How to Use Convolution Integrals to Solve Inverse Laplace Transforms

\(\Laplace\{f * g\} = F(s)G(s)\)
\( (f * g)(t) = \int_0^t f(t-\tau)g(\tau)d\tau \)
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.

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