• NoneOfUrBusiness@fedia.ioOP
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    6 hours ago

    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.

    • AliSaket@mander.xyz
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      3 hours 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.