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Mathematically, the Fourier transform is "simply" a way of representing time signals in a certain orthogonal vectorial basis. Vectors in an ordinary sense, e.g.

• weinzierl 2y ago
  Excellent answer and I am sure you are aware of this, but like to point out: "Mathematically, the Fourier transform is "simply" a way of representing time signals in a certain orthogonal vectorial basis." Not just time signals but any piecewise continuous and differentiable as well as Dirichlet integrable function. This has many applications, just a few examples from the top of my head: image processing, solving differential equations, fast multiplication. I'd also like to add that from a mathematical point of view these transforms are "lossless" in the sense that the transformed function has the exact same information as the original and you can get back the exact original even if all you have is the transform. I feel this often gets lost when people approach the Fourier transform from a more engineering perspective, not at least because we often do the transform to throw away unwanted information, like certain frequency components. In the end it really is just one of many perspectives to look at a function.
  • quibono 2y ago
    > I feel this often gets lost when people approach the Fourier transform from a more engineering perspective, not at least because we often do the transform to throw away unwanted information, like certain frequency components. That was my problem as well. My first introduction to Fourier transforms was through more of an engineering lens. I remember having trouble with the _inverse_ Fourier transform. I was OK with a Fourier inverse of an already transformed function but I wasn't quite sure what that would mean when applied to a non-transformed, "regular" function.
    • takd 2y ago
      As a related aside, the terms "cepstrum" and the "quefrencies" [1] (c.f. spectrum and frequencies) sound so hilarious that when I first heard about them I was convinced it was some kind of prank. [1] https://en.wikipedia.org/wiki/Cepstrum https://en.wikipedia.org/wiki/Cepstrum
      • cmehdy 2y ago
        This does read like a joke, I had never heard of it either and I'm wondering if many people do use this at all.. Operations on cepstra are labelled quefrency analysis (or quefrency alanysis[1]), liftering, or cepstral analysis. It may be pronounced in the two ways given, the second having the advantage of avoiding confusion with kepstrum.
        • aswanson 2y ago
          It's used in speech signal processing & seismic signal analysis.
        • ljosifov 2y ago
          Almost all speech recognizers until this latest crop (of end-to-end DL NN ASR) operated on cepstral coefficients (and their delta-s and delta-delta-s) as their feature vector.
    • pas 2y ago
• OJFord 2y ago
  > one basis could, for example, be two vectors pointing North and East; another could be a vector pointing along a certain road and one perpendicular to it. And there's no requirement that they be perpendicular is there? The second just needs 'some amount of perpendicular', North and North-East for example? Since any [n, e] can also be described as [(1-sqrt(2)*e)*n, sqrt(2)*e] in the latter. (I think that's right, but my main point is you can do it, not the particular value there, and if that's way off I'll blame the fever.)
  • Jensson 2y ago
    You typically want an orthonormal basis though, but yes you don't need it.
  • datascienced 2y ago
    If you can walk this way / and this way | then you can do / minus | to get your orthogonal -
    • OJFord 2y ago
      Yes exactly.
• eigenspace 2y ago
  I used to think of it like another basis too, but nowadays I think this basis analogy is a bit fraught, or at least not the whole story. In particular, for multidimensional spaces, the usual multidimensional Fourier transform only really works if you have a flat metric on that space (I.e. no curvature). That’s a bit of a warning signal given that our universe itself is curved. There was some very interesting work recently where it was shown how to generalize Fourier series to certain hyperbolic lattices [1], and one important outcome of that work is that the analog of the Fourier space is actually higher dimensional than the position space. Furthermore, the dimensionality of the ‘Fourier space’ in this case depends on the lattice discretization. One 2D lattice discretization may have a 4D frequency-like domain, and another 2D lattice might have a 8D frequency-like domain. [1] https://arxiv.org/abs/2108.09314 https://arxiv.org/abs/2108.09314 or https://www.pnas.org/doi/full/10.1073/pnas.2116869119 https://www.pnas.org/doi/full/10.1073/pnas.2116869119
  • Jensson 2y ago
    > That’s a bit of a warning signal given that our universe itself is curved. What does this has to do with whether they are a different basis for cases where we don't account for curvature? This seems completely irrelevant, sure the tool can't be used in some cases but it can be used as a basis change in other cases.
  • mananaysiempre 2y ago
    Not the whole story indeed, but you have to dive into representation theory somewhat to get more: the Fourier transform is more or less the representation theory of the (abelian) group of the translations of your space, thus the homogeneity requirement. The finite-lattice version[1] (a discretized torus, basically) may serve to hint what’s in stock here. [1] https://www-users.cse.umn.edu/~garrett/m/repns/notes_2014-15/01_finite_abelian.pdf https://www-users.cse.umn.edu/~garrett/m/repns/notes_2014-15... (linear algebra required at least to the degree that one is comfortable with the difference between a matrix and an operator and knows what a direct sum is)
    • eigenspace 2y ago
      If you like this topic, I strongly recommend you read the references I attached to my comment. In uniformly curved 2D hyperbolic spaces, it turns out that there is a higher dimensional non-Abelian Fuchsian translation group defined on a higher genus torus.
• enaaem 2y ago
  In the past astronomers believed in the geocentric model of the universe with epicycles. It was extremely accurate, and if more accuracy was needed they added more epicycles. It was a completely wrong model, but they unknowingly used the Fourier series as a function approximator.
  • TeMPOraL 2y ago
    It wasn't a wrong model, it was just much more complex than needed, given better understanding of physics. Viewing the universe relative to stationary Earth is a perfectly fine exercise, even if it means you have to DFT the rest of the solar system for the math to work.
• deleted 2y ago
  [deleted]
• HarHarVeryFunny 2y ago
  Sure, nothing special about sine waves as basis functions for signal decomposition. Not necessarily the best either, depending on what you want to do. Still, as pertains to whether "the frequency domain is a real place", maybe sine waves are relevant as representing resonant frequencies of physical systems. There also seems to be something fundamental about the way multiple radio frequencies can simultaneously propagate through a vacuum as long as they are different frequencies.
• BeetleB 2y ago
  Ok. Now what's the difference between the Fourier transform and the Fourier series? To me what you described sounds more like the Fourier series.
  • nextaccountic 2y ago
    the fourier transform of a periodic signal is composed of a train of dirac deltas, each multiplied by some factor the delta with smallest frequency is the fundamental frequency, and the others are harmonics when you do the inverse fourier transform on this train, each delta becomes a sinusoid that's how you can write any periodic function as a sum of sinusoids, all of them multiples of the fundamental frequency and that's the fourier series: it's just the fourier transform, followed by an inverse fourier transform, macroexpanded but the fourier series only work for periodic functions, because only periodic functions have a bunch of isolated, periodic deltas as its fourier transform so the fourier transform is only half the step of a fourier series (to write down the series you also need the inverse fourier transform) but, at the same time, the fourier transform is a generalization of the fourier series, because it works for nonperiodic functions too
    • nextaccountic 2y ago
      To elaborate more, when we say that a signal is the linear combination of infinitely many frequencies, if those frequencies are multiples of a fundamental frequency (that is, a countable set of frequencies), then we are talking about a Fourier series and the signal must be periodic But if those frequencies span a whole continuum (rather than frequencies f, 2f, 3f, 4f..), that is, an uncountable set of frequencies, then this signal is non-periodic and we can't talk about a Fourier series anymore, we must use the Fourier transform
    • BeetleB 2y ago
      > the delta with smallest frequency is the fundamental frequency, and the others are harmonics Nitpick, but this isn't true. If my signal is a linear combination of two sinusoids - one at frequency 3 and the other at frequency 5, then there is no "fundamental" frequency when you do the FT.
• woopsn 2y ago
  I agree overall. But a note - every orthonormal basis partitions the frequency spectrum. It doesn't go away, if you are using e.g. polynomials then you're building functions up out of their frequency components too. The Fourier basis has every element correspond to a specific frequency, which is special in some sense. I would say rather.. they are each designed for a purpose. A basis change can rearrange the spectrum in such a way that analysis of it is complicated. Then you're analyzing something else (eg smoothness). Most functions of interest do have distinctive spectra, even if the Fourier basis doesn't answer all the questions.