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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?

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  • $\begingroup$ Be very suspicious of that wiki material. If you go back to the definition of what is a periodic sequence and what is the definition of the DFT, you will realize that is impossible to perform the DFT on a periodic sequence. $\endgroup$ – Richard Lyons Feb 8 at 11:51
  • $\begingroup$ I think you're right, the article shouldn't have labeled the quantity over the bracket as a DFT, the DFT is a sum over $x(n)$, not $x_{N}(n)$. $\endgroup$ – Hilbert Feb 8 at 12:13
  • $\begingroup$ No, actually the article is correct. It is a DFT because as jomegA showed below, $x(n)$ is exactly $x_{N}(n)$ over ${0,1,..., N-1}$. $\endgroup$ – Hilbert Feb 8 at 12:19
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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 $$x_p(n)=\begin{cases} x(n),\;\;\;\;0\leq n \leq N-1\\ \\ 0,\;\;\;\;\;\;\;\;\;\;\; \mathrm{elsewhere}\\ \end{cases}$$

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  • $\begingroup$ Sorry but the question wasn't about the periodization equation, rather about why this equation is true:$$\sum_{n=-\infty}^\infty x[n]\cdot e^{-i 2\pi \frac{kn}{N}}= \sum_{n=0}^{N-1} \sum_{m=-\infty}^{\infty} x[n-mN] e^{-i 2\pi \frac{kn}{N}}. $$ $\endgroup$ – Hilbert Feb 8 at 10:13
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    $\begingroup$ The equations on the left and right are the same, one is periodic extension of the other. $\endgroup$ – jomegaA Feb 8 at 11:48
  • $\begingroup$ @jomegaA. Is the "sequence inside the brackets" an infinite-length sequence? $\endgroup$ – Richard Lyons Feb 9 at 16:31
  • $\begingroup$ @RichardLyons The sequence $x(n)$ is finite and $x_p(n)$ is periodic extension of it. The index $0\leq n \leq N-1$ is definitely finite. $\endgroup$ – jomegaA Feb 9 at 19:17
  • $\begingroup$ @jomegaA. Hi. I wasn't asking about $x(n)$ or $x_p(n)$. I asked if the "sequence inside the brackets", the summation of $x(n-lN)$, was an infinite-length sequence? Thanks. $\endgroup$ – Richard Lyons Feb 10 at 14:13

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