DeCarlo Ch11 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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Complex Power Prbs 2/5/00 P11-1 @ DeCarlo & P. M. Lin CHAPTER 11. PROBLEM SOLUTIONS S OLUTION 11.1. Using equation 11.3, P av = 1 2 ( e t - 1) 2 Rdt = e 2 t 2 + t - 2 e t 0 1 = 0.758 0 1 W. S OLUTION 11.2. (a) From 11.6, P av = V m 2 2 R = 50 mW for a sinusoidal input. (b) From 11.3, P av = R 2 (10cos(10 t )) 2 dt 0 /20 + ( - 10cos(10 t )) 2 dt /20 3 /20 + (10cos(10 t )) 2 dt 3 /20 2 /10 = I m 2 R 2 = 50 mW just as the previous case since the square of the absolute cos(10t) is the same as the square of cos(10t). (c) P av = 10 R 2 0.01cos 2 (10 t ) [ ] 2 dt 0 2 /10 = 10 - 3 2 R 0.5cos(20 t ) + 0.125cos(40 t ) + 0.375 [ ] dt 0 2 /10 = 1 2 0.5 20 sin(20 t ) + 0.125 40 sin(40 t ) + 0.375 t 0 2 /10 = 37.5 mW (d) t=0:1/1000:1; R=1e3; pta= (0.01*cos(10.*t)).^2.*R; ptb= (0.01*abs(cos(10.*t))).^2.*R; ptc= (0.01*cos(10.*t)).^4.*R; subplot(3,1,1); plot(t,pta); grid ylabel('W'); subplot(3,1,2); plot(t,ptb); grid ylabel('W'); subplot(3,1,3); plot(t,ptc); grid ylabel('W'); xlabel('time in seconds');
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Complex Power Prbs 2/5/00 P11-2 @ DeCarlo & P. M. Lin 0 0.2 0.4 0.6 0.8 1 0 0.05 0.1 W TextEnd 0 0.2 0.4 0.6 0.8 1 0 0.05 0.1 W TextEnd 0 0.2 0.4 0.6 0.8 1 0 0.5 1 x 10 -5 W TextEnd time in seconds S OLUTION 11.3. (a) For figure a, the period is 2, and P av = 1 2 v 2 ( t ) R 0 2 dt = 1 2 R 400 + 100 ( 29 = 25 W. In figure b, the period is 1, and P av = 20 t ( 29 2 R 0 1 dt = 400 R t 3 3 0 1 = 13.3 W. (b)
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Complex Power Prbs 2/5/00 P11-3 @ DeCarlo & P. M. Lin 0 0.5 1 1.5 2 2.5 3 3.5 4 10 15 20 25 30 35 40 Time in s Instantaneous Power in W TextEnd 0 0.5 1 1.5 2 2.5 3 3.5 4 0 5 10 15 20 25 30 35 40 Time in s Instantaneous Power in W TextEnd S OLUTION 11.4. (a) For (a), looking the definition for the effective voltage, one sees graphically that the integral over one period, 2, of the squared waveform, is 500. Dividing by the period, and taking the square root, V eff = 15.81 V. For fig. b, V eff = 20 t ( 29 2 dt 0 1 = 400 1 3 = 11.55 V (b) P av = I eff 2 R = V eff 10 2 8 = 20 W. (c) P av = V eff 10 2 8 = 10.67 W
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Complex Power Prbs 2/5/00 P11-4 @ DeCarlo & P. M. Lin S OLUTION 11.5. (a) This can be done graphically quite easily. The period of fig a, is 9s. The total area of one period of the squared waveform is 75. This yields I eff = 75 9 = 2.89 A. In fig. b, the area over one period is 25 which yields I eff = 25 3 = 2.89 A. (b) Using current division, P av = I eff 60 90 2 30 = 111.36 W. (c) The same result as (b) is obtained since the effective current is the same. S OLUTION 11.6. (a) I eff = 1 2 e t - 1 ( 29 2 0 1 dt = 0.615 A. (b) P av = I eff 2 R = 0.758 W. S OLUTION 11.7. (a) V eff 2 = 20 2 10 + 2cos(20 t ) ( 29 2 0 2 /20 dt = 20 2 100 t + 2 t + 2 40 sin(40 t ) + 2sin(20 t ) 0 2 /20 = 102.01 Hence V eff = 10.1 V. (b) V eff 2 = 1 10cos(2 t ) + 5cos(4 t ) ( 29 2 0 dt = 1 62.5 t [ ] 0 = 62.568 Hence V eff = 7.91 V. (c) Without going into detailed calculation, note the following fact about v 3 2 ( t ) . Only the product terms that have the same frequency will produce a non-zero result when integrated. Thus the integral reduces to the following: V eff 2 = 1 100cos 2 (2 t ) + 25cos 2 (4 t ) + 25cos 2 (4 t - π /4) + 50cos(4 t )cos(4 t - π /4) ( 29 0 dt Hence V eff = 50 + 12.5 + 12.5 + 25cos( / 4) [ ] = 9.627 V S OLUTION 11.8. The voltage is V = 50 0 V and the impedance Z eq = 100 - 100 j = 141.42 ∠ - 45 o . Thus I = V Z eq = 353.6 45 o mA. Hence, P av = R I eff 2 = 100 × 0.3536 2 2 = 6.2516 W.
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Complex Power Prbs 2/5/00 P11-5 @ DeCarlo & P. M. Lin S OLUTION 11.9. (a)
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