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Consider a mixture of different wavelengths being emitted from the same point (ex: a star). This light consists of a mixture of wavelengths and intensities at each wavelength. When measuring the combined signal as a time-series of amplitude values, we can then use something like Fourier Transform to get a frequency vs. power plot. Would this plot be a "physically" accurate description of what physical wavelengths are in the light, or is it somehow mathematically decomposed further.

I think... the same question - posed differently: Is it possible for the same combined waveform be generated by 2 distinct signals sources - with each source having its own set of emitted frequencies/intensities?

Suppose you have 1 star emitting at frequencies A, B, and C. Then another star emitting at frequencies J, K, and L. Is is somehow possible that the combined waveform (A+B+C) is the same as (J + K + L).

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    $\begingroup$ To get the frequency from a time series, you need to sample at a rate faster than that frequency. For visible light that is difficult. A much easier way to measure the wavelength spectrum is with the aid of a diffraction grating. $\endgroup$ Nov 19, 2023 at 5:10
  • $\begingroup$ Thank you. I think I would be approaching this from a more theoretical perspective. As to whether the there is a 1:1 relationship between the measured waveform and the original physical wavelengths present in the physical phenomena. $\endgroup$
    – codecitrus
    Nov 19, 2023 at 5:17

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Short answer (to your last question): No.

(Starting with a comment: I have no clue why it's important to you that this needs to be "light" as in optics, as it seems to me. You could have easily gotten the same answer from any wave phenomena.)

Slightly longer answer: If you completely sample your time signal, then there is no chance for confusion. However, if you start truncating your signal to sample sizes which start to be too short to properly sample your time-series, then you start getting broader and broader spectra, which then might overlap. After-all, a dirac-delta's fourier transform is a constant value across the whole reciprocal space.

EDIT: Let me put things in other words: Fourier transforms are unambiguous. If I Fourier transform a signal (be it time or frequency) and then make the opposite transformation back, I should get exactly the starting signal. Mathematically, it's a single operation. This means that it cannot be that I would get the same time-series from two different spectra: this would make it ambiguous! If I would now transform this time series back to Fourier space, which one of the two spectra should I transform to?

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  • $\begingroup$ I was wondering moreso if there is more than one single combination of signals that could combine (superimpose) to form the same waveform. In other words, if you are measuring 2 separate signals A and B, in which A is superposition of A1 + A2 + A3; then if B has the same waveform upon measurement, and you are sampling both perfectly, is it true that B has to be a superposition of equivalent signals: B1 = A1, B2 = A2, B3 = A3; Couldn't it also be produced by, say, just 2 signals that superimpose to form the same final waveform? In this case, there is a still 1:1 between the waveform and the FT $\endgroup$
    – codecitrus
    Nov 19, 2023 at 17:44
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    $\begingroup$ Re, "If I Fourier transform a signal..." Fourier Transform is only defined for periodic functions. If you sample an aperiodic signal over some time interval, $P$, and then you perform an FFT, and then you use the output of the FFT to program a synthesizer; the synthesizer will emit a periodic signal with period $P$ which, if you sample it for exactly one period, should give you back the same sequence of samples that you got when you sampled the original. In other words, the synth will keep repeating that same original clip, over and over again forever. $\endgroup$ Nov 19, 2023 at 19:04
  • $\begingroup$ @codecitrus I answered that above: If you have a signal, lets call it $A(t)$, then there is a single $\tilde{a}(f)$ that will come out after you Fourier transform it. So, If you ever measure $B(t)=A(t)$, it does not matter, you will always get $\tilde{a}(f)$. Its a mathematical operation like multiplying by 2: if my starting condition is the same, I will always get the same answer. In your case if A and B are the same, then there is only one corresponding transform. $\endgroup$ Nov 19, 2023 at 19:29
  • $\begingroup$ @SolomonSlow, well, the mathematical definition does not care if its periodic or not, you can always apply it. You are right that in a real case scenario of its application, that is what you get, though. And in general you need to understand it well to apply it correctly to your specific problems, either in the lab, or in simulation. $\endgroup$ Nov 19, 2023 at 19:35

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