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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 ?

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

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  • $\begingroup$ Oh I see thanks $\endgroup$
    – Samir
    Commented Dec 15, 2020 at 9:15

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