Question 4 Answer: For the first amplifier: V+ = 0 [0-f(t)]/4M + [0-X(t)]/1M = 1µF d(Y-0)/dt -f(t) 0.25×10-6 X(t) ×10-6 =1×10-6 dY/dt – dY(t)/dt = -0.25f(t) X(t) ---(1) – For the second amplifier: Z/Y = -2M/.1M = -20 Z(t) = -20 Y(t) ---(2) For the 3rd amplifier: V+ = 0 [0-Z(t)]/0.05M = 0.1µF d(X-0)/dt -Z(t)] 20×10-6 = 0.1×10-6 dX/dt dX(t)/dt = -200 Z(t) ---(3) Combine (1)-(3) d2x/dt2 +4000x +1000f =0
Question 4 Answer: For the first amplifier: V+ = 0 [0-f(t)]/4M + [0-X(t)]/1M = 1µF d(Y-0)/dt -f(t) 0.25×10-6 – X(t) ×10-6 =1×10-6 dY/dt dY(t)/dt = -0.25f(t) – X(t) ---(1) For the second amplifier: Z/Y = -2M/.1M = -20 Z(t) = -20 Y(t) ---(2) For the 3rd amplifier: V+ = 0 [0-Z(t)]/0.05M = 0.1µF d(X-0)/dt -Z(t)] 20×10-6 = 0.1×10-6 dX/dt dX(t)/dt = -200 Z(t) ---(3) Combine (1)-(3) d2x/dt2 +4000x +1000f =0