A causal LTI system is described $y[n] = \dfrac 34y[n-1]-\dfrac 18 y[n-2]+x[n].$Where y[n] response of the system and x[n] is excitation to the system.

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A causal LTI system is described

$y[n] = \dfrac 34y[n-1]-\dfrac 18 y[n-2]+x[n].$ Where y[n] response of the system and x[n] is excitation to the system.

written 8.5 years ago by

teamques10 ★ 69k

• modified 8.5 years ago

(i) Determine impulse response of the system.

(ii) Determine step response of the system.

(iii) Plot pole-zero pattern and state whether system is stable.

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

signals and systems

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written 8.5 years ago by

teamques10 ★ 69k

(i) Impulse response of the system:

By z-transform,

$Y(z) = \dfrac34 z^{-1} Y(z)-\dfrac18 z^{-2} Y(z)+X(z) \\ Y(z) - \dfrac34 z^{-1} Y(z)+\dfrac18 z^{-2} Y(z) = X(z) \\ Y(z)(1- \dfrac34 z^{-1}+\dfrac18 z^{-2} ) = X(z) \\ H(z) = \dfrac{Y(z)}{X(z)} =\dfrac1{1- \dfrac34 z^{-1}+\dfrac18 z^{-2}} \\ H(z) = \dfrac{z^2}{z^2- \dfrac34 z+\dfrac18} \\ H(z) = …

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