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Two LTI systems in cascaded have impulse response $h_1 [n] $ and $h_2 [n]$

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

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

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By using convolution property of z-transform,

$h[n]=h_1 [n]*h_2 [n](↔^z )h(z)=h_1 (z).h_2 (z) \\ ∴h_1 [n]=(0.9)^n u[n]-0.5(0.9)^{n-1} u[n-1]$

By z-transform,

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

$= \dfrac {z}{z-0.9}-\dfrac {0.5}{z-0.9} \\ =\dfrac {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,

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

$= \dfrac z{z-0.5}-\dfrac 1{z-0.5} \\ =\dfrac {z-0.1}{z-0.5} \\ ∴h(z)= h_1 (z) h_2 (z) \\ = \dfrac {z-0.5}{z-0.9} \times \dfrac {z-0.1}{z-0.5} \\ = \dfrac {z-1}{z-0.9} \\ =\dfrac z{z-0.9}- \dfrac {0.5}{z-0.9} \\ =\dfrac z{z-0.9} - z^{-1} \dfrac {0.5z}{z-0.9} $

By Inverse z-transform,

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

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