# Inverse discrete Fourier transform

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)

• 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 – Laurent Duval Oct 19 '19 at 17:24
• 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 – John_HB Oct 19 '19 at 18:15
• 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 – Laurent Duval Oct 19 '19 at 18:18
• ok, I'll edit the post and add my work. – John_HB Oct 19 '19 at 18:25
• You can check it now if it is correct – John_HB Oct 19 '19 at 18:36

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 :

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

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

• 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 ? – John_HB Oct 19 '19 at 20:54
• because $x[n] = 0$ for $n=N,N+1,...2N-1$... Hence $$\sum_{n=0}^{2N-1} x[n] W_N^{kn} = \sum_{n=0}^{N-1} x[n] W_N^{kn}$$ – Fat32 Oct 19 '19 at 21:09