The following question was asked : Given a Bode plot with a frequency response as illustrated by this Bode plot find the amplitude of vout for a symmetrical square wave inpute of +-10V at 2KHz. The Bode plot describes an asymptote of -40dB which crosses the 0dB line at 700Hz.
The attempt of a solution:
The fundamental of the input has pulsation of \$\omega=2\times10^{3}\times2\pi=4000\pi\$ . Which is way larger than the cutoff pulsation: \$\omega_{c}\approx300\times2pi=600\pi\$
We can therefore approximate H(s) by \$H(s)=\frac{R(s)}{E(s)}\approx\frac{1400\pi}{s^{2}}\$. .
If we analyse the response in the time domain we have \$r(t)=1400\pi\times10\int{}_{0}^{T/2}\int{}_{0}^{T/2}e(t)dt=1400\pi\times10\times\frac{T}{2}\int_{0}^{T/2}dt\$ with \$\frac{T}{2}=2.5\times10^{-4}\$
So we get \$r(t)=1400\pi\times10\times(2.5\times10^{-4})^{2}=0.00274889V \$
I saw an example for a simple integrator but I am not so sure this double integral, sould it be an indefinite integral then evaluated on the half period? I also have some intuition problem on going from the freq to the time domain, how would this work for a more complicated filter, say an elliptic one which has ripple in the band-stop. First question on the forum so sorry if some rules of etiquette have been broken here.