Determine the following

(i) The system transfer function H(z)

(ii) impulse response of the system h[n]

(iii) Step response of the system s[n].

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2015

(i) To determine the system transfer function H(z)

The given difference equation is

$y[n] =\dfrac 34 y[n - 1] – \dfrac18 y[n - 2] + x[n]$

Taking Z-transform of the difference equation

$Y(z) = \dfrac34 z^{-1} Y(z) - \dfrac18 z^{-2} Y(z) + X(z) \\ ∴[1 - \dfrac34 z^{-1} + \dfrac18 z^{-2}] Y(z) = X(z) \\ ∴H(z) = \dfrac{Y(z)}{X(z) } \\ ∴H(z) = \dfrac1{ 1 - \dfrac34 z^{-1} + \dfrac18 z^{-2}} -------------- 1 \\ = \dfrac{z^2}{ z^2 - \dfrac34 z + \dfrac18 } \\ = \dfrac{z^2}{(z - \dfrac12) (z -\dfrac 14) } \\ ∴\dfrac{H(z)} z = \dfrac z {(z - \dfrac12) (z - \dfrac14)} ------------- 2$

Expanding above equation in partial fractions

$\dfrac{H(z)}z =\dfrac{ A_1}{z - \dfrac12} + \dfrac{A_2}{z -\dfrac 14} ------------ 3 \\ ∴A_1 = (z – \dfrac12 )\dfrac{ H(z)}z |_{z = \dfrac12 } \\ = (z – \dfrac 12 ) \dfrac z{(z - \dfrac12) (z - \dfrac14)} |_{z = \dfrac12 } \\ = \dfrac z{(z - \dfrac14)} |_{z = \dfrac12 } \\ ∴A_1 = 2$

And

$∴A_2 = (z – \dfrac14 ) \dfrac{H(z)}z |_{z = \dfrac14 } \\ = (z – \dfrac14 ) \dfrac{z}{(z - \dfrac12) (z - \dfrac14)} |_{z = \dfrac14} \\ =\dfrac z{(z - \dfrac12)} |_{z = \dfrac14 } \\ ∴ A_2 = -1 $

Hence equation 3 becomes

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

This is the system transfer function.

(ii) To determine the impulse response of the system h[n]

Taking inverse Z-transform of H(z)

Since inverse Z-transform of $[\dfrac1{1 – az^{-1} }] = (a)^n u(n) \\ ∴h[n] = 2(\dfrac12)^n u(n)-(\dfrac14)^n u(n)$

This is the impulse response of the system.

(iii) To determine the step response of the system s(n)

For the step response,

$x(n) = s(n)$

Taking Z-transform,

$X(z) = \dfrac1{1 – z^{-1} }$

By convolution theorem,

$Y(z) = H(z).X(z) \\ = \dfrac1 {1 - \dfrac34 z^{-1} + \dfrac18 z^{-2}}\dfrac 1{1 – z^{-1}} \hspace {1cm} \text {From equation 1} \\ = \dfrac{z^2}{ z^2 - \dfrac34 z + \dfrac18}\dfrac z{z - 1 }\\ ∴\dfrac{Y(z)}z = \dfrac{z^2}{ (z^2 - \dfrac34 z + \dfrac18 ) (z - 1) } \\ ∴\dfrac{Y(z)}z = \dfrac{z^2}{(z - \dfrac12) (z - \dfrac14) (z - 1) }$

Expanding above equation in partial functions

$\dfrac{Y(z)}z = \dfrac{A_1}{z - \dfrac12 }+ \dfrac{A_2}{z - \dfrac14} + \dfrac{A_3}{z - 1} -------------- 4 \\ ∴A_1 = (z – \dfrac12 ) \dfrac{Y(z)}z |_{z = \dfrac12 } \\ = (z – \dfrac12 ) \dfrac{z^2}{(z - \dfrac12) (z - \dfrac14) (z - 1)} |_{z = \dfrac12 } \\ = \dfrac{z^2}{(z - \dfrac14) (z - 1)} |_{z = \dfrac12 } \\ ∴A_1 = -2 \\ ∴A_2 = (z – \dfrac14 ) \dfrac{Y(z)}z |_{z = \dfrac14 } \\ = (z – \dfrac14 ) \dfrac{z^2}{(z - \dfrac12) (z - \dfrac14) (z - 1)} |_{z = \dfrac14} \\ = \dfrac{z^2}{(z - \dfrac12) (z - 1)} |_{z = \dfrac14 } \\ ∴A_2 = \dfrac13 \\ ∴A_3 = (z – 1) \dfrac{Y(z)}z |_{z = 1} \\ = (z – 1) \dfrac{z^2}{(z - \dfrac12) (z - \dfrac14) (z - 1) }|_{z = 1} \\ =\dfrac{ z^2}{(z - \dfrac12) (z - \dfrac14)} |_{z = 1} \\ ∴A_3 = \dfrac83 $

Therefore equation 4 becomes

$\dfrac{Y(z)} z = \dfrac{-2}{z - \dfrac12} + \dfrac{\dfrac13}{z - \dfrac14 }+ \dfrac{\dfrac83}{z - 1 } \\ ∴Y(z) = \dfrac{-2}{1 - \dfrac12 z^{-1}} + \dfrac{\dfrac13}{1 - \dfrac14 z^{-1}} + \dfrac{\dfrac83}{1 - z^{-1 }}$

Taking inverse Z-transform of the above equation

Since inverse Z-transform of $[\dfrac1{1 – az^{-1}} ] = (a)^n u(n) \\ ∴y[n] = -2(\dfrac12)^n u(n)+\dfrac13 (\dfrac14)^n u(n)+\dfrac83 u(n) \\ ∴y[n] = [\dfrac83 + \dfrac13 (\dfrac14)^n-2(\dfrac12)^n ]u(n)$

This is the step response of the system.