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June 3

Laplace transform

In a Laplace transform, you take a function of time, t, and make it a function of a complex variable, s, which is defined as σ + iω. ω is the angular frequency, but what is σ? --h2g2bob (talk) 20:48, 3 June 2007 (UTC)[reply]

If you have a function f(x) that is nonzero only for x>0, then the Laplace transform G(s) (where s = σ + iω) of that function equals the Fourier transform F(ω) of the product f(x) e^{-σ x}. In other words, you multiply the function by a decaying exponential function before you take its Fourier transform, and σ is the decay constant of that exponential function. Multiplying by the exponential can make the Fourier integral convergent in cases where f itself does not have a Fourier transform. For example, f(x)=x² does not have a Fourier transform, but it does have a Laplace transform (namely G(s) = 2/s³ according to Laplace transform), because the exponential factor makes the integral converge. --mglg_(talk) 21:44, 3 June 2007 (UTC)[reply]

Many thanks --h2g2bob (talk) 10:48, 4 June 2007 (UTC)[reply]