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Comment by dustingetz | Hacker News Reader

what is an intuition for complex frequency?

Signals estimated by the FFT have two parameters: magnitude and phase. FFT results evade intuition because complex numbers are cartesian. If you convert them to polar coordinates they make more sense as magnitude and phase

https://www.gaussianwaves.com/2015/11/interpreting-fft-resul...

Note that complex numbers are merely convenient for working with two dimensional quantities. The square root of -1 is just math geek for orthogonal, and has nothing whatsoever to do with signals

(Note: GGP, not GP.) I meant complex frequency and not complex amplitude though.

From a physics perspective, complex frequency results in “evanescent waves” - ie, waves that decay rather than oscillate (technically a fully complex frequency of the form a+ib will both oscillate and decay)

Yep: e^iω (with ω real) is an oscillation, but e^σ (with σ real) is an exponential decay when σ < 1, an exponential growth with σ > 0 and constant with σ = 1

so e^(σ + iω) = e^σ * e^iω is just an exponential growth or decay modulated by a sinusoid.. or, if σ is one, is just a pure oscillation

ω is the usual frequency, but σ + iω is the complex frequency. the fourier transform deals with function that receives ω as input, and the laplace transform deals with functions that receives σ + iω instead.

so the fourier transform is just a special case of the laplace transform with σ = 0

> so the fourier transform is just a special case of the laplace transform with σ = 0

Another useful way of looking at it: Laplace transform is doing many extra Fourier-transform-but-with-decay giving you a map of which "global decay timescale" fits your data best -- since each "slice" is itself sufficient to fully describe the time series

They are all cases of integral transforms with different choices of the set of "primitive fingerprints" -- see chirp transform, wavelet transform, chirplet transform etc -- all taking advantage of the fact that if you choose one set of basis "brushes" that are not redundant with each other (e.g. having red-green-blue brushes is independent, as is magenta-green-yellow but having red-green-blue-yellow is not) then you will be able to describe your signal in terms of a composition of those kernels.

So, so I'm going with a rusty knowledge of a computer engineering course from years ago,

> a map of which "global decay timescale" fits your data best

What do you mean by this?

> -- since each "slice" is itself sufficient to fully describe the time series

But each slice is multiplied by an exponent.. which, okay, becomes a convolution that lets you recover the original function

The baseline decay describes the global dissipation or amplification over time.

While Laplace transform is most useful in more complicated systems, this concept is actually best illustrated in a damped/amplified harmonic oscillator model as it serves as the primitive archetype that more complex systems are composed from.

In a nutshell, the general solution is a linear combination of two exponential that are either pure-decay or oscillatory whose imaginary parts, if the signal is to have no imaginary parts, must cancel each other out so that the result is a real signal, which means that the two must be "synchronized" in time against each ither, i.e. having the same oscillation frequency but in the opposite rotation direction (so the imaginary part opposes each other out), and with the same decay progression. This means that the average of their complex frequencies must be real, i.e. <real mean> +/- <imag diff> for underdamped and <real mean> +/- <real diff> for overdamped, and so you can split out the mean as a common decay function, giving you decay(t)*( a*clkwise(t) + b*ccwise(t) )

    a,b:real
    E,F:real->complex
    y:real->real
    y = aE + bF
      = sum(a,b)sum(E,F)/2 + diff(a,b)diff(E,F)/2

As I meant it—of, not for.

I referred to the idea that by plugging an imaginary frequency into the Fourier transform [ETA: the grown-up Fourier transform with the complex exponent, not the schoolboy cosine kludge], you get the Laplace transform, and while that changes the inverse Fourier transform in a different way, it’s not hard to work out how specifically and obtain the inverse Laplace transform.

Why you’d want to do that, I actually don’t know how to explain convincingly. The post hoc rationalization is simple and more or less the reason people prefer the Laplace transform in signal processing: you still get a convolution theorem, but are now allowed to work with exponentially increasing functions, which standard Fourier theory (even the tempered distributions version) can’t accomodate. But while that’s useful from a toolbox standpoint, it isn’t satisfying as motivation, I think.

This is not the only way looking at the complex frequency plane turns out to be useful—there’s a whole thing about doing complex analysis to response functions aka propagators—but there too I can’t really say why you’d guess to look in that direction in the first place.

What I mentioned was that this idea of Laplace as imaginary Fourier extends beyond the reals to the group setting at least to some extent, so it’s not entirely an R-specific accident. Again, dunno why, I’ve explored this stuff a bit but am far from an expert.

> but are now allowed to work with exponentially increasing functions, which standard Fourier theory (even the tempered distributions version) can’t accomodate.

So you can't take the fourier transform of an exponential? But.. it seems you can? https://proofwiki.org/wiki/Fourier_Transform_of_Exponential_...

exp(-|x|) is not an exponential, it's just the easy, bounded, integrable, half of it :)

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