I was reading a DSP book where they had a problem on Z Transform.

Determine the z-transform and ROC for signal $\displaystyle x(n) = a^nu(n)$?

The solution states that $$ \displaystyle X(z) = \displaystyle\sum_{n=0}^\infty (az{^{-1}})^n $$ So $$ \displaystyle X(z) = \frac{ \displaystyle {z} }{z-a} $$ Hence ROC $|z| > a$

Which I understand that for all values of z > a the series $\frac{ \displaystyle {z} }{z-a} $ has finite values.

But for $|z| < a$ their exists a value : $z = 5$ and $a = 10$ the series has a finite value of -1 and so on. So why -1 is not taken as a set for which ROC exists - or rather my understanding on ROC is wrong?


1 Answer 1


The fact that evaluating the function $$ \begin{align} \frac{z}{z-a} \end{align} $$ yields a finite value for some given values of $z$ and $a$, does not mean that the original series (from $\mathcal{Z}$-transform definition) $$ \begin{align} X(z) = \sum_{n=0}^\infty(az^{-1})^n \end{align} $$ converges at all for those same values of $z$ and $a$. Before we determine the region of convergence, the only thing we can say is that if the series converges, then it will converge to $\frac{z}{z-a}$.

In the case $z=5$ and $a=10$, looking at the sequence $$ \begin{align} \left\{\sum_{n=0}^N (az^{-1})^n \right\} \end{align} $$ as $N$ increases yields: $$ \{1, 3, 7, 15, 31, 63, ...\} $$ which as you may notice does not converge to -1, and in fact diverges as $N\rightarrow\infty$. The region of convergence tells you where this sequence converges (and in this case it is $|z|>|a|$).

  • $\begingroup$ Thanks for the explanation, I get the point - only while evaluating the series should we be able to understand whether the series will diverge or converge $\endgroup$
    – Programmer
    Commented Aug 12, 2014 at 4:29

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