# Inverse Laplace help

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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