DeCarlo Ch19b Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

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Prbs Chap 19, 1/13/02 P19-1 © R. A. DeCarlo, P. M. Lin CHAPTER 19 PROBLEM SOLUTIONS S OLUTION P ROBLEM 19.40 . (a) Let Z 1 = R and Z 2 = 1/Cs or Y 2 = Cs. From problem 38, h = Z 1 1 - 1 Y 2 = R 1 - 1 Cs (b) This part is a cascade of an ideal transformer and part (a). Label the voltage and current at the port 1 of N 1 as ˆ V 1 and ˆ I 1 . From the properties of the ideal transformer, V 1 = - b ˆ V 1 and I 1 = - ˆ I 1 b . Hence ˆ V 1 I 2 = R 1 - 1 Cs ˆ I 1 V 2 - V 1 b I 2 = R 1 - 1 Cs - bI 1 V 2 Therefore V 1 I 2 = b 2 R - b b Cs I 1 V 2 From table 19.1, if h 22 = Cs 0, then the z-parameters exist and if h 11 = b 2 R 0, the y- parameters exist, i.e., if C 0 and R 0 (assuming reasonably that b 0) respectively. S OLUTION P ROBLEM 19.41 . For this solution we apply the definition of h-parameters: by inspection h 11 = V 1 I 1 V 2 = 0 = 1 2 + 2 s h 21 = I 2 I 1 V 2 = 0 = 2 V 1 - 2 sV 1 I 1 V 2 = 0 = (2 - 2 s ) V 1 (2 + 2 s ) V 1 V 2 = 0 = 1 - s s + 1 When I 1 = 0, then I 2 = 2 V 1 + 2 V 1 = 4 V 1 and V 2 = 0.5(2 V 1 ) + 1 2 s (2 V 1 ) = s + 1 s V 1 .

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Prbs Chap 19, 1/13/02 P19-2 © R. A. DeCarlo, P. M. Lin Therefore, h 12 = V 1 V 2 I 1 = 0 = s s + 1 and h 22 = I 2 V 2 I 1 = 0 = 4 V 1 V 2 I 1 = 0 = 4 s s + 1 . S OLUTION P ROBLEM 19.42 . (a) In MATLAB »h11 = 250; h12 = 0.025; h21 = 12.5; h22 = 2.25e-3; »Zs = 1e3; ZL = 500; »YL = 1/ZL YL = 2.0000e-03 »Zin = h11 - h12*h21/(h22 + (1/ZL)) Zin = 1.7647e+02 »Yout = h22 - h12*h21/(h11 + Zs) Yout = 2.0000e-03 »Zout = 1/Yout Zout = 500 (b) »% Gv1 = V1/Vs »Gv1 = Zin/(Zin + Zs) Gv1 = 1.5000e-01 »% Gv2 = V2/V1 »Gv2 = -h21/(Zin*(h22 + YL)) Gv2 = -1.6667e+01 »Gv = Gv1*Gv2 Gv = -2.5000e+00 (c) Given the above, the Thevenin equivalent seen by the capacitor is V oc = - 2.5 V in and R th = 500 . In MATLAB »Zth = ZL*Zout/(ZL + Zout) Zth = 250 »Vin = 10; »Voc = -2.5*Vin; »w = 400;
Prbs Chap 19, 1/13/02 P19-3 © R. A. DeCarlo, P. M. Lin »Zc = 1/(j*w*10e-6) Zc = 0 - 2.5000e+02i »Vc = Voc*Zc/(Zth + Zc) Vc = -1.2500e+01 + 1.2500e+01i»V2mag = abs(Vc) »V2mag = abs(Vc) V2mag = 1.7678e+01 »V2ang = angle(Vc)*180/pi V2ang = 135 From above, v 2 ( t ) = 17.678 2 cos(400 t + 135 o ) V Therefore »Pave = V2mag^2/500 Pave = 6.2500e-01 S OLUTION P ROBLEM 19.43 . (a) Using the h-parameters of stage 2 Z in2 = h 11 - h 12 h 21 h 22 +Y L = 1000+ 0.966 × 51 0.0008 + 1/64 = 4000 The load for stage 1 is the parallel combination of Z in2 and the 3 k resistance. However, because h 12 = 0, the input impedance is unaffected by the load, and hence for stage 1, Z in1 = h 11 = 2000 (b) For stage 1, because h 12 = 0, the output impedance is unaffected by the source impedance. Thus, Y out1 = h 22 = 0.05 × 10 -3 S, Z out1 =1/Y out1 = 20 × 10 3

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Prbs Chap 19, 1/13/02 P19-4 © R. A. DeCarlo, P. M. Lin For stage 2, the source impedance is the parallel combination of Z out1 and the 3 k resistance. Thus Z s2 = 20000 × 3000 20000 + 3000 =2608.7 and Y out2 = h 22 - h 12 h 21 h 11 +Z s2 = 0.0008+ 0.966 × 51 1000 + 2608.7 = 0.0145 S Z out2 =1/Y out2 = 69.19 (c) V 1 V s = Z in1 Z in1 + Z s = 2000 2000 + 2000 = 0.5 The load of stage 1 is the parallel combination of Z in2 and Z m . Thus Y L1 = Y in2 + 1/3000 = 5.834 × 10 -4 S Hence Z L1 = 1/ Y L1 = 1714 and V 2 V 1 stage 1 = 1 Z in1 × -h 21 h 22 + Y L1 = 1 2000 × -50 (0.05 + 0.5834) × 10 -3 = - 39.46 For stage 2, the load is 64 . Hence V 2 V 1 stage 2 = 1 Z in2 × -h 21 h 22 + Y L2 = 1 4000 × 51 (0.8 + 1000/64)
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