I am studying DT-FT. But I cannot figure out why we use $X(e^{j\omega})$ instead of $ X(j\omega) $ in DT FT

Thanks in advance..


1 Answer 1


The argument $e^{j\omega}$ emphasizes the $2\pi$-periodicity of the discrete-time Fourier transform (DTFT) of a sequence. Furthermore, if the $\mathcal{Z}$-transform $X(z)$ of the sequence $x[n]$ exists and if the unit circle $|z|=1$ is inside the region of convergence, then the DTFT of the sequence is simply obtained by evaluating $X(z)$ on the unit circle $z=e^{j\omega}$.

That's the same thing as with the continuous Fourier transform and the Laplace transform, referred to in your previous question.


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