The transfer function of digital causal system is given as follows: $H(z)=\dfrac {1-z^{-1}}{1-0.2z^{-1}-0.15z^{-2}} $. Draw cascade form, parallel form realization

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

$H(z)=\dfrac {1-z^{-1}}{1-0.2z^{-1}-0.15z^{-2}}$ . Draw cascade form, parallel form realization

written 8.5 years ago by

teamques10 ★ 69k

modified 3.2 years ago by

sagarkolekar ★ 11k

discrete time signal processing

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written 8.5 years ago by

teamques10 ★ 69k

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

$=\Bigg[\dfrac {(1-z^{-1})}{(1-0.5z^{-1})}\Bigg]\Bigg[\dfrac 1{1+0.3z^{-1})}\Bigg]$

Cascade form can be given as :-

enter image description here

Parallel form:-

$H(z)=\dfrac {1-2^{-1}}{1-0.2z^{-1}-0.15z^{-2}}\\ =\dfrac {z(2-1)}{z^2-0.2z-0.15} \\ \dfrac {H(z)}z=\dfrac {(z-1)}{(z-0.5)(2+0.3)}$

Now,

$\dfrac {H(2)}z=\dfrac A{z-0.5}+\dfrac {B}{z+0.3}\\ \therefore A=\dfrac {H(z)}2(z-0.5)|z=0.5 \\ =-0.625 \\ B=\dfrac {z-1}{(2-0.5)(2+0.3)}|z=-0.3 \\ =\dfrac {-0.3-1}{-0.3-0.5}\\ =1.625$

Parallel form can be given as

enter image description here

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