Question

Problem 8.61 The parameters for the circuit shown in the figure are R, 25 kQ, R2 = 100 kΩ and C2 = 1 μΡ. The power supply voltage for each op amp is 6 V. The amplifiers jumps from 0 to 250 mV att-0. No energy is stored in the feedback capacitors at the instant the Figure 〈 1of1 > R1 R2 C2 Ra Rb CC2 Voc2

Please solve parts E,F,G,H. No justification needed.

Answer 1

PART E The initial output voltage of first op-amp is, Vol (0)=-2V This is the initial voltage across capacitor Ci The initial output voltage of second op-amp is, (0) = 4 V This is the initial voltage across capacitor C, Draw the circuit in s-domain as shown below. Vcc V (s) Apply the Kirchhoff's Current Law at node 1. 100 x10500x10 0.1×10-6 S Va1(s) + 0.2x10-6 100x103 500x103 (s)| 0.01s|=V,(s)-0.02

ve(s) +0.02 oI (s) +0.01s Apply the Kirchhoffs Current Law at node 2. vo (0) 4 100x 1025 x10 10 S 有(s) 25x103 +4x10-5 V. (s) 7, (s)(-1-0.1s)-1%(s)-0.4 Substitute the expression for Voi(s) from equation (1) 100x 103 -1o6 , (s)(-1-0.1s) = _41 , (s) +0.02 +0.4 +0.01s ,(s(-0.001s -0.03s-02)-4(v (s)+0.02)+0.4 -0.001s V, (s)-0.03sV, (s)-0.2V, (s)-4g (s)-0.08+0.4 0.001sY, (s) +0.03 s, (s) + 0.2, (s)-4' (s)-0.32 s V, (s)+30sv. (s)+200v, (s)-4000, (s)-320 Apply the inverse Laplace transform +30 0+ 200v - 4000v -320 dt

For t20, V. 0.25 V Thus, 30+200v - 4000 (0.25)-320 dt +30 0+200v 1000 -320 dt di dt =-30 _ _ 200, 680 V/s* dt PART F Recall equation (2) g (S) G)-1-0.1)-G)+0.02 +0.4 0.01s For t20, V 0.25 V Apply the Laplace transform 0.25 Vz(s) Thus, 0.25 0.02 v,()-1-0.151-0.01s +0.4 0.25 -+ 0.02 0.4 v. (s) (1-0.1s)1+0.011+0.1s)

1+0.08s 0.4 (1+0.1s) 1+0.05 s(1+0.1s) 1 + 0.08s 0.4 s(1 +0.1s)(1+0.05s丿 5(1.6) (s+12.5 4 s(s+10+20) (s +10) (1 +0.1s) Apply the partial fractions 0.3 0.5 0.2 (s)=-- s+10 s+10 s+20 0.5 4.2 0.3 s S10 S20 Apply the inverse Laplace transform v.(t)-0.5-4.2 -0.3e-30rV for t20 -10t PART G Recall equation (1) (s) +0.02 0.01s -+0.01s 1.250.75 S S20 Apply the inverse Laplace transform voi (t)1.25-0.75e0 V for t2 0 1.25u(t)-0.75e 3"u(t) Vv

Differentiate vo1(t). "ol (0--1.25δ(t)-(-20)0.75e-20t u(t) dt --1 .25 δ(t) +15e-20t V/s PART H From part G, vol (t)1.25 -0.75e0 V for t2 0