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How may I deduce the time constants from these output waveforms? For \$V_{o1}\$ may I write \$\tau=1/2\pi f\$, where \$f=1/T=1/1ms=1KHz\$?

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  • \$\begingroup\$ I'd start with trying to find the transfer function of the system. From this you can usually figure out the time constant. \$\endgroup\$
    – klamb
    Commented Jun 16, 2017 at 20:56
  • \$\begingroup\$ Hit your local university's engineering library and dig out an OLD textbook on analog computers. What you have here are two absolutely classic analog computer integrators. \$\endgroup\$ Commented Jun 16, 2017 at 20:59
  • \$\begingroup\$ @klamb which is actually precisely what I did. Isn't it 1/(s*tau) where s=jw? For Vo1 that is. \$\endgroup\$
    – peripatein
    Commented Jun 16, 2017 at 21:00
  • \$\begingroup\$ Hint: Part b) question is similar to Part a) answer \$\endgroup\$ Commented Jun 16, 2017 at 21:08
  • \$\begingroup\$ If I assume R1 = 1kΩ for 0s to 0.5ms the current is equal to I = 1V/1kΩ = 1mA and the capacitor voltage need to change from 0V to -1V Hence C = Q/V = (I*t)/V = (1mA *0.5ms)/1V = 500nF, so the time constant is 0.5ms. And the time constant for a second integrator is also equal to 0.5ms. Because if you do the integration you will found that Vo2 = 0.5V at T = 0.5ms only when t2 = 0.5ms; for t2 = 1ms --->Vo2 = 0.25V; For t2 = 0.25mS ---->Vo2 = 1V. \$\endgroup\$
    – G36
    Commented Jun 17, 2017 at 14:22

2 Answers 2

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We know the inverting terminal of the opamp is going to have the same voltage as the non-inverting terminal.

Given that information, you can calculate the current across the resistor which is the current into the capacitor (since no current flows into the op-amp inverting terminal) and then you can go from there.

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  • \$\begingroup\$ you're suggesting that in order to determine the transfer function, right? \$\endgroup\$
    – peripatein
    Commented Jun 16, 2017 at 21:39
  • \$\begingroup\$ The question doesn't ask for that. It asks for time constants. I am telling you how to calculate the capacitance and resistance as to calculate that. \$\endgroup\$
    – Nino
    Commented Jun 16, 2017 at 21:43
  • \$\begingroup\$ Doing that I still get \$j\omega \tau = 1\$ \$\endgroup\$
    – peripatein
    Commented Jun 16, 2017 at 21:47
  • \$\begingroup\$ I am suggesting you solve it in the time domain, given that you are given graphs in the time domain. \$\endgroup\$
    – Nino
    Commented Jun 16, 2017 at 22:36
  • \$\begingroup\$ Well, I know that (1-0)/R1=(0-(-1))/sC1, is that better? How is that helpful? \$\endgroup\$
    – peripatein
    Commented Jun 16, 2017 at 22:39
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\$dV/dt=1V/0.5~ms~= 2V/ms~~=I_C/C\$,
If C=1uF , At t=0.5ms Vo1=1V, Ic=200uA=1V/R1, R1=5k then
R1C1 = 5ms from the 1st step.

At t=1ms Vo1 averages 0.5V and if C2 =1uF
with Vo2(1ms) = 1V Then R2= 1V/1ms = 1k so R2C2= 1ms from the 2nd input -ve step.

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