Convolution theorem and laplace transforms

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The discussion centers on the confusion surrounding the convolution theorem and Laplace transforms. The user struggles to understand why the Laplace inverse of the product of transforms, F(s)*G(s), requires integration to yield the convolution of functions f and g. They realize that the operation represented is convolution, not simple multiplication, which clarifies their misunderstanding. The user expresses frustration over their initial confusion but acknowledges the learning process involved. Ultimately, the conversation highlights the complexities of applying the convolution theorem in Laplace transforms.
indianaronald
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Okay, so this is the first time I'm encountering this theorem and I'm not very strong in calculus. But I tried to understand it myself but couldn't.

Convolution theorem is the one in the attachment as give in the book ( couldn't find a way to type that out easily). My doubt is if laplace(f) = F(s) and laplace(g) = G(s) and laplace( f*g )= F(s)*G(s), why not
laplace-inverse[ F(s)*G(s) ]=f*g, which is given but why do the integration at all after that? ( but my answers don;t match if I do it this way; that is without that final integration so I'm obviously misunderstanding it)

Thank you very much for any help.
 

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Damn. I understand. What is there to understand anyway? It is not really multiplication at all. The symbol stands for 'convolution'. I can't delete this thread. So umm...it's going to be my beacon of stupidity, I guess, thank you very much.
 

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