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Obtain Inverse Z-Transform of, X(z)=11−0.6Z−1+0.08Z−2 for i) |Z|>0.4 ii) |Z|<0.2 iii) 0.2<|Z|<0.4
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written 2.3 years ago by |
Solution:
X(z)=11−0.6Z−1+0.08Z−2
=Z2Z2−0.6Z+0.08
X(Z)Z=Z(Z−0.2)(Z−0.4)
=AZ−0.2+BZ−0.4
A=Z(Z−0.2)(Z−0.4)(Z−0.2)|Z=0.2
=−1B=Z(Z−0.2)(Z−0.4)(Z−0.4)|Z=0.4=2
∴X(Z)Z=−1Z−0.2+2Z−0.4
X(Z)=−ZZ−0.2+2ZZ−0.4
Applying inverse Z-transform,
x(n)=−(0.2)nu(n)+2(0.4)nu(n)