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$y[n] = \dfrac34 y[n-1]-\dfrac18 y[n-2]+x[n]+\dfrac12 x[n-1] $

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2014

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A system is described by the following difference equation. Find out its transfer function H(z).

written 5.0 years ago by | • modified 5.0 years ago |

$y[n] = \dfrac34 y[n-1]-\dfrac18 y[n-2]+x[n]+\dfrac12 x[n-1] $

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2014

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written 5.0 years ago by |

By using z-transform,

$Y (z) = \dfrac34 y (z)z^{-1}-\dfrac18 y(z)z^{-2} + x(z)+\dfrac12 x(z)z^{-1} \\ Y (z) (1-\dfrac34 z^{-1}+ \dfrac18 z^{-2})=x (z) (1+\dfrac12 z^{-1} ) \\ \dfrac{y (z)}{x (z)}= \dfrac {(1+\dfrac12 z^{-1})}{(1-\dfrac34 z^{-1}+\dfrac18 z^{-2})} \\ H(z)= \dfrac{z(z+1⁄2)}{(z^2-\dfrac34 z+\dfrac18)} \\ \dfrac{H (z)}z= \dfrac {(z+1⁄2)}{(z-1⁄2)(z-1⁄4) }$

By partial fraction,

$\dfrac{H (z)}z= \dfrac A{(z-1⁄2)} + \dfrac B{(z-1⁄4) } $

$(A=4)$ and $(B=-3)$

$\dfrac{H (z)}z= \dfrac 4{(z-1⁄2) }- \dfrac 3{(z-1⁄4) } \\ H (z) = \dfrac{4z}{(z-1⁄2)} - \dfrac {3z} {(z-1⁄4)} $

By Inverse z-transform,

$h[n] = 4(\dfrac 12)^n u(n)-3(\dfrac 14)^n u(n).$

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