# Why DFT exact for periodic signals?

I have in my course notes:

$$\sum_{k=0}^{N-1}y(k)e^{-j\omega_nk}\approx\sum_{k=-\infty}^{\infty}y(k)e^{-j\omega_nk}$$

Where the $\approx$ for some reason becomes $=$ when $y(k)$ is periodic. Could you please explain why this is the case?

Thank you so much!

It's not correct to say that both expressions become equal when the signal is periodic. What is correct is that - obviously - everything one can know about an $N$-periodic signal is represented by the expression on the left of your equation, and consequently, both expressions are equivalent in the sense that one can be obtained from the other.

Note that that the infinite sum on the right-hand side does not converge in the conventional sense if the signal is periodic. The Discrete-Time Fourier Transform (DTFT) - that is what the right-hand side represents - of an $N$-periodic signal can be expressed by using Dirac delta impulses:

$$X(\omega)=\frac{2\pi}{N}\sum_{k=-\infty}^{\infty}X[k]\delta\left(\omega-\frac{2\pi k}{N}\right)\tag{1}$$

The coefficients $X[k]$ can be obtained from the expression on the left-hand side of your equation be setting $\omega_k=2\pi k/N$. In that sense, both expressions convey the same information.

• Alright, awesome. That was the root of my confusion, actually. For me, the right hand side (DTFT) will give infinite amplitude at the frequencies that $x(k)$ contains - because we keep on hitting those frequencies in each period. Is that correct? – space_voyager Oct 25 '15 at 10:59
• @space_voyager: I'd put it like that: a periodic signal is made up of sinusoids, the DTFT of which is represented by Dirac impulses. – Matt L. Oct 25 '15 at 11:07
• I obtained quite a different answer over on Math.stackexchange: math.stackexchange.com/questions/1496150/… Could you explain the discrepancy between what you are saying with respect to the answer in the link above? – space_voyager Oct 25 '15 at 12:25
• @space_voyager: I only agree with the second of the 3 conclusions in that answer over at math.SE. As for the first one, I don't see how the right-hand side (the DTFT) is approximate, unless you use it to compute the FT of some continuous function, but that was never even mentioned. The third conclusion is wrong because periodic signals do have a DTFT if we allow Dirac impulses. It's not necessary and also not common to change the definition of the DTFT for periodic signals. The second conclusion is trivially true. – Matt L. Oct 25 '15 at 12:49
• This document isites.harvard.edu/fs/docs/icb.topic196302.files/9-dtft.pdf explains basically what you wrote, just thought I'd share it :) – space_voyager Oct 25 '15 at 13:06