I have learnt about the Laplace convolution theorem: $$\mathcal{L}[f(t)*g(t)]=\mathcal{L}[f(t)]\mathcal{L}[g(t)]$$ So I wonder if there any solution to $$\mathcal{L^{-1}}[F(s)*G(s)]=?$$ If there isn't, why there is in Fourier Transformation?
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1Is there any certain example in the mind? – Mikasa Dec 28 '16 at 06:45
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I made it up in my mind. I just wonder why it is not the part of the theorem. – Alex Thomson Dec 28 '16 at 06:47
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see there http://math.stackexchange.com/questions/2074411/what-is-a-general-form-for-the-laplace-transform-of-ft-ft#comment4260389_2074411 if you meant $F \ast G(s ) = \frac{1}{2i\pi} \int_{\sigma -i\infty}^{\sigma+i \infty} F(u)G(s-u)du$ then yes $\mathcal{L}^{-1}[F \ast G(s)] = f (t) g(t)$ – reuns Dec 28 '16 at 06:51
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Isn't $F(s)*G(s)=\int_0^s F(u)G(s-u),du$ – Alex Thomson Dec 28 '16 at 07:09
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1Take a look at the last five rows of this table – polfosol Dec 28 '16 at 08:31
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I know what it is now. Thank you. – Alex Thomson Dec 28 '16 at 10:48