# How to Use Convolution Integrals to Solve Inverse Laplace Transforms

fact
$$\Laplace\{f * g\} = F(s)G(s)$$
fact
$$(f * g)(t) = \int_0^t f(t-\tau)g(\tau)d\tau$$
You won't often use this theorem in the forward direction. You'll usually use it to perform inverse Laplace transforms of the form $$F(s)G(s)$$. Using the convolution theorem we can get an expression for the inverse transform but we might not be able to actually compute the integral. Of course if we're doing a simulation we can always try to solve it numerically.
example

Find the inverse Laplace transform of: $$H(s) = \frac{1}{(s^2 + 25)(s^2 + 100)}$$

Rewriting as $$H(s) = F(s)G(s) = \frac{1}{s^2 + 25} \frac{1}{s^2 + 100}$$ We can see that $$f(t) = \frac{1}{5}\sin(5t)$$, $$g(t) = \frac{1}{10}\sin(10t)$$. So $$h(t) = \frac{1}{50}\int_0^t \sin(5t - 5\tau)\sin(10\tau)d\tau$$ $$h(t) = \frac{2}{375} \sin^2(\frac{5t}{2})\sin(5t)$$
practice problems