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The transfer function of digital causal system is given as follows:

$H(z)=\frac{1-z^{-1}}{(1-0.2z^{-1} -0.15z^{-2)}}$

1.Find the difference equation

2.Draw Direct Form-I and Direct Form-II realization structure.

3.Draw cascade form, parallel form realization.

1 Answer
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The difference can be obtained by taking IZT if H(z)

$H(z)=\frac{(Y(z))}{(X(z))}$

$\frac{(Y(z))}{(X(z))}=\frac{(1-z^{-1})}{(1-0.2z^{-1}-0.15z^{-2} )}$

$Y(z)(1-0.2 z^{-1}-0.15z^{-2} )=X(z)(1-z^{-1})$

$Y(z)-0.2z^{-1} Y(z)-0.15z^{-2} Y(z)=X(z)-z^{-1} X(z)$

$Y(z)=X(z)-z^{-1} X(z)+0.2z^{-1} Y(z)+0.15z^{-2} Y(z)$

Taking inverse Z transform

$∴y(n)=x(n)-x(n-1)+0.2y(n-1)+0.15y(n-2)$

ii) Direct form-I and Direct form-II realization structure:

enter image description here

iii) Cascade form, parallel form realization cascade form:

$H(z) = \frac{1-z^{-1}}{(1-0.5z^{-1})(1+0.3z^{-1})}$

In cascade form

$H(z)=H_1 (z).H_2 (z)$

Here

$H_1 (z)=\frac{(1-z^{-1})}{(1-0.5z^{-1} )}$

$H_2 (z)=\frac{1}{(1+0.3z^{-1} )}$

enter image description here

Parallel Form:-

$H(z)=\frac{(1-z^{-1})}{(1-0.2z^{-1}-0.5z^{-2} )}=\frac{(1-z^{-1})}{((1-0.5z^{-1})(1+0.3z^{-1}))}$

$∴H(z)=\frac{(\frac{(z-1)}{z})}{(\frac{((z-0.5)(z+0.3))}{z^2 )}}=\frac{(z-1)z}{((z-0.5)(z+0.3))}$

$∴\frac{H(z)}{z} =\frac{(z-1)}{((z-0.5)(z+0.3))}$

Now, By partial fraction

$\frac{(z-1)}{((z-0.5)(z+0.3))} =\frac{A}{((z-0.5))}+\frac{B}{((z+0.3))}$

$∴z-1=A(z+0.3)+B(z-0.5)$

Put z=0.5 ; A=-0.625

Put z=-0.3 ; B=1.625

Now,

$\frac{H(z)}{z}=\frac{(-0.625)}{(z-0.5)}+\frac{1.625}{(z+0.3)}$

$H(z)=\frac{(-0.625)}{(1-0.5z^(-1) )}+\frac{1.625}{(1+0.3z^(-1) )}$

enter image description here

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