \le t < 0 \\ 2e^{-t/4} & 0 \le \frac{t}{2} < 3 \text{ or } 0 \le t < 6 \\ 0 & \text{otherwise} \end{cases} -$$ - -Observe that the instants *t* = −1.5 and 3 in *x*(*t*) correspond to the instants *t* = −3 and 6 in the expanded signal *x*(*t*/2). - -### **DR ILL 1.5 Compression and Expansion of Sinusoids** - -Show that the time compression by an integer factor *n* (*n* > 1) of a sinusoid results in a sinusoid of the same amplitude and phase, but with the frequency increased *n*-fold. Similarly, the time expansion by an integer factor *n* (*n* > 1) of a sinusoid results in a sinusoid of the same amplitude and phase, but with the frequency reduced by a factor *n*. Verify your conclusion by sketching a sinusoid sin 2*t* and the same sinusoid compressed by a factor 3 and expanded by a factor 2. - -### **[1.2-3 Time Reversal](#page-7-0)** - -Consider the signal *x*(*t*) in Fig. 1.8a. We can view *x*(*t*) as a rigid wire frame hinged at the vertical axis. To time-reverse *x*(*t*), we rotate this frame 180◦ about the vertical axis. This time reversal [the reflection of *x*(*t*) about the vertical axis] gives us the signal φ(*t*) (Fig. 1.8b). Observe that whatever happens in Fig. 1.8a at some instant *t* also happens in Fig. 1.8b at the instant −*t*, and vice versa. Therefore, - -φ(*t*) = *x*(−*t*) - -Thus, to time-reverse a signal we replace *t* with −*t*, and the time reversal of signal *x*(*t*) results in a signal *x*(−*t*). We must remember that the reversal is performed about the vertical axis, which acts as an anchor or a hinge. Recall also that the reversal of *x*(*t*) about the horizontal axis results in −*x*(*t*). - -**Figure 1.8** Time reversal of a signal. - -### **EXAMPLE 1.5 Time Reversal of a Signal** - -For the signal *x*(*t*) illustrated in Fig. 1.9a, sketch *x*(−*t*), which is time-reversed *x*(*t*). - -The instants −1 and −5 in *x*(*t*) are mapped into instants 1 and 5 in *x*(−*t*). Because *x*(*t*) = *et*/2, we have *x*(−*t*) = *e*−*t*/2. The signal *x*(−*t*) is depicted in Fig. 1.9b. We can describe *x*(*t*) and *x*(−*t*) as - -$$ -x(t) = \begin{cases} e^{t/2} & -1 \ge t > -5\\ 0 & \text{otherwise} \end{cases} -$$ - -and its time-reversed version *x*(−*t*) is obtained by replacing *t* with −*t* in *x*(*t*) as - -*x*(−*t*) = *e*−*t*/2 −1 ≥ −*t* > −5 or 1 ≤ *t* < 5 0 otherwise - -### **[1.2-4 Combined Operations](#page-7-0)** - -Certain complex operations require simultaneous use of more than one of the operations just described. The most general operation involving all the three operations is *x*(*at* − *b*), which is realized in two possible sequences of operation: - -- 1. Time-shift *x*(*t*) by *b* to obtain *x*(*t*−*b*). Now time-scale the shifted signal *x*(*t*−*b*) by *a* [i.e., replace *t* with *at*] to obtain *x*(*at* −*b*). -- 2. Time-scale *x*(*t*) by *a* to obtain *x*(*at*). Now time-shift *x*(*at*) by *b*/*a* [i.e., replace *t* with *t* − (*b*/*a*)] to obtain *x*[*a*(*t* − *b*/*a*)] = *x*(*at* − *b*). In either case, if *a* is negative, time scaling involves time reversal. - -For example, the signal *x*(2*t*−6) can be obtained in two ways. We can delay *x*(*t*) by 6 to obtain *x*(*t* − 6), and then time-compress this signal by factor 2 (replace *t* with 2*t*) to obtain *x*(2*t* − 6). Alternately, we can first time-compress *x*(*t*) by factor 2 to obtain *x*(2*t*), then delay this signal by 3 (replace *t* with *t* −3) to obtain *x*(2*t* −6). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/015_1.3 CLASSIFICATION OF SIGNALS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/015_1.3 CLASSIFICATION OF SIGNALS.md deleted file mode 100644 index 747efccce4c458b0ffd85d408ee08174735adbb1..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/015_1.3 CLASSIFICATION OF SIGNALS.md +++ /dev/null @@ -1,85 +0,0 @@ -## **[1.3 CLASSIFICATION OF](#page-7-0) SIGNALS** - -Classification helps us better understand and utilize the items around us. Cars, for example, are classified as sports, offroad, family, and so forth. Knowing you have a sports car is useful in deciding whether to drive on a highway or on a dirt road. Knowing you want to drive up a mountain, you would probably choose an offroad vehicle over a family sedan. Similarly, there are several classes of signals. Some signal classes are more suitable for certain applications than others. Further, different signal classes often require different mathematical tools. Here we shall consider only the following classes of signals, which are suitable for the scope of this book: - -- 1. Continuous-time and discrete-time signals -- 2. Analog and digital signals -- 3. Periodic and aperiodic signals -- 4. Energy and power signals -- 5. Deterministic and probabilistic signals - -### **[1.3-1 Continuous-Time and Discrete-Time Signals](#page-7-0)** - -A signal that is specified for a continuum of values of time *t* (Fig. 1.10a) is a *continuous-time signal,* and a signal that is specified only at discrete values of *t* (Fig. 1.10b) is a *discrete-time signal*. Telephone and video camera outputs are continuous-time signals, whereas the quarterly gross national product (GNP), monthly sales of a corporation, and stock market daily averages are discrete-time signals. - -### **[1.3-2 Analog and Digital Signals](#page-7-0)** - -The concept of continuous time is often confused with that of analog. The two are not the same. The same is true of the concepts of discrete time and digital. A signal whose amplitude can take on any value in a continuous range is an *analog signal*. This means that an analog signal amplitude can take on an infinite number of values. A *digital signal,* on the other hand, is one whose amplitude can take on only a finite number of values. Signals associated with a digital computer are digital because they take on only two values (binary signals). A digital signal whose amplitudes can take on *M* values is an *M*-ary signal of which binary (*M* = 2) is a special case. The terms *continuous time* and *discrete time* qualify the nature of a signal along the time (horizontal) axis. The terms *analog* and *digital,* on the other hand, qualify the nature of the signal amplitude (vertical axis). Figure 1.11 shows examples of signals of various types. It is clear that analog is not necessarily continuous-time and digital need not be discrete-time. Figure 1.11c shows an example of an analog discrete-time signal. An analog signal can be converted into a digital signal [analog-to-digital (A/D) conversion] through quantization (rounding off ), as explained in Sec. 8.3. - -**Figure 1.10 (a)** Continuous-time and **(b)** discrete-time signals. - -### **[1.3-3 Periodic and Aperiodic Signals](#page-7-0)** - -A signal *x*(*t*) is said to be *periodic* if for some positive constant *T*0 - -$$ -x(t) = x(t + T_0) \qquad \text{for all } t \tag{1.7} -$$ - -The *smallest* value of *T*0 that satisfies the periodicity condition of Eq. (1.7) is the *fundamental period* of *x*(*t*). The signals in Figs. 1.2b and 1.3e are periodic signals with periods 2 and 1, respectively. A signal is *aperiodic* if it is not periodic. Signals in Figs. 1.2a, 1.3a, 1.3b, 1.3c, and 1.3d are all aperiodic. - -By definition, a periodic signal *x*(*t*) remains unchanged when time-shifted by one period. For this reason, a periodic signal must start at *t* = −∞: if it started at some finite instant, say, *t* = 0, the time-shifted signal *x*(*t* + *T*0) would start at *t* = −*T*0 and *x*(*t* + *T*0) would not be the same as 80 CHAPTER 1 SIGNALS AND SYSTEMS - -**Figure 1.11** Examples of signals: **(a)** analog, continuous time; **(b)** digital, continuous time; **(c)** analog, discrete time; and **(d)** digital, discrete time. - -**Figure 1.12** A periodic signal of period *T*0. - -*x*(*t*). Therefore, a *periodic signal, by definition, must start at t* = −∞ *and continue forever, as illustrated in Fig. 1.12.* - -Another important property of a periodic signal *x*(*t*) is that *x*(*t*) can be generated by *periodic extension* of any segment of *x*(*t*) of duration *T*0 (the period). As a result, we can generate *x*(*t*) from any segment of *x*(*t*) having a duration of one period by placing this segment and the reproduction thereof end to end ad infinitum on either side. Figure 1.13 shows a periodic signal *x*(*t*) of period *T*0 = 6. The shaded portion of Fig. 1.13a shows a segment of *x*(*t*) starting at *t* = −1 and having a duration of one period (6 seconds). This segment, when repeated forever in either direction, results in the periodic signal *x*(*t*). Figure 1.13b shows another shaded segment of *x*(*t*) of duration *T*0 starting at *t* = 0. Again, we see that this segment, when repeated forever on either side, results in *x*(*t*). The reader can verify that this construction is possible with any segment of *x*(*t*) starting at any instant as long as the segment duration is one period. - -**Figure 1.13** Generation of a periodic signal by periodic extension of its segment of one-period duration. - -An additional useful property of a periodic signal *x*(*t*) of period *T*0 is that the area under *x*(*t*) over any interval of duration *T*0 is the same; that is, for any real numbers *a* and *b*, - -$$ -\int_{a}^{a+T_0} x(t) dt = \int_{b}^{b+T_0} x(t) dt -$$ - -This result follows from the fact that a periodic signal takes the same values at the intervals of *T*0. Hence, the values over any segment of duration *T*0 are repeated in any other interval of the same duration. For convenience, the area under *x*(*t*) over any interval of duration *T*0 will be denoted by - -$$ -\int_{T_0} x(t) \, dt -$$ - -It is helpful to label signals that start at *t* = −∞ and continue forever as *everlasting* signals. Thus, an everlasting signal exists over the entire interval −∞ < *t* < ∞. The signals in Figs. 1.1b and 1.2b are examples of everlasting signals. Clearly, a periodic signal, by definition, is an everlasting signal. - -A signal that does not start before *t* = 0 is a *causal* signal. In other words, *x*(*t*) is a causal signal if - -$$ -x(t) = 0 \qquad t < 0 -$$ - -The signals in Figs. 1.3a–1.3c are causal signals. A signal that starts before *t* = 0 is a *noncausal* signal. All the signals in Figs. 1.1 and 1.2 are noncausal. Observe that an everlasting signal is always noncausal but a noncausal signal is not necessarily everlasting. The everlasting signal in Fig. 1.2b is noncausal; however, the noncausal signal in Fig. 1.2a is not everlasting. A signal that is zero for all *t* ≥ 0 is called an *anti-causal* signal. - -**Comment.** A true everlasting signal cannot be generated in practice for obvious reasons. Why should we bother to postulate such a signal? In later chapters we shall see that certain signals - -### 82 CHAPTER 1 SIGNALS AND SYSTEMS - -(e.g., an impulse and an everlasting sinusoid) that cannot be generated in practice *do* serve a very useful purpose in the study of signals and systems. - -### **1.3-4 Energy and Power Signals** - -A signal with finite energy is an *energy signal,* and a signal with finite and nonzero power is a *power signal*. The signals in Figs. 1.2a and 1.2b are examples of energy and power signals, respectively. Observe that power is the time average of energy. Since the averaging is over an infinitely large interval, a signal with finite energy has zero power, and a signal with finite power has infinite energy. Therefore, a signal cannot be both an energy signal and a power signal. If it is one, it cannot be the other. On the other hand, there are signals that are neither energy nor power signals. The ramp signal is one such case. - -**Comments.** All practical signals have finite energies and are therefore energy signals. A power signal must necessarily have infinite duration; otherwise, its power, which is its energy averaged over an infinitely large interval, will not approach a (nonzero) limit. Clearly, it is impossible to generate a true power signal in practice because such a signal has infinite duration and infinite energy. - -Also, because of periodic repetition, periodic signals for which the area under |*x*(*t*)| 2 over one period is finite are power signals; however, not all power signals are periodic. - -### **DR ILL 1.6 Neither Energy nor Power** - -Show that an everlasting exponential *e*−*at* is neither an energy nor a power signal for any real value of *a*. However, if *a* is imaginary, it is a power signal with power *Px* = 1 regardless of the value of *a*. - -### **[1.3-5 Deterministic and Random Signals](#page-7-0)** - -A signal whose physical description is known completely, in either a mathematical form or a graphical form, is a *deterministic signal*. A signal whose values cannot be predicted precisely but are known only in terms of probabilistic description, such as mean value or mean-squared value, is a *random signal*. In this book we shall exclusively deal with deterministic signals. Random signals are beyond the scope of this study. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/016_1.4 SOME USEFUL SIGNAL MODELS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/016_1.4 SOME USEFUL SIGNAL MODELS.md deleted file mode 100644 index 65807c4915c5b398865067b49c04bcee999f23c4..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/016_1.4 SOME USEFUL SIGNAL MODELS.md +++ /dev/null @@ -1,261 +0,0 @@ -## **[1.4 SOME](#page-7-0) USEFUL SIGNAL MODELS** - -In the area of signals and systems, the step, the impulse, and the exponential functions play very important roles. Not only do they serve as a basis for representing other signals, but their use can simplify many aspects of the signals and systems. - -## **[1.4-1 The Unit Step Function](#page-7-0)** *u(t)* - -In much of our discussion, the signals begin at *t* = 0 (causal signals). Such signals can be conveniently described in terms of unit step function *u*(*t*) shown in Fig. 1.14a. This function is defined by - -$$ -u(t) = \begin{cases} 1 & t \ge 0 \\ 0 & t < 0 \end{cases} -$$ - (1.8) - -If we want a signal to start at *t* = 0 (so that it has a value of zero for *t* < 0), we need only multiply the signal by *u*(*t*). For instance, the signal *e*−*at* represents an everlasting exponential that starts at *t* = −∞. The causal form of this exponential (Fig. 1.14b) can be described as *e*−*atu*(*t*). - -The unit step function also proves very useful in specifying a function with different mathematical descriptions over different intervals. Examples of such functions appear in Fig. 1.7. These functions have different mathematical descriptions over different segments of time, as seen from Eqs. (1.5) and (1.6). Such a description often proves clumsy and inconvenient in mathematical treatment. We can use the unit step function to describe such functions by a single expression that is valid for all *t*. - -Consider, for example, the rectangular pulse depicted in Fig. 1.15a. We can express such a pulse in terms of familiar step functions by observing that the pulse *x*(*t*) can be expressed as the sum of the two delayed unit step functions, as shown in Fig. 1.15b. The unit step function *u*(*t*) delayed by *T* seconds is *u*(*t* −*T*). From Fig. 1.15b, it is clear that - -$$ -x(t) = u(t-2) - u(t-4) -$$ - -**Figure 1.14 (a)** Unit step function *u*(*t*). **(b)** Exponential *e*−*atu*(*t*). - -**Figure 1.15** Representation of a rectangular pulse by step functions. - -### **EXAMPLE 1.6 Describing a Triangle Function with the Unit Step** - -Use the unit step function to describe the signal in Fig. 1.16a. - -**Figure 1.16** Representation of a signal defined interval by interval. - -$$ -x_1(t) = t[u(t) - u(t-2)] -$$ - -The signal *x*2(*t*) can be obtained by multiplying another ramp by the gate pulse illustrated in Fig. 1.16c. This ramp has a slope −2; hence it can be described by −2*t* + *c*. Now, because the ramp has a zero value at *t* = 3, the constant *c* = 6, and the ramp can be described by −2(*t*−3). Also, the gate pulse in Fig. 1.16c is *u*(*t* −2)−*u*(*t* −3). Therefore, - -$$ -x_2(t) = -2(t-3)[u(t-2) - u(t-3)] -$$ - -The signal illustrated in Fig. 1.16a can be conveniently handled by breaking it up into the two components *x*1(*t*) and *x*2(*t*), depicted in Figs. 1.16b and 1.16c, respectively. Here, *x*1(*t*) can be obtained by multiplying the ramp *t* by the gate pulse *u*(*t*) − *u*(*t* − 2), as shown in Fig. 1.16b. Therefore, - -and - -$$ -x(t) = x_1(t) + x_2(t) -$$ - -= t[u(t) - u(t-2)] - 2(t-3)[u(t-2) - u(t-3)] -= tu(t) - 3(t-2)u(t-2) + 2(t-3)u(t-3) - -### **EXAMPLE 1.7 Describing a Piecewise Function with the Unit Step** - -Describe the signal in Fig. 1.7a by a single expression valid for all *t*. - -Over the interval from −1.5 to 0, the signal can be described by a constant 2, and over the interval from 0 to 3, it can be described by 2*e*−*t*/2. Therefore, - -$$ -x(t) = 2[u(t+1.5) - u(t)] + 2e^{-t/2}[u(t) - u(t-3)] -$$ - -constant part -$$ -= 2u(t+1.5) - 2(1 - e^{-t/2})u(t) - 2e^{-t/2}u(t-3) -$$ - -Compare this expression with the expression for the same function found in Eq. (1.6). - -### **DR ILL 1.7 Using Reflected Unit Step Functions** - -Show that the signals depicted in Figs. 1.17a and 1.17b can be described as *u*(−*t*) and *e*−*atu*(−*t*), respectively. - -### **DR ILL 1.8 Describing a Piecewise Function with the Unit Step** - -Show that the signal shown in Fig. 1.18 can be described as - -$$ -x(t) = (t-1)u(t-1) - (t-2)u(t-2) - u(t-4) -$$ - -## **[1.4-2 The Unit Impulse Function](#page-7-0)** *δ(t)* - -The unit impulse function δ(*t*) is one of the most important functions in the study of signals and systems. This function was first defined in two parts by P. A. M. Dirac as - -$$ -\delta(t) = 0 \quad t \neq 0 \quad \text{and} \quad \int_{-\infty}^{\infty} \delta(t) dt = 1 \tag{1.9} -$$ - -**Figure 1.18** Signal for Drill 1.8. - -We can visualize an impulse as a tall, narrow, rectangular pulse of unit area, as illustrated in Fig. 1.19b. The width of this rectangular pulse is a very small value → 0. Consequently, its height is a very large value 1/ → ∞. The unit impulse therefore can be regarded as a rectangular pulse with a width that has become infinitesimally small, a height that has become infinitely large, and an overall area that has been maintained at unity. Thus δ(*t*) = 0 everywhere except at *t* = 0, where it is undefined. For this reason, a unit impulse is represented by the spearlike symbol in Fig. 1.19a. - -Other pulses, such as the exponential, triangular, or Gaussian types, may also be used in impulse approximation. The important feature of the unit impulse function is not its shape but the fact that its effective duration (pulse width) approaches zero while its area remains at unity. For example, the exponential pulse α*e*−α*t u*(*t*) in Fig. 1.20a becomes taller and narrower as α increases. - -**Figure 1.19** A unit impulse and its - -**Figure 1.20** Other possible approximations to a unit impulse. - -In the limit as α → ∞, the pulse height → ∞, and its width or duration → 0. Yet, the area under the pulse is unity regardless of the value of α because - -$$ -\int_0^\infty \alpha e^{-\alpha t} dt = 1 -$$ - -The pulses in Figs. 1.20b and 1.20c behave in a similar fashion. Clearly, the exact impulse function cannot be generated in practice; it can only be approached. - -From Eq. (1.9), it follows that the function *k*δ(*t*) = 0 for all *t* = 0, and its area is *k*. Thus, *k*δ(*t*) is an impulse function whose area is *k* (in contrast to the unit impulse function, whose area is 1). - -### MULTIPLICATION OF A FUNCTION BY AN IMPULSE - -Let us now consider what happens when we multiply the unit impulse δ(*t*) by a function φ(*t*) that is known to be continuous at *t* = 0. Since the impulse has nonzero value only at *t* = 0, and the value of φ(*t*) at *t* = 0 is φ(0), we obtain - -$$ -\phi(t)\delta(t) = \phi(0)\delta(t) -$$ - -Thus, multiplication of a continuous-time function φ(*t*) with an unit impulse located at *t* = 0 results in an impulse, which is located at *t* = 0 and has strength φ(0) [the value of φ(*t*) at the location of the impulse]. Use of exactly the same argument leads to the generalization of this result, stating that provided φ(*t*) is continuous at *t* = *T*,φ(*t*) multiplied by an impulse δ(*t* − *T*) (impulse located at *t* = *T*) results in an impulse located at *t* = *T* and having strength φ(*T*) [the value of φ(*t*) at the location of the impulse]. - -$$ -\phi(t)\delta(t-T) = \phi(T)\delta(t-T) \tag{1.10} -$$ - -### SAMPLING PROPERTY OF THE UNIT IMPULSE FUNCTION - -From Eq. (1.10) it follows that - -$$ -\int_{-\infty}^{\infty} \phi(t)\delta(t-T) dt = \phi(T) \int_{-\infty}^{\infty} \delta(t) dt = \phi(T) -$$ -\n(1.11) - -provided φ(*t*) is continuous at *t* = *T*. This result means that *the area under the product of a function with an impulse* δ(*t* − *T*) *is equal to the value of that function at the instant at which the unit impulse is located.* This property is very important and useful and is known as the *sampling* or *sifting property* of the unit impulse. - -### UNIT IMPULSE AS A GENERALIZED FUNCTION - -The definition of the unit impulse function given in Eq. (1.9) is not mathematically rigorous, which leads to serious difficulties. First, the impulse function does not define a unique function: for example, it can be shown that δ(*t*)+δ(˙ *t*) also satisfies Eq. (1.9) [1]. Moreover, δ(*t*) is not even a true function in the ordinary sense. An ordinary function is specified by its values for all time *t*. The impulse function is zero everywhere except at *t* = 0, and at this, the only interesting part of its range, it is undefined. These difficulties are resolved by defining the impulse as a generalized function rather than an ordinary function. A *generalized function* is defined by its effect on other functions instead of by its value at every instant of time. - -In this approach the impulse function is defined by the sampling property [Eq. (1.11)]. We say nothing about what the impulse function is or what it looks like. Instead, the impulse function is defined in terms of its effect on a test function φ(*t*). We define a unit impulse as a function for which the area under its product with a function φ(*t*) is equal to the value of the function φ(*t*) at the instant at which the impulse is located. It is assumed that φ(*t*) is continuous at the location of the impulse. Recall that the sampling property [Eq. (1.11)] is the consequence of the classical (Dirac) definition of the unit impulse in Eq. (1.9). In contrast, *the sampling property [Eq. (1.11)] defines the impulse function in the generalized function approach.* - -We now present an interesting application of the generalized function definition of an impulse. Because the unit step function *u*(*t*) is discontinuous at *t* = 0, its derivative *du*/*dt* does not exist at *t* = 0 in the ordinary sense. We now show that this derivative *does* exist in the generalized sense, and it is, in fact, δ(*t*). As a proof, let us evaluate the integral of (*du*/*dt*)φ(*t*), using integration by parts: - -$$ -\int_{-\infty}^{\infty} \frac{du(t)}{dt} \phi(t) dt = u(t) \phi(t) \Big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} u(t) \dot{\phi}(t) dt -$$ -$$ -= \phi(\infty) - 0 - \int_{0}^{\infty} \dot{\phi}(t) dt -$$ -$$ -= \phi(\infty) - \phi(t) \Big|_{0}^{\infty} = \phi(0) -$$ - -This result shows that *du*/*dt* satisfies the sampling property of δ(*t*). Therefore it is an impulse δ(*t*) in the generalized sense—that is, - -$$ -\frac{du(t)}{dt} = \delta(t) \tag{1.12} -$$ - -Consequently, - -$$ -\int_{-\infty}^{t} \delta(\tau) d\tau = u(t) -$$ - -These results can also be obtained graphically from Fig. 1.19b. We observe that the area from −∞ to *t* under the limiting form of δ(*t*) in Fig. 1.19b is zero if *t* < −/2 and unity if *t* ≥ /2 with → 0. Consequently, - -$$ -\int_{-\infty}^{t} \delta(\tau) d\tau = \begin{cases} 0 & t < 0 \\ 1 & t \ge 0 \end{cases} -$$ -$$ -= u(t) -$$ - -This result shows that the unit step function can be obtained by integrating the unit impulse function. Similarly the unit ramp function *x*(*t*) = *tu*(*t*) can be obtained by integrating the unit step function. We may continue with unit parabolic function *t* 2/2 obtained by integrating the unit ramp, and so on. On the other side, we have derivatives of impulse function, which can be defined as generalized functions (see Prob. 1.4-12). All these functions, derived from the unit impulse function (successive derivatives and integrals), are called *singularity functions*. † - -### **DR ILL 1.9 Simplifying Expressions Containing the Unit Impulse** - -Show that - -(a) -$$ -(t^3 + 3)\delta(t) = 3\delta(t) -$$ - -\n(b) $\left[\sin\left(t^2 - \frac{\pi}{2}\right)\right] \delta(t) = -\delta(t)$ -\n(c) $e^{-2t}\delta(t) = \delta(t)$ - -(d) -$$ -\frac{\omega^2 + 1}{\omega^2 + 9} \delta(\omega - 1) = \frac{1}{5} \delta(\omega - 1) -$$ - -## **DR ILL 1.10 Simplifying Integrals Containing the Unit Impulse** - -Show that - -(a) -$$ -\int_{-\infty}^{\infty} \delta(t) e^{-j\omega t} dt = 1 -$$ - -\n(b) -$$ -\int_{-\infty}^{\infty} \delta(t-2) \cos\left(\frac{\pi t}{4}\right) dt = 0 -$$ - -\n(c) -$$ -\int_{-\infty}^{\infty} e^{-2(x-t)} \delta(2-t) dt = e^{-2(x-2)} -$$ - -## **[1.4-3 The Exponential Function](#page-7-0)** *est* - -Another important function in the area of signals and systems is the exponential signal *est*, where *s* is complex in general, given by - -``` -s = σ +jω -``` - -† Singularity functions were defined by late Prof. S. J. Mason as follows. A singularity is a point at which a function does not possess a derivative. Each of the singularity functions (or if not the function itself, then the function differentiated a finite number of times) has a singular point at the origin and is zero elsewhere [2]. - -Therefore, - -$$ -e^{st} = e^{(\sigma + j\omega)t} = e^{\sigma t} e^{j\omega t} = e^{\sigma t} (\cos \omega t + j \sin \omega t) -$$ - (1.13) - -Since *s*∗ = σ −*j*ω (the conjugate of *s*), then - -$$ -e^{s^*t} = e^{(\sigma - j\omega)t} = e^{\sigma t}e^{-j\omega t} = e^{\sigma t}(\cos \omega t - j\sin \omega t) -$$ - -and - -$$ -e^{\sigma t} \cos \omega t = \frac{1}{2} (e^{st} + e^{s^* t}) -$$ -\n(1.14) - -A comparison of Eq. (1.13) with Euler's formula shows that *est* is a generalization of the function *ej*ω*t* , where the frequency variable *j*ω is generalized to a complex variable *s* = σ + *j*ω. For this reason, we designate the variable *s* as the *complex frequency*. In fact, function *est* encompasses a large class of functions. The following functions are either special cases of or can be expressed in terms of *est*: - -- 1. A constant *k* = *ke*0*t* (*s* = 0) -- 2. A monotonic exponential *e*σ*t* (ω = 0, *s* = σ ) -- 3. A sinusoid cos ω*t* (σ = 0, *s* = ±*j*ω) -- 4. An exponentially varying sinusoid *e*σ*t* cos ω*t* (*s* = σ ±*j*ω) - -These functions are illustrated in Fig. 1.21. - -The complex frequency *s* can be conveniently represented on a *complex frequency plane* (*s* plane), as depicted in Fig. 1.22. The horizontal axis is the real axis (σ axis), and the vertical axis is the imaginary axis (ω axis). The absolute value of the imaginary part of *s* is |ω| (the - -**Figure 1.21** Sinusoids of complex frequency σ +*j*ω. - -**Figure 1.22** Complex frequency plane. - -*radian* frequency), which indicates the frequency of oscillation of *est*; the real part σ (the *neper* frequency) gives information about the rate of increase or decrease of the amplitude of *est*. For signals whose complex frequencies lie on the real axis (σ axis, where ω = 0), the frequency of oscillation is zero. Consequently these signals are monotonically increasing or decreasing exponentials (Fig. 1.21a). For signals whose frequencies lie on the imaginary axis (ω axis, where σ = 0), *e*σ*t* = 1. Therefore, these signals are conventional sinusoids with constant amplitude (Fig. 1.21b). The case *s* = 0 (σ = ω = 0) corresponds to a constant (dc) signal because *e*0*t* = 1. For the signals illustrated in Figs. 1.21c and 1.21d, both σ and ω are nonzero; the frequency *s* is complex and does not lie on either axis. The signal in Fig. 1.21c decays exponentially. Therefore, σ is negative, and *s* lies to the left of the imaginary axis. In contrast, the signal in Fig. 1.21d *grows* exponentially. Therefore, σ is positive, and *s* lies to the right of the imaginary axis. Thus the *s* plane (Fig. 1.21) can be separated into two parts: the *left half-plane* (LHP) corresponding to exponentially decaying signals and the *right half-plane* (RHP) corresponding to exponentially growing signals. The imaginary axis separates the two regions and corresponds to signals of constant amplitude. - -An exponentially growing sinusoid *e*2*t* cos 5*t*, for example, can be expressed as a linear combination of exponentials *e*(2+*j*5)*t* and *e*(2−*j*5)*t* with complex frequencies 2 + *j*5 and 2−*j*5, respectively, which lie in the RHP. An exponentially decaying sinusoid *e*−2*t* cos 5*t* can be expressed as a linear combination of exponentials *e*(−2+*j*5)*t* and *e*(−2−*j*5)*t* with complex frequencies −2 + *j*5 and −2 − *j*5, respectively, which lie in the LHP. A constant-amplitude sinusoid cos 5*t* can be expressed as a linear combination of exponentials *ej*5*t* and *e*−*j*5*t* with complex frequencies ±*j*5, which lie on the imaginary axis. Observe that the monotonic exponentials *e*±2*t* are also generalized sinusoids with complex frequencies ±2. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/017_1.5 EVEN AND ODD FUNCTIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/017_1.5 EVEN AND ODD FUNCTIONS.md deleted file mode 100644 index 52638f57be5f511474076461c73bed35180520be..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/017_1.5 EVEN AND ODD FUNCTIONS.md +++ /dev/null @@ -1,88 +0,0 @@ -## **[1.5 EVEN AND](#page-7-0) ODD FUNCTIONS** - -A function *xe*(*t*) is said to be an *even function* of *t* if it is symmetrical about the vertical axis. A function *xo*(*t*) is said to be an *odd function* of *t* if it is antisymmetrical about the vertical axis. Mathematically expressed, these symmetry conditions require - -$$ -x_e(t) = x_e(-t) -$$ - and $x_o(t) = -x_o(-t)$ (1.15) - -An even function has the same value at the instants *t* and −*t* for all values of *t*. On the other hand, the value of an odd function at the instant *t* is the negative of its value at the instant −*t*. An example even signal and an example odd signal are shown in Figs. 1.23a and 1.23b, respectively. - -### **[1.5-1 Some Properties of Even and Odd Functions](#page-7-0)** - -Even and odd functions have the following properties: - -even function ×odd function = odd function odd function ×odd function = even function even function ×even function = even function - -The proofs are trivial and follow directly from the definition of odd and even functions [Eq. (1.15)]. - -### AREA - -Because of the symmetries of even and odd functions about the vertical axis, it follows from Eq. (1.15) [or Fig. 1.23] that - -$$ -\int_{-a}^{a} x_e(t) dt = 2 \int_{0}^{a} x_e(t) dt \quad \text{and} \quad \int_{-a}^{a} x_o(t) dt = 0 \quad (1.16) -$$ - -These results are valid under the assumption that there is no impulse (or its derivatives) at the origin. The proof of these statements is obvious from the plots of even and odd functions. Formal proofs, left as an exercise for the reader, can be accomplished by using the definitions in Eq. (1.15). - -Because of their properties, study of odd and even functions proves useful in many applications, as will become evident in later chapters. - -### **[1.5-2 Even and Odd Components of a Signal](#page-7-0)** - -Every signal *x*(*t*) can be expressed as a sum of even and odd components because - -$$ -x(t) = \underbrace{\frac{1}{2}[x(t) + x(-t)]}_{\text{even}} + \underbrace{\frac{1}{2}[x(t) - x(-t)]}_{\text{odd}} -$$ -(1.17) - -From the definitions in Eq. (1.15), we can clearly see that the first component on the right-hand side is an even function, while the second component is odd. This is apparent from the fact that replacing *t* by −*t* in the first component yields the same function. The same maneuver in the second component yields the negative of that component. - -### **EXAMPLE 1.8 Finding the Even and Odd Components of a Signal** - -Find and sketch the even and odd components of *x*(*t*) = *e*−*atu*(*t*). - -Based on Eq. (1.17), we can express *x*(*t*) as a sum of the even component *xe*(*t*) and the odd component *xo*(*t*) as - -$$ -x(t) = x_e(t) + x_o(t) -$$ - -where - -$$ -x_e(t) = \frac{1}{2} [e^{-at}u(t) + e^{at}u(-t)] \quad \text{and} \quad x_o(t) = \frac{1}{2} [e^{-at}u(t) - e^{at}u(-t)] -$$ - -The function *e*−*atu*(*t*) and its even and odd components are illustrated in Fig. 1.24. - -### **EXAMPLE 1.9 Finding the Even and Odd Components of a Complex Signal** - -Find the even and odd components of *ejt*. - -From Eq. (1.17), - -*ejt* = *xe*(*t*)+*xo*(*t*) - -where - -*xe*(*t*) = 1 2 [*ejt* +*e*−*jt*] = cos *t* and *xo*(*t*) = 1 2 [*ejt* −*e*−*jt*] = *j*sin *t* - -### A MODIFICATION FOR COMPLEX SIGNALS - -While a complex signal can be decomposed into even and odd components, it is more common to decompose complex signals using conjugate symmetries. A complex signal *x*(*t*) is said to be *conjugate-symmetric* if *x*(*t*) = *x*∗(−*t*). A conjugate-symmetric signal is even in the real part and odd in the imaginary part. Thus, a real conjugate-symmetric signal is an even signal. A signal is *conjugate-antisymmetric* if *x*(*t*) = −*x*∗(−*t*). A conjugate-antisymmetric signal is odd in the real part and even in the imaginary part. A real conjugate-antisymmetric signal is an odd signal. Any signal *x*(*t*) can be decomposed into a conjugate-symmetric portion *xcs*(*t*) plus a conjugate-antisymmetric portion *xca*(*t*). That is, - -$$ -x(t) = x_{cs}(t) + x_{ca}(t) -$$ - -where - -$$ -x_{cs}(t) = \frac{x(t) + x^*(-t)}{2} -$$ - and $x_{ca}(t) = \frac{x(t) - x^*(-t)}{2}$ - -The proof is similar to the one for decomposing a signal into even and odd components. As we shall see in later chapters, conjugate symmetries commonly occur in real-world signals and their transforms. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/018_1.6 SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/018_1.6 SYSTEMS.md deleted file mode 100644 index 9a1b4942520f153fedb0dda2534daf1d63d0c121..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/018_1.6 SYSTEMS.md +++ /dev/null @@ -1,57 +0,0 @@ -## **[1.6 SYSTEMS](#page-7-0)** - -As mentioned in Sec. 1.1, systems are used to process signals to allow modification or extraction of additional information from the signals. A system may consist of physical components (hardware realization) or of an algorithm that computes the output signal from the input signal (software realization). - -Roughly speaking, a physical system consists of interconnected components, which are characterized by their terminal (input–output) relationships. In addition, a system is governed by laws of interconnection. For example, in electrical systems, the terminal relationships are the familiar voltage-current relationships for the resistors, capacitors, inductors, transformers, transistors, and so on, as well as the laws of interconnection (i.e., Kirchhoff's laws). We use these laws to derive mathematical equations relating the outputs to the inputs. These equations then represent a *mathematical model* of the system. - -A system can be conveniently illustrated by a "black box" with one set of accessible terminals where the input variables *x*1(*t*), *x*2(*t*), ..., *xj*(*t*) are applied and another set of accessible terminals where the output variables *y*1(*t*), *y*2(*t*),..., *yk*(*t*) are observed (Fig. 1.25). - -The study of systems consists of three major areas: mathematical modeling, analysis, and design. Although we shall be dealing with mathematical modeling, our main concern is with - -| x1(t) | y1(t) | | -|--------|--------|-----------------------------------------| -| x2(t) | y2(t) | | -| •
• | •
• | | -| •
• | •
• | | -| •
• | •
• | | -| xj(t) | yk(t) | Figure 1.25 Representation of a system. | - -analysis and design. The major portion of this book is devoted to the analysis problem—how to determine the system outputs for the given inputs and a given mathematical model of the system (or rules governing the system). To a lesser extent, we will also consider the problem of design or synthesis—how to construct a system that will produce a desired set of outputs for the given inputs. - -### DATA NEEDED TO COMPUTE SYSTEM RESPONSE - -To understand what data we need to compute a system response, consider a simple *RC* circuit with a current source *x*(*t*) as its input (Fig. 1.26). - -The output voltage *y*(*t*) is given by - -$$ -y(t) = Rx(t) + \frac{1}{C} \int_{-\infty}^{t} x(\tau) d\tau -$$ -\n(1.18) - -The limits of the integral on the right-hand side are from −∞ to *t* because this integral represents the capacitor charge due to the current *x*(*t*) flowing in the capacitor, and this charge is the result of the current flowing in the capacitor from −∞. Now, Eq. (1.18) can be expressed as - -$$ -y(t) = Rx(t) + \frac{1}{C} \int_{-\infty}^{0} x(\tau) d\tau + \frac{1}{C} \int_{0}^{t} x(\tau) d\tau -$$ - -The middle term on the right-hand side is *vC*(0), the capacitor voltage at *t* = 0. Therefore, - -$$ -y(t) = v_C(0) + Rx(t) + \frac{1}{C} \int_0^t x(\tau) d\tau \qquad t \ge 0 -$$ - -This equation can be readily generalized as - -$$ -y(t) = v_C(t_0) + Rx(t) + \frac{1}{C} \int_{t_0}^t x(\tau) d\tau \qquad t \ge t_0 -$$ -\n(1.19) - -From Eq. (1.18), the output voltage *y*(*t*) at an instant *t* can be computed if we know the input current flowing in the capacitor throughout its entire past (−∞ to *t*). Alternatively, if we know the input current *x*(*t*) from some moment *t*0 onward, then, using Eq. (1.19), we can still calculate *y*(*t*) for *t* ≥ *t*0 from a knowledge of the input current, provided we know *vC*(*t*0), the initial capacitor voltage (voltage at *t*0). Thus *vC*(*t*0) contains all the relevant information about the circuit's entire - -**Figure 1.26** Example of a simple electrical system. - -past (−∞ to *t*0) that we need to compute *y*(*t*) for *t* ≥ *t*0. Therefore, the response of a system at *t* ≥ *t*0 can be determined from its input(s) during the interval *t*0 to *t* and from certain *initial conditions* at *t* = *t*0. - -In the preceding example, we needed only one initial condition. However, in more complex systems, several initial conditions may be necessary. We know, for example, that in passive *RLC* networks, the initial values of all inductor currents and all capacitor voltages† are needed to determine the outputs at any instant *t* ≥ 0 if the inputs are given over the interval [0,*t*]. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/019_1.7 CLASSIFICATION OF SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/019_1.7 CLASSIFICATION OF SYSTEMS.md deleted file mode 100644 index 170322800114887471cbe81788d41e9bf3262d11..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/019_1.7 CLASSIFICATION OF SYSTEMS.md +++ /dev/null @@ -1,344 +0,0 @@ -## **[1.7 CLASSIFICATION OF](#page-7-0) SYSTEMS** - -Systems may be classified broadly in the following categories: - -- 1. Linear and nonlinear systems -- 2. Constant-parameter and time-varying-parameter systems -- 3. Instantaneous (memoryless) and dynamic (with memory) systems -- 4. Causal and noncausal systems -- 5. Continuous-time and discrete-time systems -- 6. Analog and digital systems -- 7. Invertible and noninvertible systems -- 8. Stable and unstable systems - -Other classifications, such as deterministic and probabilistic systems, are beyond the scope of this text and are not considered. - -### **[1.7-1 Linear and Nonlinear Systems](#page-7-0)** - -### THE CONCEPT OF LINEARITY - -A system whose output is proportional to its input is an *example* of a linear system. But linearity implies more than this; it also implies the *additivity property:* that is, if several inputs are acting on a system, then the total effect on the system due to all these inputs can be determined by considering one input at a time while assuming all the other inputs to be zero. The total effect is then the sum of all the component effects. This property may be expressed as follows: for a linear system, if an input *x*1 acting alone has an effect *y*1, and if another input *x*2, also acting alone, has an effect *y*2, then, with both inputs acting on the system, the total effect will be *y*1 +*y*2. Thus, if - -$$ -x_1 \longrightarrow y_1 -$$ - and $x_2 \longrightarrow y_2$ - -then for all *x*1 and *x*2 - -*x*1 +*x*2 −→ *y*1 +*y*2 (1.20) - -In addition, a linear system must satisfy the *homogeneity* or scaling property, which states that for arbitrary real or imaginary number *k*, if an input is increased *k*-fold, the effect also increases *k*-fold. Thus, if - -*x* −→ *y* - -† Strictly speaking, this means independent inductor currents and capacitor voltages. - -### 98 CHAPTER 1 SIGNALS AND SYSTEMS - -then for all real or imaginary *k* - -$$ -kx \longrightarrow ky \tag{1.21} -$$ - -Thus, linearity implies two properties: homogeneity (scaling) and additivity.† Both these properties can be combined into one property (*superposition*), which is expressed as follows: If - -$$ -x_1 \longrightarrow y_1 -$$ - and $x_2 \longrightarrow y_2$ - -then for all inputs *x*1 and *x*2 and all constants *k*1 and *k*2, - -$$ -k_1x_1 + k_2x_2 \longrightarrow k_1y_1 + k_2y_2 \tag{1.22} -$$ - -There is another useful way to view the linearity condition described in Eq. (1.22): the response of a linear system is unchanged whether the operations of summing and scaling precede the system (sum and scale act on inputs) or follow the system (sum and scale act on outputs). *Thus, linearity implies commutability between a system and the operations of summing and scaling.* It may appear that additivity implies homogeneity. Unfortunately, homogeneity does not always follow from additivity. Drill 1.11 demonstrates such a case. - -### **DR ILL 1.11 Additivity but Not Homogeneity** - -Show that a system with the input *x*(*t*) and the output *y*(*t*) related by *y*(*t*) = Re{*x*(*t*)} satisfies the additivity property but violates the homogeneity property. Hence, such a system is not linear. [*Hint:* Show that Eq. (1.21) is not satisfied when *k* is complex.] - -### RESPONSE OF A LINEAR SYSTEM - -For the sake of simplicity, we discuss only *single-input, single-output* (*SISO*) systems. But the discussion can be readily extended to *multiple-input, multiple-output* (*MIMO*) systems. - -A system's output for *t* ≥ 0 is the result of two independent causes: the initial conditions of the system (or the system state) at *t* = 0 and the input *x*(*t*) for *t* ≥ 0. If a system is to be linear, the output must be the sum of the two components resulting from these two causes: first, the *zero-input response* (ZIR) that results only from the initial conditions at *t* = 0 with the input *x*(*t*) = 0 for *t* ≥ 0, and then the *zero-state response* (ZSR) that results only from the input *x*(*t*) for *t* ≥ 0 when the initial conditions (at *t* = 0) are assumed to be zero. When all the appropriate initial conditions are zero, the system is said to be in *zero state*. The system output is zero when the input is zero only if the system is in zero state. - -In summary, a linear system response can be expressed as the sum of the zero-input and zero-state responses: - -total response = zero-input response + zero-state response - -† A linear system must also satisfy the additional condition of *smoothness,* where small changes in the system's inputs must result in small changes in its outputs [3]. - -This property of linear systems, which permits the separation of an output into components resulting from the initial conditions and from the input, is called the *decomposition property*. For the *RC* circuit of Fig. 1.26, the response *y*(*t*) was found to be [see Eq. (1.19) with *t*0 = 0] - -$$ -y(t) = \underbrace{v_C(0)}_{\text{ZIR}} + \underbrace{Rx(t) + \frac{1}{C} \int_0^t x(\tau) d\tau}_{\text{ZSR}} \tag{1.23} -$$ - -From Eq. (1.23), it is clear that if the input *x*(*t*) = 0 for *t* ≥ 0, the output *y*(*t*) = *vC*(0). Hence *vC*(0) is the zero-input response of the response *y*(*t*). Similarly, if the system state (the voltage *vC* in this case) is zero at *t* = 0, the output is given by the second component on the right-hand side of Eq. (1.23). Clearly this is the zero-state response of the response *y*(*t*). - -In addition to the decomposition property, linearity implies that both the zero-input and zero-state components must obey the principle of superposition with respect to each of their respective causes. For example, if we increase the initial condition *k*-fold, the zero-input response must also increase *k*-fold. Similarly, if we increase the input *k*-fold, the zero-state response must also increase *k*-fold. These facts can be readily verified from Eq. (1.23) for the *RC* circuit in Fig. 1.26. For instance, if we double the initial condition *vC*(0), the zero-input response doubles; if we double the input *x*(*t*), the zero-state response doubles. - -## **EXAMPLE 1.10 Linearity of Constant-Coefficient Linear Differential Equations** - -Show that the system described by the equation - -$$ -\frac{dy(t)}{dt} + 3y(t) = x(t) -$$ - (1.24) - -is linear. - -Let the system response to the inputs *x*1(*t*) and *x*2(*t*) be *y*1(*t*) and *y*2(*t*), respectively. Then - -$$ -\frac{dy_1(t)}{dt} + 3y_1(t) = x_1(t) \qquad \text{and} \qquad \frac{dy_2(t)}{dt} + 3y_2(t) = x_2(t) -$$ - -Multiplying the first equation by *k*1, the second by *k*2, and adding them yield - -$$ -\frac{d}{dt}[k_1y_1(t) + k_2y_2(t)] + 3[k_1y_1(t) + k_2y_2(t)] = k_1x_1(t) + k_2x_2(t) -$$ - -But this equation is the system equation [Eq. (1.24)] with - -*x*(*t*) = *k*1*x*1(*t*)+*k*2*x*2(*t*) and *y*(*t*) = *k*1*y*1(*t*)+*k*2*y*2(*t*) - -Therefore, when the input is *k*1*x*1(*t*) + *k*2*x*2(*t*), the system response is *k*1*y*1(*t*) + *k*2*y*2(*t*). Consequently, the system is linear. Using this argument, we can readily generalize the result to - -### 100 CHAPTER 1 SIGNALS AND SYSTEMS - -show that a system described by a differential equation of the form - -$$ -a_0 \frac{d^N y(t)}{dt^N} + a_1 \frac{d^{N-1} y(t)}{dt^{N-1}} + \dots + a_N y(t) = b_{N-M} \frac{d^M x(t)}{dt^M} + \dots + b_{N-1} \frac{dx(t)}{dt} + b_N x(t) \quad (1.25) -$$ - -is a linear system. The coefficients *ai* and *bi* in this equation can be constants or functions of time. Although here we proved only zero-state linearity, it can be shown that such systems are also zero-input linear and have the decomposition property. - -## **DR ILL 1.12 Linearity of a Differential Equation with Time-Varying Parameters** - -Show that the system described by the following equation is linear: - -$$ -\frac{dy(t)}{dt} + t^2 y(t) = (2t + 3)x(t) -$$ - -### **DR ILL 1.13 A Nonlinear Differential Equation** - -Show that the system described by the following equation is nonlinear: - -$$ -y(t)\frac{dy(t)}{dt} + 3y(t) = x(t) -$$ - -### MORE COMMENTS ON LINEAR SYSTEMS - -Almost all systems observed in practice become nonlinear when large enough signals are applied to them. However, it is possible to approximate most of the nonlinear systems by linear systems for small-signal analysis. The analysis of nonlinear systems is generally difficult. Nonlinearities can arise in so many ways that describing them with a common mathematical form is impossible. Not only is each system a category in itself, but even for a given system, changes in initial conditions or input amplitudes may change the nature of the problem. On the other hand, the superposition property of linear systems is a powerful unifying principle that allows for a general solution. The superposition property (linearity) greatly simplifies the analysis of linear systems. Because of the decomposition property, we can evaluate separately the two components of the output. The zero-input response can be computed by assuming the input to be zero, and the zero-state response can be computed by assuming zero initial conditions. Moreover, if we express an input *x*(*t*) as a sum of simpler functions, - -$$ -x(t) = a_1 x_1(t) + a_2 x_2(t) + \cdots + a_m x_m(t) -$$ - -then, by virtue of linearity, the response *y*(*t*) is given by - -$$ -y(t) = a_1 y_1(t) + a_2 y_2(t) + \cdots + a_m y_m(t) -$$ - -where *yk*(*t*) is the zero-state response to an input *xk*(*t*). This apparently trivial observation has profound implications. As we shall see repeatedly in later chapters, it proves extremely useful and opens new avenues for analyzing linear systems. - -For example, consider an arbitrary input *x*(*t*) such as the one shown in Fig. 1.27a. We can approximate *x*(*t*) with a sum of rectangular pulses of width *t* and of varying heights. The approximation improves as *t* → 0, when the rectangular pulses become impulses spaced *t* seconds apart (with *t* → 0).† Thus, an arbitrary input can be replaced by a weighted sum of impulses spaced *t* (*t* → 0) seconds apart. Therefore, if we know the system response to a unit impulse, we can immediately determine the system response to an arbitrary input *x*(*t*) by adding the system response to each impulse component of *x*(*t*). A similar situation is depicted in Fig. 1.27b, where *x*(*t*) is approximated by a sum of step functions of varying magnitude and spaced *t* seconds apart. The approximation improves as *t* becomes smaller. Therefore, if we know the system response to a unit step input, we can compute the system response to any arbitrary input *x*(*t*) with relative ease. Time-domain analysis of linear systems (discussed in Ch. 2) uses this approach. - -Chapters 4, 5, 6, and 7 employ the same approach but instead use sinusoids or exponentials as the basic signal components. We show that any arbitrary input signal can be expressed as a weighted sum of sinusoids (or exponentials) having various frequencies. Thus a knowledge of the system response to a sinusoid enables us to determine the system response to an arbitrary input *x*(*t*). - -**Figure 1.27** Signal representation in terms of impulse and step components. - -† Here, the discussion of a rectangular pulse approaching an impulse at *t* → 0 is somewhat imprecise. It is explained in Sec. 2.4 with more rigor. - -### **[1.7-2 Time-Invariant and Time-Varying Systems](#page-7-0)** - -Systems whose parameters do not change with time are *time-invariant* (also *constant-parameter*) systems. For such a system, if the input is delayed by *T* seconds, the output is the same as before but delayed by *T* (assuming initial conditions are also delayed by *T*). This property is expressed graphically in Fig. 1.28. We can also illustrate this property, as shown in Fig. 1.29. We can delay the output *y*(*t*) of a system *S* by applying the output *y*(*t*) to a *T* second delay (Fig. 1.29a). If the system is time invariant, then the delayed output *y*(*t*−*T*) can also be obtained by first delaying the input *x*(*t*) before applying it to the system, as shown in Fig. 1.29b. In other words, the system *S* and the time delay commute if the system *S* is time invariant. This would not be true for time-varying systems. Consider, for instance, a time-varying system specified by *y*(*t*) = *e*−*t x*(*t*). The output for such a system in Fig. 1.29a is *e*−(*t*−*T*) *x*(*t* − *T*). In contrast, the output for the system in Fig. 1.29b is *e*−*t x*(*t* −*T*). - -**Figure 1.28** Time-invariance property. - -**Figure 1.29** Illustration of timeinvariance property. - -It is possible to verify that the system in Fig. 1.26 is a time-invariant system. Networks composed of *RLC* elements and other commonly used active elements such as transistors are time-invariant systems. A system with an input–output relationship described by a linear differential equation of the form given in Ex. 1.10 [Eq. (1.25)] is a linear time-invariant (LTI) system when the coefficients *ai* and *bi* of such equation are constants. If these coefficients are functions of time, then the system is a linear *time-varying* system. - -The system described in Drill 1.12 is linear time varying. Another familiar example of a time-varying system is the carbon microphone, in which the resistance *R* is a function of the mechanical pressure generated by sound waves on the carbon granules of the microphone. The output current from the microphone is thus modulated by the sound waves, as desired. - -### **EXAMPLE 1.11 Assessing System Time Invariance** - -Determine the time invariance of the following systems: **(a)** *y*(*t*)=*x*(*t*)*u*(*t*) and **(b)** *y*(*t*)= *d dt x*(*t*). - -**(a)** In this case, the output equals the input for *t* ≥ 0 and is otherwise zero. Clearly, the input is being modified by a time-dependent function, so the system is likely time variant. We can prove that the system is not time invariant through a counterexample. Letting *x*1(*t*) = δ(*t*+1), we see that *y*1(*t*) = 0. However, *x*2(*t*) = *x*1(*t*−2) = δ(*t*−1) produces an output of *y*2(*t*) = δ(*t* − 1), which does equal *y*1(*t* − 2) = 0 as time-invariance would require. Thus, *y*(*t*) = *x*(*t*)*u*(*t*) is a time variant system. - -**(b)** Although it appears that *x*(*t*) is being modified by a time-dependent function, this is not the case. The output of this system is simply the slope of the input. If the input is delayed, so too is the output. Applying input *x*(*t*) to the system produces output *y*(*t*) = *d dt x*(*t*); delaying this output by *T* produces *y*(*t* − *T*) = *d d*(*t*−*T*) *x*(*t* − *T*) = *d dt x*(*t* − *T*). This is just the output of the system to a delayed input *x*(*t* − *T*). Since the *T*-delayed output of the system to input *x*(*t*) equals the output of the system to the *T*-delayed input *x*(*t* −*T*), the system is time invariant. - -### **DR ILL 1.14 A Time-Variant System** - -Show that a system described by the following equation is a time-varying-parameter system: - -$$ -y(t) = (\sin t)x(t-2) -$$ - -[*Hint:* Show that the system fails to satisfy the time-invariance property.] - -### **[1.7-3 Instantaneous and Dynamic Systems](#page-7-0)** - -As observed earlier, a system's output at any instant *t* generally depends on the entire past input. However, in a special class of systems, the output at any instant *t* depends only on its input at that - -### 104 CHAPTER 1 SIGNALS AND SYSTEMS - -instant. In resistive networks, for example, any output of the network at some instant *t* depends only on the input at the instant *t*. In these systems, past history is irrelevant in determining the response. Such systems are said to be *instantaneous* or *memoryless* systems. More precisely, a system is said to be instantaneous (or memoryless) if its output at any instant *t* depends, at most, on the strength of its input(s) at the same instant *t*, and not on any past or future values of the input(s). Otherwise, the system is said to be *dynamic* (or a system with memory). A system whose response at *t* is completely determined by the input signals over the past *T* seconds [interval from (*t*−*T*) to *t*] is a *finite-memory system* with a memory of *T* seconds. Networks containing inductive and capacitive elements generally have infinite memory because the response of such networks at any instant *t* is determined by their inputs over the entire past (−∞,*t*). This is true for the *RC* circuit of Fig. 1.26. - -### **EXAMPLE 1.12 Assessing System Memory** - -Determine whether the following systems are memoryless: **(a)** *y*(*t* − 1) = 2*x*(*t* − 1), **(b)** *y*(*t*) = *d dt x*(*t*), and **(c)** *y*(*t*) = (*t* −1)*x*(*t*). - -**(a)** In this case, the output at time *t* −1 is just twice the input at the same time *t* −1. Since the output at a particular time depends only on the strength of the input at the same time, the system is memoryless. - -**(b)** Although it appears that the output *y*(*t*) at time *t* depends on the input *x*(*t*) at the same time *t*, we know that the slope (derivative) of *x*(*t*) cannot be determined solely from a single point. There must be some memory, even if infinitesimally small, involved. This is confirmed by using the fundamental theorem of calculus to express the system as - -$$ -y(t) = \lim_{T \to 0} \frac{x(t) - x(t - T)}{T} -$$ - -Since the output at a particular time depends on more than just the input at the same time, the system is not memoryless. - -**(c)** The output *y*(*t*) at time *t* is just the input *x*(*t*) at the same time *t* multiplied by the (time-dependent) coefficient *t* − 1. Since the output at a particular time depends only on the strength of the input at the same time, the system is memoryless. - -### **[1.7-4 Causal and Noncausal Systems](#page-7-0)** - -A *causal* (also known as a *physical* or *nonanticipative*) system is one for which the output at any instant *t*0 depends only on the value of the input *x*(*t*) for *t* ≤ *t*0. In other words, the value of the output at the present instant depends only on the past and present values of the input *x*(*t*), not on its future values. To put it simply, in a causal system the output cannot start before the input is applied. If the response starts before the input, it means that the system knows the input in the - -**Figure 1.30** Input–output of a noncausal system and the causal output achieved by delay. - -future and acts on this knowledge before the input is applied. A system that violates the condition of causality is called a *noncausal* (or *anticipative*) system. - -Any practical system that operates in real time† must necessarily be causal. We do not yet know how to build a system that can respond to future inputs (inputs not yet applied). A noncausal system is a prophetic system that knows the future input and acts on it in the present. Thus, if we apply an input starting at *t* = 0 to a noncausal system, the output would begin even before *t* = 0. For example, consider the system specified by - -$$ -y(t) = x(t-2) + x(t+2) -$$ -\n(1.26) - -For the input *x*(*t*) illustrated in Fig. 1.30a, the output *y*(*t*), as computed from Eq. (1.26) (shown in Fig. 1.30b), starts even before the input is applied. Equation (1.26) shows that *y*(*t*), the output at *t*, is given by the sum of the input values 2 seconds before and 2 seconds after *t* (at *t* − 2 and *t* + 2, respectively). But if we are operating the system in real time at *t*, we do not know what the value of the input will be 2 seconds later. Thus it is impossible to implement this system in real time. For this reason, noncausal systems are unrealizable in *real time*. - -### **EXAMPLE 1.13 Assessing System Causality** - -Determine whether the following systems are causal: **(a)** *y*(*t*) = *x*(−*t*), **(b)** *y*(*t*) = *x*(*t* + 1), and **(c)** *y*(*t* +1) = *x*(*t*). - -† In real-time operations, the response to an input is essentially simultaneous (contemporaneous) with the input itself. - -**(a)** Here, the output is a reflection of the input. We can easily use a counterexample to disprove the causality of this system. The input *x*(*t*) = δ(*t* − 1), which is nonzero at *t* = 1, produces an output *y*(*t*) = δ(*t* + 1), which is nonzero at *t* = −1, a time 2 seconds earlier than the input! Clearly the system is not causal. - -**(b)** In this case, the output at time *t* depends on the input at future time of *t* + 1. Clearly the system is not causal. - -**(c)** In this case, the output at time *t* + 1 depends on the input one second in the past, at time *t*. Since the output does not depend on future values of the input, the system is causal. - -### WHY STUDY NONCAUSAL SYSTEMS? - -The foregoing discussion may suggest that noncausal systems have no practical purpose. This is not the case; they are valuable in the study of systems for several reasons. First, noncausal systems *are* realizable when the independent variable is other than "time" (e.g., *space*). Consider, for example, an electric charge of density *q*(*x*) placed along the *x* axis for *x* ≥ 0. This charge density produces an electric field *E*(*x*) that is present at every point on the *x* axis from *x* = −∞ to ∞. In this case the input [i.e., the charge density *q*(*x*)] starts at *x* = 0, but its output [the electric field *E*(*x*)] begins before *x* = 0. Clearly, this space-charge system is noncausal. This discussion shows that only temporal systems (systems with time as independent variable) must be causal to be realizable. The terms "before" and "after" have a special connection to causality only when the independent variable is time. This connection is lost for variables other than time. Nontemporal systems, such as those occurring in optics, can be noncausal and still realizable. - -Moreover, even for temporal systems, such as those used for signal processing, the study of noncausal systems is important. In such systems we may have all input data prerecorded. This often happens with speech, geophysical, and meteorological signals, and with space probes. In such cases, the input's future values are available to us. For example, suppose we had a set of input signal records available for the system described by Eq. (1.26). We can then compute *y*(*t*) since, for any *t*, we need only refer to the records to find the input's value 2 seconds before and 2 seconds after *t*. Thus, noncausal systems can be realized, although not in real time. We may therefore be able to realize a noncausal system, provided we are willing to accept a time delay in the output. Consider a system whose output *y*ˆ(*t*) is the same as *y*(*t*) in Eq. (1.26) delayed by 2 seconds (Fig. 1.30c), so that - -$$ -\hat{y}(t) = y(t-2) = x(t-4) + x(t) -$$ - -Here the value of the output *y*ˆ at any instant *t* is the sum of the values of the input *x* at *t* and at the instant 4 seconds earlier [at (*t* − 4)]. In this case, the output at any instant *t* does not depend on future values of the input, and the system is causal. The output of this system, which is *y*ˆ(*t*), is identical to that in Eq. (1.26) or Fig. 1.30b except for a delay of 2 seconds. Thus, a noncausal system may be realized or satisfactorily approximated in real time by using a causal system with a delay. - -A third reason for studying noncausal systems is that they provide an upper bound on the performance of causal systems. For example, if we wish to design a filter for separating a signal from noise, then the optimum filter is invariably a noncausal system. Although unrealizable, this - -Noncausal systems are realizable with time delay! - -noncausal system's performance acts as the upper limit on what can be achieved and gives us a standard for evaluating the performance of causal filters. - -At first glance, noncausal systems may seem to be inscrutable. Actually, there is nothing mysterious about these systems and their approximate realization through physical systems with delay. If we want to know what will happen one year from now, we have two choices: go to a prophet (an unrealizable person) who can give the answers instantly, or go to a wise man and allow him a delay of one year to give us the answer! If the wise man is truly wise, he may even be able, by studying trends, to shrewdly guess the future very closely with a delay of less than a year. Such is the case with noncausal systems—nothing more and nothing less. - -### **DR ILL 1.15 A Noncausal System** - -Show that a system described by the following equation is noncausal: - -$$ -y(t) = \int_{t-5}^{t+5} x(\tau) d\tau -$$ - -Show that this system can be realized physically if we accept a delay of 5 seconds in the output. - -### **[1.7-5 Continuous-Time and Discrete-Time Systems](#page-7-0)** - -Signals defined or specified over a continuous range of time are *continuous-time signals,* denoted by symbols *x*(*t*), *y*(*t*), and so on. Systems whose inputs and outputs are continuous-time signals are *continuous-time systems*. On the other hand, signals defined only at discrete instants of time *t*0, *t*1, *t*2,...,*tn*,... are *discrete-time signals,* denoted by the symbols *x*(*tn*), *y*(*tn*), and so on, where *n* is some integer. Systems whose inputs and outputs are discrete-time signals are *discrete-time systems*. A digital computer is a familiar example of this type of system. In practice, discrete-time signals can arise from sampling continuous-time signals. For example, when the sampling is - -#### 108 CHAPTER 1 SIGNALS AND SYSTEMS - -uniform, the discrete instants *t*0, *t*1, *t*2, ... are uniformly spaced so that - -$$ -t_{k+1} - t_k = T \qquad \text{for all } k -$$ - -In such case, the discrete-time signals represented by the samples of continuous-time signals *x*(*t*), *y*(*t*), and so on can be expressed as *x*(*nT*), *y*(*nT*), and so on; for convenience, we further simplify this notation to *x*[*n*], *y*[*n*], ..., where it is understood that *x*[*n*] = *x*(*nT*) and that *n* is some integer. A typical discrete-time signal is shown in Fig. 1.31. A discrete-time signal may also be viewed as a sequence of numbers ..., *x*[−1], *x*[0], *x*[1], *x*[2], .... Thus, a discrete-time system may be seen as processing a sequence of numbers *x*[*n*] and yielding as an output another sequence of numbers *y*[*n*]. - -Discrete-time signals arise naturally in situations that are inherently discrete time, such as population studies, amortization problems, national income models, and radar tracking. They may also arise as a result of sampling continuous-time signals in sampled data systems, digital filtering, and the like. Digital filtering is a particularly interesting application in which continuous-time signals are processed by using discrete-time systems, as shown in Fig. 1.32. A continuous-time signal *x*(*t*) is first sampled to convert it into a discrete-time signal *x*[*n*], which then is processed by the discrete-time system to yield a discrete-time output *y*[*n*]. A continuous-time signal *y*(*t*) is finally constructed from *y*[*n*]. In this manner, we can process a continuous-time signal with an appropriate discrete-time system such as a digital computer. Because discrete-time systems have several significant advantages over continuous-time systems, there is an accelerating trend toward processing continuous-time signals with discrete-time systems. - -**Figure 1.32** Processing continuous-time signals by discrete-time systems. - -### **[1.7-6 Analog and Digital Systems](#page-7-0)** - -Analog and digital signals are discussed in Sec. 1.3-2. A system whose input and output signals are analog is an *analog system;* a system whose input and output signals are digital is a *digital system*. A digital computer is an example of a digital (binary) system. Observe that a digital computer is a digital as well as a discrete-time system. - -### **[1.7-7 Invertible and Noninvertible Systems](#page-7-0)** - -A system *S* performs certain operation(s) on input signal(s). If we can obtain the input *x*(*t*) back from the corresponding output *y*(*t*) by some operation, the system *S* is said to be *invertible*. When several different inputs result in the same output (as in a rectifier), it is impossible to obtain the input from the output, and the system is *noninvertible*. Therefore, for an invertible system, it is essential that every input have a unique output so that there is a one-to-one mapping between an input and the corresponding output. The system that achieves the inverse operation [of obtaining *x*(*t*) from *y*(*t*)] is the *inverse system* for *S*. For instance, if *S* is an ideal integrator, then its inverse system is an ideal differentiator. Consider a system *S* connected in tandem with its inverse *Si*, as shown in Fig. 1.33. The input *x*(*t*) to this tandem system results in signal *y*(*t*) at the output of *S*, and the signal *y*(*t*), which now acts as an input to *Si*, yields back the signal *x*(*t*) at the output of *Si*. Thus, *Si* undoes the operation of *S* on *x*(*t*), yielding back *x*(*t*). A system whose output is equal to the input (for all possible inputs) is an *identity* system. Cascading a system with its inverse system, as shown in Fig. 1.33, results in an identity system. - -In contrast, a rectifier, specified by an equation *y*(*t*) = |*x*(*t*)|, is noninvertible because the rectification operation cannot be undone. - -Inverse systems are very important in signal processing. In many applications, the signals are distorted during the processing, and it is necessary to undo the distortion. For instance, in transmission of data over a communication channel, the signals are distorted owing to non-ideal frequency response and finite bandwidth of a channel. It is necessary to restore the signal as closely as possible to its original shape. Such equalization is also used in audio systems and photographic systems. - -inverse results in an identity system. - -### **EXAMPLE 1.14 Assessing System Invertibility** - -Determine whether the following systems are invertible: **(a)** *y*(*t*) = *x*(−*t*), **(b)** *y*(*t*) = *tx*(*t*), and **(c)** *y*(*t*) = *d dt x*(*t*). - -**(a)** Here, the output is a reflection of the input, which does not cause any loss to the input. The input can, in fact, be exactly recovered by simply reflecting the output [*x*(*t*) = *y*(−*t*)], which is to say that a reflecting system is its own inverse. Thus, *y*(*t*) = *x*(−*t*) is an invertible system. - -### 110 CHAPTER 1 SIGNALS AND SYSTEMS - -**(b)** In this case, one might be tempted to recover the input from the output as *x*(*t*) = 1 *t y*(*t*). This approach works almost everywhere, except at *t* = 0 where the input value *x*(0) cannot be recovered. Due to this single lost point, the system *y*(*t*) = *tx*(*t*) is not invertible. - -**(c)** Differentiation eliminates any dc component. For example, the inputs *x*1(*t*) = 1 and *x*2(*t*) = 2 both produce the same output *y*(*t*) = 0. Given only *y*(*t*) = 0, it is impossible to know if the original input was *x*1(*t*) = 1, *x*2(*t*) = 2, or something else entirely. Since unique inputs do produce unique outputs, we know that *y*(*t*) = *d dt x*(*t*) is not an invertible system. - -### **[1.7-8 Stable and Unstable Systems](#page-7-0)** - -Systems can also be classified as *stable* or *unstable* systems. Stability can be *internal* or *external*. If every *bounded input* applied at the input terminal results in a *bounded output,* the system is said to be stable *externally*. External stability can be ascertained by measurements at the external terminals (input and output) of the system. This type of stability is also known as the stability in the BIBO (bounded-input/bounded-output) sense. The concept of internal stability is postponed to Ch. 2 because it requires some understanding of internal system behavior, introduced in that chapter. - -### **EXAMPLE 1.15 Assessing System BIBO Stability** - -Determine whether the following systems are BIBO-stable: **(a)** *y*(*t*) = *x*2(*t*), **(b)** *y*(*t*) = *tx*(*t*), and **(c)** *y*(*t*) = *d dt x*(*t*). - -**(a)** This system squares an input to produce the output. If the input is bounded, which is to say that |*x*(*t*)| ≤ *Mx* < ∞ for all *t*, then we see that - -$$ -|y(t)| = |x^2(t)| = |x(t)|^2 \le M_x^2 < \infty -$$ - -Since the output amplitude is guaranteed to be bounded for any bounded-amplitude input, the system *y*(*t*) = *x*2(*t*) is BIBO-stable. - -**(b)** We can prove that *y*(*t*) = *tx*(*t*) is not BIBO-stable with a simple example. The bounded-amplitude input *x*(*t*) = *u*(*t*) produces the output *y*(*t*) = *tu*(*t*) whose amplitude grows to infinity as *t* → ∞. Thus, *y*(*t*) = *tx*(*t*) is a BIBO-unstable system. - -**(c)** We can prove that *y*(*t*) = *d dt x*(*t*) is not BIBO-stable with an example. The bounded-amplitude input *x*(*t*) = *u*(*t*) produces the output *y*(*t*) = δ(*t*) whose amplitude is infinite at *t* = 0. Thus, *y*(*t*) = *d dt x*(*t*) is a BIBO-unstable system. - -### **DR ILL 1.16 A Noninvertible BIBO-Stable System** - -Show that a system described by the equation *y*(*t*) = *x*2(*t*) is noninvertible but BIBO-stable. diff --git "a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/020_1.8 SYSTEM MODEL - INPUT\342\200\223OUTPUT DESCRIPTION.md" "b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/020_1.8 SYSTEM MODEL - INPUT\342\200\223OUTPUT DESCRIPTION.md" deleted file mode 100644 index 892380810555373db23e41b7b4683c84d3e2e34b..0000000000000000000000000000000000000000 --- "a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/020_1.8 SYSTEM MODEL - INPUT\342\200\223OUTPUT DESCRIPTION.md" +++ /dev/null @@ -1,249 +0,0 @@ -## **1.8 SYSTEM [MODEL: INPUT–OUTPUT](#page-8-0) DESCRIPTION** - -A system description in terms of the measurements at the input and output terminals is called the *input–output description*. As mentioned earlier, systems theory encompasses a variety of systems, such as electrical, mechanical, hydraulic, acoustic, electromechanical, and chemical, as well as social, political, economic, and biological. The first step in analyzing any system is the construction of a system model, which is a mathematical expression or a rule that satisfactorily approximates the dynamical behavior of the system. In this chapter we shall consider only continuous-time systems. Modeling of discrete-time systems is discussed in Ch. 3. - -### **[1.8-1 Electrical Systems](#page-8-0)** - -To construct a system model, we must study the relationships between different variables in the system. In electrical systems, for example, we must determine a satisfactory model for the voltage-current relationship of each element, such as Ohm's law for a resistor. In addition, we must determine the various constraints on voltages and currents when several electrical elements are interconnected. These are the laws of interconnection—the well-known Kirchhoff laws for voltage and current (KVL and KCL). From all these equations, we eliminate unwanted variables to obtain equation(s) relating the desired output variable(s) to the input(s). The following examples demonstrate the procedure of deriving input–output relationships for some LTI electrical systems. - -### 112 CHAPTER 1 SIGNALS AND SYSTEMS - -By using the voltage-current laws of each element (inductor, resistor, and capacitor), we can express this equation as - -$$ -\frac{dy(t)}{dt} + 3y(t) + 2\int_{-\infty}^{t} y(\tau) d\tau = x(t) -$$ -\n(1.27) - -Differentiating both sides of this equation, we obtain - -$$ -\frac{d^2y(t)}{dt^2} + 3\frac{dy(t)}{dt} + 2y(t) = \frac{dx(t)}{dt} -$$ -\n(1.28) - -This differential equation is the input–output relationship between the output *y*(*t*) and the input *x*(*t*). - -It proves convenient to use a compact notation *D* for the differential operator *d*/*dt*. This notation can be repeatedly applied. Thus, - -$$ -\frac{dy(t)}{dt} \equiv Dy(t), \qquad \frac{d^2y(t)}{dt^2} \equiv D^2y(t), \qquad \dots, \qquad \frac{d^Ny(t)}{dt^N} \equiv D^Ny(t) -$$ - -With this notation, Eq. (1.28) can be expressed as - -$$ -(D2 + 3D + 2)y(t) = Dx(t) -$$ -\n(1.29) - -The differential operator is the inverse of the integral operator, so we can use the operator 1/*D* to represent integration.† - -$$ -\int_{-\infty}^{t} y(\tau) d\tau \equiv \frac{1}{D} y(t) -$$ - -$$ -\frac{d}{dt} \left[ \int_{-\infty}^{t} y(\tau) d\tau \right] = y(t) -$$ - -† Use of operator 1/*D* for integration generates some subtle mathematical difficulties because the operators *D* and 1/*D* do not commute. For instance, we know that *D*(1/*D*) = 1 because - -However, (1/*D*)*D* is not necessarily unity. Use of Cramer's rule in solving simultaneous integro-differential equations will always result in cancellation of operators 1/*D* and *D*. This procedure may yield erroneous results when the factor *D* occurs in the numerator as well as in the denominator. This happens, for instance, in circuits with all-inductor loops or all-capacitor cut sets. To eliminate this problem, avoid the integral operation in system equations so that the resulting equations are differential rather than integro-differential. In electrical circuits, this can be done by using charge (instead of current) variables in loops containing capacitors and choosing current variables for loops without capacitors. In the literature this problem of commutativity of *D* and 1/*D* is largely ignored. As mentioned earlier, such a procedure gives erroneous results only in special systems, such as the circuits with all-inductor loops or all-capacitor cut sets. Fortunately such systems constitute a very small fraction of the systems we deal with. For further discussion of this topic and a correct method of handling problems involving integrals, see [4]. - -Consequently, Eq. (1.27) can be expressed as - -$$ -\left(D+3+\frac{2}{D}\right)y(t) = x(t) -$$ - -Multiplying both sides by *D* to differentiate the expression, we obtain - -$$ -(D^2 + 3D + 2)y(t) = Dx(t) -$$ - -which is identical to Eq. (1.29). - -Recall that Eq. (1.29) is not an algebraic equation, and *D*2+3*D*+2 is not an algebraic term that multiplies *y*(*t*); it is an operator that operates on *y*(*t*). It means that we must perform the following operations on *y*(*t*): take the second derivative of *y*(*t*) and add to it 3 times the first derivative of *y*(*t*) and 2 times *y*(*t*). Clearly, a polynomial in *D* multiplied by *y*(*t*) represents a certain differential operation on *y*(*t*). - -### **EXAMPLE 1.17 Input–Output Equation of a Series** *RC* **Circuit** - -Using operator notation, find the equation relating input to output for the series *RC* circuit of Fig. 1.35 if the input is the voltage *x*(*t*) and output is - -- **(a)** the loop current *i*(*t*) -- **(b)** the capacitor voltage *y*(*t*) - -**(a)** The loop equation for the circuit is - -$$ -R i(t) + \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau = x(t) -$$ - -or - -$$ -15i(t) + 5 \int_{-\infty}^{t} i(\tau) d\tau = x(t) -$$ - -With operator notation, this equation can be expressed as - -$$ -15 i(t) + \frac{5}{D} i(t) = x(t) -$$ -\n(1.30) - -**(b)** Multiplying both sides of Eq. (1.30) by *D* (i.e., differentiating the equation), we obtain - -(15*D*+5)*i*(*t*) = *Dx*(*t*) - -Using the fact that *i*(*t*) = *C dy*(*t*) *dt* = 1 5*Dy*(*t*), simple substitution yields - -$$ -(3D+1)y(t) = x(t) -$$ -\n(1.31) - -## **DR ILL 1.17 Input–Output Equation of a Series** *RLC* **Circuit with Inductor Voltage as Output** - -If the inductor voltage *vL*(*t*) is taken as the output, show that the *RLC* circuit in Fig. 1.34 has an input–output equation of (*D*2 +3*D*+2)*vL*(*t*) = *D*2*x*(*t*). - -## **DR ILL 1.18 Input–Output Equation of a Series** *RC* **Circuit with Capacitor Voltage as Output** - -If the capacitor voltage *vC*(*t*) is taken as the output, show that the *RLC* circuit in Fig. 1.34 has an input–output equation of (*D*2 +3*D*+2)*vC*(*t*) = 2*x*(*t*). - -### **[1.8-2 Mechanical Systems](#page-8-0)** - -Planar motion can be resolved into translational (rectilinear) motion and rotational (torsional) motion. Translational motion will be considered first. We shall restrict ourselves to motions in one dimension. - -### TRANSLATIONAL SYSTEMS - -The basic elements used in modeling translational systems are ideal masses, linear springs, and dashpots providing viscous damping. The laws of various mechanical elements are now discussed. - -For a *mass M* (Fig. 1.36a), a force *x*(*t*) causes a motion *y*(*t*) and acceleration *y*¨(*t*). From Newton's law of motion, - -$$ -x(t) = M\ddot{y}(t) = M\frac{d^2y(t)}{dt^2} = MD^2y(t) -$$ - -The force *x*(*t*) required to stretch (or compress) a *linear spring* (Fig. 1.36b) by an amount *y*(*t*) is given by - -$$ -x(t) = Ky(t) -$$ - -where *K* is the *stiffness* of the spring. - -**Figure 1.36** Some elements in translational mechanical systems. - -For *a linear dashpot* (Fig. 1.36c), which operates by virtue of viscous friction, the force moving the dashpot is proportional to the relative velocity *y*˙(*t*) of one surface with respect to the other. Thus - -$$ -x(t) = B\dot{y}(t) = B\frac{dy(t)}{dt} = BDy(t) -$$ - -where *B* is the *damping coefficient* of the dashpot or the viscous friction. - -### **EXAMPLE 1.18 Input–Output Equation for a Translational Mechanical System** - -Find the input–output relationship for the translational mechanical system shown in Fig. 1.37a or its equivalent in Fig. 1.37b. The input is the force *x*(*t*), and the output is the mass position *y*(*t*). - -### 116 CHAPTER 1 SIGNALS AND SYSTEMS - -In mechanical systems it is helpful to draw a free-body diagram of each junction, which is a point at which two or more elements are connected. In Fig. 1.37, the point representing the mass is a junction. The displacement of the mass is denoted by *y*(*t*). The spring is also stretched by the amount *y*(*t*), and therefore it exerts a force −*Ky*(*t*) on the mass. The dashpot exerts a force −*By*˙(*t*) on the mass, as shown in the free-body diagram (Fig. 1.37c). By Newton's second law, the net force must be *My*¨(*t*). Therefore, - -$$ -M\ddot{y}(t) = -B\dot{y}(t) - Ky(t) + x(t) -$$ - -or - -$$ -(MD2 + BD + K)y(t) = x(t) -$$ - -### ROTATIONAL SYSTEMS - -In rotational systems, the motion of a body may be defined as its motion about a certain axis. The variables used to describe rotational motion are torque (in place of force), angular position (in place of linear position), angular velocity (in place of linear velocity), and angular acceleration (in place of linear acceleration). The system elements are *rotational mass* or *moment of inertia* (in place of mass) and *torsional springs* and *torsional dashpots* (in place of linear springs and dashpots). The terminal equations for these elements are analogous to the corresponding equations for translational elements. If *J* is the moment of inertia (or rotational mass) of a rotating body about a certain axis, then the external torque required for this motion is equal to *J* (rotational mass) times the angular acceleration. If θ (*t*) is the angular position of the body, θ (¨ *t*) is its angular acceleration, and - -torque = -$$ -J\ddot{\theta}(t) = J\frac{d^2\theta(t)}{dt^2} = JD^2\theta(t) -$$ - -Similarly, if *K* is the stiffness of a torsional spring (per unit angular twist), and θ is the angular displacement of one terminal of the spring with respect to the other, then - -torque = -$$ -K\theta(t) -$$ - -Finally, the torque due to viscous damping of a torsional dashpot with damping coefficient *B* is - -torque = -$$ -B\dot{\theta}(t) -$$ - = $BD\theta(t)$ - -### **EXAMPLE 1.19 Input–Output Equation for Aircraft Roll Angle** - -The attitude of an aircraft can be controlled by three sets of surfaces (shown shaded in Fig. 1.38): elevators, rudder, and ailerons. By manipulating these surfaces, one can set the aircraft on a desired flight path. The roll angle ϕ(*t*) can be controlled by deflecting in the opposite direction the two aileron surfaces as shown in Fig. 1.38. Assuming only rolling motion, find the equation relating the roll angle ϕ(*t*) to the input (deflection) θ (*t*). - -**Figure 1.38** Attitude control of an airplane. - -*J* - -The aileron surfaces generate a torque about the roll axis proportional to the aileron deflection angle θ (*t*). Let this torque be *c*θ (*t*), where *c* is the constant of proportionality. Air friction dissipates the torque *B*ϕ(˙ *t*). The torque available for rolling motion is then *c*θ (*t*) − *B*ϕ(˙ *t*). If *J* is the moment of inertia of the plane about the *x* axis (roll axis), then - -net torque = -$$ -J\ddot{\varphi}(t) = c\theta(t) - B\dot{\varphi}(t) -$$ - -and - -$$ -J\frac{d^2\varphi(t)}{dt^2} + B\frac{d\varphi(t)}{dt} = c\theta(t) \qquad \text{or} \qquad (JD^2 + BD)\varphi(t) = c\theta(t) -$$ - -This is the desired equation relating the output (roll angle ϕ(*t*)) to the input (aileron angle θ (*t*)). - -The roll velocity ω(*t*) is ϕ(˙ *t*). If the desired output is the roll velocity ω(*t*) rather than the roll angle ϕ(*t*), then the input–output equation would be - -$$ -\frac{d\omega(t)}{dt} + B\omega(t) = c\theta(t) \qquad \text{or} \qquad (JD + B)\omega(t) = c\theta(t) -$$ - -## **DR ILL 1.19 Input–Output Equation of a Rotational Mechanical System** - -Torque *T* (*t*) is applied to the rotational mechanical system shown in Fig. 1.39a. The torsional spring stiffness is *K*; the rotational mass (the cylinder's moment of inertia about the shaft) is *J*; the viscous damping coefficient between the cylinder and the ground is *B*. Find the equation relating the output angle θ (*t*) to the input torque *T* (*t*). [*Hint:* A free-body diagram is shown in Fig. 1.39b.] - -### **[1.8-3 Electromechanical Systems](#page-8-0)** - -A wide variety of electromechanical systems is used to convert electrical signals into mechanical motion (mechanical energy) and vice versa. Here we consider a rather simple example of an armature-controlled dc motor driven by a current source *x*(*t*), as shown in Fig. 1.40a. The torque *T* (*t*) generated in the motor is proportional to the armature current *x*(*t*). Therefore, - -$$ -\mathcal{T}(t) = K_T x(t) -$$ - -where *KT* is a constant of the motor. This torque drives a mechanical load whose free-body diagram is shown in Fig. 1.40b. The viscous damping (with coefficient *B*) dissipates a torque *B*θ (˙ *t*). If *J* is the moment of inertia of the load (including the rotor of the motor), then the net torque *T* (*t*)−*B*θ (˙ *t*) must be equal to *J*θ (¨ *t*): - -$$ -J\ddot{\theta}(t) = \mathcal{T}(t) - B\dot{\theta}(t) -$$ - -Thus, - -$$ -(JD2 + BD)\theta(t) = \mathcal{T}(t) = K_T x(t) -$$ - -which in conventional form can be expressed as - -$$ -J\frac{d^2\theta(t)}{dt^2} + B\frac{d\theta(t)}{dt} = K_T x(t) -$$ -\n(1.32) - -**Figure 1.40** Armature-controlled dc motor. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/021_1.9 INTERNAL AND EXTERNAL DESCRIPTIONS OF A SYSTEM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/021_1.9 INTERNAL AND EXTERNAL DESCRIPTIONS OF A SYSTEM.md deleted file mode 100644 index 04581e9ca1f20cd03473aef0957f531d03fc6e5b..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/021_1.9 INTERNAL AND EXTERNAL DESCRIPTIONS OF A SYSTEM.md +++ /dev/null @@ -1,33 +0,0 @@ -## **1.9 INTERNAL AND EXTERNAL [DESCRIPTIONS OF A](#page-8-0) SYSTEM** - -The input–output relationship of a system is an *external description* of that system. We have found an external description (not the *internal description*) of systems in all the examples discussed so far. This may puzzle the reader because in each of these cases, we derived the input–output relationship by analyzing the internal structure of that system. Why is this not an internal description? What makes a description internal? Although it is true that we did find the input–output description by internal analysis of the system, we did so strictly for convenience. We could have obtained the input–output description by making observations at the external (input and output) terminals, for example, by measuring the output for certain inputs, such as an impulse or a sinusoid. A description that can be obtained from measurements at the external terminals (even when the rest of the system is sealed inside an inaccessible black box) is an external description. Clearly, the input–output description is an external description. What, then, is an internal description? An internal description is capable of providing complete information about all possible signals in the system. An external description may not give such complete information. An external description can always be found from an internal description, but the converse is not necessarily true. We shall now give an example to clarify the distinction between an external and an internal description. - -Let the circuit in Fig. 1.41a with the input *x*(*t*) and the output *y*(*t*) be enclosed inside a "black box" with only the input and the output terminals accessible. To determine its external description, let us apply a known voltage *x*(*t*) at the input terminals and measure the resulting output voltage *y*(*t*). - -Let us also assume that there is some initial charge *Q*0 present on the capacitor. The output voltage will generally depend on both, the input *x*(*t*) and the initial charge *Q*0. To compute the output resulting because of the charge *Q*0, assume the input *x*(*t*) = 0 (short across the input). In this case, the currents in the two 2 resistors in the upper and the lower branches at the output terminals are equal and opposite because of the balanced nature of the circuit. Clearly, the capacitor charge results in zero voltage at the output.† - -† The output voltage *y*(*t*) resulting because of the capacitor charge [assuming *x*(*t*) = 0] is the zero-input response, which, as argued above, is zero. The output component due to the input *x*(*t*) (assuming zero initial capacitor charge) is the zero-state response. Complete analysis of this problem is given later in Ex. 1.21. - -**Figure 1.41** A system that cannot be described by external measurements. - -Now, to compute the output *y*(*t*) resulting from the input voltage *x*(*t*), we assume zero initial capacitor charge (short across the capacitor terminals). The current *i*(*t*) (Fig. 1.41a), in this case, divides equally between the two parallel branches because the circuit is balanced. Thus, the voltage across the capacitor continues to remain zero. Therefore, for the purpose of computing the current *i*(*t*), the capacitor may be removed or replaced by a short. The resulting circuit is equivalent to that shown in Fig. 1.41b, which shows that the input *x*(*t*) sees a load of 5, and - -$$ -i(t) = \frac{1}{5}x(t) -$$ - -Also, because *y*(*t*) = 2*i*(*t*), - -$$ -y(t) = \frac{2}{5}x(t) -$$ - -This is the total response. Clearly, for the external description, the capacitor does not exist. No external measurement or external observation can detect the presence of the capacitor. Furthermore, if the circuit is enclosed inside a "black box" so that only the external terminals are accessible, it is impossible to determine the currents (or voltages) inside the circuit from external measurements or observations. An internal description, however, can provide every possible signal inside the system. In Ex. 1.21, we shall find the internal description of this system and show that it is capable of determining every possible signal in the system. - -For most systems, the external and internal descriptions are equivalent, but there are a few exceptions, as in the present case, where the external description gives an inadequate picture of the system. This happens when the system is *uncontrollable* and/or *unobservable*. - -Figure 1.42 shows structural representations of simple uncontrollable and unobservable systems. In Fig. 1.42a, we note that part of the system (subsystem *S*2) inside the box cannot be controlled by the input *x*(*t*). In Fig. 1.42b, some of the system outputs (those in subsystem *S*2) cannot be observed from the output terminals. If we try to describe either of these systems by applying an external input *x*(*t*) and then measuring the output *y*(*t*), the measurement will not characterize the complete system but only the part of the system (here *S*1) that is both controllable - -**Figure 1.42** Structures of uncontrollable and unobservable systems. - -and observable (linked to both the input and output). Such systems are undesirable in practice and should be avoided in any system design. The system in Fig. 1.41a can be shown to be neither controllable nor observable. It can be represented structurally as a combination of the systems in Figs. 1.42a and 1.42b. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/022_1.10 INTERNAL DESCRIPTION - THE STATE-SPACE DESCRIPTION.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/022_1.10 INTERNAL DESCRIPTION - THE STATE-SPACE DESCRIPTION.md deleted file mode 100644 index a00a53ad3f29f7f2244c2967a524bce18ba17217..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/022_1.10 INTERNAL DESCRIPTION - THE STATE-SPACE DESCRIPTION.md +++ /dev/null @@ -1,176 +0,0 @@ -## **1.10 INTERNAL [DESCRIPTION: THE](#page-8-0) STATE-SPACE DESCRIPTION** - -We shall now introduce the *state-space* description of a linear system, which is an internal description of a system. In this approach, we identify certain key variables, called the *state variables,* of the system. These variables have the property that every possible signal in the system can be expressed as a linear combination of these state variables. For example, we can show that every possible signal in a passive *RLC* circuit can be expressed as a linear combination of independent capacitor voltages and inductor currents, which, therefore, are state variables for the circuit. - -To illustrate this point, consider the network in Fig. 1.43. We identify two state variables: the capacitor voltage *q*1 and the inductor current *q*2. If the values of *q*1, *q*2, and the input *x*(*t*) are known at some instant *t*, we can demonstrate that every possible signal (current or voltage) in the circuit can be determined at *t*. For example, if *q*1 = 10, *q*2 = 1, and the input *x* = 20 at some instant, the remaining voltages and currents at that instant will be - -$$ -i_1 = (x - q_1)/1 = 20 - 10 = 10 \text{A} -$$ - -\n -$$ -v_1 = x - q_1 = 20 - 10 = 10 \text{V} -$$ - -\n -$$ -v_2 = q_1 = 10 \text{V} -$$ - -\n -$$ -i_2 = q_1/2 = 5 \text{A} -$$ - -\n -$$ -i_C = i_1 - i_2 - q_2 = 10 - 5 - 1 = 4 \text{A} -$$ - -\n -$$ -i_3 = q_2 = 1 \text{A} -$$ - -\n -$$ -v_3 = 5q_2 = 5 \text{V} -$$ - -\n -$$ -v_L = q_1 - v_3 = 10 - 5 = 5 \text{V} -$$ - -\n(1.33) - -Thus all signals in this circuit are determined. Clearly, state variables consist of the *key variables* in a system; a knowledge of the state variables allows one to determine every possible output of the system. Note that the *state-variable description is an internal description* of a system because it is capable of describing all possible signals in the system. - -**Figure 1.43** Choosing suitable initial conditions in a network. - -### **EXAMPLE 1.20 State-Space Description of a System** - -This example illustrates how state equations may be natural and easier to determine than other descriptions, such as loop or node equations. Consider again the network in Fig. 1.43 with *q*1 and *q*2 as the state variables and write the state equations. - -This can be done by simple inspection of Fig. 1.43. Since *q*˙1 is the current through the capacitor, - -$$ -\dot{q}_1 = i_C = i_1 - i_2 - q_2 -$$ - -= $(x - q_1) - 0.5 q_1 - q_2$ -= $-1.5 q_1 - q_2 + x$ - -Also 2*q*˙2, the voltage across the inductor, is given by - -$$ -2\dot{q}_2 = q_1 - v_3 -$$ -$$ -= q_1 - 5q_2 -$$ - -or - -$$ -\dot{q}_2 = 0.5 q_1 - 2.5 q_2 -$$ - -Thus, the state equations are - -$$ -\dot{q}_1 = -1.5q_1 - q_2 + x \n\dot{q}_2 = 0.5q_1 - 2.5q_2 -$$ -\n(1.34) - -This is a set of two simultaneous first-order differential equations. This set of equations comprises the *state equations*. Once these equations have been solved for *q*1 and *q*2, everything else in the circuit can be determined by using Eq. (1.33), which are known as the *output equations*. Thus, in this approach, we have two sets of equations, the state equations and the output equations. Once we have solved the state equations, all possible outputs can be obtained - -from the output equations. In the input–output description, an *N*th-order system is described by an *N*th-order equation. In the state-variable approach, the same system is described by *N* simultaneous first-order state equations.† - -### **EXAMPLE 1.21 Controllability and Observability** - -Investigate the nature of state equations and the issue of controllability and observability for the circuit in Fig. 1.41a. - -This circuit has only one capacitor and no inductors. Hence, there is only one state variable, the capacitor voltage *q*(*t*). Since *C* = 1 F, the capacitor current is *q*˙. There are two sources in this circuit: the input *x*(*t*) and the capacitor voltage *q*(*t*). The response due to *x*(*t*), assuming *q*(*t*) = 0, is the zero-state response, which can be found from Fig. 1.44a, where we have shorted the capacitor [*q*(*t*) = 0]. The response due to *q*(*t*) assuming *x*(*t*) = 0, is the zero-input response, which can be found from Fig. 1.44b, where we have shorted *x*(*t*) to ensure *x*(*t*) = 0. It is now trivial to find both the components. - -Figure 1.44a shows zero-state currents in every branch. It is clear that the input *x*(*t*) sees an effective resistance of 5 , and, hence, the current through *x*(*t*) is *x*/5 A, which divides in the two parallel branches, resulting in the current *x*/10 through each branch. - -Examining the circuit in Fig. 1.44b for the zero-input response, we note that the capacitor voltage is *q* and the current is *q*˙. We also observe that the capacitor sees two loops in parallel, each with resistance 4 and current *q*˙/2. Interestingly, the 3 branch is effectively shorted because the circuit is balanced, and thus the voltage across the terminals *cd* is zero. The total current in any branch is the sum of the currents in that branch in Figs. 1.44a and 1.44b (principle of superposition). - -Branch - -\n -$$ -c = \frac{x}{10} + \frac{\dot{q}}{2} \quad 2\left(\frac{x}{10} + \frac{\dot{q}}{2}\right) -$$ -\n -$$ -cb = \frac{x}{10} - \frac{\dot{q}}{2} \quad 2\left(\frac{x}{10} - \frac{\dot{q}}{2}\right) -$$ -\n -$$ -ad = \frac{x}{10} - \frac{\dot{q}}{2} \quad 2\left(\frac{x}{10} - \frac{\dot{q}}{2}\right) -$$ -\n -$$ -bd = \frac{x}{10} + \frac{\dot{q}}{2} \quad 2\left(\frac{x}{10} + \frac{\dot{q}}{2}\right) -$$ -\n -$$ -ec = \frac{x}{5} \quad 3\left(\frac{x}{5}\right) -$$ -\n -$$ -ed = \frac{x}{5} \quad x -$$ -\n(1.35) - -† This assumes the system to be controllable and observable. If it is not, the input–output description equation will be of an order lower than the corresponding number of state equations. - -**Figure 1.44** Analysis of a system that is neither controllable nor observable. - -To find the state equation, we note that the current in branch *ca* is (*x*/10)+ ˙*q*/2 and the current in branch *cb* is (*x*/10)− ˙*q*/2. Hence, the equation around the loop *acba* is - -$$ -q = 2\left[-\frac{x}{10} - \frac{\dot{q}}{2}\right] + 2\left[\frac{x}{10} - \frac{\dot{q}}{2}\right] = -2\dot{q} -$$ - -$$ -\dot{q} = -0.5q -$$ - (1.36) - -or - -This is the desired state equation. - -Substitution of *q*˙ = −0.5*q* in Eq. (1.35) shows that every possible current and voltage in the circuit can be expressed in terms of the state variable *q* and the input *x*, as desired. Hence, the set of Eq. (1.35) is the output equation for this circuit. Once we have solved the state equation [Eq. (1.36)] for *q*, we can determine every possible output in the circuit. - -The output *y*(*t*) is given by - -$$ -y(t) = 2\left[\frac{x}{10} - \frac{\dot{q}}{2}\right] + 2\left[\frac{x}{10} + \frac{\dot{q}}{2}\right] = \frac{2}{5}x(t) -$$ - (1.37) - -A little examination of the state and the output equations indicates the nature of this system. Equation (1.36) shows that the state *q*(*t*) is independent of the input *x*(*t*); hence the system state *q* cannot be controlled by the input. Moreover, Eq. (1.37) shows that the output *y*(*t*) does not depend on the state *q*(*t*). Thus, the system state cannot be observed from the output terminals. Hence, the system is neither controllable nor observable. Such is not the case of other systems examined earlier. Consider, for example, the circuit in Fig. 1.43. The state equations [Eq. (1.34)] show that the states are influenced by the input directly or indirectly. Hence, the system is controllable. Moreover, as Eq. (1.33) shows, every possible output is expressed in terms of the state variables and the input. Hence, the states are also observable. - -State-space techniques are useful not just because of their ability to provide internal system description, but for several other reasons, including the following. - -- 1. State equations of a system provide a mathematical model of great generality that can describe not just linear systems, but also nonlinear systems; not just time-invariant systems, but also time-varying parameter systems; not just SISO (single-input/single-output) systems, but also multiple-input/multiple-output (MIMO) systems. Indeed, state equations are ideally suited for the analysis, synthesis, and optimization of MIMO systems. -- 2. Compact matrix notation and the powerful techniques of linear algebra greatly facilitate complex manipulations. Without such features, many important results of the modern system theory would have been difficult to obtain. State equations can yield a great deal of information about a system even when they are not solved explicitly. -- 3. State equations lend themselves readily to digital computer simulation of complex systems of high order, with or without nonlinearities, and with multiple inputs and outputs. -- 4. For second-order systems (*N* = 2), a graphical method called *phase-plane analysis* can be used on state equations, whether they are linear or nonlinear. - -The real benefits of the state-space approach, however, are realized for highly complex systems of large order. Much of the book is devoted to introduction of the basic concepts of linear systems analysis, which must necessarily begin with simpler systems without using the state-space approach. Chapter 10 deals with the state-space analysis of linear, time-invariant, continuous-time, and discrete-time systems. - -### **DR ILL 1.20 State Equations for a Series** *RLC* **Circuit** - -Write the state equations for the series *RLC* circuit shown in Fig. 1.45, using the inductor current *q*1(*t*) and the capacitor voltage *q*2(*t*) as state variables. Express every voltage and current in this circuit as a linear combination of *q*1, *q*2, and *x*. - -### **ANSWERS** - -*q*1 = −3*q*1 −*q*2 +*x* and *q*2 = 2*q*1. - -**Figure 1.45** Circuit for Drill 1.20. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/023_1.11 MATLAB - WORKING WITH FUNCTIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/023_1.11 MATLAB - WORKING WITH FUNCTIONS.md deleted file mode 100644 index eb008968d4f314ac13e036752c1f95e54834d77e..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/023_1.11 MATLAB - WORKING WITH FUNCTIONS.md +++ /dev/null @@ -1,179 +0,0 @@ -## **[1.11 MATLAB: WORKING WITH](#page-8-0) FUNCTIONS** - -Working with functions is fundamental to signals and systems applications. MATLAB provides several methods of defining and evaluating functions. An understanding and proficient use of these methods are therefore necessary and beneficial. - -### **[1.11-1 Anonymous Functions](#page-8-0)** - -Many simple functions are most conveniently represented by using MATLAB anonymous functions. An anonymous function provides a symbolic representation of a function defined in terms of MATLAB operators, functions, or other anonymous functions. For example, consider defining the exponentially damped sinusoid *f*(*t*) = *e*−*t* cos(2π*t*). - -``` ->> f = @(t) exp(-t).*cos(2*pi*t); -``` - -In this context, the @ symbol identifies the expression as an anonymous function, which is assigned a name of f. Parentheses following the @ symbol are used to identify the function's independent variables (input arguments), which in this case is the single time variable t. Input arguments, such as t, are local to the anonymous function and are not related to any workspace variables with the same names. - -Once defined, *f*(*t*) can be evaluated simply by passing the input values of interest. For example, - ->> t = 0; f(t) ans = 1 - -evaluates *f*(*t*) at *t* = 0, confirming the expected result of unity. The same result is obtained by passing *t* = 0 directly. - ->> f(0) ans = 1 - -Vector inputs allow the evaluation of multiple values simultaneously. Consider the task of plotting *f*(*t*) over the interval (−2 ≤ *t* ≤ 2). Gross function behavior is clear: *f*(*t*) should oscillate four times with a decaying envelope. Since accurate hand sketches are cumbersome, MATLAB-generated plots are an attractive alternative. As the following example illustrates, care must be taken to ensure reliable results. - -Suppose vector t is chosen to include only the integers contained in (−2 ≤ *t* ≤ 2), namely, [−2,−1, 0, 1, 2]. - ->> t = (-2:2); - -This vector input is evaluated to form a vector output. - ->> f(t) ans = 7.3891 2.7183 1.0000 0.3679 0.1353 The plot command graphs the result, which is shown in Fig. 1.46. - -``` ->> plot(t,f(t)); ->> xlabel('t'); ylabel('f(t)'); grid; -``` - -Grid lines, added by using the grid command, aid feature identification. Unfortunately, the plot does not illustrate the expected oscillatory behavior. More points are required to adequately represent *f*(*t*). - -The question, then, is how many points is enough?† If too few points are chosen, information is lost. If too many points are chosen, memory and time are wasted. A balance is needed. For oscillatory functions, plotting 20 to 200 points per oscillation is normally adequate. For the present case, t is chosen to give 100 points per oscillation. - ->> t = (-2:0.01:2); - -Again, the function is evaluated and plotted. - -**Figure 1.46** *f*(*t*) = *e*−*t* cos(2π*t*) for t = (-2:2). - -**Figure 1.47** *f*(*t*) = *e*−*t* cos(2π*t*) for t = (-2:0.01:2). - -† Sampling theory, presented later, formally addresses important aspects of this question. - ->> plot(t,f(t)); >> xlabel('t'); ylabel('f(t)'); grid; - -The result, shown in Fig. 1.47, is an accurate depiction of *f*(*t*). - -### **[1.11-2 Relational Operators and the Unit Step Function](#page-8-0)** - -The unit step function *u*(*t*) arises naturally in many practical situations. For example, a unit step can model the act of turning on a system. With the help of relational operators, anonymous functions can represent the unit step function. - -In MATLAB, a relational operator compares two items. If the comparison is true, a logical true (1) is returned. If the comparison is false, a logical false (0) is returned. Sometimes called indicator functions, relational operators indicates whether a condition is true. Six relational operators are available: <, >, <=, >=, ==, and ~=. - -The unit step function is readily defined using the >= relational operator. - ->> u = @(t) 1.0.\*(t>=0); - -Any function with a jump discontinuity, such as the unit step, is difficult to plot. Consider plotting *u*(*t*) by using t = (-2:2). - ->> t = (-2:2); plot(t,u(t)); >> xlabel('t'); ylabel('u(t)'); - -Two significant problems are apparent in the resulting plot, shown in Fig. 1.48. First, MATLAB automatically scales plot axes to tightly bound the data. In this case, this normally desirable feature obscures most of the plot. Second, MATLAB connects plot data with lines, making a true jump discontinuity difficult to achieve. The coarse resolution of vector t emphasizes the effect by showing an erroneous sloping line between *t* = −1 and *t* = 0. - -The first problem is corrected by vertically enlarging the bounding box with the axis command. The second problem is reduced, but not eliminated, by adding points to vector t. - -**Figure 1.48** *u*(*t*) for t = (-2:2). - -**Figure 1.49** *u*(*t*) for t = (-2:0.01:2) with axis modification. - ->> t = (-2:0.01:2); plot(t,u(t)); >> xlabel('t'); ylabel('u(t)'); >> axis([-2 2 -0.1 1.1]); - -The four-element vector argument of axis specifies *x* axis minimum, *x* axis maximum, *y* axis minimum, and *y* axis maximum, respectively. The improved results are shown in Fig. 1.49. - -Relational operators can be combined using logical AND, logical OR, and logical negation: &, |, and ~, respectively. For example, (t>0)&(t<1) and ~((t<=0)|(t>=1)) both test if 0 < *t* < 1. To demonstrate, consider defining and plotting the unit pulse *p*(*t*) = *u*(*t*) − *u*(*t* − 1), as shown in Fig. 1.50: - ->> p = @(t) 1.0.\*((t>=0)&(t<1)); >> t = (-1:0.01:2); plot(t,p(t)); >> xlabel('t'); ylabel('p(t) = u(t)-u(t-1)'); >> axis([-1 2 -.1 1.1]); - -Since anonymous functions can be constructed using other anonymous functions, we could have used our previously defined unit step anonymous function to define *p*(*t*) as p = @(t) u(t)-u(t-1);. - -**Figure 1.50** *p*(*t*) = *u*(*t*)−*u*(*t* −1) over (−1 ≤ *t* ≤ 2). - -### 130 CHAPTER 1 SIGNALS AND SYSTEMS - -For scalar operands, MATLAB also supports two short-circuit logical constructs. A short-circuit logical AND is performed by using &&, and a short-circuit logical OR is performed by using ||. Short-circuit logical operators are often more efficient than traditional logical operators because they test the second portion of the expression only when necessary. That is, when scalar expression A is found false in (A&&B), scalar expression B is not evaluated, since a false result is already guaranteed. Similarly, scalar expression B is not evaluated when scalar expression A is found true in (A||B), since a true result is already guaranteed. - -### **[1.11-3 Visualizing Operations on the Independent Variable](#page-8-0)** - -Two operations on a function's independent variable are commonly encountered: shifting and scaling. Anonymous functions are well suited to investigate both operations. - -Consider *g*(*t*) = *f*(*t*)*u*(*t*) = *e*−*t* cos(2π*t*)*u*(*t*), a causal version of *f*(*t*). MATLAB easily multiplies anonymous functions. Thus, we create *g*(*t*) by multiplying our anonymous functions for *f*(*t*) and *u*(*t*). † - ->> g = @(t) f(t).\*u(t); - -A combined shifting and scaling operation is represented by *g*(*at* + *b*), where *a* and *b* are arbitrary real constants. As an example, consider plotting *g*(2*t* +1) over (−2 ≤ *t* ≤ 2). With *a* = 2, the function is compressed by a factor of 2, resulting in twice the oscillations per unit *t*. Adding the condition *b* > 0 shifts the waveform to the left. Given anonymous function g, an accurate plot is nearly trivial to obtain. - -``` ->> t = (-2:0.01:2); ->> plot(t,g(2*t+1)); xlabel('t'); ylabel('g(2t+1)'); grid; -``` - -Figure 1.51 confirms the expected waveform compression and left shift. As a final check, realize that function *g*(·) turns on when the input argument is zero. Therefore, *g*(2*t* + 1) should turn on when 2*t* +1 = 0 or at *t* = −0.5, a fact again confirmed by Fig. 1.51. - -**Figure 1.51** *g*(2*t* +1) over (−2 ≤ *t* ≤ 2). - -† Although we define g in terms of f and u, the function g will not change if we later change either f or u unless we subsequently redefine g as well. - -**Figure 1.52** *g*(−*t* +1) over (−2 ≤ *t* ≤ 2). - -**Figure 1.53** *h*(*t*) = *g*(2*t* +1)+*g*(−*t* +1) over (−2 ≤ *t* ≤ 2). - -Next, consider plotting *g*(−*t* + 1) over (−2 ≤ *t* ≤ 2). Since *a* < 0, the waveform will be reflected. Adding the condition *b* > 0 shifts the final waveform to the right. - ->> plot(t,g(-t+1)); xlabel('t'); ylabel('g(-t+1)'); grid; - -Figure 1.52 confirms both the reflection and the right shift. - -Up to this point, Figs. 1.51 and 1.52 could be reasonably sketched by hand. Consider plotting the more complicated function *h*(*t*) = *g*(2*t* + 1) + *g*(−*t* + 1) over (−2 ≤ *t* ≤ 2) (Fig. 1.53); an accurate hand sketch would be quite difficult. With MATLAB, the work is much less burdensome. - ->> plot(t,g(2\*t+1)+g(-t+1)); xlabel('t'); ylabel('h(t)'); grid; - -### **[1.11-4 Numerical Integration and Estimating Signal Energy](#page-8-0)** - -Interesting signals often have nontrivial mathematical representations. Computing signal energy, which involves integrating the square of these expressions, can be a daunting task. Fortunately, many difficult integrals can be accurately estimated by means of numerical integration techniques. - -### 132 CHAPTER 1 SIGNALS AND SYSTEMS - -Even if the integration appears simple, numerical integration provides a good way to verify analytical results. - -To start, consider the simple signal *x*(*t*) = *e*−*t* (*u*(*t*)−*u*(*t*−1)). The energy of *x*(*t*) is expressed as *Ex* = \$ ∞ −∞ |*x*(*t*)| 2 *dt* = \$ 1 0 *e*−2*t dt*. Integrating yields *Ex* = 0.5(1 − *e*−2) ≈ 0.4323. The energy integral can also be evaluated numerically. Figure 1.27 helps illustrate the simple method of rectangular approximation: evaluate the integrand at points uniformly separated by *t*, multiply each by *t* to compute rectangle areas, and then sum over all rectangles. First, we create function *x*(*t*). - ->> x = @(t) exp(-t).\*((t>=0)&(t<1)); - -With *t* = 0.01, a suitable time vector is created. - ->> t = (0:0.01:1); - -The final result is computed by using the sum command. - ->> E\_x = sum(x(t).\*x(t)\*0.01) E\_x = 0.4367 - -The result is not perfect, but at 1% relative error it is close. By reducing *t*, the approximation is improved. For example, *t* = 0.001 yields E\_x = 0.4328, or 0.1% relative error. - -Although simple to visualize, rectangular approximation is not the best numerical integration technique. The MATLAB function quad implements a better numerical integration technique called recursive adaptive Simpson quadrature.† To operate, quad requires a function describing the integrand, the lower limit of integration, and the upper limit of integration. Notice that no *t* needs to be specified. - -To use quad to estimate *Ex*, the integrand must first be described. - ->> x\_squared = @(t) x(t).\*x(t); - -Estimating *Ex* immediately follows. - ->> E\_x = quad(x\_squared,0,1) E\_x = 0.4323 - -In this case, the relative error is −0.0026%. - -The same techniques can be used to estimate the energy of more complex signals. Consider *g*(*t*), defined previously. Energy is expressed as *Eg* = \$ ∞ 0 *e*−2*t* cos2 (2π*t*)*dt*. A closed-form solution exists, but it takes some effort. MATLAB provides an answer more quickly. - ->> g\_squared = @(t) g(t).\*g(t); - -† A comprehensive treatment of numerical integration is outside the scope of this text. Details of this particular method are not important for the current discussion; it is sufficient to say that it is better than the rectangular approximation. - -Although the upper limit of integration is infinity, the exponentially decaying envelope ensures *g*(*t*) is effectively zero well before *t* = 100. Thus, an upper limit of *t* = 100 is used along with *t* = 0.001. - ->> t = (0:0.001:100); >> E\_g = sum(g\_squared(t)\*0.001) E\_g = 0.2567 - -A slightly better approximation is obtained with the quad function. - ->> E\_g = quad(g\_squared,0,100) E\_g = 0.2562 - -### **DR ILL 1.21 Computing Signal Energy with MATLAB** - -Use MATLAB to confirm that the energy of signal *h*(*t*), defined previously as *h*(*t*) = *g*(2*t* + 1)+*g*(−*t* +1), is *Eh* = 0.3768. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/024_1.12 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/024_1.12 SUMMARY.md deleted file mode 100644 index 1e7fbad54947f4725e9068fe231d7289f4748fe8..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/024_1.12 SUMMARY.md +++ /dev/null @@ -1,45 +0,0 @@ -## **[1.12 SUMMARY](#page-8-0)** - -A *signal* is a set of data or information. A *system* processes input signals to modify them or extract additional information from them to produce output signals (response). A system may be made up of physical components (hardware realization), or it may be an algorithm that computes an output signal from an input signal (software realization). - -A convenient measure of the size of a signal is its energy, if it is finite. If the signal energy is infinite, the appropriate measure is its power, if it exists. The signal power is the time average of its energy (averaged over the entire time interval from −∞ to ∞). For periodic signals, the time averaging need be performed over only one period in view of the periodic repetition of the signal. Signal power is also equal to the mean squared value of the signal (averaged over the entire time interval from *t* = −∞ to ∞). - -Signals can be classified in several ways. - -- 1. A *continuous-time signal* is specified for a continuum of values of the independent variable (such as time *t*). A *discrete-time signal* is specified only at a finite or a countable set of time instants. -- 2. An *analog signal* is a signal whose amplitude can take on any value over a continuum. On the other hand, a signal whose amplitudes can take on only a finite number of values is a *digital signal*. The terms *discrete-time* and *continuous-time* qualify the nature of a signal along the time axis (horizontal axis). The terms *analog* and *digital,* on the other hand, qualify the nature of the signal amplitude (vertical axis). -- 3. A *periodic signal x*(*t*) is defined by the fact that *x*(*t*) = *x*(*t* +*T*0) for some *T*0. The smallest positive value of *T*0 for which this relationship is satisfied is called the *fundamental period*. A periodic signal remains unchanged when shifted by an integer multiple of its period. A periodic signal *x*(*t*) can be generated by a periodic extension of any contiguous segment of *x*(*t*) of duration *T*0. Finally, a periodic signal, by definition, must exist over the entire time interval −∞ < *t* < ∞. A signal is *aperiodic* if it is not periodic. - -- 4. An *everlasting signal* starts at *t* = −∞ and continues forever to *t* = ∞. Hence, periodic signals are everlasting signals. A *causal signal* is a signal that is zero for *t* < 0. -- 5. A signal with finite energy is an *energy signal*. Similarly a signal with a finite and nonzero power (mean-square value) is a *power signal*. A signal can be either an energy signal or a power signal, but not both. However, there are signals that are neither energy nor power signals. -- 6. A signal whose physical description is known completely in a mathematical or graphical form is a *deterministic signal*. A *random signal* is known only in terms of its probabilistic description such as mean value or mean-square value, rather than by its mathematical or graphical form. - -A signal *x*(*t*) delayed by *T* seconds (right-shifted) can be expressed as *x*(*t* − *T*); on the other hand, *x*(*t*) advanced by *T* (left-shifted) is *x*(*t* + *T*). A signal *x*(*t*) time-compressed by a factor *a*(*a* > 1) is expressed as *x*(*at*); on the other hand, the same signal time-expanded by factor *a*(*a* > 1) is *x*(*t*/*a*). The signal *x*(*t*) when time-reversed can be expressed as *x*(−*t*). - -The unit step function *u*(*t*) is very useful in representing causal signals and signals with different mathematical descriptions over different intervals. - -In the classical (Dirac) definition, the unit impulse function δ(*t*) is characterized by unit area and is concentrated at a single instant *t* = 0. The impulse function has a sampling (or sifting) property, which states that the area under the product of a function with a unit impulse is equal to the value of that function at the instant at which the impulse is located (assuming the function to be continuous at the impulse location). In the modern approach, the impulse function is viewed as a generalized function and is defined by the sampling property. - -The exponential function *est*, where *s* is complex, encompasses a large class of signals that includes a constant, a monotonic exponential, a sinusoid, and an exponentially varying sinusoid. - -A real signal that is symmetrical about the vertical axis (*t* = 0) is an *even* function of time, and a real signal that is antisymmetrical about the vertical axis is an *odd* function of time. The product of an even function and an odd function is an odd function. However, the product of an even function and an even function or an odd function and an odd function is an even function. The area under an odd function from *t* = −*a* to *a* is always zero regardless of the value of *a*. On the other hand, the area under an even function from *t* = −*a* to *a* is two times the area under the same function from *t* = 0 to *a* (or from *t* = −*a* to 0). Every signal can be expressed as a sum of odd and even functions of time. - -A system processes input signals to produce output signals (response). The input is the cause, and the output is its effect. In general, the output is affected by two causes: the internal conditions of the system (such as the initial conditions) and the external input. - -Systems can be classified in several ways. - -- 1. Linear systems are characterized by the linearity property, which implies superposition; if several causes (such as various inputs and initial conditions) are acting on a linear system, the total output (response) is the sum of the responses from each cause, assuming that all the remaining causes are absent. A system is nonlinear if superposition does not hold. -- 2. In time-invariant systems, system parameters do not change with time. The parameters of time-varying-parameter systems change with time. -- 3. For memoryless (or instantaneous) systems, the system response at any instant *t* depends only on the value of the input at *t*. For systems with memory (also known as dynamic - -systems), the system response at any instant *t* depends not only on the present value of the input, but also on the past values of the input (values before *t*). - -- 4. In contrast, if a system response at *t* also depends on the future values of the input (values of input beyond *t*), the system is noncausal. In causal systems, the response does not depend on the future values of the input. Because of the dependence of the response on the future values of input, the effect (response) of noncausal systems occurs before the cause. When the independent variable is time (temporal systems), the noncausal systems are prophetic systems, and therefore, unrealizable, although close approximation is possible with some time delay in the response. Noncausal systems with independent variables other than time (e.g., space) are realizable. -- 5. Systems whose inputs and outputs are continuous-time signals are continuous-time systems; systems whose inputs and outputs are discrete-time signals are discrete-time systems. If a continuous-time signal is sampled, the resulting signal is a discrete-time signal. We can process a continuous-time signal by processing the samples of the signal with a discrete-time system. -- 6. Systems whose inputs and outputs are analog signals are analog systems; those whose inputs and outputs are digital signals are digital systems. -- 7. If we can obtain the input *x*(*t*) back from the output *y*(*t*) of a system *S* by some operation, the system *S* is said to be invertible. Otherwise the system is noninvertible. -- 8. A system is stable if bounded input produces bounded output. This defines external stability because it can be ascertained from measurements at the external terminals of the system. External stability is also known as the stability in the BIBO (bounded-input/bounded-output) sense. Internal stability, discussed later in Ch. 2, is measured in terms of the internal behavior of the system. - -The system model derived from a knowledge of the internal structure of the system is its internal description. In contrast, an external description is a representation of a system as seen from its input and output terminals; it can be obtained by applying a known input and measuring the resulting output. In the majority of practical systems, an external description of a system so obtained is equivalent to its internal description. At times, however, the external description fails to describe the system adequately. Such is the case with the so-called uncontrollable or unobservable systems. - -A system may also be described in terms of certain set of key variables called state variables. In this description, an *N*th-order system can be characterized by a set of *N* simultaneous first-order differential equations in *N* state variables. State equations of a system represent an internal description of that system. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/025_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/025_REFERENCES.md deleted file mode 100644 index ff6571a70b8cfd07026472736f2e7136e8df189d..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/025_REFERENCES.md +++ /dev/null @@ -1,50 +0,0 @@ -### **[REFERENCES](#page-8-0)** - -- 1. Papoulis, A., *The Fourier Integral and Its Applications*. McGraw-Hill, New York, 1962. -- 2. Mason, S. J., *Electronic Circuits, Signals, and Systems*. Wiley, New York, 1960. -- 3. Kailath, T., *Linear Systems*. Prentice-Hall, Englewood Cliffs, NJ, 1980. -- 4. Lathi, B. P., *Signals and Systems*. Berkeley-Cambridge Press, Carmichael, CA, 1987. - -## **[PROBLEMS](#page-8-0)** - -- **1.1-1** Find the energies of the signals illustrated in Fig. P1.1-1. Comment on the effect on energy of sign change, time shifting, or doubling of the signal. What is the effect on the energy if the signal is multiplied by *k*? -- **1.1-2** Repeat Prob. 1.1-1 for the signals in Fig. P1.1-2. -- **1.1-3** (a) Find the energies of the pair of signals *x*(*t*) and *y*(*t*) depicted in Figs. P1.1-3a and P1.1-3b. Sketch and find the energies of signals *x*(*t*) + *y*(*t*) and *x*(*t*) − *y*(*t*). Can you make any observation from these results? - - (b) Repeat part (a) for the signal pair illustrated in Fig. P1.1-3c. Is your observation in part (a) still valid? - -- **1.1-4** Find the power of the periodic signal *x*(*t*) shown in Fig. P1.1-4. Find also the powers and the rms values of: - - (a) −*x*(*t*) - - (b) 2*x*(*t*) - - (c) *cx*(*t*) - - Comment. -- **1.1-5** By original design, a system outputs a 10-volt pulse that is 3 seconds in duration. It is desired to upgrade the square-pulse output with a "soft-start" pulse that steps up to 10 volts in 1-volt increments spaced every 20 milliseconds. Determine the signal duration *T* so that the "soft-start" pulse has the same signal energy as the original square pulse. - -**Figure P1.1-2** - --8 - -*t* 3 - -- **1.1-6** Determine the power and the rms value for each of the following signals: - - (a) 5+10 cos(100*t* +π/3) - - (b) 10 cos(100*t* +π/3)+16 sin(150*t* +π/5) - - (c) (10+2 sin 3*t*) cos 10*t* - - (d) 10 cos 5*t* cos 10*t* - - (e) 10 sin 5*t* cos 10*t* - - (f) *ej*α*t* cosω0*t* -- **1.1-7** Figure P1.1-7 shows a periodic 50% duty cycle dc-offset sawtooth wave *x*(*t*) with peak amplitude *A*. Determine the energy and power of *x*(*t*). -- **1.1-8** Two periodic signals that differ only by a 90-degree phase shift are considered to be quadrature signals. For example, cos(2π*t*) and sin(2π*t*) are quadrature signals. Another pair of quadrature signals is *x*(*t*) = sgn[cos(2π*t*)] and - -*y*(*t*) = sgn[sin(2π*t*)], where sgn is the sign (or signum) function. - -- (a) Plot *x*(*t*) and determine its power *Px* and energy *Ex*. -- (b) Plot *y*(*t*) and determine its power *Py* and energy *Ey*. -- (c) Consider the complex function *f*(*t*) = *x*(*t*)+ *jy*(*t*). Determine the power and energy of *f*(*t*). -- (d) When real functions *x*(*t*) and *y*(*t*) are combined as *f*(*t*) = *x*(*t*) + *jy*(*t*), is it generally true that *Ef* = *Ex* + *Ey* and *Pf* = *Px* + *Py*? Prove your answer. -- **1.1-9** There are many useful properties related to signal energy. Prove each of the following statements. In each case, let energy signal *x*1(*t*) have energy *E*[*x*1(*t*)], let energy signal *x*2(*t*) have - -**Figure P1.1-7** - -energy *E*[*x*2(*t*)], and let *T* be a nonzero, finite, real-valued constant. - -- (a) Prove *E*[*Tx*1(*t*)] = *T*2*E*[*x*1(*t*)]. That is, amplitude scaling a signal by constant *T* scales the signal energy by *T*2. \ No newline at end of file diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/026_PROBLEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/026_PROBLEMS.md deleted file mode 100644 index c2c0f99452c7968ea342f1f2d7df96f17544bb23..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/026_PROBLEMS.md +++ /dev/null @@ -1,768 +0,0 @@ -- (b) Prove *E*[*x*1(*t*)] = *E*[*x*1(*t* − *T*)]. That is, shifting a signal does not affect its energy. -- (c) If (*x*1(*t*) = 0) ⇒ (*x*2(*t*) = 0) and (*x*2(*t*) = 0) ⇒ (*x*1(*t*) = 0), then prove *E*[*x*1(*t*) + *x*2(*t*)] = *E*[*x*1(*t*)] + *E*[*x*2(*t*)]. That is, the energy of the sum of two nonoverlapping signals is the sum of the two individual energies. -- (d) Prove *E*[*x*1(*Tt*)] = (1/|*T*|)*E*[*x*1(*t*)]. That is, time-scaling a signal by *T* reciprocally scales the signal energy by 1/|*T*|. -- **1.1-10** Consider the signal *x*(*t*) shown in Fig. P1.1-10. Outside the interval shown, *x*(*t*) is zero. Determine the signal energy *E*[*x*(*t*)]. [*Hint:* Use the results of Prob. 1.1-9.] - -**Figure P1.1-10** - -**1.1-11** (a) Show that the power of a signal - -$$ -x(t) = \sum_{k=m}^{n} D_k e^{j\omega_k t} -$$ - -$$ -P_x = \sum_{k=m}^{n} |D_k|^2 -$$ - -assuming all frequencies to be distinct, that is, ω*i* = ω*k* for all *i* = *k*. - -- (b) Use the result in part (a) to determine the power of each of the signals in Prob. 1.1-6. -- **1.1-12** A binary signal *x*(*t*) = 0 for *t* < 0. For positive time, *x*(*t*) toggles between one and zero as follows: one for 1 second, zero for 1 second, one for 1 second, zero for 2 seconds, one for 1 second, zero for 3 seconds, and so forth. That is, the "on" time is always 1 second, but the "off" time successively increases by 1 second between each toggle. A portion of *x*(*t*) is shown in Fig. P1.1-12. Determine the energy and power of *x*(*t*). -- **1.2-1** For the signal *x*(*t*) depicted in Fig. P1.2-1, sketch the signals - - (a) *x*(−*t*) - -is - -- (b) *x*(*t* +6) -- (c) *x*(3*t*) -- (d) *x*(*t*/2) -- **1.2-2** For the signal *x*(*t*) illustrated in Fig. P1.2-2, sketch - - (a) *x*(*t* −4) - - (b) *x*(*t*/1.5) - - (c) *x*(−*t*) - - (d) *x*(2*t* −4) - - (e) *x*(2−*t*) -- **1.2-3** In Fig. P1.2-3, express signals *x*1(*t*), *x*2(*t*), *x*3(*t*), *x*4(*t*), and *x*5(*t*) in terms of signal *x*(*t*) and its time-shifted, time-scaled, or time-reversed versions. -- **1.2-4** For an energy signal *x*(*t*) with energy *Ex*, show that the energy of any one of the signals −*x*(*t*), *x*(−*t*), and *x*(*t* − *T*) is *Ex*. Show also that the energy of *x*(*at*) as well as *x*(*at* − *b*) is *Ex*/*a*, but the energy of *ax*(*t*) is *a*2*Ex*. This shows that time - -#### **Figure P1.2-3** - -inversion and time shifting do not affect signal energy. On the other hand, time compression of a signal (*a* > 1) reduces the energy, and time expansion of a signal (*a* < 1) increases the energy. What is the effect on signal energy if the signal is multiplied by a constant *a*? - -**1.2-5** Define 2*x*(−3*t* + 1) = *t*[*u*(−*t* − 1) − *u*(−*t* + 1)], where *u*(*t*) is the unit step function. (a) Plot 2*x*(−3*t* +1) over a suitable range of *t*. - -- (b) Plot *x*(*t*) over a suitable range of *t*. -- **1.2-6** Consider the signal *x*(*t*) = 2−*tu*(*t*) , where *u*(*t*) is the unit step function. - - (a) Accurately sketch *x*(*t*) over (−1 ≤ *t* ≤ 1). - - (b) Accurately sketch *y*(*t*) = 0.5*x*(1 − 2*t*) over (−1 ≤ *t* ≤ 1). -- **1.2-7** Define signals *y*(*t*) and *z*(*t*) as in Fig. P1.2-7. - - (a) Determine constants *a*, *b*, and *c* to produce *z*(*t*) = *ax*(*bt* +*c*) in Fig. P1.2-7. - -- **Figure P1.2-7** -- (b) Determine and sketch a signal *v*(*t*) such that *z*(*t*) = \$ *t* −∞ *v*(τ )*d*τ . -- **1.3-1** Think of a real-world signal that is a personally relevant and interesting. Describe the signal and then classify it according to the six following characteristics: - - (a) continuous-time or discrete-time - - (b) analog or digital - - (c) periodic or aperiodic - - (d) energy or power - - (e) causal or noncausal - - (f) deterministic or random - -If possible, think of a second real-world signal that has the opposite six characteristics of your first signal. If such a second signal is not possible, carefully explain why that is the case. - -**1.3-2** Define signal *y*(*t*) = %∞ *k*=−∞ *x*(0.5*t* − 10*k*), where - -$$ -x(t) = \begin{cases} e^{-2t} & t \ge 1 \\ 0 & t < 1 \end{cases} -$$ - -- (a) Determine the constant *a* such that the signal *x*(−2*t* +*a*) is borderline anticausal. -- (b) Is the signal *y*(*t*) periodic? If so, determine the period *Ty*. If not, explain why *y*(*t*) is not periodic. -- **1.3-3** Determine whether each of the following statements is true or false. If the statement is false, demonstrate this by proof or example. - - (a) Every continuous-time signal is an analog signal. - - (b) Every discrete-time signal is a digital signal. - - (c) If a signal is not an energy signal, then it must be a power signal and vice versa. - - (d) An energy signal must be of finite duration. - - (e) A power signal cannot be causal. - - (f) A periodic signal cannot be anticausal. -- **1.3-4** Determine whether each of the following statements is true or false. If the statement is - -false, demonstrate by proof or example why the statement is false. - -- (a) Every bounded periodic signal is a power signal. -- (b) Every bounded power signal is a periodic signal. -- (c) If an energy signal *x*(*t*) has energy *E*, then the energy of *x*(*at*) is *E*/*a*. Assume *a* is real and positive. -- (d) If a power signal *x*(*t*) has power *P*, then the power of *x*(*at*) is *P*/*a*. Assume *a* is real and positive. -- **1.3-5** Given *x*1(*t*) = cos(*t*), *x*2(*t*) = sin(π*t*), and *x*3(*t*) = *x*1(*t*)+*x*2(*t*). - - (a) Determine the fundamental periods *T*1 and *T*2 of signals *x*1(*t*) and *x*2(*t*). - - (b) Show that *x*3(*t*) is not periodic, which requires *T*3 = *k*1*T*1 = *k*2*T*2 for some integers *k*1 and *k*2. - - (c) Determine the powers *Px*1 , *Px*2 , and *Px*3 of signals *x*1(*t*), *x*2(*t*), and *x*3(*t*). -- **1.3-6** For any constant ω, is the function *f*(*t*) = sin(ω*t*) a periodic function of the independent variable *t*? Justify your answer. -- **1.3-7** The signal shown in Fig. P1.3-7 is defined as - -$$ -x(t) = \begin{cases} t & 0 \le t < 1 \\ 0.5 + 0.5 \cos(2\pi t) & 1 \le t < 2 \\ 3 - t & 2 \le t < 3 \\ 0 & \text{otherwise} \end{cases} -$$ - -The energy of *x*(*t*) is *E* ≈ 1.0417. - -- (a) What is the energy of *y*1(*t*) = (1/3)*x*(2*t*)? -- (b) A periodic signal *y*2(*t*) is defined as - -$$ -y_2(t) = \begin{cases} x(t) & 0 \le t < 4\\ y_2(t+4) & \forall t \end{cases} -$$ - -What is the power of *y*2(*t*)? - -(c) What is the power of *y*3(*t*) = (1/3)*y*2(2*t*)? - -**Figure P1.3-7** - -- **1.3-8** Let *y*1(*t*) = *y*2(*t*) = *t* 2 over 0 ≤ *t* ≤ 1. Notice, this statement does not require *y*1(*t*) = *y*2(*t*) for all *t*. - - (a) Define *y*1(*t*) as an even, periodic signal with period *T*1 = 2. Sketch *y*1(*t*) and determine its power. - - (b) Design an odd, periodic signal *y*2(*t*) with period *T*2 = 3 and power equal to unity. Fully describe *y*2(*t*) and sketch the signal over at least one full period. [*Hint:* There are an infinite number of possible solutions to this problem—you need to find only one of them!] - - (c) We can create a complex-valued function *y*3(*t*) = *y*1(*t*) + *jy*2(*t*). Determine whether this signal is periodic. If yes, determine the period *T*3. If no, justify why the signal is not periodic. - - (d) Determine the power of *y*3(*t*) defined in part (c). The power of a complex-valued function *z*(*t*) is - -$$ -P = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} z(\tau) z^*(\tau) d\tau -$$ - -- **1.4-1** Sketch the following signals: - - (a) *u*(*t* −5)−*u*(*t* −7) - - (b) *u*(*t* −5) +*u*(*t* −7) - - (c) *t* 2[*u*(*t* −1)−*u*(*t* −2)] - - (d) (*t* −4)[*u*(*t* −2) −*u*(*t* −4)] -- **1.4-2** Express each of the signals in Fig. P1.4-2 by a single expression valid for all *t*. - -**1.4-3** Letting *w*(*t*) = *t*[*u*(*t*)−*u*(*t* −1)], define the periodic signal *x*(*t*) as - -$$ -x(t) = \sum_{k=-\infty}^{\infty} w(2t + 2k) - 0.5w(2t + 2k - 1) -$$ - -- (a) Sketch *w*(*t*) and *x*(*t*). What is the fundamental period *T*0 of signal *x*(*t*)? -- (b) Sketch *y*(*t*) = *d dt x*(1−0.5*t*). -- (c) Determine the energy *Ez* and power *Pz* of the signal *z*(*t*) = *x*(0.5 − 1.5*t*)[*u*(*t*)−*u*(*t* −1)]. Sketching *z*(*t*) should help. -- **1.4-4** Define signal *x*(*t*) = *u*(*t*−1)−*u*(*t*−2.5)−2δ(*t*− 4)+δ(*t* −6). - - (a) Sketch *y*(*t*) = \$ *t* −∞ *x*(τ )*d*τ . - - (b) Describe a simple change that can be made to the right-most delta function in *x*(*t*) so that *y*(*t*) = \$ *t* −∞ *x*(τ )*d*τ has finite energy. - - (c) Sketch *z*(*t*) = \$ ∞ *t x*(τ )*d*τ . - - (d) Determine real constants *A* and *B* so that *w*(*t*) = *x t*−*A B* has a region of support [−2, 2]. -- **1.4-5** Simplify the following expressions: - -(a) -$$ -\left(\frac{\sin t}{t^2 + 2}\right) \delta(t) -$$ - -\n(b) $\left(\frac{j\omega + 2}{\omega^2 + 9}\right) \delta(\omega)$ - -(c) [*e*−*t* cos(3*t* −60◦)]δ(*t*) - -(d) -$$ -\left(\frac{\sin\left[\frac{\pi}{2}(t-2)\right]}{t^2+4}\right)\delta(1-t) -$$ - -(e) -$$ -\left(\frac{1}{j\omega+2}\right)\delta(\omega+3) -$$ - -(f) -$$ -\left(\frac{\sin k\omega}{\omega}\right)\delta(\omega) -$$ - -[*Hint:* Use Eq. (1.10). For part (f) use L'Hôpital's rule.] - -**1.4-6** Evaluate the following integrals: - -(a) -$$ -\int_{-\infty}^{\infty} \delta(\tau) x(t-\tau) d\tau -$$ - -\n(b) -$$ -\int_{-\infty}^{\infty} x(\tau) \delta(t-\tau) d\tau -$$ - -\n(c) -$$ -\int_{-\infty}^{\infty} \delta(t) e^{-j\omega t} dt -$$ - -\n(d) -$$ -\int_{-\infty}^{\infty} \delta(2t-3) \sin \pi t dt -$$ - -\n(e) -$$ -\int_{-\infty}^{\infty} \delta(t+3) e^{-t} dt -$$ - -\n(f) -$$ -\int_{-\infty}^{\infty} (t^3+4) \delta(1-t) dt -$$ - -\n(g) -$$ -\int_{-\infty}^{\infty} x(2-t) \delta(3-t) dt -$$ - -\n(h) -$$ -\int_{-\infty}^{\infty} e^{(x-1)} \cos \left[ \frac{\pi}{2} (x-5) \right] \delta(x-3) dx -$$ - -**1.4-7** For real and positive constant *a*, evaluate the following integral: - -$$ -\int_{-\infty}^{\infty} \delta(at) \, dt -$$ - -- **1.4-8** (a) Find and sketch *dx*/*dt* for the signal *x*(*t*) shown in Fig. P1.2-2. - - (b) Find and sketch *d*2*x*/*dt*2 for the signal *x*1(*t*) depicted in Fig. P1.4-2a. -- **1.4-9** Find and sketch \$ *t* −∞ *x*(*t*)*dt* for the signals *x*(*t*) illustrated in Fig. P1.4-9. -- **1.4-10** Using the generalized function definition of impulse [Eq. (1.11) with *T* = 0], show that δ(*t*) is an even function of *t*. -- **1.4-11** Using the generalized function definition of impulse [Eq. (1.11) with *T* = 0], show that - -$$ -\delta(at) = \frac{1}{|a|} \delta(t) -$$ - -**1.4-12** Show that - -**Figure P1.4-9** - -$$ -\int_{-\infty}^{\infty} \dot{\delta}(t)\phi(t) dt = -\dot{\phi}(0) -$$ - -where φ(*t*) and φ(˙ *t*) are continuous at *t* = 0, and φ(*t*) → 0 as *t* → ±∞. This integral defines δ(˙ *t*) as a generalized function. [*Hint:* Use integration by parts.] - -- **1.4-13** A sinusoid *e*σ*t* cos ω*t* can be expressed as a sum of exponentials *est* and *e*−*st* [Eq. (1.14)] with complex frequencies *s* = σ +*j*ω and *s* = σ −*j*ω. Locate in the complex plane the frequencies of the following sinusoids: - - (a) cos 3*t* - - (b) *e*−3*t* cos 3*t* - - (c) *e*2*t* cos 3*t* - - (d) *e*−2*t* - - (e) *e*2*t* - - (f) 5 -- **1.5-1** Find and sketch the odd and the even components of the following: - - (a) *u*(*t*) - - (b) *tu*(*t*) - - (c) sinω0*t* - - (d) cosω0*t* - - (e) cos(ω0*t* +θ ) - - (f) sinω0*tu*(*t*) - - (g) cosω0*tu*(*t*) -- **1.5-2** Define *x*(*t*) = 2*u*(*t*+1)−*u*(*t*−2)−*u*(*t*−3). - - (a) Letting *xo*(*t*) designate the odd portion of *x*(*t*), accurately sketch *xo*(1−2*t*). - -- (b) Letting *xe*(*t*) designate the even portion of *x*(*t*), accurately sketch *xe*(2+*t*/3). -- **1.5-3** (a) Determine even and odd components of the signal *x*(*t*) = *e*−2*t u*(*t*). - - (b) Show that the energy of *x*(*t*) is the sum of energies of its odd and even components found in part (a). - - (c) Generalize the result in part (b) for any finite energy signal. -- **1.5-4** (a) If *xe*(*t*) and *xo*(*t*) are even and the odd components of a real signal *x*(*t*), then show that - -$$ -\int_{-\infty}^{\infty} x_e(t)x_o(t) dt = 0 -$$ - -(b) Show that - -$$ -\int_{-\infty}^{\infty} x(t) dt = \int_{-\infty}^{\infty} x_e(t) dt -$$ - -- **1.5-5** An aperiodic signal is defined as *x*(*t*) = sin(π*t*)*u*(*t*), where *u*(*t*) is the continuous-time step function. Is the odd portion of this signal, *xo*(*t*), periodic? Justify your answer. -- **1.5-6** An aperiodic signal is defined as *x*(*t*) = cos(π*t*)*u*(*t*), where *u*(*t*) is the continuous-time step function. Is the even portion of this signal, *xe*(*t*), periodic? Justify your answer. -- **1.5-7** Consider the signal *x*(*t*) shown in Fig. P1.5-7. - -- (a) Determine and carefully sketch *v*(*t*) = 3*x*(−(1/2)(*t* +1)). -- (b) Determine the energy and power of *v*(*t*). -- (c) Determine and carefully sketch the even portion of *v*(*t*), *ve*(*t*). -- (d) Let *a* = 2 and *b* = 3; sketch *v*(*at* + *b*), *v*(*at*) +*b*, *av*(*t* +*b*), and *av*(*t*)+*b*. -- (e) Let *a* = −3 and *b* = −2; sketch *v*(*at* + *b*), *v*(*at*)+*b*, *av*(*t* +*b*), and *av*(*t*) +*b*. -- **1.5-8** Consider the signal *y*(*t*) = (1/5)*x*(−2*t* − 3) shown in Fig. P1.5-8. - -**Figure P1.5-8** - -- (a) Does *y*(*t*) have an odd portion, *yo*(*t*)? If so, determine and carefully sketch *yo*(*t*). Otherwise, explain why no odd portion exists. -- (b) Determine and carefully sketch the original signal *x*(*t*). -- **1.5-9** Consider the signal −(1/2)*x*(−3*t* + 2) shown in Fig. P1.5-9. - -### **Figure P1.5-9** - -- (a) Determine and carefully sketch the original signal *x*(*t*). -- (b) Determine and carefully sketch the even portion of the original signal *x*(*t*). -- (c) Determine and carefully sketch the odd portion of the original signal *x*(*t*). -- **1.5-10** The conjugate symmetric (or Hermitian) portion of a signal is defined as *wcs*(*t*) = (*w*(*t*) + *w*∗(−*t*))/2. Show that the real portion of *wcs*(*t*) is even and that the imaginary portion of *wcs*(*t*) is odd. -- **1.5-11** The conjugate antisymmetric (or skew-Hermitian) portion of a signal is defined as *wca*(*t*) = (*w*(*t*) − *w*∗(−*t*))/2. Show that the real portion of *wca*(*t*) is odd and that the imaginary portion of *wca*(*t*) is even. - -**1.5-12** Define *w*(*t*) = *ej*(*t*+π/4) . - -- (a) Referring to the definition in Prob. 1.5-10, determine *wcs*(*t*). Express your simplified answer in standard rectangular form. -- (b) Referring to the definition in Prob. 1.5-11, determine *wca*(*t*). Express your simplified answer in standard polar form. - -### 144 CHAPTER 1 SIGNALS AND SYSTEMS - -**1.5-13** Figure P1.5-13 plots a complex signal *w*(*t*) in the complex plane over the time range (0 ≤ *t* ≤ 1). The time *t* =0 corresponds with the origin, while the time *t* = 1 corresponds with the point (2, 1). - -**Figure P1.5-13** - -- (a) In the complex plane, plot *w*(*t*) over (−1 ≤ *t* ≤ 1) if *w*(*t*) is an even signal. -- (b) In the complex plane, plot *w*(*t*) over (−1 ≤ *t* ≤ 1) if *w*(*t*) is an odd signal. -- (c) In the complex plane, plot *w*(*t*) over (−1 ≤ *t* ≤ 1) if *w*(*t*) is a conjugate symmetric signal. [*Hint:* See Prob. 1.5-10.] -- (d) In the complex plane, plot *w*(*t*) over (−1 ≤ *t* ≤ 1) if *w*(*t*) is a conjugate antisymmetric signal. [*Hint:* See Prob. 1.5-11.] -- (e) In the complex plane, plot as much of *w*(3*t*) as possible. -- **1.5-14** Define complex signal *x*(*t*) = *t* 2(1 + *j*) over interval (1 ≤ *t* ≤ 2). The remaining portion is defined such that *x*(*t*) is a minimum-energy, skew-Hermitian signal. - - (a) Fully describe *x*(*t*) for all *t*. - - (b) Sketch *y*(*t*) = Re{*x*(*t*)} versus the independent variable *t*. - - (c) Sketch *z*(*t*) = Re{*jx*(−2*t* + 1)} versus the independent variable *t*. - - (d) Determine the energy and power of *x*(*t*). - -[*Hint:* See Prob. 1.5-11 for a definition of skew-Hermitian signals.] - -- **1.6-1** Write the input–output relationship for an ideal integrator. Determine the zero-input and zero-state components of the response. -- **1.6-2** A force *x*(*t*) acts on a ball of mass *M* (Fig. P1.6-2). Show that the velocity *v*(*t*) of the ball at any instant *t* > 0 can be determined if we know the force *x*(*t*) over the interval from 0 to *t* and the ball's initial velocity *v*(0). - -### **Figure P1.6-2** - -- **1.6-3** From your personal experience, provide an example of: - - (a) a single-input, single-output (SISO) system - - (b) a multiple-input, single-output (MISO) system - - (c) a single-input, multiple-output (SIMO) system - - (d) a multiple-input, multiple-output (MIMO) system -- **1.7-1** For the systems described by the following equations, with the input *x*(*t*) and output *y*(*t*), determine which of the systems are linear and which are nonlinear. - -(a) -$$ -\frac{dy(t)}{dt} + 2y(t) = x^2(t) -$$ - -\n(b) -$$ -\frac{dy(t)}{dt} + 3ty(t) = t^2x(t) -$$ - -\n(c) -$$ -3y(t) + 2 = x(t) -$$ - -\n(d) -$$ -\frac{dy(t)}{dt} + y^2(t) = x(t) -$$ - -\n(e) -$$ -\left(\frac{dy(t)}{dt}\right)^2 + 2y(t) = x(t) -$$ - -\n(f) -$$ -\frac{dy(t)}{dt} + (\sin t)y(t) = \frac{dx(t)}{dt} + 2x(t) -$$ - -(g) -$$ -\frac{dy(t)}{dt} + 2y(t) = x(t) \frac{dx(t)}{dt} -$$ - -(h) -$$ -y(t) = \int_0^t x(\tau) d\tau -$$ - -−∞ **1.7-2** For the systems described by the following equations, with the input *x*(*t*) and output *y*(*t*), explain with reasons which of the systems are time-invariant parameter systems and which are time-varying-parameter systems. - -(a) -$$ -y(t) = x(t-2) -$$ - -\n(b) $y(t) = x(-t)$ -\n(c) $y(t) = x(at)$ -\n(d) $y(t) = tx(t-2)$ -\n(e) $y(t) = \int_{-5}^{5} x(\tau) d\tau$ -\n(f) $y(t) = \left(\frac{dx(t)}{dt}\right)^2$ - -- **1.7-3** Two inputs, temperature *T*(*t*) and wind speed *V*(*t*), produce an output, wind chill *W*(*t*), according to *W*(*t*) = 35.74 + 0.6215*T*(*t*) − 35.75{*V*(*t*)} 0.16 + 0.4275*T*(*t*){*V*(*t*)} 0.16. The independent variable here is time, *t*. Answer the following questions yes or no, and provide mathematical justification for each answer. - - (a) Is this system BIBO-stable? - - (b) Is the system memoryless? - - (c) Is the system causal? - - (d) For simplicity, let the wind speed be constant, *V*(*t*) = *kV* . Thus, *W*(*t*) = *k*1 + *k*2*T*(*t*) for some constants *k*1 and *k*2. Is this simplified system linear? - - (e) For simplicity, let the temperature be constant, *T*(*t*) = *kT* . Thus, *W*(*t*) = *k*3 + *k*4 {*V*(*t*)} 0.16 for some constants *k*3 and *k*4. Is this simplified system linear? -- **1.7-4** Input voltage *x*(*t*) applied to an inverting op-amp follower circuit produces output *y*(*t*) according to - -$$ -y(t + t_{p}) = \begin{cases} -V_{ref} & x(t) > V_{ref} \\ V_{ref} & x(t) < -V_{ref} \\ -x(t) & \text{otherwise} \end{cases} -$$ - -where op-amp reference voltage *V*ref and propagation delay *t*p are both positive constants. Answer the following questions yes or no, and provide mathematical justification for each answer. - -- (a) Is this system BIBO-stable? -- (b) Is the system causal? -- (c) Is the system invertible? -- (d) Is the system linear? -- (e) Is the system memoryless? -- (f) Is the system time invariant? -- **1.7-5** Repeat Prob. 1.7-4 for a system with input *x*(*t*) that produces output *y*(*t*) according to - -$$ -y(t+1) = \begin{cases} -2x(t) & \text{when } x(t) \ge 0\\ 0 & \text{otherwise} \end{cases} -$$ - -**1.7-6** Repeat Prob. 1.7-4 for a system with input *x*(*t*) that produces output *y*(*t*) according to - -$$ -y(t-1) = \begin{cases} x(t-1) & \text{when } \frac{d}{dt}x(t) \ge 0\\ x(t-2) & \text{otherwise} \end{cases} -$$ - -- **1.7-7** Repeat Prob. 1.7-4 for a system that multiplies a given input by a ramp function, *r*(*t*) = *tu*(*t*). That is, *y*(*t*) = *x*(*t*)*r*(*t*). -- **1.7-8** Repeat Prob. 1.7-4 for a system with input *x*(*t*) that produces output *y*(*t*) according to - -$$ -y(t) = \frac{d}{dt}x(t-1) -$$ - -**1.7-9** Repeat Prob. 1.7-4 for a system with input *x*(*t*) that produces output *y*(*t*) according to - -$$ -y(t) = \begin{cases} x(t) & \text{if } x(t) > 0\\ 0 & \text{if } x(t) \le 0 \end{cases} -$$ - -**1.7-10** A continuous-time system is given by - -$$ -y(t) = 0.5 \int_{-\infty}^{\infty} x(\tau) [\delta(t-\tau) - \delta(t+\tau)] d\tau -$$ - -Recall that δ(*t*) designates the Dirac delta function. - -- (a) Explain what this system does. -- (b) Is the system BIBO-stable? Justify your answer. -- (c) Is the system linear? Justify your answer. -- (d) Is the system memoryless? Justify your answer. -- (e) Is the system causal? Justify your answer. -- (f) Is the system time invariant? Justify your answer. -- **1.7-11** For a certain LTI system with the input *x*(*t*), the output *y*(*t*) and the two initial conditions *q*1(0) and *q*2(0), the following observations were made: - -| x(t) | q1(0) | q2(0) | y
(t) | -|------|-------|-------|--------------------| -| 0 | 1 | −1 | e−t
u(t) | -| 0 | 2 | 1 | e−t
(3t +2)u(t) | -| u(t) | −1 | −1 | 2u(t) | - -Determine *y*(*t*) when both the initial conditions are zero and the input *x*(*t*) is as shown in Fig. P1.7-11. [*Hint:* There are three causes: the input and each of the two initial conditions. Because of the linearity property, if a cause is increased by a factor *k*, the response to that cause also increases by the same factor *k*. Moreover, if causes are added, the corresponding responses add.] - -**Figure P1.7-11** - -**1.7-12** A system is specified by its input–output relationship as - -$$ -y(t) = \frac{x^2(t)}{dx(t)/dt} -$$ - -Show that the system satisfies the homogeneity property but not the additivity property. - -**1.7-13** Show that the circuit in Fig. P1.7-13 is zero-state linear but not zero-input linear. Assume all diodes to have identical (matched) characteristics. The output is the current *y*(*t*). - -**Figure P1.7-13** - -**1.7-14** The inductor *L* and the capacitor *C* in Fig. P1.7-14 are nonlinear, which makes the circuit nonlinear. The remaining three elements are linear. Show that the output *y*(*t*) of this nonlinear circuit satisfies the linearity conditions with respect to the input *x*(*t*) and the initial conditions (all the initial inductor currents and capacitor voltages). - -**1.7-15** For the systems described by the following equations, with the input *x*(*t*) and output *y*(*t*), determine which are causal and which are noncausal. - -(a) -$$ -y(t) = x(t-2) -$$ - -(b) $y(t) = x(-t)$ - -(c) *y*(*t*) = *x*(*at*) *a* > 1 - -(d) *y*(*t*) = *x*(*at*) *a* < 1 - -**1.7-16** For the systems described by the following equations, with the input *x*(*t*) and output *y*(*t*), determine which are invertible and which are noninvertible. For the invertible systems, find the input–output relationship of the inverse system. - -(a) -$$ -y(t) = \int_{-\infty}^{t} x(\tau) d\tau -$$ - -\n(b) $y(t) = x^{n}(t), x(t)$ real, *n* integer -\n(c) $y(t) = \frac{dx(t)}{dt}$ - -$$ -(c) \ y(t) = \frac{t}{dt} -$$ - -(d) -$$ -y(t) = x(3t - 6) -$$ - -(e) -$$ -y(t) = \cos [x(t)] -$$ - -(f) -$$ -y(t) = e^{x(t)} -$$ -, $x(t)$ real - -- **1.7-17** Figure P1.7-17 displays an input *x*1(*t*) to a linear time-invariant (LTI) system *H*, the corresponding output *y*1(*t*), and a second input *x*2(*t*). - - (a) Bill suggests that *x*2(*t*) = 2*x*1(3*t*)−*x*1(*t*−1). Is Bill correct? If yes, prove it. If not, correct his error. - - (b) Bill wants to know the output *y*2(*t*) in response to the input *x*2(*t*). Provide him with an expression for *y*2(*t*) in terms of *y*1(*t*). Use MATLAB to plot *y*2(*t*). - -- **1.7-18** A linear time-invariant system *H* acts on input *x*(*t*) = *u*(*t* − 0.5) − *u*(*t* − 1.5) to produce output *y*(*t*) = *H* {*x*(*t*)} = 0.5*u*(*t*) + 0.5*u*(*t* − 1) −*u*(*t* −2). - - (a) Is it possible that the system is causal? Explain your answer. If not causal, determine the shift necessary to make the system causal. - - (b) Is it possible that the system is memoryless? Explain your answer. - - (c) Suppose the output *y*(*t*) is applied to an identical system *H* to produce output - -$$ -z(t) = H\{y(t)\} = H\{H\{x(t)\}\} -$$ - -If possible, determine and sketch *z*(*t*). If not possible, explain why *z*(*t*) cannot be determined using the information given. - -**1.8-1** For the circuit depicted in Fig. P1.8-1, find the differential equations relating outputs *y*1(*t*) and *y*2(*t*) to the input *x*(*t*). - -**1.8-2** For the circuit depicted in Fig. P1.8-2, find the differential equations relating outputs *y*1(*t*) and *y*2(*t*) to the input *x*(*t*). - -**Figure P1.8-2** - -- **1.8-3** A simplified (one-dimensional) model of an automobile suspension system is shown in Fig. P1.8-3. In this case, the input is not a force but a displacement *x*(*t*) (the road contour). Find the differential equation relating the output *y*(*t*) (auto body displacement) to the input *x*(*t*) (the road contour). -- **1.8-4** A field-controlled dc motor is shown in Fig. P1.8-4. Its armature current *ia* is maintained constant. The torque generated by this motor is proportional to the field current *if* (torque= *Kf if*). Find the differential equation relating the output position θ to the input voltage *x*(*t*). The motor and load together have a moment of inertia *J*. -- **1.8-5** Water flows into a tank at a rate of *qi* units/s and flows out through the outflow valve at a rate of *q*0 units/s (Fig. P1.8-5). Determine the equation relating the outflow *q*0 to the input *qi*. The outflow rate is proportional to the head *h*. Thus *q*0 = *Rh*, where *R* is the valve resistance. Determine also the differential equation relating the head *h* to the input *qi*. [Hint: The net inflow of water in time Δ*t* is (*qi* − *q*0)Δ*t*. This inflow is also *A*Δ*h*, where *A* is the cross section of the tank.] -- **1.8-6** Consider the circuit shown in Fig. P1.8-6, with input voltage *x*(*t*) and output currents *y*1(*t*), *y*2(*t*), and *y*3(*t*). - - (a) What is the order of this system? Explain your answer. - - (b) Determine the matrix representation for this system. - - (c) Use Cramer's rule to determine the output current *y*3(*t*) for the input voltage *x*(*t*) = [2− | cos(*t*)|]*u*(*t* −1). -- **1.10-1** Write state equations for the parallel *RLC* circuit in Fig. P1.8-2. Use the capacitor voltage *q*1 and the inductor current *q*2 as your state variables. - -Show that every possible current or voltage in the circuit can be expressed in terms of *q*1, *q*2 and the input *x*(*t*). - -**1.10-2** Write state equations for the third-order circuit shown in Fig. P1.10-2, using the inductor currents *q*1, *q*2 and the capacitor voltage *q*3 as state variables. Show that every possible voltage or current in this circuit can be expressed as a linear combination of *q*1, *q*2, *q*3, and the input *x*(*t*). Also, at some instant *t*, it was found that - -*q*1 = 5, *q*2 = 1, *q*3 = 2, and *x* = 10. Determine the voltage across and the current through every element in this circuit. - -- **1.11-1** Provide MATLAB code and output that plots the odd portion *xo*(*t*) of the function *x*(*t*) = 2−*t* cos(2π*t*)*u*(*t*−π ) over a suitable-length interval using a suitable number of points. -- **1.11-2** Provide MATLAB code and output that plots the even portion *xe*(*t*) of the function *x*(*t*) = 2−*t*/2 cos(4π*t*)*u*(*t* − 0.5) over a suitable *t* using *t* = 0.002 second between points. -- **1.11-3** Define *x*(*t*) = *et*(1+*j*2π )*u*(−*t*) and *y*(*t*) = Re\* 2*x* −5−*t* 2 +. - - (a) Use MATLAB to plot Re{*x*(*t*)} versus Im{*x*(*at*)} for *a* = 0.5, 1, and 2 and −10 ≤ - -### **Figure P1.10-2** - -*t* ≤ 10. How important is the scale factor *a* on the shape of the resulting figure? - -- (b) Use MATLAB to plot *y*(*t*) over −10 ≤ *t* ≤ 10. Analytically determine the time *t*0 where *y*(*t*) has a jump discontinuity. Verify your calculation of *t*0 using the plot of *y*(*t*). -- (c) Use MATLAB and numerical integration to compute the energy *Ex* of signal *x*(*t*). -- (d) Use MATLAB and numerical integration to compute the energy *Ey* of signal *y*(*t*). -- **1.11-4** Consider the signal *x*(*t*) = *u*( *t* 2 + 1) − *u*(*t* − 1) − δ( *t* 2 ). Define *y*(*t*) = \$ *t*−3 −∞ *x*(τ )*d*τ and *z*(*t*) = \$ ∞ *t x*(τ )*d*τ . - - (a) Using MATLAB, accurately plot *y*(*t*). - - (b) Using MATLAB, accurately plot *z*(*t*). - - (c) Using MATLAB, accurately plot *w*(*t*) = *d dt y*(*t*) +*z*(*t*) . - - - -# **TIME-DOMAIN ANALYSIS OF [CONTINUOUS-TIME](#page-8-0) SYSTEMS** - -In this book we consider two methods of analysis of linear time-invariant (LTI) systems: the time-domain method and the frequency-domain method. In this chapter we discuss the *time-domain analysis* of linear, time-invariant, continuous-time (LTIC) systems. - -## **[2.1 INTRODUCTION](#page-8-0)** - -For the purpose of analysis, we shall consider *linear differential systems*. This is the class of LTIC systems introduced in Ch. 1, for which the input *x*(*t*) and the output *y*(*t*) are related by linear differential equations of the form - -$$ -\frac{d^{N}y(t)}{dt^{N}} + a_{1} \frac{d^{N-1}y(t)}{dt^{N-1}} + \dots + a_{N-1} \frac{dy(t)}{dt} + a_{N}y(t) -$$ -\n -$$ -= b_{N-M} \frac{d^{M}x(t)}{dt^{M}} + b_{N-M+1} \frac{d^{M-1}x(t)}{dt^{M-1}} + \dots + b_{N-1} \frac{dx(t)}{dt} + b_{N}x(t) \tag{2.1} -$$ - -where all the coefficients *ai* and *bi* are constants. Using operator notation *D* to represent *d*/*dt*, we can express this equation as - -$$ -(DN + a1DN-1 + ··· + aN-1D + aN)y(t) -$$ - -= (bN-MDM + bN-M+1DM-1 + ··· + bN-1D + bN)x(t) - -or - -$$ -Q(D)y(t) = P(D)x(t) -$$ -\n(2.2) - -where the polynomials *Q*(*D*) and *P*(*D*) are - -$$ -Q(D) = DN + a1DN-1 + \dots + aN-1D + aN -$$ - -$$ -P(D) = bN-MDM + bN-M+1DM-1 + \dots + bN-1D + bN -$$ - -Theoretically the powers *M* and *N* in the foregoing equations can take on any value. However, practical considerations make *M* > *N* undesirable for two reasons. In Sec. 4.3-3, we shall show that an LTIC system specified by Eq. (2.1) acts as an (*M* − *N*)th-order differentiator. A differentiator represents an unstable system because a bounded input like the step input results in an unbounded output, δ(*t*). Second, noise is enhanced by a differentiator. Noise is a wideband signal containing components of all frequencies from 0 to a very high frequency approaching ∞. † Hence, noise contains a significant amount of rapidly varying components. We know that the derivative of any rapidly varying signal is high. Therefore, any system specified by Eq. (2.1) in which *M* > *N* will magnify the high-frequency components of noise through differentiation. It is entirely possible for noise to be magnified so much that it swamps the desired system output even if the noise signal at the system's input is tolerably small. Hence, practical systems generally use *M* ≤ *N*. For the rest of this text we assume implicitly that *M* ≤ *N*. For the sake of generality, we shall assume *M* = *N* in Eq. (2.1). - -In Ch. 1, we demonstrated that a system described by Eq. (2.2) is linear. Therefore, its response can be expressed as the sum of two components: the zero-input response and the zero-state response (decomposition property).‡ Therefore, - -total response = zero-input response + zero-state response - -The zero-input response is the system output when the input *x*(*t*) = 0, and thus it is the result of internal system conditions (such as energy storages, initial conditions) alone. It is independent of the external input *x*(*t*). In contrast, the zero-state response is the system output to the external input *x*(*t*) when the system is in zero state, meaning the absence of all internal energy storages: that is, all initial conditions are zero. - -## **2.2 SYSTEM RESPONSE TO INTERNAL [CONDITIONS:](#page-8-0) THE ZERO-INPUT RESPONSE** - -The zero-input response *y*0(*t*) is the solution of Eq. (2.2) when the input *x*(*t*) = 0 so that - -$$ -Q(D)y_0(t) = 0 -$$ - -$$ -Q(D)y_0(t) = 0 -$$ - -If *y*(*t*) is the zero-state response, then *y*(*t*) is the solution of - -$$ -Q(D)y(t) = P(D)x(t) -$$ - -subject to zero initial conditions (zero-state). Adding these two equations, we have - -$$ -Q(D)[y_0(t) + y(t)] = P(D)x(t) -$$ - -Clearly, *y*0(*t*)+*y*(*t*) is the general solution of Eq. (2.2). - -† Noise is any undesirable signal, natural or manufactured, that interferes with the desired signals in the system. Some of the sources of noise are the electromagnetic radiation from stars, the random motion of electrons in system components, interference from nearby radio and television stations, transients produced by automobile ignition systems, and fluorescent lighting. - -‡ We can verify readily that the system described by Eq. (2.2) has the decomposition property. If *y*0(*t*) is the zero-input response, then, by definition, - -or - -$$ -(DN + a1DN-1 + \dots + aN-1D + aN)y0(t) = 0 -$$ -\n(2.3) - -A solution to this equation can be obtained systematically [1]. However, we will take a shortcut by using heuristic reasoning. Equation (2.3) shows that a linear combination of *y*0(*t*) and its *N* successive derivatives is zero, not at *some* values of *t*, but for all *t*. Such a result is possible *if and only if y*0(*t*) and all its *N* successive derivatives are of the same form. Otherwise their sum can never add to zero for all values of *t*. We know that only an exponential function *e*λ*t* has this property. So let us assume that - -$$ -y_0(t) = ce^{\lambda t} -$$ - -is a solution to Eq. (2.3). Then - -$$ -Dy_0(t) = \frac{dy_0(t)}{dt} = c\lambda e^{\lambda t} -$$ -$$ -D^2y_0(t) = \frac{d^2y_0(t)}{dt^2} = c\lambda^2 e^{\lambda t} -$$ -$$ -\vdots -$$ -$$ -D^Ny_0(t) = \frac{d^Ny_0(t)}{dt^N} = c\lambda^N e^{\lambda t} -$$ - -Substituting these results in Eq. (2.3), we obtain - -$$ -c(\lambda^N + a_1 \lambda^{N-1} + \dots + a_{N-1} \lambda + a_N)e^{\lambda t} = 0 -$$ - -For a nontrivial solution of this equation, - -$$ -\lambda^{N} + a_{1}\lambda^{N-1} + \dots + a_{N-1}\lambda + a_{N} = 0 -$$ -\n(2.4) - -This result means that *ce*λ*t* is indeed a solution of Eq. (2.3), provided λ satisfies Eq. (2.4). Note that the polynomial in Eq. (2.4) is identical to the polynomial *Q*(*D*) in Eq. (2.3), with λ replacing *D*. Therefore, Eq. (2.4) can be expressed as - -$$ -Q(\lambda) = 0 -$$ - -Expressing *Q*(λ) in factorized form, we obtain - -$$ -Q(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_N) = 0 -$$ -\n(2.5) - -Clearly, λ has *N* solutions: λ1, λ2, ..., λ*N*, assuming that all λ*i* are distinct. Consequently, Eq. (2.3) has *N* possible solutions: *c*1*e*λ1*t* , *c*2*e*λ2*t* , ..., *cNe*λ*Nt* , with *c*1, *c*2,..., *cN* as arbitrary constants. We - -can readily show that a general solution is given by the sum of these *N* solutions† so that - -$$ -y_0(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + \dots + c_N e^{\lambda_N t} -$$ - (2.6) - -where *c*1, *c*2, ..., *cN* are arbitrary constants determined by *N* constraints (the auxiliary conditions) on the solution. - -Observe that the polynomial *Q*(λ), which is characteristic of the system, has nothing to do with the input. For this reason the polynomial *Q*(λ) is called the *characteristic polynomial* of the system. The equation - -$$ -Q(\lambda) = 0 -$$ - -is called the *characteristic equation* of the system. Equation (2.5) clearly indicates that λ1, λ2, ..., λ*N* are the roots of the characteristic equation; consequently, they are called the *characteristic roots* of the system. The terms *characteristic values, eigenvalues,* and *natural frequencies* are also used for characteristic roots.‡ The exponentials *e*λ*it* (*i* = 1, 2,...,*n*) in the zero-input response are the *characteristic modes* (also known as *natural modes* or simply as *modes*) of the system. There is a characteristic mode for each characteristic root of the system, and the *zero-input response is a linear combination of the characteristic modes of the system*. - -An LTIC system's characteristic modes comprise its single most important attribute. Characteristic modes not only determine the zero-input response but also play an important role in determining the zero-state response. In other words, the entire behavior of a system is dictated primarily by its characteristic modes. In the rest of this chapter we shall see the pervasive presence of characteristic modes in every aspect of system behavior. - -### REPEATED ROOTS - -The solution of Eq. (2.3) as given in Eq. (2.6) assumes that the *N* characteristic roots λ1, λ2, ..., λ*N* are distinct. If there are repeated roots (same root occurring more than once), the form of the solution is modified slightly. By direct substitution we can show that the solution of the equation - -$$ -(D - \lambda)^2 y_0(t) = 0 -$$ - -is given by - -$$ -y_0(t) = (c_1 + c_2 t)e^{\lambda t} -$$ - -† To prove this assertion, assume that *y*1(*t*), *y*2(*t*), ..., *yN*(*t*) are all solutions of Eq. (2.3). Then - -$$ -Q(D)y_1(t) = 0 -$$ - -\n -$$ -Q(D)y_2(t) = 0 -$$ - -\n -$$ -\vdots -$$ - -\n -$$ -Q(D)y_N(t) = 0 -$$ - -Multiplying these equations by *c*1, *c*2, ..., *cN*, respectively, and adding them together yield - -$$ -Q(D)[c_1y_1(t) + c_2y_2(t) + \cdots + c_Ny_n(t)] = 0 -$$ - -This result shows that *c*1*y*1(*t*) + *c*2*y*2(*t*) +···+ *cNyn*(*t*) is also a solution of the homogeneous equation [Eq. (2.3)]. - -‡ *Eigenvalue* is German for "characteristic value." diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/027_2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/027_2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS.md deleted file mode 100644 index bcc699680ca008257120e742cf516abf606a5f83..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/027_2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS.md +++ /dev/null @@ -1,192 +0,0 @@ -#### 154 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -In this case the root λ repeats twice. Observe that the characteristic modes in this case are *e*λ*t* and *te*λ*t* . Continuing this pattern, we can show that for the differential equation - -$$ -(D - \lambda)^r y_0(t) = 0 -$$ - -the characteristic modes are *e*λ*t* , *te*λ*t* , *t* 2*e*λ*t* , ..., *t r*−1*e*λ*t* , and that the solution is - -$$ -y_0(t) = (c_1 + c_2t + \dots + c_rt^{r-1})e^{\lambda t} -$$ - -Consequently, for a system with the characteristic polynomial - -$$ -Q(\lambda) = (\lambda - \lambda_1)^r (\lambda - \lambda_{r+1}) \cdots (\lambda - \lambda_N) -$$ - -the characteristic modes are *e*λ1*t* , *te*λ1*t* , ..., *t r*−1*e*λ1*t* , *e*λ*r*+1*t* , ..., *e*λ*Nt* and the solution is - -$$ -y_0(t) = (c_1 + c_2t + \dots + c_rt^{r-1})e^{\lambda_1 t} + c_{r+1}e^{\lambda_{r+1} t} + \dots + c_N e^{\lambda_N t} -$$ - -### COMPLEX ROOTS - -The procedure for handling complex roots is the same as that for real roots. For complex roots, the usual procedure leads to complex characteristic modes and the complex form of solution. However, it is possible to avoid the complex form altogether by selecting a real-form of solution, as described next. - -For a real system, complex roots must occur in pairs of conjugates if the coefficients of the characteristic polynomial *Q*(λ) are to be real. Therefore, if α + *j*β is a characteristic root, α − *j*β must also be a characteristic root. The zero-input response corresponding to this pair of complex conjugate roots is - -$$ -y_0(t) = c_1 e^{(\alpha + j\beta)t} + c_2 e^{(\alpha - j\beta)t} -$$ -\n(2.7) - -For a real system, the response *y*0(*t*) must also be real. This is possible only if *c*1 and *c*2 are conjugates. Let - -$$ -c_1 = \frac{c}{2}e^{j\theta} \qquad \text{and} \qquad c_2 = \frac{c}{2}e^{-j\theta} -$$ - -This yields - -$$ -y_0(t) = \frac{c}{2}e^{i\theta}e^{(\alpha+j\beta)t} + \frac{c}{2}e^{-j\theta}e^{(\alpha-j\beta)t} -$$ - -= -$$ -\frac{c}{2}e^{\alpha t}[e^{i(\beta t + \theta)} + e^{-j(\beta t + \theta)}] -$$ - -= -$$ -ce^{\alpha t}\cos(\beta t + \theta) -$$ - (2.8) - -Therefore, the zero-input response corresponding to complex conjugate roots α ± *j*β can be expressed in a complex form [Eq. (2.7)] or a real form [Eq. (2.8)]. - -### **EXAMPLE 2.1 Finding the Zero-Input Response** - -Find *y*0(*t*), the zero-input response of the response for an LTIC system described by - -- **(a)** the simple-root system (*D*2 + 3*D* + 2)*y*(*t*) = *Dx*(*t*) with initial conditions *y*0(0) = 0 and *y*˙0(0) = −5. -- **(b)** the repeated-root system (*D*2 + 6*D* + 9)*y*(*t*) = (3*D* + 5)*x*(*t*) with initial conditions *y*0(0) = 3 and *y*˙0(0) = −7. -- **(c)** the complex-root system (*D*2 + 4*D* + 40)*y*(*t*) = (*D* + 2)*x*(*t*) with initial conditions *y*0(0) = 2 and *y*˙0(0) = 16.78. - -**(a)** Note that *y*0(*t*), being the zero-input response (*x*(*t*) = 0), is the solution of (*D*2 +3*D*+ 2)*y*0(*t*) = 0. The characteristic polynomial of the system is λ2 + 3λ + 2. The characteristic equation of the system is therefore λ2 + 3λ + 2 = (λ + 1)(λ + 2) = 0. The characteristic roots of the system are λ1 = −1 and λ2 = −2, and the characteristic modes of the system are *e*−*t* and *e*−2*t* . Consequently, the zero-input response is - -$$ -y_0(t) = c_1 e^{-t} + c_2 e^{-2t} -$$ - -Differentiating this expression, we obtain - -$$ -\dot{y}_0(t) = -c_1 e^{-t} - 2c_2 e^{-2t} -$$ - -To determine the constants *c*1 and *c*2, we set *t* = 0 in the equations for *y*0(*t*) and *y*˙0(*t*) and substitute the initial conditions *y*0(0) = 0 and *y*˙0(0) = −5, yielding - -$$ -0 = c_1 + c_2 -$$ - -$$ --5 = -c_1 - 2c_2 -$$ - -Solving these two simultaneous equations in two unknowns for *c*1 and *c*2 yields - -$$ -c_1 = -5 \qquad \text{and} \qquad c_2 = 5 -$$ - -Therefore, - -$$ -y_0(t) = -5e^{-t} + 5e^{-2t} -$$ - (2.9) - -This is the zero-input response of *y*(*t*). Because *y*0(*t*) is present at *t* = 0−, we are justified in assuming that it exists for *t* ≥ 0.† - -**(b)** The characteristic polynomial is λ2 + 6λ + 9 = (λ + 3)2, and its characteristic roots are λ1 = −3, λ2 = −3 (repeated roots). Consequently, the characteristic modes of the system are *e*−3*t* and *te*−3*t* . The zero-input response, being a linear combination of the characteristic modes, is given by - -$$ -y_0(t) = (c_1 + c_2 t)e^{-3t} -$$ - -† *y*0(*t*) may be present even before *t* = 0−. However, we can be sure of its presence only from *t* = 0− onward. - -We can find the arbitrary constants *c*1 and *c*2 from the initial conditions *y*0(0) = 3 and *y*˙0(0) = −7 following the procedure in part (a). The reader can show that *c*1 = 3 and *c*2 = 2. Hence, - -$$ -y_0(t) = (3+2t)e^{-3t} \qquad t \ge 0 -$$ - -**(c)** The characteristic polynomial is λ2 + 4λ + 40 = (λ + 2 − *j*6)(λ + 2 + *j*6). The characteristic roots are −2 ± *j*6.† The solution can be written either in the complex form [Eq. (2.7)] or in the real form [Eq. (2.8)]. The complex form is *y*0(*t*) = *c*1*e*λ1*t* + *c*2*e*λ2*t* , where λ1 = −2+*j*6 and λ2 = −2−*j*6. Since α = −2 and β =6, the real-form solution is [see Eq. (2.8)] - -$$ -y_0(t) = ce^{-2t}\cos(6t+\theta) -$$ - -Differentiating this expression, we obtain - -$$ -\dot{y}_0(t) = -2ce^{-2t}\cos{(6t + \theta)} - 6ce^{-2t}\sin{(6t + \theta)} -$$ - -To determine the constants *c* and θ, we set *t* = 0 in the equations for *y*0(*t*) and *y*˙0(*t*) and substitute the initial conditions *y*0(0) = 2 and *y*˙0(0) = 16.78, yielding - -$$ -2 = c \cos \theta -$$ - -16.78 = $-2c \cos \theta - 6c \sin \theta$ - -Solution of these two simultaneous equations in two unknowns *c*cos θ and *c*sinθ yields - -*c*cos θ = 2 and *c*sinθ = −3.463 - -Squaring and then adding these two equations yield - -$$ -c^{2} = (2)^{2} + (-3.464)^{2} = 16 \Longrightarrow c = 4 -$$ - -Next, dividing *c*sinθ = −3.463 by *c*cos θ = 2 yields - -$$ -\tan \theta = \frac{-3.463}{2} -$$ - -and - -$$ -\theta = \tan^{-1} \left( \frac{-3.463}{2} \right) = -\frac{\pi}{3} -$$ - -Therefore, - -$$ -y_0(t) = 4e^{-2t}\cos\left(6t - \frac{\pi}{3}\right) -$$ - -For the plot of *y*0(*t*), refer again to Fig. B.11c. - -$$ -\lambda^2 + 4\lambda + 40 = (\lambda^2 + 4\lambda + 4) + 36 = (\lambda + 2)^2 + (6)^2 = (\lambda + 2 - j6)(\lambda + 2 + j6) -$$ - -† The complex conjugate roots of a second-order polynomial can be determined by using the formula in Sec. B.8-10 or by expressing the polynomial as a sum of two squares. The latter can be accomplished by completing the square with the first two terms, as follows: - -### **EXAMPLE 2.2 Using MATLAB to Find Polynomial Roots** - -Find the roots λ1 and λ2 of the polynomial λ2 + 4λ + *k* for three values of *k*: **(a)** *k* = 3, **(b)** *k* = 4, and **(c)** *k* = 40. - -**(a)** >> r = roots([1 4 3]).' r = -3 -1 For *k* = 3, the polynomial roots are therefore λ1 = −3 and λ2 = −1. **(b)** >> r = roots([1 4 4]).' r = -2 -2 For *k* = 4, the polynomial roots are therefore λ1 = λ2 = −2. **(c)** >> r = roots([1 4 40]).' r = -2.00+6.00i -2.00-6.00i For *k* = 40, the polynomial roots are therefore λ1 = −2+*j*6 and λ2 = −2−*j*6. - -### **EXAMPLE 2.3 Using MATLAB to Find the Zero-Input Response** - -Consider an LTIC system specified by the differential equation - -$$ -(D^2 + 4D + k)y(t) = (3D + 5)x(t) -$$ - -Using initial conditions *y*0(0) = 3 and *y*˙0(0) = −7, apply MATLAB's dsolve command to determine the zero-input response when: **(a)** *k* = 3, **(b)** *k* = 4, and **(c)** *k* = 40. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/028_2.2 SYSTEM RESPONSE TO INTERNAL CONDITIONS - THE ZERO-INPUT RESPONSE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/028_2.2 SYSTEM RESPONSE TO INTERNAL CONDITIONS - THE ZERO-INPUT RESPONSE.md deleted file mode 100644 index bfd5a6468e14e81dc59cf559e447622f99b75c63..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/028_2.2 SYSTEM RESPONSE TO INTERNAL CONDITIONS - THE ZERO-INPUT RESPONSE.md +++ /dev/null @@ -1,203 +0,0 @@ -**(a)** - ->> y\_0 = dsolve('D2y+4\*Dy+3\*y=0','y(0)=3','Dy(0)=-7','t') y\_0 = 1/exp(t) + 2/exp(3\*t) - -For *k* = 3, the zero-input response is therefore *y*0(*t*) = *e*−*t* +2*e*−3*t* . **(b)** - ->> y\_0 = dsolve('D2y+4\*Dy+4\*y=0','y(0)=3','Dy(0)=-7','t') y\_0 = 3/exp(2\*t) - t/exp(2\*t) - -For *k* = 4, the zero-input response is therefore *y*0(*t*) = 3*e*−2*t* −*te*−2*t* . **(c)** - ->> -$$ -y_0 = dsolve('D2y+4*Dy+40*y=0', 'y(0)=3', 'Dy(0)=-7', 't') -$$ - -\n $y_0 = (3*cos(6*t))/exp(2*t) - sin(6*t)/(6*exp(2*t))$ - -For *k* = 40, the zero-input response is therefore *y*0(*t*) = 3*e*−2*t* cos(6*t*)− 1 6 *e*−2*t*sin(6*t*). - -### **DR ILL 2.1 Finding the Zero-Input Response of a First-Order System** - -Find the zero-input response of an LTIC system described by (*D* + 5)*y*(*t*) = *x*(*t*) if the initial condition is *y*(0) = 5. - -### **ANSWER** - -*y*0(*t*) = 5*e*−5*t t* ≥ 0 - -## **DR ILL 2.2 Finding the Zero-Input Response of a Second-Order System** - -Letting *y*0(0) = 1 and *y*˙0(0) = 4, solve - -(*D*2 +2*D*)*y*0(*t*) = 0 - -**ANSWER** - -*y*0(*t*) = 3−2*e*−2*t t* ≥ 0 - -### PRACTICAL INITIAL CONDITIONS AND THE MEANING OF 0− AND 0+ - -In Ex. 2.1 the initial conditions *y*0(0) and *y*˙0(0) were supplied. In practical problems, we must derive such conditions from the physical situation. For instance, in an *RLC* circuit, we may be given the conditions (initial capacitor voltages, initial inductor currents, etc.). - -From this information, we need to derive *y*0(0), *y*˙0(0), ... for the desired variable as demonstrated in the next example. - -In much of our discussion, the input is assumed to start at *t* = 0, unless otherwise mentioned. Hence, *t* = 0 is the reference point. The conditions immediately before *t* = 0 (just before the input is applied) are the conditions at *t* = 0−, and those immediately after *t* = 0 (just after the input is applied) are the conditions at *t* = 0+ (compare this with the historical time frames BCE and CE). In practice, we are likely to know the initial conditions at *t* = 0− rather than at *t* = 0+. The two sets of conditions are generally different, although in some cases they may be identical. - -The total response *y*(*t*) consists of two components: the zero-input response *y*0(*t*) [response due to the initial conditions alone with *x*(*t*) = 0] and the zero-state response resulting from the input alone with all initial conditions zero. At *t* = 0−, the total response *y*(*t*) consists solely of the zero-input response *y*0(*t*) because the input has not started yet. Hence the initial conditions on *y*(*t*) are identical to those of *y*0(*t*). Thus, *y*(0−) = *y*0(0−), *y*˙(0−) = ˙*y*0(0−), and so on. Moreover, *y*0(*t*) is the response due to initial conditions alone and does not depend on the input *x*(*t*). Hence, application of the input at *t* = 0 does not affect *y*0(*t*). This means the initial conditions on *y*0(*t*) at *t* = 0− and 0+ are identical; that is, *y*0(0−), *y*˙0(0−), ... are identical to *y*0(0+), *y*˙0(0+), ..., respectively. It is clear that for *y*0(*t*), there is no distinction between the initial conditions at *t* = 0−, 0, and 0+. They are all the same. But this is not the case with the total response *y*(*t*), which consists of both the zero-input and zero-state responses. Thus, in general, *y*(0−) = *y*(0+), *y*˙(0−) = ˙*y*(0+), and so on. - -### **EXAMPLE 2.4 Consideration of Initial Conditions** - -A voltage *x*(*t*) = 10*e*−3*t u*(*t*) is applied at the input of the *RLC* circuit illustrated in Fig. 2.2a. Find the loop current *y*(*t*) for *t* ≥ 0 if the initial inductor current is zero [*y*(0−) = 0] and the initial capacitor voltage is 5 volts [*vC*(0−) = 5]. - -The differential (loop) equation relating *y*(*t*) to *x*(*t*) was derived in Eq. (1.29) as - -$$ -(D^2 + 3D + 2)y(t) = Dx(t) -$$ - -The zero-state component of *y*(*t*) resulting from the input *x*(*t*), assuming that all initial conditions are zero, that is, *y*(0−) = *vC*(0−) = 0, will be obtained later in Ex. 2.9. In this example we shall find the zero-input reponse *y*0(*t*). For this purpose, we need two initial conditions, *y*0(0) and *y*˙0(0). These conditions can be derived from the given initial conditions, *y*(0−) = 0 and *vC*(0−) = 5, as follows. Recall that *y*0(*t*) is the loop current when the input terminals are shorted so that the input *x*(*t*) = 0 (zero-input), as depicted in Fig. 2.2b. We now compute *y*0(0) and *y*˙0(0), the values of the loop current and its derivative at *t* = 0, from the initial values of the inductor current and the capacitor voltage. Remember that the inductor current cannot change instantaneously in the absence of an impulsive voltage. Similarly, the capacitor voltage cannot change instantaneously in the absence of an impulsive current. Therefore, when the input terminals are shorted at *t* = 0, the inductor current is still zero and the capacitor voltage is still 5 volts. Thus, - -*y*0(0) = 0 - -**Figure 2.1** Circuits for Ex. 2.4. - -To determine *y*˙0(0), we use the loop equation for the circuit in Fig. 2.2b. Because the voltage across the inductor is *L*(*dy*0/*dt*) or *y*˙0(*t*), this equation can be written as follows: - -$$ -\dot{y}_0(t) + 3y_0(t) + v_C(t) = 0 -$$ - -Setting *t* = 0, we obtain - -$$ -\dot{y}_0(0) + 3y_0(0) + v_C(0) = 0 -$$ - -But *y*0(0) = 0 and *vC*(0) = 5. Consequently, - -*y*˙0(0) = −5 - -Therefore, the desired initial conditions are - -$$ -y_0(0) = 0 -$$ - and $\dot{y}_0(0) = -5$ - -Thus, the problem reduces to finding *y*0(*t*), the zero-input component of *y*(*t*) of the system specified by the equation (*D*2 +3*D*+2)*y*(*t*) = *Dx*(*t*), when the initial conditions are *y*0(0) = 0 and *y*˙0(0) = −5. We have already solved this problem in Ex. 2.1a, where we found - -$$ -y_0(t) = -5e^{-t} + 5e^{-2t} \qquad t \ge 0 -$$ - -This is the zero-input component of the loop current *y*(*t*). - -It is interesting to find the initial conditions at *t* = 0− and 0+ for the total response *y*(*t*). Let us compare *y*(0−) and *y*˙(0−) with *y*(0+) and *y*˙(0+). The two pairs can be compared by writing the loop equation for the circuit in Fig. 2.2a at *t* = 0− and *t* = 0+. The only difference between the two situations is that at *t* = 0−, the input *x*(*t*) = 0, whereas at *t* = 0+, the input *x*(*t*) = 10 [because *x*(*t*) = 10*e*−3*t* ]. Hence, the two loop equations are - -$$ -\dot{y}(0^-) + 3y(0^-) + v_C(0^-) = 0 -$$ - -$$ -\dot{y}(0^+) + 3y(0^+) + v_C(0^+) = 10 -$$ - -The loop current *y*(0+) = *y*(0−) = 0 because it cannot change instantaneously in the absence of impulsive voltage. The same is true of the capacitor voltage. Hence, *vC*(0+) = *vC*(0−) = 5. Substituting these values in the foregoing equations, we obtain *y*˙(0−) = −5 and *y*˙(0+) = 5. Thus, - -*y*(0−) = 0, *y*˙(0−) = −5 and *y*(0+) = 0, *y*˙(0+) = 5 (2.10) - -### **DR ILL 2.3 Zero-Input Response of an** *RC* **Circuit** - -In the circuit in Fig. 2.2a, the inductance *L* = 0 and the initial capacitor voltage *vC*(0) = 30 volts. Show that the zero-input component of the loop current is given by *y*0(*t*) = −10*e*−2*t*/3 for *t* ≥ 0. - -### INDEPENDENCE OF THE ZERO-INPUT AND ZERO-STATE RESPONSES - -In Ex. 2.4 we computed the zero-input component without using the input *x*(*t*). The zero-state response can be computed from the knowledge of the input *x*(*t*) alone; the initial conditions are assumed to be zero (system in zero state). The two components of the system response (the zero-input and zero-state responses) are independent of each other. *The two worlds of zero-input response and zero-state response coexist side by side, neither one knowing or caring what the other is doing. For each component, the other is totally irrelevant.* - -### ROLE OF AUXILIARY CONDITIONS IN SOLUTION OF DIFFERENTIAL EQUATIONS - -The solution of a differential equation requires additional pieces of information (the *auxiliary conditions*). Why? We now show heuristically why a differential equation does not, in general, have a unique solution unless some additional constraints (or conditions) on the solution are known. - -Differentiation operation is not invertible unless one piece of information about *y*(*t*) is given. To get back *y*(*t*) from *dy*/*dt*, we must know one piece of information, such as *y*(0). Thus, differentiation is an irreversible (noninvertible) operation during which certain information is lost. To invert this operation, one piece of information about *y*(*t*) must be provided to restore the original *y*(*t*). Using a similar argument, we can show that, given *d*2*y*/*dt*2, we can determine *y*(*t*) uniquely only if two additional pieces of information (constraints) about *y*(*t*) are given. In general, to determine *y*(*t*) uniquely from its *N*th derivative, we need *N* additional pieces of information (constraints) about *y*(*t*). These constraints are also called *auxiliary conditions*. When these conditions are given at *t* = 0, they are called *initial conditions*. - -### **[2.2-1 Some Insights into the Zero-Input Behavior of a System](#page-8-0)** - -By definition, the zero-input response is the system response to its internal conditions, assuming that its input is zero. Understanding this phenomenon provides interesting insight into system behavior. If a system is disturbed momentarily from its rest position and if the disturbance is then - -#### 162 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -removed, the system will not come back to rest instantaneously. In general, it will come back to rest over a period of time and only through a special type of motion that is characteristic of the system.† For example, if we press on an automobile fender momentarily and then release it at *t* = 0, there is no external force on the automobile for *t* > 0.‡ The auto body will eventually come back to its rest (equilibrium) position, but not through any arbitrary motion. It must do so by using only a form of response that is sustainable by the system on its own without any external source, since the input is zero. Only characteristic modes satisfy this condition. *The system uses a proper combination of characteristic modes to come back to the rest position while satisfying appropriate boundary (or initial) conditions.* - -If the shock absorbers of the automobile are in good condition (high damping coefficient), the characteristic modes will be monotonically decaying exponentials, and the auto body will come to rest rapidly without oscillation. In contrast, for poor shock absorbers (low damping coefficients), the characteristic modes will be exponentially decaying sinusoids, and the body will come to rest through oscillatory motion. When a series *RC* circuit with an initial charge on the capacitor is shorted, the capacitor will start to discharge exponentially through the resistor. This response of the *RC* circuit is caused entirely by its internal conditions and is sustained by this system without the aid of any external input. The exponential current waveform is therefore the characteristic mode of the *RC* circuit. - -Mathematically we know that *any combination of characteristic modes can be sustained by the system alone without requiring an external input*. This fact can be readily verified for the series *RL* circuit shown in Fig. 2.2. The loop equation for this system is - -$$ -(D+2)y(t) = x(t) -$$ - -It has a single characteristic root λ = −2, and the characteristic mode is *e*−2*t* . We now verify that a loop current *y*(*t*) = *ce*−2*t* can be sustained through this circuit without any input voltage. The input voltage *x*(*t*) required to drive a loop current *y*(*t*) = *ce*−2*t* is given by - -$$ -x(t) = L\frac{dy(t)}{dt} + Ry(t) -$$ - -= $\frac{d}{dt}(ce^{-2t}) + 2ce^{-2t}$ -= $-2ce^{-2t} + 2ce^{-2t} = 0$ - -**Figure 2.2** Modes always get a free ride. - -† This assumes that the system will eventually come back to its original rest (or equilibrium) position. - -‡ We ignore the force of gravity, which merely causes a constant displacement of the auto body without affecting the other motion. - -Clearly, the loop current *y*(*t*) = *ce*−2*t* is sustained by the *RL* circuit on its own, without the necessity of an external input. - -### THE RESONANCE PHENOMENON - -We have seen that any signal consisting of a system's characteristic mode is sustained by the system on its own; the system offers no obstacle to such signals. Imagine what would happen if we were to drive the system with an external input that is one of its characteristic modes. This would be like pouring gasoline on a fire in a dry forest or hiring a child to eat ice cream. A child would gladly do the job without pay. Think what would happen if he were paid by the amount of ice cream he ate! He would work overtime. He would work day and night, until he became sick. The same thing happens with a system driven by an input of the form of characteristic mode. The system response grows without limit, until it burns out.† We call this behavior the *resonance phenomenon*. An intelligent discussion of this important phenomenon requires an understanding of the zero-state response; for this reason we postpone this topic until Sec. 2.6-7. - -## **2.3 THE UNIT IMPULSE [RESPONSE](#page-8-0)** *h(t)* - -In Ch. 1 we explained how a system response to an input *x*(*t*) may be found by breaking this input into narrow rectangular pulses, as illustrated earlier in Fig. 1.27a, and then summing the system response to all the components. The rectangular pulses become impulses in the limit as their widths approach zero. Therefore, the system response is the sum of its responses to various impulse components. This discussion shows that if we know the system response to an impulse input, we can determine the system response to an arbitrary input *x*(*t*). We now discuss a method of determining *h*(*t*), the unit impulse response of an LTIC system described by the *N*th-order differential equation [Eq. (2.1)] - -$$ -\frac{d^N y(t)}{dt^N} + a_1 \frac{d^{N-1} y(t)}{dt^{N-1}} + \dots + a_{N-1} \frac{dy(t)}{dt} + a_N y(t) -$$ - -= $b_{N-M} \frac{d^M x(t)}{dt^M} + b_{N-M+1} \frac{d^{M-1} x(t)}{dt^{M-1}} + \dots + b_{N-1} \frac{dx(t)}{dt} + b_N x(t)$ - -Recall that noise considerations restrict practical systems to *M* ≤ *N*. Under this constraint, the most general case is *M* = *N*. Therefore, Eq. (2.1) can be expressed as - -$$ -(DN + a1DN-1 + \dots + aN-1D + aN)y(t) = (b0DN + b1DN-1 + \dots + bN-1D + bN)x(t) -$$ - (2.11) - -Before deriving the general expression for the unit impulse response *h*(*t*), it is illuminating to understand qualitatively the nature of *h*(*t*). The impulse response *h*(*t*) is the system response to an impulse input δ(*t*) applied at *t* = 0 with all the initial conditions zero at *t* = 0−. An impulse input δ(*t*) is like lightning, which strikes instantaneously and then vanishes. But in its wake, in that single moment, objects that have been struck are rearranged. Similarly, an impulse input δ(*t*) appears momentarily at *t* = 0, and then it is gone forever. But in that moment it generates energy storages; that is, it creates nonzero initial conditions instantaneously within the system at - -† In practice, the system in resonance is more likely to go in saturation because of high amplitude levels. - -*t* = 0+. Although the impulse input δ(*t*) vanishes for *t* > 0 so that the system has no input after the impulse has been applied, the system will still have a response generated by these newly created initial conditions. The impulse response *h*(*t*), therefore, must consist of the system's characteristic modes for *t* ≥ 0+. As a result, - -*h*(*t*) = characteristic mode terms *t* ≥ 0+ - -This response is valid for *t* > 0. But what happens at *t* = 0? At a single moment *t* = 0, there can at most be an impulse,† so the form of the complete response *h*(*t*) is - -$$ -h(t) = A_0 \delta(t) + \text{characteristic mode terms} \qquad t \ge 0 \tag{2.12} -$$ - -because *h*(*t*) is the unit impulse response. Setting *x*(*t*) = δ(*t*) and *y*(*t*) = *h*(*t*) in Eq. (2.11) yields - -$$ -(DN + a1DN-1 + \dots + aN-1D + aN)h(t) = (b0DN + b1DN-1 + \dots + bN-1D + bN)\delta(t) -$$ - -In this equation we substitute *h*(*t*) from Eq. (2.12) and compare the coefficients of similar impulsive terms on both sides. The highest order of the derivative of impulse on both sides is *N*, with its coefficient value as *A*0 on the left-hand side and *b*0 on the right-hand side. The two values must be matched. Therefore, *A*0 = *b*0 and - -$$ -h(t) = b_0 \delta(t) + \text{characteristic modes} \tag{2.13} -$$ - -In Eq. (2.11), if *M* < *N*, *b*0 = 0. Hence, the impulse term *b*0δ(*t*) exists only if *M* = *N*. The unknown coefficients of the *N* characteristic modes in *h*(*t*) in Eq. (2.13) can be determined by using the technique of impulse matching, as explained in the following example. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/029_2.3 THE UNIT IMPULSE RESPONSE h(t).md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/029_2.3 THE UNIT IMPULSE RESPONSE h(t).md deleted file mode 100644 index dca570021a9a6cf36bc8904d31c6a068b33a3521..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/029_2.3 THE UNIT IMPULSE RESPONSE h(t).md +++ /dev/null @@ -1,250 +0,0 @@ -### **EXAMPLE 2.5 Impulse Response via Impulse Matching** - -Find the impulse response *h*(*t*) for a system specified by - -$$ -(D2 + 5D + 6)y(t) = (D + 1)x(t) -$$ -\n(2.14) - -In this case, *b*0 = 0. Hence, *h*(*t*) consists of only the characteristic modes. The characteristic polynomial is λ2 + 5λ + 6 = (λ + 2)(λ + 3). The roots are −2 and −3. Hence, the impulse - -† It might be possible for the derivatives of δ(*t*) to appear at the origin. However, if *M* ≤ *N*, it is impossible for *h*(*t*) to have any derivatives of δ(*t*). This conclusion follows from Eq. (2.11) with *x*(*t*) = δ(*t*) and *y*(*t*) = *h*(*t*). The coefficients of the impulse and all its derivatives must be matched on both sides of this equation. If *h*(*t*) contains δ(1) (*t*), the first derivative of δ(*t*), the left-hand side of Eq. (2.11) will contain a term δ(*N*+1) (*t*). But the highest-order derivative term on the right-hand side is δ(*N*) (*t*). Therefore, the two sides cannot match. Similar arguments can be made against the presence of the impulse's higher-order derivatives in *h*(*t*). - -response *h*(*t*) is - -$$ -h(t) = (c_1 e^{-2t} + c_2 e^{-3t}) u(t) -$$ -\n(2.15) - -Letting *x*(*t*) = δ(*t*) and *y*(*t*) = *h*(*t*) in Eq. (2.14), we obtain - -$$ -\ddot{h}(t) + 5\dot{h}(t) + 6h(t) = \dot{\delta}(t) + \delta(t) -$$ -\n(2.16) - -Recall that initial conditions *h*(0−) and *h*˙(0−) are both zero. But the application of an impulse at *t* = 0 creates new initial conditions at *t* = 0+. Let *h*(0+) = *K*1 and *h*˙(0+) = *K*2. These jump discontinuities in *h*(*t*) and *h*˙(*t*) at *t* = 0 result in impulse terms *h*˙(0) = *K*1δ(*t*) and *h*¨(0) = *K*1δ(˙ *t*) + *K*2δ(*t*) on the left-hand side. Matching the coefficients of impulse terms on both sides of Eq. (2.16) yields - -$$ -5K_1 + K_2 = 1 -$$ -, $K_1 = 1$ $\implies$ $K_1 = 1, K_2 = -4$ - -We now use these values *h*(0+) = *K*1 = 1 and *h*˙(0+) = *K*2 = −4 in Eq. (2.15) to find *c*1 and *c*2. Setting *t* = 0+ in Eq. (2.15), we obtain *c*1 + *c*2 = 1. Also setting *t* = 0+ in *h*˙(*t*), we obtain −2*c*1 −3*c*1 = −4. These two simultaneous equations yield *c*1 = −1 and *c*2 = 2. Therefore, - -$$ -h(t) = (-e^{-2t} + 2e^{-3t})u(t) -$$ - -Although the method used in this example is relatively simple, we can simplify it still further by using a modified version of impulse matching. - -### SIMPLIFIED IMPULSE MATCHING METHOD - -The alternate technique we present now allows us to reduce the procedure to a simple routine to determine *h*(*t*). To avoid the needless distraction, the proof for this procedure is placed in Sec. 2.8. There, we show that for an LTIC system specified by Eq. (2.11), the unit impulse response *h*(*t*) is given by - -$$ -h(t) = b_0 \delta(t) + [P(D)y_n(t)]u(t) -$$ -\n(2.17) - -where *yn*(*t*) is a linear combination of the characteristic modes of the system subject to the following initial conditions: - -$$ -y_n(0) = \dot{y}_n(0) = \ddot{y}_n(0) = \dots = y_n^{(N-2)}(0) = 0 -$$ - and $y_n^{(N-1)}(0) = 1$ (2.18) - -where *y*(*k*) *n* (0) is the value of the *k*th derivative of *yn*(*t*) at *t* = 0. We can express this set of conditions for various values of *N* (the system order) as follows: - -$$ -N = 1 : y_n(0) = 1 -$$ - -\n -$$ -N = 2 : y_n(0) = 0, \dot{y}_n(0) = 1 -$$ - -\n -$$ -N = 3 : y_n(0) = \dot{y}_n(0) = 0, \ddot{y}_n(0) = 1 -$$ - -and so on. - -As stated earlier, if the order of *P*(*D*) is less than the order of *Q*(*D*), that is, if *M* < *N*, then *b*0 = 0, and the impulse term *b*0δ(*t*) in *h*(*t*) is zero. - -### **EXAMPLE 2.6 Impulse Response via Simplified Impulse Matching** - -Determine the unit impulse response *h*(*t*) for a system specified by the equation - -$$ -(D2 + 3D + 2) y(t) = Dx(t) -$$ - (2.19) - -This is a second-order system (*N* = 2) having the characteristic polynomial - -$$ -(\lambda^2 + 3\lambda + 2) = (\lambda + 1)(\lambda + 2) -$$ - -The characteristic roots of this system are λ = −1 and λ = −2. Therefore, - -$$ -y_n(t) = c_1 e^{-t} + c_2 e^{-2t} -$$ -\n(2.20) - -Differentiation of this equation yields - -$$ -\dot{y}_n(t) = -c_1 e^{-t} - 2c_2 e^{-2t} \tag{2.21} -$$ - -The initial conditions are [see Eq. (2.18)] - -$$ -\dot{y}_n(0) = 1 \qquad \text{and} \qquad y_n(0) = 0 -$$ - -Setting *t* = 0 in Eqs. (2.20) and (2.21), and substituting the initial conditions just given, we obtain - -$$ -0 = c_1 + c_2 -$$ - -$$ -1 = -c_1 - 2c_2 -$$ - -Solution of these two simultaneous equations yields - -$$ -c_1 = 1 \qquad \text{and} \qquad c_2 = -1 -$$ - -Therefore, - -$$ -y_n(t) = e^{-t} - e^{-2t} -$$ - -Moreover, according to Eq. (2.19), *P*(*D*) = *D* so that - -$$ -P(D)y_n(t) = Dy_n(t) = \dot{y}_n(t) = -e^{-t} + 2e^{-2t} -$$ - -Also in this case, *b*0 = 0 [the second-order term is absent in *P*(*D*)]. Therefore, - -$$ -h(t) = [P(D)y_n(t)]u(t) = (-e^{-t} + 2e^{-2t})u(t) -$$ - -**Comment.** In the above discussion, we have assumed *M* ≤ *N*, as specified by Eq. (2.11). Section 2.8 shows that the expression for *h*(*t*) applicable to all possible values of *M* and *N* is given by - -$$ -h(t) = P(D)[y_n(t)u(t)] -$$ - -where *yn*(*t*) is a linear combination of the characteristic modes of the system subject to initial conditions [Eq. (2.18)]. This expression reduces to Eq. (2.17) when *M* ≤ *N*. - -Determination of the impulse response *h*(*t*) using the procedures in this section is relatively simple. However, in Ch. 4 we shall discuss another, even simpler method using the Laplace transform. As the next example demonstrates, it is also possible to find *h*(*t*) using functions from MATLAB's symbolic math toolbox. - -### **EXAMPLE 2.7 Using MATLAB to Find the Impulse Response** - -Determine the impulse response *h*(*t*) for an LTIC system specified by the differential equation - -$$ -(D^2 + 3D + 2)y(t) = Dx(t) -$$ - -This is a second-order system with *b*0 = 0. First we find the zero-input component for initial conditions *y*(0−) = 0, and *y*˙(0−) = 1. Since *P*(*D*) = *D*, the zero-input response is differentiated and the impulse response immediately follows as *h*(*t*) = 0δ(*t*)+ [*Dyn*(*t*)]*u*(*t*). - ->> y\_n = dsolve('D2y+3\*Dy+2\*y=0','y(0)=0','Dy(0)=1','t'); h = diff(y\_n) h = 2/exp(2\*t) - 1/exp(t) - -Therefore, *h*(*t*) = (2*e*−2*t* −*e*−*t* )*u*(*t*). - -### **DR ILL 2.4 Finding the Impulse Response** - -Determine the unit impulse response of LTIC systems described by the following equations: - -- **(a)** (*D*+2)*y*(*t*) = (3*D*+5)*x*(*t*) -- **(b)** *D*(*D*+2)*y*(*t*) = (*D*+4)*x*(*t*) -- **(c)** (*D*2 +2*D*+1)*y*(*t*) = *Dx*(*t*) - -### **ANSWERS** - -- **(a)** 3δ(*t*)−*e*−2*t u*(*t*) -- **(b)** (2−*e*−2*t* )*u*(*t*) -- **(c)** (1−*t*)*e*−*t u*(*t*) - -### SYSTEM RESPONSE TO DELAYED IMPULSE - -If *h*(*t*) is the response of an LTIC system to the input δ(*t*), then *h*(*t*−*T*) is the response of this same system to the input δ(*t* − *T*). This conclusion follows from the time-invariance property of LTIC systems. Thus, by knowing the unit impulse response *h*(*t*), we can determine the system response to a delayed impulse δ(*t* − *T*). Next, we put this result to good use in finding an LTIC system's zero-state response. - -## **2.4 SYSTEM [RESPONSE TO](#page-8-0) EXTERNAL INPUT: THE ZERO-STATE RESPONSE** - -This section is devoted to the determination of the zero-state response of an LTIC system. This is the system response *y*(*t*) to an input *x*(*t*) when the system is in the zero state, that is, when all initial conditions are zero. *We shall assume that the systems discussed in this section are in the zero state unless mentioned otherwise.* Under these conditions, the zero-state response will be the total response of the system. - -We shall use the superposition property for finding the system response to an arbitrary input *x*(*t*). Let us define a basic pulse *p*(*t*) of unit height and width τ , starting at *t* = 0 as illustrated in Fig. 2.3a. Figure 2.3b shows an input *x*(*t*) as a sum of narrow rectangular pulses. The pulse starting at *t* = *n*τ in Fig. 2.3b has a height *x*(*n*τ ) and can be expressed as *x*(*n*τ )*p*(*t*−*n*τ ). Now, *x*(*t*) is the sum of all such pulses. Hence, - -$$ -x(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau) p(t - n\Delta \tau) = \lim_{\Delta \tau \to 0} \sum_{\tau} \left[ \frac{x(n\Delta \tau)}{\Delta \tau} \right] p(t - n\Delta \tau) \Delta \tau -$$ - -The term [*x*(*n*τ )/τ ]*p*(*t* − *n*τ ) represents a pulse *p*(*t* − *n*τ ) with height *x*(*n*τ )/τ . As τ → 0, the height of this strip → ∞, but its area remains *x*(*n*τ ). Hence, this strip approaches an impulse *x*(*n*τ )δ(*t* −*n*τ ) as τ → 0 (Fig. 2.3e). Therefore, - -$$ -x(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau) \delta(t - n\Delta \tau) \Delta \tau -$$ - (2.22) - -To find the response for this input *x*(*t*), we consider the input and the corresponding output pairs, as shown in Figs. 2.3c–2.3f and also shown by directed arrow notation as follows: - -input -$$ -\Rightarrow -$$ - output -\n -$$ -\delta(t) \Longrightarrow h(t) -$$ -\n -$$ -\delta(t - n\Delta\tau) \Longrightarrow h(t - n\Delta\tau) -$$ -\n -$$ -[x(n\Delta\tau)\Delta\tau]\delta(t - n\Delta\tau) \Longrightarrow [x(n\Delta\tau)\Delta\tau]h(t - n\Delta\tau) -$$ -\n -$$ -\lim_{\Delta\tau \to 0} \sum_{\tau} x(n\Delta\tau)\delta(t - n\Delta\tau) \Delta\tau \Longrightarrow \lim_{\Delta\tau \to 0} \sum_{\tau} x(n\Delta\tau)h(t - n\Delta\tau) \Delta\tau -$$ -\n -$$ -x(t) \quad [\text{see Eq. (2.22)}] -$$ - -**Figure 2.3** Finding the system response to an arbitrary input *x*(*t*). - -Therefore,† - -$$ -y(t) = \lim_{\Delta \tau \to 0} \sum_{\tau} x(n\Delta \tau)h(t - n\Delta \tau)\Delta \tau -$$ - -= -$$ -\int_{-\infty}^{\infty} x(\tau)h(t - \tau) d\tau -$$ - (2.23) - -This is the result we seek. We have obtained the system response *y*(*t*) to an arbitrary input *x*(*t*) in terms of the unit impulse response *h*(*t*). Knowing *h*(*t*), we can determine the response *y*(*t*) to any input. *Observe once again the all-pervasive nature of the system's characteristic modes. The system response to any input is determined by the impulse response, which, in turn, is made up of characteristic modes of the system.* - -It is important to keep in mind the assumptions used in deriving Eq. (2.23). We assumed a linear time-invariant (LTI) system. Linearity allowed us to use the principle of superposition, and time invariance made it possible to express the system's response to δ(*t* −*n*τ ) as *h*(*t* −*n*τ ). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/030_2.4 SYSTEM RESPONSE TO EXTERNAL INPUT - THE ZERO-STATE RESPONSE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/030_2.4 SYSTEM RESPONSE TO EXTERNAL INPUT - THE ZERO-STATE RESPONSE.md deleted file mode 100644 index 34099911d7703aed452cdadff3b0cd493cfde1e7..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/030_2.4 SYSTEM RESPONSE TO EXTERNAL INPUT - THE ZERO-STATE RESPONSE.md +++ /dev/null @@ -1,785 +0,0 @@ -### **[2.4-1 The Convolution Integral](#page-8-0)** - -The zero-state response *y*(*t*) obtained in Eq. (2.23) is given by an integral that occurs frequently in the physical sciences, engineering, and mathematics. For this reason this integral is given a special name: the *convolution integral*. The convolution integral of two functions *x*1(*t*) and *x*2(*t*) is denoted symbolically by *x*1(*t*) ∗ *x*2(*t*) and is defined as - -$$ -x_1(t) * x_2(t) \equiv \int_{-\infty}^{\infty} x_1(\tau) x_2(t - \tau) d\tau -$$ - (2.24) - -Some important properties of the convolution integral follow. - -### THE COMMUTATIVE PROPERTY - -Convolution operation is commutative; that is, *x*1(*t*) ∗ *x*2(*t*) = *x*2(*t*) ∗ *x*1(*t*). This property can be proved by a change of variable. In Eq. (2.24), if we let *z* = *t* − τ so that τ = *t* − *z* and *d*τ = −*dz*, we obtain - -$$ -x_1(t) * x_2(t) = -\int_{-\infty}^{-\infty} x_2(z) x_1(t-z) dz -$$ - -= -$$ -\int_{-\infty}^{\infty} x_2(z) x_1(t-z) dz -$$ - -= -$$ -x_2(t) * x_1(t) -$$ - (2.25) - -$$ -y(t) = \int_{-\infty}^{\infty} x(\tau)h(t, \tau) d\tau -$$ - -where *h*(*t*, τ ) is the system response at instant *t* to a unit impulse input located at τ . - - - -† In deriving this result we have assumed a time-invariant system. If the system is time-varying, then the system response to the input δ(*t*−*n*Δτ ) cannot be expressed as *h*(*t*−*n*Δτ ) but instead has the form *h*(*t*,*n*Δτ ). Use of this form modifies Eq. (2.23) to - -### THE DISTRIBUTIVE PROPERTY - -According to the distributive property, - -$$ -x_1(t) * [x_2(t) + x_3(t)] = x_1(t) * x_2(t) + x_1(t) * x_3(t) -$$ -\n(2.26) - -### THE ASSOCIATIVE PROPERTY - -According to the associative property, - -$$ -x_1(t) * [x_2(t) * x_3(t)] = [x_1(t) * x_2(t)] * x_3(t) -$$ -\n(2.27) - -The proofs of Eqs. (2.26) and (2.27) follow directly from the definition of the convolution integral. They are left as an exercise for the reader. - -THE SHIFT PROPERTY If - -$$ -x_1(t) * x_2(t) = c(t) -$$ - -then - -$$ -x_1(t) * x_2(t - T) = x_1(t - T) * x_2(t) = c(t - T) -$$ - -More generally, we see that - -$$ -x_1(t - T_1) * x_2(t - T_2) = c(t - T_1 - T_2) -$$ -\n(2.28) - -**Proof.** We are given - -$$ -x_1(t) * x_2(t) = \int_{-\infty}^{\infty} x_1(\tau) x_2(t - \tau) d\tau = c(t) -$$ - -Therefore, - -$$ -x_1(t) * x_2(t-T) = \int_{-\infty}^{\infty} x_1(\tau) x_2(t-T-\tau) d\tau -$$ -$$ -= c(t-T) -$$ - -The equally simple proof of Eq. (2.28) follows a similar approach. - -### CONVOLUTION WITH AN IMPULSE - -Convolution of a function *x*(*t*) with a unit impulse results in the function *x*(*t*) itself. By definition of convolution, - -$$ -x(t) * \delta(t) = \int_{-\infty}^{\infty} x(\tau) \delta(t - \tau) d\tau -$$ - -Because δ(*t* −τ ) is an impulse located at τ = *t*, according to the sampling property of the impulse [Eq. (1.11)], the integral here is just the value of *x*(τ ) at τ = *t*, that is, *x*(*t*). Therefore, - -$$ -x(t) * \delta(t) = x(t) -$$ - -Actually this result was derived earlier [Eq. (2.22)]. - -#### 172 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -### THE WIDTH PROPERTY - -If the durations (widths) of *x*1(*t*) and *x*2(*t*) are finite, given by *T*1 and *T*2, respectively, then the duration (width) of *x*1(*t*) ∗ *x*2(*t*) is *T*1 + *T*2 (Fig. 2.4). The proof of this property follows readily from the graphical considerations discussed later in Sec. 2.4-2. - -**Figure 2.4** Width property of convolution. - -### ZERO-STATE RESPONSE AND CAUSALITY - -The (zero-state) response *y*(*t*) of an LTIC system is - -$$ -y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau -$$ -\n(2.29) - -In deriving Eq. (2.29), we assumed the system to be linear and time-invariant. There were no other restrictions either on the system or on the input signal *x*(*t*). Since, in practice, most systems are causal, their response cannot begin before the input. Furthermore, most inputs are also causal, which means they start at *t* = 0. - -Causality restriction on both signals and systems further simplifies the limits of integration in Eq. (2.29). By definition, the response of a causal system cannot begin before its input begins. Consequently, the causal system's response to a unit impulse δ(*t*) (which is located at *t* = 0) cannot begin before *t* = 0. Therefore, a *causal system's unit impulse response h*(*t*) *is a causal signal*. - -It is important to remember that the integration in Eq. (2.29) is performed with respect to τ (not *t*). If the input *x*(*t*) is causal, *x*(τ ) = 0 for τ < 0. Therefore, *x*(τ ) = 0 for τ < 0, as illustrated in Fig. 2.5a. Similarly, if *h*(*t*) is causal, *h*(*t* − τ ) = 0 for *t* − τ < 0; that is, for τ > *t*, as depicted in Fig. 2.5a. Therefore, the product *x*(τ )*h*(*t* − τ ) = 0 everywhere except over the nonshaded interval 0 ≤ τ ≤ *t* shown in Fig. 2.5a (assuming *t* ≥ 0). Observe that if *t* is negative, *x*(τ )*h*(*t* − τ ) = 0 for all τ , as shown in Fig. 2.5b. Therefore, Eq. (2.29) reduces to - -$$ -y(t) = x(t) * h(t) = \begin{cases} \int_0^t x(\tau)h(t-\tau) d\tau & t \ge 0\\ 0 & t < 0 \end{cases} -$$ - (2.30) - -The lower limit of integration in Eq. (2.30) is taken as 0− to avoid the difficulty in integration that can arise if *x*(*t*) contains an impulse at the origin. This result shows that if *x*(*t*) and *h*(*t*) are both causal, the response *y*(*t*) is also causal. - -Because of the convolution's commutative property [Eq. (2.25)], we can also express Eq. (2.30) as [assuming causal *x*(*t*) and *h*(*t*)] - -$$ -y(t) = \begin{cases} \int_0^t h(\tau)x(t-\tau)d\tau & t \ge 0\\ 0 & t < 0 \end{cases} -$$ - -Hereafter, the lower limit of 0− will be implied even when we write it as 0. As in Eq. (2.30), this result assumes that both the input and the system are causal. - -### **EXAMPLE 2.8 Computing the Zero-State Response** - -For an LTIC system with the unit impulse response *h*(*t*) = *e*−2*t u*(*t*), determine the response *y*(*t*) for the input - -$$ -x(t) = e^{-t}u(t) -$$ - -Here both *x*(*t*) and *h*(*t*) are causal (Fig. 2.6). Hence, from Eq. (2.30), we obtain - -$$ -y(t) = \int_0^t x(\tau)h(t-\tau) d\tau \qquad t \ge 0 -$$ - -Because *x*(*t*) = *e*−*t u*(*t*) and *h*(*t*) = *e*−2*t u*(*t*), - -$$ -x(\tau) = e^{-\tau} u(\tau) -$$ - and $h(t - \tau) = e^{-2(t - \tau)} u(t - \tau)$ - -Remember that the integration is performed with respect to τ (not *t*), and the region of integration is 0 ≤ τ ≤ *t*. Hence, τ ≥ 0 and *t* − τ ≥ 0. Therefore, *u*(τ ) = 1 and *u*(*t* − τ ) = 1; consequently, - -$$ -y(t) = \int_0^t e^{-\tau} e^{-2(t-\tau)} d\tau \qquad t \ge 0 -$$ - -Because this integration is with respect to τ , we can pull *e*−2*t* outside the integral, giving us - -$$ -y(t) = e^{-2t} \int_0^t e^{\tau} d\tau = e^{-2t} (e^t - 1) = e^{-t} - e^{-2t} \qquad t \ge 0 -$$ - -Moreover, *y*(*t*) = 0 when *t* < 0 [see Eq. (2.30)]. Therefore, - -$$ -y(t) = (e^{-t} - e^{-2t})u(t) -$$ - -The response is depicted in Fig. 2.6c. - -### **DR ILL 2.5 Computing the Zero-State Response** - -For an LTIC system with the impulse response *h*(*t*) = 6*e*−*t u*(*t*), determine the system response to the input: **(a)** 2*u*(*t*) and **(b)** 3*e*−3*t u*(*t*). - -### **ANSWERS** - -- **(a)** 12(1−*e*−*t* )*u*(*t*) -- **(b)** 9(*e*−*t* −*e*−3*t* )*u*(*t*) - -### **DR ILL 2.6 Zero-State Response with Resonance** - -Repeat Drill 2.5 for the input *x*(*t*) = *e*−*t u*(*t*). - -### **ANSWER** - -6*te*−*t u*(*t*) - -### THE CONVOLUTION TABLE - -The task of convolution is considerably simplified by a ready-made convolution table (Table 2.1). This table, which lists several pairs of signals and their convolution, can conveniently determine *y*(*t*), a system response to an input *x*(*t*), without performing the tedious job of integration. For instance, we could have readily found the convolution in Ex. 2.8 by using pair 4 (with λ1 = −1 and λ2 = −2) to be (*e*−*t* −*e*−2*t* )*u*(*t*). The following example demonstrates the utility of this table. - -### **EXAMPLE 2.9 Convolution by Tables** - -Use Table 2.1 to compute the loop current *y*(*t*) of the *RLC* circuit in Ex. 2.4 for the input *x*(*t*) = 10*e*−3*t u*(*t*) when all the initial conditions are zero. - -The loop equation for this circuit [see Ex. 1.16 or Eq. (1.29)] is - -$$ -(D^2 + 3D + 2)y(t) = Dx(t) -$$ - -The impulse response *h*(*t*) for this system, as obtained in Ex. 2.6, is - -$$ -h(t) = (2e^{-2t} - e^{-t})u(t) -$$ - -The input is *x*(*t*) = 10*e*−3*t u*(*t*), and the response *y*(*t*) is - -$$ -y(t) = x(t) * h(t) = 10e^{-3t}u(t) * [2e^{-2t} - e^{-t}]u(t) -$$ - -Using the distributive property of the convolution [Eq. (2.26)], we obtain - -$$ -y(t) = 10e^{-3t}u(t) * 2e^{-2t}u(t) - 10e^{-3t}u(t) * e^{-t}u(t) -$$ - -= 20[e-3tu(t) \* e-2tu(t)] - 10[e-3tu(t) \* e-tu(t)] - -Now the use of pair 4 in Table 2.1 yields - -$$ -y(t) = \frac{20}{-3 - (-2)} [e^{-3t} - e^{-2t}] u(t) - \frac{10}{-3 - (-1)} [e^{-3t} - e^{-t}] u(t) -$$ - -= $-20(e^{-3t} - e^{-2t}) u(t) + 5(e^{-3t} - e^{-t}) u(t)$ -= $(-5e^{-t} + 20e^{-2t} - 15e^{-3t}) u(t)$ - -| No. | x1(t) | x2(t) | x1(t)
∗ x2(t)
= x2(t)
∗ x1(t) | -|-----|-------------------------|--------------------|-----------------------------------------------------------------------------------| -| 1 | x(t) | δ(t −T) | x(t −T) | -| 2 | eλt
u(t) | u(t) | 1−eλt
u(t)
−λ | -| 3 | u(t) | u(t) | tu(t) | -| 4 | eλ1t
u(t) | eλ2t
u(t) | eλ1t −eλ2t
u(t)
λ1
= λ2
λ1
−λ2 | -| 5 | eλt
u(t) | eλt
u(t) | teλt
u(t) | -| 6 | teλt
u(t) | eλt
u(t) | 1
2eλt
t
u(t)
2 | -| 7 | Nu(t)
t | eλt
u(t) | "N
N! eλt
N−k
N!t
λN+1 u(t) −
u(t)
λk+1(N
−k)!
k=0 | -| 8 | Mu(t)
t | Nu(t)
t | M!N!
M+N+1u(t)
t
(M +N +1)! | -| 9 | teλ1t
u(t) | eλ2t
u(t) | eλ2t −eλ1t +(λ1
−λ2)teλ1t
u(t)
−λ2)2
(λ1 | -| 10 | Meλt
t
u(t) | Neλt
t
u(t) | M!N!
M+N+1eλt
t
u(t)
(N + M +1)! | -| 11 | Meλ1t
t
u(t) | Neλ2t
t
u(t) | "M
(−1)kM!(N
M−keλ1t
+k)!t
−λ2)N+k+1 u(t)
k!(M −k)!(λ1
k=0 | -| | λ1
= λ2 | | "N
(−1)kN!(M
N−keλ2t
+k)!t
−λ1)M+k+1 u(t)
+
k!(N −k)!(λ2
k=0 | -| 12 | e−αt cos(βt
+θ )u(t) | eλt
u(t) | cos(θ −φ)eλt −e−αt cos(βt
+θ −φ)
u(t)

(α +λ)2 +β2 | -| | | | φ = tan−1[−β/(α
+λ)] | -| 13 | eλ1t
u(t) | eλ2t
u(−t) | eλ1t
u(t)+eλ2t
u(−t)
Reλ2
> Reλ1
λ2
−λ1 | -| 14 | eλ1t
u(−t) | eλ2t
u(−t) | eλ1t −eλ2t
u(−t)
λ2
−λ1 | - -**TABLE 2.1** Select Convolution Integrals - -### **DR ILL 2.7 Convolution by Tables** - -Use Table 2.1 to show *e*−2*t u*(*t*) ∗ (1−*e*−*t* )*u*(*t*) = 1 2 −*e*−*t* + 1 2 *e*−2*t u*(*t*). - -### **DR ILL 2.8 Zero-State Response by Convolution Table** - -Rework Drills 2.5 and 2.6 using Table 2.1. - -### **DR ILL 2.9 Another Zero-State Response by Convolution Table** - -For an LTIC system with the unit impulse response *h*(*t*) = *e*−2*t u*(*t*), determine the zero-state response *y*(*t*) if the input *x*(*t*) = sin 3*t u*(*t*). [*Hint:* Use pair 12 from Table 2.1.] - -### **ANSWER** - -1 13 [3*e*−2*t* + √13 cos(3*t* −146.32◦)]*u*(*t*) or 1 13 [3*e*−2*t* − √13 cos(3*t* +33.68◦)]*u*(*t*) - -### RESPONSE TO COMPLEX INPUTS - -The LTIC system response discussed so far applies to general input signals, real or complex. However, if the system is real, that is, if *h*(*t*) is real, then we shall show that the real part of the input generates the real part of the output, and a similar conclusion applies to the imaginary part. - -If the input is *x*(*t*) = *xr*(*t*) + *jxi*(*t*), where *xr*(*t*) and *xi*(*t*) are the real and imaginary parts of *x*(*t*), then for real *h*(*t*) - -$$ -y(t) = h(t) * [x_r(t) + jx_i(t)] = h(t) * x_r(t) + jh(t) * x_i(t) = y_r(t) + jy_i(t) -$$ - -where *yr*(*t*) and *yi*(*t*) are the real and the imaginary parts of *y*(*t*). Using the right-directed-arrow notation to indicate a pair of the input and the corresponding output, the foregoing result can be expressed as follows. If - -$$ -x(t) = x_r(t) + jx_i(t) \implies y(t) = y_r(t) + jy_i(t) -$$ - -then - -$$ -x_r(t) \implies y_r(t) \quad \text{and} \quad x_i(t) \implies y_i(t) \tag{2.31} -$$ - -### MULTIPLE INPUTS - -Multiple inputs to LTI systems can be treated by applying the superposition principle. Each input is considered separately, with all other inputs assumed to be zero. The sum of all these individual system responses constitutes the total system output when all the inputs are applied simultaneously. - -### **[2.4-2 Graphical Understanding of Convolution Operation](#page-8-0)** - -The convolution operation can be grasped readily through a graphical interpretation of the convolution integral. Such an understanding is helpful in evaluating the convolution integral of more complex signals. In addition, graphical convolution allows us to grasp visually or mentally the convolution integral's result, which can be of great help in sampling, filtering, and many other problems. Finally, many signals have no exact mathematical description, so they can be described only graphically. If two such signals are to be convolved, we have no choice but to perform their convolution graphically. - -We shall now explain the convolution operation by convolving the signals *x*(*t*) and *g*(*t*), illustrated in Figs. 2.7a and 2.7b, respectively. If *c*(*t*) is the convolution of *x*(*t*) with *g*(*t*), then - -$$ -c(t) = \int_{-\infty}^{\infty} x(\tau)g(t-\tau) d\tau -$$ - -One of the crucial points to remember here is that this integration is performed with respect to τ so that *t* is just a parameter (like a constant). This consideration is especially important when we sketch the graphical representations of the functions *x*(τ ) and *g*(*t*−τ ). Both these functions should be sketched as functions of τ , not of *t*. - -The function *x*(τ ) is identical to *x*(*t*), with τ replacing *t* (Fig. 2.7c). Therefore, *x*(*t*) and *x*(τ ) will have the same graphical representations. Similar remarks apply to *g*(*t*) and *g*(τ ) (Fig. 2.7d). - -To appreciate what *g*(*t* − τ ) looks like, let us start with the function *g*(τ ) (Fig. 2.7d). Time reversal of this function (reflection about the vertical axis τ = 0) yields *g*(−τ ) (Fig. 2.7e). Let us denote this function by φ(τ ): - -$$ -\phi(\tau) = g(-\tau) -$$ - -Now φ(τ ) shifted by *t* seconds is φ(τ −*t*), given by - -$$ -\phi(\tau - t) = g[-(\tau - t)] = g(t - \tau) -$$ - -Therefore, we first time-reverse *g*(τ ) to obtain *g*(−τ ) and then time-shift *g*(−τ ) by *t* to obtain *g*(*t* − τ ). For positive *t*, the shift is to the right (Fig. 2.7f); for negative *t*, the shift is to the left (Figs. 2.7g, 2.7h). - -The preceding discussion gives us a graphical interpretation of the functions *x*(τ ) and *g*(*t*−τ ). The convolution *c*(*t*) is the area under the product of these two functions. Thus, to compute *c*(*t*) at some positive instant *t* = *t*1, we first obtain *g*(−τ ) by inverting *g*(τ ) about the vertical axis. Next, we right-shift or delay *g*(−τ ) by *t*1 to obtain *g*(*t*1 − τ ) (Fig. 2.7f), and then we multiply this function by *x*(τ ), giving us the product *x*(τ )*g*(*t*1 − τ ) (shaded portion in Fig. 2.7f). The area *A*1 under this product is *c*(*t*1), the value of *c*(*t*) at *t* = *t*1. We can therefore plot *c*(*t*1) = *A*1 on a curve describing *c*(*t*), as shown in Fig. 2.7i. The area under the product *x*(τ )*g*(−τ ) in Fig. 2.7e is *c*(0), the value of the convolution for *t* = 0 (at the origin). - -**Figure 2.7** Graphical explanation of the convolution operation. - -### 180 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -A similar procedure is followed in computing the value of *c*(*t*) at *t* = *t*2, where *t*2 is negative (Fig. 2.7g). In this case, the function *g*(−τ ) is shifted by a negative amount (that is, left-shifted) to obtain *g*(*t*2−τ ). Multiplication of this function with *x*(τ ) yields the product *x*(τ )*g*(*t*2−τ ). The area under this product is *c*(*t*2) = *A*2, giving us another point on the curve *c*(*t*) at *t* = *t*2 (Fig. 2.7i). This procedure can be repeated for all values of *t*, from −∞ to ∞. The result will be a curve describing *c*(*t*) for all time *t*. Note that when *t* ≤ −3, *x*(τ ) and *g*(*t*−τ ) do not overlap (see Fig. 2.7h); therefore, *c*(*t*) = 0 for *t* ≤ −3. - -### SUMMARY OF THE GRAPHICAL PROCEDURE - -The procedure for graphical convolution can be summarized as follows: - -- 1. Keep the function *x*(τ ) fixed. -- 2. Visualize the function *g*(τ ) as a rigid wire frame, and rotate (or invert) this frame about the vertical axis (τ = 0) to obtain *g*(−τ ). -- 3. Shift the inverted frame along the τ axis by *t*0 seconds. The shifted frame now represents *g*(*t*0 −τ ). -- 4. The area under the product of *x*(τ ) and *g*(*t*0 − τ ) (the shifted frame) is *c*(*t*0), the value of the convolution at *t* = *t*0. -- 5. Repeat this procedure, shifting the frame by different values (positive and negative) to obtain *c*(*t*) for all values of *t*. - -The graphical procedure discussed here appears very complicated and discouraging at first reading. Indeed, some people claim that convolution has driven many electrical engineering undergraduates to contemplate theology either for salvation or as an alternative career (*IEEE Spectrum,* March 1991, p. 60). Actually, the bark of convolution is worse than its bite. In graphical convolution, we need to determine the area under the product *x*(τ )*g*(*t* − τ ) for all values of *t* from −∞ to ∞. However, a mathematical description of *x*(τ )*g*(*t* − τ ) is generally valid over a range - -Convolution: Its bark is worse than its bite! - -of *t*. Therefore, repeating the procedure for every value of *t* amounts to repeating it only a few times for different ranges of *t*. - -We can also use the commutative property of convolution to our advantage by computing *x*(*t*) ∗ *g*(*t*) or *g*(*t*) ∗ *x*(*t*), whichever is simpler. As a rule of thumb, *convolution computations are simplified if we choose to invert (time-reverse) the simpler of the two functions*. For example, if the mathematical description of *g*(*t*) is simpler than that of *x*(*t*), then *x*(*t*) ∗ *g*(*t*) will be easier to compute than *g*(*t*) ∗ *x*(*t*). In contrast, if the mathematical description of *x*(*t*) is simpler, the reverse will be true. - -We shall demonstrate graphical convolution with the following examples. Let us start by using this graphical method to rework Ex. 2.8. - -### **EXAMPLE 2.10 Graphical Convolution of Two Causal Functions** - -Determine graphically *y*(*t*) = *x*(*t*) ∗ *h*(*t*) for *x*(*t*) = *e*−*t u*(*t*) and *h*(*t*) = *e*−2*t u*(*t*). - -In Figs. 2.8a and 2.8b we have *x*(*t*) and *h*(*t*), respectively; and Fig. 2.8c shows *x*(τ ) and *h*(−τ ) as functions of τ . The function *h*(*t* −τ ) is now obtained by shifting *h*(−τ ) by *t*. If *t* is positive, the shift is to the right (delay); if *t* is negative, the shift is to the left (advance). Figure 2.8d shows that for negative *t*, *h*(*t* −τ ) [obtained by left-shifting *h*(−τ )] does not overlap *x*(τ ), and the product *x*(τ )*h*(*t* −τ ) = 0, so that - -$$ -y(t) = 0 \qquad t < 0 -$$ - -Figure 2.8e shows the situation for *t* ≥ 0. Here *x*(τ ) and *h*(*t* −τ ) do overlap, but the product is nonzero only over the interval 0 ≤ τ ≤ *t* (shaded interval). Therefore, - -$$ -y(t) = \int_0^t x(\tau)h(t-\tau) d\tau \qquad t \ge 0 -$$ - -All we need to do now is substitute correct expressions for *x*(τ ) and *h*(*t* − τ ) in this integral. From Figs. 2.8a and 2.8b, it is clear that the segments of *x*(*t*) and *g*(*t*) to be used in this convolution (Fig. 2.8e) are described by - -$$ -x(t) = e^{-t} \qquad \text{and} \qquad h(t) = e^{-2t} -$$ - -Therefore, - -$$ -x(\tau) = e^{-\tau} -$$ - and $h(t - \tau) = e^{-2(t - \tau)}$ - -Consequently, - -$$ -y(t) = \int_0^t e^{-\tau} e^{-2(t-\tau)} d\tau = e^{-2t} \int_0^t e^{\tau} d\tau = e^{-t} - e^{-2t} \qquad t \ge 0 -$$ - -Moreover, *y*(*t*) = 0 for *t* < 0 so that - -$$ -y(t) = (e^{-t} - e^{-2t})u(t) -$$ - -## **EXAMPLE 2.11 Graphical Convolution: Causal Function and Two-Sided Function** - -Find *c*(*t*) = *x*(*t*) ∗ *g*(*t*) for the signals depicted in Figs. 2.9a and 2.9b. - -Since *x*(*t*) is simpler than *g*(*t*), it is easier to evaluate *g*(*t*) ∗ *x*(*t*) than *x*(*t*) ∗ *g*(*t*). However, we shall intentionally take the more difficult route and evaluate *x*(*t*) ∗ *g*(*t*). - -From *x*(*t*) and *g*(*t*) (Figs. 2.9a and 2.9b, respectively), observe that *g*(*t*) is composed of two segments. As a result, it can be described as - -$$ -g(t) = \begin{cases} 2e^{-t} & \text{segment A} \\ -2e^{2t} & \text{segment B} \end{cases} -$$ - -Therefore, - -$$ -g(t - \tau) = \begin{cases} 2e^{-(t-\tau)} & \text{segment A} \\ -2e^{2(t-\tau)} & \text{segment B} \end{cases} -$$ - -The segment of *x*(*t*) that is used in convolution is *x*(*t*) = 1 so that *x*(τ ) = 1. Figure 2.9c shows *x*(τ ) and *g*(−τ ). - -To compute *c*(*t*) for *t* ≥ 0, we right-shift *g*(−τ ) to obtain *g*(*t*−τ ), as illustrated in Fig. 2.9d. Clearly, *g*(*t* − τ ) overlaps with *x*(τ ) over the shaded interval, that is, over the range τ ≥ 0; segment A overlaps with *x*(τ ) over the interval (0,*t*), while segment B overlaps with *x*(τ ) over (*t*,∞). Remembering that *x*(τ ) = 1, we have - -$$ -c(t) = \int_0^\infty x(\tau)g(t-\tau) d\tau -$$ - -= -$$ -\int_0^t 2e^{-(t-\tau)} d\tau + \int_t^\infty -2e^{2(t-\tau)} d\tau -$$ - -= -$$ -2(1 - e^{-t}) - 1 = 1 - 2e^{-t} \qquad t \ge 0 -$$ - -Figure 2.9e shows the situation for *t* < 0. Here the overlap is over the shaded interval, that is, over the range τ ≥ 0, where only the segment B of *g*(*t*) is involved. Therefore, - -$$ -c(t) = \int_0^\infty x(\tau)g(t-\tau) d\tau = \int_0^\infty -2e^{2(t-\tau)} d\tau = -e^{2t} \qquad t \le 0 -$$ - -Therefore, - -$$ -c(t) = \begin{cases} 1 - 2e^{-t} & t \ge 0\\ -e^{2t} & t \le 0 \end{cases} -$$ - -Figure 2.9f shows a plot of *c*(*t*). - -## **EXAMPLE 2.12 Graphical Convolution of Two Finite-Duration Functions** - -Find *x*(*t*) ∗ *g*(*t*) for the functions *x*(*t*) and *g*(*t*) shown in Figs. 2.10a and 2.10b. - -Here, *x*(*t*) has a simpler mathematical description than that of *g*(*t*), so it is preferable to time-reverse *x*(*t*). Hence, we shall determine *g*(*t*) ∗ *x*(*t*) rather than *x*(*t*) ∗ *g*(*t*). Thus, - -$$ -c(t) = g(t) * x(t) = \int_{-\infty}^{\infty} g(\tau) x(t - \tau) d\tau -$$ - -First, we determine the expressions for the segments of *x*(*t*) and *g*(*t*) used in finding *c*(*t*). According to Figs. 2.10a and 2.10b, these segments can be expressed as - -$$ -x(t) = 1 -$$ - and $g(t) = \frac{1}{3}t$ - -so that - -$$ -x(t - \tau) = 1 \qquad \text{and} \qquad g(\tau) = \frac{1}{3}\tau -$$ - -Figure 2.10c shows *g*(τ ) and *x*(−τ ), whereas Fig. 2.10d shows *g*(τ ) and *x*(*t* − τ ), which is *x*(−τ ) shifted by *t*. Because the edges of *x*(−τ ) are at τ = −1 and 1, the edges of *x*(*t* − τ ) are at −1 + *t* and 1 + *t*. The two functions overlap over the interval (0, 1 + *t*) (shaded interval) so that - -$$ -c(t) = \int_0^{1+t} g(\tau) x(t-\tau) d\tau = \int_0^{1+t} \frac{1}{3} \tau d\tau = \frac{1}{6} (t+1)^2 \qquad -1 \le t \le 1 \qquad (2.32) -$$ - -This situation, depicted in Fig. 2.10d, is valid only for −1 ≤ *t* ≤ 1. For *t* ≥ 1 but ≤ 2, the situation is as illustrated in Fig. 2.10e. The two functions overlap only over the range −1 + *t* to 1 + *t* (shaded interval). Note that the expressions for *g*(τ ) and *x*(*t* − τ ) do not change; only the range of integration changes. Therefore, - -$$ -c(t) = \int_{-1+t}^{1+t} \frac{1}{3}\tau \,d\tau = \frac{2}{3}t \qquad \qquad 1 \le t \le 2 \tag{2.33} -$$ - -Also note that the expressions in Eqs. (2.32) and (2.33) both apply at *t* = 1, the transition point between their respective ranges. We can readily verify that both expressions yield a value of 2/3 at *t* = 1 so that *c*(1) = 2/3. The continuity of *c*(*t*) at transition points indicates a high probability of a correct answer. Continuity of *c*(*t*) at transition points is assured as long as *x*(*t*) and *g*(*t*) contain no impulse functions. - -For *t* ≥ 2 but ≤ 4, the situation is as shown in Fig. 2.10f. The functions *g*(τ ) and *x*(*t* − τ ) overlap over the interval from −1+*t* to 3 (shaded interval) so that - -$$ -c(t) = \int_{-1+t}^{3} \frac{1}{3}\tau \,d\tau = -\frac{1}{6}(t^2 - 2t - 8) \qquad 2 \le t \le 4 \qquad (2.34) -$$ - -Both Eqs. (2.33) and (2.34) apply at the transition point *t* = 2. We can readily verify that *c*(2) = 4/3 when either of these expressions is used. - -For *t* ≥ 4, *x*(*t* − τ ) has been shifted so far to the right that it no longer overlaps with *g*(τ ) as depicted in Fig. 2.10g. Consequently, - -$$ -c(t) = 0 \qquad \qquad t \ge 4 -$$ - -We now turn our attention to negative values of *t*. We have already determined *c*(*t*) up to *t* = −1. For *t* < −1, there is no overlap between the two functions, as illustrated in Fig. 2.10h, so that - -$$ -c(t) = 0 \qquad \qquad t \le -1 -$$ - -Combining our results, we see that - -$$ -c(t) = \begin{cases} \frac{1}{6}(t+1)^2 & -1 \le t < 1 \\ \frac{2}{3}t & 1 \le t < 2 \\ -\frac{1}{6}(t^2 - 2t - 8) & 2 \le t < 4 \\ 0 & \text{otherwise} \end{cases} -$$ - -Figure 2.10i plots *c*(*t*) according to this expression. - -### THE WIDTH OF CONVOLVED FUNCTIONS - -The widths (durations) of *x*(*t*), *g*(*t*), and *c*(*t*) in Ex. 2.12 (Fig. 2.10) are 2, 3, and 5, respectively. Note that the width of *c*(*t*) in this case is the sum of the widths of *x*(*t*) and *g*(*t*). This observation is not a coincidence. Using the concept of graphical convolution, we can readily see that if *x*(*t*) and *g*(*t*) have the finite widths of *T*1 and *T*2 respectively, then the width of *c*(*t*) is equal to *T*1 +*T*2. The reason is that the time it takes for a signal of width (duration) *T*1 to completely pass another signal of width (duration) *T*2 so that they become non-overlapping is *T*1+*T*2. When the two signals become non-overlapping, the convolution goes to zero. - -### **DR ILL 2.10 Interchanging Convolution Order** - -Rework Ex. 2.11 by evaluating *g*(*t*) ∗ *x*(*t*). - -### **DR ILL 2.11 Showing Commutability Using Two Causal Signals** - -Use graphical convolution to show that *x*(*t*) ∗ *g*(*t*) = *g*(*t*) ∗ *x*(*t*) = *c*(*t*) in Fig. 2.11. - -## **DR ILL 2.12 Showing Commutability Using a Causal Signal and an Anticausal Signal** - -## **DR ILL 2.13 Showing Commutability Using Shifted Signals** - -Repeat Drill 2.11 for the functions in Fig. 2.13. - -### THE PHANTOM OF THE SIGNALS AND SYSTEMS OPERA - -In the study of signals and systems we often come across some signals such as an impulse, which cannot be generated in practice and have never been sighted by anyone.† One wonders why we even consider such idealized signals. The answer should be clear from our discussion so far in this chapter. Even if the impulse function has no physical existence, we can compute the system response *h*(*t*) to this phantom input according to the procedure in Sec. 2.3, and knowing *h*(*t*), we can compute the system response to any arbitrary input. The concept of impulse response, therefore, provides an effective intermediary for computing system response to an arbitrary input. In addition, the impulse response *h*(*t*) itself provides a great deal of information and insight about the system behavior. In Sec. 2.6 we show that the knowledge of impulse response provides much valuable information, such as the response time, pulse dispersion, and filtering properties of the system. Many other useful insights about the system behavior can be obtained by inspection of *h*(*t*). - -Similarly, in frequency-domain analysis (discussed in later chapters), we use an *everlasting exponential* (or *sinusoid)* to determine system response. An everlasting exponential (or sinusoid), too, is a phantom, which nobody has ever seen and which has no physical existence. But it provides another effective intermediary for computing the system response to an arbitrary input. Moreover, the system response to everlasting exponential (or sinusoid) provides valuable information and insight regarding the system's behavior. Clearly, idealized impulses and everlasting sinusoids are friendly and helpful spirits. - -Interestingly, the unit impulse and the everlasting exponential (or sinusoid) are the dual of each other in the time-frequency duality, to be studied in Ch. 7. Actually, the time-domain and the frequency-domain methods of analysis are the dual of each other. - -### WHY CONVOLUTION? AN INTUITIVE EXPLANATION OF SYSTEM RESPONSE - -On the surface, it appears rather strange that the response of linear systems (those gentlest of the gentle systems) should be given by such a tortuous operation of convolution, where one signal is fixed and the other is inverted and shifted. To understand this odd behavior, consider a hypothetical impulse response *h*(*t*) that decays linearly with time (Fig. 2.14a). This response is strongest at *t* =0, the moment the impulse is applied, and it decays linearly at future instants so that one second later (at *t* = 1 and beyond), it ceases to exist. This means that the closer the impulse input is to an instant *t*, the stronger is its response at *t*. - -Now consider the input *x*(*t*) shown in Fig. 2.14b. To compute the system response, we break the input into rectangular pulses and approximate these pulses with impulses. Generally, the response of a causal system at some instant *t* will be determined by all the impulse components of the input before *t*. Each of these impulse components will have different weight in determining the response at the instant *t*, depending on its proximity to *t*. As seen earlier, the closer the impulse is to *t*, the stronger is its influence at *t*. The impulse at *t* has the greatest weight (unity) in determining - -† The late Prof. S. J. Mason, the inventor of signal flow graph techniques, used to tell a story of a student frustrated with the impulse function. The student said, "The unit impulse is a thing that is so small you can't see it, except at one place (the origin), where it is so big you can't see it. In other words, you can't see it at all; at least I can't!" [2]. - -**Figure 2.14** Intuitive explanation of convolution. - -the response at *t*. The weight decreases linearly for all impulses before *t* until the instant *t* − 1. The input before *t* − 1 has no influence (zero weight). Thus, to determine the system response at *t*, we must assign a linearly decreasing weight to impulses occurring before *t*, as shown in Fig. 2.14b. This weighting function is precisely the function *h*(*t* − τ ). The system response at *t* is then determined not by the input *x*(τ ) but by the weighted input *x*(τ )*h*(*t* − τ ), and the summation of all these weighted inputs is the convolution integral. - -### **[2.4-3 Interconnected Systems](#page-8-0)** - -A larger, more complex system can often be viewed as the interconnection of several smaller subsystems, each of which is easier to characterize. Knowing the characterizations of these subsystems, it becomes simpler to analyze such large systems. We shall consider here two basic interconnections, cascade and parallel. Figure 2.15a shows *S*1 and *S*2, two LTIC subsystems connected in parallel, and Fig. 2.15b shows the same two systems connected in cascade. - -In Fig. 2.15a, the device depicted by the symbol inside a circle represents an adder, which adds signals at its inputs. Also the junction from which two (or more) branches radiate out is called the *pickoff node*. Every branch that radiates out from the pickoff node carries the same signal (the signal at the junction). In Fig. 2.15a, for instance, the junction at which the input is applied is a pickoff node from which two branches radiate out, each of which carries the input signal at the node. - -Let the impulse response of *S*1 and *S*2 be *h*1(*t*) and *h*2(*t*), respectively. Further assume that interconnecting these systems, as shown in Fig. 2.15, does not load them. This means that the impulse response of either of these systems remains unchanged whether observed when these systems are unconnected or when they are interconnected. - -To find *hp*(*t*), the impulse response of the parallel system *Sp* in Fig. 2.15a, we apply an impulse at the input of *Sp*. This results in the signal δ(*t*) at the inputs of *S*1 and *S*2, leading to their outputs *h*1(*t*) and *h*2(*t*), respectively. These signals are added by the adder to yield *h*1(*t*) + *h*2(*t*) as the output of *Sp*: - -$$ -h_p(t) = h_1(t) + h_2(t) -$$ - -To find *hc*(*t*), the impulse response of the cascade system *Sc* in Fig. 2.15b, we apply the input δ(*t*) at the input of *Sc*, which is also the input to *S*1. Hence, the output of *S*1 is *h*1(*t*), which now acts - -**Figure 2.15** Interconnected systems. - -as the input to *S*2. The response of *S*2 to input *h*1(*t*) is *h*1(*t*) ∗ *h*2(*t*). Therefore, - -$$ -h_c(t) = h_1(t) * h_2(t) -$$ - -Because of the commutative property of convolution, it follows that interchanging the systems *S*1 and *S*2, as shown in Fig. 2.15c, results in the same impulse response *h*1(*t*) ∗ *h*2(*t*). This means that when several LTIC systems are cascaded, the order of systems does not affect the impulse response of the composite system. In other words, linear operations, performed in cascade, commute. The order in which they are performed is not important, at least theoretically.† - -†Change of order, however, could affect performance because of physical limitations and sensitivities to changes in the subsystems involved. - -#### 192 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -We shall give here another interesting application of the commutative property of LTIC systems. Figure 2.15d shows a cascade of two LTIC systems: a system *S* with impulse response *h*(*t*), followed by an ideal integrator. Figure 2.15e shows a cascade of the same two systems in reverse order; an ideal integrator followed by *S*. In Fig. 2.15d, if the input *x*(*t*) to *S* yields the output *y*(*t*), then the output of the system of Fig. 2.15d is the integral of *y*(*t*). In Fig. 2.15e, the output of the integrator is the integral of *x*(*t*). The output in Fig. 2.15e is identical to the output in Fig. 2.15d. Hence, it follows that if an LTIC system response to input *x*(*t*) is *y*(*t*), then the response of the same system to the integral of *x*(*t*) is the integral of *y*(*t*). In other words, - -if -$$ -x(t) \Longrightarrow y(t) -$$ - then $\int_{-\infty}^{t} x(\tau) d\tau \Longrightarrow \int_{-\infty}^{t} y(\tau) d\tau$ - -Replacing the ideal integrator with an ideal differentiator in Figs. 2.15d and 2.15e, and following a similar argument, we conclude that - -if -$$ -x(t) \Longrightarrow y(t) -$$ - then $\frac{dx(t)}{dt} \Longrightarrow \frac{dy(t)}{dt}$ - -If we let *x*(*t*) = δ(*t*) and *y*(*t*) = *h*(*t*) in Fig. 2.15e, we find that *g*(*t*), the unit step response of an LTIC system with impulse *h*(*t*), is given by - -$$ -g(t) = \int_{-\infty}^{t} h(\tau) d\tau -$$ -\n(2.35) - -We can also show that the system response to δ(˙ *t*) is *dh*(*t*)/*dt*. These results can be extended to other singularity functions. For example, the unit ramp response of an LTIC system is the integral of its unit step response, and so on. - -### INVERSE SYSTEMS - -In Fig. 2.15b, if *S*1 and *S*2 are inverse systems with impulse response *h*(*t*) and *hi*(*t*), respectively, then the impulse response of the cascade of these systems is *h*(*t*) ∗ *hi*(*t*). But, the cascade of a system with its inverse is an identity system, whose output is the same as the input. In other words, the unit impulse response of the cascade of inverse systems is also an unit impulse δ(*t*). Hence, - -$$ -h(t) * h_i(t) = \delta(t) \tag{2.36} -$$ - -We shall give an interesting application of the commutative property. As seen from Eq. (2.36), a cascade of inverse systems is an identity system. Moreover, in a cascade of several LTIC subsystems, changing the order of the subsystems in any manner does not affect the impulse response of the cascade system. Using these facts, we observe that the two systems, shown in Fig. 2.15f, are equivalent. We can compute the response of the cascade system on the right-hand side, by computing the response of the system inside the dotted box to the input *x*˙(*t*). The impulse response of the dotted box is *g*(*t*), the integral of *h*(*t*), as given in Eq. (2.35). Hence, it follows that - -$$ -y(t) = x(t) * h(t) = \dot{x}(t) * g(t) -$$ -\n(2.37) - -Recall that *g*(*t*) is the unit step response of the system. Hence, an LTIC response can also be obtained as a convolution of *x*˙(*t*) (the derivative of the input) with the unit step response of the system. This result can be readily extended to higher derivatives of the input. An LTIC system response is the convolution of the *n*th derivative of the input with the *n*th integral of the impulse response. - -### **[2.4-4 A Very Special Function for LTIC Systems:](#page-8-0) The Everlasting Exponential** *est* - -There is a very special connection of LTIC systems with the everlasting exponential function *est*, where *s* is a complex variable, in general. We now show that the LTIC system's (zero-state) response to everlasting exponential input *est* is also the same everlasting exponential (within a multiplicative constant). Moreover, no other function can make the same claim. Such an input for which the system response is also of the same form is called the *characteristic function* (also *eigenfunction*) of the system. Because a sinusoid is a form of exponential (*s* = ±*j*ω), everlasting sinusoid is also a characteristic function of an LTIC system. Note that we are talking here of an everlasting exponential (or sinusoid), which starts at *t* = −∞. - -If *h*(*t*) is the system's unit impulse response, then system response *y*(*t*) to an everlasting exponential *est* is given by - -$$ -y(t) = h(t) * e^{st} = \int_{-\infty}^{\infty} h(\tau) e^{s(t-\tau)} d\tau = e^{st} \int_{-\infty}^{\infty} h(\tau) e^{-s\tau} d\tau -$$ - -The integral on the right-most side is a function of a complex variable *s* and a constant with respect to *t*. Let us denote this term by *H*(*s*), which is also complex, in general. Thus, - -$$ -y(t) = H(s)e^{st} -$$ -\n(2.38) - -where - -$$ -H(s) = \int_{-\infty}^{\infty} h(\tau) e^{-s\tau} d\tau -$$ - (2.39) - -Equation (2.38) is valid only for the values of *s* for which *H*(*s*) exists, that is, if \$ ∞ −∞ *h*(τ )*e*−*s*τ *d*τ exists (or converges). The region in the *s* plane for which this integral converges is called the *region of convergence* for *H*(*s*). Further elaboration of the region of convergence is presented in Ch. 4. - -For a given *s*, note that *H*(*s*) is a constant. Thus, the input and the output are the same (within a multiplicative constant) for the everlasting exponential signal. - -*H*(*s*), which is called the *transfer function* of the system, is a function of complex variable *s*. An alternate definition of the transfer function *H*(*s*) of an LTIC system, as seen from Eq. (2.38), is - -$$ -H(s) = \frac{\text{output signal}}{\text{input signal}} \bigg|_{\text{input} = \text{everlasting exponential } e^{st}} \tag{2.40} -$$ - -The transfer function is defined for, and is meaningful to, LTIC systems only. It does not exist for nonlinear or time-varying systems, in general. - -We repeat again that this discussion is about the everlasting exponential, which starts at *t* = −∞, not the causal exponential *estu*(*t*), which starts at *t* = 0. - -#### 194 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -For a system specified by Eq. (2.2), the transfer function is given by - -$$ -H(s) = \frac{P(s)}{Q(s)}\tag{2.41} -$$ - -This follows readily by considering an everlasting input *x*(*t*) = *est*. According to Eq. (2.38), the output is *y*(*t*) = *H*(*s*)*est*. Substitution of this *x*(*t*) and *y*(*t*) in Eq. (2.2) yields - -$$ -H(s)[Q(D)e^{st}] = P(D)e^{st} -$$ - -Moreover, - -$$ -D^r e^{st} = \frac{d^r e^{st}}{dt^r} = s^r e^{st} -$$ - -Hence, - -$$ -P(D)e^{st} = P(s)e^{st} \qquad \text{and} \qquad Q(D)e^{st} = Q(s)e^{st} -$$ - -Consequently, - -$$ -H(s) = \frac{P(s)}{Q(s)} -$$ - -### **DR ILL 2.14 Ideal Integrator and Differentiator Transfer Functions** - -Show that the transfer function of an ideal integrator is *H*(*s*) = 1/*s* and that of an ideal differentiator is *H*(*s*) = *s*. Find the answer in two ways: using Eq. (2.39) and using Eq. (2.41). [*Hint:* Find *h*(*t*) for the ideal integrator and differentiator. You also may need to use the result in Prob. 1.4-12.] - -### A FUNDAMENTAL PROPERTY OF LTI SYSTEMS - -We can show that Eq. (2.38) is a fundamental property of LTI systems and it follows directly as a consequence of linearity and time invariance. To show this let us assume that the response of an LTI system to an everlasting exponential *est* is *y*(*s*,*t*). If we define - -$$ -H(s,t) = \frac{y(s,t)}{e^{st}} -$$ - -then - -$$ -y(s,t) = H(s,t) e^{st} -$$ - -Because of the time-invariance property, the system response to input *es*(*t*−*T*) is *H*(*s*,*t* − *T*) *es*(*t*−*T*) , that is, - -$$ -y(s, t - T) = H(s, t - T) e^{s(t - T)} -$$ -\n(2.42) - -The delayed input *es*(*t*−*T*) represents the input *est* multiplied by a constant *e*−*sT* . Hence, according to the linearity property, the system response to *es*(*t*−*T*) must be *y*(*s*,*t*) *e*−*sT* . Hence, - -$$ -y(s,t-T) = y(s,t) e^{-sT} = H(s,t) e^{s(t-T)} -$$ - -Comparison of this result with Eq. (2.42) shows that - -$$ -H(s,t) = H(s,t-T) \qquad \text{for all } T -$$ - -This means *H*(*s*,*t*) is independent of *t*, and we can express *H*(*s*,*t*) = *H*(*s*). Hence, - -$$ -y(s,t) = H(s) e^{st} -$$ - -### **[2.4-5 Total Response](#page-8-0)** - -Assuming distinct roots, the total response of a linear system can be expressed as the sum of its zero-input response (ZIR) and its zero-state response (ZSR): - -total response = -$$ -\underbrace{\sum_{k=1}^{N} c_k e^{\lambda_k t}}_{\text{ZIR}} + \underbrace{x(t) * h(t)}_{\text{ZSR}} -$$ - -For repeated roots, the zero-input component should be appropriately modified. - -For the series *RLC* circuit in Ex. 2.4 with the input *x*(*t*) = 10*e*−3*t u*(*t*) and the initial conditions *y*(0−) = 0, *vC*(0−) = 5, we determined the zero-input response in Ex. 2.1a [Eq. (2.9)]. We found the zero-state response in Ex. 2.9. From the results in Exs. 2.1a and 2.9, we obtain - -total current = -$$ -\underbrace{(-5e^{-t} + 5e^{-2t})}_{\text{zero-input current}} + \underbrace{(-5e^{-t} + 20e^{-2t} - 15e^{-3t})}_{\text{zero-state current}} \qquad t \ge 0 -$$ - (2.43) - -Figure 2.16a shows the zero-input, zero-state, and total responses. - -**Figure 2.16** Total response and its components. - -### NATURAL AND FORCED RESPONSE - -For the *RLC* circuit in Ex. 2.4, the characteristic modes were found to be *e*−*t* and *e*−2*t* . As we expected, the zero-input response is composed exclusively of characteristic modes. Note, however, that even the zero-state response [Eq. (2.43)] contains characteristic mode terms. This observation is generally true of LTIC systems. We can now lump together all the characteristic mode terms in the total response, giving us a component known as the *natural response yn*(*t*). The remainder, consisting entirely of noncharacteristic mode terms, is known as the *forced response y*φ(*t*). The total response of the *RLC* circuit in Ex. 2.4 can be expressed in terms of natural and forced components by regrouping the terms in Eq. (2.43) as - -total current = -$$ -\underbrace{(-10e^{-t} + 25e^{-2t})}_{\text{natural response } y_n(t)} + \underbrace{(-15e^{-3t})}_{\text{forced response } y_\phi(t)} \qquad t \ge 0 -$$ - (2.44) - -Figure 2.16b shows the natural, forced, and total responses. - -The classical solution to a differential equation includes the natural (also called the *homogeneous* or *complementary*) solution and the forced (also known as the *particular*) solution; traditional courses on differential equations provide simplified procedures to determine these components. Unfortunately, the classical solution lacks the engineering intuition and utility afforded by the zero-input and zero-state responses. The classical approach cannot separate the responses arising from internal conditions and external input. While the natural and forced solutions can be obtained from the zero-input and zero-state responses, the converse is not true. Further, the classical method is unable to express the system response to an input *x*(*t*) as an explicit function of *x*(*t*). In fact, the classical method is restricted to a certain class of inputs and cannot handle arbitrary inputs, as can the method to determine the zero-state response. For these (and other) reasons, we do not further detail the classical solution of differential equations. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/031_2.5 SYSTEM STABILITY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/031_2.5 SYSTEM STABILITY.md deleted file mode 100644 index ced8e268c3637d57ec68cb5d1c36ee3b02c86a7c..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/031_2.5 SYSTEM STABILITY.md +++ /dev/null @@ -1,166 +0,0 @@ -## **[2.5 SYSTEM](#page-8-0) STABILITY** - -Stability is an important system property. Two types of system stability are generally considered: external (BIBO) stability and internal (asymptotic) stability. Let us consider both stability types in turn. - -### **[2.5-1 External \(BIBO\) Stability](#page-8-0)** - -To understand the intuitive basis for the BIBO (bounded-input/bounded-output) stability of a system introduced in Sec. 1.7, let us examine the stability concept as applied to a right circular cone. Such a cone can be made to stand forever on its circular base, on its apex, or on its side. For this reason, these three states of the cone are said to be *equilibrium states*. Qualitatively, however, the three states show very different behavior. If the cone, standing on its circular base, were to be disturbed slightly and then left to itself, it would eventually return to its original equilibrium position. In such a case, the cone is said to be in *stable equilibrium*. In contrast, if the cone stands on its apex, then the slightest disturbance will cause the cone to move farther and farther away from its equilibrium state. The cone in this case is said to be in an *unstable equilibrium*. The cone lying on its side, if disturbed, will neither go back to the original state nor continue to move farther away from the original state. Thus it is said to be in a *neutral equilibrium*. Clearly, when a system is in stable equilibrium, application of a small disturbance (input) produces a small response. In contrast, when the system is in unstable equilibrium, even a minuscule disturbance (input) produces an unbounded response. The BIBO-stability definition can be understood in the light of this concept. If every bounded input produces bounded output, the system is (BIBO) stable.† In contrast, if even one bounded input results in unbounded response, the system is (BIBO) unstable. - -For an LTIC system, - -$$ -y(t) = h(t) * x(t) = \int_{-\infty}^{\infty} h(\tau) x(t - \tau) d\tau -$$ - -Therefore, - -$$ -|y(t)| \le \int_{-\infty}^{\infty} |h(\tau)| |x(t - \tau)| d\tau -$$ - -Moreover, if *x*(*t*) is bounded, then |*x*(*t* −τ )| < *K*1 < ∞, and - -$$ -|y(t)| \leq K_1 \int_{-\infty}^{\infty} |h(\tau)| d\tau -$$ - -Hence for BIBO stability, - -$$ -\int_{-\infty}^{\infty} |h(\tau)| d\tau < \infty \tag{2.45} -$$ - -This is a sufficient condition for BIBO stability. We can show that this is also a necessary condition (see Prob. 2.5-7). Therefore, for an LTIC system, if its impulse response *h*(*t*) is absolutely integrable, the system is (BIBO) stable. Otherwise it is (BIBO) unstable. In addition, we shall show in Ch. 4 that a necessary (but not sufficient) condition for an LTIC system described by Eq. (2.1) to be BIBO-stable is *M* ≤ *N*. If *M* > *N*, the system is unstable. This is one of the reasons to avoid systems with *M* > *N*. - -Because the BIBO stability of a system can be ascertained by measurements at the external terminals (input and output), this is an external stability criterion. It is no coincidence that the BIBO criterion in Eq. (2.45) is in terms of the impulse response, which is an external description of the system. - -As observed in Sec. 1.9, the internal behavior of a system is not always ascertainable from the external terminals. Therefore, external (BIBO) stability may not be a correct indication of internal stability. Indeed, some systems that appear stable by the BIBO criterion may be internally unstable. This is like a room on fire inside a house: no trace of fire is visible from outside, but the entire house will be burned to ashes. - -The BIBO stability is meaningful only for systems in which the internal and the external description are equivalent (controllable and observable systems). Fortunately, most practical systems fall into this category, and whenever we apply this criterion, we implicitly assume that the system, in fact, belongs to this category. Internal stability is all-inclusive, and external stability can always be determined from internal stability. For this reason, we now investigate the internal stability criterion. - -† The system is assumed to be in zero state. - -### **[2.5-2 Internal \(Asymptotic\) Stability](#page-8-0)** - -Because of the great variety of possible system behaviors, there are several definitions of internal stability in the literature. Here we shall consider a definition that is suitable for causal, linear, time-invariant (LTI) systems. - -If, in the absence of an external input, a system remains in a particular state (or condition) indefinitely, then that state is said to be an *equilibrium state* of the system. For an LTI system, zero state, in which all initial conditions are zero, is an equilibrium state. Now suppose an LTI system is in zero state and we change this state by creating small nonzero initial conditions (small disturbance). These initial conditions will generate signals consisting of characteristic modes in the system. By analogy with the cone, if the system is stable, it should eventually return to zero state. In other words, when left to itself, every mode in a stable system arising as a result of nonzero initial conditions should approach 0 as *t*→∞. However, if even one of the modes grows with time, the system will never return to zero state, and the system would be identified as unstable. In the borderline case, some modes neither decay to zero nor grow indefinitely, while all the remaining modes decay to zero. This case is like the neutral equilibrium in the cone. Such a system is said to be *marginally* stable. Internal stability is also called *asymptotic* stability or stability in the sense of *Lyapunov*. - -For a system characterized by Eq. (2.1), we can restate the internal stability criterion in terms of the location of the *N* characteristic roots λ1, λ2, ..., λ*N* of the system in a complex plane. The characteristic modes are of the form *e*λ*kt* or *t r e*λ*kt* . The locations of various roots in the complex plane and the corresponding modes are shown in Fig. 2.17. These modes → 0 as *t* → ∞ if Re λ*k* < 0. In contrast, the modes → ∞ as *t* → ∞ if Reλ*k* > 0.† - -From Fig. 2.17, we see that a system is (asymptotically) stable if all its characteristic roots lie in the LHP, that is, if Reλ*k* < 0 for all *k*. If even a single characteristic root lies in the RHP, the system is (asymptotically) unstable. Modes due to roots on the imaginary axis (λ = ±*j*ω0) are of the form *e*±*j*ω0*t* . Hence, if some roots are on the imaginary axis, and all the remaining roots are in the LHP, the system is marginally stable (assuming that the roots on the imaginary axis are not repeated). If the imaginary axis roots are repeated, the characteristic modes are of the form *t r e*±*j*ω*kt* , which *do* grow with time indefinitely. Hence, the system is unstable. Figure 2.18 shows stability regions in the complex plane. - -To summarize: - -- 1. An LTIC system is asymptotically stable if, and only if, all the characteristic roots are in the LHP. The roots may be simple (unrepeated) or repeated. -- 2. An LTIC system is unstable if, and only if, one or both of the following conditions exist: (i) at least one root is in the RHP; (ii) there are repeated roots on the imaginary axis. -- 3. An LTIC system is marginally stable if, and only if, there are no roots in the RHP, and there are some unrepeated roots on the imaginary axis. - -$$ -\lim_{t \to \infty} e^{\lambda t} = \lim_{t \to \infty} e^{(\alpha + j\beta)t} = \lim_{t \to \infty} e^{\alpha t} e^{j\beta t} = \begin{cases} 0 & \alpha < 0 \\ \infty & \alpha > 0 \end{cases} -$$ - -This conclusion is also valid for the terms of the form *t r e*λ*t* . - -† This may be seen from the fact that if α and β are the real and the imaginary parts of a root λ, then - -**Figure 2.17** Location of characteristic roots and the corresponding characteristic modes. - -### **[2.5-3 Relationship Between BIBO and Asymptotic Stability](#page-8-0)** - -External stability is determined by applying an external input with zero initial conditions, while internal stability is determined by applying the nonzero initial conditions and no external input. This is why these stabilities are also called the *zero-state stability* and the *zero-input stability*, respectively. - -Recall that *h*(*t*), the impulse response of an LTIC system, is a linear combination of the system characteristic modes. For an LTIC system, specified by Eq. (2.1), we can readily show that when a characteristic root λ*k* is in the LHP, the corresponding mode *e*λ*kt* is absolutely integrable. In - -contrast, if λ*k* is in the RHP or on the imaginary axis, *e*λ*kt* is not absolutely integrable.† This means that an asymptotically stable system is BIBO-stable. Moreover, a marginally stable or asymptotically unstable system is BIBO-unstable. The converse is not necessarily true; that is, BIBO stability does not necessarily inform us about the internal stability of the system. For instance, if a system is uncontrollable and/or unobservable, some modes of the system are invisible and/or uncontrollable from the external terminals [3]. Hence, the stability picture portrayed by the external description is of questionable value. BIBO (external) stability cannot assure internal (asymptotic) stability, as the following example shows. - -### **EXAMPLE 2.13 A BIBO-Stable but Asymptotically Unstable System** - -An LTID system consists of two subsystems *S*1 and *S*2 in cascade (Fig. 2.19). The impulse response of these systems are *h*1(*t*) and *h*2(*t*), respectively, given by - -$$ -h_1(t) = \delta(t) - 2e^{-t}u(t) -$$ - and $h_2(t) = e^t u(t)$ - -Comment on the BIBO and asymptotic stability of the composite system. - -$$ -\int_{-\infty}^{\infty} |e^{\lambda \tau} u(\tau)| d\tau = \int_{0}^{\infty} e^{\alpha \tau} d\tau = \begin{cases} -1/\alpha & \alpha < 0\\ \infty & \alpha \ge 0 \end{cases} -$$ - -This conclusion is also valid when the integrand is of the form |*t ke*λ*t u*(*t*)|. - -† Consider a mode of the form *e*λ*t* , where λ = α +*j*β. Hence, *e*λ*t* = *e*α*t ej*β*t* and |*e*λ*t* | = *e*α*t* . Therefore, - -The composite system impulse response *h*(*t*) is given by - -$$ -h(t) = h_1(t) * h_2(t) = h_2(t) * h_1(t) = e^t u(t) * [\delta(t) - 2e^{-t} u(t)] -$$ - -= $e^t u(t) - 2 \left[ \frac{e^t - e^{-t}}{2} \right] u(t)$ -= $e^{-t} u(t)$ - -If the composite cascade system were to be enclosed in a black box with only the input and the output terminals accessible, any measurement from these external terminals would show that the impulse response of the system is *e*−*t u*(*t*), without any hint of the dangerously unstable system the system is harboring within. - -The composite system is BIBO-stable because its impulse response, *e*−*t u*(*t*), is absolutely integrable. Observe, however, the subsystem *S*2 has a characteristic root 1, which lies in the RHP. Hence, *S*2 is asymptotically unstable. Eventually, *S*2 will burn out (or saturate) because of the unbounded characteristic response generated by intended or unintended initial conditions, no matter how small. We shall show in Ex. 10.12 that this composite system is observable, but not controllable. If the positions of *S*1 and *S*2 were interchanged (*S*2 followed by *S*1), the system is still BIBO-stable, but asymptotically unstable. In this case, the analysis in Ex. 10.12 shows that the composite system is controllable, but not observable. - -This example shows that BIBO stability does not always imply asymptotic stability. However, asymptotic stability always implies BIBO stability. - -Fortunately, uncontrollable and/or unobservable systems are not commonly observed in practice. Henceforth, in determining system stability, we shall assume that unless otherwise mentioned, the internal and the external descriptions of a system are equivalent, implying that the system is controllable and observable. - -### **EXAMPLE 2.14 Investigating Asymptotic and BIBO Stability** - -Investigate the asymptotic and the BIBO stability of LTIC system described by the following equations, assuming that the equations are internal system descriptions: - -- **(a)** (*D*+1)(*D*2 +4*D*+8)*y*(*t*) = (*D*−3)*x*(*t*) -- **(b)** (*D*−1)(*D*2 +4*D*+8)*y*(*t*) = (*D*+2)*x*(*t*) -- **(c)** (*D*+2)(*D*2 +4)*y*(*t*) = (*D*2 +*D*+1)*x*(*t*) -- **(d)** (*D*+1)(*D*2 +4)2*y*(*t*) = (*D*2 +2*D*+8)*x*(*t*) - -The characteristic polynomials of these systems are - -- **(a)** (λ+1)(λ2 +4λ+8) = (λ+1)(λ+2−*j*2)(λ+2+*j*2) -- **(b)** (λ−1)(λ2 +4λ+8) = (λ−1)(λ+2−*j*2)(λ+2+*j*2) -- **(c)** (λ+2)(λ2 +4) = (λ+2)(λ−*j*2)(λ+*j*2) -- **(d)** (λ+1)(λ2 +4)2 = (λ+2)(λ−*j*2)2(λ+*j*2)2 - -Consequently, the characteristic roots of the systems are (see Fig. 2.20): - -- **(a)** −1, −2±*j*2 -- **(b)** 1, −2±*j*2 -- **(c)** −2, ±*j*2 -- **(d)** −1, ±*j*2, ±*j*2 - -System (a) is asymptotically stable (all roots in LHP), system (b) is unstable (one root in RHP), system (c) is marginally stable (unrepeated roots on imaginary axis) and no roots in RHP, and system (d) is unstable (repeated roots on the imaginary axis). BIBO stability is readily determined from the asymptotic stability. System (a) is BIBO-stable, system (b) is BIBO-unstable, system (c) is BIBO-unstable, and system (d) is BIBO-unstable. We have assumed that these systems are controllable and observable. - -### **DR ILL 2.15 Assessing Stability by Characteristic Roots** - -For each case, plot the characteristic roots and determine asymptotic and BIBO stabilities. Assume the equations reflect internal descriptions. - -- **(a)** *D*(*D*+2)*y*(*t*) = 3*x*(*t*) -- **(b)** *D*2(*D*+3)*y*(*t*) = (*D*+5)*x*(*t*) -- **(c)** (*D*+1)(*D*+2)*y*(*t*) = (2*D*+3)*x*(*t*) -- **(d)** (*D*2 +1)(*D*2 +9)*y*(*t*) = (*D*2 +2*D*+4)*x*(*t*) -- **(e)** (*D*+1)(*D*2 −4*D*+9)*y*(*t*) = (*D*+7)*x*(*t*) - -### **ANSWERS** - -- **(a)** Marginally stable, but BIBO-unstable -- **(b)** Unstable in both senses -- **(c)** Stable in both senses -- **(d)** Marginally stable, but BIBO-unstable -- **(e)** Unstable in both senses. - -### IMPLICATIONS OF STABILITY - -All practical signal-processing systems must be asymptotically stable. Unstable systems are useless from the viewpoint of signal processing because any set of intended or unintended initial conditions leads to an unbounded response that either destroys the system or (more likely) leads it to some saturation conditions that change the nature of the system. Even if the discernible initial conditions are zero, stray voltages or thermal noise signals generated within the system will act as initial conditions. Because of exponential growth of a mode or modes in unstable systems, a stray signal, no matter how small, will eventually cause an unbounded output. - -Marginally stable systems, though BIBO unstable, do have one important application in the oscillator, which is a system that generates a signal on its own without the application of an external input. Consequently, the oscillator output is a zero-input response. If such a response is to be a sinusoid of frequency ω0, the system should be marginally stable with characteristic roots at ±*j*ω0. Thus, to design an oscillator of frequency ω0, we should pick a system with the characteristic polynomial (λ−*j*ω0)(λ+*j*ω0) = λ2 +ω0 2. A system described by the differential equation - -$$ -(D2 + \omega_02)y(t) = x(t) -$$ - -will do the job. However, practical oscillators are invariably realized using nonlinear systems. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/032_2.6 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/032_2.6 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR.md deleted file mode 100644 index 40df9c2be10077b53f329dfe4ecc4a49a96dfd3e..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/032_2.6 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR.md +++ /dev/null @@ -1,239 +0,0 @@ -## **2.6 INTUITIVE [INSIGHTS INTO](#page-8-0) SYSTEM BEHAVIOR** - -This section attempts to provide an understanding of what determines system behavior. Because of its intuitive nature, the discussion is more or less qualitative. We shall now show that the most important attributes of a system are its characteristic roots or characteristic modes because they determine not only the zero-input response but also the entire behavior of the system. - -### **[2.6-1 Dependence of System Behavior on Characteristic Modes](#page-8-0)** - -Recall that the zero-input response of a system consists of the system's characteristic modes. For a stable system, these characteristic modes decay exponentially and eventually vanish. This behavior may give the impression that these modes do not substantially affect system behavior in general and system response in particular. This impression is totally wrong! We shall now see that the system's characteristic modes leave their imprint on every aspect of the system behavior. *We may compare the system's characteristic modes (or roots) to a seed that eventually dissolves in the* - -### 204 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -*ground; however, the plant that springs from it is totally determined by the seed. The imprint of the seed exists on every cell of the plant.* - -To understand this interesting phenomenon, recall that the characteristic modes of a system are very special to that system because it can sustain these signals without the application of an external input. In other words, the system offers a free ride and ready access to these signals. Now imagine what would happen if we actually drove the system with an input having the form of a characteristic mode! We would expect the system to respond strongly (this is, in fact, the resonance phenomenon discussed later in this section). If the input is not exactly a characteristic mode but is close to such a mode, we would still expect the system response to be strong. However, if the input is very different from any of the characteristic modes, we would expect the system to respond poorly. We shall now show that these intuitive deductions are indeed true. - -Intuition can cut the math jungle instantly! - -We have devised a measure of similarity of signals later (see in Ch. 6). Here we shall take a simpler approach. Let us restrict the system's inputs to exponentials of the form *e*ζ *t* , where ζ is generally a complex number. The similarity of two exponential signals *e*ζ *t* and *e*λ*t* will then be measured by the closeness of ζ and λ. If the difference ζ − λ is small, the signals are similar; if ζ −λ is large, the signals are dissimilar. - -Now consider a first-order system with a single characteristic mode *e*λ*t* and the input *e*ζ *t* . The impulse response of this system is then given by *Ae*λ*t* , where the exact value of *A* is not important for this qualitative discussion. The system response *y*(*t*) is given by - -$$ -y(t) = h(t) * x(t) = Ae^{\lambda t}u(t) * e^{\zeta t}u(t) -$$ - -From the convolution table (Table 2.1), we obtain - -$$ -y(t) = \frac{A}{\zeta - \lambda} [e^{\zeta t} - e^{\lambda t}] u(t) -$$ -\n(2.46) - -Clearly, if the input *e*ζ *t* is similar to *e*λ*t* , ζ −λ is small and the system response is large. *The closer the input x*(*t*) *to the characteristic mode, the stronger the system response.* In contrast, if the input is very different from the natural mode, ζ − λ is large and the system responds poorly. This is precisely what we set out to prove. - -We have proved the foregoing assertion for a single-mode (first-order) system. It can be generalized to an *N*th-order system, which has *N* characteristic modes. The impulse response *h*(*t*) of such a system is a linear combination of its *N* modes. Therefore, if *x*(*t*) is similar to any one of the modes, the corresponding response will be high; if it is similar to none of the modes, the response will be small. Clearly, the characteristic modes are very influential in determining system response to a given input. - -It would be tempting to conclude on the basis of Eq. (2.46) that if the input is identical to the characteristic mode, so that ζ = λ, then the response goes to infinity. Remember, however, that if ζ = λ, the numerator on the right-hand side of Eq. (2.46) also goes to zero. We shall study this interesting behavior (resonance phenomenon) later in this section. - -We now show that *mere inspection of the impulse response h*(*t*) *(which is composed of characteristic modes) reveals a great deal about the system behavior*. - -### **[2.6-2 Response Time of a System: The System Time Constant](#page-8-0)** - -Like human beings, systems have a certain response time. In other words, when an input (stimulus) is applied to a system, a certain amount of time elapses before the system fully responds to that input. This time lag or response time is called the system *time constant*. As we shall see, a system's time constant is equal to the width of its impulse response *h*(*t*). - -An input δ(*t*) to a system is instantaneous (zero duration), but its response *h*(*t*) has a duration *Th*. Therefore, the system requires a time *Th* to respond fully to this input, and we are justified in viewing *Th* as the system's response time or time constant. We arrive at the same conclusion via another argument. The output is a convolution of the input with *h*(*t*). If an input is a pulse of width *Tx*, then the output pulse width is *Tx* + *Th* according to the width property of convolution. This conclusion shows that the system requires *Th* seconds to respond fully to any input. *The system time constant indicates how fast the system is. A system with a smaller time constant is a faster system that responds quickly to an input. A system with a relatively large time constant is a sluggish system that cannot respond well to rapidly varying signals.* - -Strictly speaking, the duration of the impulse response *h*(*t*) is ∞ because the characteristic modes approach zero asymptotically as *t* → ∞. However, beyond some value of *t*, *h*(*t*) becomes negligible. It is therefore necessary to use some suitable measure of the impulse response's effective width. - -There is no single satisfactory definition of effective signal duration (or width) applicable to every situation. For the situation depicted in Fig. 2.21, a reasonable definition of the duration *h*(*t*) would be *Th*, the width of the rectangular pulse *h*ˆ(*t*). This rectangular pulse *h*ˆ(*t*) has an area identical to that of *h*(*t*) and a height identical to that of *h*(*t*) at some suitable instant *t* = *t*0. In Fig. 2.21, *t*0 is chosen as the instant at which *h*(*t*) is maximum. According to this definition,† - -$$ -T_h h(t_0) = \int_{-\infty}^{\infty} h(t) dt -$$ - -† This definition is satisfactory when *h*(*t*) is a single, mostly positive (or mostly negative) pulse. Such systems are lowpass systems. This definition should not be applied indiscriminately to all systems. - -or - -$$ -T_h = \frac{\int_{-\infty}^{\infty} h(t) dt}{h(t_0)} -$$ -\n(2.47) - -Now if a system has a single mode - -*h*(*t*) = *Ae*λ*t u*(*t*) - -with λ negative and real, then *h*(*t*) is maximum at *t* = 0 with value *h*(0) = *A*. Therefore, according to Eq. (2.47), - -$$ -T_h = \frac{1}{A} \int_0^\infty A e^{\lambda t} dt = -\frac{1}{\lambda} -$$ - -Thus, the time constant in this case is simply the (negative of the) reciprocal of the system's characteristic root. For the multimode case, *h*(*t*) is a weighted sum of the system's characteristic modes, and *Th* is a weighted average of the time constants associated with the *N* modes of the system. - -### **[2.6-3 Time Constant and Rise Time of a System](#page-8-0)** - -Rise time of a system, defined as the time required for the unit step response to rise from 10% to 90% of its steady-state value, is an indication of the speed of response.† The system time constant may also be viewed from a perspective of rise time. The unit step response *y*(*t*) of a system is the convolution of *u*(*t*) with *h*(*t*). Let the impulse response *h*(*t*) be a rectangular pulse of width *Th*, as shown in Fig. 2.22. This assumption simplifies the discussion, yet gives satisfactory results for qualitative discussion. The result of this convolution is illustrated in Fig. 2.22. Note that the output does not rise from zero to a final value instantaneously as the input rises; instead, the output takes *Th* seconds to accomplish this. Hence, the rise time *Tr* of the system is equal to the system time constant - -$$ -T_r = T_h -$$ - -This result and Fig. 2.22 show clearly that a system generally does not respond to an input instantaneously. Instead, it takes time *Th* for the system to respond fully. - -† Because of varying definitions of rise time, the reader may find different results in the literature. The qualitative and intuitive nature of this discussion should always be kept in mind. - -**Figure 2.22** Rise time of a system. - -### **[2.6-4 Time Constant and Filtering](#page-8-0)** - -A larger time constant implies a sluggish system because the system takes longer to respond fully to an input. Such a system cannot respond effectively to rapid variations in the input. In contrast, a smaller time constant indicates that a system is capable of responding to rapid variations in the input. Thus, there is a direct connection between a system's time constant and its filtering properties. - -A high-frequency sinusoid varies rapidly with time. A system with a large time constant will not be able to respond well to this input. Therefore, such a system will suppress rapidly varying (high-frequency) sinusoids and other high-frequency signals, thereby acting as a lowpass filter (a filter allowing the transmission of low-frequency signals only). We shall now show that a system - -**Figure 2.23** Time constant and filtering. - -with a time constant *Th* acts as a lowpass filter having a cutoff frequency of *fc* = 1/*Th* hertz, so that sinusoids with frequencies below *fc* Hz are transmitted reasonably well, while those with frequencies above *fc* Hz are suppressed. - -To demonstrate this fact, let us determine the system response to a sinusoidal input *x*(*t*) by convolving this input with the effective impulse response *h*(*t*) in Fig. 2.23a. From Figs. 2.23b and 2.23c we see the process of convolution of *h*(*t*) with the sinusoidal inputs of two different frequencies. The sinusoid in Fig. 2.23b has a relatively high frequency, while the frequency of the sinusoid in Fig. 2.23c is low. Recall that the convolution of *x*(*t*) and *h*(*t*) is equal to the area under the product *x*(τ )*h*(*t* − τ ). This area is shown shaded in Figs. 2.23b and 2.23c for the two cases. For the high-frequency sinusoid, it is clear from Fig. 2.23b that the area under *x*(τ )*h*(*t* − τ ) is very small because its positive and negative areas nearly cancel each other out. In this case the output *y*(*t*) remains periodic but has a rather small amplitude. This happens when the period of the sinusoid is much smaller than the system time constant *Th*. In contrast, for the low-frequency sinusoid, the period of the sinusoid is larger than *Th*, rendering the partial cancellation of area under *x*(τ )*h*(*t* −τ ) less effective. Consequently, the output *y*(*t*) is much larger, as depicted in Fig. 2.23c. - -Between these two possible extremes in system behavior, a transition point occurs when the period of the sinusoid is equal to the system time constant *Th*. The frequency at which this transition occurs is known as the *cutoff frequency fc* of the system. Because *Th* is the period of cutoff frequency *fc*, - -$$ -f_c = \frac{1}{T_h} -$$ - -The frequency *fc* is also known as the bandwidth of the system because the system transmits or passes sinusoidal components with frequencies below *fc* while attenuating components with frequencies above *fc*. Of course, the transition in system behavior is gradual. There is no dramatic change in system behavior at *fc* = 1/*Th*. Moreover, these results are based on an idealized (rectangular pulse) impulse response; in practice these results will vary somewhat, depending on the exact shape of *h*(*t*). Remember that the "feel" of general system behavior is more important than exact system response for this qualitative discussion. - -Since the system time constant is equal to its rise time, we have - -$$ -T_r = \frac{1}{f_c} \qquad \text{or} \qquad f_c = \frac{1}{T_r} \tag{2.48} -$$ - -Thus, a system's bandwidth is inversely proportional to its rise time. Although Eq. (2.48) was derived for an idealized (rectangular) impulse response, its implications are valid for lowpass LTIC systems, in general. For a general case, we can show that [1] - -$$ -f_c = \frac{k}{T_r} -$$ - -where the exact value of *k* depends on the nature of *h*(*t*). An experienced engineer often can estimate quickly the bandwidth of an unknown system by simply observing the system response to a step input on an oscilloscope. - -### **[2.6-5 Time Constant and Pulse Dispersion \(Spreading\)](#page-8-0)** - -In general, the transmission of a pulse through a system causes pulse dispersion (or spreading). Therefore, the output pulse is generally wider than the input pulse. This system behavior can have serious consequences in communication systems in which information is transmitted by pulse amplitudes. Dispersion (or spreading) causes interference or overlap with neighboring pulses, thereby distorting pulse amplitudes and introducing errors in the received information. - -Earlier we saw that if an input *x*(*t*) is a pulse of width *Tx*, then *Ty*, the width of the output *y*(*t*), is - -$$ -T_{y}=T_{x}+T_{h} -$$ - -This result shows that an input pulse spreads out (disperses) as it passes through a system. Since *Th* is also the system's time constant or rise time, the amount of spread in the pulse is equal to the time constant (or rise time) of the system. - -### **[2.6-6 Time Constant and Rate of Information Transmission](#page-8-0)** - -In pulse communications systems, which convey information through pulse amplitudes, the rate of information transmission is proportional to the rate of pulse transmission. We shall demonstrate that to avoid the destruction of information caused by dispersion of pulses during their transmission through the channel (transmission medium), the rate of information transmission should not exceed the bandwidth of the communications channel. - -Since an input pulse spreads out by *Th* seconds, the consecutive pulses should be spaced *Th* seconds apart to avoid interference between pulses. Thus, the rate of pulse transmission should not exceed 1/*Th* pulses/second. But 1/*Th* = *fc*, the channel's bandwidth, so that we can transmit pulses through a communications channel at a rate of *fc* pulses per second and still avoid significant interference between the pulses. The rate of information transmission is therefore proportional to the channel's bandwidth (or to the reciprocal of its time constant).† - -The discussion of Secs. 2.6-2, 2.6-3, 2.6-4, 2.6-5, and 2.6-6) shows that the system time constant determines much of a system's behavior—its filtering characteristics, rise time, pulse dispersion, and so on. In turn, the time constant is determined by the system's characteristic roots. Clearly the characteristic roots and their relative amounts in the impulse response *h*(*t*) determine the behavior of a system. - -### **EXAMPLE 2.15 Intuitive Insights into Lowpass System Behavior** - -Find the time constant *Th*, rise time *Tr*, and cutoff frequency *fc* for a lowpass system that has impulse response *h*(*t*) = *te*−*t u*(*t*). Determine the maximum rate that pulses of 1 second - -† Theoretically, a channel of bandwidth *fc* can transmit correctly up to 2*fc* pulse amplitudes per second [4]. Our derivation here, being very simple and qualitative, yields only half the theoretical limit. In practice it is not easy to attain the upper theoretical limit. - -#### 210 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -duration can be transmitted through the system so that interference is essentially avoided between adjacent pulses at the system output. - -The system impulse response *h*(*t*) = *te*−*t u*(*t*), which looks similar to the impulse response of Fig. 2.21, has a peak value of *e*−1 = 0.3679 at a time *t*0 = 1. According to Eq. (2.47) and using integration by parts, the system time constant is therefore - -$$ -T_h = \frac{\int_0^\infty t e^{-t} dt}{e^{-1}} = e^1 \left( -t e^{-t} \Big|_0^\infty + \int_0^\infty e^{-t} dt \right) = e^1 \left( 0 - e^{-t} \Big|_0^\infty \right) = e^1(1) = 2.7183 -$$ - -Thus, - -$$ -T_h = 2.7183 -$$ - s, $T_r = T_h = 2.7183$ s, and $f_c = \frac{1}{T_h} = 0.3679$ Hz - -Due to its lowpass nature, this system will spread an input pulse of 1 second to an output with width - -*Ty* = *Tx* +*Th* = 1+2.7183 = 3.7183 s - -To avoid interference between pulses at the output, the pulse transmission rate should be no more than the reciprocal of the output pulse width. That is, - -> maximum pulse transmission rate = 1 3.7183 = 0.2689 pulse/s - -By narrowing the input pulses, the pulse transmission rate could increase up to *fc* = 0.3679 pulse/s. - -### **[2.6-7 The Resonance Phenomenon](#page-8-0)** - -Finally, we come to the fascinating phenomenon of resonance. As we have already mentioned several times, this phenomenon is observed when the input signal is identical or is very close to a characteristic mode of the system. For the sake of simplicity and clarity, we consider a first-order system having only a single mode, *e*λ*t* . Let the impulse response of this system be† - -$$ -h(t) = Ae^{\lambda t} -$$ - -and let the input be - -$$ -x(t) = e^{(\lambda - \epsilon)t} -$$ - -The system response *y*(*t*) is then given by - -$$ -y(t) = Ae^{\lambda t} * e^{(\lambda - \epsilon)t} -$$ - -† For convenience, we omit multiplying *x*(*t*) and *h*(*t*) by *u*(*t*). Throughout this discussion, we assume that they are causal. - -### 2.6 Intuitive Insights into System Behavior 211 - -From the convolution table we obtain - -$$ -y(t) = \frac{A}{\epsilon} \left[ e^{\lambda t} - e^{(\lambda - \epsilon)t} \right] = A e^{\lambda t} \left( \frac{1 - e^{-\epsilon t}}{\epsilon} \right) -$$ - (2.49) - -Now, as → 0, both the numerator and the denominator of the term in the parentheses approach zero. Applying L'Hôpital's rule to this term yields - -$$ -\lim_{\epsilon \to 0} y(t) = A t e^{\lambda t} -$$ - -Clearly, the response does not go to infinity as → 0, but it acquires a factor *t*, which approaches ∞ as *t* → ∞. If λ has a negative real part (so that it lies in the LHP), *e*λ*t* decays faster than *t* and *y*(*t*) → 0 as *t* → ∞. The resonance phenomenon in this case is present, but its manifestation is aborted by the signal's own exponential decay. - -This discussion shows that*resonance is a cumulative phenomenon,* not instantaneous. It builds up linearly with *t*. † When the mode decays exponentially, the signal decays too fast for resonance to counteract the decay; as a result, the signal vanishes before resonance has a chance to build it up. However, if the mode were to decay at a rate less than 1/*t*, we should see the resonance phenomenon clearly. This specific condition would be possible if Re λ ≥ 0. For instance, when Re λ = 0 so that λ lies on the imaginary axis of the complex plane (λ = *j*ω), the output becomes - -$$ -y(t) = A t e^{j\omega t} -$$ - -Here, the response does go to infinity linearly with *t*. - -For a real system, if λ = *j*ω is a root, λ∗ = −*j*ω must also be a root; the impulse response is of the form *Aej*ω*t* + *Ae*−*j*ω*t* = 2*A*cos ω*t*. The response of this system to input *A*cosω*t* is 2*A*cosω*t* ∗ cosω*t*. The reader can show that this convolution contains a term of the form *At* cos ω*t*. The resonance phenomenon is clearly visible. The system response to its characteristic mode increases linearly with time, eventually reaching ∞, as indicated in Fig. 2.24. - -Recall that when λ = *j*ω, the system is marginally stable. As we have indicated, the full effect of resonance cannot be seen for an asymptotically stable system; only in a marginally stable system does the resonance phenomenon boost the system's response to infinity when the system's input - -**Figure 2.24** Buildup of system response in resonance. - -† If the characteristic root in question repeats *r* times, resonance effect increases as *t r*−1. However, *t r*−1*e*λ*t* →0 as *t* → ∞ for any value of *r*, provided Re λ < 0 (λ in the LHP). - -### 212 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -is a characteristic mode. But even in an asymptotically stable system, we see a manifestation of resonance if its characteristic roots are close to the imaginary axis so that Re λ is a small, negative value. We can show that when the characteristic roots of a system are σ ± *j*ω0, then the system response to the input *ej*ω0*t* or the sinusoid cosω0*t* is very large for small σ. † The system response drops off rapidly as the input signal frequency moves away from ω0. This frequency-selective behavior can be studied more profitably after an understanding of frequency-domain analysis has been acquired. For this reason we postpone full discussion of this subject until Ch. 4. - -### IMPORTANCE OF THE RESONANCE PHENOMENON - -The resonance phenomenon is very important because it allows us to design frequency-selective systems by choosing their characteristic roots properly. Lowpass, bandpass, highpass, and bandstop filters are all examples of frequency-selective networks. In mechanical systems, the inadvertent presence of resonance can cause signals of such tremendous magnitude that the system may fall apart. A musical note (periodic vibrations) of proper frequency can shatter glass if the frequency is matched to the characteristic root of the glass, which acts as a mechanical system. Similarly, a company of soldiers marching in step across a bridge amounts to applying a periodic force to the bridge. If the frequency of this input force happens to be nearer to a characteristic root of the bridge, the bridge may respond (vibrate) violently and collapse, even though it would have been strong enough to carry many soldiers marching out of step. A case in point is the Tacoma Narrows Bridge failure of 1940. This bridge was opened to traffic in July 1940. Within four months of opening (on November 7, 1940), it collapsed in a mild gale, not because of the wind's brute force but because the frequencies of wind-generated vortices, which matched the natural frequencies (characteristic roots) of the bridge, caused resonance. - -Because of the great damage that may occur, mechanical resonance is generally to be avoided, especially in structures or vibrating mechanisms. If an engine with periodic force (such as piston motion) is mounted on a platform, the platform with its mass and springs should be designed so that their characteristic roots are not close to the engine's frequency of vibration. Proper design of this platform can not only avoid resonance, but also attenuate vibrations if the system roots are placed far away from the frequency of vibration. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/033_2.7 MATLAB - M-FILES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/033_2.7 MATLAB - M-FILES.md deleted file mode 100644 index 67674992565fb9624ac1f2c2d7673d38611efec2..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/033_2.7 MATLAB - M-FILES.md +++ /dev/null @@ -1,218 +0,0 @@ -## **[2.7 MATLAB: M-FILES](#page-8-0)** - -M-files are stored sequences of MATLAB commands and help simplify complicated tasks. There are two types of M-file: script and function. Both types are simple text files and require a .m filename extension. - -Although M-files can be created by using any text editor, MATLAB's built-in editor is the preferable choice because of its special features. As with any program, comments improve the readability of an M-file. Comments begin with the % character and continue through the end of the line. - -An M-file is executed by simply typing the filename (without the .m extension). To execute, M-files need to be located in the current directory or any other directory in the MATLAB path. New directories are easily added to the MATLAB path by using the addpath command. - -† This follows directly from Eq. (2.49) with λ = σ +*j*ω0 and = σ. - -### **[2.7-1 Script M-Files](#page-8-0)** - -Script files, the simplest type of M-file, consist of a series of MATLAB commands. Script files record and automate a series of steps, and they are easy to modify. To demonstrate the utility of a script file, consider the operational amplifier circuit shown in Fig. 2.25. - -The system's characteristic modes define the circuit's behavior and provide insight regarding system behavior. Using ideal, infinite gain difference amplifier characteristics, we first derive the differential equation that relates output *y*(*t*) to input *x*(*t*). Kirchhoff's current law (KCL) at the node shared by *R*1 and *R*3 provides - -$$ -\frac{x(t) - v(t)}{R_3} + \frac{y(t) - v(t)}{R_2} + \frac{0 - v(t)}{R_1} - C_2 \dot{v}(t) = 0 -$$ - -KCL at the inverting input of the op amp gives - -$$ -\frac{v(t)}{R_1} + C_1 \dot{y}(t) = 0 -$$ - -Combining and simplifying the KCL equations yield - -$$ -\ddot{y}(t) + \frac{1}{C_2} \left\{ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right\} \dot{y}(t) + \frac{1}{R_1 R_2 C_1 C_2} y(t) = -\frac{1}{R_1 R_3 C_1 C_2} x(t) -$$ - -which is the desired constant coefficient differential equation. Thus, the characteristic equation is given by - -$$ -\lambda^2 + \frac{1}{C_2} \left\{ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right\} \lambda + \frac{1}{R_1 R_2 C_1 C_2} = (a_0 \lambda^2 + a_1 \lambda + a_2) = 0 \tag{2.50} -$$ - -The roots λ1 and λ2 of Eq. (2.50) establish the nature of the characteristic modes *e*λ1*t* and *e*λ2*t* . - -As a first case, assign nominal component values of *R*1 = *R*2 = *R*3 = 10 k and *C*1 = *C*2 = 1 µF. A series of MATLAB commands allows convenient computation of the roots λ = [λ1;λ2]. Although λ can be determined using the quadratic equation, MATLAB's roots command is more convenient. The roots command requires an input vector that contains the polynomial coefficients in descending order. Even if a coefficient is zero, it must still be included in the vector. - -**Figure 2.25** Operation-amplifier circuit. - -``` -% CH2MP1.m : Chapter 2, MATLAB Program 1 -% Script M-file determines characteristic roots of op-amp circuit. -% Set component values: -R = [1e4, 1e4, 1e4]; C = [1e-6, 1e-6]; -% Determine coefficients for characteristic equation: -A = [1, (1/R(1)+1/R(2)+1/R(3))/C(2), 1/(R(1)*R(2)*C(1)*C(2))]; -% Determine characteristic roots: -lambda = roots(A); -``` - -A script file is created by placing these commands in a text file, which in this case is named CH2MP1.m. While comment lines improve program clarity, their removal does not affect program functionality. The program is executed by typing - -``` ->> CH2MP1 -``` - -After execution, all the resulting variables are available in the workspace. For example, to view the characteristic roots, type - ->> lambda lambda = -261.8034 -38.1966 - -Thus, the characteristic modes are simple decaying exponentials: *e*−261.8034*t* and *e*−38.1966*t* . - -Script files permit simple or incremental changes, thereby saving significant effort. Consider what happens when capacitor *C*1 is changed from 1.0 µF to 1.0 nF. Changing CH2MP1.m so that C = [1e-9, 1e-6] allows computation of the new characteristic roots: - -``` ->> CH2MP1 ->> lambda - lambda = 1.0e+003 * - -0.1500 + 3.1587i - -0.1500 - 3.1587i -``` - -Perhaps surprisingly, the characteristic modes are now complex exponentials capable of supporting oscillations. The imaginary portion of λ dictates an oscillation rate of 3158.7 rad/s or about 503 Hz. The real portion dictates the rate of decay. The time expected to reduce the amplitude to 25% is approximately *t* = ln 0.25/Re(λ) ≈ 0.01 second. - -### **[2.7-2 Function M-Files](#page-8-0)** - -It is inconvenient to modify and save a script file each time a change of parameters is desired. Function M-files provide a sensible alternative. Unlike script M-files, function M-files can accept input arguments as well as return outputs. Functions truly extend the MATLAB language in ways that script files cannot. - -Syntactically, a function M-file is identical to a script M-file except for the first line. The general form of the first line is - -function [*output1, ..., outputN*] = filename(*input1, ..., inputM*) - -For example, consider modification of CH2MP1.m to make function CH2MP2.m. Component values are passed to the function as two separate inputs: a length-3 vector of resistor values and a length-2 vector of capacitor values. The characteristic roots are returned as a 2×1 complex vector. - -``` -function [lambda] = CH2MP2(R,C) -% CH2MP2.m : Chapter 2, MATLAB Program 2 -% Function M-file finds characteristic roots of op-amp circuit. -% INPUTS: R = length-3 vector of resistances -% C = length-2 vector of capacitances -% OUTPUTS: lambda = characteristic roots -% Determine coefficients for characteristic equation: -A = [1, (1/R(1)+1/R(2)+1/R(3))/C(2), 1/(R(1)*R(2)*C(1)*C(2))]; -% Determine characteristic roots: -lambda = roots(A); -``` - -As with script M-files, function M-files execute by typing the name at the command prompt. However, inputs must also be included. For example, CH2MP2 easily confirms the oscillatory modes of the preceding example. - -``` ->> lambda = CH2MP2([1e4, 1e4, 1e4],[1e-9, 1e-6]) - lambda = 1.0e+003 * - -0.1500 + 3.1587i - -0.1500 - 3.1587i -``` - -Although scripts and functions have similarities, they also have distinct differences that are worth pointing out. Scripts operate on workspace data; either functions must be supplied data through inputs or they must create their own data. Unless passed as an output, variables and data created by functions remain local to the function; variables or data generated by scripts are global and are added to the workspace. To emphasize this point, consider polynomial coefficient vector A, which is created and used in both CH2MP1.m and CH2MP2.m. Following execution of function CH2MP2, the variable A is not added to the workspace. Following execution of script CH2MP1, however, A is available in the workspace. Recall, the workspace is easily viewed by typing either who or whos. - -### **[2.7-3 For-Loops](#page-9-0)** - -Real resistors and capacitors never exactly equal their nominal values. Suppose that the circuit components are measured as *R*1 = 10.322 k, *R*2 = 9.952 k, *R*3 = 10.115 k, *C*1 = 1.120 nF, and *C*2 = 1.320 µF. These values are consistent with the 10 and 25% tolerance resistor and capacitor values commonly and readily available. CH2MP2.m uses these component values to calculate the new values of λ. - -### 216 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -``` ->> lambda = CH2MP2([10322,9592,10115],[1.12e-9, 1.32e-6]) - lambda = 1.0e+003 * - -0.1136 + 2.6113i - -0.1136 - 2.6113i -``` - -Now the natural modes oscillate at 2611.3 rad/s or about 416 Hz. Decay to 25% amplitude is expected in *t* = ln 0.25/(−113.6) ≈ 0.012 second. These values, which differ significantly from the nominal values of 503 Hz and *t* ≈ 0.01 second, warrant a more formal investigation of the effect of component variations on the locations of the characteristic roots. - -It is sensible to look at three values for each component: the nominal value, a low value, and a high value. Low and high values are based on component tolerances. For example, a 10% 1 k resistor could have an expected low value of 1000(1 − 0.1) = 900 and an expected high value of 1000(1+0.1) = 1100 . For the five passive components in the design, 35 = 243 permutations are possible. - -Using either CH2MP1.m or CH2MP2.m to solve each of the 243 cases would be very tedious and boring. For-loops help automate repetitive tasks such as this. In MATLAB, the general structure of a for statement is - -``` -for variable = expression, statement, ..., statement, end -``` - -Five nested for-loops, one for each passive component, are required for the present example. - -``` -% CH2MP3.m : Chapter 2, MATLAB Program 3 -% Script M-file determines characteristic roots over a range of component - values. -% Pre-allocate memory for all computed roots: -lambda = zeros(2,243); -% Initialize index to identify each permutation: -p=0; -for R1 = 1e4*[0.9,1.0,1.1], - for R2 = 1e4*[0.9,1.0,1.1], - for R3 = 1e4*[0.9,1.0,1.1], - for C1 = 1e-9*[0.75,1.0,1.25], - for C2 = 1e-6*[0.75,1.0,1.25], - p = p+1; - lambda(:,p) = CH2MP2([R1 R2 R3],[C1 C2]); - end - end - end - end -end -plot(real(lambda(:)),imag(lambda(:)),'kx',... - real(lambda(:,1)),imag(lambda(:,1)),'kv',... - real(lambda(:,end)),imag(lambda(:,end)),'k^') -xlabel('Real'),ylabel('Imaginary') -legend('Char. Roots','Min. Val. Roots','Max. Val. Roots','Location','West'); -``` - -**Figure 2.26** Effect of component values on characteristic root locations. - -The command lambda = zeros(2,243) preallocates a 2×243 array to store the computed roots. When necessary, MATLAB performs dynamic memory allocation, so this command is not strictly necessary. However, preallocation significantly improves script execution speed. Notice also that it would be nearly useless to call script CH2MP1 from within the nested loop; script file parameters cannot be changed during execution. - -The plot instruction is quite long. Long commands can be broken across several lines by terminating intermediate lines with three dots (...). The three dots tell MATLAB to continue the present command to the next line. Black x's locate roots of each permutation. The command lambda(:) vectorizes the 2 × 243 matrix lambda into a 486 × 1 vector. This is necessary in this case to ensure that a proper legend is generated. Because of loop order, permutation *p* = 1 corresponds to the case of all components at the smallest values and permutation *p* = 243 corresponds to the case of all components at the largest values. This information is used to separately highlight the minimum and maximum cases using down-triangles () and up-triangles (), respectively. In addition to terminating each for loop, end is used to indicate the final index along a particular dimension, which eliminates the need to remember the particular size of a variable. An overloaded function, such as end, serves multiple uses and is typically interpreted based on context. - -The graphical results provided by CH2MP3 are shown in Fig. 2.26. Between extremes, root oscillations vary from 365 to 745 Hz and decay times to 25% amplitude vary from 6.2 to 12.7 ms. Clearly, this circuit's behavior is quite sensitive to ordinary component variations. - -### **[2.7-4 Graphical Understanding of Convolution](#page-9-0)** - -MATLAB graphics effectively illustrate the convolution process. Consider the case of *y*(*t*) = *x*(*t*)∗ *h*(*t*), where *x*(*t*) = 1.5 sin(π*t*)(*u*(*t*) − *u*(*t* − 1)) and *h*(*t*) = 1.5(*u*(*t*) − *u*(*t* − 1.5)) − *u*(*t* − 2) + *u*(*t* − 2.5). Program CH2MP4 steps through the convolution over the time interval (−0.25 ≤ *t* ≤ 3.75). - -``` -% CH2MP4.m : Chapter 2, MATLAB Program 4 -% Script M-file graphically demonstrates the convolution process. -figure(1) % Create figure window and make visible on screen -u = @(t) 1.0*(t>=0); -x = @(t) 1.5*sin(pi*t).*(u(t)-u(t-1)); -h = @(t) 1.5*(u(t)-u(t-1.5))-u(t-2)+u(t-2.5); -dtau = 0.005; tau = -1:dtau:4; -ti = 0; tvec = -.25:.1:3.75; -y = NaN*zeros(1,length(tvec)); % Pre-allocate memory -for t = tvec, - ti = ti+1; % Time index - xh = x(t-tau).*h(tau); lxh = length(xh); - y(ti) = sum(xh.*dtau); % Trapezoidal approximation of convolution integral - subplot(2,1,1),plot(tau,h(tau),'k-',tau,x(t-tau),'k--',t,0,'ok'); - axis([tau(1) tau(end) -2.0 2.5]); - patch([tau(1:end-1);tau(1:end-1);tau(2:end);tau(2:end)],... - [zeros(1,lxh-1);xh(1:end-1);xh(2:end);zeros(1,lxh-1)],... - [.8 .8 .8],'edgecolor','none'); - xlabel('\tau'); title('h(\tau) [solid], x(t-\tau) [dashed], h(\tau)x(t-\tau) [gray]'); - c = get(gca,'children'); set(gca,'children',[c(2);c(3);c(4);c(1)]); - subplot(2,1,2),plot(tvec,y,'k',tvec(ti),y(ti),'ok'); - xlabel('t'); ylabel('y(t) = \int h(\tau)x(t-\tau) d\tau'); - axis([tau(1) tau(end) -1.0 2.0]); grid; - drawnow; -end -``` - -At each step, the program plots *h*(τ ), *x*(*t* − τ ), and shades the area *h*(τ )*x*(*t* − τ ) gray. This gray area, which reflects the integral of *h*(τ )*x*(*t* − τ ), is also the desired result, *y*(*t*). Figures 2.27, 2.28, and 2.29 display the convolution process at times *t* of 0.75, 2.25, and 2.85 seconds, respectively. These figures help illustrate how the regions of integration change with time. Figure 2.27 has limits of integration from 0 to (*t* = 0.75). Figure 2.28 has two regions of integration, with limits (*t* −1 = 1.25) to 1.5 and 2.0 to (*t* = 2.25). The last plot, Fig. 2.29, has limits from 2.0 to 2.5. - -Several comments regarding CH2MP4 are in order. The command figure(1) opens the first figure window and, more important, makes sure it is visible. Anonymous functions are used to represent the functions *u*(*t*), *x*(*t*), and *h*(*t*). NaN, standing for not-a-number, usually results from operations such as 0/0 or ∞−∞. MATLAB refuses to plot NaN values, so preallocating *y*(*t*) with NaNs ensures that MATLAB displays only values of *y*(*t*) that have been computed. As its name suggests, length returns the length of the input vector. The subplot(a,b,c) command partitions the current figure window into an a-by-b matrix of axes and selects axes c for use. Subplots facilitate graphical comparison by allowing multiple axes in a single figure window. The patch command is used to create the gray-shaded area for *h*(τ )*x*(*t* − τ ). In CH2MP4, the get and set commands are used to reorder plot objects so that the gray area does not obscure other lines. Details of the patch, get, and set commands, as used in CH2MP4, are somewhat advanced and are not pursued here.† MATLAB also prints most Greek letters if the Greek name is preceded by a backslash (\) character. For example, \tau in the xlabel command produces the symbol τ in the plot's axis label. Similarly, an integral sign is produced by \int. Finally, the drawnow - -† Interested students should consult the MATLAB help facilities for further information. Actually, the get and set commands are extremely powerful and can help modify plots in almost any conceivable way. - -**Figure 2.27** Graphical convolution at step *t* = 0.75 second. - -**Figure 2.28** Graphical convolution at step *t* = 2.25 seconds. - -command forces MATLAB to update the graphics window for each loop iteration. Although slow, this creates an animation-like effect. Replacing drawnow with the pause command allows users to manually step through the convolution process. The pause command still forces the graphics window to update, but the program will not continue until a key is pressed. - -**Figure 2.29** Graphical convolution at step *t* = 2.85 seconds. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/034_2.8 APPENDIX - DETERMINING THE IMPULSE RESPONSE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/034_2.8 APPENDIX - DETERMINING THE IMPULSE RESPONSE.md deleted file mode 100644 index 9848f54ff8e058a449119d9e9dd47abc05834a0e..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/034_2.8 APPENDIX - DETERMINING THE IMPULSE RESPONSE.md +++ /dev/null @@ -1,77 +0,0 @@ -## **[2.8 APPENDIX: DETERMINING THE](#page-9-0) IMPULSE RESPONSE** - -In Eq. (2.13), we showed that for an LTIC system *S* specified by Eq. (2.11), the unit impulse response *h*(*t*) can be expressed as - -$$ -h(t) = b_0 \delta(t) + \text{characteristic modes} \tag{2.51} -$$ - -To determine the characteristic mode terms in Eq. (2.51), let us consider a system *S*0 whose input *x*(*t*) and the corresponding output *w*(*t*) are related by - -$$ -Q(D)w(t) = x(t) \tag{2.52} -$$ - -Observe that both the systems *S* and *S*0 have the same characteristic polynomial; namely, *Q*(λ), and, consequently, the same characteristic modes. Moreover, *S*0 is the same as *S* with *P*(*D*)=1, that is, *b*0 = 0. Therefore, according to Eq. (2.51), the impulse response of *S*0 consists of characteristic mode terms only without an impulse at *t* = 0. Let us denote this impulse response of *S*0 by *yn*(*t*). Observe that *yn*(*t*) consists of characteristic modes of *S* and therefore may be viewed as a zero-input response of *S*. Now *yn*(*t*) is the response of *S*0 to input δ(*t*). Therefore, according to Eq. (2.52), - -$$ -Q(D)y_n(t) = \delta(t) -$$ - -or - -$$ -(D^N + a_1 D^{N-1} + \dots + a_{N1} D + a_N) y_n(t) = \delta(t) -$$ - -or - -$$ -y_n^{(N)}(t) + a_1 y_n^{(N-1)}(t) + \dots + a_{N-1} y_n^{(1)}(t) + a_N y_n(t) = \delta(t) -$$ - -where *y*(*k*) *n* (*t*) represents the *k*th derivative of *yn*(*t*). The right-hand side contains a single impulse term, δ(*t*). This is possible only if *y*(*N*−1) *n* (*t*) has a unit jump discontinuity at *t* = 0, so that *y*(*N*) *n* (*t*) = δ(*t*). Moreover, the lower-order terms cannot have any jump discontinuity because this would mean the presence of the derivatives of δ(*t*). Therefore *yn*(0) = *y*(1) *n* (0) =···= *y*(*N*−2) *n* (0) = 0 (no discontinuity at *t* = 0), and the *N* initial conditions on *yn*(*t*) are - -$$ -y_n(0) = y_n^{(1)}(0) = \dots = y_n^{(N-2)}(0) = 0 -$$ - and $y_n^{(N-1)}(0) = 1$ (2.53) - -This discussion means that *yn*(*t*) is the zero-input response of the system *S* subject to initial conditions [Eq. (2.53)]. - -We now show that for the same input *x*(*t*) to both systems, *S* and *S*0, their respective outputs *y*(*t*) and *w*(*t*) are related by - -$$ -y(t) = P(D)w(t) -$$ -\n(2.54) - -To prove this result, we operate on both sides of Eq. (2.52) by *P*(*D*) to obtain - -$$ -Q(D)P(D)w(t) = P(D)x(t) -$$ - -Comparison of this equation with Eq. (2.2) leads immediately to Eq. (2.54). - -Now if the input *x*(*t*) = δ(*t*), the output of *S*0 is *yn*(*t*), and the output of *S*, according to Eq. (2.54), is *P*(*D*)*yn*(*t*). This output is *h*(*t*), the unit impulse response of *S*. Note, however, that because it is an impulse response of a causal system *S*0, the function *yn*(*t*) is causal. To incorporate this fact we must represent this function as *yn*(*t*)*u*(*t*). Now it follows that *h*(*t*), the unit impulse response of the system *S*, is given by - -$$ -h(t) = P(D)[y_n(t)u(t)] -$$ -\n -$$ -(2.55) -$$ - -where *yn*(*t*) is a linear combination of the characteristic modes of the system subject to initial conditions (2.53). - -The right-hand side of Eq. (2.55) is a linear combination of the derivatives of *yn*(*t*)*u*(*t*). Evaluating these derivatives is clumsy and inconvenient because of the presence of *u*(*t*). The derivatives will generate an impulse and its derivatives at the origin. Fortunately when *M* ≤ *N* [Eq. (2.11)], we can avoid this difficulty by using the observation in Eq. (2.51), which asserts that at *t* = 0 (the origin), *h*(*t*) = *b*0δ(*t*). Therefore, we need not bother to find *h*(*t*) at the origin. This simplification means that instead of deriving *P*(*D*)[*yn*(*t*)*u*(*t*)], we can derive *P*(*D*)*yn*(*t*) and add to it the term *b*0δ(*t*) so that - -$$ -h(t) = b_0 \delta(t) + P(D) y_n(t) \qquad t \ge 0 -$$ - -= $b_0 \delta(t) + [P(D) y_n(t)] u(t)$ - -This expression is valid when *M* ≤ *N* [the form given in Eq. (2.11)]. When *M* > *N*, Eq. (2.55) should be used. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/035_2.9 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/035_2.9 SUMMARY.md deleted file mode 100644 index 852dbc863f58c62f655de4e02fd772438568047c..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/035_2.9 SUMMARY.md +++ /dev/null @@ -1,31 +0,0 @@ -## **[2.9 SUMMARY](#page-9-0)** - -This chapter discusses time-domain analysis of LTIC systems. The total response of a linear system is a sum of the zero-input response and zero-state response. The zero-input response is the system - -#### 222 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -response generated only by the internal conditions (initial conditions) of the system, assuming that the external input is zero; hence the adjective "zero-input." The zero-state response is the system response generated by the external input, assuming that all initial conditions are zero, that is, when the system is in zero state. - -Every system can sustain certain forms of response on its own with no external input (zero input). These forms are intrinsic characteristics of the system; that is, they do not depend on any external input. For this reason they are called characteristic modes of the system. Needless to say, the zero-input response is made up of characteristic modes chosen in a combination required to satisfy the initial conditions of the system. For an *N*th-order system, there are *N* distinct modes. - -The unit impulse function is an idealized mathematical model of a signal that cannot be generated in practice.† Nevertheless, introduction of such a signal as an intermediary is very helpful in analysis of signals and systems. The unit impulse response of a system is a combination of the characteristic modes of the system‡ because the impulse δ(*t*) = 0 for *t* > 0. Therefore, the system response for *t* > 0 must necessarily be a zero-input response, which, as seen earlier, is a combination of characteristic modes. - -The zero-state response (response due to external input) of a linear system can be obtained by breaking the input into simpler components and then adding the responses to all the components. In this chapter we represent an arbitrary input *x*(*t*) as a sum of narrow rectangular pulses [staircase approximation of *x*(*t*)]. In the limit as the pulse width → 0, the rectangular pulse components approach impulses. Knowing the impulse response of the system, we can find the system response to all the impulse components and add them to yield the system response to the input *x*(*t*). The sum of the responses to the impulse components is in the form of an integral, known as the convolution integral. The system response is obtained as the convolution of the input *x*(*t*) with the system's impulse response *h*(*t*). Therefore, the knowledge of the system's impulse response allows us to determine the system response to any arbitrary input. - -LTIC systems have a very special relationship to the everlasting exponential signal *est* because the response of an LTIC system to such an input signal is the same signal within a multiplicative constant. The response of an LTIC system to the everlasting exponential input *est* is *H*(*s*)*est*, where *H*(*s*) is the transfer function of the system. - -If every bounded input results in a bounded output, the system is stable in the bounded-input/bounded-output (BIBO) sense. An LTIC system is BIBO-stable if and only if its impulse response is absolutely integrable. Otherwise, it is BIBO-unstable. BIBO stability is a stability seen from external terminals of the system. Hence, it is also called external stability or zero-state stability. - -In contrast, internal stability (or the zero-input stability) examines the system stability from inside. When some initial conditions are applied to a system in zero state, then, if the system eventually returns to zero state, the system is said to be stable in the asymptotic or Lyapunov sense. If the system's response increases without bound, it is unstable. If the system does not go to zero state and the response does not increase indefinitely, the system is marginally stable. The internal stability criterion, in terms of the location of a system's characteristic roots, can be summarized as follows: - -† However, it can be closely approximated by a narrow pulse of unit area and having a width that is much smaller than the time constant of an LTIC system in which it is used. - -‡ There is the possibility of an impulse in addition to the characteristic modes. - -- 1. An LTIC system is asymptotically stable if, and only if, all the characteristic roots are in the LHP. The roots may be repeated or unrepeated. -- 2. An LTIC system is unstable if, and only if, either one or both of the following conditions exist: (i) at least one root is in the RHP; (ii) there are repeated roots on the imaginary axis. -- 3. An LTIC system is marginally stable if, and only if, there are no roots in the RHP, and there are some unrepeated roots on the imaginary axis. - -It is possible for a system to be externally (BIBO) stable but internally unstable. When a system is controllable and observable, its external and internal descriptions are equivalent. Hence, external (BIBO) and internal (asymptotic) stabilities are equivalent and provide the same information. Such a BIBO-stable system is also asymptotically stable, and vice versa. Similarly, a BIBO-unstable system is either marginally stable or asymptotically unstable system. - -The characteristic behavior of a system is extremely important because it determines not only the system response to internal conditions (zero-input behavior), but also the system response to external inputs (zero-state behavior) and the system stability. The system response to external inputs is determined by the impulse response, which itself is made up of characteristic modes. The width of the impulse response is called the time constant of the system, which indicates how fast the system can respond to an input. The time constant plays an important role in determining such diverse system behaviors as the response time and filtering properties of the system, dispersion of pulses, and the rate of pulse transmission through the system. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/036_REFERENCES.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/036_REFERENCES.md deleted file mode 100644 index 20aa718cc3a8140d96d407f75f451b02259f2bcf..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/036_REFERENCES.md +++ /dev/null @@ -1,6 +0,0 @@ -### **[REFERENCES](#page-9-0)** - -- 1. Lathi, B. P., *Signals and Systems*. Berkeley-Cambridge Press, Carmichael, CA, 1987. -- 2. Mason, S. J., *Electronic Circuits, Signals, and Systems*. Wiley, New York, 1960. -- 3. Kailath, T., *Linear System*. Prentice-Hall, Englewood Cliffs, NJ, 1980. -- 4. Lathi, B. P., *Modern Digital and Analog Communication Systems,* 3rd ed. Oxford University Press, New York, 1998. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/037_PROBLEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/037_PROBLEMS.md deleted file mode 100644 index e3be58f632aa8079b12f7478c2edd0f2e1134dc5..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/037_PROBLEMS.md +++ /dev/null @@ -1,517 +0,0 @@ -## **[PROBLEMS](#page-9-0)** - -- **2.2-1** Determine the constants *c*1, *c*2, λ1, and λ2 for each of the following second-order systems, which have zero-input responses of the form *y*zir(*t*) = *c*1*e*λ1*t* +*c*2*e*λ2*t* . - - (a) *y*¨(*t*) + 2*y*˙(*t*) + 5*y*(*t*) = ¨*x*(*t*) − 5*x*(*t*) with *y*zir(0) = 2 and *y*˙zir(0) = 0. - - (b) *y*¨(*t*) + 2*y*˙(*t*) + 5*y*(*t*) = ¨*x*(*t*) − 5*x*(*t*) with *y*zir(0) = 4 and *y*˙zir(0) = −1. - - (c) *d*2 *dt*2 *y*(*t*) + 2 *d dt y*(*t*) = *x*(*t*) with *y*zir(0) = 1 and *y*˙zir(0) = 2. - - (d) (*D*2 +2*D*+10){*y*(*t*)} = (*D*5 −*D*){*x*(*t*)} with *y*zir(0) = ˙*y*zir(0) = 1. - - (e) (*D*2 + 7 2*D* + 3 2 ){*y*(*t*)} = (*D* + 2){*x*(*t*)} with *y*zir(0) = 3 and *y*¨zir(0) = −8. [*Caution:* The - -second IC is given in terms of the second derivative, not the first derivative]. - -- (f) 13*y*(*t*) + 4 *d dt y*(*t*) + *d*2 *dt*2 *y*(*t*) = 2*x*(*t*) − 4 *d dt x*(*t*) with *y*zir(0) = 3 and *y*¨zir(0) = −15. [*Caution:* The second IC is given in terms of the second derivative, not the first derivative]. -- **2.2-2** Consider a linear time-invariant system with input *x*(*t*) and output *y*(*t*) that is described by the differential equation - -$$ -(D+1)(D2-1) \{y(t)\} = (D5-1) \{x(t)\} -$$ - -Furthermore, assume *y*(0) = ˙*y*(0) = ¨*y*(0) = 1. - -### 224 CHAPTER 2 TIME-DOMAIN ANALYSIS OF CONTINUOUS-TIME SYSTEMS - -- (a) What is the order of this system? -- (b) What are the characteristic roots of this system? -- (c) Determine the zero-input response *y*zir(*t*). Simplify your answer. -- **2.2-3** A real LTIC system with input *x*(*t*) and output *y*(*t*) is described by the following constant-coefficient linear differential equation: - -$$ -(D3 + 9D) \{y(t)\} = (2D3 + 1) \{x(t)\}. -$$ - -- (a) What is the characteristic equation of this system? -- (b) What are the characteristic modes of this system? -- (c) Assuming *y*zir(0) = 4, *y*˙zir(0) = −18, and *y*¨zir(0) = 0, determine this system's zero-input response *y*zir(*t*). Simplify *y*zir(*t*) to include only real terms (i.e., no *j*'s should appear in your answer). -- **2.2-4** An LTIC system is specified by the equation - -$$ -(D2 + 5D + 6)y(t) = (D + 1)x(t) -$$ - -- (a) Find the characteristic polynomial, characteristic equation, characteristic roots, and characteristic modes of this system. -- (b) Find *y*0(*t*), the zero-input component of the response *y*(*t*) for *t* ≥ 0, if the initial conditions are *y*0(0−) = 2 and *y*˙0(0−) = −1. -- **2.2-5** Repeat Prob. 2.2-4 for - -$$ -(D^2 + 4D + 4)y(t) = Dx(t) -$$ - -and *y*0(0−) = 3, *y*˙0(0−) = −4. - -**2.2-6** Repeat Prob. 2.2-4 for - -$$ -D(D+1)y(t) = (D+2)x(t) -$$ - -and *y*0(0−) = ˙*y*0(0−) = 1. - -**2.2-7** Repeat Prob. 2.2-4 for - -$$ -(D^2 + 9)y(t) = (3D + 2)x(t) -$$ - -and *y*0(0−) = 0, *y*˙0(0−) = 6. - -**2.2-8** Repeat Prob. 2.2-4 for - -$$ -(D2 + 4D + 13)y(t) = 4(D+2)x(t) -$$ - -with -$$ -y_0(0^-) = 5 -$$ -, $\dot{y}_0(0^-) = 15.98$ . - -**2.2-9** Repeat Prob. 2.2-4 for - -$$ -D^2(D+1)y(t) = (D^2+2)x(t) -$$ - -with *y*0(0−) = 4, *y*˙0(0−) = 3, and *y*¨0(0−) = −1. - -**2.2-10** Repeat Prob. 2.2-4 for - -$$ -(D+1)(D^2 + 5D + 6)y(t) = Dx(t) -$$ - -with *y*0(0−) = 2, *y*˙0(0−) = −1, and *y*¨0(0−) = 5. - -- **2.2-11** A system is described by a constant-coefficient linear differential equation and has zero-input response given by *y*0(*t*) = 2*e*−*t* +3. - - (a) Is it possible for the system's characteristic equation to be λ + 1 = 0? Justify your answer. - - (b) Is it possible for the system's characteristic equation to be √3(λ2 +λ) = 0? Justify your answer. - - (c) Is it possible for the system's characteristic equation to be λ(λ + 1)2 = 0? Justify your answer. -- **2.2-12** Consider the circuit of Fig. P2.2-12. Using operator notation, this system can be described as (*D*+*a*1){*y*(*t*)} = (*b*0*D*+*b*1){*x*(*t*)}. - - (a) Determine the constants *a*1, *b*0, and *b*1 in terms of the system components *R*, *Rf* , and *C*. - - (b) Assume that *R* = 300 k, *Rf* = 1.2 M, and *C* = 5 µF. What is the zero-input response *y*0(*t*) of this system, assuming *vC*(0) = 1 V? - -**2.3-1** Determine the characteristic equation, characteristic modes, and impulse response *h*(*t*) for each of the following real LTIC systems. Since the systems are real, express each *h*(*t*) using only real terms (i.e., no *j*'s should appear in your answers). - -Problems 225 - -- (a) (*D*2 +1){*y*(*t*)} = 2*D*{*x*(*t*)} -- (b) (*D*3 +*D*){*y*(*t*)} = (2*D*3 +1){*x*(*t*)} - -(c) -$$ -\frac{d^2}{dt^2}y(t) + 2\frac{d}{dt}y(t) + 5y(t) = 8x(t) -$$ - -**2.3-2** Find the unit impulse response of a system specified by the equation - -$$ -(D2 + 4D + 3)y(t) = (D + 5)x(t) -$$ - -**2.3-3** Repeat Prob. 2.3-2 for - -$$ -(D2 + 5D + 6)y(t) = (D2 + 7D + 11)x(t) -$$ - -**2.3-4** Repeat Prob. 2.3-2 for the first-order allpass filter specified by the equation - -$$ -(D+1)y(t) = -(D-1)x(t) -$$ - -**2.3-5** Find the unit impulse response of an LTIC system specified by the equation - -(*D*2 +6*D*+9)*y*(*t*) = (2*D*+9)*x*(*t*) - -- **2.3-6** Determine and plot the unit impulse response *h*(*t*) of the op-amp circuit of Fig. P2.2-12, assuming that *R* = 300 k, *Rf* = 1.2 M, and *C* = 5 µF. -- **2.3-7** A causal LTIC system with input *x*(*t*) and output *y*(*t*) is described by the constant coefficient integral equation - -$$ -y(t) + \int 3y(t) dt + \int \int 2y(t) dt = -$$ - -$$ -\int \int x(t) dt - \int \int \int x(t) dt. -$$ - -- (a) Express this system as a constant coefficient linear differential equation in standard operator form. -- (b) Determine the characteristic modes of this system. - -**Figure P2.4-1** - -- (c) Determine the impulse response *h*(*t*) of this system. -- **2.4-1** Let *f*(*t*) = *h*1(*t*)∗*h*2(*t*), where *h*1(*t*) and *h*2(*t*) are shown in Fig. P2.4-1. In the following, use the graphical convolution procedure where you flip and shift *h*2(*t*). - - (a) Plot *h*1(τ ) and *h*2(*t* − τ ) as functions of τ . Clearly label the plots, including necessary function parameterizations. - - (b) Determine the (piecewise) regions of *f*(*t*) and set up the corresponding integrals that describe *f*(*t*) in those regions. Do not evaluate the integrals, only set them up! - - (c) Determine *f*(1), which is *f*(*t*) evaluated at *t* = 1. Provide a number, not a formula. -- **2.4-2** Consider signals *h*(*t*) = *u*(*t* + 3) − 2*u*(*t* + 1) + *u*(*t* − 1) and *x*(*t*) = cos(*t*) *u*(*t* − π/2)− *u*(*t* −3π/2) . Let *y*(*t*) = *x*(*t*) ∗ *h*(*t*). - - (a) Determine the last time *t*last that *y*(*t*) is nonzero. That is, find the smallest value *t*last such that *y*(*t*) = 0 for all *t* > *t*last. - - (b) Determine the approximate time *t*max where *y*(*t*) is a maximum. -- **2.4-3** Consider signals *h*(*t*) = −*u*(*t* + 2) + 3*u*(*t* − 1) − 2*u*(*t* − 5 2 ) and *x*(*t*) = sin(*t*)[*u*(*t* + 2π ) −*u*(*t* +π )]. Determine the approximate time *t*min where *y*(*t*)=*x*(*t*)∗*h*(*t*) is a minimum. Note, the minimum value of *y*(*t*) = 0! -- **2.4-4** An LTIC system has impulse response *h*(*t*) = 3*u*(*t* − 2). For input *x*(*t*) shown in Fig. P2.4-4, use the graphical convolution procedure to determine *y*zsr(*t*) = *h*(*t*) ∗ *x*(*t*). Accurately sketch *y*zsr(*t*). When solving for *y*zsr(*t*), flip and shift *x*(*t*) and explicitly show all integration steps—even if apparently trivial! - -**Figure P2.4-4** - -**2.4-5** Suppose an LTIC system has impulse response *h*(*t*) and input *x*(*t*) = *u*(*t*). Figure P2.4-5 shows *x*(*t*) and *h*(*t* + 1), respectively. Be careful! Figure P2.4-5 shows *h*(*t* +1), not *h*(*t*). - -**Figure P2.4-5** - -- (a) Is system *h*(*t*) causal? Mathematically justify your answer. -- (b) Use the graphical convolution procedure to determine *y*zsr(*t*) = *x*(*t*) ∗ *h*(*t*). Accurately sketch *y*zsr(*t*). When solving for *y*zsr(*t*), flip and shift *x*(*t*) and explicitly show all integration steps. - -**2.4-6** Repeat Prob. 2.4-5 using the signals of Fig. P2.4-6, rather than those of Fig. P2.4-5. Be careful! Figure P2.4-6 shows *h*(*t* −1), not *h*(*t*). - -**Figure P2.4-6** - -- **2.4-7** Suppose an LTIC system has impulse response *h*(*t*) and input *x*(*t*), both shown in Fig. P2.4-7. Use the graphical convolution procedure to determine *y*zsr(*t*) = *x*(*t*) ∗ *h*(*t*). Accurately sketch *y*zsr(*t*). When solving for *y*zsr(*t*), flip and shift *x*(*t*) and explicitly show all integration steps. -- **2.4-8** An LTIC system has impulse response *h*(*t*), as shown in Fig. P2.4-8. Let *t* have units of seconds. Let the input be *x*(*t*) = *u*(−*t* − 2) and designate the output as *y*zsr(*t*) = *x*(*t*) ∗ *h*(*t*). - - (a) Use the graphical convolution procedure where *h*(*t*) is flipped and shifted to determine *y*zsr(*t*). Accurately plot your result. - - (b) Use the graphical convolution procedure where *x*(*t*) is flipped and shifted to determine *y*zsr(*t*). Accurately plot your result. -- **2.4-9** If *c*(*t*) = *x*(*t*) ∗ *g*(*t*), then show that *Ac* = *AxAg*, where *Ax*,*Ag*, and *Ac* are the areas under *x*(*t*), *g*(*t*), and *c*(*t*), respectively. Verify this *area property* of convolution in Exs. 2.10 and 2.12. - -**Figure P2.4-7** - -**2.4-10** If *x*(*t*) ∗ *g*(*t*) = *c*(*t*), then show that *x*(*at*) ∗ *g*(*at*) = |1/*a*|*c*(*at*). This *time-scaling property* of convolution states that if both *x*(*t*) and *g*(*t*) are time-scaled by *a*, their convolution is also time-scaled by *a* (and multiplied by |1/*a*|). - -**Figure P2.4-8** - -- **2.4-11** Show that the convolution of an odd and an even function is an odd function and the convolution of two odd or two even functions is an even function. [*Hint:* Use the time-scaling property of convolution in Prob. 2.4-10.] -- **2.4-12** Suppose an LTIC system has impulse response *h*(*t*) = (1 − *t*)[*u*(*t*) − *u*(*t* − 1)] and input *x*(*t*) = *u*(−*t* −1)+*u*(*t* −1). Use the graphical convolution procedure to determine *y*zsr(*t*) = *x*(*t*) ∗ *h*(*t*). Accurately sketch *y*zsr(*t*). When solving for *y*zsr(*t*), flip and shift *h*(*t*), explicitly show all integration steps, and simplify your answer. -- **2.4-13** Using direct integration, find *e*−*atu*(*t*) ∗ *e*−*btu*(*t*). -- **2.4-14** Using direct integration, find *u*(*t*) ∗ *u*(*t*), *e*−*atu*(*t*) ∗ *e*−*atu*(*t*), and *tu*(*t*) ∗ *u*(*t*). -- **2.4-15** Using direct integration, find sin *t u*(*t*) ∗ *u*(*t*) and cos *t u*(*t*) ∗ *u*(*t*). -- **2.4-16** The unit impulse response of an LTIC system is - -$$ -h(t) = e^{-t}u(t) -$$ - -Find this system's (zero-state) response *y*(*t*) if the input *x*(*t*) is: - -$$ -(a) u(t) -$$ - -$$ -(b) e^{-t}u(t) -$$ - -$$ -(c) e^{-2t}u(t) -$$ - -(d) sin 3*t u*(*t*) - -Use the convolution table (Table 2.1) to find your answers. - -**2.4-17** Repeat Prob. 2.4-16 for - -$$ -h(t) = [2e^{-3t} - e^{-2t}]u(t) -$$ - -and if the input -$$ -x(t) -$$ - is: -\n(a) $u(t)$ -\n(b) $e^{-t}u(t)$ -\n(c) $e^{-2t}u(t)$ - -**2.4-18** Repeat Prob. 2.4-16 for - -$$ -h(t) = (1 - 2t)e^{-2t}u(t) -$$ - -and input *x*(*t*) = *u*(*t*). - -**2.4-19** Repeat Prob. 2.4-16 for - -$$ -h(t) = 4e^{-2t}\cos 3t u(t) -$$ - -and each of the following inputs *x*(*t*): - -- (a) *u*(*t*) -- (b) *e*−*t u*(*t*) -- **2.4-20** Repeat Prob. 2.4-16 for - -$$ -h(t) = e^{-t}u(t) -$$ - -and each of the following inputs *x*(*t*): - -- (a) *e*−2*t u*(*t*) -- (b) *e*−2(*t*−3) *u*(*t*) -- (c) *e*−2*t u*(*t* −3) -- (d) The gate pulse depicted in Fig. P2.4-20—and provide a sketch of *y*(*t*). - -**Figure P2.4-20** - -**2.4-21** A first-order allpass filter impulse response is given by - -$$ -h(t) = -\delta(t) + 2e^{-t}u(t) -$$ - -- (a) Find the zero-state response of this filter for the input *et u*(−*t*). -- (b) Sketch the input and the corresponding zero-state response. -- **2.4-22** Figure P2.4-22 shows the input *x*(*t*) and the impulse response *h*(*t*) for an LTIC system. Let the output be *y*(*t*). - - (a) By inspection of *x*(*t*) and *h*(*t*), find *y*(−1), *y*(0), *y*(1), *y*(2), *y*(3), *y*(4), *y*(5), and - -### **Figure P2.4-22** - -*y*(6). Thus, by merely examining *x*(*t*) and *h*(*t*), you are required to see what the result of convolution yields at *t* = −1, 0, 1, 2, 3, 4, 5, and 6. - -- (b) Find the system response to the input *x*(*t*). -- **2.4-23** The zero-state response of an LTIC system to an input *x*(*t*) = 2*e*−2*t u*(*t*) is *y*(*t*) = [4*e*−2*t* + 6*e*−3*t* ]*u*(*t*). Find the impulse response of the system. [*Hint:* We have not yet developed a method of finding *h*(*t*) from the knowledge of the input and the corresponding output. Knowing the form of *x*(*t*) and *y*(*t*), you will have to make the best guess of the general form of *h*(*t*).] -- **2.4-24** Sketch the functions *x*(*t*) = 1/(*t* 2 +1) and *u*(*t*). Now find *x*(*t*) ∗ *u*(*t*) and sketch the result. -- **2.4-25** Figure P2.4-25 shows *x*(*t*) and *g*(*t*). Find and sketch *c*(*t*) = *x*(*t*) ∗ *g*(*t*). -- **2.4-26** Find and sketch *c*(*t*) = *x*(*t*) ∗ *g*(*t*) for the functions depicted in Fig. P2.4-26. -- **2.4-27** Find and sketch *c*(*t*) = *x*1(*t*) ∗ *x*2(*t*) for the pairs of functions illustrated in Fig. P2.4-27. -- **2.4-28** Use Eq. (2.37) to find the convolution of *x*(*t*) and *w*(*t*), shown in Fig. P2.4-28. - -- **2.4-29** Determine *H*(*s*), the transfer function of an ideal time delay of *T* seconds. Find your answer by two methods: using Eq. (2.39) and using Eq. (2.40). -- **2.4-30** Determine *y*(*t*) = *x*(*t*) ∗ *h*(*t*) for the signals depicted in Fig. P2.4-30. -- **2.4-31** Two linear time-invariant systems, each with impulse response *h*(*t*), are connected in cascade. Refer to Fig. P2.4-31. Given input *x*(*t*) = *u*(*t*), determine *y*(1). That is, determine the step response at time *t* = 1 for the cascaded system shown. -- **2.4-32** Consider the electric circuit shown in Fig. P2.4-32. - - (a) Determine the differential equation that relates the input *x*(*t*) to output *y*(*t*). Recall that *iC*(*t*) = *CdvC*(*t*) *dt* and *vL*(*t*) = *LdiL*(*t*) *dt* . - - (b) Find the characteristic equation for this circuit, and express the root(s) of the characteristic equation in terms of *L* and *C*. - - (c) Determine the zero-input response given an initial capacitor voltage of one volt and an initial inductor current of zero amps. That is, find *y*0(*t*) given *vC*(0) = 1 V and - -**Figure P2.4-26** - -Problems 229 - -**Figure P2.4-27** - -*iL*(0) = 0 A. [*Hint:* The coefficient(s) in *y*0(*t*) are independent of *L* and *C*.] - -- (d) Plot *y*0(*t*) for *t* ≥ 0. Does the zero-input response, which is caused solely by initial conditions, ever "die out"? -- (e) Determine the total response *y*(*t*) to the input *x*(*t*) = *e*−*t u*(*t*). Assume an initial inductor current of *iL*(0−) = 0 A, an initial capacitor voltage of *vC*(0−) = 1 V, *L* = 1 H, and *C* = 1 F. - -**Figure P2.4-33** - -- **2.4-33** Two LTIC systems have impulse response functions given by *h*1(*t*) = (1−*t*)[*u*(*t*)−*u*(*t*−1)] and *h*2(*t*) = *t*[*u*(*t* +2)−*u*(*t* −2)]. - - (a) Carefully sketch the functions *h*1(*t*) and *h*2(*t*). - - (b) Assume that the two systems are connected in parallel, as shown in Fig. P2.4-33a. Carefully plot the equivalent impulse response function, *hp*(*t*). - - (c) Assume that the two systems are connected in cascade, as shown in Fig. P2.4-33b. Carefully plot the equivalent impulse response function, *hs*(*t*). - -- **2.4-34** Consider the circuit shown in Fig. P2.4-34. - - (a) Find the output *y*(*t*) given an initial capacitor voltage of *y*(0) = 2 volts and an input *x*(*t*) = *u*(*t*). - - (b) Given an input *x*(*t*) = *u*(*t* − 1), determine the initial capacitor voltage *y*(0) so that the output *y*(*t*) is 0.5 volt at *t* = 2 seconds. - -**Figure P2.4-34** - -**2.4-35** An analog signal is given by *x*(*t*) = *t*[*u*(*t*) − *u*(*t* − 1)], as shown in Fig. P2.4-35. Determine and plot *y*(*t*) = *x*(*t*) ∗ *x*(2*t*). - -**Figure P2.4-35** - -- **2.4-36** Consider the electric circuit shown in Fig. P2.4-36. - - (a) Determine the differential equation that relates the input current *x*(*t*) to output current *y*(*t*). Recall that - -$$ -v_L(t) = L \frac{di_L(t)}{dt} -$$ - -- (b) Find the characteristic equation for this circuit, and express the root(s) of the characteristic equation in terms of *L*1, *L*2, and *R*. -- (c) Determine the zero-input response given initial inductor currents of one ampere each. That is, find *y*0(*t*) given *iL*1 (0) = *iL*2 (0) = 1 A. - -**Figure P2.4-36** - -**Figure P2.4-38** - -- **2.4-37** An LTI system has step response given by *g*(*t*) = *e*−*t u*(*t*) − *e*−2*t u*(*t*). Determine the output of this system *y*(*t*) given an input *x*(*t*) = δ(*t* −π ) −cos( √3)*u*(*t*). -- **2.4-38** The periodic signal *x*(*t*) shown in Fig. P2.4-38 is input to a system with impulse response function *h*(*t*) = *t*[*u*(*t*) − *u*(*t* − 1.5)], also shown in Fig. P2.4-38. Use convolution to determine the output *y*(*t*) of this system. Plot *y*(*t*) over (−3 ≤ *t* ≤ 3). -- **2.4-39** Consider the electric circuit shown in Fig. P2.4-39. - - (a) Determine the differential equation relating input *x*(*t*) to output *y*(*t*). - - (b) Determine the output *y*(*t*) in response to the input *x*(*t*) = 4*te*−3*t*/2*u*(*t*). Assume component values of *R* = 1 , *C*1 = 1 F, and *C*2 = 2 F, and initial capacitor voltages of *VC*1 = 2 V and *VC*2 = 1 V. - -**Figure P2.4-39** - -- **2.4-40** An LTIC system has impulse response *h*(*t*) = 3*e*−|*t*| . - - (a) Is the system causal? Mathematically justify your answer. - - (b) Determine the zero-state response of this system if the input is *x*(*t*) = *u*(2−*t*). -- **2.4-41** A cardiovascular researcher is attempting to model the human heart. He has recorded ventricular pressure, which he believes corresponds to the heart's impulse response - -function *h*(*t*), as shown in Fig. P2.4-41. Comment on the function *h*(*t*) shown in Fig. P2.4-41. Can you establish any system properties, such as causality or stability? Do the data suggest any reason to suspect that the measurement is not a true impulse response? - -- **2.4-42** Consider an integrator system, *y*(*t*) = \$ *t* −∞ *x*(τ )*d*τ . - - (a) What is the unit impulse response *h*i(*t*) of this system? - - (b) If two such integrators are put in parallel, what is the resulting impulse response *h*p(*t*)? - - (c) If two such integrators are put in series, what is the resulting impulse response *h*s(*t*)? -- **2.4-43** The autocorrelation of a function *x*(*t*) is given by *rxx*(*t*) = \$ ∞ −∞ *x*(τ )*x*(τ − *t*)*d*τ . This equation is computed in a manner nearly identical to convolution. - - (a) Show *rxx*(*t*) = *x*(*t*) ∗ *x*(−*t*). - - (b) Determine and plot *rxx*(*t*) for the signal *x*(*t*) depicted in Fig. P2.4-43. [*Hint: rxx*(*t*) = *rxx*(−*t*).] - -**Figure P2.4-43** - -**2.4-44** Consider the circuit shown in Fig. P2.4-44. This circuit functions as an integrator. Assume ideal op-amp behavior and recall that - -*dVC*(*t*) - -- (a) Determine the differential equation that relates the input *x*(*t*) to the output *y*(*t*). -- (b) This circuit does not behave well at dc. Demonstrate this by computing the zero-state response *y*(*t*) for a unit step input *x*(*t*) = *u*(*t*). -- **2.4-45** Derive the result in Eq. (2.37) in another way. As mentioned in Ch. 1 (Fig. 1.27b), it is possible to express an input in terms of its step components, as shown in Fig. P2.4-45. Find the system response as a sum of the responses to the step components of the input. - -**2.4-46** Show that an LTIC system response to an everlasting sinusoid cosω0*t* is given by - -$$ -y(t) = |H(j\omega_0)| \cos [\omega_0 t + \angle H(j\omega_0)] -$$ - -where - -$$ -H(j\omega) = \int_{-\infty}^{\infty} h(t)e^{-j\omega t} dt -$$ - -assuming the integral on the right-hand side exists. - -**2.4-47** A line charge is located along the *x* axis with a charge density *Q*(*x*) coulombs per meter. Show that the electric field *E*(*x*) produced by this line charge at a point *x* is given by - -$$ -E(x) = Q(x) * h(x) -$$ - -where *h*(*x*) = 1/4π *x*2. [*Hint:* The charge over an interval τ located at τ = *n*τ is *Q*(*n*τ )τ . Also by Coulomb's law, the electric field *E*(*r*) at a distance *r* from a charge *q* coulombs is given by *E*(*r*) = *q*/4π *r*2.] - -- **2.4-48** A system is called complex if a real-valued input can produce a complex-valued output. Suppose a linear time-invariant complex system has impulse response *h*(*t*) = *j*[*u*(−*t* + 2) − *u*(−*t*)]. - - (a) Is this system causal? Explain. - - (b) Use convolution to determine the zero-state response *y*1(*t*) of this system in response to the unit-duration pulse *x*1(*t*) = *u*(*t*) − *u*(*t* − 1). - - (c) Using the result from part (a), determine the zero-state response *y*2(*t*) in response to *x*2(*t*) = 2*u*(*t* −1)−*u*(*t* −2)−*u*(*t* −3). -- **2.5-1** Explain, with reasons, whether the LTIC systems described by the following equations are (i) stable or unstable in the BIBO sense; (ii) asymptotically stable, unstable, or marginally stable. Assume that the systems are controllable and observable. - - (a) (*D*2 +8*D*+12)*y*(*t*) = (*D*−1)*x*(*t*) - - (b) *D*(*D*2 +3*D*+2)*y*(*t*) = (*D*+5)*x*(*t*) - - (c) *D*2(*D*2 +2)*y*(*t*) = *x*(*t*) - - (d) (*D*+1)(*D*2 −6*D*+5)*y*(*t*) = (3*D*+1)*x*(*t*) -- **2.5-2** Repeat Prob. 2.5-1 for the following: - - (a) (*D*+1)(*D*2 +2*D*+5)2*y*(*t*) = *x*(*t*) - - (b) (*D*+1)(*D*2 +9)*y*(*t*) = (2*D*+9)*x*(*t*) - - (c) (*D*+1)(*D*2 +9)2*y*(*t*) = (2*D*+9)*x*(*t*) - - (d) (*D*2 +1)(*D*2 +4)(*D*2 +9)*y*(*t*) = 3*Dx*(*t*) -- **2.5-3** Consider an LTIC system with unit impulse response *h*(*t*) = *et* 2 3 cos( 3 2 *t*) + 1 3 sin(π*t*) *u*(123 − *t*). Is this system BIBO-stable? Mathematically justify your answer. - -- **2.5-4** Consider an LTIC system with unit impulse response *h*(*t*) = 1 *t u*(*t* −*T*). - - (a) Determine, if possible, the value(s) of *T* for which this system is causal. - - (b) Determine, if possible, the value(s) of *T* for which this system is BIBO-stable. Justify all answers mathematically. -- **2.5-5** You are given the choice of a system that is guaranteed internally stable or a system that is guaranteed externally stable. Which do you choose? Why? -- **2.5-6** For a certain LTIC system, the impulse response *h*(*t*) = *u*(*t*). - - (a) Determine the characteristic root(s) of this system. - - (b) Is this system asymptotically or marginally stable, or is it unstable? - - (c) Is this system BIBO-stable? - - (d) What can this system be used for? -- **2.5-7** In Sec. 2.5 we demonstrated that for an LTIC system, the condition of Eq. (2.45) is sufficient for BIBO stability. Show that this is also a necessary condition for BIBO stability in such systems. In other words, show that if Eq. (2.45) is not satisfied, then there exists a bounded input that produces an unbounded output. [Hint: Assume that a system exists for which *h*(*t*) violates Eq. (2.45) and yet produces an output that is bounded for every bounded input. Establish the contradiction in this statement by considering an input *x*(*t*) defined by *x*(*t*1−τ )=1 when *h*(τ ) ≥ 0 and *x*(*t*1 − τ ) = −1 when *h*(τ ) < 0, where *t*1 is some fixed instant.] -- **2.5-8** An analog LTIC system with impulse response function *h*(*t*) = *u*(*t* + 2) − *u*(*t* − 2) is presented with an input *x*(*t*) = *t*(*u*(*t*) −*u*(*t* −2)). - - (a) Determine and plot the system output *y*(*t*) = *x*(*t*) ∗ *h*(*t*). - - (b) Is this system stable? Is this system causal? Justify your answers. -- **2.5-9** A system has an impulse response function shaped like a rectangular pulse, *h*(*t*) = *u*(*t*) − *u*(*t* − 1). Is the system stable? Is the system causal? -- **2.5-10** A continuous-time LTI system has impulse response function *h*(*t*) =%∞ *i*=0(0.5)*i* δ(*t*−*i*). - - (a) Is the system causal? Prove your answer. - - (b) Is the system stable? Prove your answer. - -- **2.6-1** Data at a rate of 1 million pulses per second are to be transmitted over a certain communications channel. The unit step response *g*(*t*) for this channel is shown in Fig. P2.6-1. - - (a) Can this channel transmit data at the required rate? Explain your answer. - - (b) Can an audio signal consisting of components with frequencies up to 15 kHz be transmitted over this channel with reasonable fidelity? - -**Figure P2.6-1** - -- **2.6-2** Determine a frequency ω that will cause the input *x*(*t*) = cos(ω*t*) to produce a strong response when applied to the system described by (*D*2 + 2*D* + 13/4){*y*(*t*)} = *x*(*t*). Carefully explain your choice. -- **2.6-3** Figure P2.6-3 shows the impulse response *h*(*t*) of a lowpass LTIC system. Determine the peak amplitude *A* and time constant *Th* so that rectangular impulse response *h*ˆ(*t*) is an appropriate approximation of *h*(*t*). The two graphs of Fig. P2.6-3 are not necessarily drawn to the same scale. - -**Figure P2.6-3** - -- **2.6-4** A certain communication channel has a bandwidth of 10 kHz. A pulse of 0.5 ms duration is transmitted over this channel. - - (a) Determine the width (duration) of the received pulse. - - (b) Find the maximum rate at which these pulses can be transmitted over this channel without interference between the successive pulses. - -- **2.6-5** A first-order LTIC system has a characteristic root λ = −104. - - (a) Determine *Tr*, the rise time of its unit step input response. - - (b) Determine the bandwidth of this system. - - (c) Determine the rate at which the information pulses can be transmitted through this system. -- **2.6-6** A lowpass system with a 6 MHz cutoff frequency needs to transmit data pulse that are 500 6 ns wide. Determine a suitable transmission rate *F*rate (pulses/s) for this system. -- **2.6-7** Sketch an impulse response *h*(*t*) of a non-causal LP system that has an approximate cutoff frequency of 5 kHz. Since many solutions are possible, be sure to properly justify your answer. -- **2.6-8** Two LTIC transmission channels are available: the first has impulse response *h*1(*t*) = *u*(*t*) − *u*(*t* − 1) and the second has impulse response *h*2(*t*) = δ(*t*) + 0.5δ(*t* − 1) + 0.25δ(*t* − 2). Explain which channel is better suited for the transmission of high-speed digital data (pulses). -- **2.6-9** Consider a linear time-invariant system with impulse response *h*(*t*) shown in Fig. P2.6-9. Outside the interval shown, *h*(*t*) = 0. - -### **Figure P2.6-9** - -- (a) What is the rise time *Tr* of this system? Remember, rise time is the time between the application of a unit step and the moment at which the system has "fully" responded. -- (b) Suppose *h*(*t*) represents the response of a communication channel. What conditions might cause the channel to have such an impulse response? What is the maximum - -average number of pulses per unit time that can be transmitted without causing interference? Justify your answer. - -- (c) Determine the system output *y*(*t*) = *x*(*t*) ∗ *h*(*t*) for *x*(*t*) = [*u*(*t* − 2) − *u*(*t*)]. Accurately sketch *y*(*t*) over (0 ≤ *t* ≤ 10). -- **2.6-10** A lowpass LTIC system has impulse response *h*(*t*) = −*te*−*t u*(*t*). - - (a) Accurately sketch *h*(*t*). - - (b) Describe a rectangular impulse response *h*ˆ(*t*) as an appropriate approximation of *h*(*t*). What is the approximate cutoff frequency of this system? -- **2.6-11** A lowpass LTIC system has impulse response *h*(*t*), as shown in Fig. P2.4-8. - - (a) As discussed in Sec. 2.6-2, determine a rectangular approximation *h*ˆ(*t*) to *h*(*t*). - - (b) Using *h*ˆ(*t*), what is the time constant *Th* of this lowpass system? - - (c) Using *h*ˆ(*t*), what is the approximate radian cutoff frequency ω*c* of this lowpass system? - - (d) Assuming a frequency ω0 ω*c*, what is the system response *y*(*t*) to the input *x*(*t*) = sin(ω0*t* +π/3)? -- **2.6-12** A first CT lowpass system with time constant *T*1 = 4 µs is put in series with a second CT lowpass system with time constant *T*2 = 2 µs. Make an educated sketch of the overall impulse response function *h*series(*t*). What is the time constant *T*series of the overall series-connected system? -- **2.7-1** An LTIC system with input *x*(*t*) and output *y*(*t*) is described by the following constant coefficient linear differential equation: - -$$ -(D4 - 16) \{y(t)\} = (D - 2) \{x(t)\}. -$$ - -- (a) What are the 4 characteristic roots of this system (λ1, λ2, λ3, and λ4)? Determine the roots by hand and then verify your answers using MATLAB's roots command. -- (b) From Eq. (2.17), computing *h*(*t*) requires a signal *y*˜*n*(*t*) = %4 *k*=1 *cke*λ*kt* . First, determine a matrix representation of the system of equations needed to solve for the four coefficients *ck*. Second, write MATLAB code that computes the length-4 column vector of coefficients *ck*. - -**2.7-2** Define *x*(*t*) = 2*u*(*t* + 2 3 ) − 2*u*(*t*). Further, define the periodic signal *h*1(*t*) as - -$$ -h_1(t) = \begin{cases} t & 0 \le t < 1 \\ h_1(t+1) & \forall t \end{cases} -$$ - -Lastly, define the aperiodic signal *h*2(*t*) in terms of *h*1(*t*) as - -$$ -h_2(t) = h_1(t)[u(t-1) - u(t-2)] -$$ - -- (a) Use MATLAB to plot *x*(*t*), *h*1(*t*), and *h*2(*t*) over the interval −2.5 ≤ *t* ≤ 3.5. -- (b) Using the graphical convolution procedure, compute *y*2(*t*) = *x*(*t*) ∗ *h*2(*t*). -- (c) Compute by hand and then MATLAB plot *y*1(*t*) = *x*(*t*) ∗ *h*1(*t*). Modify program CH2MP4.m in Sec. 2.7-4 to validate your analytical result. -- **2.7-3** Consider the circuit shown in Fig. P2.7-3. Assume ideal op-amp behavior and recall that - -$$ -i_C(t) = C \frac{dV_C(t)}{dt} -$$ - -Without a feedback resistor *Rf* , the circuit functions as an integrator and is unstable, particularly at dc. A feedback resistor *Rf* corrects this problem and results in a stable circuit that functions as a "lossy" integrator. - -**Figure P2.7-3** - -- (a) Determine the differential equation that relates the input *x*(*t*) to the output *y*(*t*). What is the corresponding characteristic equation? -- (b) To demonstrate that this "lossy" integrator is well behaved at dc, determine the zero-state - -response *y*(*t*) given a unit step input *x*(*t*) = *u*(*t*). - -- (c) Investigate the effect of 10% resistor and 25% capacitor tolerances on the system's characteristic root(s). -- **2.7-4** Consider the electric circuit shown in Fig. P2.7-4. Let *C*1 = *C*2 = 10 µF, *R*1 = *R*2 = 100 k, and *R*3 = 50 k. - - (a) Determine the corresponding differential equation describing this circuit. Is the circuit BIBO-stable? - - (b) Determine the zero-input response *y*0(*t*) if the output of each op amp initially reads one volt. - -- (c) Determine the zero-state response *y*(*t*) to a step input *x*(*t*) = *u*(*t*). -- (d) Investigate the effect of 10% resistor and 25% capacitor tolerances on the system's characteristic roots. -- **2.7-5** Input *x*(*t*)=3[*u*(*t*)−*u*(*t*−1)]+2[*u*(*t*−2)−*u*(*t*− 3)] is applied to a lowpass LTIC system with impulse response *h*(*t*) = (4 − *t*)[*u*(*t*) − *u*(*t* − 2)] to produce output *y*(*t*) = *x*(*t*) ∗ *h*(*t*). Modify program CH2MP4.m in Sec. 2.7-4 to perform the graphical convolution procedure to produce a plot of *y*(*t*). - - diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/038_3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/038_3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS.md deleted file mode 100644 index 77c602c46ffdf18daa919c5615fdc1803e567d39..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/038_3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS.md +++ /dev/null @@ -1,3 +0,0 @@ -# **TIME-DOMAIN ANALYSIS OF [DISCRETE-TIME](#page-9-0) SYSTEMS** - -In this chapter we introduce the basic concepts of discrete-time signals and systems. Furthermore, we explore the time-domain analysis of linear, time-invariant, discrete-time (LTID) systems. We show how to compute the zero-input response, determine the unit impulse response, and use convolution to evaluate the zero-state response. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/039_3.1 INTRODUCTION.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/039_3.1 INTRODUCTION.md deleted file mode 100644 index f68838901edc2790b65199fb915de36ac188068b..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/039_3.1 INTRODUCTION.md +++ /dev/null @@ -1,66 +0,0 @@ -## **[3.1 INTRODUCTION](#page-9-0)** - -A *discrete-time signal* is basically a sequence of numbers. Such signals arise naturally in inherently discrete-time situations such as population studies, amortization problems, national income models, and radar tracking. They may also arise as a result of sampling continuous-time signals in sampled data systems and digital filtering. Such signals can be denoted by *x*[*n*], *y*[*n*], and so on, where the variable *n* takes integer values, and *x*[*n*] denotes the *n*th number in the sequence labeled *x*. In this notation, the discrete-time variable *n* is enclosed in square brackets instead of parentheses, which we have reserved for enclosing continuous-time variables, such as *t*. - -Systems whose inputs and outputs are discrete-time signals are called *discrete-time systems*. A digital computer is a familiar example of this type of system. A discrete-time signal is a sequence of numbers, and a discrete-time system processes a sequence of numbers *x*[*n*] to yield another sequence *y*[*n*] as the output.† - -A discrete-time signal, when obtained by uniform sampling of a continuous-time signal *x*(*t*), can also be expressed as *x*(*nT*), where *T* is the sampling interval and *n*, the discrete variable taking on integer values. Thus, *x*(*nT*) denotes the value of the signal *x*(*t*) at *t* = *nT*. The signal *x*(*nT*) is a sequence of numbers (sample values), and hence, by definition, is a discrete-time signal. Such a signal can also be denoted by the customary discrete-time notation *x*[*n*], where *x*[*n*] = *x*(*nT*). A typical discrete-time signal is depicted in Fig. 3.1, which shows both forms of notation. By way of an example, a continuous-time exponential *x*(*t*) = *e*−*t* , when sampled every *T* = 0.1 seconds, results in a discrete-time signal *x*(*nT*) given by - -$$ -x(nT) = e^{-nT} = e^{-0.1n} -$$ - -† There may be more than one input and more than one output. - -**Figure 3.2** Processing a continuous-time signal by means of a discrete-time system. - -Clearly, this signal is a function of *n* and may be expressed as *x*[*n*]. Such representation is more convenient and will be followed throughout this book, even for signals resulting from sampling continuous-time signals. - -Digital filters can process continuous-time signals by discrete-time systems, using appropriate interfaces at the input and the output, as illustrated in Fig. 3.2. A continuous-time signal *x*(*t*) is first sampled to convert it into a discrete-time signal *x*[*n*], which is then processed by a discrete-time system to yield the output *y*[*n*]. A continuous-time signal *y*(*t*) is finally constructed from *y*[*n*]. We shall use the notations C/D and D/C for conversion from continuous to discrete time and from discrete to continuous time. By using the interfaces in this manner, we can use an appropriate discrete-time system to process a continuous-time signal. As we shall see later in our discussion, discrete-time systems have several advantages over continuous-time systems. For this reason, there is an accelerating trend toward processing continuous-time signals with discrete-time systems. - -### **[3.1-1 Size of a Discrete-Time Signal](#page-9-0)** - -Arguing along the lines similar to those used for continuous-time signals, the size of a discrete-time signal *x*[*n*] will be measured by its energy *Ex*, defined by - -$$ -E_x = \sum_{n=-\infty}^{\infty} |x[n]|^2 -$$ -\n(3.1) - -This definition is valid for real or complex *x*[*n*]. For this measure to be meaningful, the energy of a signal must be finite. A necessary condition for the energy to be finite is that the signal amplitude must → 0 as |*n*|→∞. Otherwise the sum in Eq. (3.1) will not converge. If *Ex* is finite, the signal is called an *energy signal*. - -In some cases, for instance, when the amplitude of *x*[*n*] does not → 0 as |*n*|→∞, then the signal energy is infinite, and a more meaningful measure of the signal in such a case would be the time average of the energy (if it exists), which is the signal power *Px*, defined by - -$$ -P_{x} = \lim_{N \to \infty} \frac{1}{2N + 1} \sum_{-N}^{N} |x[n]|^{2} -$$ - -In this equation, the sum is divided by 2*N* + 1 because there are 2*N* + 1 samples in the interval from −*N* to *N*. For periodic signals, the time averaging need be performed over only one period in view of the periodic repetition of the signal. If *Px* is finite and nonzero, the signal is called a *power signal*. As in the continuous-time case, a discrete-time signal can either be an energy signal or a power signal, but cannot be both at the same time. Some signals are neither energy nor power signals. - -### **EXAMPLE 3.1 Computing DT Energy and Power** - -Find the energy of the signal *x*[*n*] = *n*(*u*[*n*] − *u*[*n* − 6]), shown in Fig. 3.3a and the power for the periodic signal *y*[*n*] in Fig. 3.3b. - -By definition, - -$$ -E_x = \sum_{n=0}^{5} n^2 = 55 -$$ - -A periodic signal *x*[*n*] with period *N*0 is characterized by the fact that - -$$ -x[n] = x[n+N_0] -$$ - -The smallest value of *N*0 for which the preceding equation holds is the *fundamental period*. Such a signal is called *N*0 *periodic*. Figure 3.3b shows an example of a periodic signal *y*[*n*] of period *N*0 = 6 because each period contains 6 samples. Note that if the first sample is taken at *n* = 0, the last sample is at *n* = *N*0 − 1 = 5, not at *n* = *N*0 = 6. Because the signal *y*[*n*] is periodic, its power *Py* can be found by averaging its energy over one period. Averaging the energy over one period, we obtain - -$$ -P_{y} = \frac{1}{6} \sum_{n=0}^{5} n^{2} = \frac{55}{6} -$$ - -**Figure 3.3 (a)** Energy and **(b)** power computations for a signal. - -### **DR ILL 3.1 DT Signal Classification: Energy, Power, and Neither** - -Show that the signal *x*[*n*] = *anu*[*n*] is an energy signal of energy *Ex* = 1/(1 − |*a*| 2) if |*a*| < 1, that it is a power signal of power *Px* = 0.5 if |*a*| = 1, and that it is neither an energy signal nor a power signal if |*a*| > 1. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/040_3.2 USEFUL SIGNAL OPERATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/040_3.2 USEFUL SIGNAL OPERATIONS.md deleted file mode 100644 index bb2339360f801fef21188ed107d7f5cb53aed778..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/040_3.2 USEFUL SIGNAL OPERATIONS.md +++ /dev/null @@ -1,87 +0,0 @@ -## **3.2 USEFUL SIGNAL [OPERATIONS](#page-9-0)** - -Signal operations for *shifting,* and *scaling,* as discussed for continuous-time signals also apply, with some modifications, to discrete-time signals. - -### SHIFTING - -Consider a signal *x*[*n*] (Fig. 3.4a) and the same signal delayed (right-shifted) by 5 units (Fig. 3.4b), which we shall denote by *xs*[*n*]. † Using the argument employed for a similar operation in continuous-time signals (Sec. 1.2), we obtain - -$$ -x_s[n] = x[n-5] -$$ - -Therefore, to shift a sequence by *M* units (*M* integer), we replace *n* with *n* − *M*. Thus *x*[*n* − *M*] represents *x*[*n*] shifted by *M* units. If *M* is positive, the shift is to the right (delay). If *M* is negative, the shift is to the left (advance). Accordingly, *x*[*n* − 5] is *x*[*n*] delayed (right-shifted) by 5 units, and *x*[*n*+5] is *x*[*n*] advanced (left-shifted) by 5 units. - -† The terms "delay" and "advance" are meaningful only when the independent variable is time. For other independent variables, such as frequency or distance, it is more appropriate to refer to the "right shift" and "left shift" of a sequence. - -**Figure 3.4** Shifting and time reversal of a signal. - -## **DR ILL 3.2 Left-Shift Operation** - -Show that *x*[*n*] in Fig. 3.4a left-shifted by 3 units can be expressed as 0.729(0.9)*n* for 0 ≤ *n* ≤ 7, and zero otherwise. Sketch the shifted signal. - -### **DR ILL 3.3 Right-Shift Operation** - -Show that *x*[−*k* − *n*] can be obtained from *x*[*n*] by first right-shifting *x*[*n*] by *k* units and then time-reversing this shifted signal. - -### TIME REVERSAL - -To time-reverse *x*[*n*] in Fig. 3.4a, we rotate *x*[*n*] about the vertical axis to obtain the time-reversed signal *xr*[*n*] shown in Fig. 3.4c. Using the argument employed for a similar operation in continuous-time signals (Sec. 1.2), we obtain - -$$ -x_r[n] = x[-n] -$$ - -Therefore, to time-reverse a signal, we replace *n* with −*n* so that *x*[−*n*] is the time-reversed *x*[*n*]. For example, if *x*[*n*] = (0.9)*n* for 3 ≤ *n* ≤ 10, then *xr*[*n*] = (0.9)−*n* for 3 ≤ −*n* ≤ 10; that is, −3 ≥ *n* ≥ −10, as shown in Fig. 3.4c. - -The origin *n* = 0 is the anchor point, which remains unchanged under time-reversal operation because at *n* = 0, *x*[*n*] = *x*[−*n*] = *x*[0]. Note that while the reversal of *x*[*n*] about the vertical axis is *x*[−*n*], the reversal of *x*[*n*] about the horizontal axis is −*x*[*n*]. - -### **EXAMPLE 3.2 Time Reversal and Shifting** - -In the convolution operation, discussed later, we need to find the function *x*[*k* −*n*] from *x*[*n*]. - -This can be done in two steps: (i) time-reverse the signal *x*[*n*] to obtain *x*[−*n*]; (ii) now, right-shift *x*[−*n*] by *k*. Recall that right-shifting is accomplished by replacing *n* with *n* − *k*. Hence, right-shifting *x*[−*n*] by *k* units is *x*[−(*n* − *k*)] = *x*[*k* − *n*]. Figure 3.4d shows *x*[5 − *n*], obtained this way. We first time-reverse *x*[*n*] to obtain *x*[−*n*] in Fig. 3.4c. Next, we shift *x*[−*n*] by *k* = 5 to obtain *x*[*k* −*n*] = *x*[5−*n*], as shown in Fig. 3.4d. - -In this particular example, the order of the two operations employed is interchangeable. We can first left-shift *x*[*k*] to obtain *x*[*n* + 5]. Next, we time-reverse *x*[*n* + 5] to obtain *x*[−*n* + 5] = *x*[5 − *n*]. The reader is encouraged to verify that this procedure yields the same result, as in Fig. 3.4d. - -### **DR ILL 3.4 Time Reversal** - -Sketch the signal *x*[*n*] = *e*−0.5*n* for −3 ≤ *n* ≤ 2, and zero otherwise. Sketch the corresponding time-reversed signal and show that it can be expressed as *xr*[*n*] = *e*0.5*n* for −2 ≤ *n* ≤ 3. - -### SAMPLING RATE ALTERATION: DOWNSAMPLING, UPSAMPLING, AND INTERPOLATION - -Alteration of the sampling rate is somewhat similar to time-scaling in continuous-time signals. Consider a signal *x*[*n*] compressed by factor *M*. Compressing a signal *x*[*n*] by factor *M* yields *xd*[*n*] given by - -$$ -x_d[n] = x[Mn] -$$ - -Because of the restriction that discrete-time signals are defined only for integer values of the argument, we must restrict *M* to integer values. The values of *x*[*Mn*] at *n* = 0, 1, 2, 3,... are *x*[0], *x*[*M*], *x*[2*M*], *x*[3*M*], ... . This means *x*[*Mn*] selects every *M*th sample of *x*[*n*] and deletes all the samples in between. It reduces the number of samples by factor *M*. If *x*[*n*] is obtained by sampling a continuous-time signal, this operation implies reducing the sampling rate by factor *M*. For this reason, this operation is commonly called *downsampling*. Figure 3.5a shows a signal *x*[*n*] and Fig. 3.5b shows the signal *x*[2*n*], which is obtained by deleting odd-numbered samples of *x*[*n*]. † - -In the continuous-time case, time compression merely speeds up the signal without loss of any data. In contrast, downsampling *x*[*n*] generally causes loss of data. Under certain conditions—for example, if *x*[*n*] is the result of oversampling some continuous-time signal—then *xd*[*n*] may still retain the complete information about *x*[*n*]. - -An *interpolated* signal is generated in two steps; first, we expand *x*[*n*] by an integer factor *L* to obtain the expanded signal *xe*[*n*], as - -$$ -x_e[n] = \begin{cases} x[n/L] & n = 0, \pm L \pm 2L, \dots, \\ 0 & \text{otherwise} \end{cases} \tag{3.2} -$$ - -To understand this expression, consider a simple case of expanding *x*[*n*] by a factor 2 (*L* = 2). When *n* is odd, *n*/2 is noninteger, and *xe*[*n*] = 0. That is, *xe*[1] = *xe*[3] = *xe*[5],... are all zero, as depicted in Fig. 3.5c. Moreover, *n*/2 is integer for even *n*, and the values of *xe*[*n*] = *x*[*n*/2] for *n* = 0, 2, 4, 6,..., are *x*[0], *x*[1], *x*[2], *x*[3], ... , as shown in Fig. 3.5c. In general, for *n* = 0, 1, 2,..., *xe*[*n*] is given by the sequence - -$$ -x[0], \underbrace{0,0,\ldots,0,0}_{L-1 \text{ zeros}}, x[1], \underbrace{0,0,\ldots,0,0}_{L-1 \text{ zeros}}, x[2], \underbrace{0,0,\ldots,0,0}_{L-1 \text{ zeros}},\ldots -$$ - -Thus, the sampling rate of *xe*[*n*] is *L* times that of *x*[*n*]. Hence, this operation is commonly called *upsampling*. The upsampled signal *xe*[*n*] contains all the data of *x*[*n*], although in an expanded form. - -In the expanded signal in Fig. 3.5c, the missing (zero-valued) odd-numbered samples can be reconstructed from the non-zero-valued samples by using some suitable interpolation formula. Figure 3.5d shows such an interpolated signal *xi*[*n*], where the missing samples are constructed by using an interpolating filter. The optimum interpolating filter is usually an ideal lowpass - -† Odd-numbered samples of *x*[*n*] can be retained (and even-numbered samples deleted) by using the transformation *xd*[*n*] = *x*[2*n*+1]. - -**Figure 3.5** Compression (downsampling) and expansion (upsampling, interpolation) of a signal. - -filter, which is realizable only approximately. In practice, we may use an interpolation that is nonoptimum but realizable. The process of filtering to interpolate the zero-valued samples is called *interpolation*. Since the interpolated data are computed from the existing data, interpolation does not result in gain of information. While further discussion of interpolation is beyond our scope, Drill 3.5 and Prob. 3.11-10 introduce the idea of linear interpolation. - -### **DR ILL 3.5 Expansion and Interpolation** - -A signal *x*[*n*] is expanded by factor 2 to obtain signal *x*[*n*/2]. The odd-numbered samples (*n* odd) in this signal have zero value. Show that the linearly interpolated odd-numbered samples are given by *xi*[*n*] = (1/2){*x*[*n*−1] +*x*[*n*+1]}. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/041_3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/041_3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS.md deleted file mode 100644 index d0b3c5b6e99d4e9d05e197877b309ce2f3b25b9f..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/041_3.3 SOME USEFUL DISCRETE-TIME SIGNAL MODELS.md +++ /dev/null @@ -1,181 +0,0 @@ -## **3.3 SOME USEFUL [DISCRETE-TIME](#page-9-0) SIGNAL MODELS** - -We now discuss some important discrete-time signal models that are encountered frequently in the study of discrete-time signals and systems. - -## **[3.3-1 Discrete-Time Impulse Function](#page-9-0)** *δ***[***n***]** - -The discrete-time counterpart of the continuous-time impulse function δ(*t*) is δ[*n*], a Kronecker delta function, defined by - -$$ -\delta[n] = \begin{cases} 1 & n = 0 \\ 0 & n \neq 0 \end{cases} -$$ - -This function, also called the unit impulse sequence, is shown in Fig. 3.6a. The shifted impulse sequence δ[*n* − *m*] is depicted in Fig. 3.6b. Unlike its continuous-time counterpart δ(*t*) (the Dirac delta), the Kronecker delta is a very simple function, requiring no special esoteric knowledge of distribution theory. - -**Figure 3.6** Discrete-time impulse function: **(a)** unit impulse sequence and **(b)** shifted impulse sequence. - -## **[3.3-2 Discrete-Time Unit Step Function](#page-9-0)** *u***[***n***]** - -The discrete-time counterpart of the unit step function *u*(*t*) is *u*[*n*] (Fig. 3.7a), defined by - -$$ -u[n] = \begin{cases} 1 & \text{for } n \ge 0 \\ 0 & \text{for } n < 0 \end{cases} -$$ - -If we want a signal to start at *n* = 0 (so that it has a zero value for all *n* < 0), we need only multiply the signal by *u*[*n*]. - -**Figure 3.7 (a)** A discrete-time unit step function *u*[*n*] and **(b)** its application. - -### **EXAMPLE 3.3 Describing Signals with Unit Step and Unit Impulse Functions** - -Describe the signal *x*[*n*] shown in Fig. 3.7b by a single expression valid for all *n*. - -The signal *x*[*n*] can be broken into three components: (1) a ramp component *x*1[*n*] from *n* = 0 to 4, (2) a scaled step component *x*2[*n*] from *n* = 5 to 10, and (3) an impulse component *x*3[*n*] represented by the negative spike at *n* = 8. Let us consider each one separately. - -We express *x*1[*n*] = *n*(*u*[*n*]−*u*[*n*−5]) to account for the signal from *n* = 0 to 4. Assuming that the spike at *n* = 8 does not exist, we can express *x*2[*n*] = 4(*u*[*n*−5] −*u*[*n*−11]) to account for the signal from *n* = 5 to 10. Once these two components have been added, the only part that is unaccounted for is a spike of amplitude −2 at *n* = 8, which can be represented by - -There are many different ways of viewing *x*[*n*]. Although each way of viewing yields a different expression, they are all equivalent. We shall consider here just one possible expression. - -*x*3[*n*]=−2δ[*n*−8]. Hence, - -$$ -x[n] = x_1[n] + x_2[n] + x_3[n] -$$ - -= $n(u[n] - u[n-5]) + 4(u[n-5] - u[n-11]) - 2\delta[n-8]$ for all *n* - -We stress again that the expression is valid for all values of *n*. The reader can find several other equivalent expressions for *x*[*n*]. For example, one may consider a scaled step function from *n* = 0 to 10, subtract a ramp over the range *n* = 0 to 3, and subtract the spike. You can also play with breaking *n* into different ranges for your expression. - -## **[3.3-3 Discrete-Time Exponential](#page-9-0)** *γ n* - -A continuous-time exponential *e*λ*t* can be expressed in an alternate form as - -$$ -e^{\lambda t} = \gamma^t \qquad (\gamma = e^{\lambda} \text{ or } \lambda = \ln \gamma) -$$ - -For example, *e*−0.3*t* = (0.7408)*t* because *e*−0.3 = 0.7408. Conversely, 4*t* = *e*1.386*t* because *e*1.386 = 4, that is, ln 4 = 1.386. In the study of continuous-time signals and systems, we prefer the form *e*λ*t* rather than γ *t* . In contrast, the exponential form γ *n* is preferable in the study of discrete-time signals and systems, as will become apparent later. The discrete-time exponential γ *n* can also be expressed by using a natural base, as - -$$ -e^{\lambda n} = \gamma^n \qquad (\gamma = e^{\lambda} \text{ or } \lambda = \ln \gamma) -$$ - -Because of unfamiliarity with exponentials with bases other than *e*, exponentials of the form γ *n* may seem inconvenient and confusing at first. The reader is urged to plot some exponentials to acquire a sense of these functions. Also observe that γ −*n* = 1 γ *n* . - -### **DR ILL 3.6 Equivalent Forms of DT Exponentials** - -**(a)** Show that (i) (0.25)−*n* = 4*n*, (ii) 4−*n* = (0.25)*n*, (iii) *e*2*t* = (7.389)*t* , (iv) *e*−2*t* = (0.1353)*t* = (7.389)−*t* , (v) *e*3*n* = (20.086)*n*, and (vi) *e*−1.5*n* = (0.2231)*n* = (4.4817)−*n*. **(b)** Show that (i) 2*n* = *e*0.693*n*, (ii) (0.5)*n* = *e*−0.693*n*, and (iii) (0.8)−*n* = *e*0.2231*n*. - -**Nature of** *γ n***.** The signal *e*λ*n* grows exponentially with *n* if Reλ > 0 (λ in the RHP), and decays exponentially if Reλ < 0 (λ in the LHP). It is constant or oscillates with constant amplitude if Reλ = 0 (λ on the imaginary axis). Clearly, the location of λ in the complex plane indicates whether the signal *e*λ*n* will grow exponentially, decay exponentially, or oscillate with constant - -**Figure 3.8** The λ plane, the γ plane, and their mapping. - -amplitude (Fig. 3.8a). A constant signal (λ = 0) is also an oscillation with zero frequency. We now find a similar criterion for determining the nature of γ *n* from the location of γ in the complex plane. - -Figure 3.8a shows a complex plane (λ plane). Consider a signal *ejn*. In this case, λ = *j* lies on the imaginary axis (Fig. 3.8a), and therefore is a constant-amplitude oscillating signal. This signal *ejn* can be expressed as γ *n*, where γ = *ej*. Because the magnitude of *ej* is unity, |γ | = 1. Hence, when λ lies on the imaginary axis, the corresponding γ lies on a circle of unit radius, centered at the origin (the *unit circle* illustrated in Fig. 3.8b). Therefore, a signal γ *n* oscillates with constant amplitude if γ lies on the unit circle. Thus, the imaginary axis in the λ plane maps into the unit circle in the γ plane. - -Next consider the signal *e*λ*n*, where λ lies in the left half-plane in Fig. 3.8a. This means λ = *a* + *jb*, where *a* is negative (*a* < 0). In this case, the signal decays exponentially. This signal can be expressed as γ *n*, where - -$$ -\gamma = e^{\lambda} = e^{a+jb} = e^a e^{jb} -$$ - -and - -$$ -|\gamma| = |e^a| \, |e^{ib}| = e^a -$$ - because $|e^{ib}| = 1$ - -Also, *a* is negative (*a* < 0). Hence, |γ | = *ea* < 1. This result means that the corresponding γ lies inside the unit circle. Therefore, a signal γ *n* decays exponentially if γ lies within the unit circle (Fig. 3.8b). If, in the preceding case we select *a* to be positive (λ in the right half-plane), then |γ | > 1, and γ lies outside the unit circle. Therefore, a signal γ *n* grows exponentially if γ lies outside the unit circle (Fig. 3.8b). - -To summarize, the imaginary axis in the λ plane maps into the unit circle in the γ plane. The left half-plane in the λ plane maps into the inside of the unit circle and the right half of the λ plane maps into the outside of the unit circle in the γ plane, as depicted in Fig. 3.8. - -**Figure 3.9** Discrete-time exponentials γ *n*. - -Plots of (0.8)*n* and (−0.8)*n* appear in Figs. 3.9a and 3.9b, respectively. Plots of (0.5)*n* and (1.1)*n* appear in Figs. 3.9c and 3.9d, respectively. These plots verify our earlier conclusions about the location of γ and the nature of signal growth. Observe that a signal (−|γ |)*n* alternates sign successively (is positive for even values of *n* and negative for odd values of *n*, as depicted in Fig. 3.9b). Also, the exponential (0.5)*n* decays faster than (0.8)*n* because 0.5 is closer to the origin than 0.8. The exponential (0.5)*n* can also be expressed as 2−*n* because (0.5)−1 = 2. - -### **DR ILL 3.7 Sketching DT Exponentials** - -Sketch the following signals: **(a)** (1)*n*, **(b)** (−1)*n*, **(c)** (0.5)*n*, **(d)** (−0.5)*n*, **(e)** (0.5)−*n*, **(f)** 2−*n*, and **(g)** (−2)*n*. Express these exponentials as γ *n*, and plot γ in the complex plane for each case. Verify that γ *n* decays exponentially with *n* if γ lies inside the unit circle and that γ *n* grows with *n* if γ is outside the unit circle. If γ is on the unit circle, γ *n* is constant or oscillates with a constant amplitude. - -### 250 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Accurately hand-sketching DT signals can be tedious and difficult. As the next example shows, MATLAB is particularly well suited to plot DT signals, including exponentials. - -### **EXAMPLE 3.4 Plotting DT Exponentials with MATLAB** - -Use MATLAB to plot the following discrete-time signals over (0 ≤ *n* ≤ 8): **(a)** *xa*[*n*] = (0.8)*n*, **(b)** *xb*[*n*] = (−0.8)*n*, **(c)** *xc*[*n*] = (0.5)*n*, and **(d)** *xd*[*n*] = (1.1)*n*. - -To begin, we use anonymous functions to represent each of the four signals. Next, we plot these functions over the desired range of *n*. The results, shown in Fig. 3.10, match the earlier Fig. 3.9 plots of the same signals. - -``` ->> n = (0:8); x_a = @(n) (0.8).^n; x_b = @(n) (-0.8).^(n); ->> x_c = @(n) (0.5).^n; x_d = @(n) (1.1).^n; ->> subplot(2,2,1); stem(n,x_a(n),'k'); ylabel('x_a[n]'); xlabel('n'); ->> subplot(2,2,2); stem(n,x_b(n),'k'); ylabel('x_b[n]'); xlabel('n'); ->> subplot(2,2,3); stem(n,x_c(n),'k'); ylabel('x_c[n]'); xlabel('n'); ->> subplot(2,2,4); stem(n,x_d(n),'k'); ylabel('x_d[n]'); xlabel('n'); -``` - -## **[3.3-4 Discrete-Time Sinusoid](#page-9-0) cos***(n* **+***θ )* - -A general discrete-time sinusoid can be expressed as *C*cos(*n*+θ ), where *C* is the *amplitude,* and θ is the *phase* in radians. Also, *n* is an angle in radians. Hence, the dimensions of the frequency are *radians per sample*. This sinusoid may also be expressed as - -$$ -C\cos\left(\Omega n + \theta\right) = C\cos\left(2\pi\mathcal{F}n + \theta\right) -$$ - -where *F* = /2π. Therefore, the dimensions of the discrete-time frequency *F* are (radians/2π) per sample, which is equal to *cycles per sample*. This means if *N*0 is the period (samples/cycle) of the sinusoid, then the frequency of the sinusoid *F* = 1/*N*0 (samples/cycle). - -Figure 3.11 shows a discrete-time sinusoid cos( π 12 *n* + π 4 ). For this case, the frequency is = π/12 radians/sample. Alternately, the frequency is *F* = 1/24 cycles/sample. In other words, there are 24 samples in one cycle of the sinusoid. - -Because cos(−*x*) = cos(*x*), - -$$ -\cos\left(-\Omega n + \theta\right) = \cos\left(\Omega n - \theta\right) -$$ - -This shows that both cos(*n*+θ ) and cos(−*n*+θ ) have the same frequency (). Therefore, *the frequency of* cos(*n*+θ ) *is* ||. - -**Figure 3.11** A discrete-time sinusoid cos( π 12 *n*+ π 4 ). - -### SAMPLED CONTINUOUS-TIME SINUSOID YIELDS A DISCRETE-TIME SINUSOID - -A continuous-time sinusoid cosω*t* sampled every *T* seconds yields a discrete-time sequence whose *n*th element (at *t* = *nT*) is cosω*nT*. Thus, the sampled signal *x*[*n*] is given by - -$$ -x[n] = \cos \omega nT = \cos \Omega n \quad \text{where } \Omega = \omega T -$$ - -### 252 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Thus, a continuous-time sinusoid cosω*t* sampled every *T* seconds yields a discrete-time sinusoid cos*n*, where = ω*T*. † - -## **[3.3-5 Discrete-Time Complex Exponential](#page-9-0)** *ejn* - -Using Euler's formula, we can express an exponential *ejn* in terms of sinusoids as - -*ejn* = (cos*n*+*j*sin*n*) and *e*−*jn* = (cos*n*−*j*sin*n*) - -These equations show that *the frequency of both ejn and e*−*jn is* (radians/sample). Therefore, the frequency of *ejn* is ||. - -Observe that for *r* = 1 and θ = *n*, - -$$ -e^{j\Omega n}=re^{j\theta} -$$ - -This equation shows that the magnitude and angle of *ejn* are 1 and *n*, respectively. In the complex plane, *ejn* is a point on a unit circle at an angle *n*. - -### **EXAMPLE 3.5 Plotting a DT Sinusoid with MATLAB** - -Using MATLAB, plot the discrete-time sinusoid *x*[*n*] = cos π 12 *n*+ π 4 . - -We represent the desired sinusoid using an anonymous function. Next, we plot this function over the desired range of *n*. The result, shown in Fig. 3.12, matches the plot of the same signal shown in Fig. 3.11. - ->> n = (-30:30); x = @(n) cos(n\*pi/12+pi/4); >> clf; stem(n,x(n),'k'); ylabel('x[n]'); xlabel('n'); - -† Superficially, it may appear that a discrete-time sinusoid is a continuous-time sinusoid's cousin in a striped suit. However, some of the properties of discrete-time sinusoids are very different from those of continuous-time sinusoids. For instance, not every discrete-time sinusoid is periodic. A sinusoid cos*n* is periodic only if is a rational multiple of 2π. Also, discrete-time sinusoids are bandlimited to = π. Any sinusoid with ≥ π can always be expressed as a sinusoid of some frequency ≤ π. These peculiar properties are the direct consequence of the fact that the period of a discrete-time sinusoid must be an integer. These topics are discussed in Chs. 5 and 9. - - diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/042_3.4 EXAMPLES OF DISCRETE-TIME SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/042_3.4 EXAMPLES OF DISCRETE-TIME SYSTEMS.md deleted file mode 100644 index bdc4cbfab1184e2297a124c7b45f7126d00ae673..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/042_3.4 EXAMPLES OF DISCRETE-TIME SYSTEMS.md +++ /dev/null @@ -1,370 +0,0 @@ -## **[3.4 EXAMPLES OF](#page-9-0) DISCRETE-TIME SYSTEMS** - -We shall give here four examples of discrete-time systems. In the first two examples, the signals are inherently of the discrete-time variety. In the third and fourth examples, a continuous-time signal is processed by a discrete-time system, as illustrated in Fig. 3.2, by discretizing the signal through sampling. - -### **EXAMPLE 3.6 Savings Account** - -A person makes a deposit (the input) in a bank regularly at an interval of *T* (say, 1 month). The bank pays a certain interest on the account balance during the period *T* and mails out a periodic statement of the account balance (the output) to the depositor. Find the equation relating the output *y*[*n*] (the balance) to the input *x*[*n*] (the deposit). - -In this case, the signals are inherently discrete time. Let - -*x*[*n*] = deposit made at the *n*th discrete instant - -*y*[*n*] = account balance at the *n*th instant computed - -immediately after receipt of the *n*th deposit *x*[*n*] - -*r* = interest per dollar per period *T* - -The balance *y*[*n*] is the sum of (i) the previous balance *y*[*n* − 1], (ii) the interest on *y*[*n* − 1] during the period *T*, and (iii) the deposit *x*[*n*] - -$$ -y[n] = y[n-1] + ry[n-1] + x[n] -$$ - -= (1+r)y[n-1] + x[n] - -or - -$$ -y[n] - ay[n-1] = x[n] \qquad a = 1+r -$$ -\n(3.3) - -In this example the deposit *x*[*n*] is the input (cause) and the balance *y*[*n*] is the output (effect). - -### 254 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -A withdrawal from the account is a negative deposit. Therefore, this formulation can handle deposits as well as withdrawals. It also applies to a loan payment problem with the initial value *y*[0]=−*M*, where *M* is the amount of the loan. A loan is an initial deposit with a negative value. Alternately, we may treat a loan of *M* dollars taken at *n* = 0 as an input of −*M* at *n* = 0 (see Prob. 3.8-23). - -We can express Eq. (3.3) in an alternate form. The choice of index *n* in Eq. (3.3) is completely arbitrary, so we can substitute *n*+1 for *n* to obtain - -$$ -y[n+1] - ay[n] = x[n+1] -$$ -\n(3.4) - -We also could have obtained Eq. (3.4) directly by realizing that *y*[*n*+1], the balance at instant (*n* + 1), is the sum of *y*[*n*] plus *ry*[*n*] (the interest on *y*[*n*]) plus the deposit (input) *x*[*n* + 1] at instant (*n*+1). - -The difference equation in Eq. (3.3) uses delays, whereas the form in Eq. (3.4) uses advances. Thus, Eq. (3.3) is said to be in *delay form* and Eq. (3.4) is said to be in *advance form*. The delay form is more natural because operation of delay is causal, hence realizable. In contrast, advance operation, being noncausal, is unrealizable. We use the advance form primarily for its mathematical convenience over the delay form.† - -We shall now represent this system in a block diagram form, which is basically a road map to a hardware (or software) realization of the system. For this purpose, the causal (realizable) delay form in Eq. (3.3) will be used. There are three basic operations in this equation: *addition, scalar multiplication,* and *delay*. Figure 3.13 shows their schematic representation. In addition, we also have a *pickoff* node (Fig. 3.13d), which is used to provide multiple copies of a signal at its input. - -**Figure 3.13** Schematic representations of basic operations on sequences. - -† Use of the advance form results in discrete-time system equations that are identical in form to those for continuous-time systems. This will become apparent later. In transform analysis, advance form leads to the more convenient variable *z* instead of the clumsy *z*−1 that arises from delay form. - -**Figure 3.14** Realization of the savings account system. - -Figure 3.14 shows in block diagram form a system represented by Eq. (3.3). To understand this realization, it is helpful to rewrite Eq. (3.3) as *y*[*n*] = *ay*[*n* − 1] + *x*[*n*] (*a* = 1 + *r*). Now, assume that the output *y*[*n*] is available at the pickoff node *N*. Unit delay of *y*[*n*] results in *y*[*n* − 1], which is multiplied by a scalar of value *a* to yield *ay*[*n* − 1]. Next, we generate *y*[*n*] by adding the input *x*[*n*] and *ay*[*n* − 1]. † Observe that node *N* is a pickoff node, from which two copies of the output signal flow out: one as the feedback signal and the other as the output signal. - -### **EXAMPLE 3.7 Sales Estimate** - -During semester *n*, *x*[*n*] students enroll in a course requiring a certain textbook while the publisher sells *y*[*n*] new copies of the same book. On the average, one-quarter of students with books in salable condition resell the texts at the end of the semester, and the book life is three semesters. Write the equation relating *y*[*n*], the new books sold by the publisher, to *x*[*n*], the number of students enrolled in the *n*th semester, assuming that every student buys a book. - -In the *n*th semester, the total books *x*[*n*] sold to students must be equal to *y*[*n*] (new books from the publisher) plus the used books from students enrolled in the preceding two semesters (because the book life is only three semesters). There are *y*[*n* − 1] new books sold in semester (*n* − 1), and one-quarter of these books, that is, (1/4)*y*[*n* − 1], will be resold in the *n*th semester. Also, *y*[*n* − 2] new books are sold in semester *n* − 2, and one-quarter of these, that is, (1/4)*y*[*n* − 2], will be resold in semester (*n* − 1). Again, a quarter of these, that is, (1/16)*y*[*n* − 2], will be resold in the *n*th semester. Therefore, *x*[*n*] must be equal to the sum of *y*[*n*], (1/4)*y*[*n*−1], and (1/16)*y*[*n*−2]. - -$$ -y[n] + \frac{1}{4}y[n-1] + \frac{1}{16}y[n-2] = x[n] -$$ -\n(3.5) - -Equation (3.5) can also be expressed in an alternative form by realizing that this equation is valid for any value of *n*. Therefore, replacing *n* by *n*+2, we obtain - -$$ -y[n+2] + \frac{1}{4}y[n+1] + \frac{1}{16}y[n] = x[n+2] -$$ -\n(3.6) - -This is the alternative form of Eq. (3.5). - -† A unit delay represents 1 unit of time delay. In this example, 1 unit of delay in the output corresponds to period *T* for the actual output. - -### 256 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -To facilitate a realization of a system with this input–output equation, we rewrite the delay-form Eq. (3.5) as *y*[*n*]=−1 4 *y*[*n* − 1] − 1 16 *y*[*n* − 2] + *x*[*n*]. Figure 3.15 shows a corresponding hardware realization using two unit delays in cascade.† - -**Figure 3.15** Realization of the system representing sales estimate in Ex. 3.7. - -### **EXAMPLE 3.8 Digital Differentiator** - -Design a discrete-time system, like the one in Fig. 3.2, to differentiate continuous-time signals. This differentiator is used in an audio system having an input signal bandwidth below 20 kHz. - -In this case, the output *y*(*t*) is required to be the derivative of the input *x*(*t*). The discrete-time processor (system) *G* processes the samples of *x*(*t*) to produce the discrete-time output *y*[*n*]. Let *x*[*n*] and *y*[*n*] represent the samples *T* seconds apart of the signals *x*(*t*) and *y*(*t*), respectively, that is, - -$$ -x[n] = x(nT) \qquad \text{and} \qquad y[n] = y(nT) \tag{3.7} -$$ - -The signals *x*[*n*] and *y*[*n*] are the input and the output for the discrete-time system *G*. Now, we require that - -$$ -y(t) = \frac{dx(t)}{dt} -$$ - -Therefore, at *t* = *nT* (see Fig. 3.16a), - -$$ -y(nT) = \frac{dx(t)}{dt}\bigg|_{t=nT} = \lim_{T \to 0} \frac{1}{T} [x(nT) - x[(n - 1)T]] -$$ - -† The comments in the preceding footnote apply here also. Although 1 unit of delay in this example is one semester, we need not use this value in the hardware realization. Any value other than one semester results in a time-scaled output. - -**Figure 3.16** Digital differentiator and its realization. - -By using the notation in Eq. (3.7), the foregoing equation can be expressed as - -$$ -y[n] = \lim_{T \to 0} \frac{1}{T} \{x[n] - x[n-1]\} -$$ - -This is the input–output relationship for *G* required to achieve our objective. In practice, the sampling interval *T* cannot be zero. Assuming *T* to be sufficiently small, the equation just given can be expressed as - -$$ -y[n] = \frac{1}{T} \{x[n] - x[n-1]\} -$$ -\n(3.8) - -The approximation improves as *T* approaches 0. A discrete-time processor *G* to realize Eq. (3.8) is shown inside the shaded box in Fig. 3.16b. The system in Fig. 3.16b acts as a differentiator. This example shows how a continuous-time signal can be processed by a - -### 258 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -discrete-time system. The considerations for determining the sampling interval *T* are discussed in Chs. 5 and 8, where it is shown that to process frequencies below 20 kHz, the proper choice is - -$$ -T \le \frac{1}{2 \times \text{highest frequency}} = \frac{1}{40,000} = 25 \,\mu\text{s} -$$ - -To see how well this method works, let us consider the differentiator in Fig. 3.16b with a ramp input *x*(*t*) = *t*, depicted in Fig. 3.16c. If the system were to act as a differentiator, then the output *y*(*t*) of the system should be the unit step function *u*(*t*). Let us investigate how the system performs this particular operation and how well the system achieves the objective. - -The samples of the input *x*(*t*) = *t* at the interval of *T* seconds act as the input to the discrete-time system *G*. These samples, denoted by a compact notation *x*[*n*], are, therefore, - -$$ -x[n] = x(t)|_{t=nT} = t|_{t=nT} \qquad t \ge 0 -$$ - -= nT \qquad n \ge 0 - -Figure 3.16d shows the sampled signal *x*[*n*]. This signal acts as an input to the discrete-time system *G*. Figure 3.16b shows that the operation of *G* consists of subtracting a sample from the preceding (delayed) sample and then multiplying the difference with 1/*T*. From Fig. 3.16d, it is clear that the difference between the successive samples is a constant *nT* −(*n*−1)*T* = *T* for all samples, except for the sample at *n* = 0 (because there is no preceding sample at *n* = 0). The output of *G* is 1/*T* times the difference *T*, which is unity for all values of *n*, except *n* = 0, where it is zero. Therefore, the output *y*[*n*] of *G* consists of samples of unit values for *n* ≥ 1, as illustrated in Fig. 3.16e. The D/C (discrete-time to continuous-time) converter converts these samples into a continuous-time signal *y*(*t*), as shown in Fig. 3.16f. Ideally, the output should have been *y*(*t*) = *u*(*t*). This deviation from the ideal is due to our use of a nonzero sampling interval *T*. As *T* approaches zero, the output *y*(*t*) approaches the desired output *u*(*t*). - -The digital differentiator in Eq. (3.8) is an example of what is known as the *backward difference* system. The reason for calling it so is obvious from Fig. 3.16a. To compute the derivative of *y*(*t*), we are using the difference between the present sample value and the preceding (backward) sample value. If we use the difference between the next (forward) sample at *t* = (*n* + 1)*T* and the present sample at *t* = *nT*, we obtain a forward difference form of differentiator as - -$$ -y[n] = \frac{1}{T} \{x[n+1] - x[n]\} -$$ -\n(3.9) - -### **EXAMPLE 3.9 Digital Integrator** - -Design a digital integrator along the same lines as the digital differentiator in Ex. 3.8. - -For an integrator, the input *x*(*t*) and the output *y*(*t*) are related by - -$$ -y(t) = \int_{-\infty}^{t} x(\tau) d\tau -$$ - -### 3.4 Examples of Discrete-Time Systems 259 - -Therefore, at *t* = *nT* (see Fig. 3.16a), - -$$ -y(nT) = \lim_{T \to 0} \sum_{k=-\infty}^{n} x(kT)T -$$ - -Using the usual notation *x*(*kT*) = *x*[*k*], *y*(*nT*) = *y*[*n*], and so on, this equation can be expressed as - -$$ -y[n] = \lim_{T \to 0} T \sum_{k=-\infty}^{n} x[k] -$$ - -Assuming that *T* is small enough to justify the assumption *T* → 0, we have - -$$ -y[n] = T \sum_{k=-\infty}^{n} x[k] -$$ - (3.10) - -This equation represents an example of *accumulator* system. This digital integrator equation can be expressed in an alternate form. From Eq. (3.10), it follows that - -$$ -y[n] - y[n-1] = Tx[n] -$$ -\n(3.11) - -This is an alternate description for the digital integrator. Equations (3.10) and (3.11) are equivalent; the one can be derived from the other. Observe that the form of Eq. (3.11) is similar to that of Eq. (3.3). Hence, the block diagram representation of a digital integrator in the form of Eq. (3.11) is identical to that in Fig. 3.14 with *a* = 1 and the input multiplied by *T*. - -## RECURSIVE AND NONRECURSIVE FORMS OF DIFFERENCE EQUATION - -If Eq. (3.11) expresses Eq. (3.10) in another form, what is the difference between these two forms? Which form is preferable? To answer these questions, let us examine how the output is computed by each of these forms. In Eq. (3.10), the output *y*[*n*] at any instant *n* is computed by adding all the past input values till *n*. This can mean a large number of additions. In contrast, Eq. (3.11) can be expressed as *y*[*n*] = *y*[*n*−1] +*Tx*[*n*]. Hence, computation of *y*[*n*] involves addition of only two values: the preceding output value *y*[*n* − 1] and the present input value *x*[*n*]. The computations are done recursively by using the preceding output values. For example, if the input starts at *n* = 0, we first compute *y*[0]. Then we use the computed value *y*[0] to compute *y*[1]. Knowing *y*[1], we compute *y*[2], and so on. The computations are recursive. This is why the form of Eq. (3.11) is called *recursive* form and the form of Eq. (3.10) is called *nonrecursive* form. Clearly, "recursive" and "nonrecursive" describe two different ways of presenting the same information. Equations (3.3), (3.5), and (3.11) are examples of recursive form, and Eqs. (3.8) and (3.10) are examples of nonrecursive form. - -### KINSHIP OF DIFFERENCE EQUATIONS TO DIFFERENTIAL EQUATIONS - -We now show that a digitized version of a differential equation results in a difference equation. Let us consider a simple first-order differential equation - -$$ -\frac{dy(t)}{dt} + cy(t) = x(t) -$$ -\n(3.12) - -Consider uniform samples of *x*(*t*) at intervals of *T* seconds. As usual, we use the notation *x*[*n*] to denote *x*(*nT*), the *n*th sample of *x*(*t*). Similarly, *y*[*n*] denotes *y*[*nT*], the *n*th sample of *y*(*t*). From the basic definition of a derivative, we can express Eq. (3.12) at *t* = *nT* as - -$$ -\lim_{T \to 0} \frac{y[n] - y[n-1]}{T} + cy[n] = x[n] -$$ - -Clearing the fractions and rearranging the terms yield (assuming nonzero, but very small *T*) - -$$ -y[n] + \alpha y[n-1] = \beta x[n] -$$ -\n(3.13) - -where - -$$ -\alpha = \frac{-1}{1 + cT} \quad \text{and} \quad \beta = \frac{T}{1 + cT} -$$ - -We can also express Eq. (3.13) in advance form as - -$$ -y[n+1] + \alpha y[n] = \beta x[n+1] -$$ - -It is clear that a differential equation can be approximated by a difference equation of the same order. In this way, we can approximate an *n*th-order differential equation by a difference equation of *n*th order. Indeed, a digital computer solves differential equations by using an equivalent difference equation, which can be solved by means of simple operations of addition, multiplication, and shifting. Recall that a computer can perform only these simple operations. It must necessarily approximate complex operation like differentiation and integration in terms of such simple operations. The approximation can be made as close to the exact answer as possible by choosing sufficiently small value for *T*. - -At this stage, we have not developed tools required to choose a suitable value of the sampling interval *T*. This subject is discussed in Ch. 5 and also in Ch. 8. In Sec. 5.7, we shall discuss a systematic procedure (impulse invariance method) for finding a discrete-time system with which to realize an *N*th-order LTIC system. - -### ORDER OF A DIFFERENCE EQUATION - -Equations (3.3), (3.5), (3.9), (3.11), and (3.13) are examples of difference equations. The highest-order difference of the output signal or the input signal, whichever is higher, represents the *order* of the difference equation. Hence, Eqs. (3.3), (3.9), (3.11), and (3.13) are first-order difference equations, whereas Eq. (3.5) is of the second order. - -### **DR ILL 3.8 Digital Integrator Design** - -Design a digital integrator in Ex. 3.9 using the fact that for an integrator, the output *y*(*t*) and the input *x*(*t*) are related by *dy*(*t*)/*dt* = *x*(*t*). Approximation (similar to that in Ex. 3.8) of this equation at *t* = *nT* yields the recursive form in Eq. (3.11). - -### ANALOG, DIGITAL, CONTINUOUS-TIME, AND DISCRETE-TIME SYSTEMS - -The basic difference between continuous-time systems and analog systems, as also between discrete-time and digital systems, is fully explained in Secs. 1.7-5 and 1.7-6.† Historically, discrete-time systems have been realized with digital computers, where continuous-time signals are processed through digitized samples rather than unquantized samples. Therefore, the terms *digital filters* and *discrete-time systems* are used synonymously in the literature. This distinction is irrelevant in the analysis of discrete-time systems. For this reason, we follow this loose convention in this book, where the term *digital filter* implies a *discrete-time system,* and *analog filter* means *continuous-time system*. Moreover, the terms C/D (continuous-to-discrete-time ) and D/C will occasionally be used interchangeably with terms A/D (analog-to-digital) and D/A, respectively. - -### ADVANTAGES OF DIGITAL SIGNAL PROCESSING - -- 1. Digital systems operation can tolerate considerable variation in signal values, and hence are less sensitive to changes in the component parameter values due to temperature variation, aging, and other factors. This results in greater degree of precision and stability. Since digital systems are binary circuits, their accuracy can be increased by using more complex circuitry to increase word length, subject to cost limitations. -- 2. Digital systems do not require any factory adjustment and can be easily duplicated in volume without having to worry about precise component values. They can be fully integrated, and even highly complex systems can be placed on a single chip by using *VLSI* (very-large-scale integrated) circuits. -- 3. Digital filters are more flexible. Their characteristics can be easily altered simply by changing the program. Digital hardware implementation permits the use of microprocessors, miniprocessors, digital switching, and large-scale integrated circuits. -- 4. A greater variety of filters can be realized by digital systems. -- 5. Digital signals can be stored easily and inexpensively on various media (e.g., magnetic, optical, and solid state) without deterioration of signal quality. It is also possible (and increasingly popular) to search and select information from distant electronic storehouses, such as the cloud. -- 6. Digital signals can be coded to yield extremely low error rates and high fidelity, as well as privacy. Also, more sophisticated signal-processing algorithms can be used to process digital signals. - -† The terms *discrete-time* and *continuous-time* qualify the nature of a signal along the time axis (horizontal axis). The terms *analog* and *digital,* in contrast, qualify the nature of the signal amplitude (vertical axis). - -### 262 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -- 7. Digital filters can be easily time-shared and therefore can serve a number of inputs simultaneously. Moreover, it is easier and more efficient to multiplex several digital signals on the same channel. -- 8. Reproduction with digital messages is extremely reliable without deterioration. Analog messages such as photocopies and films, for example, lose quality at each successive stage of reproduction and have to be transported physically from one distant place to another, often at relatively high cost. - -One must weigh these advantages against such disadvantages as increased system complexity due to use of A/D and D/A interfaces, limited range of frequencies available in practice (affordable rates are gigahertz or less), and use of more power than is needed for the passive analog circuits. Digital systems use power-consuming active devices. - -### **[3.4-1 Classification of Discrete-Time Systems](#page-9-0)** - -Before examining the nature of discrete-time system equations, let us consider the concepts of linearity, time invariance (or shift invariance), and causality, which apply to discrete-time systems also. - -### LINEARITY AND TIME INVARIANCE - -For discrete-time systems, the definition of *linearity* is identical to that for continuous-time systems, as given in Eq. (1.22). We can show that the systems in Exs. 3.6, 3.7, 3.8, and 3.9 are all linear. - -Time invariance (or *shift invariance*) for discrete-time systems is also defined in a way similar to that for continuous-time systems. Systems whose parameters do not change with time (with *n*) are *time-invariant* or shift-invariant (also *constant-parameter*) systems. For such a system, if the input is delayed by *k* units or samples, the output is the same as before but delayed by *k* samples (assuming the initial conditions also are delayed by *k*). The systems in Exs. 3.6, 3.7, 3.8, and 3.9 are time-invariant because the coefficients in the system equations are constants (independent of *n*). If these coefficients were functions of *n* (time), then the systems would be linear *time-varying* systems. Consider, for example, a system described by - -$$ -y[n] = e^{-n}x[n] -$$ - -For this system, let a signal *x*1[*n*] yield the output *y*1[*n*], and another input *x*2[*n*] yield the output *y*2[*n*]. Then - -$$ -y_1[n] = e^{-n}x_1[n] -$$ - and $y_2[n] = e^{-n}x_2[n]$ - -If we let *x*2[*n*] = *x*1[*n*−*N*0], then - -$$ -y_2[n] = e^{-n}x_2[n] = e^{-n}x_1[n - N_0] \neq y_1[n - N_0] -$$ - -Clearly, this is a time-varying parameter system. - -### CAUSAL AND NONCAUSAL SYSTEMS - -A *causal* (also known as a *physical* or *nonanticipative*) system is one for which the output at any instant *n* = *k* depends only on the value of the input *x*[*n*] for *n* ≤ *k*. In other words, the value of the output at the present instant depends only on the past and present values of the input *x*[*n*], not on its future values. As we shall see, the systems in Exs. 3.6, 3.7, 3.8, and 3.9 are all causal. - -### INVERTIBLE AND NONINVERTIBLE SYSTEMS - -A discrete-time system *S* is invertible if an inverse system *Si* exists such that the cascade of *S* and *Si* results in an *identity* system. An identity system is defined as one whose output is identical to the input. In other words, for an invertible system, the input can be uniquely determined from the corresponding output. For every input there is a unique output. When a signal is processed through such a system, its input can be reconstructed from the corresponding output. There is no loss of information when a signal is processed through an invertible system. - -A cascade of a unit delay with a unit advance results in an identity system because the output of such a cascaded system is identical to the input. Clearly, the inverse of an ideal unit delay is ideal unit advance, which is a noncausal (and unrealizable) system. In contrast, a compressor *y*[*n*] = *x*[*Mn*] is not invertible because this operation loses all but every *M*th sample of the input, and, generally, the input cannot be reconstructed. Similarly, operations, such as *y*[*n*] = cos *x*[*n*] or *y*[*n*]=|*x*[*n*]|, are not invertible. - -### **DR ILL 3.9 Invertibility** - -Show that a system specified by equation *y*[*n*] = *ax*[*n*] + *b* is invertible but that the system *y*[*n*]=|*x*[*n*]|2 is noninvertible. - -### STABLE AND UNSTABLE SYSTEMS - -The concept of stability is similar to that in continuous-time systems. Stability can be *internal* or *external*. If every *bounded input* applied at the input terminal results in a *bounded output,* the system is said to be stable *externally*. External stability can be ascertained by measurements at the external terminals of the system. This type of stability is also known as the stability in the BIBO (bounded-input/bounded-output) sense. Both internal and external stability are discussed in greater detail in Sec. 3.9. - -### MEMORYLESS SYSTEMS AND SYSTEMS WITH MEMORY - -The concepts of memoryless (or instantaneous) systems and those with memory (or dynamic) are identical to the corresponding concepts of the continuous-time case. A system is memoryless if its response at any instant *n* depends at most on the input at the same instant *n*. The output at any instant of a system with memory generally depends on the past, present, and future values of the input. For example, *y*[*n*] =sin*x*[*n*] is an example of instantaneous system, and *y*[*n*]−*y*[*n*−1] =*x*[*n*] is an example of a dynamic system or a system with memory. - -### **EXAMPLE 3.10 Investigating DT System Properties** - -Consider a DT system described as *y*[*n* + 1] = *x*[*n* + 1]*x*[*n*]. Determine whether the system is **(a)** linear, **(b)** time-invariant, **(c)** causal, **(d)** invertible, **(e)** BIBO-stable, and **(f)** memoryless. - -Let us delay the input–output equation by one to obtain the equivalent but more convenient representation of *y*[*n*] = *x*[*n*]*x*[*n*−1]. - -**(a)** Linearity requires both homogeneity and additivity. Let us first investigate homogeneity. Assuming *x*[*n*] -⇒ *y*[*n*], we see that - -$$ -ax[n] \Longrightarrow (ax[n])(ax[n-1]) = a^2y[n] \neq ay[n] -$$ - -Thus, the system does not satisfy the homogeneity property. - -The system also does not satisfy the additivity property. Assuming *x*1[*n*] -⇒ *y*1[*n*] and *x*2[*n*] -⇒ *y*2[*n*], we see that input *x*[*n*] = *x*1[*n*] +*x*2[*n*] produces output *y*[*n*] as - -$$ -y[n] = (x_1[n] + x_2[n])(x_1[n-1] + x_2[n-1]) -$$ - -= $x_1[n]x_1[n-1] + x_2[n]x_2[n-1] + x_1[n]x_2[n-1] + x_2[n]x_1[n-1]$ -= $y_1[n] + y_2[n] + x_1[n]x_2[n-1] + x_2[n]x_1[n-1]$ - $\neq y_1[n] + y_2[n]$ - -Clearly, additivity is not satisfied. - -Since the system does not satisfy both the homogeneity and additivity properties, we conclude that the system is not linear. - -**(b)** To be time-invariant, a shift in any input should cause a corresponding shift in respective output. Assume that *x*[*n*] -⇒ *y*[*n*]. Applying a delay version of this input to the system yields - -$$ -x[n - N] \Longrightarrow x[n - N]x[n - 1 - N] = x[(n - N)]x[(n - N) - 1] = y[n - N] -$$ - -Since shifting an input causes a corresponding shift in the output, we conclude that the system is time-invariant. - -**(c)** To be causal, an output value cannot depend on any future input values. The output *y* at time *n* depends on the input *x* at present and past times *n* and *n*−1. Since the current output does not depend on future input values, the system is causal. - -**(d)** For a system to be invertible, every input must generate a unique output, which allows exact recovery of the input from the output. Consider two inputs to this system: *x*1[*n*] = 1 and *x*2[*n*]=−1. Both inputs generate the same output: *y*1[*n*] = *y*2[*n*] = 1. Since unique inputs do not always generate unique outputs, we conclude that the system is not invertible. - -**(e)** To be BIBO-stable, any bounded input must generate a bounded output. A bounded input satisfies |*x*[*n*]| ≤ *Mx* < ∞ for all *n*. Given this condition, the system output magnitude behaves as - -> |*y*[*n*]| = |*x*[*n*]*x*[*n*−1]| = |*x*[*n*]||*x*[*n*−1]| ≤ *M*2 *x* < ∞ - -Since any bounded input is guaranteed to produce a bounded output, it follows that the system is BIBO-stable. - -**(f)** To be memoryless, a system's output can only depend on the strength of the current input. Since the output *y* at time *n* depends on the input *x* not only at present time *n* but also on past time *n*−1, we see that the system is not memoryless. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/043_3.5 DISCRETE-TIME SYSTEM EQUATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/043_3.5 DISCRETE-TIME SYSTEM EQUATIONS.md deleted file mode 100644 index 40d13b2704523060f26d6a8a1a5aafb0a3977bc4..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/043_3.5 DISCRETE-TIME SYSTEM EQUATIONS.md +++ /dev/null @@ -1,245 +0,0 @@ -## **[3.5 DISCRETE-TIME](#page-9-0) SYSTEM EQUATIONS** - -In this section we discuss time-domain analysis of LTID (linear, time-invariant, discrete-time systems). With minor differences, the procedure is parallel to that for continuous-time systems. - -### DIFFERENCE EQUATIONS - -Equations (3.3), (3.5), (3.8), and (3.13) are examples of difference equations. Equations (3.3), (3.8), and (3.13) are first-order difference equations, and Eq. (3.5) is a second-order difference equation. All these equations are linear, with constant (not time-varying) coefficients.† Before giving a general form of an *N*th-order linear difference equation, we recall that a difference equation can be written in two forms: the first form uses delay terms such as *y*[*n* − 1], *y*[*n* − 2], *x*[*n*−1], *x*[*n*−2], and so on; and the alternate form uses advance terms such as *y*[*n*+1], *y*[*n*+2], and so on. Although the delay form is more natural, we shall often prefer the advance form, not just for the general notational convenience, but also for resulting notational uniformity with the operator form for differential equations. This facilitates the commonality of the solutions and concepts for continuous-time and discrete-time systems. - -We start here with a general difference equation, written in advance form as - -$$ -y[n+N] + a_1y[n+N-1] + \cdots + a_{N-1}y[n+1] + a_Ny[n] = -$$ - -\n -$$ -b_{N-M}x[n+M] + b_{N-M+1}x[n+M-1] + \cdots + b_{N-1}x[n+1] + b_Nx[n] -$$ - (3.14) - -This is a linear difference equation whose order is max(*N*,*M*). We have assumed the coefficient of *y*[*n* + *N*] to be unity (*a*0 = 1) without loss of generality. If *a*0 = 1, we can divide the equation throughout by *a*0 to normalize the equation to have *a*0 = 1. - -### CAUSALITY CONDITION - -For a causal system, the output cannot depend on future input values. This means that when the system equation is in the advance form of Eq. (3.14), causality requires *M* ≤ *N*. If *M* were to be greater than *N*, then *y*[*n*+*N*], the output at *n*+*N* would depend on *x*[*n*+ *M*], which is the input at the later instant *n*+ *M*. For a general causal case, *M* = *N*, and Eq. (3.14) can be expressed as - -$$ -y[n+N] + a_1y[n+N-1] + \cdots + a_{N-1}y[n+1] + a_Ny[n] = -$$ - -\n -$$ -b_0x[n+N] + b_1x[n+N-1] + \cdots + b_{N-1}x[n+1] + b_Nx[n] -$$ -\n(3.15) - -† Equations such as (3.3), (3.5), (3.8), and (3.13) are considered to be linear according to the classical definition of linearity. Some authors label such equations as *incrementally linear*. We prefer the classical definition. It is just a matter of individual choice and makes no difference in the final results. - -where some of the coefficients on either side can be zero. In this *N*th-order equation, *a*0, the coefficient of *y*[*n*+*N*], is normalized to unity. Equation (3.15) is valid for all values of *n*. Therefore, it is still valid if we replace *n* by *n* − *N* throughout the equation [see Eqs. (3.3) and (3.4)]. Such replacement yields a delay-form alternative: - -$$ -y[n] + a_1y[n-1] + \dots + a_{N-1}y[n-N+1] + a_Ny[n-N] = -$$ - -\n -$$ -b_0x[n] + b_1x[n-1] + \dots + b_{N-1}x[n-N+1] + b_Nx[n-N] -$$ - (3.16) - -### **[3.5-1 Recursive \(Iterative\) Solution of Difference Equation](#page-9-0)** - -Equation (3.16) can be expressed as - -$$ -y[n] = -a_1y[n-1] - a_2y[n-2] - \cdots - a_Ny[n-N] + b_0x[n] + b_1x[n-1] + \cdots + b_Nx[n-N] -$$ -(3.17) - -In Eq. (3.17), *y*[*n*] is computed from 2*N* + 1 pieces of information; the preceding *N* values of the output: *y*[*n* − 1], *y*[*n* − 2], ... , *y*[*n* − *N*], and the preceding *N* values of the input: *x*[*n* − 1], *x*[*n* − 2], ... , *x*[*n* − *N*], and the present value of the input *x*[*n*]. Initially, to compute *y*[0], the *N* initial conditions *y*[−1], *y*[−2], ... , *y*[−*N*] serve as the preceding *N* output values. Hence, knowing the *N* initial conditions and the input, we can determine recursively the entire output *y*[0], *y*[1], *y*[2], *y*[3], ... , one value at a time. For instance, to find *y*[0] we set *n* = 0 in Eq. (3.17). The left-hand side is *y*[0], and the right-hand side is expressed in terms of *N* initial conditions *y*[−1], *y*[−2], ... , *y*[−*N*] and the input *x*[0] if *x*[*n*] is causal (because of causality, other input terms *x*[−*n*] = 0). Similarly, knowing *y*[0] and the input, we can compute *y*[1] by setting *n* = 1 in Eq. (3.17). Knowing *y*[0] and *y*[1], we find *y*[2], and so on. Thus, we can use this recursive procedure to find the complete response *y*[0], *y*[1], *y*[2], .... For this reason, this equation is classed as a recursive form. This method basically reflects the manner in which a computer would solve a recursive difference equation, given the input and initial conditions. Equation (3.17) [or Eq. (3.16)] is nonrecursive if all the *N* − 1 coefficients *ai* = 0 (*i* = 1, 2,...,*N* − 1). In this case, it can be seen that *y*[*n*] is computed only from the input values and without using any previous outputs. Generally speaking, the recursive procedure applies only to equations in the recursive form. The recursive (iterative) procedure is demonstrated by the following examples. - -### **EXAMPLE 3.11 Iterative Solution to a First-Order Difference Equation** - -Solve iteratively - -$$ -y[n] - 0.5y[n-1] = x[n] -$$ - -with initial condition *y*[−1] = 16 and causal input *x*[*n*] = *n*2*u*[*n*]. This equation can be expressed as - -*y*[*n*] = 0.5*y*[*n*−1] +*x*[*n*] (3.18) - -If we set *n* = 0 in Eq. (3.18), we obtain - -$$ -y[0] = 0.5y[-1] + x[0] -$$ - -= 0.5(16) + 0 = 8 - -Now, setting *n* = 1 in Eq. (3.18) and using the value *y*[0] = 8 (computed in the first step) and *x*[1] = (1)2 = 1, we obtain - -$$ -y[1] = 0.5(8) + (1)^2 = 5 -$$ - -Next, setting *n* = 2 in Eq. (3.18) and using the value *y*[1] = 5 (computed in the previous step) and *x*[2] = (2)2, we obtain - -$$ -y[2] = 0.5(5) + (2)^2 = 6.5 -$$ - -Continuing in this way iteratively, we obtain - -$$ -y[3] = 0.5(6.5) + (3)^{2} = 12.25 -$$ - -\n -$$ -y[4] = 0.5(12.25) + (4)^{2} = 22.125 -$$ - -\n -$$ -\vdots -$$ - -The output *y*[*n*] is depicted in Fig. 3.17. - -We now present one more example of iterative solution—this time for a second-order equation. The iterative method can be applied to a difference equation in delay form or advance form. In Ex. 3.11 we considered the former. Let us now apply the iterative method to the advance form. - -### **EXAMPLE 3.12 Iterative Solution to a Second-Order Difference Equation** - -Solve iteratively - -$$ -y[n+2] - y[n+1] + 0.24y[n] = x[n+2] - 2x[n+1] -$$ - -### 268 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -with initial conditions *y*[−1] = 2, *y*[−2] = 1 and a causal input *x*[*n*] = *nu*[*n*]. The system equation can be expressed as - -$$ -y[n+2] = y[n+1] - 0.24y[n] + x[n+2] - 2x[n+1] -$$ -\n(3.19) - -Setting *n* = −2 in Eq. (3.19) and then substituting *y*[−1] = 2, *y*[−2] = 1, *x*[0] = *x*[−1] = 0, we obtain - -$$ -y[0] = 2 - 0.24(1) + 0 - 0 = 1.76 -$$ - -Setting *n* = −1 in Eq. (3.19) and then substituting *y*[0] = 1.76, *y*[−1] = 2, *x*[1] = 1, *x*[0] = 0, we obtain - -$$ -y[1] = 1.76 - 0.24(2) + 1 - 0 = 2.28 -$$ - -Setting *n* = 0 in Eq. (3.19) and then substituting *y*[0] = 1.76, *y*[1] = 2.28, *x*[2] = 2, and *x*[1] = 1 yield - -$$ -y[2] = 2.28 - 0.24(1.76) + 2 - 2(1) = 1.8576 -$$ - -and so on. - -With MATLAB, we can readily verify and extend these recursive calculations. - -``` ->> n = -2:5; y = [1,2,zeros(1,length(n)-2)]; x = [0,0,n(3:end)]; ->> for k = 1:length(n)-2, ->> y(k+2) = y(k+1)-0.24*y(k)+x(k+2)-2*x(k+1); ->> end ->> n,y - n = -2 -1 0 1 2 3 4 5 - y = 1.0000 2.0000 1.7600 2.2800 1.8576 0.3104 -2.1354 -5.2099 -``` - -Note carefully the recursive nature of the computations. From the *N* initial conditions (and the input), we obtained *y*[0] first. Then, using this value of *y*[0] and the preceding *N* − 1 initial conditions (along with the input), we find *y*[1]. Next, using *y*[0], *y*[1] along with the past *N* − 2 initial conditions and input, we obtained *y*[2], and so on. This method is general and can be applied to a recursive difference equation of any order. It is interesting that the hardware realization of Eq. (3.18) depicted in Fig. 3.14 (with *a* = 0.5) generates the solution precisely in this (iterative) fashion. - -### **DR ILL 3.10 Iterative Solution to a Difference Equation** - -Using the iterative method, find the first three terms of *y*[*n*] for - -*y*[*n*+1] −2*y*[*n*] = *x*[*n*] - -The initial condition is *y*[−1] = 10 and the input *x*[*n*] = 2 starting at *n* = 0. - -**ANSWER** *y*[0] = 20, *y*[1] = 42, and *y*[2] = 86 - -We shall see in the future that the solution of a difference equation obtained in this direct (iterative) way is useful in many situations. Despite the many uses of this method, a closed-form solution of a difference equation is far more useful in the study of system behavior and its dependence on the input and various system parameters. For this reason we shall develop a systematic procedure to analyze discrete-time systems along lines similar to those used for continuous-time systems. - -### OPERATOR NOTATION - -In difference equations, it is convenient to use operator notation similar to that used in differential equations for the sake of compactness. In continuous-time systems, we used the operator *D* to denote the operation of differentiation. For discrete-time systems, we shall use the operator *E* to denote the operation for advancing a sequence by one time unit. Thus, - -$$ -Ex[n] \equiv x[n+1] -$$ - -$$ -E^{2}x[n] \equiv x[n+2] -$$ - -$$ -\vdots -$$ - -$$ -E^{N}x[n] \equiv x[n+N] -$$ - -Let us use this advance operator notation to represent several systems investigated earlier. The first-order difference equation of a savings account is [see Eq. (3.4)] - -$$ -y[n+1] - ay[n] = x[n+1] -$$ - -Using the operator notation, we can express this equation as - -$$ -E\mathbf{y}[n] - a\mathbf{y}[n] = Ex[n] \qquad \text{or} \qquad (E - a)\mathbf{y}[n] = Ex[n] -$$ - -Similarly, the second-order book sales estimate described by Eq. (3.6) as - -$$ -y[n+2] + \frac{1}{4}y[n+1] + \frac{1}{16}y[n] = x[n+2] -$$ - -can be expressed in operator notation as - -$$ -(E^2 + \frac{1}{4}E + \frac{1}{16})y[n] = E^2x[n] -$$ - -The general *N*th-order advance-form difference equation of Eq. (3.15) can be expressed as - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y[n] = (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n] -$$ - -or - -$$ -Q[E]y[n] = P[E]x[n] \tag{3.20} -$$ - -where *Q*[*E*] and *P*[*E*] are *N*th-order polynomial operators - -$$ -Q[E] = EN + a1EN-1 + \dots + aN-1E + aN -$$ - -$$ -P[E] = b0EN + b1EN-1 + \dots + bN-1E + bN -$$ - -### RESPONSE OF LINEAR DISCRETE-TIME SYSTEMS - -Following the procedure used for continuous-time systems, we can show that Eq. (3.20) is a linear equation (with constant coefficients). A system described by such an equation is a linear, time-invariant, discrete-time (LTID) system. We can verify, as in the case of LTIC systems (see the footnote on page 151), that the general solution of Eq. (3.20) consists of zero-input and zero-state components. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/044_3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS - THE ZERO-INPUT RESPONSE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/044_3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS - THE ZERO-INPUT RESPONSE.md deleted file mode 100644 index 97ed67f3fbe1087d139540182d5325bda5da7fad..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/044_3.6 SYSTEM RESPONSE TO INTERNAL CONDITIONS - THE ZERO-INPUT RESPONSE.md +++ /dev/null @@ -1,336 +0,0 @@ -## **3.6 SYSTEM RESPONSE TO INTERNAL [CONDITIONS:](#page-9-0) THE ZERO-INPUT RESPONSE** - -The zero-input response *y*0[*n*] is the solution of Eq. (3.20) with *x*[*n*] = 0; that is, - -$$ -Q[E]y_0[n] = 0 -$$ - -or - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y0[n] = 0 -$$ -\n(3.21) - -Although we can solve this equation systematically, even a cursory examination points to the solution. This equation states that a linear combination of *y*0[*n*] and advanced *y*0[*n*] is zero, *not for some values of n, but for all n.* Such a situation is possible *if and only if y*0[*n*] and advanced *y*0[*n*] have the same form. Only an exponential function γ *n* has this property, as the following equation indicates: - -$$ -E^k\{\gamma^n\} = \gamma^{n+k} = \gamma^k\gamma^n -$$ - -This expression shows that γ *n* advanced by *k* units is a constant (γ *k*) times γ *n*. Therefore, the solution of Eq. (3.21) must be of the form† - -$$ -y_0[n] = c\gamma^n \tag{3.22} -$$ - -To determine *c* and γ , we substitute this solution in Eq. (3.21). Since *Eky*0[*n*] = *y*0[*n*+*k*] = *c*γ *n*+*k*, this produces - -$$ -c(\gamma^{N} + a_1 \gamma^{N-1} + \cdots + a_{N-1} \gamma + a_N) \gamma^{n} = 0 -$$ - -For a nontrivial solution of this equation, - -$$ -\gamma^{N} + a_{1}\gamma^{n-1} + \dots + a_{N-1}\gamma + a_{N} = 0 -$$ -\n(3.23) - -or - -*Q*[γ ] = 0 - -Our solution *c*γ *n* [Eq. (3.22)] is correct, provided γ satisfies Eq. (3.23). Now, *Q*[γ ] is an *N*th-order polynomial and can be expressed in the factored form (assuming all distinct roots): - -$$ -(\gamma - \gamma_1)(\gamma - \gamma_2) \cdots (\gamma - \gamma_N) = 0 -$$ - -Clearly, γ has *N* solutions γ1, γ2, ... , γ*N* and, therefore, Eq. (3.21) also has *N* solutions *c*1γ *n* 1 , *c*2γ *n* 2 , ... , *cn*γ *n N*. In such a case, we have shown that the general solution is a linear combination - -† A signal of the form *nm*γ *n* also satisfies this requirement under certain conditions (repeated roots), discussed later. - -#### 3.6 System Response to Internal Conditions: The Zero-Input Response 271 - -of the *N* solutions (see the footnote on page 153). Thus, - -$$ -y_0[n] = c_1 \gamma_1^n + c_2 \gamma_2^n + \cdots + c_n \gamma_N^n -$$ - -where γ1, γ2, ... , γ*n* are the roots of Eq. (3.23) and *c*1, *c*2, ... , *cn* are arbitrary constants determined from *N* auxiliary conditions, generally given in the form of initial conditions. The polynomial *Q*[γ ] is called the *characteristic polynomial* of the system, and *Q*[γ ] = 0 [Eq. (3.23)] is the *characteristic equation* of the system. Moreover, γ1, γ2, ... , γ*N*, the roots of the characteristic equation, are called *characteristic roots* or *characteristic values* (also *eigenvalues*) of the system. The exponentials γ *n i* (*i* = 1, 2,...,*N*) are the *characteristic modes* or *natural modes* of the system. A characteristic mode corresponds to each characteristic root of the system, and the *zero-input response is a linear combination of the characteristic modes of the system*. - -### **EXAMPLE 3.13 Zero-Input Response of a Second-Order System with Real Roots** - -The LTID system described by the difference equation - -$$ -y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2] -$$ - -has input *x*[*n*] = 4−*nu*[*n*] and initial conditions *y*[−1] = 0 and *y*[−2] = 25/4. Determine the zero-input response *y*0[*n*]. The zero-state response of this system is considered later, in Ex. 3.21. - -The system equation in operator notation is - -$$ -(E^2 - 0.6E - 0.16)y[n] = 5E^2x[n] -$$ - -The characteristic polynomial is - -$$ -\gamma^2 - 0.6\gamma - 0.16 = (\gamma + 0.2)(\gamma - 0.8) -$$ - -The characteristic equation is - -$$ -(\gamma + 0.2)(\gamma - 0.8) = 0 -$$ - -The characteristic roots are γ1 = −0.2 and γ2 = 0.8. The zero-input response is - -$$ -y_0[n] = c_1(-0.2)^n + c_2(0.8)^n \tag{3.24} -$$ - -To determine arbitrary constants *c*1 and *c*2, we set *n* = −1 and −2 in Eq. (3.24), then substitute *y*0[−1] = 0 and *y*0[−2] = 25/4 to obtain† - -$$ -\begin{array}{c}\n0 = -5c_1 + \frac{5}{4}c_2 \\ -\frac{25}{4} = 25c_1 + \frac{25}{16}c_2\n\end{array}\n\right\} \quad \Longrightarrow \quad c_1 = \frac{1}{5} -$$ -\n -$$ -c_2 = \frac{4}{5} -$$ - -† The initial conditions *y*[−1] and *y*[−2] are the conditions given on the total response. But because the input does not start until *n* = 0, the zero-state response is zero for *n* < 0. Hence, at *n* = −1 and −2 the total response consists of the zero-input component only so that *y*[−1] = *y*0[−1] and *y*[−2] = *y*0[−2]. - -### 272 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Therefore, - -$$ -y_0[n] = \frac{1}{5}(-0.2)^n + \frac{4}{5}(0.8)^n -$$ - $n \ge 0$ - -The reader can verify this solution by computing the first few terms using the iterative method (see Exs. 3.11 and 3.12). - -### **DR ILL 3.11 Zero-Input Response of First-Order Systems** - -Find and sketch the zero-input response for the systems described by the following equations: - -- **(a)** *y*[*n*+1] −0.8*y*[*n*] = 3*x*[*n*+1] -- **(b)** *y*[*n*+1] +0.8*y*[*n*] = 3*x*[*n*+1] - -In each case the initial condition is *y*[−1] = 10. Verify the solutions by computing the first three terms using the iterative method. - -### **ANSWERS** - -- **(a)** 8(0.8)*n* -- **(b)** −8(−0.8)*n* - -### **DR ILL 3.12 Zero-Input Response of a Second-Order System with Real Roots** - -Find the zero-input response of a system described by the equation - -$$ -y[n] + 0.3y[n-1] - 0.1y[n-2] = x[n] + 2x[n-1] -$$ - -The initial conditions are *y*0[−1] = 1 and *y*0[−2] = 33. Verify the solution by computing the first three terms iteratively. - -### **ANSWER** - -*y*0[*n*] = (0.2)*n* +2(−0.5)*n* - -Section 3.5-1 introduced the method of recursion to solve difference equations. As the next example illustrates, the zero-input response can likewise be found through recursion. Since it does not provide a closed-form solution, recursion is generally not the preferred method of solving difference equations. - -### REPEATED ROOTS - -So far we have assumed the system to have *N* distinct characteristic roots γ1, γ2, ... , γ*N* with corresponding characteristic modes γ *n* 1 , γ *n* 2 , ... , γ *n N*. If two or more roots coincide (repeated roots), the form of characteristic modes is modified. Direct substitution shows that if a root γ repeats *r* times (root of multiplicity *r*), the corresponding characteristic modes for this root are γ *n*, *n*γ *n*, *n*2γ *n*, ... , *nr*−1γ *n*. Thus, if the characteristic equation of a system is - -$$ -Q[\gamma]=(\gamma-\gamma_1)^r(\gamma-\gamma_{r+1})(\gamma-\gamma_{r+2})\cdots(\gamma-\gamma_N) -$$ - -then the zero-input response of the system is - -$$ -y_0[n] = (c_1 + c_2n + c_3n^2 + \dots + c_rn^{r-1})\gamma_1^n + c_{r+1}\gamma_{r+1}^n + c_{r+2}\gamma_{r+2}^n + \dots + c_n\gamma_N^n -$$ - -### **EXAMPLE 3.15 Zero-Input Response of a Second-Order System with Repeated Roots** - -Consider a second-order difference equation with repeated roots: - -$$ -(E2 + 6E + 9)y[n] = (2E2 + 6E)x[n] -$$ - -Determine the zero-input response *y*0[*n*] if the initial conditions are *y*0[−1]=−1/3 and *y*0[−2]=−2/9. - -The characteristic polynomial is γ 2 +6γ +9 = (γ +3)2, and we have a repeated characteristic root at γ = −3. The characteristic modes are (−3)*n* and *n*(−3)*n*. Hence, the zero-input response is - -> *y*0[*n*] = (*c*1 +*c*2*n*)(−3) *n* - -Although we can determine the constants *c*1 and *c*2 from the initial conditions following a procedure similar to Ex. 3.13, we instead use MATLAB to perform the needed calculations. - ->> c = inv([(-3)^(-1) -1\*(-3)^(-1);(-3)^(-2) -2\*(-3)^(-2)])\*[-1/3;-2/9] c=4 3 - -Thus, the zero-input response is - -$$ -y_0[n] = (4+3n)(-3)^n -$$ - $n \ge 0$ - -### COMPLEX ROOTS - -As in the case of continuous-time systems, the complex roots of a discrete-time system will occur in pairs of conjugates if the system equation coefficients are real. Complex roots can be treated exactly as we would treat real roots. However, just as in the case of continuous-time systems, we can also use the real form of solution as an alternative. - -First we express the complex conjugate roots γ and γ ∗ in polar form. If |γ | is the magnitude and β is the angle of γ , then - -$$ -\gamma = |\gamma|e^{i\beta} -$$ - and $\gamma^* = |\gamma|e^{-j\beta}$ - -The zero-input response is given by - -$$ -y_0[n] = c_1 \gamma^n + c_2(\gamma^*)^n = c_1 |\gamma|^n e^{i\beta n} + c_2 |\gamma|^n e^{-j\beta n} -$$ - -For a real system, *c*1 and *c*2 must be conjugates so that *y*0[*n*] is a real function of *n*. Let - -$$ -c_1 = \frac{c}{2}e^{j\theta} \qquad \text{and} \qquad c_2 = \frac{c}{2}e^{-j\theta} -$$ - -Then - -$$ -y_0[n] = \frac{c}{2} |\gamma|^n \left[ e^{j(\beta n + \theta)} + e^{-j(\beta n + \theta)} \right] = c |\gamma|^n \cos(\beta n + \theta) -$$ -\n(3.25) - -where *c* and θ are arbitrary constants determined from the auxiliary conditions. This is the solution in real form, which avoids dealing with complex numbers. - -### **EXAMPLE 3.16 Zero-Input Response of a Second-Order System with Complex Roots** - -Consider a second-order difference equation with complex-conjugate roots: - -$$ -(E2 - 1.56E + 0.81)y[n] = (E + 3)x[n] -$$ - -Determine the zero-input response *y*0[*n*] if the initial conditions are *y*0[−1] = 2 and *y*0[−2] = 1. - -The characteristic polynomial is (γ 2 − 1.56γ + 0.81) = (γ − 0.78 − *j*0.45)(γ − 0.78 + *j*0.45). The characteristic roots are 0.78 ± *j*0.45; that is, 0.9*e*±*j*(π/6) . We could immediately write the solution as - -$$ -y_0[n] = c(0.9)^n e^{j\pi n/6} + c^*(0.9)^n e^{-j\pi n/6} -$$ - -Setting *n* = −1 and −2 and using the initial conditions *y*0[−1] = 2 and *y*0[−2] = 1, we find *c* = 1.1550−*j*0.2025 = 1.1726*e*−*j*0.1735 and *c*∗ = 1.1550+*j*0.2025 = 1.1726*ej*0.1735. - ->> gamma = roots([1 -1.56 0.81]); >> c = inv([gamma(1)^(-1) gamma(2)^(-1);gamma(1)^(-2) gamma(2)^(-2)])\*[2;1] c = 1.1550 - 0.2025i 1.1550 + 0.2025i - -Alternately, we could also find the unknown coefficient by using the real form of the solution, as given in Eq. (3.25). In the present case, the roots are 0.9*e*±*j*(π/6) . Hence, |γ | = 0.9 and β = π/6, and the zero-input response, according to Eq. (3.25), is given by - -$$ -y_0[n] = c(0.9)^n \cos\left(\frac{\pi}{6}n + \theta\right) -$$ - -To determine the constants *c* and θ, we set *n* = −1 and −2 in this equation and substitute the initial conditions *y*0[−1] = 2 and *y*0[−2] = 1 to obtain - -$$ -2 = \frac{c}{0.9} \cos\left(-\frac{\pi}{6} + \theta\right) = \frac{c}{0.9} \left[\frac{\sqrt{3}}{2} \cos\theta + \frac{1}{2} \sin\theta\right] -$$ -$$ -1 = \frac{c}{(0.9)^2} \cos\left(-\frac{\pi}{3} + \theta\right) = \frac{c}{0.81} \left[\frac{1}{2} \cos\theta + \frac{\sqrt{3}}{2} \sin\theta\right] -$$ - -or - -$$ -\frac{\sqrt{3}}{1.8}c\cos\theta + \frac{1}{1.8}c\sin\theta = 2 -$$ -$$ -\frac{1}{1.62}c\cos\theta + \frac{\sqrt{3}}{1.62}c\sin\theta = 1 -$$ - -These are two simultaneous equations in two unknowns *c*cos θ and *c*sin θ. Solution of these equations yields - -$$ -c \cos \theta = 2.308 -$$ - -$$ -c \sin \theta = -0.397 -$$ - -Dividing *c*sin θ by *c*cos θ yields - -$$ -\tan \theta = \frac{-0.397}{2.308} = \frac{-0.172}{1} -$$ -$$ -\theta = \tan^{-1}(-0.172) = -0.17 \text{ rad} -$$ - -Substituting θ = −0.17 radian in *c*cos θ = 2.308 yields *c* = 2.34 and - -$$ -y_0[n] = 2.34(0.9)^n \cos\left(\frac{\pi}{6}n - 0.17\right) -$$ - $n \ge 0$ - -Observe that here we have used radian units for both β and θ. We also could have used the degree unit, although this practice is not recommended. The important consideration is to be consistent and to use the same units for both β and θ. - -### **DR ILL 3.13 Zero-Input Response of a Second-Order System with Complex Roots** - -Find the zero-input response of a system described by the equation - -$$ -y[n] + 4y[n-2] = 2x[n] -$$ - -The initial conditions are *y*0[−1]=−1/(2 √2) and *y*0[−2] = 1/(4 √ 2). Verify the solution by computing the first three terms iteratively. - -**ANSWER** - -*y*0[*n*] = (2)*n* cos π 2 *n*− 3π 4 - -# **3.7 THE UNIT IMPULSE [RESPONSE](#page-9-0)** *h***[***n***]** - -Consider an *n*th-order system specified by the equation - -$$ -(EN + a1EN-1 + \dots + aN-1E + aN)y[n] = (b0EN + b1EN-1 + \dots + bN-1E + bN)x[n] -$$ - -or - -$$ -Q[E]y[n] = P[E]x[n] -$$ - -The unit impulse response *h*[*n*] is the solution of this equation for the input δ[*n*] with all the initial conditions zero; that is, - -$$ -Q[E]h[n] = P[E]\delta[n] \tag{3.26} -$$ - -subject to initial conditions - -$$ -h[-1] = h[-2] = \cdots = h[-N] = 0 -$$ - -Equation (3.26) can be solved to determine *h*[*n*] iteratively or in a closed form. The following example demonstrates the iterative solution. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/045_3.7 THE UNIT IMPULSE RESPONSE h[n].md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/045_3.7 THE UNIT IMPULSE RESPONSE h[n].md deleted file mode 100644 index 6955e5bfe41c8fa38a7a7a6b3785560f84743891..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/045_3.7 THE UNIT IMPULSE RESPONSE h[n].md +++ /dev/null @@ -1,130 +0,0 @@ -### **EXAMPLE 3.17 Iterative Determination of the Impulse Response** - -Iteratively compute the first two values of the impulse response *h*[*n*] of a system described by the equation - -*y*[*n*] −0.6*y*[*n*−1] −0.16*y*[*n*−2] = 5*x*[*n*] - -To determine the unit impulse response, we let the input *x*[*n*] = δ[*n*] and the output *y*[*n*] = *h*[*n*] in the system's difference equation to obtain - -$$ -h[n] - 0.6h[n-1] - 0.16h[n-2] = 5\delta[n] -$$ - -subject to zero initial state; that is, *h*[−1] = *h*[−2] = 0. Setting *n* = 0 in this equation yields - -*h*[0] −0.6(0)−0.16(0) = 5(1) ⇒ *h*[0] = 5 - -Setting *n* = 1 in the same equation and using *h*[0] = 5, we obtain - -*h*[1] −0.6(5)−0.16(0) = 5(0) ⇒ *h*[1] = 3 - -Continuing this way, we can determine any number of terms of *h*[*n*]. Unfortunately, such a solution does not yield a closed-form expression for *h*[*n*]. Nevertheless, determining a few values of *h*[*n*] can be useful in determining the closed-form solution, as the following development shows. - -## **[3.7-1 The Closed-Form Solution of](#page-9-0)** *h***[***n***]** - -Recall that *h*[*n*] is the system response to input δ[*n*], which is zero for *n* > 0. We know that when the input is zero, only the characteristic modes can be sustained by the system. Therefore, *h*[*n*] must be made up of characteristic modes for *n* > 0. At *n* = 0, it may have some nonzero value *A*0 so that a general form of *h*[*n*] can be expressed as† - -$$ -h[n] = A_0 \delta[n] + y_c[n]u[n] -$$ -\n(3.27) - -where *yc*[*n*] is a linear combination of the characteristic modes. We now substitute Eq. (3.27) in Eq. (3.26) to obtain *Q*[*E*](*A*0δ[*n*] +*yc*[*n*]*u*[*n*]) = *P*[*E*]δ[*n*]. Because *yc*[*n*] is made up of characteristic modes, *Q*[*E*]*yc*[*n*]*u*[*n*] = 0, and we obtain *A*0*Q*[*E*]δ[*n*] = *P*[*E*]δ[*n*], that is, - -$$ -A_0 (\delta[n+N] + a_1 \delta[n+N-1] + \cdots + a_N \delta[n]) = b_0 \delta[n+N] + \cdots + b_N \delta[n] -$$ - -Setting *n* = 0 in this equation and using the fact that δ[*m*] = 0 for all *m* = 0, and δ[0] = 1, we obtain - -$$ -A_0 a_N = b_N \quad \Longrightarrow \quad A_0 = \frac{b_N}{a_N} \tag{3.28} -$$ - -Hence,‡ - -$$ -h[n] = \frac{b_N}{a_N} \delta[n] + y_c[n]u[n] \tag{3.29} -$$ - -The *N* unknown coefficients in *yc*[*n*] (on the right-hand side) can be determined from a knowledge of *N* values of *h*[*n*]. Fortunately, it is a straightforward task to determine values of *h*[*n*] iteratively, as demonstrated in Ex. 3.17. We compute *N* values *h*[0], *h*[1], *h*[2], ... , *h*[*N* −1] iteratively. Now, setting *n* = 0, 1, 2, ... , *N* −1 in Eq. (3.29), we can determine the *N* unknowns in *yc*[*n*]. This point will become clear in the following example. - -### **EXAMPLE 3.18 Closed-Form Determination of the Impulse Response** - -Determine the unit impulse response *h*[*n*] for a system in Ex. 3.17 specified by the equation - -$$ -y[n] - 0.6y[n-1] - 0.16y[n-2] = 5x[n] -$$ - -† We assume that the term *yc*[*n*] consists of characteristic modes for *n* &gt; 0 only. To reflect this behavior, the characteristic terms should be expressed in the form γ *n j u*[*n* − 1]. But because *u*[*n* − 1] = *u*[*n*] − δ[*n*], *cj*γ *n j u*[*n* − 1] = *cj*γ *n j u*[*n*] − *cj*δ[*n*], and *yc*[*n*] can be expressed in terms of exponentials γ *n j u*[*n*] (which start at *n* = 0), plus an impulse at *n* = 0. - -‡ If *aN* = 0, then *A*0 cannot be determined by Eq. (3.28). In such a case, we show in Sec. 3.12 that *h*[*n*] is of the form *A*0δ[*n*] + *A*1δ[*n* − 1] + *yc*[*n*]*u*[*n*]. We have here *N* + 2 unknowns, which can be determined from *N* +2 values *h*[0],*h*[1],...,*h*[*N* +1] found iteratively. - -This equation can be expressed in the advance form as - -$$ -y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2] -$$ - -or in advance operator form as - -$$ -(E^2 - 0.6E - 0.16)y[n] = 5E^2x[n] -$$ - -The characteristic polynomial is - -$$ -\gamma^2 - 0.6\gamma - 0.16 = (\gamma + 0.2)(\gamma - 0.8) -$$ - -The characteristic modes are (−0.2)*n* and (0.8)*n*. Therefore, - -$$ -y_c[n] = c_1(-0.2)^n + c_2(0.8)^n -$$ - -Inspecting the system difference equation, we see that *aN* = −0.16 and *bN* = 0. Therefore, according to Eq. (3.29), - -$$ -h[n] = [c_1(-0.2)^n + c_2(0.8)^n]u[n] -$$ - -To determine *c*1 and *c*2, we need to find two values of *h*[*n*] iteratively. From Ex. 3.17, we know that *h*[0] = 5 and *h*[1] = 3. Setting *n* = 0 and 1 in our expression for *h*[*n*] and using the fact that *h*[0] = 5 and *h*[1] = 3, we obtain - -$$ -\begin{array}{c}\n5 = c_1 + c_2 \\ -3 = -0.2c_1 + 0.8c_2\n\end{array}\n\right\} \implies c_1 = 1\n\begin{array}{c}\nc_1 = 1 \\ -c_2 = 4\n\end{array} -$$ - -Therefore, - -$$ -h[n] = [(-0.2)^n + 4(0.8)^n]u[n] -$$ - -### **DR ILL 3.14 Closed-Form Determination of the Impulse Response** - -Find *h*[*n*], the unit impulse response of the LTID systems specified by the following equations: - -- **(a)** *y*[*n*+1] −*y*[*n*] = *x*[*n*] -- **(b)** *y*[*n*] −5*y*[*n*−1] +6*y*[*n*−2] = 8*x*[*n*−1] −19*x*[*n*−2] -- **(c)** *y*[*n*+2] −4*y*[*n*+1] +4*y*[*n*] = 2*x*[*n*+2] −2*x*[*n*+1] -- **(d)** *y*[*n*] = 2*x*[*n*] −2*x*[*n*−1] - -### **ANSWERS** - -- **(a)** *h*[*n*] = *u*[*n*−1] -- **(b)** *h*[*n*]=−19 6 δ[*n*] + 3 2 (2)*n* + 5 3 (3)*n u*[*n*] -- **(c)** *h*[*n*] = (2+*n*)2*nu*[*n*] -- **(d)** *h*[*n*] = 2δ[*n*] −2δ[*n*−1] - -### **EXAMPLE 3.19 Filtering Perspective of the Unit Impulse Response** - -Use the MATLAB filter command to solve Ex. 3.18. - -There are several ways to find the impulse response using MATLAB. In this method, we first specify the unit impulse function, which will serve as our input. Vectors a and b are created to specify the system. The filter command is then used to determine the impulse response. In fact, this method can be used to determine the zero-state response for any input. - -**Comment.** Although it is relatively simple to determine the impulse response *h*[*n*] by using the procedure in this section, in Ch. 5 we shall discuss the much simpler method of the *z*-transform. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/046_3.8 SYSTEM RESPONSE TO EXTERNAL INPUT - THE ZERO-STATE RESPONSE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/046_3.8 SYSTEM RESPONSE TO EXTERNAL INPUT - THE ZERO-STATE RESPONSE.md deleted file mode 100644 index 82d90e804d4e1729bcfe6bb9fb3d82fff4a895c4..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/046_3.8 SYSTEM RESPONSE TO EXTERNAL INPUT - THE ZERO-STATE RESPONSE.md +++ /dev/null @@ -1,620 +0,0 @@ -## **3.8 SYSTEM [RESPONSE TO](#page-9-0) EXTERNAL INPUT: THE ZERO-STATE RESPONSE** - -The zero-state response *y*[*n*] is the system response to an input *x*[*n*] when the system is in the zero state. In this section we shall assume that systems are in the zero state unless mentioned otherwise, so that the zero-state response will be the total response of the system. Here we follow the procedure parallel to that used in the continuous-time case by expressing an arbitrary input *x*[*n*] as a sum of impulse components. A signal *x*[*n*] in Fig. 3.20a can be expressed as a sum of impulse components, such as those depicted in Figs. 3.20b–3.20f. The component of *x*[*n*] at *n* = *m* is *x*[*m*]δ[*n*−*m*], and *x*[*n*] is the sum of all these components summed from *m* = −∞ to ∞. - -Therefore, - -$$ -x[n] = x[0]\delta[n] + x[1]\delta[n-1] + x[2]\delta[n-2] + \cdots -$$ - -+ x[-1]\delta[n+1] + x[-2]\delta[n+2] + \cdots -= -$$ -\sum_{m=-\infty}^{\infty} x[m]\delta[n-m] -$$ -(3.30) - -### 282 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -For a linear system, if we know the system response to impulse δ[*n*], we can obtain the system response to any arbitrary input by summing the system response to various impulse components. Let *h*[*n*] be the system response to impulse input δ[*n*]. We shall use the notation - -$$ -x[n] \Longrightarrow y[n] -$$ - -to indicate the input and the corresponding response of the system. Thus, if - -$$ -\delta[n] \Longrightarrow h[n] -$$ - -then because of time invariance - -$$ -\delta[n-m] \Longrightarrow h[n-m] -$$ - -and because of linearity - -$$ -x[m]\delta[n-m] \Longrightarrow x[m]h[n-m] -$$ - -and again because of linearity - -$$ -\underbrace{\sum_{m=-\infty}^{\infty} x[m]\delta[n-m]}_{x[n]} \quad \Longrightarrow \quad \underbrace{\sum_{m=-\infty}^{\infty} x[m]h[n-m]}_{y[n]} -$$ - -The left-hand side is *x*[*n*] [see Eq. (3.30)], and the right-hand side is the system response *y*[*n*] to input *x*[*n*]. Therefore,† - -$$ -y[n] = \sum_{m = -\infty}^{\infty} x[m]h[n-m] -$$ -\n(3.31) - -The summation on the right-hand side is known as the *convolution sum* of *x*[*n*] and *h*[*n*], and is represented symbolically by *x*[*n*] ∗ *h*[*n*] - -$$ -x[n] * h[n] = \sum_{m=-\infty}^{\infty} x[m]h[n-m] -$$ - -### PROPERTIES OF THE CONVOLUTION SUM - -The structure of the convolution sum is similar to that of the convolution integral. Moreover, the properties of the convolution sum are similar to those of the convolution integral. We shall enumerate these properties here without proof. The proofs are similar to those for the convolution integral and may be derived by the reader. - -$$ -y[n] = \sum_{m=-\infty}^{\infty} x[m]h[n,m] -$$ - -† In deriving this result, we have assumed a time-invariant system. The system response to input δ[*n* − *m*] for a time-varying system cannot be expressed as *h*[*n* − *m*]; instead, it has the form *h*[*n*, *m*]. Using this form, Eq. (3.31) is modified as follows: - -**The Commutative Property.** - -$$ -x_1[n] * x_2[n] = x_2[n] * x_1[n] -$$ - -**The Distributive Property.** - -$$ -x_1[n] * (x_2[n] + x_3[n]) = x_1[n] * x_2[n] + x_1[n] * x_3[n] -$$ - -**The Associative Property.** - -$$ -x_1[n] * (x_2[n] * x_3[n]) = (x_1[n] * x_2[n]) * x_3[n] -$$ - -**The Shifting Property.** If - -$$ -x_1[n] * x_2[n] = c[n] -$$ - -then - -$$ -x_1[n-m]*x_2[n-p] = c[n-m-p] -$$ -\n(3.32) - -**The Convolution with an Impulse.** - -$$ -x[n] * \delta[n] = x[n] -$$ - -**The Width Property.** If *x*1[*n*] and *x*2[*n*] have finite widths of *W*1 and *W*2, respectively, then the width of *x*1[*n*] ∗ *x*2[*n*] is *W*1 + *W*2. The width of a signal is 1 less than the number of its elements (length). Thus the signal in Fig. 3.22h has six elements (length of 6) but a width of only 5. Alternately, the property may be stated in terms of lengths as follows: if *x*1[*n*] and *x*2[*n*] have finite lengths of *L*1 and *L*2 elements, respectively, then the length of *x*1[*n*] ∗ *x*2[*n*] is *L*1 + *L*2 − 1 elements. - -### CAUSALITY AND ZERO-STATE RESPONSE - -In deriving Eq. (3.31), we assumed the system to be linear and time-invariant. There were no other restrictions on either the input signal or the system. In our applications, almost all the input signals are causal, and a majority of the systems are also causal. These restrictions further simplify the limits of the sum in Eq. (3.31). If the input *x*[*n*] is causal, *x*[*m*] = 0 for *m* < 0. Similarly, if the system is causal (i.e., if *h*[*n*] is causal), then *h*[*x*] = 0 for negative *x* so that *h*[*n* − *m*] = 0 when *m* > *n*. Therefore, if *x*[*n*] and *h*[*n*] are both causal, the product *x*[*m*]*h*[*n*−*m*] = 0 for *m* < 0 and for *m* > *n*, and it is nonzero only for the range 0 ≤ *m* ≤ *n*. Therefore, Eq. (3.31) in this case reduces to - -$$ -y[n] = \sum_{m=0}^{n} x[m]h[n-m] -$$ -\n(3.33) - -We shall evaluate the convolution sum first by an analytical method and later with graphical aid. - -### **EXAMPLE 3.20 Convolution of Causal Signals** - -Determine *c*[*n*] = *x*[*n*] ∗ *g*[*n*] for - -$$ -x[n] = (0.8)^n u[n] -$$ - and $g[n] = (0.3)^n u[n]$ - -We have - -$$ -c[n] = \sum_{m=-\infty}^{\infty} x[m]g[n-m] -$$ - -Note that - -$$ -x[m] = (0.8)^m u[m] -$$ - and $g[n-m] = (0.3)^{n-m} u[n-m]$ - -Both *x*[*n*] and *g*[*n*] are causal. Therefore [see Eq. (3.33)], - -$$ -c[n] = \sum_{m=0}^{n} x[m]g[n-m] = \sum_{m=0}^{n} (0.8)^m u[m] (0.3)^{n-m} u[n-m] -$$ - -In this summation, *m* lies between 0 and *n* (0 ≤ *m* ≤ *n*). Therefore, if *n* ≥ 0, then both *m* and *n*−*m* ≥ 0 so that *u*[*m*] = *u*[*n*−*m*] = 1. If *n* < 0, *m* is negative because *m* lies between 0 and *n*, and *u*[*m*] = 0. Therefore, - -$$ -c[n] = \begin{cases} \sum_{m=0}^{n} (0.8)^m (0.3)^{n-m} & n \ge 0\\ 0 & n < 0 \end{cases} -$$ - -or - -$$ -c[n] = (0.3)^n \sum_{m=0}^n \left(\frac{0.8}{0.3}\right)^m u[n] -$$ - -This is a geometric progression with common ratio (0.8/0.3). From Sec. B.8-3 we have - -$$ -c[n] = (0.3)^n \frac{(0.8)^{n+1} - (0.3)^{n+1}}{(0.3)^n (0.8 - 0.3)} u[n] -$$ - -= 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n] - -### **DR ILL 3.15 Convolution of Causal Signals** - -Show that (0.8)*nu*[*n*] ∗ *u*[*n*] = 5[1−(0.8)*n*+1]*u*[*n*]. - -### CONVOLUTION SUM FROM A TABLE - -Just as in the continuous-time case, we have prepared a table (Table 3.1) from which convolution sums may be determined directly for a variety of signal pairs. For example, the convolution in Ex. 3.20 can be read directly from this table (pair 4) as - -$$ -(0.8)^n u[n] * (0.3)^n u[n] = \frac{(0.8)^{n+1} - (0.3)^{n+1}}{0.8 - 0.3} u[n] = 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n] -$$ - -We shall demonstrate the use of the convolution table in the following example. - -| No. | x1[n] | x2[n] | x1[n]
∗ x2[n]
= x2[n]
∗ x1[n] | -|-----|-----------------------------|---------------|---------------------------------------------------------------------------------------------------------| -| 1 | δ[n−k] | x[n] | x[n−k] | -| 2 | γ nu[n] | u[n] | 1−γ n+1
!
u[n]
1−γ | -| 3 | u[n] | u[n] | (n+1)u[n] | -| 4 | γ n
1 u[n] | γ n
2 u[n] | γ n+1
−γ n+1
1
2
u[n]
γ1
= γ2
γ1
−γ2 | -| 5 | u[n] | nu[n] | n(n+1)
u[n]
2 | -| 6 | γ nu[n] | nu[n] | γ (γ n −1)
!
+n(1−γ )
u[n]
(1−γ )2 | -| 7 | nu[n] | nu[n] | 1
6 n(n−1)(n+1)u[n] | -| 8 | γ nu[n] | γ nu[n] | (n+1)γ nu[n] | -| 9 | nγ n
1 u[n] | γ n
2 u[n] | !
γ1γ2
γ1
−γ2
γ n
2 −γ n
nγ n
u[n]
γ1
= γ2
1 +
1
−γ2)2
(γ1
γ2 | -| 10 | n cos(βn+θ
γ1
)u[n] | nu[n]
γ2 | 1
n+1 cos[β(n+1)+θ−φ]− γ2
n+1 cos(θ−φ)]u[n]
R[ γ1 | -| | | | 1/2
R =
2 +
2 −2 γ1 γ2
γ1
γ2
cosβ | -| | | | !
( γ1 sinβ)
φ = tan−1
( γ1 cosβ − γ2 ) | -| 11 | γ n
1 u[−(n+1)] | γ n
2 u[n] | γ2
γ1
γ n
γ n
2 u[n] +
1 u[−(n+1)]
γ1 > γ2
γ1
−γ2
γ1
−γ2 | - -**TABLE 3.1** Select Convolution Sums - -### **EXAMPLE 3.21 Convolution by Tables** - -Using Table 3.1, find the (zero-state) response *y*[*n*] of an LTID system described by the equation - -$$ -y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2] -$$ - -if the input *x*[*n*] = 4−*nu*[*n*]. - -The input can be expressed as *x*[*n*] = 4−*nu*[*n*] = (1/4)*nu*[*n*] = (0.25)*nu*[*n*]. The unit impulse response of this system, obtained in Ex. 3.18, is - -$$ -h[n] = [(-0.2)^n + 4(0.8)^n]u[n] -$$ - -Therefore, - -$$ -y[n] = x[n] * h[n] -$$ - -= (0.25)nu[n] \* [(-0.2)nu[n] + 4(0.8)nu[n] -= (0.25)nu[n] \* (-0.2)nu[n] + (0.25)nu[n] \* 4(0.8)nu[n] - -We use pair 4 (Table 3.1) to find the foregoing convolution sums. - -$$ -y[n] = \left[\frac{(0.25)^{n+1} - (-0.2)^{n+1}}{0.25 - (-0.2)} + 4 \frac{(0.25)^{n+1} - (0.8)^{n+1}}{0.25 - 0.8}\right] u[n] -$$ - -= $(2.22[(0.25)^{n+1} - (-0.2)^{n+1}] - 7.27[(0.25)^{n+1} - (0.8)^{n+1}])u[n]$ -= $[-5.05(0.25)^{n+1} - 2.22(-0.2)^{n+1} + 7.27(0.8)^{n+1}]u[n]$ - -Recognizing that - -$$ -\gamma^{n+1} = \gamma(\gamma)^n -$$ - -we can express *y*[*n*] as - -$$ -y[n] = [-1.26(0.25)^{n} + 0.444(-0.2)^{n} + 5.81(0.8)^{n}]u[n] -$$ - -= [-1.26(4)-n + 0.444(-0.2)n + 5.81(0.8)n]u[n] - -### **DR ILL 3.16 Convolution by Tables** - -Use Table 3.1 to show that - -(a) -$$ -(0.8)^{n+1}u[n] * u[n] = 4[1 - 0.8(0.8)^n]u[n] -$$ - -**(b)** *n*3−*nu*[*n*] ∗ (0.2)*nu*[*n*] = 15 4 (0.2)*n* − 1− 2 3 *n* 3−*n u*[*n*] - -(c) -$$ -e^{-n}u[n] * 2^{-n}u[n] = \frac{2}{2-e} \left[ e^{-n} - \frac{e}{2}2^{-n} \right] u[n] -$$ - -### **EXAMPLE 3.22 Filtering Perspective of the Zero-State Response** - -Use the MATLAB filter command to compute and sketch the zero-state response for the system described by (*E*2 +0.5*E* −1)*y*[*n*] = (2*E*2 +6*E*)*x*[*n*] and the input *x*[*n*] = 4−*nu*[*n*]. - -We solve this problem using the same approach as Ex. 3.19. Although the input is bounded and quickly decays to zero, the system itself is unstable and an unbounded output results. - ->> n = (0:11); x = @(n) 4.^(-n).\*(n>=0); >> a = [1 0.5 -1]; b = [2 6 0]; y = filter(b,a,x(n)); >> clf; stem(n,y,'k'); xlabel('n'); ylabel('y[n]'); axis([-0.5 11.5 -20 25]); - -### RESPONSE TO COMPLEX INPUTS - -As in the case of real continuous-time systems, we can show that for an LTID system with real *h*[*n*], if the input and the output are expressed in terms of their real and imaginary parts, then the real part of the input generates the real part of the response and the imaginary part of the input generates the imaginary part. Thus, if - -$$ -x[n] = x_r[n] + jx_i[n] -$$ - and $y[n] = y_r[n] + jy_i[n]$ - -using the right-directed arrow to indicate the input–output pair, we can show that - -$$ -x_r[n] \Longrightarrow y_r[n] -$$ - and $x_i[n] \Longrightarrow y_i[n]$ (3.34) - -The proof is similar to that used to derive Eq. (2.31) for LTIC systems. - -### MULTIPLE INPUTS - -Multiple inputs to LTI systems can be treated by applying the superposition principle. Each input is considered separately, with all other inputs assumed to be zero. The sum of all these individual system responses constitutes the total system output when all the inputs are applied simultaneously. - -### **DR ILL 3.17 Response to Multiple Inputs** - -Show that the system described by *y*[*n*] − 0.6*y*[*n* − 1] − 0.16*y*[*n* − 2] = 5*x*[*n*] responds to input *x*[*n*] = δ[*n*] +4−*nu*[*n*] with output *y*[*n*] = [−1.26(4)−*n* +1.444(−0.2)*n* +9.81(0.8)*n*]*u*[*n*]. [*Hint:* Use the results of Exs. 3.18 and 3.21.] - -### **[3.8-1 Graphical Procedure for the Convolution Sum](#page-9-0)** - -The steps in evaluating the convolution sum are parallel to those followed in evaluating the convolution integral. The convolution sum of causal signals *x*[*n*] and *g*[*n*] is given by - -$$ -c[n] = \sum_{m=0}^{n} x[m]g[n-m] -$$ - -We first plot *x*[*m*] and *g*[*n* − *m*] as functions of *m* (not *n*), because the summation is over *m*. Functions *x*[*m*] and *g*[*m*] are the same as *x*[*n*] and *g*[*n*], plotted, respectively, as functions of *m* (see Fig. 3.22). The convolution operation can be performed as follows: - -- 1. Invert *g*[*m*] about the vertical axis (*m* = 0) to obtain *g*[−*m*] (Fig. 3.22d). Figure 3.22e shows both *x*[*m*] and *g*[−*m*]. -- 2. Shift *g*[−*m*] by *n* units to obtain *g*[*n* − *m*]. For *n* > 0, the shift is to the right (delay); for *n* < 0, the shift is to the left (advance). Figure 3.22f shows *g*[*n* − *m*] for *n* > 0; for *n* < 0, see Fig. 3.22g. -- 3. Next we multiply *x*[*m*] and *g*[*n*−*m*] and add all the products to obtain *c*[*n*]. The procedure is repeated for each value of *n* over the range −∞ to ∞. - -We shall demonstrate by an example the graphical procedure for finding the convolution sum. Although both the functions in this example are causal, this procedure is applicable to the general case. - -### **EXAMPLE 3.23 Graphical Procedure for the Convolution Sum** - -Find *c*[*n*] = *x*[*n*] ∗*g*[*n*], where *x*[*n*] and *g*[*n*] are depicted in Figs. 3.22a and 3.22b, respectively. - -We are given - -*x*[*n*] = (0.8) *n* and *g*[*n*] = (0.3) *n* - -Therefore, - -*x*[*m*] = (0.8) *m* and *g*[*n*−*m*] = (0.3) *n*−*m* - -**Figure 3.22** Graphical procedure to convolve *x*[*n*] and *g*[*n*]. - -### 290 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Figure 3.22f shows the general situation for *n* ≥ 0. The two functions *x*[*m*] and *g*[*n*−*m*] overlap over the interval 0 ≤ *m* ≤ *n*. Therefore, - -$$ -c[n] = \sum_{m=0}^{n} x[m]g[n-m] -$$ - -= $\sum_{m=0}^{n} (0.8)^m (0.3)^{n-m}$ -= $(0.3)^n \sum_{m=0}^{n} \left(\frac{0.8}{0.3}\right)^m$ -= $2[(0.8)^{n+1} - (0.3)^{n+1}]$ $n \ge 0$ (see Sec. B.8-3) - -For *n* < 0, there is no overlap between *x*[*m*] and *g*[*n*−*m*], as shown in Fig. 3.22g, so that - -*c*[*n*] = 0 *n* < 0 - -Combining pieces, we see that - -$$ -c[n] = 2[(0.8)^{n+1} - (0.3)^{n+1}]u[n] -$$ - -which agrees with the result found earlier in Ex. 3.20. - -### **DR ILL 3.18 Graphical Procedure for the Convolution Sum** - -Find (0.8)*nu*[*n*] ∗ *u*[*n*] graphically and sketch the result. - -### **ANSWER** - -5(1−(0.8)*n*+1)*u*[*n*] - -### AN ALTERNATIVE FORM OF GRAPHICAL PROCEDURE: THE SLIDING-TAPE METHOD - -This algorithm is convenient when the sequences *x*[*n*] and *g*[*n*] are short or when they are available only in graphical form. The algorithm is basically the same as the graphical procedure in Fig. 3.22. The only difference is that instead of presenting the data as graphical plots, we display it as a sequence of numbers on tapes. Otherwise the procedure is the same, as will become clear in the following example. - -### **EXAMPLE 3.24 Sliding-Tape Method for the Convolution Sum** - -Use the sliding-tape method to convolve the two sequences *x*[*n*] and *g*[*n*] depicted in Figs. 3.23a and 3.23b, respectively. - -In this procedure we write the sequences *x*[*n*] and*g*[*n*] in the slots of two tapes: *x* tape and *g* tape (Fig. 3.23c). Now leave the *x* tape stationary (to correspond to *x*[*m*]). The *g*[−*m*] tape is obtained by inverting the *g*[*m*] tape about the origin (*m* = 0) so that the slots corresponding to *x*[0] and *g*[0] remain aligned (Fig. 3.23d). We now shift the inverted tape by *n* slots, multiply values on two tapes in adjacent slots, and add all the products to find *c*[*n*]. Figures 3.23d–3.23i show the cases for *n* = 0–5. Figures 3.23j, 3.23k, and 3.23l show the cases for *n* = −1,−2, and −3, respectively. - -For the case of *n* = 0, for example (Fig. 3.23d), - -$$ -c[0] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) = -3 -$$ - -For *n* = 1 (Fig. 3.23e), - -$$ -c[1] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) = -2 -$$ - -Similarly, - -$$ -c[2] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) = 0 -$$ - -\n -$$ -c[3] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) = 3 -$$ - -\n -$$ -c[4] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) + (4 \times 1) = 7 -$$ - -\n -$$ -c[5] = (-2 \times 1) + (-1 \times 1) + (0 \times 1) + (1 \times 1) + (2 \times 1) + (3 \times 1) + (4 \times 1) = 7 -$$ - -Figure 3.23i shows that *c*[*n*] = 7 for *n* ≥ 4. - -Similarly, we compute *c*[*n*] for negative *n* by sliding the tape backward, one slot at a time, as shown in the plots corresponding to *n* = −1, −2, and −3, respectively (Figs. 3.23j, 3.23k, and 3.23l). - -$$ -c[-1] = (-2 \times 1) + (-1 \times 1) = -3 -$$ - -\n -$$ -c[-2] = (-2 \times 1) = -2 -$$ - -\n -$$ -c[-3] = 0 -$$ - -Figure 3.23l shows that *c*[*n*] = 0 for *n* ≤ 3. Figure 3.23m shows the plot of *c*[*n*]. - -Rotate the *g* tape about the vertical axis as shown in (d) - -### **DR ILL 3.19 Sliding-Tape Method for the Convolution Sum** - -Use the graphical procedure of Ex. 3.24 (sliding-tape technique) to show that *x*[*n*] ∗ *g*[*n*] = *c*[*n*] in Fig. 3.24. Verify the width property of convolution. - -## **EXAMPLE 3.25 Convolution of Two Finite-Duration Signals Using MATLAB** - -For the signals *x*[*n*] and *g*[*n*] depicted in Fig. 3.24, use MATLAB to compute and plot *c*[*n*] = *x*[*n*] ∗ *g*[*n*]. - -### **[3.8-2 Interconnected Systems](#page-9-0)** - -As with continuous-time case, we can determine the impulse response of systems connected in parallel (Fig. 3.26a) and cascade (Figs. 3.26b, 3.26c). We can use arguments identical to those used for the continuous-time systems in Sec. 2.4-3 to show that if two LTID systems *S*1 and *S*2 with impulse responses *h*1[*n*] and *h*2[*n*], respectively, are connected in parallel, the composite parallel system impulse response is *h*1[*n*] + *h*2[*n*]. Similarly, if these systems are connected in cascade, the impulse response of the composite system is *h*1[*n*] ∗ *h*2[*n*]. Moreover, because *h*1[*n*] ∗ *h*2[*n*] = *h*2[*n*] ∗ *h*1[*n*], linear systems commute. Their orders can be interchanged without affecting the composite system behavior. - -**Figure 3.26** Interconnected systems. - -### INVERSE SYSTEMS - -If the two systems in cascade are the inverse of each other, with impulse responses *h*[*n*] and *hi*[*n*], respectively, then the impulse response of the cascade of these systems is *h*[*n*] ∗ *hi*[*n*]. But, the cascade of a system with its inverse is an identity system, whose output is the same as the input. Hence, the unit impulse response of an identity system is δ[*n*]. Consequently, - -$$ -h[n] * h_i[n] = \delta[n] -$$ - -As an example, we show that an accumulator system and a backward difference system are the inverse of each other. An accumulator system is specified by† - -$$ -y[n] = \sum_{k=-\infty}^{n} x[k] \tag{3.35} -$$ - -The backward difference system is specified by - -$$ -y[n] = x[n] - x[n-1] \tag{3.36} -$$ - -From Eq. (3.35), we find *h*acc[*n*], the impulse response of the accumulator, as - -$$ -h_{\text{acc}}[n] = \sum_{k=-\infty}^{n} \delta[k] = u[n] -$$ - -Similarly, from Eq. (3.36), *h*bdf[*n*], the impulse response of the backward difference system is given by - -$$ -h_{\text{bdf}}[n] = \delta[n] - \delta[n-1] -$$ - -We can verify that - -$$ -h_{\text{acc}} * h_{\text{bdf}} = u[n] * {\delta[n] - \delta[n-1]} = u[n] - u[n-1] = \delta[n] -$$ - -Roughly speaking, a discrete-time accumulator is analogous to a continuous-time integrator, and a backward difference system is analogous to a differentiator. We have already encountered examples of these systems in Exs. 3.8 and 3.9 (digital differentiator and integrator). - -### SYSTEM RESPONSE TO %*n k*=−∞ *x*[*k*] - -Figure 3.26d shows a cascade of two LTID systems: a system *S* with impulse response *h*[*n*], followed by an accumulator. Figure 3.26e shows a cascade of the same two systems in reverse order: an accumulator followed by *S*. In Fig. 3.26d, if the input *x*[*n*] to *S* results in the output *y*[*n*], then the output of the system in Fig. 3.26d is the %*y*[*k*]. In Fig. 3.26e, the output of the accumulator is the sum %*x*[*k*]. Because the output of the system in Fig. 3.26e is identical to that of system Fig. 3.26d, it follows that - -if -$$ -x[n] \Longrightarrow y[n] -$$ -, then $\sum_{k=-\infty}^{n} x[k] \Longrightarrow \sum_{k=-\infty}^{n} y[k]$ - -† Equations (3.35) and (3.36) are identical to Eqs. (3.10) and (3.8), respectively, with *T* = 1. - -#### 296 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -If we let *x*[*n*] = δ[*n*] and *y*[*n*] = *h*[*n*], we find that *g*[*n*], the unit step response of an LTID system with impulse response *h*[*n*], is given by - -$$ -g[n] = \sum_{k=-\infty}^{n} h[k] \tag{3.37} -$$ - -The reader can readily prove the inverse relationship - -$$ -h[n] = g[n] - g[n-1] -$$ - -### A VERY SPECIAL FUNCTION FOR LTID SYSTEMS: THE EVERLASTING EXPONENTIAL *zn* - -In Sec. 2.4-4, we showed that there exists one signal for which the response of an LTIC system is the same as the input within a multiplicative constant. The response of an LTIC system to an everlasting exponential input *est* is *H*(*s*)*est*, where *H*(*s*) is the system transfer function. We now show that for an LTID system, the same role is played by an everlasting exponential *zn*. The system response *y*[*n*] in this case is given by - -$$ -y[n] = h[n] * zn -$$ - -= -$$ -\sum_{m=-\infty}^{\infty} h[m] z^{n-m} -$$ - -= -$$ -zn \sum_{m=-\infty}^{\infty} h[m] z^{-m} -$$ - -For causal *h*[*n*], the limits on the sum on the right-hand side would range from 0 to ∞. In any case, this sum is a function of *z*. Assuming that this sum converges, let us denote it by *H*[*z*]. Thus, - -$$ -y[n] = H[z]z^n \tag{3.38} -$$ - -where - -$$ -H[z] = \sum_{m=-\infty}^{\infty} h[m]z^{-m} -$$ -\n(3.39) - -Equation (3.38) is valid only for values of *z* for which the sum on the right-hand side of Eq. (3.39) exists (converges). Note that *H*[*z*] is a constant for a given *z*. Thus, the input and the output are the same (within a multiplicative constant) for the everlasting exponential input *zn*. - -*H*[*z*], which is called the *transfer function* of the system, is a function of the complex variable *z*. An alternate definition of the transfer function *H*[*z*] of an LTID system from Eq. (3.38) is - -$$ -H[z] = \frac{\text{output signal}}{\text{input signal}} \bigg|_{\text{input} = \text{everlasting exponential } z^n} -$$ - (3.40) - -The transfer function is defined for, and is meaningful to, LTID systems only. It does not exist for nonlinear or time-varying systems in general. - -We repeat again that in this discussion we are talking of the everlasting exponential, which starts at *n* = −∞, not the causal exponential *znu*[*n*], which starts at *n* = 0. - -For a system specified by Eq. (3.20), the transfer function is given by - -$$ -H[z] = \frac{P[z]}{Q[z]} \tag{3.41} -$$ - -This follows readily by considering an everlasting input *x*[*n*] = *zn*. According to Eq. (3.40), the output is *y*[*n*] = *H*[*z*]*zn*. Substitution of this *x*[*n*] and *y*[*n*] in Eq. (3.20) yields - -$$ -H[z]\{Q[E]z^n\} = P[E]z^n -$$ - -Moreover, - -$$ -E^k z^n = z^{n+k} = z^k z^n -$$ - -Hence, - -$$ -P[E]z^n = P[z]z^n \qquad \text{and} \qquad Q[E]z^n = Q[z]z^n -$$ - -Consequently, - -$$ -H[z] = \frac{P[z]}{Q[z]} -$$ - -### **DR ILL 3.20 DT System Transfer Function** - -Show that the transfer function of the digital differentiator in Ex. 3.8 (big shaded block in Fig. 3.16b) is given by *H*[*z*] = (*z*−1)/*Tz*, and the transfer function of an unit delay, specified by *y*[*n*] = *x*[*n*−1], is given by 1/*z*. - -### **[3.8-3 Total Response](#page-9-0)** - -The total response of an LTID system can be expressed as a sum of the zero-input and zero-state responses: - -total response = -$$ -\underbrace{\sum_{j=1}^{N} c_j \gamma_j^n}_{\text{ZIR}} + \underbrace{x[n] * h[n]}_{\text{ZSR}} -$$ - -In this expression, the zero-input response should be appropriately modified for the case of repeated roots. We have developed procedures to determine these two components. From the system equation, we find the characteristic roots and characteristic modes. The zero-input response is a linear combination of the characteristic modes. From the system equation, we also determine *h*[*n*], the impulse response, as discussed in Sec. 3.7. Knowing *h*[*n*] and the input *x*[*n*], we find the zero-state response as the convolution of *x*[*n*] and *h*[*n*]. The arbitrary constants *c*1, *c*2,..., *cn* in the zero-input response are determined from the *n* initial conditions. For the system described by the equation - -$$ -y[n+2] - 0.6y[n+1] - 0.16y[n] = 5x[n+2] -$$ - -#### 298 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -with initial conditions *y*[−1] = 0, *y*[−2] = 25/4 and input *x*[*n*] = (4)−*nu*[*n*], we have determined the two components of the response in Exs. 3.13 and 3.21, respectively. From the results in these examples, the total response for *n* ≥ 0 is - -total response = -$$ -\underbrace{0.2(-0.2)^n + 0.8(0.8)^n}_{\text{ZIR}} + \underbrace{0.444(-0.2)^n + 5.81(0.8)^n - 1.26(4)^{-n}}_{\text{ZSR}} -$$ -(3.42) - -### NATURAL AND FORCED RESPONSE - -The characteristic modes of this system are (−0.2)*n* and (0.8)*n*. The zero-input response is made up of characteristic modes exclusively, as expected, but the characteristic modes also appear in the zero-state response. When all the characteristic mode terms in the total response are lumped together, the resulting component is the *natural response*. The remaining part of the total response that is made up of noncharacteristic modes is the *forced response*. For the present case, Eq. (3.42) yields - -total response = -$$ -\underbrace{0.644(-0.2)^n + 6.61(0.8)^n}_{\text{natural response}} + \underbrace{-1.26(4)^{-n}}_{\text{forced response}} \qquad n \ge 0 -$$ - -Just like differential equations, the classical solution to difference equations includes the natural and forced responses, a decomposition that lacks the engineering intuition and utility afforded by the zero-input and zero-state responses. The classical approach cannot separate the responses arising from internal conditions and external input. While the natural and forced solutions can be obtained from the zero-input and zero-state responses, the converse is not true. Further, the classical method is unable to express the system response to an input *x*[*n*] as an explicit function of *x*[*n*]. In fact, the classical method is restricted to a certain class of inputs and cannot handle arbitrary inputs as can the method to determine the zero-state response. For these (and other) reasons, we do not further detail the classical approach and its direct calculation of the forced and natural responses. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/047_3.9 SYSTEM STABILITY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/047_3.9 SYSTEM STABILITY.md deleted file mode 100644 index ba3d8d1f3576c5bb6e7093d94f12ad6470ed193f..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/047_3.9 SYSTEM STABILITY.md +++ /dev/null @@ -1,176 +0,0 @@ -## **[3.9 SYSTEM](#page-9-0) STABILITY** - -The concepts and criteria for the BIBO (external) stability and internal (asymptotic) stability for discrete-time systems are identical to those corresponding to continuous-time systems. The comments in Sec. 2.5 for LTIC systems concerning the distinction between external and internal stability are also valid for LTID systems. Let us begin with external (BIBO) stability. - -### **[3.9-1 External \(BIBO\) Stability](#page-9-0)** - -Recall that - -$$ -y[n] = h[n] * x[n] = \sum_{m=-\infty}^{\infty} h[m]x[n-m] -$$ - -and - -$$ -|y[n]| = \left|\sum_{m=-\infty}^{\infty} h[m]x[n-m]\right| \le \sum_{m=-\infty}^{\infty} |h[m]| |x[n-m]| -$$ - -If *x*[*n*] is bounded, then |*x*[*n*−*m*]| < *K*1 < ∞, and - -$$ -|y[n]| \leq K_1 \sum_{m=-\infty}^{\infty} |h[m]| -$$ - -Clearly the output is bounded if the summation on the right-hand side is bounded; that is, if - -$$ -\sum_{n=-\infty}^{\infty} |h[n]| < K_2 < \infty \tag{3.43} -$$ - -This is a sufficient condition for BIBO stability. We can show that this is also a necessary condition (see Prob. 3.9-1). Therefore, if the impulse response *h*[*n*] of an LTID system is absolutely summable, the system is (BIBO) stable. Otherwise it is unstable. - -All the comments about the nature of external and internal stability in Ch. 2 apply to discrete-time case. We shall not elaborate them further. - -### **[3.9-2 Internal \(Asymptotic\) Stability](#page-9-0)** - -For LTID systems, as in the case of LTIC systems, internal stability, called asymptotical stability or stability in the sense of Lyapunov (also the zero-input stability), is defined in terms of the zero-input response of a system. - -For an LTID system specified by a difference equation in the form of Eq. (3.15) [or Eq. (3.20)], the zero-input response consists of the characteristic modes of the system. The mode corresponding to a characteristic root γ is γ *n*. To be more general, let γ be complex so that - -$$ -\gamma = |\gamma|e^{i\beta} -$$ - and $\gamma^n = |\gamma|^n e^{i\beta n}$ - -Since the magnitude of *ej*β*n* is always unity regardless of the value of *n*, the magnitude of γ *n* is |γ | *n*. Therefore, - -if -$$ -|\gamma| < 1 -$$ -, then $\gamma^n \to 0$ as $n \to \infty$ -if $|\gamma| > 1$ , then $\gamma^n \to \infty$ as $n \to \infty$ -and if $|\gamma| = 1$ , then $|\gamma|^n = 1$ for all *n* - -The characteristic modes corresponding to characteristic roots at various locations in the complex plane appear in Fig. 3.27. - -These results can be grasped more effectively in terms of the location of characteristic roots in the complex plane. Figure 3.28 shows a circle of unit radius, centered at the origin in a complex plane. Our discussion shows that if all characteristic roots of the system lie inside the *unit circle*, |γ*i*| < 1 for all *i* and the system is asymptotically stable. On the other hand, even if one characteristic root lies outside the unit circle, the system is unstable. If none of the characteristic - -**Figure 3.27** Characteristic roots locations and the corresponding characteristic modes. - -(d) - -(a) - -(b) - -roots lie outside the unit circle, but some simple (unrepeated) roots lie on the circle itself, the system is marginally stable. If two or more characteristic roots coincide on the unit circle (repeated roots), the system is unstable. The reason is that for repeated roots, the zero-input response is of the form *nr*−1γ *n*, and if |γ | = 1, then |*nr*−1γ *n*| = *nr*−1 → ∞ as *n* → ∞. † Note, however, that repeated roots inside the unit circle do not cause instability. - -**Figure 3.28** Characteristic root locations and system stability. - -To summarize: - -- 1. An LTID system is asymptotically stable if, and only if, all the characteristic roots are inside the unit circle. The roots may be simple or repeated. -- 2. An LTID system is unstable if, and only if, either one or both of the following conditions exist: (i) at least one root is outside the unit circle; (ii) there are repeated roots on the unit circle. -- 3. An LTID system is marginally stable if and only if there are no roots outside the unit circle and there are some unrepeated roots on the unit circle. - -### **[3.9-3 Relationship Between BIBO and Asymptotic Stability](#page-9-0)** - -For LTID systems, the relation between the two types of stability is similar to those in LTIC systems. For a system specified by Eq. (3.15), we can readily show that if a characteristic root γ*k* - -† If the development of discrete-time systems is parallel to that of continuous-time systems, we wonder why the parallel breaks down here. Why, for instance, are LHP and RHP not the regions demarcating stability and instability? The reason lies in the form of the characteristic modes. In continuous-time systems, we chose the form of characteristic mode as *e*λ*it* . In discrete-time systems, for computational convenience, we choose the form to be γ *n i* . Had we chosen this form to be *e*λ*in* where γ*i* = *e*λ*i* , then the LHP and RHP (for the location of λ*i*) again would demarcate stability and instability. The reason is that if γ = *e*λ, |γ | = 1 implies |*e*λ| = 1, and therefore λ = *j*ω. This shows that the unit circle in γ plane maps into the imaginary axis in the λ plane. - -### 302 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -is inside the unit circle, the corresponding mode γ *n k* is absolutely summable. In contrast, if γ*k* lies outside the unit circle, or on the unit circle, γ *n k* is not absolutely summable.† - -This means that an asymptotically stable system is BIBO-stable. Moreover, a marginally stable or asymptotically unstable system is BIBO-unstable. The converse is not necessarily true. The stability picture portrayed by the external description is of questionable value. BIBO (external) stability cannot ensure internal (asymptotic) stability, as the following example shows. - -### **EXAMPLE 3.26 A BIBO-Stable but Asymptotically Unstable System** - -An LTID systems consists of two subsystems *S*1 and *S*2 in cascade (Fig. 3.29). The impulse response of these systems are *h*1[*n*] and *h*2[*n*], respectively, given by - -> *h*1[*n*] = 4δ[*n*] −3(0.5) *n u*[*n*] and *h*2[*n*] = 2*n u*[*n*] - -Investigate the BIBO and asymptotic stability of the composite system. - -The composite system impulse response *h*[*n*] is given by - -$$ -h[n] = h_1[n] * h_2[n] = h_2[n] * h_1[n] = 2nu[n] * (4\delta[n] - 3(0.5)nu[n]) -$$ - -= 4(2)nu[n] - 3\left[\frac{2^{n+1} - (0.5)n+1}{2 - 0.5}\right]u[n] -= (0.5)nu[n] - -If the composite cascade system were to be enclosed in a black box with only the input and the output terminals accessible, any measurement from these external terminals would show that the impulse response of the system is (0.5)*nu*[*n*], without any hint of the unstable system sheltered inside the composite system. - -† This conclusion follows from the fact that (see Sec. B.8-3) - -$$ -\sum_{n=-\infty}^{\infty} |\gamma_k^n| u[n] = \sum_{n=0}^{\infty} |\gamma_k|^n = \frac{1}{1 - |\gamma_k|} \qquad |\gamma_k| < 1 -$$ - -Moreover, if |γ | ≥ 1, the sum diverges and goes to ∞. These conclusions are valid also for the modes of the form *nr* γ *n k* . - -The composite system is BIBO-stable because its impulse response (0.5)*nu*[*n*] is absolutely summable. However, the system *S*2 is asymptotically unstable because its characteristic root, 2, lies outside the unit circle. This system will eventually burn out (or saturate) because of the unbounded characteristic response generated by intended or unintended initial conditions, no matter how small. - -The system is asymptotically unstable, though BIBO-stable. This example shows that BIBO stability does not necessarily ensure asymptotic stability when a system is uncontrollable, unobservable, or both. The internal and the external descriptions of a system are equivalent only when the system is controllable and observable. In such a case, BIBO stability means the system is asymptotically stable, and vice versa. - -Fortunately, uncontrollable or unobservable systems are not common in practice. Henceforth, in determining system stability, we shall assume that unless otherwise mentioned, the internal and the external descriptions of the system are equivalent, implying that the system is controllable and observable. - -### **EXAMPLE 3.27 Investigating Asymptotic and BIBO Stability** - -Determine the internal and external stability of systems specified by the following equations. In each case plot the characteristic roots in the complex plane. - -- **(a)** *y*[*n*+2] +2.5*y*[*n*+1] +*y*[*n*] = *x*[*n*+1] −2*x*[*n*] -- **(b)** *y*[*n*] −*y*[*n*−1] +0.21*y*[*n*−2] = 2*x*[*n*−1] +3*x*[*n*−2] -- **(c)** *y*[*n*+3] +2*y*[*n*+2] + 3 2 *y*[*n*+1] + 1 2 *y*[*n*] = *x*[*n*+1] -- **(d)** (*E*2 −*E* +1)2*y*[*n*] = (3*E* +1)*x*[*n*] - -**(a)** The characteristic polynomial is - -$$ -\gamma^2 + 2.5\gamma + 1 = (\gamma + 0.5)(\gamma + 2) -$$ - -The characteristic roots are −0.5 and −2. Because | − 2| > 1 (−2 lies outside the unit circle), the system is BIBO-unstable and also asymptotically unstable (Fig. 3.30a). - -**(b)** The characteristic polynomial is - -$$ -\gamma^2 - \gamma + 0.21 = (\gamma - 0.3)(\gamma - 0.7) -$$ - -The characteristic roots are 0.3 and 0.7, both of which lie inside the unit circle. The system is BIBO-stable and asymptotically stable (Fig. 3.30b). - -**(c)** The characteristic polynomial is - -$$ -\gamma^{3} + 2\gamma^{2} + \frac{3}{2}\gamma + \frac{1}{2} = (\gamma + 1)(\gamma^{2} + \gamma + \frac{1}{2}) = (\gamma + 1)(\gamma + 0.5 - j0.5)(\gamma + 0.5 + j0.5) -$$ - -### 304 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -The characteristic roots are −1, −0.5 ± *j*0.5 (Fig. 3.30c). One of the characteristic roots is on the unit circle and the remaining two roots are inside the unit circle. The system is BIBO-unstable but marginally stable. - -**(d)** The characteristic polynomial is - -$$ -(\gamma^2 - \gamma + 1)^2 = \left(\gamma - \frac{1}{2} - j\frac{\sqrt{3}}{2}\right)^2 \left(\gamma - \frac{1}{2} + j\frac{\sqrt{3}}{2}\right)^2 -$$ - -The characteristic roots are (1/2)±*j*( √3/2) = 1*e*±*j*(π/3) repeated twice, and they lie on the unit circle (Fig. 3.30d). The system is BIBO-unstable and asymptotically unstable. - -### **DR ILL 3.21 Assessing Stability by Characteristic Roots** - -Using the complex plane, locate the characteristic roots of the following systems, and use the characteristic root locations to determine external and internal stability of each system. - -**(a)** (*E* +1)(*E*2 +6*E* +25)*y*[*n*] = 3*Ex*[*n*] - -**(b)** (*E* −1)2(*E* +0.5)*y*[*n*] = (*E*2 +2*E* +3)*x*[*n*] - -### **ANSWERS** - -Both systems are BIBO-and asymptotically unstable. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/048_3.10 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/048_3.10 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR.md deleted file mode 100644 index 1e104e92bfba95a32aac49730fc24abbe706ac00..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/048_3.10 INTUITIVE INSIGHTS INTO SYSTEM BEHAVIOR.md +++ /dev/null @@ -1,41 +0,0 @@ -## **[3.10 INTUITIVE](#page-9-0) INSIGHTS INTO SYSTEM BEHAVIOR** - -The intuitive insights into the behavior of continuous-time systems and their qualitative proofs, discussed in Sec. 2.6, also apply to discrete-time systems. For this reason, we shall merely mention here without discussion some of the insights presented in Sec. 2.6. - -The system's entire (zero-input and zero-state) behavior is strongly influenced by the characteristic roots (or modes) of the system. The system responds strongly to input signals similar to its characteristic modes and poorly to inputs very different from its characteristic modes. In fact, when the input is a characteristic mode of the system, the response goes to infinity, provided the mode is a nondecaying signal. This is the resonance phenomenon. The width of an impulse response *h*[*n*] indicates the response time (time required to respond fully to an input) of the system. It is the time constant of the system.† Discrete-time pulses are generally dispersed when passed through a discrete-time system. The amount of dispersion (or spreading out) is equal to the system time constant (or width of *h*[*n*]). The system time constant also determines the rate at which the system can transmit information. A smaller time constant corresponds to a higher rate of information transmission, and vice versa. We keep in mind that concepts such as time constant and pulse dispersion only coarsely illustrate system behavior. Let us illustrate these ideas with an example. - -### **EXAMPLE 3.28 Intuitive Insights into Lowpass DT System Behavior** - -Determine the time constant, rise time, pulse dispersion, and filter characteristics of a lowpass DT system with impulse response *h*[*n*] = 2(0.6)*nu*[*n*]. - -† This part of the discussion applies to systems with impulse response *h*[*n*] that is a mostly positive (or mostly negative) pulse. - -### 306 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -Since *h*[*n*] resembles a single, mostly positive pulse, we know that the DT system is lowpass. Similar to the CT case shown in Sec. 2.6, we can determine the time constant *Th* as the width of a rectangle that approximates *h*[*n*]. This rectangle possesses the same peak height and total sum (area), as does *h*[*n*]. The peak of *h*[*n*] is 2, and the total sum (area) is - -$$ -\sum_{n=0}^{\infty} 2(0.6)^n = 2\frac{1-0}{1-0.6} = 5 -$$ - -Since the width of a DT signal is 1 less than its length, we see that the time constant *Th* (rectangle width) is - -$$ -T_h = \text{rectangle width} = \frac{\text{area}}{\text{height}} - 1 = \frac{5}{2} - 1 = 1.5 \text{ samples} -$$ - -Since time constant, rise time, pulse dispersion are all given by the same value, we see that - -time constant = rise time = pulse dispersion = *Th* = 1.5 samples - -The approximate cutoff frequency of our DT system can be determined as the frequency of a DT sinusoid whose period equals the length of the rectangle approximation to *h*[*n*]. That is, - -cutoff frequency = -$$ -\frac{1}{T_h + 1} = \frac{2}{5} -$$ - cycles/sample - -Equivalently, we can express the cutoff frequency as 4π/5 radians/sample. - -Notice that *Th* is not an integer and thus lacks a clear physical meaning for our DT system. How, for example, can it take 1.5 samples for our DT system to fully respond to an input? We can put our minds at ease by remembering the approximate nature of *Th*, which is meant to provide only a rough understanding of system behavior. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/049_3.11 MATLAB - DISCRETE-TIME SIGNALS AND SYSTEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/049_3.11 MATLAB - DISCRETE-TIME SIGNALS AND SYSTEMS.md deleted file mode 100644 index 669dfaf8484e7379afc7c6b31c065a21732b0cd2..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/049_3.11 MATLAB - DISCRETE-TIME SIGNALS AND SYSTEMS.md +++ /dev/null @@ -1,175 +0,0 @@ -## **[3.11 MATLAB: DISCRETE-TIME](#page-9-0) SIGNALS AND SYSTEMS** - -MATLAB is naturally and ideally suited to discrete-time signals and systems. Many special functions are available for discrete-time data operations, including the stem, filter, and conv commands. In this section, we investigate and apply these and other commands. - -### **[3.11-1 Discrete-Time Functions and Stem Plots](#page-9-0)** - -Consider the discrete-time function *f*[*n*] = *e*−*n*/5 cos(π*n*/5)*u*[*n*]. In MATLAB, there are many ways to represent *f*[*n*] including M-files or, for particular *n*, explicit command line evaluation. In this example, however, we use an anonymous function. - ->> f = @(n) exp(-n/5).\*cos(pi\*n/5).\*(n>=0); - -A true discrete-time function is undefined (or zero) for noninteger *n*. Although anonymous function f is intended as a discrete-time function, its present construction does not restrict *n* to be integer, and it can therefore be misused. For example, MATLAB dutifully returns 0.8606 to f(0.5) when a NaN (not-a-number) or zero is more appropriate. The user is responsible for appropriate function use. - -Next, consider plotting the discrete-time function *f*[*n*] over (−10 ≤ *n* ≤ 10). The stem command simplifies this task. - -``` ->> n = (-10:10)'; ->> stem(n,f(n),'k'); ->> xlabel('n'); ylabel('f[n]'); -``` - -Here, stem operates much like the plot command: dependent variable f(n) is plotted against independent variable n with black lines. The stem command emphasizes the discrete-time nature of the data, as Fig. 3.31 illustrates. - -For discrete-time functions, the operations of shifting, inversion, and scaling can have surprising results. Compare *f*[−2*n*] with *f*[−2*n* + 1]. Contrary to the continuous case, the second is not a shifted version of the first. We can use separate subplots, each over (−10 ≤ *n* ≤ 10), to help illustrate this fact. Notice that unlike the plot command, the stem command cannot simultaneously plot multiple functions on a single axis; overlapping stem lines would make such plots difficult to read anyway. - -``` ->> subplot(2,1,1); stem(n,f(-2*n),'k'); ylabel('f[-2n]'); ->> subplot(2,1,2); stem(n,f(-2*n+1),'k'); ylabel('f[-2n+1]'); xlabel('n'); -``` - -The results are shown in Fig. 3.32. Interestingly, the original function *f*[*n*] can be recovered by interleaving samples of *f*[−2*n*] and *f*[−2*n*+1] and then time-reflecting the result. - -Care must always be taken to ensure that MATLAB performs the desired computations. Our anonymous function f is a case in point: although it correctly downsamples, it does not properly upsample (see Prob. 3.11-2). MATLAB does what it is told, but it is not always told how to do everything correctly! - -**Figure 3.31** *f*[*n*] over (−10 ≤ *n* ≤ 10). - -**Figure 3.32** *f*[−2*n*] and *f*[−2*n*+1] over (−10 ≤ *n* ≤ 10). - -### **[3.11-2 System Responses Through Filtering](#page-9-0)** - -MATLAB's filter command provides an efficient way to evaluate the system response of a constant coefficient linear difference equation represented in delay form as - -$$ -\sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{N} b_k x[n-k] -$$ -\n(3.44) - -In the simplest form, filter requires three input arguments: a length-(*N* + 1) vector of feedforward coefficients [*b*0,*b*1,...,*bN*], a length-(*N* + 1) vector of feedback coefficients [*a*0,*a*1,...,*aN*], and an input vector.† Since no initial conditions are specified, the output corresponds to the system's zero-state response. - -To serve as an example, consider a system described by *y*[*n*]−*y*[*n*−1]+*y*[*n*−2] = *x*[*n*]. When *x*[*n*] = δ[*n*], the zero-state response is equal to the impulse response *h*[*n*], which we compute over (0 ≤ *n* ≤ 30). - ->> b = [1 0 0]; a = [1 -1 1]; >> n = (0:30)'; delta = @(n) 1.0.\*(n==0); >> h = filter(b,a,delta(n)); >> clf; stem(n,h,'k'); axis([-.5 30.5 -1.1 1.1]); >> xlabel('n'); ylabel('h[n]'); - -† It is important to pay close attention to the inevitable notational differences found throughout engineering documents. In MATLAB help documents, coefficient subscripts begin at 1 rather than 0 to better conform with MATLAB indexing conventions. That is, MATLAB labels *a*0 as a(1), *b*0 as b(1), and so forth. - -**Figure 3.33** *h*[*n*] for *y*[*n*] −*y*[*n*−1] +*y*[*n*−2] = *x*[*n*]. - -**Figure 3.34** Resonant zero-state response *y*[*n*] for *x*[*n*] = cos(2π*n*/6)*u*[*n*]. - -As shown in Fig. 3.33, *h*[*n*] appears to be (*N*0 = 6)-periodic for *n* ≥ 0. Since periodic signals are not absolutely summable, %∞ *n*=−∞ |*h*[*n*]| is not finite and the system is not BIBO-stable. Furthermore, the sinusoidal input *x*[*n*] = cos(2π*n*/6)*u*[*n*], which is (*N*0 = 6)-periodic for *n* ≥ 0, should generate a resonant zero-state response. - -``` ->> x = @(n) cos(2*pi*n/6).*(n>=0); ->> y = filter(b,a,x(n)); ->> stem(n,y,'k'); xlabel('n'); ylabel('y[n]'); -``` - -The response's linear envelope, shown in Fig. 3.34, confirms a resonant response. The characteristic equation of the system is γ 2 − γ + 1, which has roots γ = *e*±*j*π/3. Since the input *x*[*n*] = cos(2π*n*/6)*u*[*n*] = (1/2)(*ej*π*n*/3 + *e*−*j*π*n*/3)*u*[*n*] coincides with the characteristic roots, a resonant response is guaranteed. - -By adding initial conditions, the filter command can also compute a system's zero-input response and total response. Continuing the preceding example, consider finding the zero-input response for *y*[−1] = 1 and *y*[−2] = 2 over (0 ≤ *n* ≤ 30). - -``` ->> z_i = filtic(b,a,[1 2]); ->> y_0 = filter(b,a,zeros(size(n)),z_i); ->> stem(n,y_0,'k'); xlabel('n'); ylabel('y_{0} [n]'); ->> axis([-0.5 30.5 -2.1 2.1]); -``` - -**Figure 3.35** Zero-input response *y*0[*n*] for *y*[−1] = 1 and *y*[−2] = 2. - -There are many physical ways to implement a particular equation. MATLAB implements Eq. (3.44) by using the popular direct form II transposed structure.† Consequently, initial conditions must be compatible with this implementation structure. The signal-processing toolbox function filtic converts the traditional *y*[−1], *y*[−2], ..., *y*[−*N*] initial conditions for use with the filter command. An input of zero is created with the zeros command. The dimensions of this zero input are made to match the vector n by using the size command. Finally, \_{ } forces subscript text in the graphics window, and ^{ } forces superscript text. The results are shown in Fig. 3.35. - -Given *y*[−1] = 1 and *y*[−2] = 2 and an input *x*[*n*] = cos(2π*n*/6)*u*[*n*], the total response is easy to obtain with the filter command. - -$$ -\Rightarrow y\_total = filter(b,a,x(n),z_i); -$$ - -Summing the zero-state and zero-input response gives the same result. Computing the total absolute error provides a check. - -``` ->> sum(abs(y_total-(y + y_0))) - ans = 1.8430e-014 -``` - -Within computer round-off, both methods return the same sequence. - -### **[3.11-3 A Custom Filter Function](#page-9-0)** - -The filtic command is available only if the signal-processing toolbox is installed. To accommodate installations without the signal-processing toolbox and to help develop your MATLAB skills, consider writing a function similar in syntax to filter that directly uses the ICs *y*[−1], *y*[−2], ..., *y*[−*N*]. Normalizing *a*0 = 1 and solving Eq. (3.44) for *y*[*n*] yield - -$$ -y[n] = \sum_{k=0}^{N} b_k x[n-k] - \sum_{k=1}^{N} a_k y[n-k] -$$ - -This recursive form provides a good basis for our custom filter function. - -† Implementation structures, such as direct form II transposed, are discussed in Ch. 4. - -``` -function [y] = CH3MP1(b,a,x,yi); -% CH3MP1.m : Chapter 3, MATLAB Program 1 -% Function M-file filters data x to create y -% INPUTS: b = vector of feedforward coefficients -% a = vector of feedback coefficients -% x = input data vector -% yi = vector of initial conditions [y[-1], y[-2], ...] -% OUTPUTS: y = vector of filtered output data -yi = flipud(yi(:)); % Properly format IC's. -y = [yi;zeros(length(x),1)]; % Preinitialize y, beginning with IC's. -x = [zeros(length(yi),1);x(:)]; % Append x with zeros to match size of y. -b = b/a(1);a = a/a(1); % Normalize coefficients. -for n = length(yi)+1:length(y), - for nb = 0:length(b)-1, - y(n) = y(n) + b(nb+1)*x(n-nb); % Feedforward terms. - end - for na = 1:length(a)-1, - y(n) = y(n) - a(na+1)*y(n-na); % Feedback terms. - end -end -y = y(length(yi)+1:end); % Strip off IC's for final output. -``` - -Most instructions in CH3MP1 have been discussed; now we turn to the flipud instruction. The flip up-down command flipud reverses the order of elements in a column vector. Although not used here, the flip left-right command fliplr reverses the order of elements in a row vector. Note that typing help *filename* displays the first contiguous set of comment lines in an M-file. Thus, it is good programming practice to document M-files, as in CH3MP1, with an initial block of clear comment lines. - -As an exercise, the reader should verify that CH3MP1 correctly computes the impulse response *h*[*n*], the zero-state response *y*[*n*], the zero-input response *y*0[*n*], and the total response *y*[*n*]+*y*0[*n*]. - -### **[3.11-4 Discrete-Time Convolution](#page-9-0)** - -Convolution of two finite-duration discrete-time signals is accomplished by using the conv command. For example, the discrete-time convolution of two length-4 rectangular pulses, *g*[*n*] = (*u*[*n*]−*u*[*n*−4])∗(*u*[*n*]−*u*[*n*−4]), is a length-(4+4−1=7) triangle. Representing *u*[*n*]−*u*[*n*−4] by the vector [1, 1, 1, 1], the convolution is computed by - ->> conv([1 1 1 1],[1 1 1 1]) ans = 1 2 3 4 3 2 1 - -Notice that (*u*[*n*+4] −*u*[*n*]) ∗ (*u*[*n*] −*u*[*n*−4]) is also computed by conv([1 1 1 1],[1 1 1 1]) and obviously yields the same result. The difference between these two cases is the regions of support: (0 ≤ *n* ≤ 6) for the first and (−4 ≤ *n* ≤ 2) for the second. Although the conv command - -### 312 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -does not compute the region of support, it is relatively easy to obtain. If vector w begins at *n* = *nw* and vector v begins at *n* = *nv*, then conv(w,v) begins at *n* = *nw* +*nv*. - -In general, the conv command cannot properly convolve infinite-duration signals. This is not too surprising, since computers themselves cannot store an infinite-duration signal. For special cases, however, conv can correctly compute a portion of such convolution problems. Consider the common case of convolving two causal signals. By passing the first *N* samples of each, conv returns a length-(2*N* − 1) sequence. The first *N* samples of this sequence are valid; the remaining *N* −1 samples are not. - -To illustrate this point, reconsider the zero-state response *y*[*n*] over (0 ≤ *n* ≤ 30) for system *y*[*n*]−*y*[*n*−1]+*y*[*n*−2] = *x*[*n*] given input *x*[*n*] = cos(2π*n*/6)*u*[*n*]. The results obtained by using a filtering approach are shown in Fig. 3.34. - -The response can also be computed using convolution according to *y*[*n*] = *h*[*n*] ∗ *x*[*n*]. The impulse response of this system is† - -$$ -h[n] = \left\{ \cos\left(\frac{\pi n}{3}\right) + \frac{1}{\sqrt{3}} \sin\left(\frac{\pi n}{3}\right) \right\} u[n] -$$ - -Both *h*[*n*] and *x*[*n*] are causal and have infinite duration, so conv can be used to obtain a portion of the convolution. - -``` ->> u = @(n) 1.0.*(n>=0); h = @(n) (cos(pi*n/3)+sin(pi*n/3)/sqrt(3)).*u(n); ->> y = conv(h(n),x(n)); ->> stem([0:60],y,'k'); xlabel('n'); ylabel('y[n]'); -``` - -The conv output is fully displayed in Fig. 3.36. As expected, the results are correct over (0 ≤ *n* ≤ 30). The remaining values are clearly incorrect; the output envelope should continue to grow, not decay. Normally, these incorrect values are not displayed. - ->> stem(n,y(1:31),'k'); xlabel('n'); ylabel('y[n]'); - -The resulting plot is identical to Fig. 3.34. - -**Figure 3.36** *y*[*n*] for *x*[*n*] = cos(2π*n*/6)*u*[*n*] computed with conv. - -† Techniques to analytically determine *h*[*n*] are presented in Ch. 5. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/050_3.12 APPENDIX - IMPULSE RESPONSE FOR A SPECIAL CASE.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/050_3.12 APPENDIX - IMPULSE RESPONSE FOR A SPECIAL CASE.md deleted file mode 100644 index 59d3a438497f70a22fbcc2a2d0bd2fa57a1bbb09..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/050_3.12 APPENDIX - IMPULSE RESPONSE FOR A SPECIAL CASE.md +++ /dev/null @@ -1,21 +0,0 @@ -## **[3.12 APPENDIX: IMPULSE](#page-9-0) RESPONSE FOR A SPECIAL CASE** - -When *aN* =0, *A*0 =*bN*/*aN* becomes indeterminate, and the procedure needs to be modified slightly. When *aN* = 0, *Q*[*E*] can be expressed as *EQ*ˆ [*E*], and Eq. (3.26) can be expressed as - -$$ -E\tilde{Q}[E]h[n] = P[E]\delta[n] = P[E]\{E\delta[n-1]\} = EP[E]\delta[n-1] -$$ - -Hence, - -$$ -\hat{Q}[E]h[n] = P[E]\delta[n-1] -$$ - -In this case the input vanishes not for *n* ≥ 1, but for *n* ≥ 2. Therefore, the response consists not only of the zero-input term and an impulse *A*0δ[*n*] (at *n* = 0), but also of an impulse *A*1δ[*n*−1] (at *n* = 1). Therefore, - -$$ -h[n] = A_0 \delta[n] + A_1 \delta[n-1] + y_c[n]u[n] -$$ - -We can determine the unknowns *A*0, *A*1, and the *N* − 1 coefficients in *yc*[*n*] from the *N* + 1 number of initial values *h*[0], *h*[1], ... , *h*[*N*], determined as usual from the iterative solution of the equation *Q*[*E*]*h*[*n*] = *P*[*E*]δ[*n*]. † Similarly, if *aN* = *aN*−1 = 0, we need to use the form *h*[*n*] = *A*0δ[*n*]+*A*1δ[*n*−1]+*A*2δ[*n*−2]+*yc*[*n*]*u*[*n*]. The *N* +1 unknown constants are determined from the *N* +1 values *h*[0], *h*[1], ... , *h*[*N*], determined iteratively, and so on. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/051_3.13 SUMMARY.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/051_3.13 SUMMARY.md deleted file mode 100644 index 9cf57cf973202ddcded64cccc87e87ddf11be92b..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/051_3.13 SUMMARY.md +++ /dev/null @@ -1,25 +0,0 @@ -## **[3.13 SUMMARY](#page-9-0)** - -This chapter discusses time-domain analysis of LTID (linear, time-invariant, discrete-time) systems. The analysis is parallel to that of LTIC systems, with some minor differences. Discrete-time systems are described by difference equations. For an *N*th-order system, *N* auxiliary conditions must be specified for a unique solution. Characteristic modes are discrete-time exponentials of the form γ *n* corresponding to an unrepeated root γ , and the modes are of the form *ni* γ *n* corresponding to a repeated root γ . - -The unit impulse function δ[*n*] is a sequence of a single number of unit value at *n* = 0. The unit impulse response *h*[*n*] of a discrete-time system is a linear combination of its characteristic modes.‡ - -The zero-state response (response due to external input) of a linear system is obtained by breaking the input into impulse components and then adding the system responses to all the impulse components. The sum of the system responses to the impulse components is in the form of a sum, known as the convolution sum, whose structure and properties are similar to the convolution integral. The system response is obtained as the convolution sum of the input *x*[*n*] with the system's impulse response *h*[*n*]. Therefore, the knowledge of the system's impulse response allows us to determine the system response to any arbitrary input. - -LTID systems have a very special relationship to the everlasting exponential signal *zn* because the response of an LTID system to such an input signal is the same signal within a multiplicative - -† *Q*ˆ [γ ] is now an (*N* −1)-order polynomial. Hence there are only *N* −1 unknowns in *yc*[*n*]. ‡ There is a possibility of an impulse δ[*n*] in addition to characteristic modes. - -### 314 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -constant. The response of an LTID system to the everlasting exponential input *zn* is *H*[*z*]*zn*, where *H*[*z*] is the transfer function of the system. - -The external stability criterion, the bounded-input/bounded-output (BIBO) stability criterion, states that a system is stable if and only if every bounded input produces a bounded output. Otherwise the system is unstable. - -The internal stability criterion can be stated in terms of the location of characteristic roots of the system as follows: - -- 1. An LTID system is asymptotically stable if and only if all the characteristic roots are inside the unit circle. The roots may be repeated or unrepeated. -- 2. An LTID system is unstable if and only if either one or both of the following conditions exist: (i) at least one root is outside the unit circle; (ii) there are repeated roots on the unit circle. -- 3. An LTID system is marginally stable if and only if there are no roots outside the unit circle and some unrepeated roots on the unit circle. - -An asymptotically stable system is always BIBO-stable. The converse is not necessarily true. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/052_PROBLEMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/052_PROBLEMS.md deleted file mode 100644 index 942181533f5c05f1006deb3ca4a5e38646f7d766..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/052_PROBLEMS.md +++ /dev/null @@ -1,965 +0,0 @@ -## **[PROBLEMS](#page-9-0)** - -- **3.1-1** Find the energy of the signals depicted in Fig. P3.1-1. -- **3.1-2** Find the power of the signals illustrated in Fig. P3.1-2. -- **3.1-3** Show that the power of a signal *Dej*(2π/*N*0)*n* is |*D*| 2. Hence, show that the power of a signal *x*[*n*] =%*N*0−1 *r*=0 *Drejr*(2π/*N*0)*n* is *Px* =%*N*0−1 *r*=0 |*Dr*| 2. Use the fact that - -$$ -\sum_{k=0}^{N_0-1} e^{j(r-m)2\pi k/N_0} = \begin{cases} N_0 & r=m\\ 0 & \text{otherwise} \end{cases} -$$ - -- **3.1-4** (a) Determine even and odd components of the signal *x*[*n*] = (0.8)*nu*[*n*]. - - (b) Show that the energy of *x*[*n*] is the sum of energies of its odd and even components found in part (a). - - (c) Generalize the result in part (b) for any finite energy signal. -- **3.1-5** (a) If *xe*[*n*] and *xo*[*n*] are the even and the odd components of causal energy signal *x*[*n*], then determine *Exe* and *Ex*0 , and show that *Exe* +*Ex*0 = *Ex*. - - (b) Show that the cross-energy of *xe* and *xo* is zero, that is, - -$$ -\sum_{n=-\infty}^{\infty} x_e[n]x_o[n] = 0 -$$ - -**3.1-6** Define - -$$ -x[n] = \begin{cases} \left(\frac{1}{3}\right)^n & n \ge 0\\ A^n & n < 0 \end{cases} -$$ - -- (a) Determine the energy *Ex* and power *Px* of *x*[*n*] if *A* = 1 2 . -- (b) Determine the energy *Ex* and power *Px* of *x*[*n*] if *A* = 1. -- (c) Determine the energy *Ex* and power *Px* of *x*[*n*] if *A* = 2. -- **3.1-7** Determine the energy *Ex* and power *Px* of the complex DT signal *x*[*n*] = Re\* 3(*ej*π/4)*n* + -- **3.2-1** If the energy of a signal *x*[*n*] is *Ex*, then find the energy of the following: - - (a) *x*[−*n*] - - (b) *x*[*n*−*m*] - - (c) *x*[*m*−*n*] - - (d) *Kx*[*n*] (*m* integer and *K* constant) -- **3.2-2** If the power of a periodic signal *x*[*n*] is *Px*, find and comment on the powers and the rms values of the following: - - (a) −*x*[*n*] - - (b) *x*[−*n*] - - (c) *x*[*n*−*m*] (*m* integer) - - (d) *cx*[*n*] - - (e) *x*[*m*−*n*] (*m* integer) - -Problems 315 - -**Figure P3.1-2** - -- **3.2-3** Letting ↓ identify *n* = 0, define the nonzero values of signal *x*[*n*] as [−1, 2,−3, 4,−5, 4,−3, ↓ 2,−1]. - - (a) Using vector form, represent signal *y*[*n*] = *x*[−3*n* + 2]. Be sure to identify the *n* = 0 element. - - (b) Using vector form, represent signal *z*[*n*] = *x*[*n*/2 − 3]. Be sure to identify the *n* = 0 element. -- **3.2-4** Letting ↓ identify *n* = 0, define the nonzero values of signal *x*[*n*] as [−1, ↓ 2, - - −3, 4,−5, 4,−3, 2,−1]. (a) Determine the energy *Ex* and power *Px* of the signal *x*[*n*]. - - (b) Using vector form, represent signal *y*[*n*] = *x*[2(*n* + 2)]. Be sure to identify the *n* = 0 element. - - (c) Using vector form, represent signal *z*[*n*] = *x*[−*n*−6 3 ]. Be sure to identify the *n* = 0 element. -- **3.2-5** Letting ↓ identify the *n* = 0 value, describe a 4-periodic signal *w*[*n*] using vector notation as - - [· · · , 1, 2, 3, 4, ↓ 1, 2, 3, 4, 1, 2, 3, 4,···]. - - (a) Determine the energy *Ex* and power *Px* of signal *x*[*n*] = *w*[2*n*]. - - (b) Determine the energy *Ey* and power *Py* of signal *y*[*n*] = *w*[2− *n* 3 ]. -- **3.2-6** Let DT signal *x*[*n*] have values [1, 2, 3, 4, 5, 6] for 0 ≤ *n* ≤ 5 and let DT signal *y*[*n*] have values [5, 0, 0, 3, 0, 0, 1] for 0 ≤ *n* ≤ 6. Both signals are zero outside the ranges given. Further, define a 6-periodic replication of *y*[*n*] as *y*˜[*n*] = %∞ *k*=−∞ *y*[*n*−6*k*] . - - (a) Determine the energy *Ex* and power *Px* of signal *x*[*n*]. - - (b) Determine the smallest-magnitude integers *N*1, *N*2, and *N*3 such that *y*[*n*] = *x*[ *N*1*n N*2 +*N*3]. - - (c) Determine the energy *Ey*˜ and power *Py*˜ of *y*˜[*n*]. -- **3.2-7** For the signal shown in Fig. P3.1-1b, sketch the following signals: - - (a) *x*[−*n*] - - (b) *x*[*n*+6] - - (c) *x*[*n*−6] - - (d) *x*[3*n*] - - (e) *x* ' *n* ( - - 3 (f) *x*[3−*n*] - -- **3.2-8** Repeat Prob. 3.2-7 for the signal depicted in Fig. P3.1-1c. -- **3.2-9** Letting ↓ identify *n* = 0, consider a DT signal *x*[*n*] whose nonzero values are given as *x*[*n*]=[1, −3, 2, 2, 3, −2, −1, 1, 2, −3, ↓ 3, 3, −2, 1, −3, 2, 3, −1]. Accurately sketch *y*[*n*] = *x*[−1 − 2*n*] and *z*[*n*] = *x*[−2 + *n*/3] over −5 ≤ *n* ≤ 4. -- **3.2-10** Define - -$$ -x[n] = \begin{cases} \left(\frac{1}{2}\right)^n & n \ge 0\\ 0 & n < 0 \end{cases} -$$ - -Determine and locate the two largest non-zero values of: - -- (a) *y*a[*n*] = *x*[2*n*] -- (b) *y*b[*n*] = *x*[*n*/3] -- (c) *y*c[*n*] = *x*[3*n*+1] -- (d) *y*d[*n*] = *x*[−2*n*+5] -- (e) *y*e[*n*] = *x*[−(*n*+8)/2] -- **3.3-1** Sketch, and find the power of, the following signals: - - (a) (1)*n* - - (b) (−1)*n* - - (c) *u*[*n*] - - (d) (−1)*nu*[*n*] - - (e) cos'π 3 *n*+ π 6 ( -- **3.3-2** Show that - - (a) δ[*n*] +δ[*n*−1] = *u*[*n*] −*u*[*n*−2] - - (b) 2*n*−1 sin π*n* 3 *u*[*n*]= 1 2 2*n* sin π*n* 3 *u*[*n*−1] - - (c) *n*(*n*−1)γ *nu*[*n*] = *n*(*n*−1)γ *nu*[*n*−2] - - (d) (*u*[*n*] +(−1)*nu*[*n*])sinπ*n* = 0 for all *n* - -(e) -$$ -(u[n] + (-1)^{n+1}u[n])\cos(\frac{\pi n}{2}) = 0 -$$ - for all *n* - -- **3.3-3** Sketch the following signals: - - (a) *u*[*n*−2] −*u*[*n*−6] - - (b) *n*{*u*[*n*] −*u*[*n*−7]} - - (c) (*n*−2){*u*[*n*−2] −*u*[*n*−6]} - - (d) (−*n*+8){*u*[*n*−6] −*u*[*n*−9]} - - (e) (*n*−2){*u*[*n*−2]−*u*[*n*−6]}+(−*n*+8){*u*[*n*− 6] −*u*[*n*−9]} -- **3.3-4** Describe each of the signals in Fig. P3.1-1 by a single expression valid for all *n*. -- **3.3-5** Why are DT signals of the form *zn* so important to the study of LTID systems? -- **3.3-6** Explain the similarities and differences between the Kronecker delta function δ[*n*] and the Dirac delta function δ(*t*). - -- **3.3-7** The following signals are in the form *e*λ*n*. Express them in the form γ *n*: - - (a) *e*−0.5*n* - - (b) *e*0.5*n* - - (c) *e*−*j*π*n* - - (d) *ej*π*n* - -In each case show the locations of λ and γ in the complex plane. Verify that an exponential is growing if γ lies outside the unit circle (or if λ lies in the RHP), is decaying if γ lies within the unit circle (or if λ lies in the LHP), and has a constant amplitude if γ lies on the unit circle (or if λ lies on the imaginary axis). - -- **3.3-8** Express the following signals, which are in the form *e*λ*n*, in the form γ *n*: - - (a) *e*−(1+*j*π )*n* - - (b) *e*−(1−*j*π )*n* - - (c) *e*(1+*j*π )*n* - - (d) *e*(1−*j*π )*n* - - (e) *e*−[1+*j*(π/3)]*n* - - (f) *e*[1−*j*(π/3)]*n* -- **3.3-9** The concepts of even and odd functions for discrete-time signals are identical to those of the continuous-time signals discussed in Sec. 1.5. Using these concepts, find and sketch the odd and the even components of the following: - - (a) *u*[*n*] - - (b) *nu*[*n*] - - (c) sinπ*n* 4 - - (d) cosπ*n* 4 -- **3.4-1** A cash register output *y*[*n*] represents the total cost of *n* items rung up by a cashier. The input *x*[*n*] is the cost of the *n*th item. - - (a) Write the difference equation relating *y*[*n*] to *x*[*n*]. - - (b) Realize this system using a time-delay element. -- **3.4-2** Let *p*[*n*] be the population of a certain country at the beginning of the *n*th year. The birth and death rates of the population during any year are 3.3 and 1.3%, respectively. If *i*[*n*] is the total number of immigrants entering the country during the *n*th year, write the difference equation relating *p*[*n* + 1], *p*[*n*], and *i*[*n*]. Assume that the immigrants enter the country throughout the year at a uniform rate. - -- **3.4-3** A moving average is used to detect a trend of a rapidly fluctuating variable, such as the stock market average. A variable may fluctuate (up and down) daily, masking its long-term (secular) trend. We can discern the long-term trend by smoothing or averaging the past *N* values of the variable. For the stock market average, we may consider a 5-day moving average *y*[*n*] to be the mean of the past 5 days' market closing values *x*[*n*], *x*[*n*−1],..., *x*[*n*−4]. - - (a) Write the difference equation relating *y*[*n*] to the input *x*[*n*]. - - (b) Use time-delay elements to realize the 5-day moving-average filter. -- **3.4-4** The digital integrator in Ex. 3.9 is specified by - -$$ -y[n] - y[n-1] = Tx[n] -$$ - -If an input *u*[*n*] is applied to such an integrator, show that the output is (*n* + 1)*Tu*[*n*], which approaches the desired ramp *nTu*[*n*] as *T* → 0. - -**3.4-5** Approximate the following second-order differential equation with a difference equation. - -$$ -\frac{d^2y(t)}{dt^2} + a_1 \frac{dy(t)}{dt} + a_0 y(t) = x(t) -$$ - -- **3.4-6** Letting ↓ identify *n* = 0, define the nonzero values of signal *g*[*n*] in vector form as [1, 2, 3, 4, 5, 4, 3, ↓ 2, 1]. The impulse response of an LTID system is defined in terms of *g*[*n*] as *h*[*n*] = *g*[−2*n*−1]. - - (a) Express the nonzero values of *h*[*n*] in vector form, taking care to identify the *n* = 0 point. - - (b) Write a constant-coefficient linear difference equation (input *x*[*n*] and output *y*[*n*]) that has impulse response *h*[*n*]. - - (c) Show that the system is both linear and time-invariant. - - (d) Determine, if possible, whether the system is BIBO-stable. - - (e) Determine, if possible, whether the system is memoryless. - - (f) Determine, if possible, whether the system is causal. -- **3.4-7** An LTID system has an impulse response function *h*[*n*] = *u*[−(5−*n*)/3]. - - (a) Using an accurate sketch or vector representation, graphically depict *h*[*n*]. - -- (b) Determine, if possible, whether the system is BIBO-stable. -- (c) Determine, if possible, whether the system is memoryless. -- (d) Determine, if possible, whether the system is causal. -- **3.4-8** The voltage at the *n*th node of a resistive ladder in Fig. P3.4-8 is *v*[*n*], (*n* = 0, 1, 2,...,*N*). Show that *v*[*n*] satisfies the second-order difference equation - -$$ -v[n+2] - Av[n+1] + v[n] = 0 \quad A = 2 + \frac{1}{a} -$$ - -[*Hint:* Consider the node equation at the *n*th node with voltage *v*[*n*].] - -- **3.4-9** Determine whether each of the following statements is true or false. If the statement is false, demonstrate by proof or example why the statement is false. If the statement is true, explain why. - - (a) A discrete-time signal with finite power cannot be an energy signal. - -- (b) A discrete-time signal with infinite energy must be a power signal. -- (c) The system described by *y*[*n*] = (*n*+1)*x*[*n*] is causal. -- (d) The system described by *y*[*n* − 1] = *x*[*n*] is causal. -- (e) If an energy signal *x*[*n*] has energy *E*, then the energy of *x*[*an*] is *E*/|*a*|. -- **3.4-10** A linear time-invariant system produces output *y*1[*n*] in response to input *x*1[*n*], as shown in Fig. P3.4-10. Determine and sketch the output *y*2[*n*] that results when input *x*2[*n*] is applied to the same system. -- **3.4-11** A system is described by - -$$ -y[n] = \frac{1}{2} \sum_{k=-\infty}^{\infty} x[k] (\delta[n-k] + \delta[n+k]) -$$ - -- (a) Explain what this system does. -- (b) Is the system BIBO-stable? Justify your answer. -- (c) Is the system linear? Justify your answer. - -**Figure P3.4-8** - -**Figure P3.4-10** - -- (d) Is the system memoryless? Justify your answer. -- (e) Is the system causal? Justify your answer. -- (f) Is the system time-invariant? Justify your answer. -- **3.4-12** A discrete-time system is given by - -$$ -y[n+1] = \frac{x[n]}{x[n+1]} -$$ - -- (a) Is the system BIBO-stable? Justify your answer. -- (b) Is the system memoryless? Justify your answer. -- (c) Is the system causal? Justify your answer. -- **3.4-13** Explain why the continuous-time system *y*(*t*) = *x*(2*t*) is always invertible and yet the corresponding discrete-time system *y*[*n*] = *x*[2*n*] is not invertible. -- **3.4-14** Consider the input–output relationships of two similar discrete-time systems: - -$$ -y_1[n] = \sin\left(\frac{\pi}{2}n + 1\right)x[n] -$$ - -and - -$$ -y_2[n] = \sin\left(\frac{\pi}{2}(n+1)\right) x[n] -$$ - -Explain why *x*[*n*] can be recovered from *y*1[*n*] yet *x*[*n*] cannot be recovered from *y*2[*n*]. - -- **3.4-15** Consider a system that multiplies a given input by a ramp function, *r*[*n*]. That is, *y*[*n*] = *x*[*n*]*r*[*n*]. - - (a) Is the system BIBO-stable? Justify your answer. - - (b) Is the system linear? Justify your answer. - - (c) Is the system memoryless? Justify your answer. - - (d) Is the system causal? Justify your answer. - - (e) Is the system time-invariant? Justify your answer. -- **3.4-16** A jet-powered car is filmed using a camera operating at 60 frames per second. Let variable *n* designate the film frame, where *n* = 0 corresponds to engine ignition (film before ignition is discarded). By analyzing each frame of the film, it is possible to determine the car position *x*[*n*], measured in meters, from the original starting position *x*[0] = 0. - -From physics, we know that velocity is the time derivative of position: - -$$ -v(t) = \frac{d}{dt}x(t) -$$ - -Furthermore, we know that acceleration is the time derivative of velocity: - -$$ -a(t) = \frac{d}{dt}v(t) -$$ - -We can estimate the car velocity from the film data by using a simple difference equation *v*[*n*] = *k*(*x*[*n*] −*x*[*n*−1]). - -- (a) Determine the appropriate constant *k* to ensure *v*[*n*] has units of meters per second. -- (b) Determine a standard-form constant coefficient difference equation that outputs an estimate of acceleration, *a*[*n*], using an input of position, *x*[*n*]. Identify the advantages and shortcomings of estimating acceleration *a*(*t*) with *a*[*n*]. What is the impulse response *h*[*n*] for this system? -- **3.5-1** An LTID system is described by a constant coefficient linear difference equation 2*y*[*n*] + 2*y*[*n*−1] = *x*[*n*−1]. - - (a) Express this system in standard advance operator form. - - (b) Using recursion, determine the first 5 values of the system impulse response *h*[*n*]. - - (c) Using recursion, determine the first 5 values of the system zero-state response to input *x*[*n*] = 2*u*[*n*]. - - (d) Using recursion, determine for (0 ≤ *n* ≤ 4) the system zero-input response if *y*[−1] = 1. -- **3.5-2** Solve recursively (first three terms only): - - (a) *y*[*n*+1] −0.5*y*[*n*] = 0, with *y*[−1] = 10 - - (b) *y*[*n* + 1] + 2*y*[*n*] = *x*[*n* + 1], with *x*[*n*] = *e*−*nu*[*n*] and *y*[−1] = 0 -- **3.5-3** Solve the following equation recursively (first three terms only): - -$$ -y[n] - 0.6y[n-1] - 0.16y[n-2] = 0 -$$ - -with - -$$ -y[-1] = -25, y[-2] = 0. -$$ - -**3.5-4** Solve recursively the second-order difference Eq. (3.6) for sales estimate (first three terms only), assuming *y*[−1] = *y*[−2] = 0 and *x*[*n*] = 100*u*[*n*]. - -**3.5-5** Solve the following equation recursively (first three terms only): - -$$ -y[n+2] + 3y[n+1] + 2y[n] = x[n+2] + 3x[n+1] + 3x[n] -$$ - -with *x*[*n*] = (3)*nu*[*n*], *y*[−1] = 3, and *y*[−2] = 2 - -**3.5-6** Repeat Prob. 3.5-5 for - -$$ -y[n] + 2y[n-1] + y[n-2] = 2x[n] - x[n-1] -$$ - -with *x*[*n*] = (3)−*nu*[*n*], *y*[−1] = 2, and *y*[−2] = 3. - -- **3.6-1** Given *y*0[−1] = 3 and *y*0[−2]=−1, determine the closed-form expression of the zero-input response *y*0[*n*] of an LTID system described by the equation *y*[*n*]+ 1 6 *y*[*n*−1]− 1 6 *y*[*n*−2] = 1 3 *x*[*n*] +2 3 *x*[*n*−2]. -- **3.6-2** Solve - -$$ -y[n+2] + 3y[n+1] + 2y[n] = 0 -$$ - -if -$$ -y[-1] = 0 -$$ - and $y[-2] = 1$ . - -**3.6-3** Solve - -$$ -y[n+2] + 2y[n+1] + y[n] = 0 -$$ - -if -$$ -y[-1] = 1 -$$ - and $y[-2] = 1$ . - -**3.6-4** Solve - -$$ -y[n+2] - 2y[n+1] + 2y[n] = 0 -$$ - -if *y*[−1] = 1 and *y*[−2] = 0. - -**3.6-5** For the general *N*th-order difference Eq. (3.16), letting - -*a*1 = *a*2 =···= *aN*−1 = 0 - -results in a general causal *N*th-order LTI *nonrecursive* difference equation - -$$ -y[n] = b_0 x[n] + b_1 x[n-1] + \cdots + b_N x[n-N] -$$ - -Show that the characteristic roots for this system are zero—hence, that the zero-input response is zero. Consequently, the total response consists of the zero-state component only. - -**3.6-6** Leonardo Pisano Fibonacci, a famous thirteenthcentury mathematician, generated the sequence of integers - -$$ -\{0,1,1,2,3,5,8,13,21,34,\dots\} -$$ - -while addressing, oddly enough, a problem involving rabbit reproduction. An element of the Fibonacci sequence is the sum of the previous two. - -- (a) Find the constant-coefficient difference equation whose zero-input response *f*[*n*] with auxiliary conditions *f*[1] = 0 and *f*[2] = 1 is a Fibonacci sequence. Given *f*[*n*] is the system output, what is the system input? -- (b) What are the characteristic roots of this system? Is the system stable? -- (c) Designating 0 and 1 as the first and second Fibonacci numbers, determine the fiftieth Fibonacci number. Determine the one thousandth Fibonacci number. -- **3.6-7** Find *v*[*n*], the voltage at the *n*th node of the resistive ladder depicted in Fig. P3.4-8, if *V* = 100 volts and *a* = 2. [*Hint* 1: Consider the node equation at the *n*th node with voltage *v*[*n*]. *Hint* 2: See Prob. 3.4-8 for the equation for *v*[*n*]. The auxiliary conditions are *v*[0] = 100 and *v*[*N*] = 0.] -- **3.6-8** Consider the discrete-time system *y*[*n*] + *y*[*n* − 1] + 0.25*y*[*n* − 2] = √3*x*[*n* − 8]. Find the zero input response, *y*0[*n*], if *y*0[−1] = 1 and *y*0[1] = 1. -- **3.6-9** Provide a standard-form polynomial *Q*(*X*) such that *Q*(*E*){*y*[*n*]} = *x*[*n*] corresponds to a marginally stable third-order LTID system and *Q*(*D*){*y*(*t*)} = *x*(*t*) corresponds to a stable third-order LTIC system. -- **3.7-1** Find the unit impulse response *h*[*n*] of systems specified by the following equations: (a) *y*[*n*+1] +2*y*[*n*] = *x*[*n*] (b) *y*[*n*] +2*y*[*n*−1] = *x*[*n*] -- **3.7-2** Determine the unit impulse response *h*[*n*] of the following systems. In each case, use recursion to verify the *n* = 3 value of the closed-form expression of *h*[*n*]. - - (a) (*E*2 +1){*y*[*n*]} = (*E* +0.5){*x*[*n*]} - - (b) *y*[*n*] −*y*[*n*−1] +0.25*y*[*n*−2] = *x*[*n*] - - (c) *y*[*n*] − 1 6 *y*[*n*−1] − 1 6 *y*[*n*−2] = 1 3 *x*[*n*−2] - -- (d) *y*[*n*] + 1 6 *y*[*n*−1] − 1 6 *y*[*n*−2] = 1 3 *x*[*n*] -- (e) *y*[*n*] + 1 4 *y*[*n*−2] = *x*[*n*] -- (f) (*E*2 − 4 9 ){*y*[*n*]} = (*E*2 +1){*x*[*n*]} -- (g) (*E*2 − 1 4 )(*E* + 1 2 ){*y*[*n*]} = *E*3{*x*[*n*]} -- (h) (*E* − 1 2 )2{*y*[*n*]} = *x*[*n*] -- **3.7-3** Consider a DT system with input *x*[*n*] and output *y*[*n*] described by the difference equation - -$$ -4y[n+1] + y[n-1] = 8x[n+1] + 8x[n] -$$ - -- (a) What is the order of this system? -- (b) Determine the characteristic mode(s) of the system. -- (c) Determine a closed-form expression for the system's impulse response *h*[*n*]. -- **3.7-4** Repeat Prob. 3.7-3 for a system described by the difference equation - -$$ -y[n+3] - \frac{3}{10}y[n+2] - \frac{1}{10}y[n+1] = 2x[n+1] -$$ - -**3.7-5** Repeat Prob. 3.7-1 for - -$$ -(E2 - 6E + 9)y[n] = Ex[n] -$$ - -**3.7-6** Repeat Prob. 3.7-1 for - -$$ -y[n] - 6y[n-1] + 25y[n-2] = 2x[n] - 4x[n-1] -$$ - -**3.7-7** (a) For the general *N*th-order difference Eq. (3.16), letting - -$$ -a_0 = a_1 = a_2 = \cdots = a_{N-1} = 0 -$$ - -results in a general causal *N*th-order LTI *nonrecursive* difference equation - -$$ -y[n] = \sum_{i=0}^{N} b_i x[n-i] -$$ - -Find the impulse response *h*[*n*] for this system. [*Hint:* The characteristic equation for this case is γ *n* = 0. Hence, all the characteristic roots are zero. In this case, *yc*[*n*] = 0, and the approach in Sec. 3.7 does not work. Use a direct method to find *h*[*n*] by realizing that *h*[*n*] is the response to unit impulse input.] - -(b) Find the impulse response of a nonrecursive LTID system described by the equation - -$$ -y[n] = 3x[n] - 5x[n-1] - 2x[n-3] -$$ - -Observe that the impulse response has only a finite (*N*) number of nonzero elements. For this reason, such systems are called *finite-impulse response* (FIR) systems. For a general recursive case [Eq. (3.20)], the impulse response has an infinite number of nonzero elements, and such systems are called *infinite-impulse response* (IIR) systems. - -**3.8-1** The convolution *y*[*n*] = 5 2*n u*[*n*+5] ∗ (3*nu*[−*n*−2]) can be represented as - -$$ -y[n] = \begin{cases} C_1(\gamma_1)^n & n < N \\ C_2(\gamma_2)^n & n \ge N \end{cases} -$$ - -Using the graphical convolution procedure, determine constants *C*1, *C*2, γ1, γ2, and *N*. - -- **3.8-2** Use the graphical convolution procedure to determine the following: - - (a) *y*a[*n*] = *u*[*n*] ∗ (*u*[*n* − 5] − *u*[*n* − 9] + (0.5)(*n*−8) *u*[*n*−9]) - - (b) *y*b[*n*] = ( 1 2 )|*n*| ∗ *u*[−*n*+5] -- **3.8-3** Let *x*[*n*] = (0.5)*n* (*u*[*n*+4] −*u*[*n*−4]) be input into an LTID system with an impulse response given by - -$$ -h[n] = \begin{cases} 2 & \text{[(}n \text{ mod } 6) < 4 \text{]} \text{ and } [n \ge 0] \\ 0 & \text{otherwise} \end{cases} -$$ - -Recall, (*n* mod *p*) is the remainder of the division *n*/*p*. The system is described according to the difference equation *y*[*n*]−*y*[*n*−6] = 2*x*[*n*]+ 2*x*[*n*−1] +2*x*[*n*−2] +2*x*[*n*−3]. - -- (a) Determine the six characteristic roots (γ1 through γ6) of the system. -- (b) Determine the value of *y*[10], the zero-state output of system *h*[*n*] in response to *x*[*n*] at time *n* = 10. Express your result in decimal form to at least three decimal places (e.g., *y*[10] = 3.142). -- **3.8-4** An LTID system has impulse response *h*[*n*] = (0.5)(*n*+3) (*u*[*n*] −*u*[*n*+6]). A 6-periodic DT - -input signal *x*[*n*] is given by - -$$ -x[n] = \begin{cases} 1 & n = 0, \pm 3, \pm 6, \pm 9, \pm 12, \dots \\ 2 & n = 1, 1 \pm 6, 1 \pm 12, \dots \\ 3 & n = 2, 2 \pm 6, 2 \pm 12, \dots \\ 0 & \text{otherwise} \end{cases} -$$ - -- (a) Is system *h*[*n*] causal? Mathematically justify your answer. -- (b) Determine the value of *y*[12], the zero-state output of system *h*[*n*] in response to *x*[*n*] at time *n* = 12. Express your result in decimal form to at least three decimal places (e.g., *y*[12] = 1.234). -- **3.8-5** Find the (zero-state) response *y*[*n*] of an LTID system whose unit impulse response is - -$$ -h[n] = (-2)^n u[n-1] -$$ - -and the input is *x*[*n*] = *e*−*nu*[*n* + 1]. Find your answer by computing the convolution sum and also by using Table 3.1. - -**3.8-6** Find the (zero-state) response *y*[*n*] of an LTID system if the input is *x*[*n*] = 3*n*−1*u*[*n*+2], and - -$$ -h[n] = \frac{1}{2} [\delta[n-2] - (-2)^{n+1}] u[n-3] -$$ - -**3.8-7** Find the (zero-state) response *y*[*n*] of an LTID system if the input *x*[*n*] = (3)*n*+2*u*[*n*+1], and - -$$ -h[n] = [(2)^{n-2} + 3(-5)^{n+2}]u[n-1] -$$ - -**3.8-8** Find the (zero-state) response *y*[*n*] of an LTID system if the input *x*[*n*] = (3)−*n*+2*u*[*n*+3], and - -$$ -h[n] = 3(n-2)(2)^{n-3}u[n-4] -$$ - -**3.8-9** Find the (zero-state) response *y*[*n*] of an LTID system if its input *x*[*n*] = (2)*nu*[*n*−1], and - -$$ -h[n] = (3)^n \cos\left(\frac{\pi}{3}n - 0.5\right) u[n] -$$ - -Find your answer using only Table 3.1. - -- **3.8-10** Consider an LTID system ("system 1") described by (*E* − 1 2 ){*y*[*n*]} = *x*[*n*]. - - (a) Determine the impulse response *h*1[*n*] for system 1. Simplify your answer. - - (b) Determine the step response *s*[*n*] for system 1 (the step response is the output in response to a unit step input). Simplify your answer. - -- (c) Determine the impulse response *h*cascade[*n*] of system 1 cascaded with an LTID system with impulse response *h*2[*n*]=−3*u*[*n*−13]. Simplify your answer. -- **3.8-11** Derive the results in entries 1, 2, and 3 in Table 3.1. [*Hint:* You may need to use the information in Sec. B.8-3.] -- **3.8-12** Derive the results in entries 4, 5, and 6 in Table 3.1. -- **3.8-13** Derive the results in entries 7 and 8 in Table 3.1. [*Hint:* You may need to use the information in Sec. B.8-3.] -- **3.8-14** Derive the results in entries 9 and 11 in Table 3.1. [*Hint:* You may need to use the information in Sec. B.8-3.] -- **3.8-15** Find the total response of a system specified by the equation - -$$ -y[n+1] + 2y[n] = x[n+1] -$$ - -if *y*[−1] = 10, and the input *x*[*n*] = *e*−*nu*[*n*]. - -- **3.8-16** Find an LTID system (zero-state) response if its impulse response *h*[*n*] = (0.5)*nu*[*n*], and the input *x*[*n*] is (a) 2*nu*[*n*] (b) 2*n*−3*u*[*n*] - - (c) 2*nu*[*n*−2] - -[*Hint:* You may need to use the convolution shift property of Eq. (3.32).] - -**3.8-17** For a system specified by equation - -$$ -y[n] = x[n] - 2x[n-1] -$$ - -Find the system response to input *x*[*n*] = *u*[*n*]. What is the order of the system? What type of system (recursive or nonrecursive) is this? Is the knowledge of initial condition(s) necessary to find the system response? Explain. - -- **3.8-18** (a) A discrete-time LTI system is shown in Fig. P3.8-18. Express the overall impulse response of the system, *h*[*n*], in terms of *h*1[*n*], *h*2[*n*], *h*3[*n*], *h*4[*n*], and *h*5[*n*]. - - (b) Two LTID systems in cascade have impulse response *h*1[*n*] and *h*2[*n*], respectively. Show that if *h*1[*n*] = (0.9)*nu*[*n*] − 0.5(0.9)*n*−1*u*[*n* − 1] and *h*2[*n*] = (0.5)*nu*[*n*] − 0.9(0.5)*n*−1*u*[*n* − 1], the cascade system is an identity system. - -**Figure P3.8-18** - -**3.8-19** (a) Show that for a causal system, Eq. (3.37) can also be expressed as - -$$ -g[n] = \sum_{k=0}^{n} h[n-k] -$$ - -- (b) How would the expressions in part (a) change if the system is not causal? -- **3.8-20** An LTID system with input *x*[*n*] and output *y*[*n*] has impulse response *h*[*n*] = 2(*u*[*n* + 2] − *u*[*n* − 3]). - - (a) Write a constant-coefficient linear difference equation that has the given impulse response. [*Hint:* First express *h*[*n*] in terms of delta functions δ[*n*].] - - (b) Using graphical convolution, determine the zero-state output of this system in response to the anticausal input *x*[*n*] = 2*nu*[−*n*]. A simplified closed-form solution is required. -- **3.8-21** Consider three LTID systems: system 1 has impulse response *h*1[*n*]=[ ↓ 2, −3, 4], system 2 has impulse response *h*2[*n*]=[ ↓ 0, 0, −6, − 9, 3], and system 3 is an identity system (output equals input). - - (a) Determine the overall impulse response *h*[*n*] if system 1 is connected in cascade with a parallel connection of systems 2 and 3. - - (b) For input *x*[*n*] = *u*[−*n*], determine the zero-state response *y*zsr[*n*] of system 2. -- **3.8-22** In the savings account problem described in Ex. 3.6, a person deposits \$500 at the beginning of every month, starting at *n* = 0 with the exception at *n* = 4, when instead of depositing \$500, she withdraws \$1000. Find *y*[*n*] if the interest rate is 1% per month (*r* = 0.01). - -**3.8-23** To pay off a loan of *M* dollars in *N* number of payments using a fixed monthly payment of *P* dollars, show that - -$$ -P = \frac{rM}{1 - (1 + r)^{-N}} -$$ - -where *r* is the interest rate per dollar per month. [*Hint:* This problem can be modeled by Eq. (3.3) with the payments of *P* dollars starting at *n* = 1. The problem can be approached in two ways. First, consider the loan as the initial condition *y*0[0]=−*M*, and the input *x*[*n*] = *Pu*[*n* − 1]. The loan balance is the sum of the zero-input component (due to the initial condition) and the zero-state component *h*[*n*] ∗ *x*[*n*]. Second, consider the loan as an input −*M* at *n* = 0 along with the input due to payments. The loan balance is now exclusively a zero-state component *h*[*n*] ∗ *x*[*n*]. Because the loan is paid off in *N* payments, set *y*[*N*] = 0.] - -- **3.8-24** A person receives an automobile loan of \$10,000 from a bank at the interest rate of 1.5% per month. His monthly payment is \$500, with the first payment due one month after he receives the loan. Compute the number of payments required to pay off the loan. Note that the last payment may not be exactly \$500. [*Hint:* Follow the procedure in Prob. 3.8-23 to determine the balance *y*[*n*]. To determine *N*, the number of payments, set *y*[*N*] = 0. In general, *N* will not be an integer. The number of payments *K* is the largest integer ≤ *N*. The residual payment is |*y*[*K*]|.] -- **3.8-25** Letting ↓ identify the *n* = 0 values, use the sliding-tape method to determine the following: - - (a) *y*a = [ ↓ 2, 3,−2,−3] ∗ [−10, ↓ 0,−5] (b) *y*b = [2, ↓ −1, 3,−2] ∗ [−1,−4, 1, ↓ −2] (c) *y*c = [ ↓ 0, 0, 3, 2, 1, 2, 3]∗[2, 3,−2, ↓ 1] - - (d) *y*d = [5, 0, 0, ↓ −2, 8] ∗ [−1, 1, ↓ 3, 3,−2, 3] - - (e) *y*e = ([1, ↓ −1]∗[ ↓ 1,−1]) ∗ ([ ↓ 1,−1]∗[1, ↓ −1]) - - (f) *y*f = ([2, ↓ −1]∗[ ↓ 1,−2]) ∗ ([ ↓ 1,−2]∗[2, ↓ −1]) Outside the values shown, assume all signals are zero. - -### 324 CHAPTER 3 TIME-DOMAIN ANALYSIS OF DISCRETE-TIME SYSTEMS - -- **3.8-26** Let ↓ identify the *n* = 0 and consider DT signals *x*[*n*] and *h*[*n*] whose nonzero values are given as *x*[*n*]=[1, 2, 3, ↓ 4, 5] and *h*[*n*] = [−2, −1, 1, ↓ 2]. Use DT convolution (any method) to determine *y*[*n*] = (2*x*[*n* − 30]) ∗ −3 2 *h*[*n*−10] . Express your result in vector notation, making sure to indicate the time index of the leftmost (nonzero) element. -- **3.8-27** Using the sliding-tape algorithm, show that (a) *u*[*n*] ∗ *u*[*n*] = (*n*+1)*u*[*n*] (b) (*u*[*n*] −*u*[*n*−*m*])∗*u*[*n*] = (*n*+1)*u*[*n*]−(*n*− *m*+1)*u*[*n*−*m*] -- **3.8-28** Using the sliding-tape algorithm, find *x*[*n*] ∗*g*[*n*] for the signals shown in Fig. P3.8-28. - -- **3.8-29** Repeat Prob. 3.8-28 for the signals shown in Fig. P3.8-29. -- **3.8-30** Repeat Prob. 3.8-28 for the signals shown in Fig. P3.8-30. -- **3.8-31** Letting ↓ identify *n* = 0, define the nonzero values of signal *x*[*n*] as [1, 2, ↓ 2]. Similarly, define the non-zero values of signal *y*[*n*] as [3, 4, 6, 6, 11, ↓ 2,−2]. Using the sliding-tape algorithm as the basis for your work, determine the signal *h*[*n*] so that *y*[*n*] = *x*[*n*] ∗ *h*[*n*]. -- **3.8-32** The convolution sum in Eq. (3.33) can be expressed in a matrix form as *y* = *Hx*, where *y* is a column vector containing *y*[0], *y*[1],... *y*[*n*]; *x* is a column vector containing *x*[0], *x*[1],... *x*[*n*]; - -**Figure P3.8-30** - -and *H* is a lower triangular matrix defined as - -| | ⎡
h[0] | 0 | 0 | | 0 | ⎤ | -|--------|-----------|--------|---|---------|------|--------| -| | h[1]
⎢ | h[0] | 0 | | 0 | ⎥ | -| H
= | ⎢
⎢ | | | | | ⎥
⎥ | -| | ⎣
h[n] | h[n−1] | | ···
| h[0] | ⎦ | -| | | | | | | | - -Knowing *h*[*n*] and the output *y*[*n*], we can determine the input *x*[*n*] according to *x* = *H*−1*y*. This operation is the reverse of convolution and is known as *deconvolution*. Moreover, knowing *x*[*n*] and *y*[*n*], we can determine *h*[*n*]. This can be done by expressing the foregoing matrix equation as *n* + 1 simultaneous equations in terms of *n* + 1 unknowns *h*[0], *h*[1], ... , *h*[*n*]. These equations can readily be solved iteratively. Thus, we can synthesize a system that yields a certain output *y*[*n*] for a given input *x*[*n*]. - -- (a) Design a system (i.e., determine *h*[*n*]) that will yield the output sequence (8, 12, 14, 15, 15.5, 15.75, ...) for the input sequence (1, 1, 1, 1, 1, 1, ...). -- (b) For a system with the impulse response sequence (1, 2, 4, ...), the output sequence was (1, 7/3, 43/9, ...). Determine the input sequence. -- **3.8-33** A second-order LTID system has zero-input response - -$$ -y_0[n] = [3, 2\frac{1}{3}, 2\frac{1}{9}, 2\frac{1}{27}, \dots] -$$ - -= -$$ -\sum_{k=0}^{\infty} \left\{ 2 + \left(\frac{1}{3}\right)^k \right\} \delta[n-k] -$$ - -- (a) Determine the characteristic equation of this system, *a*0γ 2 +*a*1γ +*a*2 = 0. -- (b) Find a bounded, causal input with infinite duration that would cause a strong response from this system. Justify your choice. - -**Figure P3.8-34** - -- (c) Find a bounded, causal input with infinite duration that would cause a weak response from this system. Justify your choice. -- **3.8-34** An LTID filter has an impulse response function given by *h*1[*n*] = δ[*n* + 2] − δ[*n* − 2]. A second LTID system has an impulse response function given by *h*2[*n*] = *n*(*u*[*n*+4] −*u*[*n*−4]). - - (a) Carefully sketch the functions *h*1[*n*] and *h*2[*n*] over (−10 ≤ *n* ≤ 10). - - (b) Assume that the two systems are connected in parallel, as shown in Fig. P3.8-34a. Determine the impulse response *hp*[*n*] for the parallel system in terms of *h*1[*n*] and *h*2[*n*]. Sketch *hp*[*n*] over (−10 ≤ *n* ≤ 10). - - (c) Assume that the two systems are connected in cascade, as shown in Fig. P3.8-34b. Determine the impulse response *hs*[*n*] for the cascade system in terms of *h*1[*n*] and *h*2[*n*]. Sketch *hs*[*n*] over (−10 ≤ *n* ≤ 10). -- **3.8-35** This problem investigates an interesting application of discrete-time convolution: the expansion of certain polynomial expressions. - - (a) By hand, expand (*z*3+*z*2+*z*+1)2. Compare the coefficients to [1, 1, 1, 1]∗[1, 1, 1, 1]. - - (b) Formulate a relationship between discretetime convolution and the expansion of constant-coefficient polynomial expressions. - - (c) Use convolution to expand (*z*−4 − 2*z*−3 + 3*z*−2)4. - - (d) Use convolution to expand (*z*5 +2*z*4 +3*z*2 + 5)2(*z*−4 −5*z*−2 +13). -- **3.8-36** Joe likes coffee, and he drinks his coffee according to a very particular routine. He begins by adding two teaspoons of sugar to his mug, which he then fills to the brim with hot coffee. He drinks 2/3 of the mug's contents, adds another two teaspoons of sugar, and tops the mug off with steaming hot coffee. This refill procedure - -continues, sometimes for many, many cups of coffee. Joe has noted that his coffee tends to taste sweeter with the number of refills. - -Let independent variable *n* designate the coffee refill number. In this way, *n* = 0 indicates the first cup of coffee, *n* = 1 is the first refill, and so forth. Let *x*[*n*] represent the sugar (measured in teaspoons) added into the system (a coffee mug) on refill *n*. Let *y*[*n*] designate the amount of sugar (again, teaspoons) contained in the mug on refill *n*. - -- (a) The sugar (teaspoons) in Joe's coffee can be represented using a standard second-order constant coefficient difference equation *y*[*n*] + *a*1*y*[*n* − 1] + *a*2*y*[*n* − 2] = *b*0*x*[*n*] + *b*1*x*[*n* − 1] + *b*2*x*[*n* − 2]. Determine the constants *a*1, *a*2, *b*0, *b*1, and *b*2. -- (b) Determine *x*[*n*], the driving function to this system. -- (c) Solve the difference equation for *y*[*n*]. This requires finding the total solution. Joe always starts with a clean mug from the dishwasher, so *y*[−1] (the sugar content before the first cup) is zero. -- (d) Determine the steady-state value of *y*[*n*]. That is, what is *y*[*n*] as *n* → ∞? If possible, suggest a way of modifying *x*[*n*] so that the sugar content of Joe's coffee remains a constant for all nonnegative *n*. -- **3.8-37** A system is called complex if a real-valued input can produce a complex-valued output. Consider a causal complex system described by a first-order constant coefficient linear difference equation: - -(*jE* +0.5)*y*[*n*] = (−5*E*)*x*[*n*] - -- (a) Determine the impulse response function *h*[*n*] for this system. -- (b) Given input *x*[*n*] = *u*[*n* − 5] and initial condition *y*0[−1] = *j*, determine the system's total output *y*[*n*] for *n* ≥ 0. -- **3.8-38** A discrete-time LTI system has impulse response function *h*[*n*] = *n*(*u*[*n* − 2] − *u*[*n*+2]). - - (a) Carefully sketch the function *h*[*n*] over (−5 ≤ *n* ≤ 5). - - (b) Determine the difference equation representation of this system, using *y*[*n*] to designate the output and *x*[*n*] to designate the input. - -- **3.8-39** Consider three discrete-time signals: *x*[*n*], *y*[*n*], and *z*[*n*]. Denoting convolution as ∗, identify the expression(s) that is(are) equivalent to *x*[*n*](*y*[*n*] ∗ *z*[*n*]): - - (a) (*x*[*n*] ∗ *y*[*n*])*z*[*n*] (b) (*x*[*n*]*y*[*n*]) ∗ (*x*[*n*]*z*[*n*]) - - (c) (*x*[*n*]*y*[*n*]) ∗ *z*[*n*] - - (d) none of the above - - Justify your answer! -- **3.8-40** A causal system with input *x*[*n*] and output *y*[*n*] is described by - -$$ -y[n] - ny[n-1] = x[n] -$$ - -- (a) By recursion, determine the first six nonzero values of *h*[*n*], the response to *x*[*n*] = δ[*n*]. Do you think this system is BIBO-stable? Why? -- (b) Compute *y*R[4] recursively from *y*R[*n*] − *ny*R[*n*−1] = *x*[*n*], assuming all initial conditions are zero and *x*[*n*] = *u*[*n*]. The subscript R is only used to emphasize a recursive solution. -- (c) Define *y*C[*n*] = *x*[*n*]∗*h*[*n*]. Using *x*[*n*] = *u*[*n*] and *h*[*n*] from part (a), compute *y*C[4]. The subscript C is only used to emphasize a convolution solution. -- (d) In this chapter, both recursion and convolution are presented as potential methods to compute the zero-state response (ZSR) of a discrete-time system. Comparing parts (b) and (c), we see that *y*R[4] = *y*C[4]. Why are the two results not the same? Which method, if any, yields the correct ZSR value? -- **3.9-1** In Sec. 3.9-1 we showed that for BIBO stability in an LTID system, it is sufficient for its impulse response *h*[*n*] to satisfy Eq. (3.43). Show that this is also a necessary condition for the system to be BIBO-stable. In other words, show that if Eq. (3.43) is not satisfied, there exists a bounded input that produces unbounded output. [*Hint:* Assume that a system exists for which *h*[*n*] violates Eq. (3.43), yet its output is bounded for every bounded input. Establish the contradiction in this statement by considering an input *x*[*n*] defined by *x*[*n*1 − *m*] = 1 when *h*[*m*] > 0 and *x*[*n*1−*m*]=−1 when *h*[*m*]<0, where *n*1 is some fixed integer.] - -- **3.9-2** Each of the following equations specifies an LTID system. Determine whether each of these systems is BIBO-stable or -unstable. Determine also whether each is asymptotically stable, unstable, or marginally stable. - - (a) *y*[*n* + 2] + 0.6*y*[*n* + 1] − 0.16*y*[*n*] = *x*[*n* + 1] −2*x*[*n*] - - (b) *y*[*n*] + 3*y*[*n* − 1] + 2*y*[*n* − 2] = *x*[*n* − 1] + 2*x*[*n*−2] - - (c) (*E* −1)2 *E* + 1 2 *y*[*n*] = *x*[*n*] - - (d) *y*[*n*] +2*y*[*n*−1] +0.96*y*[*n*−2] = *x*[*n*] - - (e) *y*[*n*]+*y*[*n*−1]−2*y*[*n*−2] = *x*[*n*]+2*x*[*n*−1] - - (f) (*E*2 −1)(*E*2 +1)*y*[*n*] = *x*[*n*] -- **3.9-3** Consider two LTIC systems in cascade, as illustrated in Fig. 3.29. The impulse response of the system *S*1 is *h*1[*n*] = 2*n u*[*n*] and the impulse response of the system *S*2 is *h*2[*n*] = δ[*n*] − 2δ[*n*−1]. Is the cascaded system asymptotically stable or unstable? Determine the BIBO stability of the composite system. -- **3.9-4** Figure P3.9-4 locates the characteristic roots of ten causal, LTID systems, labeled A through J. Each system has only two roots and is described using operator notation as *Q*(*E*)*y*[*n*] = *P*(*E*)*x*[*n*]. All plots are drawn to scale, with the unit circle shown for reference. For each of the following parts, identify all the answers that are correct. - - (a) Identify all systems that are unstable. - - (b) Assuming all systems have *P*(*E*) = *E*2, identify all systems that are real. Recall that a real system always generates a real-valued response to a real-valued input. - - (c) Identify all systems that support oscillatory natural modes. - -- (d) Identify all systems that have at least one mode whose envelop decays at a rate of 2−*n*. -- (e) Identify all systems that have only one mode. -- **3.9-5** A discrete-time LTI system has impulse response given by - -$$ -h[n] = \delta[n] + \left(\frac{1}{3}\right)^n u[n-1] -$$ - -- (a) Is the system stable? Is the system causal? Justify your answers. -- (b) Plot the signal *x*[*n*] = *u*[*n*−3] −*u*[*n*+3]. -- (c) Determine the system's zero-state response *y*[*n*] to the input *x*[*n*] = *u*[*n* − 3] − *u*[*n* + 3]. Plot *y*[*n*] over (−10 ≤ *n* ≤ 10). -- **3.9-6** An LTID system has an impulse response given by - -$$ -h[n] = \left(\frac{1}{2}\right)^{|n|} -$$ - -- (a) Is the system causal? Justify your answer. -- (b) Compute %∞ *n*=−∞ |*h*[*n*]|. Is this system BIBO-stable? -- (c) Compute the energy and power of input signal *x*[*n*] = 3*u*[*n*−5]. -- (d) Using input *x*[*n*] = 3*u*[*n* − 5], determine the zero-state response of this system at time *n* = 10. That is, determine *y*zsr[10]. -- **3.10-1** Determine a constant coefficient linear difference equation that describes a system for which the input *x*[*n*] = 2( 1 3 )*nu*[−*n* − 4] causes resonance. -- **3.10-2** If one exists, determine a real input *x*[*n*] that will cause resonance in the causal LTID system described by (*E*2 +1){*y*[*n*]} = (*E*+0.5){*x*[*n*]}. If no such input exists, explain why not. - -**Figure P3.9-4** - -- **3.10-3** Consider two lowpass LTID systems, one with infinite-duration impulse response *h*1[*n*] = −(0.5)*nu*[*n*] and the other with finite-duration impulse response *h*2[*n*] = 2(*u*[*n*] − *u*[*n* − 4]). Which system (1, 2, both, or neither) would more efficiently transmit a binary communication signal? Carefully justify your result. -- **3.11-1** Write a MATLAB program that recursively computes and then plots the solution to *y*[*n*] − 1 3 *y*[*n* − 1] + 1 2 *y*[*n* − 2] = *x*[*n*] for (0 ≤ *n* ≤ 100) given *x*[*n*] = δ[*n*] + *u*[*n* − 50] and *y*[−2] = *y*[−1] = 2. -- **3.11-2** Consider the discrete-time function *f*[*n*] = *e*−*n*/5 cos(π*n*/5)*u*[*n*]. Section 3.11 uses anonymous functions in describing DT signals. - - f = @(n) exp(-n/5).\*cos(pi\*n/5).\*(n>=0); - -While this anonymous function operates correctly for a downsampling operation such as f[2n], it does not operate correctly for an upsampling operation, such as f[n/2]. Modify the anonymous function f so that it also correctly accommodates upsampling operations. Test your code by computing and plotting f(n/2) over (−10 ≤ *n* ≤ 10). - -- **3.11-3** Write MATLAB code to compute and plot the DT convolutions of Prob. 3.8-25. -- **3.11-4** An indecisive student contemplates whether he should stay home or take his final exam, which is being held 2 miles away. Starting at home, the student travels half the distance to the exam location before changing his mind. The student turns around and travels half the distance between his current location and his home before changing his mind again. This process of changing direction and traveling half the remaining distance continues until the student either reaches a destination or dies from exhaustion. - - (a) Determine a suitable difference equation description of this system. - - (b) Use MATLAB to simulate the difference equation in part (a). Where does the student end up as *n* → ∞? How does your answer change if the student goes two-thirds the way each time, rather than halfway? - - (c) Determine a closed-form solution to the equation in part (a). Use this solution to verify the results in part (b). - -**3.11-5** The cross-correlation function between *x*[*n*] and *y*[*n*] is given as - -$$ -r_{xy}[k] = \sum_{n=-\infty}^{\infty} x[n]y[n-k] -$$ - -Notice that *rxy*[*k*] is quite similar to the convolution sum. The independent variable *k* corresponds to the relative *shift* between the two inputs. - -- (a) Express *rxy*[*k*] in terms of convolution. Is *rxy*[*k*] = *ryx*[*k*]? -- (b) Cross-correlation is said to indicate similarity between two signals. Do you agree? Why or why not? -- (c) If *x*[*n*] and *y*[*n*] are both finite duration, MATLAB's conv command is well suited to compute *rxy*[*k*]. Write a MATLAB function that computes the cross-correlation function using the conv command. Four vectors are passed to the function (x, y, nx, and ny) corresponding to the inputs *x*[*n*], *y*[*n*], and their respective time vectors. Notice that x and y are not necessarily the same length. Two outputs should be created (rxy and k) corresponding to *rxy*[*k*] and its shift vector. -- (d) Test your code from part (c) using *x*[*n*] = *u*[*n* − 5] − *u*[*n* − 10] over (0 ≤ *n* = *nx* ≤ 20) and *y*[*n*] = *u*[−*n*−15]−*u*[−*n*−10]+δ[*n*− 2] over (−20 ≤ *n* = *ny* ≤ 10). Plot the result rxy as a function of the shift vector k. What shift *k* gives the largest magnitude of *rxy*[*k*]? Does this make sense? -- **3.11-6** Suppose a vector x exists in the MATLAB workspace, corresponding to a finite-duration DT signal *x*[*n*] - - (a) Write a MATLAB function that, when passed vector x, computes and returns Ex, the energy of *x*[*n*]. - - (b) Write a MATLAB function that, when passed vector x, computes and returns Px, the power of *x*[*n*]. Assume that *x*[*n*] is periodic and that vector x contains data for an integer number of periods of *x*[*n*]. -- **3.11-7** A causal *N*-point max filter assigns *y*[*n*] to the maximum of {*x*[*n*],..., *x*[*n*−(*N* −1)]}. - - (a) Write a MATLAB function that performs *N*-point max filtering on a length-*M* input vector x. The two function inputs are vector x and scalar N. To create the length-*M* output - -vector y, initially pad the input vector with *N* − 1 zeros. The MATLAB command max may be helpful. - -- (b) Test your filter and MATLAB code by filtering a length-45 input defined as *x*[*n*] = cos(π*n*/5) + δ[*n* − 30] − δ[*n* − 35]. Separately plot the results for *N* = 4, *N* = 8, and *N* = 12. Comment on the filter behavior. -- **3.11-8** A causal *N*-point min filter assigns *y*[*n*] to the minimum of {*x*[*n*],..., *x*[*n*−(*N* −1)]}. - - (a) Write a MATLAB function that performs *N*-point min filtering on a length-*M* input vector x. The two function inputs are vector x and scalar N. To create the length-*M* output vector y, initially pad the input vector with *N* − 1 zeros. The MATLAB command min may be helpful. - - (b) Test your filter and MATLAB code by filtering a length-45 input defined as *x*[*n*] = cos(π*n*/5) + δ[*n* − 30] − δ[*n* − 35]. Separately plot the results for *N* = 4,*N* = 8, and *N* = 12. Comment on the filter behavior. -- **3.11-9** A causal *N*-point median filter assigns *y*[*n*] to the median of {*x*[*n*],..., *x*[*n*−(*N* −1)]}. The median is found by sorting sequence {*x*[*n*],..., *x*[*n* − (*N* − 1)]} and choosing the middle value (odd *N*) or the average of the two middle values (even *N*). - - (a) Write a MATLAB function that performs *N*-point median filtering on a length-*M* input vector x. The two function inputs are vector x and scalar N. To create the length-*M* output vector y, initially pad the input vector with *N* − 1 zeros. The MATLAB command sort or median may be helpful. - - (b) Test your filter and MATLAB code by filtering a length-45 input defined as *x*[*n*] = cos(π*n*/5) + δ[*n* − 30] − δ[*n* − 35]. Separately plot the results for *N* = 4, *N* = 8, and *N* = 12. Comment on the filter behavior. -- **3.11-10** Recall that *y*[*n*] = *x*[*n*/*N*] represents an upsample by *N* operation. An interpolation filter replaces the inserted zeros with more realistic values. A linear interpolation filter has impulse response - -$$ -h[n] = \sum_{k=-(N-1)}^{N-1} \left(1 - \left|\frac{k}{N}\right|\right) \delta(n-k) -$$ - -- (a) Determine a constant coefficient difference equation that has impulse response *h*[*n*]. -- (b) The impulse response *h*[*n*] is noncausal. What is the smallest time shift necessary to make the filter causal? What is the effect of this shift on the behavior of the filter? -- (c) Write a MATLAB function that will compute the parameters necessary to implement an interpolation filter using MATLAB's filter command. That is, your function should output filter vectors b and a given an input scalar *N*. -- (d) Test your filter and MATLAB code. To do this, create *x*[*n*] = cos(*n*) for (0 ≤ *n* ≤ 9). Upsample *x*[*n*] by *N* = 10 to create a new signal *xup*[*n*]. Design the corresponding *N* = 10 linear interpolation filter, filter *xup*[*n*] to produce *y*[*n*], and plot the results. -- **3.11-11** A causal *N*-point moving-average filter has impulse response *h*[*n*] = (*u*[*n*] − *u*[*n* − *N*])/*N*. - - (a) Determine a constant-coefficient difference equation that has impulse response *h*[*n*]. - - (b) Write a MATLAB function that will compute the parameters necessary to implement an *N*-point moving-average filter using MATLAB's filter command. That is, your function should output filter vectors b and a given a scalar input *N*. - - (c) Test your filter and MATLAB code by filtering a length-45 input defined as *x*[*n*] = cos(π*n*/5) + δ[*n* − 30] − δ[*n* − 35]. Separately plot the results for *N* = 4, *N* = 8, and *N* = 12. Comment on the filter behavior. - - (d) Problem 3.11-10 introduces linear interpolation filters, for use following an upsample by *N* operation. Within a scale factor, show that a cascade of two *N*-point moving-average filters is equivalent to the linear interpolation filter. What is the scale factor difference? Test this idea with MAT-LAB. Create *x*[*n*] = cos(*n*) for (0 ≤ *n* ≤ 9). Upsample *x*[*n*] by *N* = 10 to create a new signal *xup*[*n*]. Design an *N* = 10 moving-average filter. Filter *xup*[*n*] twice and scale to produce *y*[*n*]. Plot the results. Does the output from the cascaded pair of moving-average filters linearly interpolate the upsampled data? - -# **[CONTINUOUS-TIME](#page-10-0) SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM** - -Because of the linearity (superposition) property of linear time-invariant systems, we can find the response of these systems by breaking the input *x*(*t*) into several components and then summing the system response to all the components of *x*(*t*). We have already used this procedure in time-domain analysis, in which the input *x*(*t*) is broken into impulsive components. In the *frequency-domain analysis* developed in this chapter, we break up the input *x*(*t*) into exponentials of the form *est*, where the parameter *s* is the complex frequency of the signal *est*, as explained in Sec. 1.4-3. This method offers an insight into the system behavior complementary to that seen in the time-domain analysis. In fact, the time-domain and the frequency-domain methods are duals of each other. - -The tool that makes it possible to represent arbitrary input *x*(*t*) in terms of exponential components is the *Laplace transform*, which is discussed in the following section. - -## **4.1 THE LAPLACE [TRANSFORM](#page-10-0)** - -For a signal *x*(*t*), its Laplace transform *X*(*s*) is defined by - -$$ -X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt -$$ -\n(4.1) - -The signal *x*(*t*) is said to be the *inverse Laplace transform* of *X*(*s*). It can be shown that - -$$ -x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds -$$ -\n(4.2) - -where *c* is a constant chosen to ensure the convergence of the integral in Eq. (4.1), as explained later. See also [1]. - -This pair of equations is known as the *bilateral Laplace transform pair*, where *X*(*s*) is the direct Laplace transform of *x*(*t*) and *x*(*t*) is the inverse Laplace transform of *X*(*s*). Symbolically, - -> *X*(*s*) = *L*[*x*(*t*)] and *x*(*t*) = *L*−1 [*X*(*s*)] - -Note that - -CHAPTER - -**4** - -$$ -\mathcal{L}^{-1}\{\mathcal{L}[x(t)]\} = x(t) \qquad \text{and} \qquad \mathcal{L}\{\mathcal{L}^{-1}[X(s)]\} = X(s) -$$ - -It is also common practice to use a bidirectional arrow to indicate a Laplace transform pair, as follows: - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -The Laplace transform, defined in this way, can handle signals existing over the entire time interval from −∞ to ∞ (causal and noncausal signals). For this reason it is called the *bilateral* (or *two-sided*) Laplace transform. Later we shall consider a special case—the *unilateral* or *one-sided* Laplace transform—which can handle only causal signals. - -### LINEARITY OF THE LAPLACE TRANSFORM - -We now prove that the Laplace transform is a linear operator by showing that the principle of superposition holds, implying that if - -$$ -x_1(t) \Longleftrightarrow X_1(s) -$$ - and $x_2(t) \Longleftrightarrow X_2(s)$ - -then - -$$ -a_1x_1(t) + a_2x_2(t) \Longleftrightarrow a_1X_1(s) + a_2X_2(s) -$$ - -The proof is simple. By definition, - -$$ -\mathcal{L}[a_1x_1(t) + a_2x_2(t)] = \int_{-\infty}^{\infty} [a_1x_1(t) + a_2x_2(t)]e^{-st} dt -$$ - -\n -$$ -= a_1 \int_{-\infty}^{\infty} x_1(t)e^{-st} dt + a_2 \int_{-\infty}^{\infty} x_2(t)e^{-st} dt -$$ - -\n -$$ -= a_1X_1(s) + a_2X_2(s) -$$ -\n(4.3) - -This result can be extended to any finite sum. - -### THE REGION OF CONVERGENCE (ROC) - -The *region of convergence* (ROC), also called the region of existence, for the Laplace transform, *X*(*s*), is the set of values of *s* (the region in the complex plane) for which the integral in Eq. (4.1) converges. This concept will become clear in the following example. - -### **EXAMPLE 4.1 Laplace Transform and ROC of a Causal Exponential** - -For a signal *x*(*t*) = *e*−*atu*(*t*), find the Laplace transform *X*(*s*) and its ROC. - -By definition, - -$$ -X(s) = \int_{-\infty}^{\infty} e^{-at} u(t) e^{-st} dt -$$ - -Because *u*(*t*) = 0 for *t* < 0 and *u*(*t*) = 1 for *t* ≥ 0, - -$$ -X(s) = \int_0^\infty e^{-at} e^{-st} dt = \int_0^\infty e^{-(s+a)t} dt = -\frac{1}{s+a} e^{-(s+a)t} \Big|_0^\infty \tag{4.4} -$$ - -### 332 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Note that *s* is complex and as *t* → ∞, the term *e*−(*s*+*a*)*t* does not necessarily vanish. Here we recall that for a complex number *z* = α +*j*β, - -$$ -e^{-zt} = e^{-(\alpha+j\beta)t} = e^{-\alpha t}e^{-j\beta t} -$$ - -Now |*e*−*j*β*t* | = 1 regardless of the value of β*t*. Therefore, as *t* → ∞, *e*−*zt* → 0 only if α > 0, and *e*−*zt* → ∞ if α < 0. Thus, - -$$ -\lim_{t \to \infty} e^{-zt} = \begin{cases} 0 & \text{Re } z > 0 \\ \infty & \text{Re } z < 0 \end{cases} -$$ - (4.5) - -Clearly, - -$$ -\lim_{t \to \infty} e^{-(s+a)t} = \begin{cases} 0 & \text{Re}(s+a) > 0\\ \infty & \text{Re}(s+a) < 0 \end{cases} -$$ - -Use of this result in Eq. (4.4) yields - -$$ -X(s) = \frac{1}{s+a} \qquad \text{Re}(s+a) > 0 -$$ -$$ -e^{-at}u(t) \Longleftrightarrow \frac{1}{s+a} \qquad \text{Re } s > -a \tag{4.6} -$$ - -or - -The ROC of -$$ -X(s) -$$ - is Re $s > -a$ , as shown in the shaded area in Fig. 4.1a. This fact means that the integral defining $X(s)$ in Eq. (4.4) exists only for the values of $s$ in the shaded region in Fig. 4.1a. For other values of $s$ , the integral in Eq. (4.4) does not converge. For this reason, the shaded region is called the *ROC* (or the *region of existence*) for $X(s)$ . - -**Figure 4.1** Signals **(a)** *e*−*atu*(*t*) and **(b)** −*e*−*atu*(−*t*) have the same Laplace transform but different regions of convergence. - -### REGION OF CONVERGENCE FOR FINITE-DURATION SIGNALS - -A finite-duration signal *xf*(*t*) is a signal that is nonzero only for *t*1 ≤ *t* ≤ *t*2, where both *t*1 and *t*2 are finite numbers and *t*2 > *t*1. For a finite-duration, absolutely integrable signal, the ROC is the entire *s* plane. This is clear from the fact that if *xf*(*t*) is absolutely integrable and a finite-duration signal, then *x*(*t*)*e*−σ*t* is also absolutely integrable for any value of σ because the integration is over the finite range of *t* only. Hence, the Laplace transform of such a signal converges for every value of *s*. This means that the ROC of a general signal *x*(*t*) remains unaffected by the addition of any absolutely integrable, finite-duration signal *xf*(*t*) to *x*(*t*). In other words, if *R* represents the ROC of a signal *x*(*t*), then the ROC of a signal *x*(*t*)+*xf*(*t*) is also *R*. - -### ROLE OF THE REGION OF CONVERGENCE - -The ROC is required for evaluating the inverse Laplace transform *x*(*t*) from *X*(*s*), as defined by Eq. (4.2). The operation of finding the inverse transform requires an integration in the complex plane, which needs some explanation. The path of integration is along *c*+*j*ω, with ω varying from −∞ to ∞. † Moreover, the path of integration must lie in the ROC (or existence) for *X*(*s*). For the signal *e*−*atu*(*t*), this is possible if *c* > −*a*. One possible path of integration is shown (dotted) in Fig. 4.1a. Thus, to obtain *x*(*t*) from *X*(*s*), the integration in Eq. (4.2) is performed along this path. When we integrate [1/(*s* + *a*)]*est* along this path, the result is *e*−*atu*(*t*). Such integration in the complex plane requires a background in the theory of functions of complex variables. We can avoid this integration by compiling a table of Laplace transforms (Table 4.1), where the Laplace transform pairs are tabulated for a variety of signals. To find the inverse Laplace transform of, say, 1/(*s* + *a*), instead of using the complex integral of Eq. (4.2), we look up the table and find the inverse Laplace transform to be *e*−*atu*(*t*) (assuming that the ROC is Re *s* > −*a*). Although the table given here is rather short, it comprises the functions of most practical interest. A more comprehensive table appears in Doetsch [2]. - -### THE UNILATERAL LAPLACE TRANSFORM - -To understand the need for defining unilateral transform, let us find the Laplace transform of signal *x*(*t*) illustrated in Fig. 4.1b: - -$$ -x(t) = -e^{-at}u(-t) -$$ - -The Laplace transform of this signal is - -$$ -X(s) = \int_{-\infty}^{\infty} -e^{-at}u(-t)e^{-st}dt -$$ - -Because *u*(−*t*) = 1 for *t* < 0 and *u*(−*t*) = 0 for *t* > 0, - -$$ -X(s) = \int_{-\infty}^{0} -e^{-at}e^{-st} dt = -\int_{-\infty}^{0} e^{-(s+a)t} dt = \frac{1}{s+a}e^{-(s+a)t} \Big|_{-\infty}^{0} -$$ - -† The discussion about the path of convergence is rather complicated, requiring the concepts of contour integration and understanding of the theory of complex variables. For this reason, the discussion here is somewhat simplified. - -| No. | x(t) | X(s) | -|-----|-------------------------------------------------------|-----------------------------------------------------------| -| 1 | δ(t) | 1 | -| 2 | u(t) | 1
s | -| 3 | tu(t) | 1
s2 | -| 4 | nu(t)
t | n!
sn+1 | -| 5 | eλt
u(t) | 1
s−λ | -| 6 | teλt
u(t) | 1
(s−λ)2 | -| 7 | neλt
t
u(t) | n!
(s−λ)n+1 | -| 8a | cos bt u(t) | s
s2 +b2 | -| 8b | sin bt u(t) | b
s2 +b2 | -| 9a | e−at cos
bt u(t) | s+a
(s+a)2 +b2 | -| 9b | e−atsin
bt u(t) | b
(s+a)2 +b2 | -| 10a | re−at cos(bt
+θ )u(t) | (r cos θ )s +(ar cos θ −brsin θ )
s2 +2as
+(a2 +b2) | -| 10b | re−at cos(bt
+θ )u(t) | 0.5rejθ
0.5re−jθ
s+a−jb +
s +a+jb | -| 10c | re−at cos(bt
+θ )u(t) | As+B
s2 +2as+c | -| |
A2c+B2 −2ABa
r =
c−a2 | | -| | Aa−B
θ = tan−1
√
c−a2
A | | -| | b = √
c−a2 | | -| 10d | e−at
!
B−Aa
Acos bt +
sin bt
u(t)
b | As+B
s2 +2as+c | -| | b = √
c−a2 | | - -**TABLE 4.1** Select (Unilateral) Laplace Transform Pairs - -Equation (4.5) shows that - -lim *t*→−∞*e*−(*s*+*a*)*t* = 0 Re (*s*+*a*) < 0 - -Hence, - -$$ -X(s) = \frac{1}{s+a} \qquad \text{Re } s < -a -$$ - -The signal −*e*−*atu*(−*t*) and its ROC (Re *s* < −*a*) are depicted in Fig. 4.1b. Note that the Laplace transforms for the signals *e*−*atu*(*t*) and −*e*−*atu*(−*t*) are identical except for their regions of convergence. Therefore, for a given *X*(*s*), there may be more than one inverse transform, depending on the ROC. In other words, unless the ROC is specified, there is no one-to-one correspondence between *X*(*s*) and *x*(*t*). This fact increases the complexity in using the Laplace transform. The complexity is the result of trying to handle causal as well as noncausal signals. If we restrict all our signals to the causal type, such an ambiguity does not arise. There is only one inverse transform of *X*(*s*) = 1/(*s* + *a*), namely, *e*−*atu*(*t*). To find *x*(*t*) from *X*(*s*), we need not even specify the ROC. In summary, if all signals are restricted to the causal type, then, for a given *X*(*s*), there is only one inverse transform *x*(*t*). † - -The unilateral Laplace transform is a special case of the bilateral Laplace transform in which all signals are restricted to being causal; consequently, the limits of integration for the integral in Eq. (4.1) can be taken from 0 to ∞. Therefore, the unilateral Laplace transform *X*(*s*) of a signal *x*(*t*) is defined as - -$$ -X(s) = \int_{0^{-}}^{\infty} x(t)e^{-st} dt -$$ - (4.7) - -We choose 0− (rather than 0+ used in some texts) as the lower limit of integration. This convention not only ensures inclusion of an impulse function at *t* = 0, but also allows us to use initial conditions at 0− (rather than at 0+) in the solution of differential equations via the Laplace transform. In practice, we are likely to know the initial conditions before the input is applied (at 0−), not after the input is applied (at 0+). Indeed, the very meaning of the term "initial conditions" implies conditions at *t* = 0− (conditions before the input is applied). Detailed analysis of desirability of using *t* = 0− appears in Sec. 4.3. - -The unilateral Laplace transform simplifies the system analysis problem considerably because of its *uniqueness property*, which says that for a given *X*(*s*), there is a unique inverse transform. But there is a price for this simplification: we cannot analyze noncausal systems or use noncausal inputs. However, in most practical problems, this restriction is of little consequence. For this reason, we shall first consider the unilateral Laplace transform and its application to system analysis. (The bilateral Laplace transform is discussed later, in Sec. 4.11.) - -Basically there is no difference between the unilateral and the bilateral Laplace transform. The unilateral transform is the bilateral transform that deals with a subclass of signals starting at *t* = 0 (causal signals). Therefore, the expression [Eq. (4.2)] for the inverse Laplace transform remains unchanged. In practice, the term *Laplace transform* means *the unilateral Laplace transform*. - -† Actually, *X*(*s*) specifies *x*(*t*) within a null function *n*(*t*), which has the property that the area under |*n*(*t*)| 2 is zero over any finite interval 0 to *t* (*t* > 0) (Lerch's theorem). For example, if two functions are identical everywhere except at finite number of points, they differ by a null function. - -### EXISTENCE OF THE LAPLACE TRANSFORM - -The variable *s* in the Laplace transform is complex in general, and it can be expressed as *s* =σ +*j*ω. By definition, - -$$ -X(s) = \int_{0^-}^{\infty} x(t)e^{-st} dt = \int_{0^-}^{\infty} [x(t)e^{-\sigma t}]e^{-j\omega t} dt -$$ - -Because |*ej*ω*t* | = 1, the integral on the right-hand side of this equation converges if - -$$ -\int_{0^{-}}^{\infty} \left| x(t)e^{-\sigma t} \right| dt < \infty \tag{4.8} -$$ - -Hence the existence of the Laplace transform is guaranteed if the integral in Eq. (4.8) is finite for some value of σ. Any signal that grows no faster than an exponential signal *Me*σ0*t* for some *M* and σ0 satisfies the condition of Eq. (4.8). Thus, if for some *M* and σ0, - -$$ -|x(t)| \le Me^{\sigma_0 t} \tag{4.9} -$$ - -we can choose σ>σ0 to satisfy Eq. (4.8).† The signal *et* 2 , in contrast, grows at a rate faster than *e*σ0*t* , and consequently is not Laplace–transformable.‡ Fortunately such signals (which are not Laplace–transformable) are of little consequence from either a practical or a theoretical viewpoint. If σ0 is the smallest value of σ for which the integral in Eq. (4.8) is finite, σ0 is called the *abscissa of convergence* and the ROC of *X*(*s*) is Re *s* > σ0. The abscissa of convergence for *e*−*atu*(*t*) is −*a* (the ROC is Re *s* > −*a*). - -### **EXAMPLE 4.2 Bilateral Laplace Transform of Common Causal Signals** - -Determine the Laplace transform of the following: **(a)** δ(*t*), **(b)** *u*(*t*), and **(c)** cos ω0*t u*(*t*). diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/053_4 CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/053_4 CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM.md deleted file mode 100644 index a14a38e7eb4936f1515d8cf5e6bfedc876b83b93..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/053_4 CONTINUOUS-TIME SYSTEM ANALYSIS USING THE LAPLACE TRANSFORM.md +++ /dev/null @@ -1,591 +0,0 @@ -**(a)** - -$$ -\mathcal{L}[\delta(t)] = \int_{0^-}^{\infty} \delta(t) e^{-st} dt -$$ - -Using the sampling property [Eq. (1.11) with *T* = 0], we obtain - -$$ -\mathcal{L}[\delta(t)] = 1 \qquad \text{for all } s -$$ - -that is, - -$$ -\delta(t) \Longleftrightarrow 1 \qquad \text{for all } s -$$ - -† The condition of Eq. (4.9) is sufficient but not necessary for the existence of the Laplace transform. For example, *x*(*t*) = 1/ √*t* is infinite at *t* = 0, and Eq. (4.9) cannot be satisfied; but the transform of 1/ √*t* exists and is given by √π/*s*. - -‡ However, if we consider a truncated (finite-duration) signal *et* 2 , the Laplace transform exists. - -**(b)** To find the Laplace transform of *u*(*t*), recall that *u*(*t*) = 1 for *t* ≥ 0. Therefore, - -$$ -\mathcal{L}[u(t)] = \int_{0^{-}}^{\infty} u(t)e^{-st} dt = \int_{0^{-}}^{\infty} e^{-st} dt = -\frac{1}{s}e^{-st} \Big|_{0^{-}}^{\infty} -$$ - -= $\frac{1}{s}$ Re $s > 0$ - -We also could have obtained this result from Eq. (4.6) by letting *a* = 0. **(c)** Because cos ω0*t u*(*t*) = 1 2 [*ej*ω0*t* +*e*−*j*ω0*t* ]*u*(*t*), we know that - -$$ -\mathcal{L}[\cos \omega_0 t u(t)] = \frac{1}{2} \mathcal{L}[e^{j\omega_0 t} u(t) + e^{-j\omega_0 t} u(t)] -$$ - -From Eq. (4.6), it follows that - -$$ -\mathcal{L}[\cos \omega_0 t u(t)] = \frac{1}{2} \left[ \frac{1}{s - j\omega_0} + \frac{1}{s + j\omega_0} \right] \qquad \text{Re}\,(s \pm j\omega) = \text{Re}\,s > 0 -$$ -\n -$$ -= \frac{s}{s^2 + \omega_0^2} \qquad \text{Re}\,s > 0 \tag{4.10} -$$ - -For the unilateral Laplace transform, there is a unique inverse transform of *X*(*s*); consequently, there is no need to specify the ROC explicitly. For this reason, we shall generally ignore any mention of the ROC for unilateral transforms. Recall, also, that in the unilateral Laplace transform it is understood that every signal *x*(*t*) is zero for *t* < 0, and it is appropriate to indicate this fact by multiplying the signal by *u*(*t*). - -### **DR ILL 4.1 Bilateral Laplace Transform of Gate Functions** - -By direct integration, find the Laplace transform *X*(*s*) and the region of convergence of *X*(*s*) for the gate functions shown in Fig. 4.2. - -**Figure 4.2** Gate functions for Drill 4.1. - -### **ANSWERS** - -(a) -$$ -\frac{1}{s}(1-e^{-2s}) -$$ - for all *s* -\n(b) $\frac{1}{s}(1-e^{-2s})e^{-2s}$ for all *s* - -### **[4.1-1 Finding the Inverse Transform](#page-10-0)** - -Finding the inverse Laplace transform by using Eq. (4.2) requires integration in the complex plane, a subject beyond the scope of this book (but see, e.g., [3]). For our purpose, we can find the inverse transforms from Table 4.1. All we need is to express *X*(*s*) as a sum of simpler functions of the forms listed in the table. Most of the transforms *X*(*s*) of practical interest are *rational functions*, that is, ratios of polynomials in *s*. Such functions can be expressed as a sum of simpler functions by using partial fraction expansion (see Sec. B.5). - -Values of *s* for which *X*(*s*) = 0 are called the *zeros* of *X*(*s*); the values of *s* for which *X*(*s*)→∞ are called the *poles* of *X*(*s*). If *X*(*s*) is a rational function of the form *P*(*s*)/*Q*(*s*), the roots of *P*(*s*) are the zeros and the roots of *Q*(*s*) are the poles of *X*(*s*). - -### **EXAMPLE 4.3 Inverse Unilateral Laplace Transform** - -Find the inverse unilateral Laplace transforms of - -(a) -$$ -\frac{7s-6}{s^2-s-6} -$$ - -\n(b) -$$ -\frac{2s^2+5}{s^2+3s+2} -$$ - -\n(c) -$$ -\frac{6(s+34)}{s(s^2+10s+34)} -$$ - -\n(d) -$$ -\frac{8s+10}{(s+1)(s+2)^3} -$$ - -In no case is the inverse transform of these functions directly available in Table 4.1. Rather, we need to expand these functions into partial fractions, as discussed in Sec. B.5-1. Today, it is very easy to find partial fractions via software such as MATLAB. However, just as the availability of a calculator does not obviate the need for learning the mechanics of arithmetical operations (addition, multiplication, etc.), the widespread availability of computers does not eliminate the need to learn the mechanics of partial fraction expansion. - -$$ -\left( \mathbf{a}\right) -$$ - -$$ -X(s) = \frac{7s - 6}{(s + 2)(s - 3)} = \frac{k_1}{s + 2} + \frac{k_2}{s - 3} -$$ - -To determine *k*1, corresponding to the term (*s* + 2), we cover up (conceal) the term (*s* + 2) in *X*(*s*) and substitute *s* = −2 (the value of *s* that makes *s* + 2 = 0) in the remaining expression (see Sec. B.5-2): - -$$ -k_1 = \frac{7s - 6}{(s + 2)(s - 3)}\bigg|_{s = -2} = \frac{-14 - 6}{-2 - 3} = 4 -$$ - -Similarly, to determine *k*2 corresponding to the term (*s* − 3), we cover up the term (*s* − 3) in *X*(*s*) and substitute *s* = 3 in the remaining expression - -$$ -k_2 = \frac{7s - 6}{(s + 2)(s - 3)}\bigg|_{s=3} = \frac{21 - 6}{3 + 2} = 3 -$$ - -Therefore, - -$$ -X(s) = \frac{7s - 6}{(s + 2)(s - 3)} = \frac{4}{s + 2} + \frac{3}{s - 3} -$$ -(4.11) - -### CHECKING THE ANSWER - -It is easy to make a mistake in partial fraction computations. Fortunately it is simple to check the answer by recognizing that *X*(*s*) and its partial fractions must be equal for every value of *s* if the partial fractions are correct. Let us verify this assertion in Eq. (4.11) for some convenient value, say, *s* = 0. Substitution of *s* = 0 in Eq. (4.11) yields† - -$$ -1 = 2 - 1 = 1 -$$ - -We can now be sure of our answer with a high margin of confidence. Using pair 5 of Table 4.1 in Eq. (4.11), we obtain - -$$ -x(t) = \mathcal{L}^{-1}\left(\frac{4}{s+2} + \frac{3}{s-3}\right) = (4e^{-2t} + 3e^{3t})u(t) -$$ - -**(b)** - -$$ -X(s) = \frac{2s^2 + 5}{s^2 + 3s + 2} = \frac{2s^2 + 5}{(s+1)(s+2)} -$$ - -Observe that *X*(*s*) is an improper function with *M* = *N*. In such a case, we can express *X*(*s*) as a sum of the coefficient of the highest power in the numerator plus partial fractions corresponding to the poles of *X*(*s*) (see Sec. B.5-5). In the present case, the coefficient of the highest power in the numerator is 2. Therefore, - -$$ -X(s) = 2 + \frac{k_1}{s+1} + \frac{k_2}{s+2} -$$ - -where - -$$ -k_1 = \frac{2s^2 + 5}{(s+1)(s+2)}\bigg|_{s=-1} = \frac{2+5}{-1+2} = 7 -$$ - -and - -$$ -k_2 = \frac{2s^2 + 5}{(s+1)(s+2)}\bigg|_{s=-2} = \frac{8+5}{-2+1} = -13 -$$ - -Therefore, - -$$ -X(s) = 2 + \frac{7}{s+1} - \frac{13}{s+2} -$$ - -From Table 4.1, pairs 1 and 5, we obtain - -$$ -x(t) = 2\delta(t) + (7e^{-t} - 13e^{-2t})u(t) -$$ - -† Because *X*(*s*) = ∞ at its poles, we should avoid the pole values (−2 and 3 in the present case) for checking. The answers may check even if partial fractions are wrong. This situation can occur when two or more errors cancel their effects. But the chances of this problem arising for randomly selected values of *s* are extremely small. - -**(c)** - -$$ -X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{6(s+34)}{s(s+5-j3)(s+5+j3)} -$$ -$$ -= \frac{k_1}{s} + \frac{k_2}{s+5-j3} + \frac{k_2^*}{s+5+j3} -$$ - -Note that the coefficients (*k*2 and *k*∗ 2) of the conjugate terms must also be conjugate (see Sec. B.5). Now - -$$ -k_1 = \frac{6(s+34)}{s(s^2+10s+34)}\bigg|_{s=0} = \frac{6 \times 34}{34} = 6 -$$ - -$$ -k_2 = \frac{6(s+34)}{s(s+5-j3)(s+5+j3)}\bigg|_{s=-5+j3} = \frac{29+j3}{-3-j5} = -3+j4 -$$ - -Therefore, - -$$ -k_2^* = -3 - j4 -$$ - -To use pair 10b of Table 4.1, we need to express *k*2 and *k*∗ 2 in polar form. - -$$ --3 + j4 = (\sqrt{3^2 + 4^2}) e^{j \tan^{-1}(4/4)} = 5 e^{j \tan^{-1}(4/4)} -$$ - -Observe that tan−1(4/−3) = tan−1(−4/3). This fact is evident in Fig. 4.3. For further discussion of this topic, see Ex. B.1. - -**Figure 4.3** Visualizing tan−1(−4/3) = tan−1(4/−3). - -From Fig. 4.3, we observe that - -$$ -k_2 = -3 + j4 = 5e^{j126.9^{\circ}} -$$ - -so - -$$ -k_2^* = 5e^{-j126.9^\circ} -$$ - -Therefore, - -$$ -X(s) = \frac{6}{s} + \frac{5e^{j126.9^{\circ}}}{s+5-j3} + \frac{5e^{-j126.9^{\circ}}}{s+5+j3} -$$ - -From Table 4.1 (pairs 2 and 10b), we obtain - -$$ -x(t) = [6 + 10e^{-5t}\cos(3t + 126.9^\circ)]u(t) -$$ - -### ALTERNATIVE METHOD USING QUADRATIC FACTORS - -The foregoing procedure involves considerable manipulation of complex numbers. Pair 10c (Table 4.1) indicates that the inverse transform of quadratic terms (with complex conjugate poles) can be found directly without having to find first-order partial fractions. We discussed such a procedure in Sec. B.5-2. For this purpose, we shall express *X*(*s*) as - -$$ -X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{k_1}{s} + \frac{As+B}{s^2+10s+34} -$$ - -We have already determined that *k*1 = 6 by the (Heaviside) "cover-up" method. Therefore, - -$$ -\frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{As+B}{s^2+10s+34} -$$ - -Clearing the fractions by multiplying both sides by *s*(*s*2 +10*s*+34) yields - -$$ -6(s+34) = (6+A)s2 + (60+B)s + 204 -$$ - -Now, equating the coefficients of *s*2 and *s* on both sides yields - -$$ -A = -6 \qquad \text{and} \qquad B = -54 -$$ - -and - -$$ -X(s) = \frac{6}{s} + \frac{-6s - 54}{s^2 + 10s + 34} -$$ - -We now use pairs 2 and 10c to find the inverse Laplace transform. The parameters for pair 10c are *A* = −6, *B* = −54, *a* = 5, *c* = 34, *b* = √ *c*−*a*2 = 3, and - -$$ -r = \sqrt{\frac{A^2c + B^2 - 2ABA}{c - a^2}} = 10 \qquad \theta = \tan^{-1} \frac{Aa - B}{A\sqrt{c - a^2}} = 126.9^{\circ} -$$ - -Therefore, - -$$ -x(t) = [6 + 10e^{-5t}\cos(3t + 126.9^\circ)]u(t) -$$ - -which agrees with the earlier result. - -### SHORTCUTS - -The partial fractions with quadratic terms also can be obtained by using shortcuts. We have - -$$ -X(s) = \frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{As+B}{s^2+10s+34} -$$ - -### 342 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -We can determine *A* by eliminating *B* on the right-hand side. This step can be accomplished by multiplying both sides of the equation for *X*(*s*) by *s* and then letting *s*→ ∞. This procedure yields - -$$ -0 = 6 + A \quad \Longrightarrow \quad A = -6 -$$ - -Therefore, - -$$ -\frac{6(s+34)}{s(s^2+10s+34)} = \frac{6}{s} + \frac{-6s+16}{s^2+10s+34} -$$ - -To find *B*, we let *s* take on any convenient value, say, *s* = 1, in this equation to obtain - -$$ -\frac{210}{45} = 6 + \frac{B - 6}{45} \implies B = -54 -$$ - -a result that agrees with the answer found earlier. - -**(d)** - -$$ -X(s) = \frac{8s+10}{(s+1)(s+2)^3} = \frac{k_1}{s+1} + \frac{a_0}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2} -$$ - -where - -$$ -k_1 = \frac{8s + 10}{(s+1)(s+2)^3}\Big|_{s=-1} = 2 -$$ - -\n -$$ -a_0 = \frac{8s + 10}{(s+1)(s+2)^3}\Big|_{s=-2} = 6 -$$ - -\n -$$ -a_1 = \left\{\frac{d}{ds}\left[\frac{8s + 10}{(s+1)(s+2)^3}\right]\right\}_{s=-2} = -2 -$$ - -\n -$$ -a_2 = \frac{1}{2}\left\{\frac{d^2}{ds^2}\left[\frac{8s + 10}{(s+1)(s+2)^3}\right]\right\}_{s=-2} = -2 -$$ - -Therefore, - -$$ -X(s) = \frac{2}{s+1} + \frac{6}{(s+2)^3} - \frac{2}{(s+2)^2} - \frac{2}{s+2} -$$ - -and - -$$ -x(t) = [2e^{-t} + (3t^2 - 2t - 2)e^{-2t}]u(t) -$$ - -### ALTERNATIVE METHOD:AHYBRID OF HEAVISIDE AND CLEARING FRACTIONS - -In this method, the simpler coefficients *k*1 and *a*0 are determined by the Heaviside "cover-up" procedure, as discussed earlier. To determine the remaining coefficients, we use the clearing-fraction method. Using the values *k*1 = 2 and *a*0 = 6 obtained earlier by the Heaviside "cover-up" method, we have - -$$ -\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2} -$$ - -We now clear fractions by multiplying both sides of the equation by (*s* + 1)(*s* + 2)3. This procedure yields† - -$$ -8s + 10 = 2(s + 2)3 + 6(s + 1) + a1(s + 1)(s + 2) + a2(s + 1)(s + 2)2 -$$ - -= (2 + a2)s3 + (12 + a1 + 5a2)s2 + (30 + 3a1 + 8a2)s + (22 + 2a1 + 4a2) - -Equating coefficients of *s*3 and *s*2 on both sides, we obtain - -$$ -0 = (2 + a_2) \implies a_2 = -2 0 = 12 + a_1 + 5a_2 = 2 + a_1 \implies a_1 = -2 -$$ - -We can stop here if we wish, since the two desired coefficients *a*1 and *a*2 have already been found. However, equating the coefficients of *s*1 and *s*0 serves as a check on our answers. This step yields - -$$ -8 = 30 + 3a_1 + 8a_2 -$$ - -$$ -10 = 22 + 2a_1 + 4a_2 -$$ - -Substitution of *a*1 = *a*2 = −2, obtained earlier, satisfies these equations. This step confirms the correctness of our answers. - -### ANOTHER ALTERNATIVE:AHYBRID OF HEAVISIDE AND SHORTCUTS - -In this method, the simpler coefficients *k*1 and *a*0 are determined by the Heaviside "cover-up" procedure, as discussed earlier. The usual shortcuts are then used to determine the remaining coefficients. Using the values *k*1 = 2 and *a*0 = 6, determined earlier by the Heaviside method, we have - -$$ -\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} + \frac{a_2}{s+2} -$$ - -There are two unknowns, *a*1 and *a*2. If we multiply both sides by *s* and then let *s* → ∞, we eliminate *a*1. This procedure yields - -0 = 2+*a*2 ⇒ *a*2 = −2 - -Therefore, - -$$ -\frac{8s+10}{(s+1)(s+2)^3} = \frac{2}{s+1} + \frac{6}{(s+2)^3} + \frac{a_1}{(s+2)^2} - \frac{2}{s+2} -$$ - -There is now only one unknown, *a*1. This value can be determined readily by setting *s* equal to any convenient value, say, *s* = 0. This step yields - -$$ -\frac{10}{8} = 2 + \frac{3}{4} + \frac{a_1}{4} - 1 \implies a_1 = -2 -$$ - -† We could have cleared fractions without finding *k*1 and *a*0. This alternative, however, proves more laborious because it increases the number of unknowns to 4. By predetermining *k*1 and *a*0, we reduce the unknowns to 2. Moreover, this method provides a convenient check on the solution. This hybrid procedure achieves the best of both methods. - -### **EXAMPLE 4.4 Inverse Laplace Transform with MATLAB** - -Using the MATLAB residue command, determine the inverse Laplace transform of each of the following functions: - -(a) -$$ -X_a(s) = \frac{2s^2 + 5}{s^2 + 3s + 2} -$$ - -\n(b) $X_b(s) = \frac{2s^2 + 7s + 4}{(s+1)(s+2)^2}$ -\n(c) $X_c(s) = \frac{8s^2 + 21s + 19}{(s+2)(s^2 + s + 7)}$ - -In each case, we use the MATLAB residue command to perform the necessary partial fraction expansions. The inverse Laplace transform follows using Table 4.1. - -``` -(a) ->> num = [2 0 5]; den = [1 3 2]; ->> [r, p, k] = residue(num,den) - r = -13 - 7 - p = -2 -``` - -Therefore, *Xa*(*s*) = −13/(*s*+2) +7/(*s*+1) +2 and *xa*(*t*) = (−13*e*−2*t* +7*e*−*t* )*u*(*t*)+2δ(*t*). - -``` -(b) -``` - --1 k= 2 - -``` ->> num = [2 7 4]; den = [conv([1 1],conv([1 2],[1 2]))]; ->> [r, p, k] = residue(num,den) - r= 3 - 2 - -1 - p = -2 - -2 - -1 - k = [] -``` - -Therefore, *Xb*(*s*) = 3/(*s*+2) +2/(*s*+2)2 −1/(*s*+1) and *xb*(*t*) = (3*e*−2*t* +2*te*−2*t* −*e*−*t* )*u*(*t*). - -**(c)** In this case, a few calculations are needed beyond the results of the residue command so that pair 10b of Table 4.1 can be utilized. - ->> num = [8 21 19]; den = [conv([1 2],[1 1 7])]; >> [r, p, k]= residue(num,den) - -``` -r = 3.5000-0.48113i - 3.5000+0.48113i - 1.0000 - p = -0.5000+2.5981i - -0.5000-2.5981i - -2.0000 - k = [] ->> ang = angle(r), mag = abs(r) - ang = -0.13661 - 0.13661 - 0 - mag = 3.5329 - 3.5329 - 1.0000 -``` - -Thus, - -$$ -X_c(s) = \frac{1}{s+2} + \frac{3.5329e^{-j0.13661}}{s+0.5-j2.5981} + \frac{3.5329e^{j0.13661}}{s+0.5+j2.5981} -$$ - -and - -$$ -x_c(t) = [e^{-2t} + 1.7665e^{-0.5t}\cos(2.5981t - 0.1366)]u(t). -$$ - -## **EXAMPLE 4.5 Symbolic Laplace and Inverse Laplace Transforms with MATLAB** - -Using MATLAB's symbolic math toolbox, determine the following: - -- **(a)** the direct unilateral Laplace transform of *xa*(*t*) = sin(*at*)+cos(*bt*) -- **(b)** the inverse unilateral Laplace transform of *Xb*(*s*) = *as*2/(*s*2 +*b*2) - -**(a)** Here, we use the sym command to symbolically define our variables and expression for *xa*(*t*), and then we use the laplace command to compute the (unilateral) Laplace transform. - ->> syms a b t; x\_a = sin(a\*t)+cos(b\*t); >> X\_a = laplace(x\_a); X\_a = a/(a^2 + s^2) + s/(b^2 + s^2) - -Therefore, *Xa*(*s*) = *a s*2+*a*2 + *s s*2+*b*2 . It is also easy to use MATLAB to determine *Xa*(*s*) in standard rational form. - ->> X\_a = collect(X\_a) X\_a = (a^2\*s + a\*b^2 + a\*s^2 + s^3)/(s^4 + (a^2 + b^2)\*s^2 + a^2\*b^2) - -Thus, we also see that *Xa*(*s*) = *s*3+*as*2+*a*2*s*+*ab*2 *s*4+(*a*2+*b*2)*s*2+*a*2*b*2 - -**(b)** A similar approach is taken for the inverse Laplace transform, except that the ilaplace command is used rather than the laplace command. - -``` ->> syms a b s; X_b = (a*s^2)/(s^2+b^2); ->> x_b = ilaplace(X_b) - x_b = a*dirac(t) - a*b*sin(b*t) -``` - -Therefore, *xb*(*t*) = *a*δ(*t*)−*ab*sin(*bt*)*u*(*t*). - -### **DR ILL 4.2 Laplace Transform** - -Show that the Laplace transform of 10*e*−3*t* cos (4*t* + 53.13◦) is (6*s* − 14)/(*s*2 + 6*s* + 25). Use Table 4.1. - -### **DR ILL 4.3 Inverse Laplace Transform** - -Find the inverse Laplace transform of the following: - -(a) -$$ -\frac{s+17}{s^2+4s-5} -$$ - -\n(b) -$$ -\frac{3s-5}{(s+1)(s^2+2s+5)} -$$ - -\n(c) -$$ -\frac{16s+43}{(s-2)(s+3)^2} -$$ - -### **ANSWERS** - -(a) -$$ -(3e^t - 2e^{-5t})u(t) -$$ - -- **(b)** −2*e*−*t* + 5 2 *e*−*t* cos(2*t* −36.87◦) *u*(*t*) -- **(c)** [3*e*2*t* +(*t* −3)*e*−3*t* ]*u*(*t*) - -## A HISTORICAL NOTE: MARQUIS PIERRE-SIMON DE LAPLACE (1749–1827) - -The Laplace transform is named after the great French mathematician and astronomer Laplace, who first presented the transform and its applications to differential equations in a paper published in 1779. - -Laplace developed the foundations of potential theory and made important contributions to special functions, probability theory, astronomy, and celestial mechanics. In his *Exposition du système du monde* (1796), Laplace formulated a nebular hypothesis of cosmic origin and tried to explain the universe as a pure mechanism. In his *Traité de mécanique céleste* (*celestial mechanics*), which completed the work of Newton, Laplace used mathematics and physics to subject the solar system and all heavenly bodies to the laws of motion and the principle of gravitation. Newton had - -Pierre-Simon de Laplace and Oliver Heaviside - -been unable to explain the irregularities of some heavenly bodies; in desperation, he concluded that God himself must intervene now and then to prevent such catastrophes as Jupiter eventually falling into the sun (and the moon into the earth), as predicted by Newton's calculations. Laplace proposed to show that these irregularities would correct themselves periodically and that a little patience—in Jupiter's case, 929 years—would see everything returning automatically to order; thus there was no reason why the solar and the stellar systems could not continue to operate by the laws of Newton and Laplace to the end of time [4]. - -Laplace presented a copy of *Mécanique céleste* to Napoleon, who, after reading the book, took Laplace to task for not including God in his scheme: "You have written this huge book on the system of the world without once mentioning the author of the universe." "Sire," Laplace retorted, "I had no need of that hypothesis." Napoleon was not amused, and when he reported this reply to another great mathematician-astronomer, Louis de Lagrange, the latter remarked, "Ah, but that is a fine hypothesis. It explains so many things" [5]. - -Napoleon, following his policy of honoring and promoting scientists, made Laplace the minister of the interior. To Napoleon's dismay, however, the new appointee attempted to bring "the spirit of infinitesimals" into administration, and so Laplace was transferred hastily to the Senate. - -### OLIVER HEAVISIDE (1850–1925) - -Although Laplace published his transform method to solve differential equations in 1779, the method did not catch on until a century later. It was rediscovered independently in a rather awkward form by an eccentric British engineer, Oliver Heaviside (1850–1925), one of the tragic figures in the history of science and engineering. Despite his prolific contributions to electrical engineering, he was severely criticized during his lifetime and was neglected later to the point that - -### 348 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -hardly a textbook today mentions his name or credits him with contributions. Nevertheless, his studies had a major impact on many aspects of modern electrical engineering. It was Heaviside who made transatlantic communication possible by inventing cable loading, but few mention him as a pioneer or an innovator in telephony. It was Heaviside who suggested the use of inductive cable loading, but the credit is given to M. Pupin, who was not even responsible for building the first loading coil.† In addition, Heaviside was [6]: - -- The first to find a solution to the distortionless transmission line. -- The innovator of lowpass filters. -- The first to write Maxwell's equations in modern form. -- The codiscoverer of rate energy transfer by an electromagnetic field. -- An early champion of the now-common phasor analysis. -- An important contributor to the development of vector analysis. In fact, he essentially created the subject independently of Gibbs [7]. -- An originator of the use of operational mathematics used to solve linear integro-differential equations, which eventually led to rediscovery of the ignored Laplace transform. -- The first to theorize (along with Kennelly of Harvard) that a conducting layer (the Kennelly–Heaviside layer) of atmosphere exists, which allows radio waves to follow earth's curvature instead of traveling off into space in a straight line. -- The first to posit that an electrical charge would increase in mass as its velocity increases, an anticipation of an aspect of Einstein's special theory of relativity [8]. He also forecast the possibility of superconductivity. - -Heaviside was a self-made, self-educated man. Although his formal education ended with elementary school, he eventually became a pragmatically successful mathematical physicist. He began his career as a telegrapher, but increasing deafness forced him to retire at the age of 24. He then devoted himself to the study of electricity. His creative work was disdained by many professional mathematicians because of his lack of formal education and his unorthodox methods. - -Heaviside had the misfortune to be criticized both by mathematicians, who faulted him for lack of rigor, and by men of practice, who faulted him for using too much mathematics and thereby confusing students. Many mathematicians, trying to find solutions to the distortionless transmission line, failed because no rigorous tools were available at the time. Heaviside succeeded because he used mathematics not with rigor, but with insight and intuition. Using his much maligned operational method, Heaviside successfully attacked problems that the rigid mathematicians could not solve, problems such as the flow-of-heat in a body of spatially varying conductivity. Heaviside brilliantly used this method in 1895 to demonstrate a fatal flaw in Lord Kelvin's determination of the geological age of the earth by secular cooling; he used the same flow-of-heat theory as for his cable analysis. Yet the mathematicians of the Royal Society remained unmoved and were not the least impressed by the fact that Heaviside had found the answer to problems no one else could solve. Many mathematicians who examined his work dismissed it - -† Heaviside developed the theory for cable loading, George Campbell built the first loading coil, and the telephone circuits using Campbell's coils were in operation before Pupin published his paper. In the legal fight over the patent, however, Pupin won the battle: he was a shrewd self-promoter, and Campbell had poor legal support. - -with contempt, asserting that his methods were either complete nonsense or a rehash of known ideas [6]. - -Sir William Preece, the chief engineer of the British Post Office, a savage critic of Heaviside, ridiculed Heaviside's work as too theoretical and, therefore, leading to faulty conclusions. Heaviside's work on transmission lines and loading was dismissed by the British Post Office and might have remained hidden, had not Lord Kelvin himself publicly expressed admiration for it [6]. - -Heaviside's operational calculus may be formally inaccurate, but in fact it anticipated the operational methods developed in more recent years [9]. Although his method was not fully understood, it provided correct results. When Heaviside was attacked for the vague meaning of his operational calculus, his pragmatic reply was, "Shall I refuse my dinner because I do not fully understand the process of digestion?" - -Heaviside lived as a bachelor hermit, often in near-squalid conditions, and died largely unnoticed, in poverty. His life demonstrates the persistent arrogance and snobbishness of the intellectual establishment, which does not respect creativity unless it is presented in the strict language of the establishment. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/054_4.2 SOME PROPERTIES OF THE LAPLACE TRANSFORM.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/054_4.2 SOME PROPERTIES OF THE LAPLACE TRANSFORM.md deleted file mode 100644 index 31dab8d144e386cb5de7bfec969f24f96fc7a816..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/054_4.2 SOME PROPERTIES OF THE LAPLACE TRANSFORM.md +++ /dev/null @@ -1,544 +0,0 @@ -## **4.2 SOME [PROPERTIES OF THE](#page-10-0) LAPLACE TRANSFORM** - -Properties of the Laplace transform are useful not only in the derivation of the Laplace transform of functions but also in the solutions of linear integro-differential equations. A glance at Eqs. (4.2) and (4.1) shows that there is a certain measure of symmetry in going from *x*(*t*) to *X*(*s*), and vice versa. This symmetry or duality is also carried over to the properties of the Laplace transform. This fact will be evident in the following development. - -We are already familiar with two properties: linearity [Eq. (4.3)] and the uniqueness property of the Laplace transform discussed earlier. - -### **[4.2-1 Time Shifting](#page-10-0)** - -The time-shifting property states that if - -*x*(*t*) ⇐⇒ *X*(*s*) - -then for *t*0 ≥ 0 - -$$ -x(t - t_0) \Longleftrightarrow X(s)e^{-st_0} \tag{4.12} -$$ - -Observe that *x*(*t*) starts at *t* = 0, and, therefore, *x*(*t* − *t*0) starts at *t* = *t*0. This fact is implicit, but is not explicitly indicated in Eq. (4.12). This often leads to inadvertent errors. To avoid such a pitfall, we should restate the property as follows. If - -$$ -x(t)u(t) \Longleftrightarrow X(s) -$$ - -then - -$$ -x(t-t_0)u(t-t_0) \Longleftrightarrow X(s)e^{-st_0} \qquad t_0 \ge 0 -$$ - -**Proof.** - -$$ -\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_0^\infty x(t-t_0)u(t-t_0)e^{-st}dt -$$ - -Setting *t* −*t*0 = τ , we obtain - -$$ -\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_{-t_0}^{\infty} x(\tau)u(\tau)e^{-s(\tau+t_0)}d\tau -$$ - -Because *u*(τ ) = 0 for τ < 0 and *u*(τ ) = 1 for τ ≥ 0, the limits of integration can be taken from 0 to ∞. Thus, - -$$ -\mathcal{L}[x(t-t_0)u(t-t_0)] = \int_0^\infty x(\tau)e^{-s(\tau+t_0)}d\tau -$$ -$$ -= e^{-st_0} \int_0^\infty x(\tau)e^{-s\tau}d\tau -$$ -$$ -= X(s)e^{-st_0} -$$ - -Note that *x*(*t* − *t*0)*u*(*t* − *t*0) is the signal *x*(*t*)*u*(*t*) delayed by *t*0 seconds. The time-shifting property states that *delaying a signal by t*0 *seconds amounts to multiplying its transform e*−*st*0 . - -This property of the unilateral Laplace transform holds only for positive *t*0 because if *t*0 were negative, the signal *x*(*t* −*t*0)*u*(*t* −*t*0) may not be causal. - -We can readily verify this property in Drill 4.1. If the signal in Fig. 4.2a is *x*(*t*)*u*(*t*), then the signal in Fig. 4.2b is *x*(*t* − 2)*u*(*t* − 2). The Laplace transform for the pulse in Fig. 4.2a is (1/*s*)(1−*e*−2*s* ). Therefore, the Laplace transform for the pulse in Fig. 4.2b is (1/*s*)(1−*e*−2*s* )*e*−2*s* . - -The time-shifting property proves very convenient in finding the Laplace transform of functions with different descriptions over different intervals, as the following example demonstrates. - -### **EXAMPLE 4.6 Laplace Transform and the Time-Shifting Property** - -Find the Laplace transform of *x*(*t*) depicted in Fig. 4.4a. - -Describing mathematically a function such as the one in Fig. 4.4a is discussed in Sec. 1.4. The function *x*(*t*) in Fig. 4.4a can be described as a sum of two components shown in Fig. 4.4b. The equation for the first component is *t*−1 over 1 ≤ *t* ≤ 2 so that this component can be described by (*t* −1)[*u*(*t* −1)−*u*(*t* −2)]. The second component can be described by *u*(*t* −2)−*u*(*t* −4). Therefore, - -$$ -x(t) = (t-1)[u(t-1) - u(t-2)] + [u(t-2) - u(t-4)] -$$ - -= $(t-1)u(t-1) - (t-1)u(t-2) + u(t-2) - u(t-4)$ (4.13) - -**Figure 4.4** Finding a piecewise representation of a signal *x*(*t*). - -The first term on the right-hand side is the signal *tu*(*t*) delayed by 1 second. Also, the third and fourth terms are the signal *u*(*t*) delayed by 2 and 4 seconds, respectively. The second term, however, cannot be interpreted as a delayed version of any entry in Table 4.1. For this reason, we rearrange it as - -$$ -(t-1)u(t-2) = (t-2+1)u(t-2) = (t-2)u(t-2) + u(t-2) -$$ - -We have now expressed the second term in the desired form as *tu*(*t*) delayed by 2 seconds plus *u*(*t*) delayed by 2 seconds. With this result, Eq. (4.13) can be expressed as - -$$ -x(t) = (t-1)u(t-1) - (t-2)u(t-2) - u(t-4) -$$ - -Application of the time-shifting property to *tu*(*t*) ⇐⇒ 1/*s*2 yields - -$$ -(t-1)u(t-1) \Longleftrightarrow \frac{1}{s^2}e^{-s} \qquad \text{and} \qquad (t-2)u(t-2) \Longleftrightarrow \frac{1}{s^2}e^{-2s} -$$ - -Also - -$$ -u(t) \Longleftrightarrow \frac{1}{s} -$$ - and $u(t-4) \Longleftrightarrow \frac{1}{s}e^{-4s}$ - -Therefore, - -$$ -X(s) = \frac{1}{s^2}e^{-s} - \frac{1}{s^2}e^{-2s} - \frac{1}{s}e^{-4s} -$$ - -### **EXAMPLE 4.7 Inverse Laplace Transform and the Time-Shifting Property** - -Find the inverse Laplace transform of - -$$ -X(s) = \frac{s+3+5e^{-2s}}{(s+1)(s+2)} -$$ - -### 352 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Observe the exponential term *e*−2*s* in the numerator of *X*(*s*), indicating time delay. In such a case, we should separate *X*(*s*) into terms with and without a delay factor, as - -$$ -X(s) = \underbrace{\frac{s+3}{(s+1)(s+2)}}_{X_1(s)} + \underbrace{\frac{5e^{-2s}}{(s+1)(s+2)}}_{X_2(s)e^{-2s}} -$$ - -where - -$$ -X_1(s) = \frac{s+3}{(s+1)(s+2)} = \frac{2}{s+1} - \frac{1}{s+2} -$$ -$$ -X_2(s) = \frac{5}{(s+1)(s+2)} = \frac{5}{s+1} - \frac{5}{s+2} -$$ - -Therefore, - -$$ -x_1(t) = (2e^{-t} - e^{-2t})u(t) -$$ - -$$ -x_2(t) = 5(e^{-t} - e^{-2t})u(t) -$$ - -Also, because - -$$ -X(s) = X_1(s) + X_2(s)e^{-2s} -$$ - -we can write - -$$ -x(t) = x_1(t) + x_2(t-2) -$$ - -= $(2e^{-t} - e^{-2t})u(t) + 5[e^{-(t-2)} - e^{-2(t-2)}]u(t-2)$ - -## **DR ILL 4.4 Laplace Transform and the Time-Shifting Property** - -Find the Laplace transform of the signal illustrated in Fig. 4.5. - -## **DR ILL 4.5 Inverse Laplace Transform and the Time-Shifting Property** - -Find the inverse Laplace transform of *X*(*s*) = 3*e*−2*s* (*s*−1)(*s*+2) . - -**ANSWER** *et*−2 −*e*−2(*t*−2) *u*(*t* −2) - -### **[4.2-2 Frequency Shifting](#page-10-0)** - -The frequency-shifting property states that if - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -then - -$$ -x(t)e^{s_0t} \Longleftrightarrow X(s-s_0) \tag{4.14} -$$ - -Observe the symmetry (or duality) between this property and the time-shifting property of Eq. (4.12). - -**Proof.** - -$$ -\mathcal{L}[x(t)e^{s_0t}] = \int_{0^-}^{\infty} x(t)e^{s_0t}e^{-st} dt = \int_{0^-}^{\infty} x(t)e^{-(s-s_0)t} dt = X(s-s_0) -$$ - -### **EXAMPLE 4.8 Frequency-Shifting Property** - -Derive pair 9a in Table 4.1 from pair 8a and the frequency-shifting property. - -Pair 8a is - -$$ -\cos btu(t) \Longleftrightarrow \frac{s}{s^2 + b^2} -$$ - -From the frequency-shifting property [Eq. (4.14)] with *s*0 = −*a*, we obtain - -$$ -e^{-at}\cos btu(t) \Longleftrightarrow \frac{s+a}{(s+a)^2 + b^2} -$$ - -### **DR ILL 4.6 Frequency-Shifting Property** - -Derive pair 6 in Table 4.1 from pair 3 and the frequency-shifting property. - -### 354 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -We are now ready to consider the two most important properties of the Laplace transform: time differentiation and time integration. - -### **[4.2-3 The Time-Differentiation Property](#page-10-0)** - -The time-differentiation property states that if† - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -then - -$$ -\frac{dx(t)}{dt} \Longleftrightarrow sX(s) - x(0^-) -$$ - -Repeating this property a second time (differentiating twice) yields - -$$ -\frac{d^2x(t)}{dt^2} \Longleftrightarrow s^2X(s) - sx(0^-) - \dot{x}(0^-) -$$ - -Repeated differentiation yields - -$$ -\frac{d^n x(t)}{dt^n} \Longleftrightarrow s^n X(s) - s^{n-1} x(0^-) - s^{n-2} \dot{x}(0^-) - \dots - x^{(n-1)}(0^-) -$$ - -= $s^n X(s) - \sum_{k=1}^n s^{n-k} x^{(k-1)}(0^-)$ (4.15) - -where *x*(*r*) (0−) is *dr x*/*dtr* at *t* = 0−. - -**Proof.** - -$$ -\mathcal{L}\left[\frac{dx(t)}{dt}\right] = \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt -$$ - -Integrating by parts, we obtain - -$$ -\mathcal{L}\left[\frac{dx(t)}{dt}\right] = x(t)e^{-st}\Big|_{0^{-}}^{\infty} + s\int_{0^{-}}^{\infty} x(t)e^{-st}dt -$$ - -For the Laplace integral to converge [i.e., for *X*(*s*) to exist], it is necessary that *x*(*t*)*e*−*st* → 0 as *t* → ∞ for the values of *s* in the ROC for *X*(*s*). Thus, - -$$ -\mathcal{L}\left[\frac{dx(t)}{dt}\right] = -x(0^-) + sX(s) -$$ - -Repeated application of this procedure yields Eq. (4.15). - -$$ -tx(t) \Longleftrightarrow -\frac{d}{ds}X(s) -$$ - -† The dual of the time-differentiation property is the frequency-differentiation property, which states that - -Find the Laplace transform of the signal *x*(*t*) in Fig. 4.6a by using Table 4.1 and the time-differentiation and time-shifting properties of the Laplace transform. - -Figures 4.6b and 4.6c show the first two derivatives of *x*(*t*). Recall that the derivative at a point of jump discontinuity is an impulse of strength equal to the amount of jump [see Eq. (1.12)]. Therefore, - -$$ -\frac{d^2x(t)}{dt^2} = \delta(t) - 3\delta(t-2) + 2\delta(t-3) -$$ - -#### 356 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -The Laplace transform of this equation yields - -$$ -\mathcal{L}\left(\frac{d^2x(t)}{dt^2}\right) = \mathcal{L}\left[\delta(t) - 3\delta(t-2) + 2\delta(t-3)\right] -$$ - -Using the time-differentiation property of Eq. (4.15), the time-shifting property of Eq. (4.12), and the facts that *x*(0−) = ˙*x*(0−) = 0, and δ(*t*) ⇐⇒ 1, we obtain - -$$ -s^2 X(s) - 0 - 0 = 1 - 3e^{-2s} + 2e^{-3s} -$$ - -Therefore, - -$$ -X(s) = \frac{1}{s^2} (1 - 3e^{-2s} + 2e^{-3s}) -$$ - -which confirms the earlier result in Drill 4.4. - -### **[4.2-4 The Time-Integration Property](#page-10-0)** - -The time-integration property states that if† - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -then - -$$ -\int_{0^{-}}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s} \quad \text{and} \quad \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s} + \frac{\int_{-\infty}^{0^{-}} x(\tau) d\tau}{s} \quad (4.16) -$$ - -**Proof.** To prove the first part of Eq. (4.16), we define - -$$ -g(t) = \int_{0^-}^{t} x(\tau) d\tau -$$ - -so that - -$$ -\frac{d}{dt}g(t) = x(t) \qquad \text{and} \qquad g(0^-) = 0 -$$ - -Now, if - -$$ -g(t) \Longleftrightarrow G(s) -$$ - -then - -$$ -X(s) = \mathcal{L}\left[\frac{d}{dt}g(t)\right] = sG(s) - g(0^{-}) = sG(s) -$$ - -$$ -\frac{x(t)}{t} \Longleftrightarrow \int_{s}^{\infty} X(z) dz -$$ - -† The dual of the time-integration property is the frequency-integration property, which states that - -### 4.2 Some Properties of the Laplace Transform 357 - -Therefore, - -$$ -G(s) = \frac{X(s)}{s} -$$ - -or - -$$ -\int_{0^-}^t x(\tau) d\tau \Longleftrightarrow \frac{X(s)}{s} -$$ - -To prove the second part of Eq. (4.16), observe that - -$$ -\int_{-\infty}^{t} x(\tau) d\tau = \int_{-\infty}^{0^-} x(\tau) d\tau + \int_{0^-}^{t} x(\tau) d\tau -$$ - -Note that the first term on the right-hand side is a constant for *t* ≥ 0. Taking the Laplace transform of the foregoing equation and using the first part of Eq. (4.16), we obtain - -$$ -\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{\int_{-\infty}^{0^-} x(\tau) d\tau}{s} + \frac{X(s)}{s} -$$ - -### **[4.2-5 The Scaling Property](#page-10-0)** - -The scaling property states that if - -$$ -x(t) \Longleftrightarrow X(s) -$$ - -then for *a* > 0 - -$$ -x(at) \Longleftrightarrow \frac{1}{a}X\left(\frac{s}{a}\right) -$$ - -The proof is given in Ch. 7. Note that *a* is restricted to positive values because if *x*(*t*) is causal, then *x*(*at*) is anticausal (is zero for *t* ≥ 0) for negative *a*, and anticausal signals are not permitted in the (unilateral) Laplace transform. - -Recall that *x*(*at*) is the signal *x*(*t*) time-compressed by the factor *a*, and *X*( *s a* ) is *X*(*s*) expanded along the *s* scale by the same factor *a* (see Sec. 1.2-2). The scaling property states that *time compression of a signal by a factor a causes expansion of its Laplace transform in the s scale by the same factor. Similarly, time expansion x*(*t*) *causes compression of X*(*s*) *in the s scale by the same factor.* - -### **[4.2-6 Time Convolution and Frequency Convolution](#page-10-0)** - -Another pair of properties states that if - -$$ -x_1(t) \Longleftrightarrow X_1(s) -$$ - and $x_2(t) \Longleftrightarrow X_2(s)$ - -then (*time-convolution property*) - -$$ -x_1(t) * x_2(t) \Longleftrightarrow X_1(s)X_2(s) \tag{4.17} -$$ - -### 358 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -and (*frequency-convolution property*) - -$$ -x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi j}[X_1(s) * X_2(s)] -$$ - -Observe the symmetry (or duality) between the two properties. Proofs of these properties are postponed to Ch. 7. - -Equation (2.39) indicates that *H*(*s*), the transfer function of an LTIC system, is the Laplace transform of the system's impulse response *h*(*t*); that is, - -$$ -h(t) \Longleftrightarrow H(s) -$$ - -If the system is causal, *h*(*t*) is causal, and, according to Eq. (2.39), *H*(*s*) is the unilateral Laplace transform of *h*(*t*). Similarly, if the system is noncausal, *h*(*t*) is noncausal, and *H*(*s*) is the bilateral transform of *h*(*t*). - -We can apply the time-convolution property to the LTIC input–output relationship *y*(*t*) = *x*(*t*) ∗ *h*(*t*) to obtain - -$$ -Y(s) = X(s)H(s) \tag{4.18} -$$ - -The response *y*(*t*) is the zero-state response of the LTIC system to the input *x*(*t*). From Eq. (4.18), it follows that - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{\mathcal{L}[zero-state response]}{\mathcal{L}[input]} -$$ -(4.19) - -This may be considered an alternate definition of the LTIC system transfer function *H*(*s*). It is the *ratio of the transform of zero-state response to the transform of the input.* - -### **EXAMPLE 4.10 Time-Convolution Property** - -Use the time-convolution property of the Laplace transform to determine *c*(*t*)=*eatu*(*t*)∗*ebtu*(*t*). - -From Eq. (4.17), it follows that - -$$ -C(s) = \frac{1}{(s-a)(s-b)} = \frac{1}{a-b} \left[ \frac{1}{s-a} - \frac{1}{s-b} \right] -$$ - -The inverse transform of this equation yields - -$$ -c(t) = \frac{1}{a-b}(e^{at} - e^{bt})u(t) -$$ - -### INITIAL AND FINAL VALUES - -In certain applications, it is desirable to know the values of *x*(*t*) as *t* → 0 and *t* → ∞ [initial and final values of *x*(*t*)] from the knowledge of its Laplace transform *X*(*s*). Initial and final value theorems provide such information. - -*The initial value theorem* states that if *x*(*t*) and its derivative *dx*/*dt* are both Laplace transformable, then - -$$ -x(0^+) = \lim_{s \to \infty} sX(s) \tag{4.20} -$$ - -provided the limit on the right-hand side of Eq. (4.20) exists. - -*The final value theorem* states that if both *x*(*t*) and *dx*/*dt* are Laplace transformable, then - -$$ -\lim_{t \to \infty} x(t) = \lim_{s \to 0} sX(s) \tag{4.21} -$$ - -provided *sX*(*s*) has no poles in the RHP or on the imaginary axis. To prove these theorems, we begin by setting *n* = 1 in Eq. (4.15). Using the definition of the Laplace transform, we see that - -$$ -sX(s) - x(0^{-}) = \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt -$$ - -= $\int_{0^{-}}^{0^{+}} \frac{dx(t)}{dt} e^{-st} dt + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$ -= $x(t) \Big|_{0^{-}}^{0^{+}} + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$ -= $x(0^{+}) - x(0^{-}) + \int_{0^{+}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt$ - -Therefore, - -$$ -sX(s) = x(0^+) + \int_{0^+}^{\infty} \frac{dx(t)}{dt} e^{-st} dt -$$ - -and - -$$ -\lim_{s \to \infty} sX(s) = x(0^+) + \lim_{s \to \infty} \int_{0^+}^{\infty} \frac{dx(t)}{dt} e^{-st} dt -$$ -$$ -= x(0^+) + \int_{0^+}^{\infty} \frac{dx(t)}{dt} \left( \lim_{s \to \infty} e^{-st} \right) dt -$$ -$$ -= x(0^+) -$$ - -**Comment.** The initial value theorem applies only if *X*(*s*) is strictly proper (*M* < *N*), because for *M* ≥ *N*, lim*s*→∞ *sX*(*s*) does not exist, and the theorem does not apply. In such a case, we can still find the answer by using long division to express *X*(*s*) as a polynomial in *s* plus a strictly proper fraction, where *M* < *N*. For example, by using long division, we can express - -$$ -\frac{s^3 + 3s^2 + s + 1}{s^2 + 2s + 1} = (s + 1) - \frac{2s}{s^2 + 2s + 1} -$$ - -The inverse transform of the polynomial in *s* is in terms of δ(*t*), and its derivatives, which are zero at *t* = 0+. In the foregoing case, the inverse transform of *s* + 1 is δ(˙ *t*) + δ(*t*). Hence, the desired - -### 360 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -*x*(0+) is the value of the remainder (strictly proper) fraction, for which the initial value theorem applies. In the present case, - -$$ -x(0^+) = \lim_{s \to \infty} \frac{-2s^2}{s^2 + 2s + 1} = -2 -$$ - -To prove the final value theorem, we let *n* = 1 and *s* → 0 in Eq. (4.15) to obtain - -$$ -\lim_{s \to 0} [sX(s) - x(0^{-})] = \lim_{s \to 0} \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} e^{-st} dt = \int_{0^{-}}^{\infty} \frac{dx(t)}{dt} dt -$$ -$$ -= x(t)|_{0^{-}}^{\infty} = \lim_{t \to \infty} x(t) - x(0^{-}) -$$ - -a deduction that leads to the desired result, Eq. (4.21). - -**Comment.** The final value theorem applies only if the poles of *X*(*s*) are in the LHP (including *s* = 0). If *X*(*s*) has a pole in the RHP, *x*(*t*) contains an exponentially growing term and *x*(∞) does not exist. If there is a pole on the imaginary axis, then *x*(*t*) contains an oscillating term and *x*(∞) does not exist. However, if there is a pole at the origin, then *x*(*t*) contains a constant term, and hence, *x*(∞) exists and is a constant. - -### **EXAMPLE 4.11 Initial and Final Values** - -Determine the initial and final values of *y*(*t*) if its Laplace transform *Y*(*s*) is given by - -$$ -Y(s) = \frac{10(2s+3)}{s(s^2+2s+5)} -$$ - -Equations (4.20) and (4.21) yield - -$$ -y(0^+) = \lim_{s \to \infty} sY(s) = \lim_{s \to \infty} \frac{10(2s+3)}{(s^2+2s+5)} = 0 -$$ - -$$ -y(\infty) = \lim_{s \to 0} sY(s) = \lim_{s \to 0} \frac{10(2s+3)}{(s^2+2s+5)} = 6 -$$ - -Table 4.2 summarizes the most important unilateral Laplace transform properties. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/055_4.3 SOLUTION OF DIFFERENTIAL AND INTEGRO-DIFFERENTIAL EQUATIONS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/055_4.3 SOLUTION OF DIFFERENTIAL AND INTEGRO-DIFFERENTIAL EQUATIONS.md deleted file mode 100644 index 2e8f5c95d455b3ff554ec84907f8558611d7cf1f..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/055_4.3 SOLUTION OF DIFFERENTIAL AND INTEGRO-DIFFERENTIAL EQUATIONS.md +++ /dev/null @@ -1,533 +0,0 @@ -## **4.3 SOLUTION OF DIFFERENTIAL AND [INTEGRO-DIFFERENTIAL](#page-10-0) EQUATIONS** - -The time-differentiation property of the Laplace transform has set the stage for solving linear differential (or integro-differential) equations with constant coefficients. Because *dky*/*dtk* ⇐⇒ *skY*(*s*), the Laplace transform of a differential equation is an algebraic equation that can be readily - -| Operation | x(t) | X(s) | -|------------------------------|------------------------|-------------------------------------------------------| -| Addition | x1(t)+x2(t) | X1(s)+X2(s) | -| Scalar multiplication | kx(t) | kX(s) | -| Time differentiation | dx(t)
dt | sX(s)−x(0−) | -| | d2x(t)
dt2 | s2X(s)−sx(0−)−
˙x(0−) | -| | d3x(t)
dt3 | s3X(s)−s2x(0−)−sx˙(0−)−
¨x(0−) | -| | dnx(t)
dtn | %n
snX(s)−
sn−kx(k−1)
(0−)
k=1 | -| Time integration | # t
x(τ )dτ
0− | 1
X(s)
s | -| | # t
x(τ )dτ | # 0−
1
1
X(s)+
x(t)dt
s
s | -| Time shifting | −∞
x(t −t0)u(t −t0) | −∞
X(s)e−st0
t0
≥ 0 | -| Frequency shifting | x(t)es0t | X(s−s0) | -| Frequency
differentiation | −tx(t) | dX(s)
ds | -| Frequency integration | x(t)
t | # ∞
X(z)dz
s | -| Scaling | x(at),a ≥ 0 | s
1
X
a
a | -| Time convolution | x1(t) ∗ x2(t) | X1(s)X2(s) | -| Frequency convolution | x1(t)x2(t) | 1
X1(s) ∗ X2(s)
2πj | -| Initial value | x(0+) | sX(s)
(n > m)
lim | -| Final value | x(∞) | s→∞
sX(s)
[poles of sX(s) in LHP]
lim
s→0 | - -**TABLE 4.2** Unilateral Laplace Transform Properties - -solved for *Y*(*s*). Next we take the inverse Laplace transform of *Y*(*s*) to find the desired solution *y*(*t*). The following examples demonstrate the Laplace transform procedure for solving linear differential equations with constant coefficients. - -## **EXAMPLE 4.12 Laplace Transform to Solve a Second-Order Linear Differential Equation** - -Solve the second-order linear differential equation - -$$ -(D2 + 5D + 6)y(t) = (D + 1)x(t) -$$ - -for the initial conditions *y*(0−) = 2 and *y*˙(0−) = 1 and the input *x*(*t*) = *e*−4*t u*(*t*). 362 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -The equation is - -$$ -\frac{d^2y(t)}{dt^2} + 5\frac{dy(t)}{dt} + 6y(t) = \frac{dx(t)}{dt} + x(t) -$$ -\n(4.22) - -Let - -$$ -y(t) \Longleftrightarrow Y(s) -$$ - -Then from Eq. (4.15), - -$$ -\frac{dy(t)}{dt} \Longleftrightarrow sY(s) - y(0^-) = sY(s) - 2 -$$ - -and - -$$ -\frac{d^2y(t)}{dt^2} \Longleftrightarrow s^2Y(s) - sy(0^-) - \dot{y}(0^-) = s^2Y(s) - 2s - 1 -$$ - -Moreover, for *x*(*t*) = *e*−4*t u*(*t*), - -$$ -X(s) = \frac{1}{s+4} \quad \text{and} \quad \frac{dx(t)}{dt} \Longleftrightarrow sX(s) - x(0^-) = \frac{s}{s+4} - 0 = \frac{s}{s+4} -$$ - -Taking the Laplace transform of Eq. (4.22), we obtain - -$$ -[s^{2}Y(s) - 2s - 1] + 5[sY(s) - 2] + 6Y(s) = \frac{s}{s+4} + \frac{1}{s+4} -$$ - -Collecting all the terms of *Y*(*s*) and the remaining terms separately on the left-hand side, we obtain - -$$ -(s2 + 5s + 6)Y(s) - (2s + 11) = \frac{s+1}{s+4} -$$ - (4.23) - -Therefore, - -$$ -(s2 + 5s + 6)Y(s) = (2s + 11) + \frac{s+1}{s+4} = \frac{2s2 + 20s + 45}{s+4} -$$ - -and - -$$ -Y(s) = \frac{2s^2 + 20s + 45}{(s^2 + 5s + 6)(s + 4)} = \frac{2s^2 + 20s + 45}{(s + 2)(s + 3)(s + 4)} -$$ - -Expanding the right-hand side into partial fractions yields - -$$ -Y(s) = \frac{13/2}{s+2} - \frac{3}{s+3} - \frac{3/2}{s+4} -$$ - -The inverse Laplace transform of this equation yields - -$$ -y(t) = \left(\frac{13}{2}e^{-2t} - 3e^{-3t} - \frac{3}{2}e^{-4t}\right)u(t) -$$ -\n(4.24) - -Example 4.12 demonstrates the ease with which the Laplace transform can solve linear differential equations with constant coefficients. The method is general and can solve a linear differential equation with constant coefficients of any order. - -### ZERO-INPUT AND ZERO-STATE COMPONENTS OF RESPONSE - -The Laplace transform method gives the total response, which includes zero-input and zero-state components. It is possible to separate the two components if we so desire. The initial condition terms in the response give rise to the zero-input response. For instance, in Ex. 4.12, the terms attributable to initial conditions *y*(0−) = 2 and *y*˙(0−) = 1 in Eq. (4.23) generate the zero-input response. These initial condition terms are −(2*s* + 11), as seen in Eq. (4.23). The terms on the right-hand side are exclusively due to the input. Equation (4.23) is reproduced below with the proper labeling of the terms - -$$ -(s2 + 5s + 6) Y(s) - (2s + 11) = \frac{s+1}{s+4} -$$ - -so that - -$$ -(s2 + 5s + 6) Y(s) = \underbrace{(2s + 11)}_{initial condition terms} + \underbrace{\underbrace{s + 1}_{s + 4}}_{input terms} -$$ - -Therefore, - -$$ -Y(s) = \underbrace{\frac{2s+11}{s^2+5s+6}}_{\text{ZIR}} + \underbrace{\frac{s+1}{(s+4)(s^2+5s+6)}}_{\text{ZSR}} \\ -= \left[ \frac{7}{s+2} - \frac{5}{s+3} \right] + \left[ \frac{-1/2}{s+2} + \frac{2}{s+3} - \frac{3/2}{s+4} \right] -$$ - -Taking the inverse transform of this equation yields - -$$ -y(t) = \underbrace{\left(7e^{-2t} - 5e^{-3t}\right)u(t)}_{\text{ZIR}} + \underbrace{\left(-\frac{1}{2}e^{-2t} + 2e^{-3t} - \frac{3}{2}e^{-4t}\right)u(t)}_{\text{ZSR}} -$$ - -### **[4.3-1 Comments on Initial Conditions at](#page-10-0) 0− and at 0+** - -The initial conditions in Ex. 4.12 are *y*(0−) = 2 and *y*˙(0−) = 1. If we let *t* = 0 in the total response in Eq. (4.24), we find *y*(0) = 2 and *y*˙(0) = 2, which is at odds with the given initial conditions. Why? Because the initial conditions are given at *t* = 0− (just before the input is applied), when only the zero-input response is present. The zero-state response is the result of the input *x*(*t*) applied at *t* = 0. Hence, this component does not exist at *t* = 0−. Consequently, the initial conditions at *t* = 0− are satisfied by the zero-input response, not by the total response. We can readily verify in this example that the zero-input response does indeed satisfy the given initial conditions at *t* = 0−. It is the total response that satisfies the initial conditions at *t* = 0+, which are generally different from the initial conditions at 0−. - -There also exists a *L*+ version of the Laplace transform, which uses the initial conditions at *t* = 0+ rather than at 0− (as in our present *L*− version). The *L*+ version, which was in vogue till the early 1960s, is identical to the *L*− version except the limits of Laplace integral [Eq. (4.7)] are from 0+ to ∞. Hence, by definition, the origin *t* = 0 is excluded from the domain. This version, still used in some math books, has some serious difficulties. For instance, the Laplace transform of δ(*t*) is zero because δ(*t*) = 0 for *t* ≥ 0+. Moreover, this approach is rather clumsy in the theoretical study - -### 364 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -of linear systems because the response obtained cannot be separated into zero-input and zero-state components. As we know, the zero-state component represents the system response as an explicit function of the input, and without knowing this component, it is not possible to assess the effect of the input on the system response in a general way. The *L*+ version can separate the response in terms of the natural and the forced components, which are not as interesting as the zero-input and the zero-state components. Note that we can always determine the natural and the forced components from the zero-input and the zero-state components [e.g., Eq. (2.44) from Eq. (2.43)], but the converse is not true. Because of these and some other problems, electrical engineers (wisely) started discarding the *L*+ version in the early 1960s. - -It is interesting to note the time-domain duals of these two Laplace versions. The classical method is the dual of the *L*+ method, and the convolution (zero-input/zero-state) method is the dual of the *L*− method. The first pair uses the initial conditions at 0+, and the second pair uses those at *t* = 0−. The first pair (the classical method and the *L*+ version) is awkward in the theoretical study of linear system analysis. It was no coincidence that the *L*− version was adopted immediately after the introduction to the electrical engineering community of state-space analysis (which uses zero-input/zero-state separation of the output). - -## **DR ILL 4.7 Laplace Transform to Solve a Second-Order Linear Differential Equation** - -Solve - -$$ -\frac{d^2y(t)}{dt^2} + 4\frac{dy(t)}{dt} + 3y(t) = 2\frac{dx(t)}{dt} + x(t) -$$ - -for the input *x*(*t*) = *u*(*t*). The initial conditions are *y*(0−) = 1 and *y*˙(0−) = 2. - -### **ANSWER** - -*y*(*t*) = 1 3 (1+9*e*−*t* −7*e*−3*t* )*u*(*t*) - -### **EXAMPLE 4.13 Laplace Transform to Solve an Electric Circuit** - -In the circuit of Fig. 4.7a, the switch is in the closed position for a long time before *t* = 0, when it is opened instantaneously. Find the inductor current *y*(*t*) for *t* ≥ 0. - -When the switch is in the closed position (for a long time), the inductor current is 2 amperes and the capacitor voltage is 10 volts. When the switch is opened, the circuit is equivalent to that depicted in Fig. 4.7b, with the initial inductor current *y*(0−) = 2 and the initial capacitor voltage *vC*(0−) = 10. The input voltage is 10 volts, starting at *t* = 0, and, therefore, can be represented by 10*u*(*t*). - -**Figure 4.7** Analysis of a network with a switching action. - -The loop equation of the circuit in Fig. 4.7b is - -$$ -\frac{dy(t)}{dt} + 2y(t) + 5 \int_{-\infty}^{t} y(\tau) d\tau = 10u(t) -$$ -\n(4.25) - -If - -*y*(*t*) ⇐⇒ *Y*(*s*) - -then - -$$ -\frac{dy(t)}{dt} \Longleftrightarrow sY(s) - y(0^-) = sY(s) - 2 -$$ - -and [see Eq. (4.16)] - -$$ -\int_{-\infty}^{t} y(\tau) d\tau \Longleftrightarrow \frac{Y(s)}{s} + \frac{\int_{-\infty}^{0} y(\tau) d\tau}{s} -$$ - -Because *y*(*t*) is the capacitor current, the integral \$ 0− −∞ *y*(τ )*d*τ is *qC*(0−), the capacitor charge at *t* = 0−, which is given by *C* times the capacitor voltage at *t* = 0−. Therefore, - -$$ -\int_{-\infty}^{0^-} y(\tau) d\tau = q_C(0^-) = C v_C(0^-) = \frac{1}{5}(10) = 2 -$$ - -and - -$$ -\int_{-\infty}^{t} y(\tau) d\tau \Longleftrightarrow \frac{Y(s)}{s} + \frac{2}{s} -$$ - -#### 366 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Using these results, the Laplace transform of Eq. (4.25) is - -$$ -sY(s) - 2 + 2Y(s) + \frac{5Y(s)}{s} + \frac{10}{s} = \frac{10}{s} -$$ - -or - -$$ -\[s+2+\frac{5}{s}\]Y(s) = 2 -$$ - -and - -$$ -Y(s) = \frac{2s}{s^2 + 2s + 5} -$$ - -To find the inverse Laplace transform of *Y*(*s*), we use pair 10c (Table 4.1) with values *A* = 2, *B* = 0, *a* = 1, and *c* = 5. This yields - -$$ -r = \sqrt{\frac{20}{4}} = \sqrt{5} -$$ -, $b = \sqrt{c - a^2} = 2$ and $\theta = \tan^{-1}(\frac{2}{4}) = 26.6^{\circ}$ - -Therefore, - -$$ -y(t) = \sqrt{5}e^{-t}\cos(2t + 26.6^{\circ})u(t) -$$ - -This response is shown in Fig. 4.7c. - -**Comment.** In our discussion so far, we have multiplied input signals by *u*(*t*), implying that the signals are zero prior to *t* = 0. This is needlessly restrictive. These signals can have any arbitrary value prior to *t* = 0. As long as the initial conditions at *t* = 0 are specified, we need only the knowledge of the input for *t* ≥ 0 to compute the response for *t* ≥ 0. Some authors use the notation 1(*t*) to denote a function that is equal to *u*(*t*) for *t* ≥ 0 and that has arbitrary value for negative *t*. We have abstained from this usage to avoid needless confusion caused by the introduction of a new function, which is very similar to *u*(*t*). - -### **[4.3-2 Zero-State Response](#page-10-0)** - -Consider an *N*th-order LTIC system specified by the equation - -$$ -Q(D)y(t) = P(D)x(t) -$$ - -or - -$$ -(DN + a1DN-1 + \dots + aN-1D + aN)y(t) = (b0DN + b1DN-1 + \dots + bN-1D + bN)x(t) -$$ - (4.26) - -We shall now find the general expression for the zero-state response of an LTIC system. Zero-state response *y*(*t*), by definition, is the system response to an input when the system is initially relaxed (in zero state). Therefore, *y*(*t*) satisfies Eq. (4.26) with zero initial conditions - -$$ -y(0^-) -$$ - = $\dot{y}(0^-)$ = $\ddot{y}(0^-)$ = $\dots$ = $y^{(N-1)}(0^-)$ = 0 - -Moreover, the input *x*(*t*) is causal so that - -$$ -x(0^-) = \dot{x}(0^-) = \ddot{x}(0^-) = \cdots = x^{(N-1)}(0^-) = 0 -$$ - -Let - -$$ -y(t) \Longleftrightarrow Y(s) -$$ - and $x(t) \Longleftrightarrow X(s)$ - -Because of zero initial conditions, - -$$ -D^{r}y(t) = \frac{d^{r}}{dt^{r}}y(t) \Longleftrightarrow s^{r}Y(s) -$$ -$$ -D^{k}x(t) = \frac{d^{k}}{dt^{k}}x(t) \Longleftrightarrow s^{k}X(s) -$$ - -Therefore, the Laplace transform of Eq. (4.26) yields - -$$ -(sN + a1sN-1 + \dots + aN-1s + aN)Y(s) = (b0sN + b1sN-1 + \dots + bN-1s + bN)X(s) -$$ - -or - -$$ -Y(s) = \frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N} X(s) = \frac{P(s)}{Q(s)} X(s) -$$ - -But we have shown in Eq. (4.18) that *Y*(*s*) = *H*(*s*)*X*(*s*). Consequently, - -$$ -H(s) = \frac{P(s)}{Q(s)}\tag{4.27} -$$ - -This is the transfer function of a linear differential system specified in Eq. (4.26). The same result has been derived earlier in Eq. (2.41) using an alternate (time-domain) approach. - -We have shown that *Y*(*s*), the Laplace transform of the zero-state response *y*(*t*), is the product of *X*(*s*) and *H*(*s*), where *X*(*s*) is the Laplace transform of the input *x*(*t*) and *H*(*s*) is the system transfer function [relating the particular output *y*(*t*) to the input *x*(*t*)]. - -### INTUITIVE INTERPRETATION OF THE LAPLACE TRANSFORM - -So far we have treated the Laplace transform as a machine that converts linear integro-differential equations into algebraic equations. There is no physical understanding of how this is accomplished or what it means. We now discuss a more intuitive interpretation and meaning of the Laplace transform. - -In Ch. 2, Eq. (2.38), we showed that LTI system response to an everlasting exponential *est* is *H*(*s*)*est*. If we could express every signal as a linear combination of everlasting exponentials of the form *est*, we could readily obtain the system response to any input. For example, if - -$$ -x(t) = \sum_{k=1}^{K} X(s_i) e^{s_i t} -$$ - -the response of an LTIC system to such input *x*(*t*) is given by - -$$ -y(t) = \sum_{k=1}^{K} X(s_i) H(s_i) e^{s_i t} -$$ - -### 368 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Unfortunately, the class of signals that can be expressed in this form is very small. However, we can express almost all signals of practical utility as a sum of everlasting exponentials over a continuum of frequencies. This is precisely what the Laplace transform in Eq. (4.2) does. - -$$ -x(t) = \frac{1}{2\pi j} \int_{c-j\infty}^{c+j\infty} X(s)e^{st} ds -$$ -\n(4.28) - -Invoking the linearity property of the Laplace transform, we can find the system response *y*(*t*) to input *x*(*t*) in Eq. (4.28) as† - -$$ -y(t) = \frac{1}{2\pi j} \int_{c'-j\infty}^{c'+j\infty} X(s)H(s)e^{st} ds = \mathcal{L}^{-1}X(s)H(s) -$$ -(4.29) - -Clearly, - -*Y*(*s*) = *X*(*s*)*H*(*s*) - -We can now represent the transformed version of the system, as depicted in Fig. 4.8a. The input *X*(*s*) is the Laplace transform of *x*(*t*), and the output *Y*(*s*) is the Laplace transform of (the zero-input response) *y*(*t*). The system is described by the transfer function *H*(*s*). The output *Y*(*s*) is the product *X*(*s*)*H*(*s*). - -Recall that *s* is the complex frequency of *est*. This explains why the Laplace transform method is also called the *frequency-domain* method. Note that *X*(*s*),*Y*(*s*), and *H*(*s*) are the frequency-domain representations of *x*(*t*), *y*(*t*), and *h*(*t*), respectively. We may view the boxes marked *L* and *L*−1 in Fig. 4.8a as the interfaces that convert the time-domain entities into the corresponding frequency-domain entities, and vice versa. All real-life signals begin in the time domain, and the final answers must also be in the time domain. First, we convert the time-domain input(s) into the frequency-domain counterparts. The problem itself is solved in the frequency domain, resulting in the answer *Y*(*s*), also in the frequency domain. Finally, we convert *Y*(*s*) to *y*(*t*). Solving the problem is relatively simpler in the frequency domain than in the time domain. Henceforth, we shall omit the explicit representation of the interface boxes *L* and *L*−1, representing signals and systems in the frequency domain, as shown in Fig. 4.8b. - -**Figure 4.8** Alternate interpretation of the Laplace transform. - -† Recall that *H*(*s*) has its own region of validity. Hence, the limits of integration for the integral in Eq. (4.28) are modified in Eq. (4.29) to accommodate the region of existence (validity) of *X*(*s*) as well as *H*(*s*). - -### **EXAMPLE 4.14 Laplace Transform to Find the Zero-State Response** - -Find the response *y*(*t*) of an LTIC system described by the equation - -$$ -\frac{d^2y(t)}{dt^2} + 5\frac{dy(t)}{dt} + 6y(t) = \frac{dx(t)}{dt} + x(t) -$$ - -if the input *x*(*t*) = 3*e*−5*t u*(*t*) and all the initial conditions are zero; that is, the system is in the zero state. - -The system equation is - -$$ -\underbrace{(D^2 + 5D + 6)}_{Q(D)} y(t) = \underbrace{(D + 1)}_{P(D)} x(t) -$$ - -Therefore, - -$$ -H(s) = \frac{P(s)}{Q(s)} = \frac{s+1}{s^2 + 5s + 6} -$$ - -Also, - -$$ -X(s) = \mathcal{L}[3e^{-5t}u(t)] = \frac{3}{s+5} -$$ - -and - -$$ -Y(s) = X(s)H(s) = \frac{3(s+1)}{(s+5)(s^2+5s+6)} -$$ - -= -$$ -\frac{3(s+1)}{(s+5)(s+2)(s+3)} = \frac{-2}{s+5} - \frac{1}{s+2} + \frac{3}{s+3} -$$ - -The inverse Laplace transform of this equation is - -$$ -y(t) = (-2e^{-5t} - e^{-2t} + 3e^{-3t})u(t) -$$ - -### **EXAMPLE 4.15 Laplace Transform to Find System Transfer Functions** - -Show that the transfer function of: - -- **(a)** an ideal delay of *T* seconds is *e*−*sT* -- **(b)** an ideal differentiator is *s* -- **(c)** an ideal integrator is 1/*s* - -**(a) Ideal Delay.** For an ideal delay of *T* seconds, the input *x*(*t*) and output *y*(*t*) are related by - -*y*(*t*) = *x*(*t* −*T*) and *Y*(*s*) = *X*(*s*)*e*−*sT* [see Eq. (4.12)] - -### 370 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -Therefore, - -$$ -H(s) = \frac{Y(s)}{X(s)} = e^{-sT} -$$ -\n(4.30) - -**(b) Ideal Differentiator.** For an ideal differentiator, the input *x*(*t*) and the output *y*(*t*) are related by - -$$ -y(t) = \frac{dx(t)}{dt} -$$ - -The Laplace transform of this equation yields - -$$ -Y(s) = sX(s) \qquad [x(0^-) = 0 \text{ for a causal signal}] -$$ - -and - -$$ -H(s) = \frac{Y(s)}{X(s)} = s -$$ -\n(4.31) - -**(c) Ideal Integrator.** For an ideal integrator with zero initial state, that is, *y*(0−) = 0, - -$$ -y(t) = \int_0^t x(\tau) d\tau \quad \text{and} \quad Y(s) = \frac{1}{s}X(s) -$$ - -Therefore, - -$$ -H(s) = \frac{1}{s} \tag{4.32} -$$ - -### **DR ILL 4.8 Differential Equation and Zero-State Response from a System Transfer Function** - -For an LTIC system with transfer function, - -$$ -H(s) = \frac{s+5}{s^2 + 4s + 3} -$$ - -- **(a)** Describe the differential equation relating the input *x*(*t*) and output *y*(*t*). -- **(b)** Find the system response *y*(*t*) to the input *x*(*t*) = *e*−2*t u*(*t*) if the system is initially in zero state. - -**ANSWERS** - -(a) -$$ -\frac{d^2y(t)}{dt^2} + 4\frac{dy(t)}{dt} + 3y(t) = \frac{dx(t)}{dt} + 5x(t) -$$ - -**(b)** -$$ -y(t) = (2e^{-t} - 3e^{-2t} + e^{-3t})u(t) -$$ - -### **[4.3-3 Stability](#page-10-0)** - -Equation (4.27) shows that the denominator of *H*(*s*) is *Q*(*s*), which is apparently identical to the characteristic polynomial *Q*(λ) defined in Ch. 2. Does this mean that the denominator of *H*(*s*) is the characteristic polynomial of the system? This may or may not be the case, since if *P*(*s*) and *Q*(*s*) in Eq. (4.27) have any common factors, they cancel out, and the effective denominator of *H*(*s*) is not necessarily equal to *Q*(*s*). Recall also that the system transfer function *H*(*s*), like *h*(*t*), is defined in terms of measurements at the external terminals. Consequently, *H*(*s*) and *h*(*t*) are both external descriptions of the system. In contrast, the characteristic polynomial *Q*(*s*) is an internal description. Clearly, we can determine only external stability, that is, BIBO stability, from *H*(*s*). If all the poles of *H*(*s*) are in LHP, all the terms in *h*(*t*) are decaying exponentials, and *h*(*t*) is absolutely integrable [see Eq. (2.45)].† Consequently, the system is BIBO-stable. Otherwise the system is BIBO-unstable. - -Beware of right half-plane poles! - -So far, we have assumed that *H*(*s*) is a proper function, that is, *M* ≤ *N*. We now show that if *H*(*s*) is improper, that is, if *M* > *N*, the system is BIBO-unstable. In such a case, using long division, we obtain *H*(*s*) = *R*(*s*) + *H* (*s*), where *R*(*s*) is an (*M* − *N*)th-order polynomial and *H* (*s*) is a proper transfer function. For example, - -$$ -H(s) = \frac{s^3 + 4s^2 + 4s + 5}{s^2 + 3s + 2} = s + \frac{s^2 + 2s + 5}{s^2 + 3s + 2} -$$ - -As shown in Eq. (4.31), the term *s* is the transfer function of an ideal differentiator. If we apply step function (bounded input) to this system, the output will contain an impulse (unbounded output). Clearly, the system is BIBO-unstable. Moreover, such a system greatly amplifies noise because differentiation enhances higher frequencies, which generally predominate in a noise signal. These - -† Values of *s* for which *H*(*s*) is ∞ are the *poles* of *H*(*s*). Thus, poles of *H*(*s*) are the values of *s* for which the denominator of *H*(*s*) is zero. - -### 372 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -are two good reasons to avoid improper systems (*M* > *N*). In our future discussion, we shall implicitly assume that the systems are proper, unless stated otherwise. - -If *P*(*s*) and *Q*(*s*) do not have common factors, then the denominator of *H*(*s*) is identical to *Q*(*s*), the characteristic polynomial of the system. In this case, we can determine internal stability by using the criterion described in Sec. 2.5. Thus, if *P*(*s*) and *Q*(*s*) have no common factors, the asymptotic stability criterion in Sec. 2.5 can be restated in terms of the poles of the transfer function of a system, as follows: - -- 1. An LTIC system is asymptotically stable if and only if all the poles of its transfer function *H*(*s*) are in the LHP. The poles may be simple or repeated. -- 2. An LTIC system is unstable if and only if either one or both of the following conditions exist: (i) at least one pole of *H*(*s*) is in the RHP; (ii) there are repeated poles of *H*(*s*) on the imaginary axis. -- 3. An LTIC system is marginally stable if and only if there are no poles of *H*(*s*) in the RHP and some unrepeated poles on the imaginary axis. - -The locations of zeros of *H*(*s*) have no role in determining the system stability. - -### **EXAMPLE 4.16 BIBO and Asymptotic Stability** - -Figure 4.9a shows a cascade connection of two LTIC systems *S*1 followed by *S*2. The transfer functions of these systems are *H*1(*s*) = 1/(*s* − 1) and *H*2(*s*) = (*s* − 1)/(*s* + 1), respectively. Determine the BIBO and asymptotic stability of the composite (cascade) system. - -**Figure 4.9** Distinction between BIBO and asymptotic stability. - -If the impulse responses of *S*1 and *S*2 are *h*1(*t*) and *h*2(*t*), respectively, then the impulse response of the cascade system is *h*(*t*) = *h*1(*t*)∗*h*2(*t*). Hence, *H*(*s*) = *H*1(*s*)*H*2(*s*). In the present case, - -$$ -H(s) = \left(\frac{1}{s-1}\right)\left(\frac{s-1}{s+1}\right) = \frac{1}{s+1} -$$ - -The pole of *S*1 at *s* = 1 cancels with the zero at *s* = 1 of *S*2. This results in a composite system having a single pole at *s* = −1. If the composite cascade system were to be enclosed inside a black box with only the input and the output terminals accessible, any measurement from these external terminals would show that the transfer function of the system is 1/(*s*+1), without any hint of the fact that the system is housing an unstable system (Fig. 4.9b). - -The impulse response of the cascade system is *h*(*t*) = *e*−*t u*(*t*), which is absolutely integrable. Consequently, the system is BIBO-stable. - -To determine the asymptotic stability, we note that *S*1 has one characteristic root at 1, and *S*2 also has one root at −1. Recall that the two systems are independent (one does not load the other), and the characteristic modes generated in each subsystem are independent of the other. Clearly, the mode *et* will not be eliminated by the presence of *S*2. Hence, the composite system has two characteristic roots, located at ±1, and the system is asymptotically unstable, though BIBO-stable. - -Interchanging the positions of *S*1 and *S*2 makes no difference in this conclusion. This example shows that BIBO stability can be misleading. If a system is asymptotically unstable, it will destroy itself (or, more likely, lead to saturation condition) because of unchecked growth of the response due to intended or unintended stray initial conditions. BIBO stability is not going to save the system. Control systems are often compensated to realize certain desirable characteristics. One should never try to stabilize an unstable system by canceling its RHP pole(s) with RHP zero(s). Such a misguided attempt will fail, not because of the practical impossibility of exact cancellation but for the more fundamental reason, as just explained. - -### **DR ILL 4.9 BIBO and Asymptotic Stability** - -Show that an ideal integrator is marginally stable but BIBO-unstable. - -### **[4.3-4 Inverse Systems](#page-10-0)** - -If *H*(*s*) is the transfer function of a system *S*, then *Si*, its inverse system has a transfer function *Hi*(*s*) given by - -$$ -H_i(s) = \frac{1}{H(s)} -$$ - -This follows from the fact the cascade of *S* with its inverse system *Si* is an identity system, with impulse response δ(*t*), implying *H*(*s*)*Hi*(*s*) = 1. For example, an ideal integrator and its inverse, an ideal differentiator, have transfer functions 1/*s* and *s*, respectively, leading to *H*(*s*)*Hi*(*s*) = 1. diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/056_4.4 ANALYSIS OF ELECTRICAL NETWORKS - THE TRANSFORMED NETWORK.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/056_4.4 ANALYSIS OF ELECTRICAL NETWORKS - THE TRANSFORMED NETWORK.md deleted file mode 100644 index 79d7b45eb92bbbaa3c0df9b23e99066301694357..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/056_4.4 ANALYSIS OF ELECTRICAL NETWORKS - THE TRANSFORMED NETWORK.md +++ /dev/null @@ -1,394 +0,0 @@ -## **4.4 ANALYSIS OF ELECTRICAL NETWORKS: THE TRANSFORMED NETWORK** - -Example 4.12 shows how electrical networks may be analyzed by writing the integro-differential equation(s) of the system and then solving these equations by the Laplace transform. We now show that it is also possible to analyze electrical networks directly without having to write the - -### 374 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -integro-differential equations. This procedure is considerably simpler because it permits us to treat an electrical network as if it were a resistive network. For this purpose, we need to represent a network in the "frequency domain" where all the voltages and currents are represented by their Laplace transforms. - -For the sake of simplicity, let us first discuss the case with zero initial conditions. If *v*(*t*) and *i*(*t*) are the voltage across and the current through an inductor of *L* henries, then - -$$ -v(t) = L \frac{di(t)}{dt} -$$ - -The Laplace transform of this equation (assuming zero initial current) is - -$$ -V(s) = LsI(s) -$$ - -Similarly, for a capacitor of *C* farads, the voltage-current relationship is *i*(*t*) = *C*(*dv*/*dt*) and its Laplace transform, assuming zero initial capacitor voltage, yields *I*(*s*) = *CsV*(*s*); that is, - -$$ -V(s) = \frac{1}{Cs}I(s) -$$ - -For a resistor of *R* ohms, the voltage-current relationship is *v*(*t*) = *Ri*(*t*), and its Laplace transform is - -$$ -V(s) = RI(s) -$$ - -Thus, in the "frequency domain," the voltage-current relationships of an inductor and a capacitor are algebraic; these elements behave like resistors of "resistance" *Ls* and 1/*Cs*, respectively. The generalized "resistance" of an element is called its *impedance* and is given by the ratio *V*(*s*)/*I*(*s*) for the element (under zero initial conditions). The impedances of a resistor of *R* ohms, an inductor of *L* henries, and a capacitance of *C* farads are *R*, *Ls*, and 1/*Cs*, respectively. - -Also, the interconnection constraints (Kirchhoff's laws) remain valid for voltages and currents in the frequency domain. To demonstrate this point, let *vj*(*t*) (*j* = 1, 2,..., *k*) be the voltages across *k* elements in a loop and let *ij*(*t*)(*j* = 1, 2,...,*m*) be the *j* currents entering a node. Then - -$$ -\sum_{j=1}^{k} v_j(t) = 0 \quad \text{and} \quad \sum_{j=1}^{m} i_j(t) = 0 -$$ - -Now if - -$$ -v_j(t) \Longleftrightarrow V_j(s) -$$ - and $i_j(t) \Longleftrightarrow I_j(s)$ - -then - -$$ -\sum_{j=1}^{k} V_j(s) = 0 \quad \text{and} \quad \sum_{j=1}^{m} I_j(s) = 0 -$$ - -This result shows that if we represent all the voltages and currents in an electrical network by their Laplace transforms, we can treat the network as if it consisted of the "resistances" *R*, *Ls*, and 1/*Cs* corresponding to a resistor *R*, an inductor *L*, and a capacitor *C*, respectively. The system equations (loop or node) are now algebraic. Moreover, the simplification techniques that have been developed for resistive circuits—equivalent series and parallel impedances, voltage and current divider rules, Thévenin and Norton theorems—can be applied to general electrical networks. The following examples demonstrate these concepts. - -**Figure 4.10 (a)** A circuit and **(b)** its transformed version. - -In the first step, we represent the circuit in the frequency domain, as illustrated in Fig. 4.10b. All the voltages and currents are represented by their Laplace transforms. The voltage 10*u*(*t*) is represented by 10/*s* and the (unknown) current *i*(*t*) is represented by its Laplace transform *I*(*s*). All the circuit elements are represented by their respective impedances. The inductor of 1 henry is represented by *s*, the capacitor of 1/2 farad is represented by 2/*s*, and the resistor of 3 ohms is represented by 3. We now consider the frequency-domain representation of voltages and currents. The voltage across any element is *I*(*s*) times its impedance. Therefore, the total voltage drop in the loop is *I*(*s*) times the total loop impedance, and it must be equal to *V*(*s*), (transform of) the input voltage. The total impedance in the loop is - -$$ -Z(s) = s + 3 + \frac{2}{s} = \frac{s^2 + 3s + 2}{s} -$$ - -The input"voltage" is *V*(*s*) = 10/*s*. Therefore, the "loop current" *I*(*s*) is - -$$ -I(s) = \frac{V(s)}{Z(s)} = \frac{10/s}{(s^2 + 3s + 2)/s} = \frac{10}{s^2 + 3s + 2} = \frac{10}{(s+1)(s+2)} = \frac{10}{s+1} - \frac{10}{s+2} -$$ - -The inverse transform of this equation yields the desired result: - -$$ -i(t) = 10(e^{-t} - e^{-2t})u(t) -$$ - -### 376 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -### INITIAL CONDITION GENERATORS - -The discussion in which we assumed zero initial conditions can be readily extended to the case of nonzero initial conditions because the initial condition in a capacitor or an inductor can be represented by an equivalent source. We now show that a capacitor *C* with an initial voltage *v*(0−) (Fig. 4.11a) can be represented in the frequency domain by an uncharged capacitor of impedance 1/*Cs* in series with a voltage source of value *v*(0−)/*s* (Fig. 4.11b) or as the same uncharged capacitor in parallel with a current source of value *Cv*(0−) (Fig. 4.11c). Similarly, an inductor *L* with an initial current *i*(0−) (Fig. 4.11d) can be represented in the frequency domain by an inductor of impedance *Ls* in series with a voltage source of value *Li*(0−) (Fig. 4.11e) or by the same inductor in parallel with a current source of value *i*(0−)/*s* (Fig. 4.11f). - -**Figure 4.11** Initial condition generators for a capacitor and an inductor. - -To prove this point, consider the terminal relationship of the capacitor in Fig. 4.11a: - -$$ -i(t) = C \frac{dv(t)}{dt} -$$ - -The Laplace transform of this equation yields - -$$ -I(s) = C[sV(s) - v(0^-)] -$$ - -### 4.4 Analysis of Electrical Networks: The Transformed Network 377 - -This equation can be rearranged as - -$$ -V(s) = \frac{1}{Cs}I(s) + \frac{v(0^{-})}{s} -$$ -\n(4.33) - -Observe that *V*(*s*) is the voltage (in the frequency domain) across the charged capacitor and *I*(*s*)/*Cs* is the voltage across the same capacitor without any charge. Therefore, the charged capacitor can be represented by the uncharged capacitor in series with a voltage source of value *v*(0−)/*s*, as depicted in Fig. 4.11b. Equation (4.33) can also be rearranged as - -$$ -V(s) = \frac{1}{Cs}[I(s) + Cv(0^{-})] -$$ - -This equation shows that the charged capacitor voltage *V*(*s*) is equal to the uncharged capacitor voltage caused by a current *I*(*s*) + *Cv*(0−). This result is reflected precisely in Fig. 4.11c, where the current through the uncharged capacitor is *I*(*s*)+*Cv*(0−). † - -For the inductor in Fig. 4.11d, the terminal equation is - -$$ -v(t) = L \frac{di(t)}{dt} -$$ - -and - -$$ -V(s) = L[sI(s) - i(0^{-})] = LsI(s) - Li(0^{-}) -$$ -\n(4.34) - -This expression is consistent with Fig. 4.11e. We can rearrange Eq. (4.34) as - -$$ -V(s) = Ls \left[ I(s) - \frac{i(0^{-})}{s} \right] -$$ - -This expression is consistent with Fig. 4.11f. - -Let us rework Ex. 4.13 using these concepts. Figure 4.12a shows the circuit in Fig. 4.7b with the initial conditions *y*(0−) = 2 and *vC*(0−) = 10. Figure 4.12b shows the frequency-domain representation (transformed circuit) of the circuit in Fig. 4.12a. The resistor is represented by its impedance 2; the inductor with initial current of 2 amperes is represented according to the arrangement in Fig. 4.11e with a series voltage source *Ly*(0−) = 2. The capacitor with initial voltage of 10 volts is represented according to the arrangement in Fig. 4.11b with a series voltage source *v*(0−)/*s* = 10/*s*. Note that the impedance of the inductor is *s* and that of the capacitor is 5/*s*. The input of 10*u*(*t*) is represented by its Laplace transform 10/*s*. - -The total voltage in the loop is (10/*s*) + 2 − (10/*s*) = 2, and the loop impedance is (*s*+2+(5/*s*)). Therefore, - -$$ -Y(s) = \frac{2}{s + 2 + 5/s} = \frac{2s}{s^2 + 2s + 5} -$$ - -which confirms our earlier result in Ex. 4.13. - -† In the time domain, a charged capacitor *C* with initial voltage *v*(0−) can be represented as the same capacitor uncharged in series with a voltage source *v*(0−)*u*(*t*), or in parallel with a current source *Cv*(0−)δ(*t*). Similarly, an inductor *L* with initial current *i*(0−) can be represented by the same inductor with zero initial current in series with a voltage source *Li*(0−)δ(*t*) or with a parallel current source *i*(0−)*u*(*t*). - -**Figure 4.12** A circuit and its transformed version with initial-condition generators. - -**Figure 4.13** Using initial condition generators and Thévenin equivalent representation. - -Inspection of this circuit shows that when the switch is closed and the steady-state conditions are reached, the capacitor voltage *vC* = 16 volts, and the inductor current *y*2 = 4 amperes. Therefore, when the switch is opened (at *t* = 0), the initial conditions are *vC*(0−) = 16 and *y*2(0−) = 4. Figure 4.13b shows the transformed version of the circuit in Fig. 4.13a. We have used equivalent sources to account for the initial conditions. The initial capacitor voltage of 16 volts is represented by a series voltage of 16/*s* and the initial inductor current of 4 amperes is represented by a source of value *Ly*2(0−) = 2. - -From Fig. 4.13b, the loop equations can be written directly in the frequency domain as - -$$ -\frac{Y_1(s)}{s} + \frac{1}{5}[Y_1(s) - Y_2(s)] = \frac{4}{s} -$$ -$$ --\frac{1}{5}Y_1(s) + \frac{6}{5}Y_2(s) + \frac{s}{2}Y_2(s) = 2 -$$ -$$ -\begin{bmatrix} \frac{1}{s} + \frac{1}{5} & -\frac{1}{5} \\ -\frac{1}{5} & \frac{6}{5} + \frac{s}{2} \end{bmatrix} \begin{bmatrix} Y_1(s) \\ Y_2(s) \end{bmatrix} = \begin{bmatrix} \frac{4}{5} \\ \frac{5}{2} \end{bmatrix} -$$ - -Application of Cramer's rule to this equation yields - -$$ -Y_1(s) = \frac{24(s+2)}{s^2 + 7s + 12} = \frac{24(s+2)}{(s+3)(s+4)} = \frac{-24}{s+3} + \frac{48}{s+4} -$$ - -and - -$$ -y_1(t) = (-24e^{-3t} + 48e^{-4t})u(t) -$$ - -Similarly, we obtain - -$$ -Y_2(s) = \frac{4(s+7)}{s^2 + 7s + 12} = \frac{16}{s+3} - \frac{12}{s+4} -$$ - -and - -$$ -y_2(t) = (16e^{-3t} - 12e^{-4t})u(t) -$$ - -We also could have used Thévenin's theorem to compute *Y*1(*s*) and *Y*2(*s*) by replacing the circuit to the right of the capacitor (right of terminals *ab*) with its Thévenin equivalent, as shown in Fig. 4.13c. Figure 4.13b shows that the Thévenin impedance *Z*(*s*) and the Thévenin source *V*(*s*) are - -$$ -Z(s) = \frac{\frac{1}{5} \left(\frac{s}{2} + 1\right)}{\frac{1}{5} + \frac{s}{2} + 1} = \frac{s+2}{5s+12} -$$ -$$ -V(s) = \frac{-\frac{1}{5}}{\frac{1}{5} + \frac{s}{2} + 1} = \frac{-4}{5s+12} -$$ - -According to Fig. 4.13c, the current *Y*1(*s*) is given by - -$$ -Y_1(s) = \frac{\frac{4}{s} - V(s)}{\frac{1}{s} + Z(s)} = \frac{24(s+2)}{s^2 + 7s + 12} -$$ - -which confirms the earlier result. We may determine *Y*2(*s*) in a similar manner. - -### **EXAMPLE 4.19 Transformed Analysis of a Coupled Inductive Network** - -The switch in the circuit in Fig. 4.14a is at position a for a long time before *t* = 0, when it is moved instantaneously to position b. Determine the current *y*1(*t*) and the output voltage *v*0(*t*) for *t* ≥ 0. - -Just before switching, the values of the loop currents are 2 and 1, respectively, that is, *y*1(0−) = 2 and *y*2(0−) = 1. - -The equivalent circuits for two types of inductive coupling are illustrated in Figs. 4.14b and 4.14c. For our situation, the circuit in Fig. 4.14c applies. Figure 4.14d shows the transformed version of the circuit in Fig. 4.14a after switching. Note that the inductors *L*1 + *M*, *L*2 + *M*, and −*M* are 3, 4, and −1 henries with impedances 3*s*, 4*s*, and −*s* respectively. The initial condition voltages in the three branches are (*L*1 + *M*)*y*1(0−) = 6, (*L*2 + *M*)*y*2(0−) = 4, and −*M*[*y*1(0−)−*y*2(0−)]=−1, respectively. The two loop equations of the circuit are - -$$ -(2s+3)Y1(s) + (s-1)Y2(s) = \frac{10}{s} + 5 -$$ -$$ -(s-1)Y1(s) + (3s+2)Y2(s) = 5 -$$ - -or - -$$ -\begin{bmatrix} 2s+3 & s-1 \ s-1 & 3s+2 \end{bmatrix} \begin{bmatrix} Y_1(s) \\ Y_2(s) \end{bmatrix} \begin{bmatrix} \frac{5s+10}{s} \\ 5 \end{bmatrix} -$$ - -Solving for *Y*1(*s*), we obtain - -$$ -Y_1(s) = \frac{2s^2 + 9s + 4}{s(s^2 + 3s + 1)} = \frac{4}{s} - \frac{1}{s + 0.382} - \frac{1}{s + 2.618} -$$ - -Therefore, - -$$ -y_1(t) = (4 - e^{-0.382t} - e^{-2.618t})u(t) -$$ - -Similarly, - -$$ -Y_2(s) = \frac{s^2 + 2s + 2}{s(s^2 + 3s + 1)} = \frac{2}{s} - \frac{1.618}{s + 0.382} + \frac{0.618}{s + 2.618} -$$ - -and - -$$ -y_2(t) = (2 - 1.618e^{-0.382t} + 0.618e^{-2.618t})u(t) -$$ - -The output voltage is therefore - -$$ -v_0(t) = y_2(t) = (2 - 1.618e^{-0.382t} + 0.618e^{-2.618t})u(t) -$$ - -## **DR ILL 4.10 Transformed Analysis of an** *RLC* **Circuit with a Switch** - -For the *RLC* circuit in Fig. 4.15, the input is switched on at *t* = 0. The initial conditions are *y*(0−) = 2 amperes and *vC*(0−) = 50 volts. Find the loop current *y*(*t*) and the capacitor voltage *vC*(*t*) for *t* ≥ 0. - -### **ANSWERS** - - - -### **[4.4-1 Analysis of Active Circuits](#page-10-0)** - -Although we have considered examples of only passive networks so far, the circuit analysis procedure using the Laplace transform is also applicable to active circuits. All that is needed is to replace the active elements with their mathematical models (or equivalent circuits) and proceed as before. - -The operational amplifier (depicted by the triangular symbol in Fig. 4.16a) is a well-known element in modern electronic circuits. The terminals with the positive and the negative signs correspond to noninverting and inverting terminals, respectively. This means that the polarity of the output voltage *v*2 is the same as that of the input voltage at the terminal marked by the positive sign (noninverting). The opposite is true for the inverting terminal, marked by the negative sign. - -Figure 4.16b shows the model (equivalent circuit) of the operational amplifier (op amp) in Fig. 4.16a. A typical op amp has a very large gain. The output voltage *v*2 = −*Av*1, where *A* is typically 105 to 106. The input impedance is very high, of the order of 1012 , and the output impedance is very low (50–100 ). For most applications, we are justified in assuming the gain *A* and the input impedance to be infinite and the output impedance to be zero. For this reason we see an ideal voltage source at the output. - -Consider now the operational amplifier with resistors *Ra* and *Rb* connected, as shown in Fig. 4.16c. This configuration is known as the *noninverting amplifier*. Observe that the input polarities in this configuration are inverted in comparison to those in Fig. 4.16a. We now show that the output voltage *v*2 and the input voltage *v*1 in this case are related by - -$$ -v_2 = Kv_1, \qquad \text{where } K = 1 + \frac{R_b}{R_a} -$$ - -First, we recognize that because the input impedance and the gain of the operational amplifier approach infinity, the input current *ix* and the input voltage *vx* in Fig. 4.16c are infinitesimal and may be taken as zero. The dependent source in this case is *Avx* instead of −*Avx* because of the input polarity inversion. The dependent source *Avx* (see Fig. 4.16b) at the output will generate current *io*, as illustrated in Fig. 4.16c. Now - -$$ -v_2 = (R_b + R_a)i_o -$$ - -**Figure 4.16** Operational amplifier and its equivalent circuit. - -and also - -$$ -v_1 = v_x + R_a i_o = R_a i_o -$$ - -Therefore, - -$$ -\frac{v_2}{v_1} = \frac{R_b + R_a}{R_a} = 1 + \frac{R_b}{R_a} = K -$$ - -or - -$$ -v_2(t) = Kv_1(t) -$$ - -The equivalent circuit of the noninverting amplifier is depicted in Fig. 4.16d. - -### **EXAMPLE 4.20 Transform Analysis of a Sallen–Key Circuit** - -The circuit in Fig. 4.17a is called the *Sallen–Key* circuit, which is frequently used in filter design. Find the transfer function *H*(*s*) relating the output voltage *vo*(*t*) to the input voltage *vi*(*t*). - -**Figure 4.17 (a)** Sallen–Key circuit and **(b)** its equivalent. - -We are required to find - -$$ -H(s) = \frac{V_o(s)}{V_i(s)} -$$ - -assuming all initial conditions to be zero. - -Figure 4.17b shows the transformed version of the circuit in Fig. 4.17a. The noninverting amplifier is replaced by its equivalent circuit. All the voltages are replaced by their Laplace transforms, and all the circuit elements are shown by their impedances. All the initial conditions are assumed to be zero, as required for determining *H*(*s*). - -We shall use node analysis to derive the result. There are two unknown node voltages, *Va*(*s*) and *Vb*(*s*), requiring two node equations. - -At node *a*, *IR*1 (*s*), the current in *R*1 (leaving the node *a*), is [*Va*(*s*) − *Vi*(*s*)]/*R*1. Similarly, *IR*2 (*s*), the current in *R*2 (leaving the node *a*), is [*Va*(*s*) − *Vb*(*s*)]/*R*2, and *IC*1 (*s*), the current in capacitor *C*1 (leaving the node *a*), is [*Va*(*s*)−*Vo*(*s*)]*C*1*s* = [*Va*(*s*)−*KVb*(*s*)]*C*1*s*. - -The sum of all the three currents is zero. Therefore, - -$$ -\frac{V_a(s) - V_i(s)}{R_1} + \frac{V_a(s) - V_b(s)}{R_2} + [V_a(s) - KV_b(s)]C_1s = 0 -$$ - -or - -$$ -\left(\frac{1}{R_1} + \frac{1}{R_2} + C_1 s\right) V_a(s) - \left(\frac{1}{R_2} + KC_1 s\right) V_b(s) = \frac{1}{R_1} V_i(s) -$$ - -Similarly, the node equation at node *b* yields - -$$ -\frac{V_b(s) - V_a(s)}{R_2} + C_2 s V_b(s) = 0 -$$ - -or - -$$ --\frac{1}{R_2}V_a(s) + \left(\frac{1}{R_2} + C_2s\right)V_b(s) = 0 -$$ - -The two node equations in two unknown node voltages *Va*(*s*) and *Vb*(*s*) can be expressed in matrix form as - -$$ -\begin{bmatrix} G_1 + G_2 + C_1 s & -(G_2 + KC_1 s) \ -G_2 & (G_2 + C_2 s) \end{bmatrix} \begin{bmatrix} V_a(s) \ V_b(s) \end{bmatrix} = \begin{bmatrix} G_1 V_i(s) \ 0 \end{bmatrix} -$$ - -where - -$$ -G_1 = \frac{1}{R_1} \quad \text{and} \quad G_2 = \frac{1}{R_2} -$$ - -Application of Cramer's rule yields - -$$ -\frac{V_b(s)}{V_i(s)} = \frac{G_1 G_2}{C_1 C_2 s^2 + [G_1 C_2 + G_2 C_2 + G_2 C_1 (1 - K)]s + G_1 G_2} -$$ - -= -$$ -\frac{\omega_0^2}{s^2 + 2\alpha s + \omega_0^2} -$$ - -where - -$$ -K = 1 + \frac{R_b}{R_a} \quad \text{and} \quad \omega_0^2 = \frac{G_1 G_2}{C_1 C_2} = \frac{1}{R_1 R_2 C_1 C_2} -$$ -$$ -2\alpha = \frac{G_1 C_2 + G_2 C_2 + G_2 C_1 (1 - K)}{C_1 C_2} = \frac{1}{R_1 C_1} + \frac{1}{R_2 C_1} + \frac{1}{R_2 C_2} (1 - K) -$$ - -Now - -$$ -V_o(s) = KV_b(s) -$$ - -Therefore, - -$$ -H(s) = \frac{V_o(s)}{V_i(s)} = K \frac{V_b(s)}{V_i(s)} = \frac{K\omega_0^2}{s^2 + 2\alpha s + \omega_0} -$$ - -2 diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/057_4.5 BLOCK DIAGRAMS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/057_4.5 BLOCK DIAGRAMS.md deleted file mode 100644 index 1091e1b0784ba70d8afd4dc7f875a582d25dc9a0..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/057_4.5 BLOCK DIAGRAMS.md +++ /dev/null @@ -1,76 +0,0 @@ -## **4.5 BLOCK [DIAGRAMS](#page-10-0)** - -Large systems may consist of an enormous number of components or elements. As anyone who has seen the circuit diagram of a radio or a television receiver can appreciate, analyzing such systems all at once could be next to impossible. In such cases, it is convenient to represent a system by suitably interconnected subsystems, each of which can be readily analyzed. Each subsystem can be characterized in terms of its input–output relationships. A linear system can be characterized by its transfer function *H*(*s*). Figure 4.18a shows a block diagram of a system with a transfer function *H*(*s*) and its input and output *X*(*s*) and *Y*(*s*), respectively. - -Subsystems may be interconnected by using cascade, parallel, and feedback interconnections (Figs. 4.18b, 4.18c, 4.18d), the three elementary types. When transfer functions appear in cascade, as depicted in Fig. 4.18b, then, as shown earlier, the transfer function of the overall system is the product of the two transfer functions. This result can also be proved by observing that in Fig. 4.18b - -$$ -\frac{Y(s)}{X(s)} = \frac{W(s)}{X(s)} \frac{Y(s)}{W(s)} = H_1(s)H_2(s) -$$ - -**Figure 4.18** Elementary connections of blocks and their equivalents. - -We can extend this result to any number of transfer functions in cascade. It follows from this discussion that the subsystems in cascade can be interchanged without affecting the overall transfer function. This commutation property of LTI systems follows directly from the commutative (and associative) property of convolution. We have already proved this property in Sec. 2.4-3. Every possible ordering of the subsystems yields the same overall transfer function. However, there may be practical consequences (such as sensitivity to parameter variation) affecting the behavior of different ordering. - -Similarly, when two transfer functions, *H*1(*s*) and *H*2(*s*), appear in parallel, as illustrated in Fig. 4.18c, the overall transfer function is given by *H*1(*s*) + *H*2(*s*), the sum of the two transfer functions. The proof is trivial. This result can be extended to any number of systems in parallel. - -When the output is fed back to the input, as shown in Fig. 4.18d, the overall transfer function *Y*(*s*)/*X*(*s*) can be computed as follows. The inputs to the adder are *X*(*s*) and −*H*(*s*)*Y*(*s*). Therefore, *E*(*s*), the output of the adder, is - -$$ -E(s) = X(s) - H(s)Y(s) -$$ - -But - -$$ -Y(s) = G(s)E(s) -$$ - -= $G(s)[X(s) - H(s)Y(s)]$ - -Therefore, - -$$ -Y(s)[1 + G(s)H(s)] = G(s)X(s) -$$ - -so that - -$$ -\frac{Y(s)}{X(s)} = \frac{G(s)}{1 + G(s)H(s)} -$$ -(4.35) - -Therefore, the feedback loop can be replaced by a single block with the transfer function shown in Eq. (4.35) (see Fig. 4.18d). - -In deriving these equations, we implicitly assume that when the output of one subsystem is connected to the input of another subsystem, the latter does not load the former. For example, the transfer function *H*1(*s*) in Fig. 4.18b is computed by assuming that the second subsystem *H*2(*s*) was not connected. This is the same as assuming that *H*2(*s*) does not load *H*1(*s*). In other words, the input–output relationship of *H*1(*s*) will remain unchanged regardless of whether *H*2(*s*) is connected. Many modern circuits use op amps with high input impedances, so this assumption is justified. When such an assumption is not valid, *H*1(*s*) must be computed under operating conditions [i.e., with *H*2(*s*) connected]. - -### **EXAMPLE 4.21 Transfer Functions of Feedback Systems Using MATLAB** - -Consider the feedback system of Fig. 4.18d with *G*(*s*) = *K*/(*s*(*s* + 8)) and *H*(*s*) = 1. Use MATLAB to determine the transfer function for each of the following cases: **(a)** *K* = 7, **(b)** *K* = 16, and **(c)** *K* = 80. - -We solve these cases using the control system toolbox function feedback. - -``` -(a) ->> H = tf(1,1); K = 7; G = tf([0 0 K],[1 8 0]); TFa = feedback(G,H) - Ha = - 7 - ------------- - s^2 + 8 s + 7 -Thus, Ha(s) = 7/(s2 +8s+7). -(b) ->> H = tf(1,1); K = 16; G = tf([0 0 K],[1 8 0]); TFb = feedback(G,H) - Hb = - 16 - -------------- - s^2 + 8 s + 16 -Thus, Hb(s) = 16/(s2 +8s+16). -(c) ->> H = tf(1,1); K = 80; G = tf([0 0 K],[1 8 0]); TFc = feedback(G,H) - Hc = - 80 - -------------- - s^2 + 8 s + 80 -Thus, Hc(s) = 80/(s2 +8s+80). -``` diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/058_4.6 SYSTEM REALIZATION.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/058_4.6 SYSTEM REALIZATION.md deleted file mode 100644 index dd6261a58d24de16d32382f22939f63ddc844df7..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/058_4.6 SYSTEM REALIZATION.md +++ /dev/null @@ -1,448 +0,0 @@ -## **4.6 SYSTEM [REALIZATION](#page-10-0)** - -We now develop a systematic method for realization (or implementation) of an arbitrary *N*th-order transfer function. The most general transfer function with *M* = *N* is given by - -$$ -H(s) = \frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N} -$$ -(4.36) - -Since realization is basically a synthesis problem, there is no unique way of realizing a system. A given transfer function can be realized in many different ways. A transfer function *H*(*s*) can be realized by using integrators or differentiators along with adders and multipliers. We avoid use of differentiators for practical reasons discussed in Secs. 2.1 and 4.3-3. Hence, in our implementation, we shall use integrators along with scalar multipliers and adders. We are already familiar with representation of all these elements except the integrator. The integrator can be represented by a box with integral sign (time-domain representation, Fig. 4.19a) or by a box with transfer function 1/*s* (frequency-domain representation, Fig. 4.19b). - -**Figure 4.19 (a)** Time-domain and **(b)** frequency-domain representations of an integrator. - -### **[4.6-1 Direct Form I Realization](#page-10-0)** - -Rather than realize the general *N*th-order system described by Eq. (4.36), we begin with a specific case of the following third-order system and then extend the results to the *N*th-order case: - -$$ -H(s) = \frac{b_0 s^3 + b_1 s^2 + b_2 s + b_3}{s^3 + a_1 s^2 + a_2 s + a_3} = \frac{b_0 + \frac{b_1}{s} + \frac{b_2}{s^2} + \frac{b_3}{s^3}}{1 + \frac{a_1}{s} + \frac{a_2}{s^2} + \frac{a_3}{s^3}} -$$ - -We can express *H*(*s*) as - -$$ -H(s) = \underbrace{\left(b_0 + \frac{b_1}{s} + \frac{b_2}{s^2} + \frac{b_3}{s^3}\right)}_{H_1(s)} \underbrace{\left(\frac{1}{1 + \frac{a_1}{s} + \frac{a_2}{s^2} + \frac{a_3}{s^3}\right)}_{H_2(s)}} -$$ - -We can realize *H*(*s*) as a cascade of transfer function *H*1(*s*) followed by *H*2(*s*), as depicted in Fig. 4.20a, where the output of *H*1(*s*) is denoted by *W*(*s*). Because of the commutative property of LTI system transfer functions in cascade, we can also realize *H*(*s*) as a cascade of *H*2(*s*) followed by *H*1(*s*), as illustrated in Fig. 4.20b, where the (intermediate) output of *H*2(*s*) is denoted by *V*(*s*). - -**Figure 4.20** Realization of a transfer function in two steps. - -The output of *H*1(*s*) in Fig. 4.20a is given by *W*(*s*) = *H*1(*s*)*X*(*s*). Hence, - -$$ -W(s) = \left(b_0 + \frac{b_1}{s} + \frac{b_2}{s^2} + \frac{b_3}{s^3}\right)X(s) -$$ -\n(4.37) - -Also, the output *Y*(*s*) and the input *W*(*s*) of *H*2(*s*) in Fig. 4.20a are related by *Y*(*s*) = *H*2(*s*)*W*(*s*). Hence, - -$$ -W(s) = \left(1 + \frac{a_1}{s} + \frac{a_2}{s^2} + \frac{a_3}{s^3}\right)Y(s) -$$ -\n(4.38) - -**Figure 4.21** Direct form I realization of an LTIC system: **(a)** third-order and **(b)** *N*th-order. - -We shall first realize *H*1(*s*). Equation (4.37) shows that the output *W*(*s*) can be synthesized by adding the input *b*0*X*(*s*) to *b*1(*X*(*s*)/*s*),*b*2(*X*(*s*)/*s*2), and *b*3(*X*(*s*)/*s*3). Because the transfer function of an integrator is 1/*s*, the signals *X*(*s*)/*s*,*X*(*s*)/*s*2, and *X*(*s*)/*s*3 can be obtained by successive integration of the input *x*(*t*). The left-half section of Fig. 4.21a shows how *W*(*s*) can be synthesized from *X*(*s*), according to Eq. (4.37). Hence, this section represents a realization of *H*1(*s*). - -To complete the picture, we shall realize *H*2(*s*), which is specified by Eq. (4.38). We can rearrange Eq. (4.38) as - -$$ -Y(s) = W(s) - \left(\frac{a_1}{s} + \frac{a_2}{s^2} + \frac{a_3}{s^3}\right) Y(s) -$$ -\n(4.39) - -Hence, to obtain *Y*(*s*), we subtract *a*1*Y*(*s*)/*s*, *a*2*Y*(*s*)/*s*2, and *a*3*Y*(*s*)/*s*3 from *W*(*s*). We have already obtained *W*(*s*) from the first step [output of *H*1(*s*)]. To obtain signals *Y*(*s*)/*s*, *Y*(*s*)/*s*2, and *Y*(*s*)/*s*3, we assume that we already have the desired output *Y*(*s*). Successive integration of *Y*(*s*) yields the needed signals *Y*(*s*)/*s*, *Y*(*s*)/*s*2, and *Y*(*s*)/*s*3. We now synthesize the final output *Y*(*s*) according to Eq. (4.39), as seen in the right-half section of Fig. 4.21a.† The left-half section in Fig. 4.21a represents *H*1(*s*) and the right-half is *H*2(*s*). We can generalize this procedure, known as the *direct form I* (DFI) realization, for any value of *N*. This procedure requires 2*N* integrators to realize an *N*th-order transfer function, as shown in Fig. 4.21b. - -### **[4.6-2 Direct Form II Realization](#page-10-0)** - -In the direct form I, we realize *H*(*s*) by implementing *H*1(*s*) followed by *H*2(*s*), as shown in Fig. 4.20a. We can also realize *H*(*s*), as shown in Fig. 4.20b, where *H*2(*s*) is followed by *H*1(*s*). - -† It may seem odd that we first assumed the existence of *Y*(*s*), integrated it successively, and then in turn generated *Y*(*s*) from *W*(*s*) and the three successive integrals of signal *Y*(*s*). This procedure poses a dilemma similar to "Which came first, the chicken or the egg?" The problem here is satisfactorily resolved by writing the expression for *Y*(*s*) at the output of the right-hand adder (at the top) in Fig. 4.21a and verifying that this expression is indeed the same as Eq. (4.38). - -**Figure 4.22** Direct form II realization of an *N*th-order LTIC system. - -This procedure is known as the *direct form II* realization. Figure 4.22a shows direct form II realization, where we have interchanged sections representing *H*1(*s*) and *H*2(*s*) in Fig. 4.21b. The output of *H*2(*s*) in this case is denoted by *V*(*s*). ‡ - -An interesting observation in Fig. 4.22a is that the input signal to both the chains of integrators is *V*(*s*). Clearly, the outputs of integrators in the left-side chain are identical to the corresponding outputs of the right-side integrator chain, thus making the right-side chain redundant. We can eliminate this chain and obtain the required signals from the left-side chain, as shown in Fig. 4.22b. This implementation halves the number of integrators to *N*, and, thus, is more efficient in hardware utilization than either Figs. 4.21b or 4.22a. This is the *direct form II* (DFII) realization. - -An *N*th-order differential equation with *N* = *M* has a property that its implementation requires a minimum of *N* integrators. A realization is *canonic* if the number of integrators used in the realization is equal to the order of the transfer function realized. Thus, canonic realization has no redundant integrators. The DFII form in Fig. 4.22b is a canonic realization, and is also called the *direct canonic* form. Note that the DFI is noncanonic. - -The direct form I realization (Fig. 4.22b) implements zeros first [the left-half section represented by *H*1(*s*)] followed by realization of poles [the right-half section represented by *H*2(*s*)] of *H*(*s*). In contrast, canonic direct implements poles first followed by zeros. Although both these realizations result in the same transfer function, they generally behave differently from the viewpoint of sensitivity to parameter variations. - -$$ -V(s) = X(s) - \left(\frac{a_1}{s} + \frac{a_2}{s^2} + \dots + \frac{a_N}{s^N}\right) V(s) -$$ - -and - -$$ -Y(s) = \left(b_0 + \frac{b_1}{s} + \frac{b_2}{s^2} + \dots + \frac{b_N}{s^N}\right) V(s) -$$ - -‡ The reader can show that the equations relating *X*(*s*),*V*(*s*), and *Y*(*s*) in Fig. 4.22a are - -### **EXAMPLE 4.22 Canonic Direct Form Realizations** - -Find the canonic direct form realization of the following transfer functions: - -(a) -$$ -\frac{5}{s+7} -$$ - -\n(b) $\frac{s}{s+7}$ -\n(c) $\frac{s+5}{s+7}$ -\n(d) $\frac{4s+28}{s^2+6s+5}$ - -All four of these transfer functions are special cases of *H*(*s*) in Eq. (4.36). - -**(a)** The transfer function 5/(*s*+7) is of the first order (*N* = 1); therefore, we need only one integrator for its realization. The feedback and feedforward coefficients are - -$$ -a_1 = 7 -$$ - and $b_0 = 0$ , $b_1 = 5$ - -The realization is depicted in Fig. 4.23a. Because *N* = 1, there is a single feedback connection from the output of the integrator to the input adder with coefficient *a*1 = 7. For *N* = 1, generally, there are *N* + 1 = 2 feedforward connections. However, in this case, *b*0 = 0, and there is only one feedforward connection with coefficient *b*1 = 5 from the output of the integrator to the output adder. Because there is only one input signal to the output adder, we can do away with the adder, as shown in Fig. 4.23a. - -**(b)** - -$$ -H(s) = \frac{s}{s+7} -$$ - -In this first-order transfer function, *b*1 = 0. The realization is shown in Fig. 4.23b. Because there is only one signal to be added at the output adder, we can discard the adder. - -**(c)** - -$$ -H(s) = \frac{s+5}{s+7} -$$ - -The realization appears in Fig. 4.23c. Here *H*(*s*) is a first-order transfer function with *a*1 = 7 and *b*0 = 1, *b*1 = 5. There is a single feedback connection (with coefficient 7) from the integrator output to the input adder. There are two feedforward connections (Fig. 4.23c).† - -$$ -H(s) = 1 - \frac{2}{s+7} -$$ - -We now realize *H*(*s*) as a parallel combination of two transfer functions, as indicated by this equation. - -† When *M* = *N* (as in this case), *H*(*s*) can also be realized in another way by recognizing that - -$$ -f_{\rm{max}} -$$ - -**(d)** - -$$ -H(s) = \frac{4s + 28}{s^2 + 6s + 5} -$$ - -This is a second-order system with *b*0 = 0, *b*1 = 4, *b*2 = 28, *a*1 = 6, and *a*2 = 5. Figure 4.23d shows a realization with two feedback connections and two feedforward connections. - -## **DR ILL 4.11 Canonic Direct Form Realization** - -Give the canonic direct realization of - -$$ -H(s) = \frac{2s}{s^2 + 6s + 25} -$$ - -### **[4.6-3 Cascade and Parallel Realizations](#page-10-0)** - -An *N*th-order transfer function *H*(*s*) can be expressed as a product or a sum of *N* first-order transfer functions. Accordingly, we can also realize *H*(*s*) as a cascade (series) or parallel form of these *N* first-order transfer functions. Consider, for instance, the transfer function in part (d) of Ex. 4.22. - -$$ -H(s) = \frac{4s + 28}{s^2 + 6s + 5} -$$ - -We can express *H*(*s*) as - -$$ -H(s) = \frac{4s + 28}{(s+1)(s+5)} = \underbrace{\left(\frac{4s + 28}{s+1}\right)}_{H_1(s)} \underbrace{\left(\frac{1}{s+5}\right)}_{H_2(s)} -$$ - -We can also express *H*(*s*) as a sum of partial fractions as - -$$ -H(s) = \frac{4s + 28}{(s+1)(s+5)} = \underbrace{\frac{6}{s+1}}_{H_3(s)} - \underbrace{\frac{2}{s+5}}_{H_4(s)} -$$ - -These equations give us the option of realizing *H*(*s*) as a cascade of *H*1(*s*) and *H*2(*s*), as shown in Fig. 4.24a, or a parallel of *H*3(*s*) and *H*4(*s*), as depicted in Fig. 4.24b. Each of the first-order transfer functions in Fig. 4.24 can be implemented by using canonic direct realizations, discussed earlier. - -This discussion by no means exhausts all the possibilities. In the cascade form alone, there are different ways of grouping the factors in the numerator and the denominator of *H*(*s*), and each grouping can be realized in DFI or canonic direct form. Accordingly, several cascade forms are possible. In Sec. 4.6-4, we shall discuss yet another form that essentially doubles the numbers of realizations discussed so far. - -From a practical viewpoint, parallel and cascade forms are preferable because parallel and certain cascade forms are numerically less sensitive than canonic direct form to small parameter variations in the system. Qualitatively, this difference can be explained by the fact that in a canonic realization all the coefficients interact with each other, and a change in any coefficient will be magnified through its repeated influence from feedback and feedforward connections. In a parallel realization, in contrast, the change in a coefficient will affect only a localized segment; the case with a cascade realization is similar. - -In the examples of cascade and parallel realization, we have separated *H*(*s*) into first-order factors. For *H*(*s*) of higher orders, we could group *H*(*s*) into factors, not all of which are necessarily of the first order. For example, if *H*(*s*) is a third-order transfer function, we could realize this function as a cascade (or a parallel) combination of a first-order and a second-order factor. - -**Figure 4.24** Realization of (4*s* +28)/[(*s*+1)(*s*+5)]: **(a)** cascade form and **(b)** parallel form. - -### REALIZATION OF COMPLEX CONJUGATE POLES - -The complex poles in *H*(*s*) should be realized as a second-order (quadratic) factor because we cannot implement multiplication by complex numbers. Consider, for example, - -$$ -H(s) = \frac{10s + 50}{(s+3)(s^2+4s+13)} -$$ - -= -$$ -\frac{10s + 50}{(s+3)(s+2-j3)(s+2+j3)} -$$ - -= -$$ -\frac{2}{s+3} - \frac{1+j2}{s+2-j3} - \frac{1-j2}{s+2+j3} -$$ - -We cannot realize first-order transfer functions individually with the poles −2 ± *j*3 because they require multiplication by complex numbers in the feedback and the feedforward paths. Therefore, we need to combine the conjugate poles and realize them as a second-order transfer function.† In the present example, we can create a cascade realization from *H*(*s*) expressed in product form as - -$$ -H(s) = \left(\frac{10}{s+3}\right) \left(\frac{s+5}{s^2+4s+13}\right) -$$ - -Similarly, we can create a parallel realization from *H*(*s*) expressed in sum form as - -$$ -H(s) = \frac{2}{s+3} - \frac{2s-8}{s^2+4s+13} -$$ - -### REALIZATION OF REPEATED POLES - -When repeated poles occur, the procedure for canonic and cascade realization is exactly the same as before. For a parallel realization, however, the procedure requires special handling, as explained in Ex. 4.23. - -### **EXAMPLE 4.23 Parallel Realization** - -Determine the parallel realization of - -$$ -H(s) = \frac{7s^2 + 37s + 51}{(s+2)(s+3)^2} = \frac{5}{s+2} + \frac{2}{s+3} - \frac{3}{(s+3)^2} -$$ - -This third-order transfer function should require no more than three integrators. But if we try to realize each of the three partial fractions separately, we require four integrators because of the one second-order term. This difficulty can be avoided by observing that the terms 1/(*s*+3) and - -† It is possible to realize complex, conjugate poles indirectly by using a cascade of two first-order transfer functions and feedback. A transfer function with poles −*a* ± *jb* can be realized by using a cascade of two identical first-order transfer functions, each having a pole at −*a* (see Prob. 4.6-15). - -**Figure 4.25** Parallel realization of (7*s*2 +37*s* +51)/((*s* +2)(*s* +3)2). - -1/(*s* + 3)2 can be realized with a cascade of two subsystems, each having a transfer function 1/(*s* + 3), as shown in Fig. 4.25. Each of the three first-order transfer functions in Fig. 4.25 may now be realized as in Fig. 4.23. - -### **DR ILL 4.12 Canonic, Cascade, and Parallel Realizations** - -Find the canonic, cascade, and parallel realization of - -$$ -H(s) = \frac{s+3}{s^2 + 7s + 10} = \left(\frac{s+3}{s+2}\right)\left(\frac{1}{s+5}\right) -$$ - -### **[4.6-4 Transposed Realization](#page-10-0)** - -Two realizations are said to be *equivalent* if they have the same transfer function. A simple way to generate an equivalent realization from a given realization is to use its *transpose*. To generate a transpose of any realization, we change the given realization as follows: - -- 1. Reverse all the arrow directions without changing the scalar multiplier values. -- 2. Replace pickoff nodes by adders and vice versa. -- 3. Replace the input *X*(*s*) with the output *Y*(*s*) and vice versa. - -Figure 4.26a shows the transposed version of the canonic direct form realization in Fig. 4.22b found according to the rules just listed. Figure 4.26b is Fig. 4.26a reoriented in the conventional form so that the input *X*(*s*) appears at the left and the output *Y*(*s*) appears at the right. Observe that this realization is also canonic. - -Rather than prove the theorem on equivalence of the transposed realizations, we shall verify that the transfer function of the realization in Fig. 4.26b is identical to that in Eq. (4.36). - -**Figure 4.26** Realization of an *N*th-order LTI transfer function in the transposed form. - -Figure 4.26b shows that *Y*(*s*) is being fed back through *N* paths. The fed-back signal appearing at the input of the top adder is - -$$ -\left(\frac{-a_1}{s} + \frac{-a_2}{s^2} + \cdots + \frac{-a_{N-1}}{s^{N-1}} + \frac{-a_N}{s^N}\right)Y(s) -$$ - -The signal *X*(*s*), fed to the top adder through *N* +1 forward paths, contributes - -$$ -\left(b_0 + \frac{b_1}{s} + \cdots + \frac{b_{N-1}}{s^{N-1}} + \frac{b_N}{s^N}\right)X(s) -$$ - -The output *Y*(*s*) is equal to the sum of these two signals (feed forward and feed back). Hence, - -$$ -Y(s) = \left(\frac{-a_1}{s} + \frac{-a_2}{s^2} + \dots + \frac{-a_{N-1}}{s^{N-1}} + \frac{-a_N}{s^N}\right)Y(s) + \left(b_0 + \frac{b_1}{s} + \dots + \frac{b_{N-1}}{s^{N-1}} + \frac{b_N}{s^N}\right)X(s) -$$ - -Transporting all the *Y*(*s*) terms to the left side and multiplying throughout by *sN*, we obtain - -$$ -(sN + a1sN-1 + \dots + aN-1s + aN)Y(s) = (b0sN + b1sN-1 + \dots + bN-1s + bN)X(s) -$$ - -Consequently, - -$$ -H(s) = \frac{Y(s)}{X(s)} = \frac{b_0 s^N + b_1 s^{N-1} + \dots + b_{N-1} s + b_N}{s^N + a_1 s^{N-1} + \dots + a_{N-1} s + a_N} -$$ - -Hence, the transfer function *H*(*s*) is identical to that in Eq. (4.36). - -We have essentially doubled the number of possible realizations. Every realization that was found earlier has a transpose. Note that the transpose of a transpose results in the same realization. - -### **EXAMPLE 4.24 Transposed Realizations** - -Find the transpose canonic direct realizations for parts (a) and (d) of Ex. 4.22 (Figs. 4.23c and 4.23d). The transfer functions are: - -(a) -$$ -\frac{s+5}{s+7} -$$ - -\n(b) $\frac{4s+28}{s^2+6s+5}$ - -Both these realizations are special cases of the one in Fig. 4.26b. - -**(a)** In this case, *N* = 1 with *a*1 = 7,*b*0 = 1,*b*1 = 5. The desired realization can be obtained by transposing Fig. 4.23c. However, we already have the general model of the transposed realization in Fig. 4.26b. The desired solution is a special case of Fig. 4.26b with *N* = 1 and *a*1 = 7,*b*0 = 1,*b*1 = 5, as shown in Fig. 4.27a. - -**(b)** In this case, *N* = 2 with *b*0 = 0, *b*1 = 4, *b*2 = 28, *a*1 = 6, *a*2 = 5. Using the model of Fig. 4.26b, we obtain the desired realization, as shown in Fig. 4.27b. - -**Figure 4.27** Transposed canonic direct form realizations of **(a)** (*s*+5)/(*s*+7) and **(b)** (4*s*+28)/(*s*2 +6*s*+5). - -### **DR ILL 4.13 Transposed Realizations** - -Find the transposed DFI and transposed canonic direct (TDFII) realizations of *H*(*s*) in Drill 4.11 - -### **[4.6-5 Using Operational Amplifiers for System Realization](#page-10-0)** - -In this section, we discuss practical implementation of the realizations described in Sec. 4.6-4. Earlier we saw that the basic elements required for the synthesis of an LTIC system (or a given transfer function) are (scalar) multipliers, integrators, and adders. All these elements can be realized by operational amplifier (op-amp) circuits. - -### OPERATIONAL AMPLIFIER CIRCUITS - -Figure 4.28 shows an op-amp circuit in the frequency domain (the transformed circuit). Because the input impedance of the op amp is infinite (very high), all the current *I*(*s*) flows in the feedback path, as illustrated. Moreover *Vx*(*s*), the voltage at the input of the op amp, is zero (very small) because of the infinite (very large) gain of the op amp. Therefore, for all practical purposes, - -$$ -Y(s) = -I(s)Z_f(s) -$$ - -Moreover, because *vx* ≈ 0, - -$$ -I(s) = \frac{X(s)}{Z(s)} -$$ - -Substitution of the second equation in the first yields - -$$ -Y(s) = -\frac{Z_f(s)}{Z(s)}X(s) -$$ - -Therefore, the op-amp circuit in Fig. 4.28 has the transfer function - -$$ -H(s) = -\frac{Z_f(s)}{Z(s)} -$$ - -By properly choosing *Z*(*s*) and *Zf*(*s*), we can obtain a variety of transfer functions, as the following development shows. - -**Figure 4.28** A basic inverting configuration op-amp circuit. - -### THE SCALAR MULTIPLIER - -If we use a resistor *Rf* in the feedback and a resistor *R* at the input (Fig. 4.29a), then *Zf*(*s*) = *Rf* , *Z*(*s*) = *R*, and - -$$ -H(s) = -\frac{R_f}{R} -$$ - -The system acts as a scalar multiplier (or an amplifier) with a negative gain *Rf* /*R*. A positive gain can be obtained by using two such multipliers in cascade or by using a single noninverting amplifier, as depicted in Fig. 4.16c. Figure 4.29a also shows the compact symbol used in circuit diagrams for a scalar multiplier. - -### THE INTEGRATOR - -If we use a capacitor *C* in the feedback and a resistor *R* at the input (Fig. 4.29b), then *Zf*(*s*) = 1/*Cs*, *Z*(*s*) = *R*, and - -$$ -H(s) = \left(-\frac{1}{RC}\right)\frac{1}{s} -$$ - -The system acts as an ideal integrator with a gain −1/*RC*. Figure 4.29b also shows the compact symbol used in circuit diagrams for an integrator. - -**Figure 4.29 (a)** Op-amp inverting amplifier. **(b)** Integrator. - -**Figure 4.30** Op-amp summing and amplifying circuit. - -### THE ADDER - -Consider now the circuit in Fig. 4.30a with *r* inputs *X*1(*s*), *X*2(*s*), ... , *Xr*(*s*). As usual, the input voltage *Vx*(*s*) 0 because the op-amp gain → ∞. Moreover, the current going into the op amp is very small ( 0) because the input impedance → ∞. Therefore, the total current in the feedback resistor *Rf* is *I*1(*s*)+*I*2(*s*)+···+*Ir*(*s*). Moreover, because *Vx*(*s*) = 0, - -$$ -I_j(s) = \frac{X_j(s)}{R_j} -$$ - $j = 1, 2, ..., r$ - -Also, - -$$ -Y(s) = -R_f[I_1(s) + I_2(s) + \dots + I_r(s)] -$$ - -= -$$ --\left[\frac{R_f}{R_1}X_1(s) + \frac{R_f}{R_2}X_2(s) + \dots + \frac{R_f}{R_r}X_r(s)\right] -$$ - -= $k_1X_1(s) + k_2X_2(s) + \dots + k_rX_r(s)$ - -where - -$$ -k_i = \frac{-R_f}{R_i} -$$ - -Clearly, the circuit in Fig. 4.30 serves an adder and an amplifier with any desired gain for each of the input signals. Figure 4.30b shows the compact symbol used in circuit diagrams for an adder with *r* inputs. - -### **EXAMPLE 4.25 Op-Amp Realization** - -Use op-amp circuits to realize the canonic direct form of the transfer function - -$$ -H(s) = \frac{2s+5}{s^2+4s+10} -$$ - -**Figure 4.31** Op-amp realization of a second-order transfer function (2*s*+5)/(*s*2 +4*s* +10). - -The basic canonic realization is shown in Fig. 4.31a. The same realization with horizontal reorientation is shown in Fig. 4.31b. Signals at various points are also indicated in the realization. For convenience, we denote the output of the last integrator by *W*(*s*). Consequently, the signals at the inputs of the two integrators are *sW*(*s*) and *s*2*W*(*s*), as shown in Figs. 4.31a and 4.31b. Op-amp elements (multipliers, integrators, and adders) change the polarity of the output signals. To incorporate this fact, we modify the canonic realization in Fig. 4.31b to that depicted in Fig. 4.31c. In Fig. 4.31b, the successive outputs of the adder and the integrators are *s*2*W*(*s*),*sW*(*s*), and *W*(*s*), respectively. Because of polarity reversals in op-amp circuits, these outputs are −*s*2*W*(*s*),*sW*(*s*), and −*W*(*s*), respectively, in Fig. 4.31c. This polarity reversal requires corresponding modifications in the signs of feedback and feedforward gains. According to Fig. 4.31b, - -$$ -s^2W(s) = X(s) - 4sW(s) - 10W(s) -$$ - -Therefore, - -$$ --s^2W(s) = -X(s) + 4sW(s) + 10W(s) -$$ - -Because the adder gains are always negative (see Fig. 4.30b), we rewrite the foregoing equation as - -$$ --s2W(s) = -1[X(s)] - 4[-sW(s)] - 10[-W(s)] -$$ - -Figure 4.31c shows the implementation of this equation. The hardware realization appears in Fig. 4.31d. Both integrators have a unity gain, which requires *RC* = 1. We have used *R* = 100 k and *C* = 10 µF. The gain of 10 in the outer feedback path is obtained in the adder by choosing the feedback resistor of the adder to be 100 k and an input resistor of 10 k. Similarly, the gain of 4 in the inner feedback path is obtained by using the corresponding input resistor of 25 k. The gains of 2 and 5, required in the feedforward connections, are obtained by using a feedback resistor of 100 k and input resistors of 50 and 20 k, respectively.† - -The op-amp realization in Fig. 4.31 is not necessarily the one that uses the fewest op amps. This example is given just to illustrate a systematic procedure for designing an op-amp circuit of an arbitrary transfer function. There are more efficient circuits (such as Sallen–Key or biquad) that use fewer op amps to realize a second-order transfer function. - -### **DR ILL 4.14 Transfer Functions of Op-Amp Circuits** - -Show that the transfer functions of the op-amp circuits in Figs. 4.32a and 4.32b are *H*1(*s*) and *H*2(*s*), respectively, where - -$$ -H_1(s) = \frac{-R_f}{R} \left( \frac{a}{s+a} \right) \qquad a = \frac{1}{R_f C_f} -$$ - -$$ -H_2(s) = -\frac{C}{C_f} \left( \frac{s+b}{s+a} \right) \qquad a = \frac{1}{R_f C_f} \qquad b = \frac{1}{RC} -$$ - -† It is possible to avoid the two inverting op amps (with gain −1) in Fig. 4.31d by adding signal *sW*(*s*) to the input and output adders directly, using the noninverting amplifier configuration in Fig. 4.16d. - - diff --git a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/059_4.7 APPLICATION TO FEEDBACK AND CONTROLS.md b/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/059_4.7 APPLICATION TO FEEDBACK AND CONTROLS.md deleted file mode 100644 index 63f8376c75f600fba41b74d4278c94423aedf4ca..0000000000000000000000000000000000000000 --- a/engineering/linear-systems-and-signals-3rd-edition-bp-lathi/059_4.7 APPLICATION TO FEEDBACK AND CONTROLS.md +++ /dev/null @@ -1,188 +0,0 @@ -## **[4.7 APPLICATION TO](#page-10-0) FEEDBACK AND CONTROLS** - -Generally, systems are designed to produce a desired output *y*(*t*) for a given input *x*(*t*). Using the given performance criteria, we can design a system, as shown in Fig. 4.33a. Ideally, such an open-loop system should yield the desired output. In practice, however, the system characteristics change with time, as a result of aging or replacement of some components, or because of changes in the operating environment. Such variations cause changes in the output for the same input. Clearly, this is undesirable in precision systems. - -**Figure 4.33 (a)** Open-loop and **(b)** closed-loop (feedback) systems. - -A possible solution to this problem is to add a signal component to the input that is not a predetermined function of time but will change to counteract the effects of changing system characteristics and the environment. In short, we must provide a correction at the system input to account for the undesired changes just mentioned. Yet since these changes are generally unpredictable, it is not clear how to preprogram appropriate corrections to the input. However, the difference between the actual output and the desired output gives an indication of the suitable correction to be applied to the system input. It may be possible to counteract the variations by feeding the output (or some function of output) back to the input. - -We unconsciously apply this principle in daily life. Consider an example of marketing a certain product. The optimum price of the product is the value that maximizes the profit of a merchant. The output in this case is the profit, and the input is the price of the item. The output (profit) can be controlled (within limits) by varying the input (price). The merchant may price the product too high initially, in which case, he will sell too few items, reducing the profit. Using feedback of the profit (output), he adjusts the price (input), to maximize his profit. If there is a sudden or unexpected change in the business environment, such as a strike-imposed shutdown of a large factory in town, the demand for the item goes down, thus reducing his output (profit). He adjusts his input (reduces price) using the feedback of the output (profit) in a way that will optimize his profit in the changed circumstances. If the town suddenly becomes more prosperous because a new factory opens, he will increase the price to maximize the profit. Thus, by continuous feedback of the output to the input, he realizes his goal of maximum profit (optimum output) in any given circumstances. We observe thousands of examples of feedback systems around us in everyday life. Most social, economical, educational, and political processes are, in fact, feedback processes. A block diagram of such a system, called the *feedback* or *closed-loop* system, is shown in Fig. 4.33b. - -A feedback system can address the problems arising because of unwanted disturbances such as random-noise signals in electronic systems, a gust of wind affecting a tracking antenna, a meteorite hitting a spacecraft, and the rolling motion of antiaircraft gun platforms mounted on ships or moving tanks. Feedback may also be used to reduce nonlinearities in a system or to control its rise time (or bandwidth). Feedback is used to achieve, with a given system, the desired objective within a given tolerance, despite partial ignorance of the system and the environment. A feedback system, thus, has an ability for supervision and self-correction in the face of changes in the system parameters and external disturbances (change in the environment). - -Consider the feedback amplifier in Fig. 4.34. Let the forward amplifier gain *G* = 10,000. One-hundredth of the output is fed back to the input (*H* = 0.01). The gain *T* of the feedback amplifier is obtained by [see Eq. (4.35)] - -$$ -T = \frac{G}{1 + GH} = \frac{10,000}{1 + 100} = 99.01 -$$ - -Suppose that because of aging or replacement of some transistors, the gain *G* of the forward amplifier changes from 10,000 to 20,000. The new gain of the feedback amplifier is given by - -$$ -T = \frac{G}{1 + GH} = \frac{20,000}{1 + 200} = 99.5 -$$ - -Surprisingly, 100% variation in the forward gain *G* causes only 0.5% variation in the feedback amplifier gain *T*. Such reduced sensitivity to parameter variations is a must in precision amplifiers. In this example, we reduced the sensitivity of gain to parameter variations at the cost of forward - -*H* **Figure 4.34** Effects of negative and positive feedback. - -#### 406 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -gain, which is reduced from 10,000 to 99. There is no dearth of forward gain (obtained by cascading stages). But low sensitivity is extremely precious in precision systems. - -Now, consider what happens when we add (instead of subtract) the signal fed back to the input. Such addition means the sign on the feedback connection is + instead of − (which is same as changing the sign of *H* in Fig. 4.34). Consequently, - -$$ -T = \frac{G}{1 - GH} -$$ - -If we let *G* = 10,000 as before and *H* = 0.9×10−4, then - -$$ -T = \frac{10,000}{1 - 0.9(10^{4})(10^{-4})} = 100,000 -$$ - -Suppose that because of aging or replacement of some transistors, the gain of the forward amplifier changes to 11,000. The new gain of the feedback amplifier is - -$$ -T = \frac{11,000}{1 - 0.9(11,000)(10^{-4})} = 1,100,000 -$$ - -Observe that in this case, a mere 10% increase in the forward gain *G* caused 1000% increase in the gain *T* (from 100,000 to 1,100,000). Clearly, the amplifier is very sensitive to parameter variations. This behavior is exactly opposite of what was observed earlier, when the signal fed back was subtracted from the input. - -What is the difference between the two situations? Crudely speaking, the former case is called the *negative feedback* and the latter is the *positive feedback*. The positive feedback increases system gain but tends to make the system more sensitive to parameter variations. It can also lead to instability. In our example, if *G* were to be 111,111, then *GH* = 1, *T* = ∞, and the system would become unstable because the signal fed back was exactly equal to the input signal itself, since *GH* = 1. Hence, once a signal has been applied, no matter how small and how short in duration, it comes back to reinforce the input undiminished, which further passes to the output, and is fed back again and again and again. In essence, the signal perpetuates itself forever. This perpetuation, even when the input ceases to exist, is precisely the symptom of instability. - -Generally speaking, a feedback system cannot be described in black and white terms, such as positive or negative. Usually *H* is a frequency-dependent component, more accurately represented by *H*(*s*); hence it varies with frequency. Consequently, what was negative feedback at lower frequencies can turn into positive feedback at higher frequencies and may give rise to instability. This is one of the serious aspects of feedback systems, which warrants a designer's careful attention. - -### **[4.7-1 Analysis of a Simple Control System](#page-10-0)** - -Figure 4.35a represents an automatic position control system, which can be used to control the angular position of a heavy object (e.g., a tracking antenna, an anti-aircraft gun mount, or the position of a ship). The input θ*i* is the desired angular position of the object, which can be set at any given value. The actual angular position θ*o* of the object (the output) is measured by a potentiometer whose wiper is mounted on the output shaft. The difference between the input θ*i* - -(set at the desired output position) and the output θ*o* (actual position) is amplified; the amplified output, which is proportional to θ*i* − θ*o*, is applied to the motor input. If θ*i* − θ*o* = 0 (the output being equal to the desired angle), there is no input to the motor, and the motor stops. But if θ*o* = θ*i*, there will be a nonzero input to the motor, which will turn the shaft until θ*o* = θ*i*. It is evident that by setting the input potentiometer at a desired position in this system, we can control the angular position of a heavy remote object. - -The block diagram of this system is shown in Fig. 4.35b. The amplifier gain is *K*, where *K* is adjustable. Let the motor (with load) transfer function that relates the output angle θ*o* to the motor input voltage be *G*(*s*) [for a starting point, see Eq. (1.32)]. This feedback arrangement is identical to that in Fig. 4.18d with *H*(*s*) = 1. Hence, *T*(*s*), the (closed-loop) system transfer function relating the output θ*o* to the input θ*i*, is - -$$ -\frac{\Theta_o(s)}{\Theta_i(s)} = T(s) = \frac{KG(s)}{1 + KG(s)} -$$ - -From this equation, we shall investigate the behavior of the automatic position control system in Fig. 4.35a for a step and a ramp input. - -### STEP INPUT - -If we desire to change the angular position of the object instantaneously, we need to apply a step input. We may then want to know how long the system takes to position itself at the desired angle, whether it reaches the desired angle, and whether it reaches the desired position smoothly (monotonically) or oscillates about the final position. If the system oscillates, we may want to know how long it takes for the oscillations to settle down. All these questions can be readily answered by finding the output θ*o*(*t*) when the input θ*i*(*t*) = *u*(*t*). A step input implies instantaneous change in the angle. This input would be one of the most difficult to follow; if the system can perform well for this input, it is likely to give a good account of itself under most other expected situations. This is why we test control systems for a step input. - -For the step input θ*i*(*t*) = *u*(*t*), *i*(*s*) = 1/*s* and - -$$ -\Theta_o(s) = \frac{1}{s}T(s) = \frac{KG(s)}{s[1+KG(s)]} -$$ - -Let the motor (with load) transfer function relating the load angle θ*o*(*t*) to the motor input voltage be *G*(*s*) = 1/(*s*(*s*+8)). This yields - -$$ -\Theta_o(s) = \frac{\frac{K}{s(s+8)}}{s\left[1 + \frac{K}{s(s+8)}\right]} = \frac{K}{s(s^2 + 8s + K)} -$$ - -Let us investigate the system behavior for three different values of gain *K*. For *K* = 7, - -$$ -\Theta_o(s) = \frac{7}{s(s^2 + 8s + 7)} = \frac{7}{s(s+1)(s+7)} = \frac{1}{s} - \frac{\frac{7}{6}}{s+1} + \frac{\frac{1}{6}}{s+7} -$$ - -**Figure 4.35 (a)** An automatic position control system. **(b)** Its block diagram. **(c)** The unit step response. **(d)** The unit ramp response. - -and - -$$ -\theta_o(t) = \left(1 - \frac{7}{6}e^{-t} + \frac{1}{6}e^{-7t}\right)u(t) -$$ - -This response, illustrated in Fig. 4.35c, shows that the system reaches the desired angle, but at a rather leisurely pace. To speed up the response let us increase the gain to, say, 80. - -For *K* = 80, - -$$ -\Theta_o(s) = \frac{80}{s(s^2 + 8s + 80)} = \frac{80}{s(s + 4 - j8)(s + 4 + j8)} -$$ -$$ -= \frac{1}{s} + \frac{\frac{\sqrt{5}}{4}e^{j153^\circ}}{s + 4 - j8} + \frac{\frac{\sqrt{5}}{4}e^{-j153^\circ}}{s + 4 + j8} -$$ - -and - -$$ -\theta_o(t) = \left[1 + \frac{\sqrt{5}}{2}e^{-4t}\cos{(8t + 153^\circ)}\right]u(t) -$$ - -This response, also depicted in Fig. 4.35c, achieves the goal of reaching the final position at a faster rate than that in the earlier case (*K* = 7). Unfortunately the improvement is achieved at the cost of ringing (oscillations) with high overshoot. In the present case, the *percent overshoot* (PO) is 21%. The response reaches its peak value at *peak time tp* = 0.393 second. The *rise time*, defined as the time required for the response to rise from 10% to 90% of its steady-state value, indicates the speed of response.† In the present case *tr* = 0.175 second. The steady-state value of the response is unity so that the *steady-state error* is zero. Theoretically it takes infinite time for the response to reach the desired value of unity. In practice, however, we may consider the response to have settled to the final value if it closely approaches the final value. A widely accepted measure of closeness is within 2% of the final value. The time required for the response to reach and stay within 2% of the final value is called the settling time *ts*. ‡ In Fig. 4.35c, we find *ts* ≈ 1 second (when *K* = 80). A good system has a small overshoot, small *tr* and *ts* and a small steady-state error. - -A large overshoot, as in the present case, may be unacceptable in many applications. Let us try to determine *K* (the gain) that yields the fastest response without oscillations. Complex characteristic roots lead to oscillations; to avoid oscillations, the characteristic roots should be real. In the present case, the characteristic polynomial is *s*2 +8*s*+*K*. For *K* > 16, the characteristic roots are complex; for *K* < 16, the roots are real. The fastest response without oscillations is obtained by choosing *K* = 16. We now consider this case. - -For *K* = 16, - -$$ -\Theta_o(s) = \frac{16}{s(s^2 + 8s + 16)} = \frac{16}{s(s+4)^2} = \frac{1}{s} - \frac{1}{s+4} - \frac{4}{(s+4)^2} -$$ - -and - -$$ -\theta_o(t) = [1 - (4t + 1)e^{-4t}]u(t) -$$ - -This response also appears in Fig. 4.35c. The system with *K* > 16 is said to be *underdamped* (oscillatory response), whereas the system with *K* < 16 is said to be *overdamped*. For *K* = 16, the system is said to be *critically damped*. - -† *Delay time td*, defined as the time required for the response to reach 50% of its steady-state value, is another indication of speed. For the present case, *td* = 0.141 second. - -‡ Typical percentage values used are 2 to 5% for *ts*. - -### 410 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -There is a trade-off between undesirable overshoot and rise time. Reducing overshoots leads to higher rise time (sluggish system). In practice, a small overshoot, which is still faster than the critical damping, may be acceptable. Note that percent overshoot PO and peak time *tp* are meaningless for the overdamped or critically damped cases. In addition to adjusting gain *K*, we may need to augment the system with some type of compensator if the specifications on overshoot and the speed of response are too stringent. - -### RAMP INPUT - -If the anti-aircraft gun in Fig. 4.35a is tracking an enemy plane moving with a uniform velocity, the gun-position angle must increase linearly with *t*. Hence, the input in this case is a ramp; that is, θ*i*(*t*) = *tu*(*t*). Let us find the response of the system to this input when *K* = 80. In this case, *i*(*s*) = 1/*s*2, and - -$$ -\Theta_o(s) = \frac{80}{s^2(s^2 + 8s + 80)} = -\frac{0.1}{s} + \frac{1}{s^2} + \frac{0.1(s - 2)}{s^2 + 8s + 80} -$$ - -Use of Table 4.1 yields - -$$ -\theta_o(t) = \left[ -0.1 + t + \frac{1}{8}e^{-8t}\cos\left(8t + 36.87^\circ\right) \right] u(t) -$$ - -This response, sketched in Fig. 4.35d, shows that there is a steady-state error *er* = 0.1 radian. In many cases such a small steady-state error may be tolerable. If, however, a zero steady-state error to a ramp input is required, this system in its present form is unsatisfactory. We must add some form of compensator to the system. - -### **EXAMPLE 4.26 Step and Ramp Responses of Feedback Systems Using MATLAB** - -Using the feedback system of Fig. 4.18d with *G*(*s*) = *K*/(*s*(*s* + 8)) and *H*(*s*) = 1, determine the step response for each of the following cases: **(a)** *K* = 7, **(b)** *K* = 16, and **(c)** *K* = 80. Additionally, find the unit ramp response when **(d)** *K* = 80. - -Example 4.21 computes the transfer functions of these feedback systems in a simple way. In this example, the conv command is used to demonstrate polynomial multiplication of the two denominator factors of *G*(*s*). Step responses are computed by using the step command. **(a–c)** - ->> H = tf(1,1); K = 7; G = tf([K],conv([1 0],[1 8])); Ha = feedback(G,H); - -&gt;> H = tf(1,1); K = 16; G = tf([K],conv([1 0],[1 8])); Hb = feedback(G,H); - -&gt;> H = tf(1,1); K = 80; G = tf([K],conv([1 0],[1 8])); Hc = feedback(G,H); - -&gt;> clf; step(Ha,'k-',Hb,'k--',Hc,'k-.'); - -&gt;> legend('K = 7','K = 16','K = 80','Location','best'); - -**Figure 4.36** Step responses for Ex. 4.26. - -**(d)** The unit ramp response is equivalent to the integral of the unit step response. We can obtain the ramp response by taking the step response of the system in cascade with an integrator. To help highlight waveform detail, we compute the ramp response over the short time interval of 0 ≤ *t* ≤ 1.5. - ->> t = 0:.001:1.5; Hd = series(Hc,tf([1],[1 0])); >> step(Hd,'k-',t); title('Unit Ramp Response'); - -### DESIGN SPECIFICATIONS - -Now the reader has some idea of the various specifications a control system might require. Generally, a control system is designed to meet given transient specifications, steady-state error specifications, and sensitivity specifications. Transient specifications include overshoot, rise time, and settling time of the response to step input. The steady-state error is the difference between - -### 412 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS - -the desired response and the actual response to a test input in steady state. The system should also satisfy a specified sensitivity specifications to some system parameter variations, or to certain disturbances. Above all, the system must remain stable under operating conditions. Discussion of design procedures used to realize given specifications is beyond the scope of this book.