written 8.1 years ago by | • modified 8.1 years ago |
Mumbai University > EXTC > Sem 4 > Signals and Systems
Marks : 05
Year : MAY 2014
written 8.1 years ago by | • modified 8.1 years ago |
Mumbai University > EXTC > Sem 4 > Signals and Systems
Marks : 05
Year : MAY 2014
written 8.1 years ago by |
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,
$x(n) = 2(1)^n u(n)+n(1)^n u(n).$