Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3

Marks : 06

Years : MAY 2016

By convolution theorem

$L^{-1} [f\bar{(s)} .g\bar { (s)}] =\int\limits_{u=0}^{u=t} f(u) g(t-u) du \\ where \space f\bar{(s)} =\dfrac s{s^2+1}, g\bar {(s)} =\dfrac 1{s^2+4} \\ L^{-1} [f\bar{(s)}] =f(t) =\cos t, L^{-1} [g\bar{(s)}]=g(t) =\dfrac {\sin 2t}2\\ \therefore L^{-1}[\dfrac {s}{(s^2+1)(s^2+4)}]=\int\limits_{u=0}^{u=t}\cos u. \dfrac {\sin2(t-u)}2 du =\dfrac 14 \int\limits_0^t\sin (2t-u)+\sin(2t-3u) \\ =\dfrac 14 [\dfrac {-\cos(2t-u)}{(-2)}-\dfrac {\cos(2t-3u)}{(-3)}]_0^{u=t} =\dfrac 14[\cos t+\dfrac {\cos t}3-\cos 2t -\dfrac {\cos 2t}3] \\ =\dfrac 14 [\dfrac {4\cos t-4\cos 2t}3] =\dfrac 4{12} [\cos t-\cos 2t]= [\dfrac {\cos t-\cos 2t}3]$