0
180views
Obtain Inverse Z-Transform of, $ X(z)=\frac{1}{1-0.6 Z^{-1}+0.08 Z^{-2}} $ $$ \text { for i) }|Z|>0.4 \text { ii) }|Z|<0.2 \text { iii) } 0.2<|Z|<0.4 $$
1 Answer
0
3views

Solution:

$ X(z)=\frac{1}{1-0.6 Z^{-1}+0.08 Z^{-2}}\\ $

$ =\frac{Z^2}{Z^2-0.6 Z+0.08}\\ $

$ \frac{X(Z)}{Z}=\frac{Z}{(Z-0.2)(Z-0.4)}\\ $

$ =\frac{A}{Z-0.2}+\frac{B}{Z-0.4}\\ $

$ A=\left.\frac{Z}{(Z-0.2)(Z-0.4)}(Z-0.2)\right|_{Z=0.2}\\ $

$ =-1 \quad B=\left.\frac{Z}{(Z-0.2)(Z-0.4)}(Z-0.4)\right|_{Z=0.4}=2\\ $

$ \therefore \frac{X(Z)}{Z}=\frac{-1}{Z-0.2}+\frac{2}{Z-0.4}\\ $

$ X(Z)=\frac{-Z}{Z-0.2}+\frac{2 Z}{Z-0.4}\\ $

Applying inverse Z-transform,

$ x(n)=-(0.2)^n u(n)+2(0.4)^n u(n)\\ $

Please log in to add an answer.