Convolution Theorem

$if\ f_1(t)\ and\ f_2(t)\ are\ two\ functions\ then\ the\ following\ Integral\\ \int_0^t f_1(u) f_2(t-u)du\\ is\ called \ the \ convolution\ and\ is\ denoted \ by\\ f_1(t).f_2(t).Thus\\ f_1(t).f_2(t).=\int_0^t f_1(u) f_2(t-u)du\\ Theorem\\ Let\ Lf_1(t)=\Phi_1(s) and \ Lf_1(t)=\Phi_1(s),then\\ L^{-1}[\Phi_1(s).\Phi_2(s)]=\int_0^t f_1(u) f_2(t-u)du\\ $

$L^{-1}(\dfrac{s}{(s^2+1)(s^2+4)})\\ L^{-1}(\dfrac{1}{(s^2+1)}.\dfrac{s}{(s^2+4)} )\\ \\ \\ \phi_1(s) =\dfrac{1}{(s^2+1)} ,\phi_12(s) =\dfrac{s}{(s^2+4)} $

$$L^{-1}\phi_2(s) =L^{-1} \dfrac{1}{(s^2+1) }\ =\ sint.\ L^{-1}\phi_1(s) =L^{-1} \dfrac{s}{(s^2+4) }\ =cos2t. $ $As\ per\ convolution\ theorem\ If\ Lf_1(t)=\Phi_1(s) and \ Lf_1(t)=\Phi_1(s),then\ L^{-1}[\Phi_1(s).\Phi_2(s)]=\int_0^t f_1(u) f_2(t-u)du\ $ $L^{-1}[\phi(s)]=\int_0^t cos2u.sint(t-u).du\ = =\dfrac{1}{2}\ \int{ sin(2-1)u+2t}- sin(1+2)u-2t\ =\dfrac{1}{2}\int sin(t+u)-sin (3u-t).du\= \dfrac{1}{2}[-\dfrac{cos(t+u)}{1}+\dfrac{cos(3u- t)}3]_0^t\= \dfrac{1}{2}[-cos2t+cos2t/3+ cost-cost/3]\= \dfrac{1}{2}[ \dfrac{2}{3}cost-\dfrac{2}{3}cos2t]\= \dfrac{cos2t-cost}{3}$