Using laplace trasnform find the convulation of (f*g)(t).

Question

$let \;f(t)=\sin (3t)$ and $g(t)=e^{-2t}$

Using laplace trasnform find the convulation of (f*g)(t).

convulation theorem: $h(t)=(f*g)(t)= \int ^t_0 f(u) g(t-u) du$

$L(sin (3t)=\frac{3}{s^2+9}\\ L(e^{-2t})=\frac{1}{s+2}$

then how we processed for this problem

Answer 1

Let $h(t)=(f*g)(t)$. We have $$\mathcal{L}\{f*g\} = \mathcal{L}\{f\} \cdot \mathcal{L}\{g\}$$ that is $$H(s)=F(s)G(s)= \frac{3}{s^2+9}\cdot\frac{1}{s+2} =-\frac{9}{13}\cdot\frac{s}{s^2+9}+\frac{6}{13}\cdot\frac{3}{s^2+9}+\frac{9}{13}\cdot\frac{1}{s+2} $$ and then $$h(t)=\mathcal L^{-1}\{F(s)G(s)\}=\left[ -\frac{9}{13}\cos(3t)+\frac{6}{13}\sin(3t)+\frac{9}{13}\mathrm e^{-2t} \right]u(t)$$