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by krackers 808 days ago
>Fourier basis

Technically it's a specialized case of the laplace basis, right? I was always surprised that lots of courses jump directly from the (bilateral) fourier transform to the unilateral laplace transform without proper analysis of the most general case that is the bilateral laplace transform: https://en.wikipedia.org/wiki/Two-sided_Laplace_transform

1 comments

That's true, Laplace corresponds to a basis of complex exponentials that can grow or decay in time instead purely imaginary exponentials. We restrict the Ae^[(a+jb)t] domain just to Ae^(jbt) for Fourier.

From an circuit analysis standpoint (your problem may be different), but exponentials that decay over time ("a" is negative) corresponds to loss in a circuit, whereas exponentials that grow over time ("a" positive) correspond to something blowing up (this is really a nonphysical result but generally means a circuit is going to oscillate on its own, without a source driving that response). I mostly do electromagnetics/passive RF types of problems, in which you generally want everything to be low-loss. In that case Fourier is perfect, especially since I typically care most about steady-state behavior.