Question

The block diagram represantation of a DT system is shown below :

condition y[-1] = 0, the system response is,

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

Answer 1

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

The diffrence equation is,
$y\left[n\right]+0.5y[n-1]=x\left[n\right]$

Characteristic equation,
$\lambda +0.5=0,$ root, $\lambda =-0.5$

$y_H\left[n\right]=C_1(-0.5{)}^n$

$y_P\left[n\right]=K$

$K+0.5K=1$

$\Rightarrow K=\frac{2}{3}$

So, $y_p\left[n\right]=\frac{2}{3}u\left[n\right]$

Complete response,

$y\left[n\right]=y_h\left[n\right]+y_p\left[n\right]$

$=C_1(-0.5{)}^n+\frac{2}{3}$

Using given initial conditions,

$y[-1]=C_1(-0.5{)}^{-1}+\frac{2}{3}=0$

$\Rightarrow C_1=\frac{1}{3}$

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

Alternative Solution ::

$x\left[n\right]=0.5y[n-1]+y\left[n\right]$

Taking z-transform,

$X\left(z\right)=Y\left(z\right)+0.5\left[z^{-1}Y\right(z)+0]$

$\Rightarrow Put,X\left(z\right)=\frac{z}{z-1}$ $[\because x[n]=u(n\left)\right]$

$Y\left(z\right)=\frac{z}{(z-1)(z+0.5)}$

by partial fraction method,

$\frac{A}{(z+0.5)}+\frac{B}{(z-1)}=\frac{z}{(z-1)(z+0.5)}$

Solving, $A=\frac{1}{3}$

$B=\frac{2}{3}$

Hence, $y\left[n\right]=\left[\frac{1}{3}\right[-0.5{]}^n+\frac{2}{3}]u\left[n\right]$