# Proof of DTFT equal to DFT when signal is periodic?

I was using the Wikipedia page on the discrete time Fourier transform to understand the connection between DFT and DTFT. The following is claimed in the article - I was wondering if anyone had a proof or a source for the proof for the highlighted part? Many thanks!

• It's not exactly equal. The DTFT of a periodic signal will have dirac delta impulses in it. The weighting factors on the dirac impulses are the numbers that come outa the DFT (perhaps with a scaling constant tossed in). Mar 5 at 17:02

Understanding the premise of the Fourier Series Expansion is a good starting point: Any single valued analytic function from $$0$$ to $$T$$ in time can be decomposed into an infinite series of sinusoids each with a frequency of $$n/T$$ where $$n$$ is any integer. Here is the key point:
The result of summing all those components will result in a waveform that matches the original function as expected from $$0$$ to $$T$$, but also since each of those components repeat themselves over that interval, then the original function will also repeat over every additional interval of $$T$$ in time! Consider that any other frequency other than these provided will not repeat as such, then they would disrupt the resulting repetition so therefore cannot exist.