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

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

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 $$z$$

• You mean, substitute $z$ with $e^{j\omega}$? – angie Sep 2 '15 at 6:44
• @angie Yes. Not sure why i'm downvoted though – usfmohy Sep 2 '15 at 13:53
• I didn't downvote you. – angie Sep 3 '15 at 20:46

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 $$z=e^{j\omega}$$.

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

• I can see they're similar – angie Sep 1 '15 at 20:04
• Does this help? – Peter K. Sep 1 '15 at 20:06
• Not really. If you could show me an example with numbers , it would be great. – angie Sep 1 '15 at 20:16
• Try to think a bit more. Looking to the equations that @PeterK. posted, what should be equal do z in order to make the Z-transform equation be equal to the Fourier Transform. In other others, you can write z = "something", and when you put this value in the z-transform, it is goingo to become the Fourier Transform. – JohnMarvin Sep 1 '15 at 20:44
• You mean $z=e^{jw}$ ? – angie Sep 1 '15 at 20:50