ECE 201 Spring 2010Homework 29 SolutionsProblem 40(a)The following differential equation can be written using KCL at the invertingterminal of the op amp and using the virtual ground concept,vs(t)R1+vout(t)R2+ Cdvout(t)dt= 0vs(t) = −100u(t)= −100, t ≥ 0⇒ vout(t) =100R2R1(1 − e−t/R2C)= 400(1 − e−2t) mV, t ≥ 0Problem not from the bookIf we denote the voltage at the non-inverting terminal of the op amp asvL(t), then using the virtual ground concept, the output volta ge is givenby Vout(t) = 10vL(t). Also, vL(t) = VIN(t) − 1000iL(t). The problem thenreduces to finding iL(t). D ue to the square wave shape of VIN(t), the graphfor iL(t) will also be periodic with maximum and minimum values of x and−x (say). The task is to find the value of x, which can be done using thefollowing equation,iL(t) = iL(∞) + [iL(0) − iL(∞)]e−t/ττ =LR= 2 µsx = 2 × 10−4+ [−x − 2 × 10−4]e−0.5⇒ x = 0.489 × 10−41The plot of Vout(t) for 0 ≤ t ≤ 2µs is drawn below. Note that Vout(t) will beperiodic with period 2µs.0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2x 10−6−2.5−2−1.5−1−0.500.511.522.5t(seconds)Vout(t)(Volts)Plot of Vout(t) for 0<=t<=2us. Note that it is periodic with period
View Full Document