Question 3.SGPYQ.39: If x[n] = (1/3)^|n| - (1/2)^n u[n] , then the region of conv......
If x[n] = (1/3)^{|n|} – (1/2)^n u[n] , then the region of convergence (ROC) of its Z-transform in the z-plane will be
(a) \frac{1}{3}<|z|<3 (b) \frac{1}{3}<|z|<\frac{1}{2}
(c) \frac{1}{2}<|z|<3 (d) \frac{1}{3}<|z|
Given that
\begin{aligned} x(n) & =\left\lgroup \frac{1}{3} \right\rgroup^{|n|}-\left\lgroup \frac{1}{3} \right\rgroup^n u(n) \\ & =\left\lgroup \frac{1}{3} \right\rgroup^n u(n)+\left\lgroup \frac{1}{2} \right\rgroup^{-n} u(-n-1)-\left\lgroup \frac{1}{2} \right\rgroup^n u(n) \end{aligned}
Talking Z-transform,
\begin{aligned} X(z)= & \sum_{n=-\infty}^{\infty}\left\lgroup\frac{1}{3} \right\rgroup^n z^{-n} u(n) \\ & +\sum_{n=-\infty}^{\infty}\left\lgroup\frac{1}{3} \right\rgroup^{-n} z^{-n} u(-n-1) \\ & -\sum_{n=-\infty}^{\infty}\left\lgroup\frac{1}{2} \right\rgroup^n z^{-n} u(n) \\ = & \sum_{n=\infty}^{\infty}\left\lgroup\frac{1}{3} \right\rgroup^n z^{-n}+\sum_{n=-\infty}^{-1}\left\lgroup\frac{1}{3} \right\rgroup^{-n} z^{-n} \\ & -\sum_{n=0}^{\infty}\left\lgroup\frac{1}{2} \right\rgroup^n z^{-n} \end{aligned}
Let m = -n, then
\begin{gathered} X(z)=\sum_{n=0}^{\infty}\left\lgroup\frac{1}{3 z} \right\rgroup^n+\sum_{m=1}^{\infty}\left\lgroup\frac{1}{3} z \right\rgroup^m-\sum_{n=0}^{\infty}\left\lgroup\frac{1}{2 z} \right\rgroup^n \\ \sum_{n=0}^{\infty}\left(\frac{1}{3 z}\right)^n ; \text { converges if }\left|\frac{1}{3 z}\right|<1 \text { or }|z|>\frac{1}{3} \\ \sum_{m=1}^{\infty}\left(\frac{1}{3} z\right)^m ; \text { converges if }\left|\frac{1}{3} z\right|<1 \text { or }|z|<3 \\ \sum_{n=0}^{\infty}\left(\frac{1}{2 z}\right)^n ; \text { converges if }\left|\frac{1}{2 z}\right|<1 \text { or }|z|>\frac{1}{2} \end{gathered}
Therefore, region of convergence =\frac{1}{2}<|z|<3