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

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.

• Hmm, We would need to see the bode plot and know the frequency of the square wave. – George Herold Feb 19 '15 at 19:21
• Your description of the Bode plot doesn't make sense to me. Do you mean that there's a horizontal asymptote at -40 dB, or that the slope is -40 dB/decade at high frequencies? What happens before and after 700 Hz? Are you using a linear approximation? If you can add a link to a picture of the Bode plot, somebody can edit it into your post. – Greg d'Eon Feb 19 '15 at 19:41
• Indeed the frequency of the square wave and my description of the bode plot wasn't too useful, I added both to the question. The reasoning behind just stating the asymptote was that at frequencies which are much higher than the cutoff frequency the transfer function can be approximated by a simple integrator. Kynit is that what you mean by linear approximation? – SolipsistElvis Feb 19 '15 at 20:31
• I think you are using 'asymptote' synonymously with 'rolloff'. The filter is two-poles so it rolls off at -40 dB per decade. Your approach is somewhat sound, but rather than using the integral in the time domain, just substitute $$s = j\omega_o$$ into your approximation for the rolloff portion of the Bode plot. $$\omega_o$$ being the frequency of your input signal. Then take the magnitude to get the gain at that frequency. – docscience Feb 20 '15 at 19:58

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 ?).

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)