Question

The inverse Z-transform of X(z) is x(n).
$X\left(z\right)=\log (1+a{z}^{-1})$
If a = 2 then the value of x(2) is _______.

• A. -2

Answer 1

The correct option is A -2
$log(1+x)=\sum _{n=1}^{\infty }\frac{-1}{n}(-x{)}^n=x-\frac{x^2}{2}+\frac{x^3}{3}-.......$

$\log (1+a{z}^{-1})=a{z}^{-1}-\frac{(a{z}^{-1}{)}^2}{2}+\frac{(a{z}^{-1}{)}^3}{3}$.......

$=\sum _{n=1}^{\infty }\frac{-1}{n}(-a{z}^{-1}{)}^n$ $|z|>|a|$

$=\sum _{n=1}^{\infty }\frac{-1}{n}(-a{)}^n{z}^{-n}$, $|z|>|a|$

$=\sum _{n=1}^{\infty }\frac{-1}{n}(-1{)}^n{a}^n{z}^{-n}$, $|z|>|a|$

$=\sum _{n=1}^{\infty }\frac{1}{n}(-1{)}^{n+1}{a}^n{z}^{-n}$, $|z|>|a|$

$X\left(z\right)=\sum _{n=1}^{\infty }\left[\frac{1}{n}\right(-1{)}^{n+1}{a}^{n}u(n-1)]z^{-n}$

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

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