Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 05

Year : MAY 2014

Given $X [z] = \dfrac {z^2+z}{z^2-2z+1} $

By using partial fraction,

$ = \dfrac {z(z+1)}{(z-1)^2 } \\ \dfrac {x [z]}z = \dfrac A{(z-1)} + \dfrac B{(z-1)^2}$

put $z=2,$ we get $A+B=3$

put $z=0,$ we get $A-B=1$

$(A= 2) $ and $(B= 1) \\ \dfrac {x [z]}z = \dfrac 2{(z-1)} + \dfrac 1{(z-1)^2} \\ x[z] = \dfrac 2z{(z-1)} + \dfrac {1z}{(z-1)^2 } \\ x[z] = \dfrac 2{(1-z^{-1})} + \dfrac {1z^{-1}}{(1-z^{-1})^2}$

ROC is $|z| \gt 1$ so by using the formula $a^n u(n)(↔^z ) \dfrac 1{1-az^{-1}}\space for\space |z| \gt |a|. $

By Inverse z-transform,

$x(n) = 2(1)^n u(n)+n(1)^n u(n).$