Square Wave Fourier Series

[Jay-qu]

So I did an experiment today at uni that involved integrating a square wave AC to become a triangle wave AC.

 

The integration was done by using a low-pass filter. The explanation given was: A square wave is a Fourier series of sin and/or cos waves and by filtering out all the high frequencies from this series it is the equivalent of integrating the square wave and hence a triangular wave will be produced.

 

Me been the keen and astute physics student asked 'but why?' and the only answer i got was 'because' not a word that i like to hear in physics! I demand mathematical proof - so I have set out to find some, but dont really know where to start - any volunteers :)

 

J

[Qfwfq]

I demand mathematical proof - so I have set out to find some, but dont really know where to start - any volunteers :D

Try starting with:

 

[math]\frac{d}{dt}\sin\omega t=\omega\sin\omega t[/math]

 

How can the filter be approximated? ;)

[Jay-qu]

sorry Q but doesnt a sin differentiate to a cos ;)

 

this makes sense though, since a square wave odd function consisting of sine wave componants will integrate to an even function composed only of cos componants.

 

Fourier series for a square wave is:

 

[math] V(t) = \frac{4V_o}{\pi}(\sin\omega t + \frac{1}{3}\sin3\omega t + \frac{1}{5}\sin5\omega t + ... )[/math]

 

integrating this:

 

[math]\int{V(t)dt} = \frac{4V_o}{\pi}\int{(\sin\omega t + \frac{1}{3}\sin3\omega t + \frac{1}{5}\sin5\omega t + ... )dt}[/math]

 

[math]= -\frac{4V_o}{\pi}(\frac{1}{\omega}\cos\omega t + \frac{1}{9 \omega}\cos3\omega t + \frac{1}{25 \omega}\cos5\omega t + ... )[/math]

[Qfwfq]

:doh: alright, alright! :hihi:

 

[math]\frac{d}{dt}\sin\omega t=\omega\cos\omega t[/math]

 

I missed the detail on the spur of the moment! How about:

 

[math]\frac{d}{dt}\sin\omega t=\omega\sin(\omega t+\frac{\pi}{2})[/math]

 

[math]\frac{d}{dt}e^{i\omega t}=i\omega e^{i\omega t}[/math]

 

or whatever...

[Jay-qu]

sorry Ive been updating my above post, still playing around with the latex and math tags

[Qfwfq]

Right, it might however be easier if you keep the factors separate:

 

[math]3\cdot3\omega,\: 5\cdot5\omega[/math] etc. or even just call them [math]\omega_i[/math] in the sum.

[Jay-qu]

Right, it might however be easier if you keep the factors separate:

 

[math]3\cdot3\omega,\: 5\cdot5\omega[/math] etc. or even just call them [math]\omega_i[/math] in the sum.

ok, thanks.

 

So by truncating the series and cutting off the higher frequency sine terms we are effectively approximating the triangular wave with a finite number of cos terms..

 

Is that right, is that what I should be proving?

[Qfwfq]

Actually, truncating the series isn't necessary, it's only a matter of the filter's behaviour and from the above you can easily deduce what behaviour would ideally be required to exactly give an integral.

 

A simple low-pass given by an RC cell will never be perfect but you could work out why it'll be a good approximation if the sqare-wave's period is brief compared with the RC product. A much better approximation can be made with an op-amp, the configuration is even called "integrator".

[Jay-qu]

Yeah the circuit was called an integrator, I see the damn thing working - but mathematically 'how' its working is proving elusive to me :)

 

So its not the fact that the filter cuts out high frequency, its just the by product of the circuit doing so is to also smooth out voltage changes?

[Erasmus00]

Yeah the circuit was called an integrator, I see the damn thing working - but mathematically 'how' its working is proving elusive to me :)

 

So its not the fact that the filter cuts out high frequency, its just the by product of the circuit doing so is to also smooth out voltage changes?

 

Yes indeed. Electronically what is happening is that the capacitor in the circuit (along with your feedback) integrates the current form being fed to it. This integration smooths out rapid fluctuations.

 

Obviously, the integral of a square wave is a triangle wave (the integral of a constant is a line).

 

Its worth noting that you can build the opposite circuit by putting the capacitor in an integrating circuit BEFORE the op-amp. This differentiating circuit is worthless in practice as transient noise spikes quickly swamp any real signal.

-Will