# Convolution theorem and laplace transforms

1. Jan 3, 2013

### indianaronald

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.

File size:
12.8 KB
Views:
2,109
2. Jan 3, 2013

### indianaronald

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 gonna be my beacon of stupidity, I guess, thank you very much.