Learning DSP on my own time. Can't figure out the proof for DFT shift theorem which states the following:

Given, $x[n]$ to be a periodic with period $N$, $\text{DFT}\{x[n]\} = X[k]$, then $$ DFT\{x[n-a]\} = e^{-j\frac{2\pi}{N}a}X[k] $$

I found a proof here, but I can't figure out how did they leap from $$\sum_{m = -\Delta}^{N-1-\Delta}{e^{-j\frac{2\pi}{N}k(m+\Delta)}x[m]}$$ to $$ \sum_{m = 0}^{N-1}{e^{-j\frac{2\pi}{N}km} e^{-j\frac{2\pi}{N}k\Delta} x[m]} $$

Specifically how did they change the limits of the summation without changing the variable?

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    $\begingroup$ By the way, you haven't accepted a single answer to any of the questions you've asked so far on this site. Why is that? $\endgroup$
    – Matt L.
    Jan 26, 2018 at 8:27
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    $\begingroup$ @MattL. For some questions I can't accept an answer because it doesn't the answer doesn't fully satisfy me however when I add answers from other people I do get a full picture. However, I did upvote the answers that I liked $\endgroup$
    – flashburn
    Jan 26, 2018 at 20:18

1 Answer 1


For an $N$-periodic sequence $y[n]$, the sum over one period is always the same, no matter how you choose the first index:


for any integer $\Delta$. Try to verify that the sequence you sum over is indeed $N$-periodic.

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    $\begingroup$ i'm glad someone else here seems to agree with me that the DFT and the Discrete Fourier Series are the same damn thing. $\endgroup$ Jan 27, 2018 at 17:52

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