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Why can one ignore the real part of a complex number because of Fourier transform "only using the rotational part"? Particularly, what property of Fourier transform is this?

Like given in this thread:

https://electronics.stackexchange.com/questions/78638/why-do-we-use-s-j-omega-in-ac-analysis-instead-of-s-sigmaj-omega


Essentially they say that (quoting clabacchio):

The reason why S=jω is chosen to evaluate AC signals is that it allows to convert the Laplace transform into Fourier transform.

The reason is that while S is a complex variable, what's used in the Fourier representation is just the rotational (imaginary) component, hence σ=0.

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  • $\begingroup$ Quoting the accepted answer by Kaz: "What's happening is that $\sigma$ is being ignored because it is assumed to be zero. The reason for it is that we are looking at the response of the system to periodic (and thus non-decaying) sinusoidal signals, whereby Laplace conveniently reduces to Fourier along the imaginary axis. The real axis in the Laplace domain represents exponential decay/growth factors that pure signals do not have, and which Fourier does not model." $\endgroup$ – Saucy O'Path Aug 17 '18 at 16:16
  • $\begingroup$ Physically $\omega$ encodes frequency of pure oscillation while $\sigma$ encodes dissipation. A common simplifying assumption in electronics is that the dissipation is negligible so you can deal with pure oscillation. $\endgroup$ – Ian Aug 17 '18 at 16:16
  • $\begingroup$ @SaucyO'Path Okay but why exactly is the real axis in Laplace domain "exponential decay/growth" factors, because I'm not familiar with Laplace transform? Or perhaps it's a common property? $\endgroup$ – mavavilj Aug 17 '18 at 16:18
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    $\begingroup$ @mavavilj More or less what you are doing in both cases is writing a function as a "linear combination" of a continuum of exponentials with different coefficients. The Laplace transform with a positive real part of $s$ corresponds to finding the coefficients of exponentials which are decaying (and oscillating) as you go forward in time. The Fourier transform with real $\omega$ corresponds to exponentials with no real part, which are purely oscillating. $\endgroup$ – Ian Aug 17 '18 at 16:20
  • $\begingroup$ @Ian I see (by looking at the definition: en.wikipedia.org/wiki/Laplace_transform#Formal_definition), and why does Fourier transform ignore those? $\endgroup$ – mavavilj Aug 17 '18 at 16:30

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