Comment on Breaking functions down to their constituent parts is nice and all, but this is a step too far.
NoneOfUrBusiness@fedia.io 2 days agoIt 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.
AliSaket@mander.xyz 2 days ago
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.