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Convolution theorem and laplace transforms

  1. Jan 3, 2013 #1
    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.
     

    Attached Files:

  2. jcsd
  3. Jan 3, 2013 #2
    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.
     
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