ECE 201 Spring 2010Homework 21 SolutionsProblem 31(a)Using voltage division,vC(0−) =20R50RV0= 0.4V0= vC(0+)(b)The Thevenin equivalent is given byVoc= V0R3R3+ (R1||R2)= 0.8V0Rth= R1||R2||R3= 4R(c)vC(∞) = 0.8V0vC(0+) = 0.4V0⇒ vC(t) = vC(∞) + [vC(t0+) − vC(∞)]e−t−t0RthC= 0.8V0− 0.4V0e−t4RC1(d)vC(T −) = vC(T +) = 0.8V0− 0.4V0e−1.5= 0.71075V0(e)τ = RthC= 4RC(f)Using the same relation as in part (c), with t0= T ,vC(t) = 0.71075V0e−t−T4RC(g)For simplicity, let all quantities be normalized to 1. Then the plot for vC(t)would look as shown on the next page.Problem 33(a)iL(0+) = iL(0−) =−1080 + 20= −0.1 A20 5 10 15 20 2500.10.20.30.40.50.60.70.8tvC(t)Plot of vC(t) for 0<=t<=4TFigure 1: Plot of vC(t) for problem 31(b)Here we have to consider two intervals, 0 ≤ t < 80 ms and 80 ms ≤ t <160 ms. For 0 ≤ t < 80 ms,iL(t) = iL(∞) + [iL(0+) − iL(∞)]e−tR/L=20100+ [−0.1 − 0.2]e−25t= 0.2 − 0.3e−25tiL(80−) = iL(80+) = 0.2 − 0.3e−2= 0.1594 AAgain, for the interval 80 ms ≤ t < 160 ms,iL(t) = iL(∞) + [iL(80+) − iL(∞)]e−(t−80)R/L= −0.1 + [0.1594 + 0.1]e−25(t−80)= −0.1 + 0.2594e−25(t−80)We can write vout(t) as the following,vout(t) = vin(t) − R1iL(t)3= 4 + 24e−25t, 0 ≤ t < 80 ms= −2 − 20.752e−25(t−80ms), 80 ms ≤ t < 160 ms0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16−30−20−100102030t(seconds)vout(t)(volts)Plot of vout(t) for 0<=t<=160msFigure 2: Plot of vout(t) for problem
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