About the proof of an equality related to the DFT [sampling the DTFT to obtain the DFT]

Question

This wiki page about the DTFT says that the DFT can be obtained from the DTFT by sampling the latter in one cycle at $N$ points:

When the DTFT is continuous, a common practice is to compute an arbitrary number of samples (${{N}}$) of one cycle of the periodic function (${X_{1/T}}$):\begin{align} \underbrace{X_{1/T}\left(\frac{k}{NT}\right)}_{X_k} &= \sum_{n=-\infty}^\infty x[n]\cdot e^{-i 2\pi \frac{kn}{N}} \quad \quad k = 0, \dots, N-1 \\ &= \underbrace{\sum_{N} x_{_N}[n]\cdot e^{-i 2\pi \frac{kn}{N}},}_{DFT}\quad \scriptstyle{\text{(sum over any }n\text{-sequence of length }N)} \end{align} where $x_{_N}$ is a periodic summation: $$x_{_N}[n]\ \triangleq\ \sum_{m=-\infty}^{\infty} x[n-mN]$$

So we essentially have the following equality $$\sum_{n=-\infty}^\infty x[n]\cdot e^{-i 2\pi \frac{kn}{N}}= \sum_{n=0}^{N-1} \sum_{m=-\infty}^{\infty} x[n-mN] e^{-i 2\pi \frac{kn}{N}}. $$

I didn't manage to prove it, is this a known identity?

Answer 1

The right hand side is the periodic repetition of the left, tells the last equation.

Starting from Fourier transform $$X(\omega)=\sum_{n=-\infty}^{\infty}x(n)e^{-j\omega n}$$ and $X(\omega)$ is periodic with period $2\pi$. So only the samples in the fundamental frequency range are sufficient.

Now we take $N$ equi-distance samples in the iterval $0\leq\omega\leq 2\pi$, so the spacing will be $\dfrac{2\pi}{N}$

Let $k$ be the sample index and $k=0,1,\dots,N-1$ and we evaluate $\omega$ at $\left(\dfrac{2\pi k}{N}\right)$

$$X\left(\dfrac{2\pi}{N}k\right)=\sum_{n=-\infty}^{\infty}x(n)e^{-j2\pi kn/N}$$

which can be written as

$$X\left(\dfrac{2\pi}{N}k\right)=\sum_{l=-\infty}^{\infty}\sum_{n=lN}^{lN+N-1}x(n)e^{-j2\pi kn/N}$$

Now change the index $n$ to $n-lN$ which results in

$$X\left(\dfrac{2\pi}{N}k\right)=\sum_{n=0}^{N-1}\left[\sum_{l=-\infty}^{\infty}x(n-lN)\right]e^{-j2\pi kn/N}$$

The sequence inside the square brackets is the periodic repetition of $x(n)$ of period $N$

$$x_p(n)=\sum_{l=-\infty}^{\infty}x(n-Nl)$$

Thus $$x_p(n)=\begin{cases} x(n),\;\;\;\;0\leq n \leq N-1\\ \\ 0,\;\;\;\;\;\;\;\;\;\;\; \mathrm{elsewhere}\\ \end{cases}$$