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Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

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Determine the inverse z transform of the function using Residue method $X(z)=\dfrac {3-2z^{-1}+z^{-2}}{1-3z^{-1}+2z^{-2}}$

written 5.0 years ago by | • modified 5.0 years ago |

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

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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} \\ = z^{n-1} \dfrac {3z^2-2z+1}{(z-1)} |_{z=2} \\ = 2^{n-1} \dfrac {3(2^2)-2(2)+1}{(2-1)} |_{z=2} \\ = 2^{n-1}\times 9 \\ =\dfrac 92 (2)^n $

Similarly residue at z=1 is,

$R(z=1) = (z-1) z^{n-1} \dfrac {3z^2-2z+1}{(z-2)(z-1)} |_{z=1} \\ = z^{n-1} \dfrac {3z^2-2z+1}{(z-2)} |_{z=1} \\ = 1^{n-1} \dfrac {3(1)-2(1)+1}{(1-2)} \\ = 1^{n-1} \times2 \\ R(z=1) = 2(1)^n \\ ∴, x [n] = res(z=2) + res(z=1) \\ x[n] = \dfrac 92 (2)^n+2(1)^n\hspace {1cm} for\space n\gt0.$

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