Can I assume that the R.O.C. from a Z-Transform will always start from or end in a pole or 0/infinity?

Using an example:

$$H(z)=\frac{(z+1)(z-1)(z+j)(z-j)}{z^4}\space,\space\space j=\sqrt{(-1)}$$

So we have 4 zeros in the unit circle equally spaced by a $\frac{\pi}{2}$ angle and 4 poles in the origin ($z=0$). If not specified, can I assume the the ROC wil necessarily start from exclusive zero and go towards infinity? Or maybe possible to exist a ROC which is a ring in the Z-plane (which will not start/end in a pole)?


The region of convergence of a power series of a meromorphic function $f:\mathbb{C}\to\mathbb{C}$ will always be a open disk with an essential singularity on the boundary. The z-transform of an LTE system is essentially an expansion of $H(z^{-1})$ into a power series of $z^{-1}$. $H$ is meromorphic and therefore the region of convergence of the power series will always end at a pole of $H(z^{-1})$.

  • $\begingroup$ "...will always be a open disk...", with exception of the cases where the boundary is at the origin or at infinity which may be included, right? $\endgroup$ May 27 '14 at 0:26
  • $\begingroup$ @FELIPE_RIBAS, no there are no exceptions. If the function converges on all $\mathbb{C}$ then that's an open set too. However note that I'm talking about regions of $\mathbb{C}$ for $z^{-1}$. If you want to express that same result in terms of $z$ then you have to apply a complex inversion to the regions, which can give you other shapes. $\endgroup$
    – Jazzmaniac
    May 27 '14 at 11:44

Since the value of Z-transform tends to infinity at poles, poles can not be a part of ROC. So ROC either starts from a pole or end in a pole. For example,

  1. If $x[n]$ is a right-sided sequence, then the ROC extends outward from the outermost pole in $X(z)$
  2. If $x[n]$ is a left-sided sequence, then the ROC extends inward from the innermost pole in $X(z)$
  3. If $x[n]$ is a two-sided sequence, the ROC will be a ring in the z-plane that is bounded on the interior and exterior by a pole
  4. If $x[n]$ is a finite-duration sequence, then the ROC is the entire z-plane, except possibly $z=0$ or $|z|=\infty$

From all these cases, one can say that poles decides the boundary of ROC.

The example you provided is a finite-duration sequence and its ROC is entire z-plane except at z =0.


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.