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Can i say that the laplace transform of a function convoluted with itself $\left[\,\mathrm{f}\left(\,t\,\right)*\mathrm{f}\left(\,t\,\right)\,\right]$ is the square of the Laplace Transform of that function $\left[\,\mathrm{F}\left(\,s\,\right)\,\right]^{\,2}$ ?.

Felix Marin
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F.Abdullah
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    Exactly it! That's the cool thing about the Laplace Transform. – Sean Roberson Jun 19 '17 at 21:52
  • okay....what if i wanted to evaluate laplace transform of fk(t) say for s=T/r where fk(t)=the convolution [f(t)*f(t)]; can i say that that equals the square of the laplace transform of f(t) at s=T/r?!! – F.Abdullah Jun 19 '17 at 23:11
  • It looks like it. I'll have an expanded answer in about an hour. – Sean Roberson Jun 19 '17 at 23:12
  • okay.. i will be waiting for your answer...Thank you so much – F.Abdullah Jun 19 '17 at 23:14
  • (assuming everything converges absolutely) $$F(s) = \int_0^\infty \int_0^\infty f(t) f(u) e^{-s(u+t)}du dt = \int_0^\infty \int_0^\infty f(\tau-u) f(u) e^{-s(u+\tau-u)} du d\tau$$ Can you finish from there ? – reuns Jun 19 '17 at 23:22
  • yes thank you... no problem for this ....but can you help me in my second question (the one in the comments) where i want to evaluate laplace transform of fk(t) at a specific value of s say s=T/r or whatever where fk(t)=[f(t)*f(t)] does that equal the square of the laplace transform of f(t) with s evaluated at T/r?!!!.......i don't know if I've explained it well but please feel free to ask for any further explanations – F.Abdullah Jun 20 '17 at 00:01
  • i mean; is this true ?! Fk(s)|s=T/r= [F(s)|s=T/r]^2

    Where fk(t)=f(t)*f(t)

     – F.Abdullah Jun 20 '17 at 01:54

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Suppose $f$ and $g$ admit Laplace transforms. Then their convolution

$$ (f * g)(t) = \int_0 ^\infty f(\tau)g(t-\tau) \ d\tau $$

has Laplace transform $\mathcal{L}(f * g) = \mathcal{L}(f) \mathcal{L}(g)$. Of course, if $f = g$ then $\mathcal{L}(f * f) = (\mathcal{L}(f))^2 $. If you wish to evaluate the transform at a specific value then you can do so.

I suggest you expand on @user1952009's comment and prove the convolution theorem, that the transform of a convolution is the product of transforms.

Sean Roberson
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