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I was trying to derive fourier transform for impulse train :

enter image description here

I know how to solve for this using using properties of fourier transform. But now I wanted to use a brute force approach to it so I did the following enter image description here

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  • $\begingroup$ Search for Fourier Transform of comb function. There are answers to that question. $\endgroup$ – Stanley Pawlukiewicz Oct 23 at 13:00
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What makes you think it's wrong? You're just not done yet. In your final expression, if $\omega n T $ is a multiple of $2\pi$, you'll sum "infinitely many ones" which gives you "infinite", whereas for other values, you're going to some numbers "equally spaced" over the complex unit circle, which gives you zero. This is exactly how a train of deltas behaves.

Of course this argument is very handwavy and somewhat dangerous as these intuitions may mislead us when making statements about infinite sums.

As a suggestion: try with a pulse train of the form $\sum_{n=-N}^N \delta(t-nT)$. Compute its Fourier transform. Then consider the limit $N\rightarrow \infty$. You should get something like $\sum_{n=-N}^N {\rm e}^{-j\omega n T}$, which you can simplify using the geometric series. Then, forming the limit over $N$ should be easier.

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  • $\begingroup$ Thank you so much for your prompt and intuitive answer Florian. I got confused looking at the result since it does not look like the well known answer given in my books :P. But I get it now. And as you suggested , i'll try the (N-> inf) way and do some simplifications. $\endgroup$ – Adarsh Oct 23 at 13:39
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One simple (simplistic) way to see that your result is correct is to realize that

$$\mathcal{F}\big\{\delta(t-nT)\big\}=e^{-j\omega nT}\tag{1}$$

So you could argue that

$$\mathcal{F}\left\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\right\}=\sum_{n=-\infty}^{\infty}e^{-j\omega nT}\tag{2}$$

which actually turns out to be correct.

The result you're looking for is

$$\mathcal{F}\left\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\right\}=\frac{2\pi}{T}\sum_{n=-\infty}^{\infty}\delta\left(\omega-n\frac{2\pi}{T}\right)\tag{3}$$

If you compute the Fourier series of the periodic function on the right-hand side of $(3)$ you end up with the right-hand side of $(2)$, which is the result you came up with.

A dual way of deriving the result is to first compute the Fourier series of the given Dirac comb:

$$\sum_{n=-\infty}^{\infty}\delta(t-nT)=\frac{1}{T}\sum_{n=-\infty}^{\infty}e^{jn2\pi t/T}\tag{4}$$

Computing the Fourier transform of the right-hand side of $(4)$ by transforming each element in the sum directly gives the expression on the right-hand side of $(3)$.

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  • $\begingroup$ Thanks for the answer Matt. I can certainly see the parallels between (2) and (3). But I'm not quite clear about the argument you make about getting fourier series of the RHS of (3). Why would i want to find fourier series of a result that is already in fourier transform form ? Sorry if i'm being silly :P $\endgroup$ – Adarsh Oct 23 at 14:08
  • $\begingroup$ @Adarsh: You can write down the Fourier series of (almost) any periodic function. If the function is in the time domain, so is it's Fourier series, if it's in the frequency domain, then its Fourier series is also in the frequency domain. It's just a different representation of the same function. I just wanted to point out that the result you came up with (RHS of Eq. 2) is the Fourier series of the result you were looking for (RHS of Eq. 3). So it's just a different representation of the same function. $\endgroup$ – Matt L. Oct 23 at 14:20
  • $\begingroup$ Oh, now i get it. Thanks for the clarification .Frankly i would have accepted both the answers if i could :P. Both of them are great :) $\endgroup$ – Adarsh Oct 24 at 4:20
  • $\begingroup$ I tried, but i don't have enough reputation. I'll do it once i have enough :) $\endgroup$ – Adarsh Oct 24 at 10:17
  • $\begingroup$ @Adarsh: OK, thanks! $\endgroup$ – Matt L. Oct 24 at 10:26

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