this post was submitted on 29 Nov 2025
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It is, but conceptually it's a lot weirder than the Fourier transform, whose idea at least is very straightforward. I mean, when doing Laplace transforms you do have to assume that int(e^tdt){0}{∞}=-1. I'd definitely rather use the Laplace transform, but you couldn't pay me to explain how that shit actually works to an undergrad student.
Basically the assumption is that the signal x(t) is equal to 0 for all t < 0 and that the integral converges. And what is a bit counter-intuitive: Laplace transformations can be regarded as generalizations of Fourier transformations, since the variable s is not only imaginary but fully complex. But yeah... I would have to brush up on it again, before explaining it as well. It's... been a while.