
We
designed our integrator to accept a 2V peak to peak square wave. It
will swing up and down by 2V peak to peak (unity gain) at a frequency
of 10kHz. Plugging our values into the relationship Vc = 1/C Integrate[
Ic * dt], where Ic = Vin/R, we can set the capacitor to a particular
value to solve for the R which we need. Initially choosing a C to be
1uF led to the use of a 100Ohm resistor (wayyy to low). So I went an
order of magnitude lower to allow us to use a 50kOhm resistor instead
with a much smaller capacitor. Our output below came close to what we
designed. If we spent a lot more time messing with different R and C
values, we could have gotten exactly 4V, but we felt 3.8 or so was
accurate enough for this.