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If anyone can help solving this exercise I'll be grateful. It's urgent. (I've added my answer, but I think it's wrong)

  • 4
    $\begingroup$ We are glad to help when the question is completed with the first steps of the OP and where he/she is stopped. Being urgent (esp. for an homework job) might not be sufficient $\endgroup$ Oct 19, 2019 at 17:24
  • $\begingroup$ it's my first time posting such a qst, didn't know that my steps are required and even if i post them i knew that they are wrong $\endgroup$
    – John_HB
    Oct 19, 2019 at 18:15
  • $\begingroup$ If we understand where your steps go on the wrong direction, it's easier to answer. Wrong steps don't matter that much in questions $\endgroup$ Oct 19, 2019 at 18:18
  • 1
    $\begingroup$ ok, I'll edit the post and add my work. $\endgroup$
    – John_HB
    Oct 19, 2019 at 18:25
  • 1
    $\begingroup$ You can check it now if it is correct $\endgroup$
    – John_HB
    Oct 19, 2019 at 18:36

1 Answer 1


No it's not correct.

Derive it like this :

1-) $2N$-point DFT of $x[n]$ is: $X_{2N}[k] = \sum_{n=0}^{2N-1} x[n] e^{-j \frac{2\pi}{2N} k n }$

2-) $Y[k] = X_{2N}[2k+1]$ be the odd-indexed samples of $X_{2N}[k]$

3-) We are looking for $y[n] = \text{N-point IDFT}\{ Y[k]\}$.

4-) Elaborate on step-2 and step-1 to see that $Y[k] = \text{N-point DFT}\{ x[n] e^{-j \frac{\pi}{N}n}\}$

5-) Then from steps 3 and 4 we get :

$$y[n] = \text{N-point IDFT}\{ \text{N-point DFT}\{ x[n] e^{-j \frac{\pi}{N}n}\} \} $$

$$ y[n] = x[n] e^{-j \frac{\pi}{N}n} $$

  • $\begingroup$ I just have one more problem after developing the eqts, how can i get rid of the interval 0< n < 2N-1 to get N-point DFT so that 0<n<N-1 ? $\endgroup$
    – John_HB
    Oct 19, 2019 at 20:54
  • $\begingroup$ because $x[n] = 0$ for $n=N,N+1,...2N-1$... Hence $$\sum_{n=0}^{2N-1} x[n] W_N^{kn} = \sum_{n=0}^{N-1} x[n] W_N^{kn} $$ $\endgroup$
    – Fat32
    Oct 19, 2019 at 21:09

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