I How to derive the Fourier transform of a comb function

[arcTomato]

TL;DR Summary

Fourier transform

Dear all.
I'm learning about the discrete Fourier transform.

$I(\nu) \equiv \int_{-\infty}^{\infty} i(t) e^{2 \pi \nu i t} d t=\frac{N}{T} \sum_{\ell=-\infty}^{\infty} \delta\left(\nu-\ell \frac{N}{T}\right)$

this $i(t)$ is comb function
$i(t)=\sum_{k=-\infty}^{\infty} \delta\left(t-\frac{k T}{N}\right)$.

I would like to see how to derive $I(ν)$.(Especially the part about transformation to ##lN/T from kT/N)
If you can teach me, please.
Thank you.

 

[mitochan]

Hi.
I(\nu)=\sum^\infty_{k=-\infty} \int^\infty_{-\infty}\delta(t-\frac{kT}{N})e^{2\pi\nu it}dt=\sum^\infty_{k=-\infty}e^{2\pi\nu i kT/N}=\sum^\infty_{l=-\infty}\delta(\nu T/N-l)=\frac{N}{T}\sum^\infty_{l=-\infty}\delta(\nu -\frac{lN}{T})

 

[arcTomato]

Thanks for reply @mitochan.
I cannnot understand what is going on this part

=\sum^\infty_{k=-\infty}e^{2\pi\nu i kT/N}=\sum^\infty_{l=-\infty}\delta(\nu T/N-l)

Could you teach me about this detail??

 

[mitochan]

RHS says $\nu T/N$ must be an integer. If not LHS =0 due to summation of various phase numbers of magnitude 1. Sumamtion in RHS says any integer is OK.

 

[arcTomato]

ok thank you.
I think I got it ;>