Sunday, July 15, 2018

Inverse Laplace Transform of e^(2 s)/((s^2 + 1)

I recently needed to compute the inverse Laplace transform of: $$\frac{e^{-2s}}{s^2 + 1}$$ For something like this we'd use the second time shifting theorem: $$ \mathcal[L] f(t-a) u(t-a) = F(s) e^{-as} $$ where in this case $ F(s) = 1/(s^2 + 1) $ We just need to take the inverse transform of $ 1/(s^2+1) $ which is $ \sin (t) $ and then substitute in the time delay so that we arrive at: $ \mathcal[L] f(t-a) u(t-a) = F(s) e^{-as} = u(t-2) \sin (t-2) $ Not too difficult. Out of curiously, I tried to use Mathematica to compute the inverse transform and most surprisingly it failed.