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I have a problem with expressing odd samples of X2 in terms of X1. I understand that the resulting DFT will be more precise in terms of expressing the exact spectrum of signal x[n], due to more samples. Moreover I know that even samples of X2(k) are copy of the spectrum of X1(k), however I do not know how to mathematically compute the odd ones.

The task is as follows:

Let $x[n]$ be a periodic sequence with period $N_1$ . Thus $x[n]$ is also periodic for period $N_2 = 2 N_1$ . We may compute $X_1[k]$: $N$-point DFT of $x[n]$ and $X_2[k]$: $2N$-point DFT of $x[n]$.

  • Express $X_2$in terms of $X_1$ Hint: it is easy with even samples $X_2[2m]$ , harder with odd ones $X2_[2m+ 1]$ , for $m$ integer.
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Consider a sequence $x[n]$ of length $N$ whose $N$-point DFT is $X[k]$. Then let $X_2[k]$ be the $2N$-point DFT of $x[n]$.

As you have stated, the even indexed samples of $X_2[k]$ will be easily shown to be: $$X_2[k] = X[k/2] ~~~,~~~k = 2m, m=0,1,...,N-1$$

Then the odd indexed samples of $X_2[k]$ will be given by the $N$-point DFT of the signal $x[n] e^{-j \frac{\pi}{N} n } $, $n=0,1,...,N-1$.

$$X_{2}[k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{\pi}{N} n} e^{-j \frac{2\pi}{N} k n} = X[k + 0.5] ~~~~ ,~~~ k =2m+1, m=0,1,...,N-1 $$

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    $\begingroup$ Thank you, I forgot about the shift in time property of DFT for solving this one. $\endgroup$ – Celeborth Nov 25 '18 at 12:08
  • $\begingroup$ Can you explain why we use here $$e^{-j\frac{\pi}{N}n}$$ and not $$e^{-j\frac{2\pi}{N}n}$$? $\endgroup$ – Celeborth Nov 25 '18 at 12:18
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    $\begingroup$ By elaborating on the DFT expression you can get it... $\endgroup$ – Fat32 Nov 25 '18 at 12:21

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