0
$\begingroup$

What are difference between finite duration and infinite duration sequences? As far as i am able to google and know only one difference that is finite duration sequence ROC exist on xy plane while infinite duration sequence ROC exist on unit circle plane

I have also attached two snapshots in this regard enter image description here

enter image description here

$\endgroup$
1
$\begingroup$

There is a slight misunderstanding here. There is no such plane called unit circle plane. Unit circle is in the xy plane (or we call it z plane). The difference is, rather, in the shape of the ROC. For infinite duration sequences, the ROC is in the form of a circular strip in the xy plane (as shown in the second figure).

In the case of finite duration sequences, ROC is the entire xy plane except at z=0 or at z = $\infty $ or at z=0 and $\infty$.

e.g. $x(n) = \{1,\underset{\uparrow}{2},3\} $

$X(z) = \sum \limits_{-\infty}^{+\infty} x(n)z^{-n}$

$ = x(-1)z^1+x(0)z^0+x(1)z^{-1} = z+2+\frac{3}{z}$

This Z transform will not exist only if $z=\infty$ (for the first term) and $z=0$ (for the last term). This can be generalised for all finite duration signals that their ROC consists of entire z plane (xy plane) except at z=0 or z=$\infty$ or at both.

$\endgroup$
7
  • $\begingroup$ Do we also have a circular strip ROC in s domain just like we have in z domain as you told? $\endgroup$ – engr Mar 3 '20 at 6:28
  • $\begingroup$ Yes. In s domain, we have a strip that is parallel to the y axis of the s plane. $\endgroup$ – Shehin Mar 3 '20 at 7:10
  • $\begingroup$ "strip that is parallel to the y axis of the s plane" Will that strip be circular/ring shaped? or you mean that strip is imaginary(y axis) itself? $\endgroup$ – engr Mar 3 '20 at 9:57
  • $\begingroup$ I couldn't understand :"This can be generalised for all finite duration signals that their ROC consists of entire z plane (xy plane) except at z=0 or z=∞ or at both." If ROC consists of either z=0 or z=∞ ,but not both,will our signal be considered finite duration or infinite duration in that case?? $\endgroup$ – engr Mar 3 '20 at 10:08
  • $\begingroup$ For a right sided finite signal, ROC is the entire z plane except z=0 and for a left sided finite signal, ROC will be entire z plane except z=$\infty $. For a two sided finite signal, ROC will be the entire z plane except both z=0 and z=$\infty$ . The example I have given is a two sided case. $\endgroup$ – Shehin Mar 3 '20 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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