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July 18th, 2014, 01:56 PM  #1 
Newbie Joined: Jul 2014 From: Marshall Islands Posts: 2 Thanks: 0  fourier transform of a square wave and rectangle function
Hello, I'm trying to work something out. If you take the Fourier transform of a rectangle pulse, say amplitude of 1 from 1/2 to 1/2, you get a sinc function (not normalized) with zero crossings at pi*k where k is an integer, right? (I'm using time domain here.) Now, if you want to do the Fourier transform of a square wave, my approach is to use the convolution integral on the above rectangle pulse by an impulse train spaced at 2 to create the square wave. Correct? dirac(t2k)? Then Fourier transform of the impulse train gives another pulse train spaced at integer multiples of 2k*pi/2 = k*pi??? I keep getting this answer, but it seems to me, I should be getting pulses at k*pi/2, since if you are convolving in the time domain to turn the rectangle into a square wave, you should be sampling in the frequency domain at odd multiple frequencies. The even multiples should fall on the sinc zero crossings...just like a square wave is created by adding an infinite amount of odd harmonics. The fact that all my harmonics keep falling on the zero crossings is obviously wrong. I just can’t figure what I’m doing. I’m thinking, if the impulse train were just as wide as the rectangle function, it would turn the function into a constant amplitude of 1 which would be the Dirac function at 0 in the frequency domain, and that’s when you would have all your impulse train harmonics falling on the zero crossings. Am I missing something or just getting something basic wrong about pulse and Dirac spacing? Thanks! 
July 18th, 2014, 10:21 PM  #2  
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications 
Hi islandboy, and welcome to the forums. Quote:
Quote:
The good news is that your reasoning is exactly right; since convolution in the time domain is multiplication in the frequency domain (and viceversa), the zeroes of the sinc function should fall on the even multiples of the impulse train, leaving no even harmonics. So let's work the problem using $\omega$ throughout. The unit height rectangular pulse has a FT of: $\displaystyle \large T \frac{\sin \left(\frac{\omega T}{2}\right)}{\frac{\omega T}{2}}$ and since $T=1$, this simplifies to: $\displaystyle \large \frac{\sin \left(\frac{\omega}{2}\right)}{\frac{\omega}{2}}$ We can see that the zeroes occur at $\omega=2k\pi$ due to the 2 in the denominator. For the impulse train: $\displaystyle \huge \mathcal{F} \large \left(\sum_k \delta (tkT)\right) = \frac{2\pi}{T} \sum_k \delta \left(\omega\frac{2k\pi}{T}\right)$ Since T=2 we get: $\displaystyle \huge \mathcal{F} \large \left(\sum_k \delta (t2k)\right) = \pi \sum_k \delta \left(\omegak\pi \right)$ And we see that the impulses occur every $k\pi$ as you stated. So the sinc function zeroes are at $\omega=2k\pi$ and the impulses are every $\omega=k\pi$. So there are no even harmonics (other than DC). _______________________________ I think that in the rectangular pulse case, you substituted $2\pi f$ for $\omega$: $\displaystyle \large \frac{\sin \left(\frac{\omega}{2}\right)}{\frac{\omega}{2}}$ became $\displaystyle \large \frac{\sin \left(\frac{2\pi f}{2}\right)}{\frac{2\pi f}{2}}=\frac{\sin(\pi f)}{\pi f}$ And the zeroes are at $f=1,2,\ldots$ so your k's in this case refer to harmonics of $f$ instead of $\omega$. I hope this helps.  
July 21st, 2014, 01:29 PM  #3 
Newbie Joined: Jul 2014 From: Marshall Islands Posts: 2 Thanks: 0  Ok, I see!
Thanks JKS! You are right about the f versus 'w'. I'll practice more with the difference. I knew what the answer had to be, just couldn't get there.


Tags 
fourier, function, rectangle, square, transform, wave 
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