Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : DEC 2015

The given difference equation is

$y[n] – 2y[n - 1] + y[n - 2] = x[n] + 3x[n - 3]$

Taking Z-transform of the difference equation

$Y(z) – 2z^{-1} Y(z) + z^{-2} Y(z) = X(z) + 3z^{-3} X(z) ----------- 1 \\ [1 - 2z^{-1} + z^{-2}] Y(z) = [1 + 3z^{-3}] X(z) \\ \dfrac {Y(z)}{X(z)} = \dfrac {1 + 3z^{-3}}{ 1 - 2z^{-1} + z^{-2}} \\ ∴H(z) = \dfrac {Y(z)}{X(z)} \\ = \dfrac {z^3 + 3}{ z(z^2 - 2z + 1) } \\ H(z) = \dfrac {z^3 + 3}{ (z^3 - 2z^2 + z) } $

Since powers of both numerator and denominator are same, the above equation is not in proper form.

Let us convert it in proper form

$∴H(z) = 1 + \dfrac {2z^2 – z + 3}{ z^3 – 2z^2 + z } \\ H(z) = 1 + \dfrac {2z^2 – z + 3}{ z(z^2 – 2z + 1) } \\ ∴H(z) = 1 + \dfrac {2z^2 – z + 3}{ z(z – 1)^2} ----------- 2 \\ Let\space H_1(z) = \dfrac {2z^2 – z + 3}{ z(z – 1)^2 } $

Let us write above equation as

$\dfrac {H_1(z)}z =\dfrac { 2z^2 – z + 3}{ z^2 (z – 1)^2} ---------- 3$

Expanding the above equation in partial fractions

$\dfrac {H_1(z)}z = \dfrac {A_1}z + \dfrac {A_2}{z^2} + \dfrac {A_3}{(z - 1)} + \dfrac {A_4}{(z - 1)^2} ----------- 4 \\ \text {Here}\space A_1 =\dfrac d{dz} [z^2\dfrac { H_1(z)}z] |_{z = 0} \\ =\dfrac d{dz} [z^2 \dfrac { 2z^2 – z + 3}{ z^2 (z – 1)^2} ] |_{z = 0} \\ = \dfrac {(z - 1)^2 (4z - 1) – (2z^2 – z + 3) (2(z - 1))}{(z - 1)^4} |_{z = 0} \\ = \dfrac {(1) (-1) – (3) (-2)}1 \\ ∴A_1 = 5 \\ A_2 = z^2 \dfrac {H_1(z)}z |_{z = 0} \\ = z^2 \dfrac { 2z^2 – z + 3}{ z^2 (z – 1)^2} |_{z = 0} \\ = \dfrac 31 \\ ∴A_2 = 3 \\ A_3 = \dfrac d{dz} [(z - 1)^2 \dfrac{H_1(z)}z ] |_{z = 1} \\ =\dfrac d{dz} [(z - 1)^2 \dfrac{2z^2 – z + 3}{ z^2 (z – 1)^2} ] |_{z = 1} \\ =\dfrac { (z^2) (4z - 1) – (2z^2 – z + 3) (2z)}{z^4} |_{z = 1} \\ = \dfrac {1(4 - 1) – (2 – 1 + 3) (2)}1 \\ ∴A_3 = -5 \\ A_4 = (z - 1)^2 \dfrac {H_1(z)}z |_{z = 1} \\ = (z - 1)^2 \dfrac {2z^2 – z + 3}{ z^2 (z – 1)^2} |_{z = 1} \\ = \dfrac {2 -1 + 3}1 \\ ∴A_4 = 4 $

Hence equation 4 becomes

$\dfrac {H_1(z)}z = \dfrac 5z + \dfrac3{z^2} + \dfrac {-5}{(z - 1)} +\dfrac 4{(z - 1)^2 } \\ H_1(z) = 5 + \dfrac 3z - 5 \dfrac z{z - 1} + 4 \dfrac z{(z - 1)^2 } $

Therefore equation 2 becomes

$H(z) = 1 + 5 + \dfrac 3z - 5\dfrac z{z - 1} + 4 \dfrac z{(z - 1)^2 } \\ H(z) = 6(1) + 3z^{-1} -5 \dfrac 1{1 – z^{-1}} + 4 \dfrac{z^{-1}}{(1 – z^{-1})^2} ------------ 5 $

Inverse Z–transform of $1 = δ(n)$

Inverse Z–transform of $z^{-nO} = δ(n - n_O)$

Inverse Z–transform of $\dfrac 1{1 – z^{-1}} = u(n)$

Inverse Z–transform of $\dfrac {az^{-1}}{(1 – az^{-1})^2} = na^n u(n)$

Taking inverse Z-transform of equation 5

$∴h(n) = 6 δ(n) + 3 δ(n - 1) -5 u(n) + 4 n u(n)$