Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2015

The given difference equation is

$y[n] – 4y[n – 1] + 4y[n – 2] = x[n] – x[n – 1]$

Taking the unilateral Z-transform of above equation, $Y(z) – 4[z^{-1} Y(z) + y(-1)] + 4[z^{-2} Y(z) + y(-1)z^{-1} + y(-2)] = X(z) – [z^{-1} X(z) + x(-1)] ---------- 1 $

Putting the initial conditions,

$y(-1) = y(-2) = 0 and \space\space x(-1) = 0$

Therefore equation 1 becomes

$Y(z) – 4z^{-1} Y(z) + 4z^{-2} Y(z) = X(z) – z^{-1} X(z)\\ [1 – 4z^{-1} + 4z^{-2}] Y(z) = [1 – z^{-1}] X(z) ----------- 2 $

Suppose input x(n) = u(n)

Then $X(z) = \dfrac 1{1 – z^{-1} }$

Therefore equation 2 becomes

$[1 – 4z^{-1} + 4z^{-2}] Y(z) = [1 – z^{-1}] \dfrac 1{1 – z^{-1}} \\ [1 – 4z^{-1} + 4z^{-2}] Y(z) = 1 \\ ∴Y(z) = \dfrac 1 {1 – 4z^{-1} + 4z^{-2} } \\ Y(z) = \dfrac {z^2}{z^2 – 4z + 4 } \\ ∴\dfrac {Y(z)}z = \dfrac z{z^2 – 4z + 4 } \\ ∴\dfrac {Y(z)}z = \dfrac z{(z - 2)^2 }$

Expanding the equation in partial fractions

$\dfrac {Y(z)}z = \dfrac {A_1}{(z - 2)} + \dfrac {A_2}{(z - 2)^2} ----------- 3 \\ \text {Here} \space A_1 = \dfrac d{dz} [(z - 2)^2 \dfrac {Y(z)}z ] |_{z = 2} \\ = \dfrac d{dz} [(z - 2)^2 z(z - 2)^2 |_{z = 2} \\ ∴A_1 = 1 \\ A_2 = (z - 2)^2 \dfrac {Y(z)}z |_{z = 2} \\ = (z - 2)^2 z(z - 2)^2 |_{z = 2} \\ ∴A_2 = 2 $

Hence equation 3 becomes

$\dfrac {Y(z)}z = \dfrac 1{(z - 2)} + \dfrac 2{(z - 2)^2 } \\ Y(z) = \dfrac z{(z - 2)} + \dfrac {2z}{(z - 2)^2 } \\ ∴Y(z) = \dfrac 1{(1 - 2z^{-1})} + \dfrac {2z^{-1}}{(1 - 2z^{-1})^2} ------------ 4$

Inverse Z-transform of $[\dfrac 1{(1 - az^{-1})} ] = a^n u(n)$

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

Therefore inverse Z-transform of equation 4 becomes

$y(n) = 2^n u(n) + n 2^n u(n) \\ ∴y(n) = (n + 1) 2^n u(n)$