(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

(i) Impulse response of the system:

By z-transform,

$Y(z) = \dfrac34 z^{-1} Y(z)-\dfrac18 z^{-2} Y(z)+X(z) \\ Y(z) - \dfrac34 z^{-1} Y(z)+\dfrac18 z^{-2} Y(z) = X(z) \\ Y(z)(1- \dfrac34 z^{-1}+\dfrac18 z^{-2} ) = X(z) \\ H(z) = \dfrac{Y(z)}{X(z)} =\dfrac1{1- \dfrac34 z^{-1}+\dfrac18 z^{-2}} \\ H(z) = \dfrac{z^2}{z^2- \dfrac34 z+\dfrac18} \\ H(z) = \dfrac{z^2}{(z-\dfrac12)(z-\dfrac14) } \\ \dfrac{H(z)}z = \dfrac z{(z-\dfrac12)(z-\dfrac14) }$

Solving the above equation using partial fractions

$\dfrac{H(z)}z = \dfrac A{(z-\dfrac12)}+ \dfrac B{(z-\dfrac14)}$

$∴ \dfrac {H(z)}z = \dfrac2{(z-\dfrac12) } -\dfrac1{(z-\dfrac14)} \\ ∴ H(z) = \dfrac 2{(1-\dfrac12 z^{-1})} -\dfrac1{(1-\dfrac14 z^{-1} ) } $

By Inverse z-transform,

Step response of the system: $x(n) = u(n)$ …………………(step response)

$X(z) = \dfrac1{1-z^{-1}}$ …………………(By z-transform)

we know, $H(z) = \dfrac{Y(z)}{X(z) } \\ ∴Y(z) = H(z).X(z) \\ =\dfrac 1{(1- \dfrac34 z^{-1}+\dfrac18 z^{-2} }*\dfrac1{1-z^{-1}} \\ =\dfrac {z^2}{z^2- \dfrac34 z+\dfrac18} *\dfrac z{z-1} \\ \dfrac{Y(z)}z = \dfrac{z^2}{z-1⁄2)(z-1⁄4)(z-1) } $

By partial fraction,

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

$\dfrac{Y(z)}z= \dfrac{-2}{(z-1⁄2)}+ \dfrac{1/3}{(z-1⁄4)}+ \dfrac{8/3}{(z-1)} \\ Y(z) = \dfrac{-2z}{(z-1⁄2)} + \dfrac{z/3}{(z-1⁄4)}+ \dfrac{8z/3}{(z-1)} $

By Inverse z-transform,

$y(n)=-2(\dfrac12)^n u(n)+\dfrac13 (\dfrac14)^n u(n)+\dfrac83 (1)^n u(n)$

(iii) The system function is given by,

$H(z) = \dfrac{z^2}{(z-\dfrac12)(z-\dfrac14) } $

Poles at: z=1/2, z=1/4.

Zeros at: z=0, z=0.

Pole zero diagrams:

Since all the poles lie inside the unit circle, the system is stable.