Question

Find the inverse z- transform of
$X\left(z\right)=\frac{z}{(z-1)(z-2{)}^2}|z|>2$

• A. $x\left[n\right]=(1-2^n-n{2}^{n-1})u\left[n\right]$
• B. $x\left[n\right]=(1+2^n-n{2}^{n-1})u\left[n\right]$
• C. $x\left[n\right]=(1-2^n+n{2}^{n-1})u\left[n\right]$
• D. $x\left[n\right]=(1+2^n+n{2}^{n-1})u\left[n\right]$

Answer 1

The correct option is C $x\left[n\right]=(1-2^n+n{2}^{n-1})u\left[n\right]$

Using partial fraction expansion,

$\frac{X\left(z\right)}{z}=\frac{a}{z-1}+\frac{b}{z-2}+\frac{c}{(z-2{)}^2}$

$\text{where, a}=\frac{1}{(z-2{)}^2}{|}_{z=1}=1$

$c=\frac{1}{(z-1)}{|}_{z=2}=1$

$b=\frac{d}{dz}\left(\frac{1}{(z-1)}\right){|}_{z=2}$

$=\frac{-1}{(z-1{)}^2}=-1$

$\therefore X\left(z\right)=\frac{z}{z-1}-\frac{z}{z-2}+\frac{z}{(z-2{)}^2},|z|>2$

Since the ROC is |z| >2, hence the sequence will be a right sided sequence

$\therefore$ we get $x\left[n\right]=(1-2^n+n{2}^{n-1})u\left[n\right]$