Chapter 11_Network Functions - The Analysis and Design of...

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The Analysis and Design of Linear Circuits Seventh Edition 11 Network Functions 11.1 Exercise Solutions Exercise 11–1. The network function for a circuit is T ( s ) = 10 s s + 100 Find the zero-state response v 2 ( t ) when the input waveform is v 1 ( t ) = cos(50 t ) V. Find the response in the s domain V 2 ( s ) = T ( s ) V 1 ( s ) = parenleftbigg 10 s s + 100 parenrightbigg parenleftbigg s s 2 + 50 2 parenrightbigg = 10 s 2 ( s + 100)( s 2 + 50 2 ) = A s + 100 + Bs + C s 2 + 50 2 = 8 s + 100 + 2 s 200 s 2 + 50 2 = 8 s + 100 + (2) s s 2 + 50 2 (4) 50 s 2 + 50 2 v 2 ( t ) = bracketleftbig 8 e 100 t + 2 cos(50 t ) 4 sin(50 t ) bracketrightbig u ( t ) V Exercise 11–2. For the circuit of Figure 11–8, find the voltage transfer function T V ( s ) = V 2 ( s ) /V 1 ( s ) and the driving-point impedance Z ( s ). Apply voltage division to find the transfer function. Sum the impedances in series to determine the driving-point impedance. T V ( s ) = V 2 ( s ) V 1 ( s ) = R R + Ls + 1 /Cs = RCs LCs 2 + RCs + 1 Z ( s ) = R + Ls + 1 Cs = LCs 2 + RCs + 1 Cs Exercise 11–3. For the circuit of Figure 11–10, (a). Find the voltage transfer function T V ( s ) = V 2 ( s ) /V 1 ( s ) and the driving-point impedance Z ( s ). Apply voltage division to find the transfer function. Sum the impedances in series to determine the driving-point impedance. Z 1 = Ls + R 1 Z 2 = 1 Cs + R 2 = R 2 Cs + 1 Cs T V ( s ) = Z 2 Z 1 + Z 2 = R 2 Cs + 1 Cs Ls + R 1 + R 2 Cs + 1 Cs = R 2 Cs + 1 LCs 2 + R 1 Cs + R 2 Cs + 1 Z ( s ) = Ls + R 1 + 1 Cs + R 2 = LCs 2 + ( R 1 + R 2 ) Cs + 1 Cs (b). Locate the poles and zeros of the transfer function when R 1 = R 2 = 1 kΩ, L = 10 mH, and C = 0 . 1 μ F. Compute the transfer function. T V ( s ) = 10 4 s + 1 10 9 s 2 + (2 × 10 4 ) s + 1 = 10 5 ( s + 10 4 ) s 2 + 200000 s + 10 9 Solution Manual Chapter 11 Page 11-1
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The Analysis and Design of Linear Circuits Seventh Edition The zeros are located at infinity and s = 10 4 rad/s. The poles are located at s = 194868 rad/s and s = 5132 rad/s. Exercise 11–4. Suppose that capacitor C 1 in the circuit of Figure 11–11 suddenly became shorted. What effect would it have on the circuit’s voltage transfer function? The transfer function would become T V ( s ) = Z 2 ( s ) R 1 = R 2 R 1 R 2 C 2 s + 1 The transfer function no longer has a pole at s = 1 /R 1 C 1 . Exercise 11–5. Suppose that capacitor C 2 in the circuit of Figure 11–11 suddenly became open-circuited. What effect would it have on the circuit’s voltage transfer function? The transfer function would become T V ( s ) = R 2 Z 1 ( s ) = R 2 C 1 s R 1 C 1 s + 1 The transfer function no longer has a pole at s = 1 /R 2 C 2 . Exercise 11–6. For the circuit shown in Figure 11–12, insert a follower at point A and find the transfer function T V ( s ) = V 2 ( s ) /V 1 ( s ). Compare your result with that found in Example 11–5 for the same transfer function. The follower effectively separates the two RC circuits. Find the transfer function on each side of the follower and multiply the results together. T 1 ( s ) = R R + 1 /Cs = RCs RCs + 1 T 2 ( s ) = 1 /Cs R + 1 /Cs = 1 RCs + 1 T V ( s ) = T 1 ( s ) T 2 ( s ) = RCs ( RCs + 1) 2 = RCs ( RCs ) 2 + 2 RCs + 1 The difference appears in the middle term in the denominator, which is now a two instead of three. The change would affect the location of the poles and the nature of the circuit response. The follower eliminates
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