Given the signal $\tilde{x}(n)$, obtained by [[periodic repetition]] of the [[finite-duration discrete-time signal]] $x(n)$, it is possible to calculate its Fourier series coefficients: $\tilde{X}(k) = \sum_{n=0}^{N-1} \tilde{x}(n) e^{-jk(2\pi/N)n}$ Since $x(n)$ is a single period of $\tilde{x}(n)$: $\begin{aligned} \tilde{X}(k) &= \sum_{n=N_1}^{N_2} x(n) e^{-jk(2\pi/N)n}\\ &= \sum_{n=-\infty}^{+\infty} x(n) e^{-jk(2\pi/N)n} \end{aligned}$ The Fourier series coefficients are samples of the Fourier transform $X(e^{j\omega})$ at frequencies $\omega_{k}=k \frac{2\pi}{N} = k \omega_{0}$: $ \tilde{X}(k) = X(e^{j\omega})|_{\omega= k\frac{2\pi}{N}}$ What corresponds to spaced samples of $\omega_0=\frac{2\pi}{N}$ of the function of $\omega \in \mathbb{R}$: $X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty} x(n) e^{-j\omega n}$ This equation is called the continuous-time Fourier transform analysis equation. Note that, due to the complex exponential, $X(e^{j\omega})$ is periodic in $\omega$ with period $2\pi$.