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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

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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.

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  • $\begingroup$ Do we also have a circular strip ROC in s domain just like we have in z domain as you told? $\endgroup$
    – DSP_CS
    Mar 3, 2020 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, 2020 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? $\endgroup$
    – DSP_CS
    Mar 3, 2020 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?? $\endgroup$
    – DSP_CS
    Mar 3, 2020 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, 2020 at 10:29

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