# Find the period of a signal with the DTFT plot

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

$$\textrm{DTFT}\{e^{jn\omega_0}\}=2\pi\delta(\omega-\omega_0)\tag{1}$$
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.
• 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 ? – Yoann A. Aug 4 at 15:03
• @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. – Matt L. Aug 4 at 15:31
• 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. – Yoann A. Aug 4 at 16:38