DeCarlo Ch19a Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

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Prbs Chap 19, 1/7/02 P19-1 © R. A. DeCarlo, P. M. Lin CHAPTER 19 PROBLEM SOLUTIONS S OLUTION P ROBLEM 19.1. Refer to figure 19.3. »Vs = 100; ZL = 20; Rs = 1e3; »beta = 149; »Zinbox = (beta + 1)*ZL Zinbox = 3000 »% By voltage division »V1 = Vs*Zinbox/(Zinbox + Rs) V1 = 75 »% To obtain the power delivered by the source »I1 = Vs/(Zinbox +Rs) I1 = 2.5000e-02 »Psource = I1*Vs Psource = 2.5000e+00 S OLUTION P ROBLEM 19.2. Refer to figure 19.4. »Vs = 100; Z1 = 30e3; Rs = 50; beta = 149; »Zboxin = Z1/(beta+1) Zboxin = 200 »V1 = Vs*Zboxin/(Zboxin + Rs) V1 = 80 »Psource = Vs^2/(Rs + Zboxin) Psource = 40 S OLUTION P ROBLEM 19.3. Refer to figure 19.5. (a) »C = 0.1e-3; vc0 = 10; Z1 = 300; Z2 = 1e3; »Z3 = 1e3; gm = 9e-3; » »Zin = Z1 + (1 + gm*Z1)*Z2 Zin = 4.0000e+03 (b) »tau = Zin*C

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Prbs Chap 19, 1/7/02 P19-2 © R. A. DeCarlo, P. M. Lin tau = 4.0000e-01 Hence, v C ( t ) = v C (0) e - t / = 10 e - 2.5 t V. S OLUTION P ROBLEM 19.4. (a) First observe that since no current can flow into the secondary we have V oc = aV pri = aRI in = 800 Vrms Now Z th = 1 j C + a 2 R = 640 - j 360 (b) Z L = Z th ( 29 * = 640 + j 360 . »Voc = 800; Rth = 640; »Pmax = Voc^2/(4*Rth) Pmax = 250 (c) By inspection the circuit is a 640 resistor in series with a 3.6 H inductor. S OLUTION P ROBLEM 19.5. Because the output is open circuited, no current flows into the secondary of the transformer, hence v oc = v sec + sin(3 t ) u ( t ) = 2 v pri + sin(3 t ) u ( t ) = cos(3 t ) + sin(3 t ) [ ] u ( t ). Additionally Z th ( s ) = 10 s + s + 4 + 4 2.5 s + 0.25 s + 9 = 20 s + 2 s + 40 S OLUTION P ROBLEM 19.6. Using Cramer's rule,
Prbs Chap 19, 1/7/02 P19-3 © R. A. DeCarlo, P. M. Lin I 1 = det V 1 1 - a 0 0.5 0 0 0 0.5 1.5 a = 0.25 V 1 1.5 a = 1 6 a V 1 Therefore R in = 6 a . To compute the average power, V 1, eff = 10 V. Hence P ave = V 1, eff 2 R in = 100 6 a watts. S OLUTION P ROBLEM 19.7. As per the hint, we write loop equations as follows: - V out 10 s - 40 s = s + 1 - 1 0 - 1 6 - 2 0 - 2 4 I out I 2 I 3 Using equation 19.6, - V out = s + 1 - - 1 0 [ ] 0.2 0.1 0.1 0.3 - 1 0 I out + - 1 0 [ ] 0.2 0.1 0.1 0.3 10 s - 40 s = ( s + 0.8) I out + 2 s Therefore V out = ( s + 0.8) - I out ( 29 - 2 s = Z th - I out ( 29 + V oc i.e., Z th = s + 0.8, and V oc = - 2 s . S OLUTION P ROBLEM 19.8. (a) Let the node voltages from left to right be V 1 , V 2 , and V out . Also inject a current I 3 into node 3. Writing nodal equations by inspection we have:

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Prbs Chap 19, 1/7/02 P19-4 © R. A. DeCarlo, P. M. Lin I in 0 I 3 = 1.5 - 1 - 0.25 - 1 2 - 0.5 - 0.25 - 0.5 0.0625 s + 0.75 V 1 V 2 V out Using equation 19.11, we have I 3 = W 22 - W 21 W 11 - 1 W 12 ( 29 V out + W 21 W 11 - 1 I in 0 = 0.0625 s + 0.75 - 0.25 0.5 [ ] 1 0.5 0.5 0.75 0.25 0.5 V out - 0.25 0.5 [ ] 1 0.5 0.5 0.75 I in 0 Thus I 3 = 0.0625 s + 0.375 ( 29 V out - 0.5 I in Therefore I sc = - I 3 ] V out = 0 = 0.5 I in . Further Z th ( s ) = 1 0.0625 s + 0.375 = 16 s + 6 .
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