What are the conditions that must be satisfied to be able to invert the $\mathcal Z$-transform?

  • $\begingroup$ If you have some function in the $z$-domain and it has a ROC, then you can always find the corresponding sequence in the time domain by performing some complex calculus (although it might be pretty laborious if not impossible to get to a closed solution). What's your point? Please clarify. $\endgroup$ – Tendero Apr 28 '17 at 16:00
  • $\begingroup$ Thanks for your reply. So you're saying when the ROC exists (systems is stable) the inverse Z-transform can always be computed? I need to know if/when then Z-transform cannot be inverted. $\endgroup$ – Marco Datola Apr 28 '17 at 17:06
  • $\begingroup$ An existing ROC doesn't imply stability. A ROC containing the unit circle does. $\endgroup$ – Tendero Apr 28 '17 at 18:36
  • $\begingroup$ Sure, in case of a causal system... I don't see how your last comment helps though? $\endgroup$ – Marco Datola Apr 28 '17 at 19:01

If I understand your question correctly then you're asking under which conditions the inverse $\mathcal{Z}$-transform of a given function $F(z)$ exists. Since the inverse $\mathcal{Z}$-transform is just a Laurent expansion of $F(z)$ around $z=0$, the inverse transform exists if $F(z)$ can be developed in a Laurent series about $z=0$. This is the case if $F(z)$ is analytic in an annulus centered at $z=0$.


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