Analysing system response in the time domain from Bode plot (intuition)

Question

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.

Answer 1

A rough calculation for the first harmonic at 2kHz - assuming a gain decrease of 40dB/dec above 700 Hz - results in a damping factor of app. 9.44 (19.5dB). Hence, the first harmonioc at 2kHz should have an amplitude of 10/9.44=1.06 volts.

Visual justification: At f=1kHz we have a damping of 7.5dB (factor 2.37)

EDIT1: Due to some incorrect numbers (obtained from the diagram) I have corrected the damping values slightly.

EDIT2: Simulation of a corresponding 2nd-order lowpass with a squarewave input results in sinusoidal signal (good quality) with an amplitude of app. 1.3 V. This is more than roughly calculated above - however, the filter data couldn`t be quite exact because the evaluation of the given magnitude response is not error-free (pole frequency?, 0 dB crossing of the asymptote or the actual curve ?).

Answer 2

Yes the data is approximate because all that was given was the bode plot shown above. The solution was provided today and I still have some trouble wrapping my head around it. Sorry it is in french and I am a little too lazy to retype and translate but the math should be self explanatory.

1) I don't understand how we can assume vout(T/2)=0

2) H(700) should be approximately 1 since the gain is zero around that frequency (crête means peak btw, so we are talking in terms of amplitude here)