Considering a [[finite-duration discrete-time signal]] $x(n)$ with Fourier transform $X(e^{j\omega})$: $x(n) \xrightarrow[\cal DTFT]{}X(e^{j\omega})$ Sampling the [[discrete-time Fourier transform (DTFT)]] at frequencies $\omega_{k}=k \frac{2\pi}{N}$ produces the coefficients $\tilde{X}(k)$: $ \begin{align} \tilde{X}(k) &= X(e^{j\omega})|_{\omega= k\frac{2\pi}{N}} \\ &= \sum_{n=-\infty}^{+\infty} x(n) e^{-j(2\pi/N)kn} \end{align} $ The [[discrete Fourier series (DFS)]] synthesis equation can be used to compute the periodic signal $\tilde{x}(n)$ corresponding to the Fourier series coefficients $\tilde{X}(k)$ $ \tilde{x}(n)=\frac{1}{N}\sum_{k=0}^{N-1} \tilde{X}(k) e^{j\frac{2\pi}{N}kn} $ that is: $ \begin{align} \tilde{x}(n) &= \frac{1}{N} \sum_{k=0}^{N-1} \left[ \sum_{m=-\infty}^{+\infty}x(m)e^{-j(2\pi/N)km}\right] e^{j(2\pi/N)kn}\\ &= \sum_{m=-\infty}^{+\infty}x(m) \left[ \frac{1}{N} \sum_{k=0}^{N-1} e^{j(2\pi/N)k(n-m)}\right] \\ & = \sum_{m=-\infty}^{+\infty}x(m) \tilde{p}(n-m) \end{align} $ Given the property of the [[sum of equally spaced complex exponentials]]: $ \tilde{p}(n-m) = \sum_{r=-\infty}^{+\infty} \delta(n-m-rN) $ That results in $ \tilde{x}(n) = x(n) \ast \tilde{p}(n) = \sum_{r=-\infty}^{+\infty} x(n-rN) $ So $\tilde{x}(n)$ is formed by adding together an infinite number of shifted replicas of $x(n)$. The original signal can be recovered if $N$ is greater that duration of the [[finite-duration discrete-time signal]] $x(n)$. The finite-lenght sequence can be recovered by: $ x(n) = \begin{cases} \tilde{x}(n), \, 0 \le n \le N-1 \\ 0, \, \text{otherwise} \end{cases} $ The sequence of samples of the DTFT corresponds to the DFS coefficients of $\tilde{x}(n)$. To maintain the duality between time and frequency domains we will also use just one period of $\tilde{X}(k)$: $ X(k) = \begin{cases} \tilde{X}(k), \, 0 \le k \le N-1 \\ 0, \, \text{otherwise} \end{cases} $ The finite-duration sequence $X(k)$ is called the [[discrete Fourier transform (DFT)]] of the [[finite-duration discrete-time signal]] $x(n)$.