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)
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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 $\endgroup$– Laurent DuvalOct 19, 2019 at 17:24
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$\begingroup$ 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_HBOct 19, 2019 at 18:15
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$\begingroup$ 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 $\endgroup$– Laurent DuvalOct 19, 2019 at 18:18
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1$\begingroup$ ok, I'll edit the post and add my work. $\endgroup$– John_HBOct 19, 2019 at 18:25
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1$\begingroup$ You can check it now if it is correct $\endgroup$– John_HBOct 19, 2019 at 18:36
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1 Answer
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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 :
$$y[n] = \text{N-point IDFT}\{ \text{N-point DFT}\{ x[n] e^{-j \frac{\pi}{N}n}\} \} $$
$$ y[n] = x[n] e^{-j \frac{\pi}{N}n} $$
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$\begingroup$ 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_HBOct 19, 2019 at 20:54
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$\begingroup$ because $x[n] = 0$ for $n=N,N+1,...2N-1$... Hence $$\sum_{n=0}^{2N-1} x[n] W_N^{kn} = \sum_{n=0}^{N-1} x[n] W_N^{kn} $$ $\endgroup$– Fat32Oct 19, 2019 at 21:09