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

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Obtain Inverse Z-Transform of,

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

$\text { for i) }|Z|>0.4 \text { ii) }|Z|<0.2 \text { iii) } 0.2<|Z|<0.4$

written 2.3 years ago by

prajapatijaimin • 3.3k

• modified 2.3 years ago

discrete time signal processing

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written 2.3 years ago by

prajapatijaimin • 3.3k

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,

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

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