Question

Consider the system shown below,



If x[n] is a discrete time signal given to a system with its impulse response as h[n] and its transformation
$H\left(z\right)=\frac{z^2}{(z-\frac{1}{2})(z-\frac{1}{4})}$
The step response of the system is

• A. $y\left[n\right]=[8-2{\left(\frac{1}{2}\right)}^n+\frac{1}{3}{\left(\frac{1}{4}\right)}^n]u\left[n\right]$
• B. $y\left[n\right]=[\frac{8}{3}-3{\left(\frac{1}{2}\right)}^n+\frac{1}{3}{\left(\frac{1}{4}\right)}^n]u\left[n\right]$
• C. $y\left[n\right]=\frac{1}{3}[8-6{\left(\frac{1}{2}\right)}^n+{\left(\frac{1}{4}\right)}^n]u\left[n\right]$
• D. $y\left[n\right]=[\frac{8}{3}-\frac{2}{3}{\left(\frac{1}{2}\right)}^n+{\left(\frac{1}{4}\right)}^n]u\left[n\right]$

Answer 1

The correct option is C $y\left[n\right]=\frac{1}{3}[8-6{\left(\frac{1}{2}\right)}^n+{\left(\frac{1}{4}\right)}^n]u\left[n\right]$

$Y\left(z\right)=X\left(z\right).H\left(z\right)$

The unit step input, $X\left(z\right)=\frac{1}{1-z^{-1}}$ $(\because \text{x[n]=unit step)}$

$Y\left(z\right)=\frac{z^3}{(z-1)(z-\frac{1}{2})(z-\frac{1}{4})}$

$\frac{Y\left(z\right)}{z}=\frac{C_1}{z-1}+\frac{C_2}{z-\frac{1}{2}}+\frac{C_3}{z-\frac{1}{4}}$

$C_1=\frac{z^2}{(z-\frac{1}{4})(z-\frac{1}{2})}{\mid }_{z=1}=\frac{8}{3}$

$C_2=\frac{z^2}{(z-1)(z-\frac{1}{4})}{\mid }_{z=\frac{1}{2}}=-2$

$C_3=\frac{z^2}{(z-1)(z-\frac{1}{2})}{\mid }_{z=\frac{1}{4}}=\frac{1}{3}$

$Y\left(z\right)=\frac{8}{3}\frac{z}{z-1}-2\frac{z}{z-\frac{1}{2}}+\frac{1}{3}\frac{z}{z-\frac{1}{4}}$

taking inverse z-transform,

$y\left[n\right]=\frac{1}{3}[8-6{\left(\frac{1}{2}\right)}^n+{\left(\frac{1}{4}\right)}^n]u\left[n\right]$