# Two Miller integrator in series

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

• I'd start with trying to find the transfer function of the system. From this you can usually figure out the time constant. Commented Jun 16, 2017 at 20:56
• 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. Commented Jun 16, 2017 at 20:59
• @klamb which is actually precisely what I did. Isn't it 1/(s*tau) where s=jw? For Vo1 that is. Commented Jun 16, 2017 at 21:00
• Hint: Part b) question is similar to Part a) answer Commented Jun 16, 2017 at 21:08
• 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.
– G36
Commented Jun 17, 2017 at 14:22

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.

• you're suggesting that in order to determine the transfer function, right? Commented Jun 16, 2017 at 21:39
• 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.
– Nino
Commented Jun 16, 2017 at 21:43
• Doing that I still get $j\omega \tau = 1$ Commented Jun 16, 2017 at 21:47
• I am suggesting you solve it in the time domain, given that you are given graphs in the time domain.
– Nino
Commented Jun 16, 2017 at 22:36
• Well, I know that (1-0)/R1=(0-(-1))/sC1, is that better? How is that helpful? Commented Jun 16, 2017 at 22:39

$$\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.