Question

$\text{If x(n) =}{\left(\frac{1}{5}\right)}^{|n|}+\left(3{)}^{n}u\right(n)\text{, then the region of convergence (ROC) of its Z-transform is}$

• A. $\frac{1}{5}<|z|\le \frac{1}{3}$
• B. $3<|z|<5$
• C. $\frac{1}{5}\le |z|<3$
• D. $3\le |z|\le 5$

Answer 1

The correct option is B $3<|z|<5$

$x\left(n\right)={\left(\frac{1}{5}\right)}^{|n|}+\left(3{)}^{n}u\right(n)$

${\left(\frac{1}{5}\right)}^{|n|}={\left(\frac{1}{5}\right)}^{n}u\left(n\right)+{\left(\frac{1}{5}\right)}^{-n}u(-n-1)$

$X\left(z\right)=(\frac{z}{z-1/5}-\frac{z}{z-5}+\frac{z}{z-3})$

$3^{n}u\left(n\right)\stackrel{Z.T}{⟷}\frac{z}{z-3}$, ROC is |z|>3

${\left(\frac{1}{5}\right)}^{|n|}\stackrel{Z.T}{⟷}\frac{z}{z-1/5}-\frac{z}{z-5},ROC\frac{1}{5}<|z|<5$

So intersection of these two ROC will give ROC of X(z) is 3 < |z| < 5.