Question

Inverse z-transform of
x(z) = log [$1+a{z}^{-1}$], z > a

• A. $n(-1{)}^{n+1}{a}^n.u(n-1)$
• B. $\frac{1}{n}(-1{)}^{n-1}{a}^n.u(n+1)$
  
• C. $\frac{1}{n}(-1{)}^{a+1}{a}^{\pi }.u(n-1)$
• D. $n(-1{)}^{n-1}{a}^n.u(n+1)$

Answer 1

The correct option is C $\frac{1}{n}(-1{)}^{a+1}{a}^{\pi }.u(n-1)$

$\frac{dX\left(z\right)}{dz}=-\frac{a{z}^{-2}}{1+a{z}^{-1}}$

$-z^{-1}[-z\frac{dX\left(z\right)}{dz}]=z^{-1}\left[\frac{-a{z}^{-1}}{1+a{z}^{-1}}\right]$

Taking the inverse z-transform, we get

$z^{-1}[-z\frac{dX\left(z\right)}{dz}]=z^{-1}\left[\frac{-a{z}^{-1}}{1+a{z}^{-1}}\right]$

$n.x\left(n\right)=z^{-1}\left[\frac{-a{z}^{-1}}{1+a{z}^{-1}}\right]$

We know that,

$a^n.u\left(n\right)\stackrel{}{⟷}\frac{1}{1-a{z}^{-1}}|z|>|a|$

$(-a{)}^a.u(n)\stackrel{}{⟷}\frac{1}{1+a{z}^{-1}}|z|>|a|$

$a(-a{)}^n.u(n)\stackrel{}{⟷}\frac{a}{1+a{z}^{-1}},|z|>|a|$

Using the time shift property, we obtain

$a(-a{)}^{n-1}.u(n-1)\stackrel{}{⟷}\frac{a{z}^{-1}}{1+a{z}^{-1}},|z|>|a|$

$-(-a{)}^{n}u(n-1)\stackrel{}{⟷}\frac{a{z}^{-1}}{1+a{z}^{-1}}|z|>|a|$

Consequently,

$n.x\left(n\right)=-(-a{)}^{n}u(n-1)$

$x\left(n\right)=\frac{-(-a{)}^n}{n}u(n-1)$

$=\frac{1}{n}(-1{)}^{n+1}{a}^n.u(n-1)$