2
$\begingroup$

I keep hearing that the DFT assumes a periodic signal. E.g. the first answer in this MATLAB Q&A site. This doesn't make any sense to me. According to the derivations I've seen of the DFT one approximates the continuous FT by a Riemann sum as in $$H(f_n)=\int_{-\infty}^{\infty} h(t) e^{2\pi i f_n t} dt \approx \Delta \sum_{k=0}^{N-1} h_k e^{2\pi i k n }{N}$$

Which to me pretty much implies that the signal dies off and isn't repeating itself indefinitely.

So my question is: does the DFT assume signals that die off? Does it assume a continuous signal (if yes, why?) and reflect this assumption in $H(f_n)$?

If it dies off then why would the whole discussion on aliasing occur? Namely, it wouldn't make much sense for that discussion to arise since what determines the precision of the DFT would be doing a good Riemann sum, not the sampling rate and its relation to the bandwith of the signal.

Thanks!

$\endgroup$
  • $\begingroup$ The derivation begins with the Fourier expansion of a function $f$ on $[0,2\pi)$. See my answer here for the full details. So right off the bat, this assumes periodicity. $\endgroup$ – Chester Nov 3 '15 at 20:13
0
$\begingroup$

It shouldn't really matter. The DFT approximates the function by a finite number of points, and you can't tell from a finite number of points whether a function is periodic or decays.

$\endgroup$
  • $\begingroup$ You mean it approximates the integral? (As opposed to the function). I think it does matter since according to what I've programmed so far, if I transform $\cos(t)$ by assuming it exists in a big interval it gives me something different than with a small interval (perhaps I'm not recalling correctly). $\endgroup$ – DLV Nov 2 '15 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.