# DFT shift theorem proof

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?

• 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? – Matt L. Jan 26 '18 at 8:27
• @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 – flashburn Jan 26 '18 at 20:18

For an $N$-periodic sequence $y[n]$, the sum over one period is always the same, no matter how you choose the first index:
$$\sum_{n=0}^{N-1}y[n]=\sum_{n=\Delta}^{N-1+\Delta}y[n]$$
for any integer $\Delta$. Try to verify that the sequence you sum over is indeed $N$-periodic.