I have an exercise and I'm struggling to resolve it. Here it is :

enter image description here

My problem is about the DTFT. I've always been taught that we use DTFT for infinite-lenght signal that are not periodic (if the signal were periodic I would have used DFS)... So, how am I supposed to find the period of x[n] and y[n] ? I must have missed or misunderstood something. Can somebody explain it to me ? (obviously without giving the explicit answer). Thanks !


1 Answer 1


The discrete-time Fourier transform (DTFT) can be used for general infinite length signals, and it can also be used for periodic signals if we allow Dirac delta impulses in the expression for the DTFT. You have the correspondence


where it is understood that the expression $(1)$ is valid in the interval $[ -\pi,\pi)$, because the DTFT is always $2\pi$-periodic.

An $N$-periodic signal has Dirac impulses at multiples of $2\pi/N$. Note that not all of these Dirac impulses must be present, some of them can have a weight of zero. I'm sure you can take it from here.

  • $\begingroup$ Thank you for the answer ! If I follow what you've explained I find for the first one : the pic at 0 means that we have a constant component in the signal. Then, for the pic at $\pi/2$ : $\pi/2 = 2\pi / 4$. So, for the first one, I've answered 4. Following the same logic for the second example, I've answered 3 ($4\pi/6 = 2\pi/3$). However, this is not the good answer ... where am I doing a mistake ? $\endgroup$
    – Yoann A.
    Aug 4, 2019 at 15:03
  • $\begingroup$ @YoannA.: For the second example, the Dirac impulses are not at integer multiples of $2\pi /3$, right? So the period cannot be $3$. I guess you'll be able to figure out the solution now. $\endgroup$
    – Matt L.
    Aug 4, 2019 at 15:31
  • $\begingroup$ I think i got it ! Pics are at multiple of $\pi / 3 = 2\pi / 6$ so the answer is 6. Is that right ? PS : I'm sorry I can't up vote you on this stackExchange because I don't have enough points. $\endgroup$
    – Yoann A.
    Aug 4, 2019 at 16:38

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