Question

If $x\left[n\right]={\left(\frac{1}{3}\right)}^{|n|}-{\left(\frac{1}{2}\right)}^{n}u\left[n\right]$, then the region of convergence (ROC) of its $Z-$ transform in the $Z-$ plane will be

• A. $\frac{1}{3}<|z|<3$
• B. $\frac{1}{2}<|z|<3$
• C. $\frac{1}{3}<|z|<\frac{1}{2}$
• D. $\frac{1}{3}<|z|$

Answer 1

The correct option is B $\frac{1}{2}<|z|<3$

$x\left[n\right]={\left(\frac{1}{3}\right)}^{|n|}-{\left(\frac{1}{2}\right)}^{n}u\left[n\right]$

$x\left[n\right]={\left(\frac{1}{3}\right)}^{n}u\left[n\right]+{\left(\frac{1}{3}\right)}^{-n}u[-n-1]-{\left(\frac{1}{2}\right)}^{n}u\left[n\right]$

${\left(\frac{1}{3}\right)}^{n}u\left[n\right]⟷\frac{1}{1-\frac{1}{3}{z}^{-1}}$ ; ROC $|z|>\frac{1}{3}$

${\left(\frac{1}{3}\right)}^{n}u[-n-1]⟷\frac{-1}{1-3z^{-1}}$ ; ROC $|z|<3$
${\left(\frac{1}{2}\right)}^{n}u\left[n\right]⟷\frac{1}{1-\frac{1}{2}{z}^{-1}}$ ; ROC $|z|>\frac{1}{2}$
So overall ROC will be intersection of there ROC's i.e.,
$\frac{1}{2}<|z|<3$