Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

Given: $X(z)=\dfrac {3-2z^{-1}+z^{-2}}{1-3z^{-1}+2z^{-2}}$

By rearranging x (z), we get

$X(z)= \dfrac {3z^2-2z+1}{1z^2-3z+2} \\ = \dfrac {3z^2-2z+1}{(z-2)(z-1)} $

Multiplying both sides by $z^{n-1}$, we get

$X(z) z^{n-1}=z^{n-1}\dfrac {3z^2-2z+1}{(z-2)(z-1) }$

There are two poles at z=2 and z=1, so the residue at z=2 is

$R(z=2) = (z-2) z^{n-1}\dfrac {3z^2-2z+1}{(z-2)(z-1)} |_{z=2} \\ = …