Let $X(\omega)$ be the DTFT of the sequence $x[n]$ given by: $$ x[n] = \{4, 2, -1, 5, -3, 1, -2, 4, 2\},\quad\text{for}\quad n \in [-6, 2] $$ I do want to compute

  1. $X(0)$
  2. $X(\pi)$
  3. $\displaystyle\int_{-\pi}^{\pi} X(\omega) d{\omega}$
  4. $\displaystyle\int_{-\pi}^{\pi}|X(\omega)|^2 d{\omega}$
  5. $\displaystyle\int_{-\pi}^{\pi} \bigg\lvert\frac{dX(\omega)}{d{\omega}}\bigg\rvert^2 d{\omega}$

What I have tried

  • For $X(0)$, I computed the sum and so it was 12.
  • For $X(\pi)$ I did $X(\pi) = x[9] = \sum_{n = 0}^{8} x[n]e^{-i\omega n} = (-1)^9 \times 12 = -12$, because 0 corresponds to 0, 2 to $\pi/8$, -1 to $\pi/4$, ..., 2 to $\pi$.
  • For $\int_{-\pi}^{\pi} X(\omega) d{\omega}$, I tried by Parseval Theorem, so it was $2\pi x[n] x[n]*$, but I don't know if it's correct. Last two I don't have any idea of how to do it.

I also have a similar exercise, but it is with DFT instead of DTFT, but it is pretty the same, isn't it?

Thanks in advance.

  • $\begingroup$ Can you include in your question (meaning edit your question instead of answering this comment with another comment) what exactly you know to be $X(\omega)$? You have told us what $x[n]$ is but I I am having a hard time figuring out what the DTFT of the sequence is. $\endgroup$ Dec 30, 2020 at 15:55
  • $\begingroup$ I edited it. Sorry, it must have been because of my English. @DilipSarwate $\endgroup$
    – Anonymous
    Dec 30, 2020 at 16:28

1 Answer 1


For the first two questions you just need to use the definition of the DTFT


and use $\omega=0$ and $\omega=\pi$, respectively.

For 3. you just need to use the definition of the inverse DTFT:


Finally, for 4. and 5. you need Parseval's theorem. In addition, for 5. you need to express the sequence corresponding to the derivative of $X(\omega)$ in terms of $x[n]$. You should be able to either derive this yourself or look it up in a table.

  • $\begingroup$ I wrote it wrong, X(0) = 12 and X(\pi) = -12. Thanks. $\endgroup$
    – Anonymous
    Dec 30, 2020 at 17:09
  • $\begingroup$ @Danny: That's right. $\endgroup$
    – Matt L.
    Dec 30, 2020 at 17:11
  • $\begingroup$ @Danny: If the answer was helpful you can accept it by clicking on the green checkmark, thanks! $\endgroup$
    – Matt L.
    Dec 30, 2020 at 21:42

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