I am facing problems related to evaluation of inverse Z-Transform using Complex Integral Method;
Consider $X(z)=\frac{z}{z-2} $ and $ROC: |z|>2$
then, $$x(n)= \frac{1}{2\pi j}\oint_c X(z) z^{n-1} \, dz$$ $$\implies x(n)= \frac{1}{2\pi j}\oint_c \frac{z^n}{z-2} \, dz \quad \dots (1)$$ $$\implies x(-1)=\frac{1}{2\pi j}\oint_c \frac{1}{z(z-2)} \, dz$$ $$=z\frac{1}{z(z-2)}|_{z=0} + (z-2)\frac{1}{z(z-2)}|_{z=2}$$ $$=-\frac{1}{2} + \frac{1}{2} =0$$ which is obvious as $x(n)=2^n u(n)$ (by inspection), $\implies x(-1)=0$

But for the same $X(z)$ , if $ROC: |z|<2$
then, from $eq(1)$ , we get: $$x(1)=\frac{1}{2\pi j}\oint_c \frac{z}{z-2} \, dz \quad \dots (2)$$ $$=2 \quad \text{(from Cauchy's Integral Theorem)}$$ but $x(n)=-2^n u(-n-1)$ (by inspection), $\implies x(1)=0$


  1. How $c$ is defined in Complex Integral Method for a given $ROC$ ?
  2. How can we evaluate $eq(2)$ ?

The closed contour $C$ must lie inside the region of convergence, so for the ROC $|z|<2$ you have no poles inside $C$ for $n\ge 0$, hence $x[n]=0$ for $n\ge 0$.

  • $\begingroup$ Sir, then can we say that for $ROC : |z| > 2$ , $C$ is a circle given by the equation $|z|=2+\epsilon$ and for $ROC : |z| < 2$ , $C$ is $|z|=2-\epsilon$ (where $\epsilon$ is an infinitesimal value) ? $\endgroup$ – Suresh Dec 30 '20 at 11:49
  • 1
    $\begingroup$ @Suresh: Yes, these are two out of infinitely many options. $\endgroup$ – Matt L. Dec 30 '20 at 11:54

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.