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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.