Finding Fourier transform of a discrete signal from its Z-transform

Question

Is it possible to find the Fourier transform of a discrete signal if you know its $\mathcal{Z}$-transform of?

Answer 1

DTFT is the Z-transform at the unit circle. So if $z=re^{j\omega}$ then for DTFT $r = 1$.

i.e If you have the Z-transform of a signal then plug-in $e^{j\omega}$ for every $z$

Answer 2

You can't always find the discrete-time Fourier transform (DTFT) from a given $\mathcal{Z}$-transform. It could be that the series

$$X(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{1}$$

doesn't converge on the unit circle $|z|=1$. In that case, if you replace $z$ by $e^{j\omega}$ in $(1)$, then the result is not the DTFT of $x[n]$, because that DTFT doesn't even exist.

In sum, the DTFT can only be obtained from the $\mathcal{Z}$-transform evaluated at $z=e^{j\omega}$ if the region of convergence (ROC) of $(1)$ includes the unit circle $|z|=1$.

Take as an example the sequence $x[n]=2^nu[n]$, where $u[n]$ is the unit step sequence. Its $\mathcal{Z}$-transform is given by

$$X(z)=\sum_{n=0}^{\infty}2^nz^{-n}=\frac{1}{1-2z^{-1}},\qquad |z|>2\tag{2}$$

Note that the ROC $|z|>2$ doesn't include the unit circle. Consequently, the function

$$X(e^{j\omega})=\frac{1}{1-2e^{-j\omega}}\tag{3}$$

cannot be the DTFT of $x[n]$, because the DTFT of $x[n]$ doesn't exist. However, $(3)$ is a valid DTFT, but of a different sequence. It is the DTFT of the sequence

$$\hat{x}[n]=-2^nu[-n-1]\tag{4}$$

which has the same $\mathcal{Z}$-transform as the original sequence $x[n]$, but with a different ROC:

$$\hat{X}(z)=-\sum_{n=-\infty}^{-1}2^nz^{-n}=\frac{1}{1-2z^{-1}},\qquad |z|<2\tag{5}$$

The ROC of $(5)$ includes the unit circle, and, consequently, the DTFT can be obtained by setting $z=e^{j\omega}$.

Answer 3

Compare the $z$ transform: $$ H(z) = \sum_{k=-\infty}^{+\infty} h[k] z^{-k} $$ to the Fourier series: $$ H_{2\pi}(\omega) = \sum_{k=-\infty}^{+\infty} h[k] e^{-j\omega k } $$