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Obtain Inverse Z-Transform of, X(z)=110.6Z1+0.08Z2  for i) |Z|>0.4 ii) |Z|<0.2 iii) 0.2<|Z|<0.4
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Solution:

X(z)=110.6Z1+0.08Z2

=Z2Z20.6Z+0.08

X(Z)Z=Z(Z0.2)(Z0.4)

=AZ0.2+BZ0.4

A=Z(Z0.2)(Z0.4)(Z0.2)|Z=0.2

=1B=Z(Z0.2)(Z0.4)(Z0.4)|Z=0.4=2

X(Z)Z=1Z0.2+2Z0.4

X(Z)=ZZ0.2+2ZZ0.4

Applying inverse Z-transform,

x(n)=(0.2)nu(n)+2(0.4)nu(n)

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