New answers tagged dtft
0
The discrete-time Fourier transform (DTFT) can be used for general infinite length signals, and it can also be used for periodic signals if we allow Dirac delta impulses in the expression for the DTFT. You have the correspondence
$$\textrm{DTFT}\{e^{jn\omega_0}\}=2\pi\delta(\omega-\omega_0)\tag{1}$$
where it is understood that the expression $(1)$ is valid ...
2
This is pretty straightforward to show. Let $x[n]$ be a sampled version of a continuous-time signal $x_c(t)$:
$$x[n]=x_c(nT)\tag{1}$$
The DTFT of $x[n]$ is defined by
$$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{2}$$
The CTFT of the signal
$$x_c(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT)=\sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT)\...
Top 50 recent answers are included
