226
Guess I'll just burn (sh.itjust.works)
submitted 5 months ago by RmDebArc_5@sh.itjust.works to c/memes@lemmy.ml
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[-] captainlezbian@lemmy.world 9 points 5 months ago
[-] chemical_cutthroat@lemmy.world 17 points 5 months ago

Only if you are blowing shit up.

[-] PM_ME_VINTAGE_30S@lemmy.sdf.org 5 points 5 months ago* (last edited 5 months ago)

If your signal looks like f(t) = K•u(t)e^at with u(t) = {1 if t≥0, 0 else}:

  • If Real(a) > 0, then your signal will eventually blow up.
  • If Real(a) < 0, then you signal will not blow up. In fact, your signal will have a maximum absolute value of |K|, and it will approach zero as time goes on.
  • If Real(a) = 0, it is either a complex sinusoid or a constant. In either case, it is bounded with maximum absolute value of |K|. It very much does not blow up.

So e pops up all the time in stable systems and bounded signals because the function e^at solves the common differential equation dx/dt = ax(t) with x(0)=1 regardless of the value of a, particularly regardless of whether or not the real part of a causes the solution to blow up.

[-] sharkfucker420@lemmy.ml 5 points 5 months ago* (last edited 5 months ago)

Usually it is e^t or something similar

this post was submitted on 22 May 2024
226 points (100.0% liked)

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