Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 05

Year : DEC 2014

By partial fraction,

$x (s)= \dfrac As +\dfrac B{(s+1)} +\dfrac C{(s+1)^2} + \dfrac D{(s+1)^3} \\ A = x (s) \times \space s \\ =\dfrac {s-2}{(s+1)^3} |s=0 = -2 \\ D = x (s) \times (s+1)^3 \\ = \dfrac{s-2}s|s=-1 = 3 \\ C = x (s) \times (s+1)^3 $

Differentiating w.r.t s

$\dfrac = d{ds} (\dfrac {s-2}s) \\ =\dfrac 2{s^2} |s=-1 =2 \\ B = x (s) \times (s+1)^3$

Differentiating w.r.t s two times

$ =\dfrac 1{2!} \dfrac { d^2}{ds^2} (\dfrac {s-2}s) \\ = \dfrac {-2}{s^3} |s=-1 = 2 \\ x (s)= \dfrac{-2}s + \dfrac 2{(s+1)} + \dfrac 2{(s+1)^2} +\dfrac 3{(s+1)^3 } $

By Inverse Laplace transform,

$x(t) = -2u(t)+2e^{-t} u(t)+2e^{-t} (t)+\dfrac 32 e^{-t} u(t).$