# Convolution Laplace transform

Find the inverse Laplace transform of the giveb function by using the
convolution theorem.

$$F(x) = \frac{s}{(s+1)(s^2+4)}$$

If I use partial fractions I get: $$\frac{s+4}{5(s^2+4)} – \frac{1}{5(x+1)}$$

which gives me Laplace inverses:

$$\frac{1}{5}(\cos2t + \sin2t) -\frac{1}{5} e^{-t}$$

$$f(t) = \int^t_0 e^{-(t -\tau)}\cos(2\tau) d\tau$$

How did they get that?

#### Solutions Collecting From Web of "Convolution Laplace transform"

Related techniques (I), (II). Using the fact about the Laplace transform $L$ that

$$L(f*g)=L(f)L(g)=F(s)G(s)\implies (f*g)(t)=L^{-1}(F(s)G(s)) .$$

In our case, given $H(s)=\frac{1}{(s+1)}\frac{s}{(s^2+4)}$

$$F(s)=\frac{1}{s+1}\implies f(t)=e^{-t},\quad G(s)=\frac{s}{s^2+4}\implies g(t)=\cos(2t).$$

Now, you use the convolution as

$$h(t) = \int_{0}^{t} e^{-(t-\tau)}\cos(2\tau) d\tau .$$