Confusion in deriving formula for fourier tansform of impulse train

Question

I was trying to derive fourier transform for impulse train :

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

Answer 1

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.

Answer 2

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 $(3)$.