# Inverse Laplace transform is not giving the same result with 2 different theorems

I'm trying to transform this image function back to its original :

$$F(s) = \frac{5}{s^2 -4s-32} (I)$$

I tried first to use the convolution theorem : $$\mathcal{F}(f*g)=\mathcal{F}(f)\mathcal{F}(g)$$

$$=> 5 . \mathcal{F}^{-1}(\frac{5}{s^2 -4s-32}) = 5\mathcal{F}^{-1}(\frac{1}{(s-8)(s+4)}) = 5 e^{4t} (*)$$(Inverse Laplace transform)

With the partial fraction decomposition I got a different result :

$$\frac{A}{s-8} + \frac{B}{s-4} => A = 5/12$$ and $$B=-5/12$$ which will lead to a different result $$F(s) = \frac{5}{12(s-8)} -\frac{5}{12(s+4)}$$ => $$\mathcal{F}^{-1}$$(II)= $$\frac{5}{12}(e^{8t} - e^{-4t} ) (**)$$(Inverse Laplace transform)

What am I doing wrong here ? Why are $$(*)$$ and $$(**)$$ not having the same result ?

## 1 Answer

You mistakenly write that f*g is just (f times g), when instead you meant to use $$f*g = \int_0^t f(z)g(t-z) dz$$.

• Oh I see thanks Commented Dec 15, 2020 at 9:15