$h_1 [n]=(0.9)^n u[n]-0.5(0.9)^{(n-1)} u[n-1]$ ; $h_2 [n]=(0.5)^n u[n]-(0.5)^{(n-1)} u[n-1]$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

By using convolution property of z-transform,

$h[n]=h_1 [n]*h_2 [n] ↔h(z) = h_1 (z) . h_2 (z) $

∴ $h_1 [n] = (0.9)^n u[n] - 0.5 (0.9)^{(n-1)} u[n-1]$

By z-transform,

$h_1 (z) = \frac{z}{z-0.9} - 0.5z^{-1} \frac{z}{z-0.9}$ (shifting property)

= $\frac{z}{z-0.9}$ - $\frac{0.5}{z-0.9}$

= $\frac{z-0.5}{z-0.9}$

Now,

∴ $h_2 [n] = (0.5)^n u[n] - (0.5)^{n-1} u[n-1]$

By z-transform,

$h_2 (z) = \frac{z}{z-0.5} - z^{-1} \frac{z}{z-0.5}$ ….. (shifting property)

= $\frac{z}{z-0.5} - \frac{1}{z-0.5}$

= $\frac{z-0.1}{z-0.5}$

∴ $h(z)= h_1 (z) . h_2 (z)$

= $\frac{z-0.5}{z-0.9} X \frac{z-0.1}{z-0.5}$

= $\frac{z-1}{z-0.9}$

= $\frac{z}{z-0.9} - \frac{0.5}{z-0.9}$

= $\frac{z}{z-0.9} - z^{-1} \frac{0.5z}{z-0.9}$

By Inverse z-transform ,

$h [n] = (0.9)^n u(n) - (0.9)^{n-1} u(n-1)$