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July 18th, 2014, 01:56 PM   #1
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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(t-2k)?

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!
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July 18th, 2014, 10:21 PM   #2
jks
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Hi islandboy, and welcome to the forums.

Quote:
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.)
Correct, except that last statement in parenthesis. By taking the FT you convert from the time domain to the frequency domain. What I think that you meant was that you are working with $f$ (in Hz) rather than $\omega$ (in radians/sec). In other words, k is a harmonic of $f=1$ Hz (you performed the steps below the line further down in this post).

Quote:
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(t-2k)?

Then Fourier transform of the impulse train gives another pulse train spaced at integer multiples of 2k*pi/2 = k*pi???
Correct, where k is a harmonic of $\omega$. So from what I can tell, the problem is that for the rectangular pulse you are working with $f$ and for the impulse train you are working with $\omega$.

The good news is that your reasoning is exactly right; since convolution in the time domain is multiplication in the frequency domain (and vice-versa), 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:

$\displaystyle \huge \mathcal{F} \large \left(\sum_k \delta (t-kT)\right) = \frac{2\pi}{T} \sum_k \delta \left(\omega-\frac{2k\pi}{T}\right)$

Since T=2 we get:

$\displaystyle \huge \mathcal{F} \large \left(\sum_k \delta (t-2k)\right) = \pi \sum_k \delta \left(\omega-k\pi \right)$

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.
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July 21st, 2014, 01:29 PM   #3
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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.
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