$H(s) = \dfrac 1{s+5} R_e(s)>-5 \ \text {Input } x (t) = e^{-t} u(t)+e^{-2t} u(t)$

Mumbai University > EXTC > Sem 4 > Signals and Systems

Marks : 10

Year : MAY 2014

By Laplace transform,

$X (s) = \dfrac 1{(s+1)} + \dfrac 1{(s+2)} \\ = \dfrac {2s+3}{(s+1)(s+2)} \\ Y (s) = x (s) \times H (s) \\ = \dfrac {2s+3}{(s+5)(s+1)(s+2)}$

By partial fraction,

$Y (s) = \dfrac A{s+1} + \dfrac B{s+2}+ \dfrac C{s+5}$

On solving we get,

$A+B+C=0…… (1) \\ 7A+6B+3C=2….. (2)\\ 10A+5B+2C=3…… (3)$

$Y (s) = \dfrac {-0.583}{s+5} + \dfrac {0.25}{s+1} + \dfrac {0.33}{s+2}$

By Inverse Laplace transform,

$y(t) = -0.583e^{-5t} u(t)+0.25e^{-t} u(t)+0.33u(t)\space\space \text {for }\space ROC\space\space R_e (S) \gt -5.$