# Inverse Laplace help

Tags:
1. Dec 2, 2016

### zachdr1

1. The problem statement, all variables and given/known data
$$F(s) = \frac{s^2e^{-4s}}{s^2+9}$$

2. Relevant equations
$$\mathcal{L} ({f(t-a)\mathcal{H}(t-a)}) = e^{-as}F(s)$$

3. The attempt at a solution

$$F(s) = e^{-4s}(\frac{s^2}{s^2+9})$$
using $$\mathcal{L} {f(t-a)\mathcal{H}(t-a)} = e^{-as}F(s)$$
$$f(t) = cos(3t-12)\mathcal{H}(t-4)$$

I do not understand why this is incorrect. My professor got $$f(t) = \delta(t - 4) - 3sin(3t - 12)\mathcal{H}(t-4)$$

EDIT: Wow I get it now, it's because $$\mathcal{L}(cos(kt)) = \frac{s}{s^2+k}$$ not $$\mathcal{L}(cos(kt)) = \frac{s^2}{s^2+k}$$ stupid mistake lol

Last edited: Dec 2, 2016
2. Dec 8, 2016

### Greg Bernhardt

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.