4 ms·

I don’t like the casting of the frequency domain view as the ‘recipe’ and the time domain view as the ‘product’. The point of Fourier is that you can switch bet

I don’t like the casting of the frequency domain view as the ‘recipe’ and the time domain view as the ‘product’. The point of Fourier is that you can switch between these perspectives losslessly - they contain equivalent information. The ‘smoothie’ metaphor of ‘unmixing’ the smoothie to get the ingredients, and then blending it to get the smoothie back conjures the impression that Fourier transformation is some sort of entropy reversing process, which is misleading. A time domain function has a frequency domain interpretation - that doesn’t mean the frequency domain function is what ‘made’ the time domain function. It’s a chicken and an egg - both make each other. Just like a function has a derivative, and you can recover the original function (modulo it’s y-offset) by taking its antiderivative - that doesn’t make the derivative the ‘recipe’ for the function.

diracs_stache 3y ago
Agreed, in image processing the time-domain representation doesn't always carry significance when spatial extent is pertinent
throwaway3091 3y ago
[dead]
nextaccountic 3y ago
the best analogy is a basis change in a vector space: you can have the same data (the same vector) viewed in different ways if you look at it using different a base. for example, in physics, the numerical value of the coordinates of an object change in different frames of reference and this is not just an analogy: in the vector space of functions the fourier transform is indeed a basis change (to be more precise, a rotation). and from this arises the fractional fourier transform, which is a halfway change: if the fourier transform is a 180 degrees rotation, a fractional transform is something in between https://en.m.wikipedia.org/wiki/Fractional_Fourier_transform https://en.m.wikipedia.org/wiki/Fractional_Fourier_transform and this also explains why the fourier transform is the inverse of itself: two 180 degrees rotations gets you to the same place you were before
- nicwilson 3y ago
  FT is a 90 degree rotation not a 180, the FT of the FT of a function is the mirror image about the origin, not the function itself.
  - meindnoch 3y ago
    90 degree rotation? That would imply that the fourier transform is orthogonal to the original function.
    - nicwilson 3y ago
      It is, because wavenumber and position are distinct variables and are orthogonal to each other. FT turn position into wavenumber (positional frequency) and wavenumber into negative position: [ 0 1] [x] [ ω] = [-1 0] [ω] [-x] see also https://en.wikipedia.org/wiki/Linear_canonical_transformation#Fourier_transform https://en.wikipedia.org/wiki/Linear_canonical_transformatio... the rotation matrix [[ 0 1], [-1 0]] is a 90 degree rotation.
      - meindnoch 3y ago
        Ok, now I see it!
  - nextaccountic 3y ago
    Ok you're right! I originally wrote down 90 degrees but then I had a conflicting view about being the inverse and then reasoned it must be 180 degrees So the fourier transform of the fourier transform isn't the same as the inverse fourier transform? (ignoring the scaling bits that can be normalized I think), so I've been lied to? Anyway here is a funny pair of questions https://math.stackexchange.com/questions/1472528/why-is-the-fourier-transform-self-inverse https://math.stackexchange.com/questions/1472528/why-is-the-... https://math.stackexchange.com/questions/3922412/why-isnt-the-fourier-transform-self-inverse https://math.stackexchange.com/questions/3922412/why-isnt-th...
- meindnoch 3y ago

diracs_stache 3y ago

Agreed, in image processing the time-domain representation doesn't always carry significance when spatial extent is pertinent

throwaway3091 3y ago

[dead]

nextaccountic 3y ago

the best analogy is a basis change in a vector space: you can have the same data (the same vector) viewed in different ways if you look at it using different a base. for example, in physics, the numerical value of the coordinates of an object change in different frames of reference and this is not just an analogy: in the vector space of functions the fourier transform is indeed a basis change (to be more precise, a rotation). and from this arises the fractional fourier transform, which is a halfway change: if the fourier transform is a 180 degrees rotation, a fractional transform is something in between https://en.m.wikipedia.org/wiki/Fractional_Fourier_transform https://en.m.wikipedia.org/wiki/Fractional_Fourier_transform and this also explains why the fourier transform is the inverse of itself: two 180 degrees rotations gets you to the same place you were before

nicwilson 3y ago
FT is a 90 degree rotation not a 180, the FT of the FT of a function is the mirror image about the origin, not the function itself.
- meindnoch 3y ago
  90 degree rotation? That would imply that the fourier transform is orthogonal to the original function.
  - nicwilson 3y ago
    It is, because wavenumber and position are distinct variables and are orthogonal to each other. FT turn position into wavenumber (positional frequency) and wavenumber into negative position: [ 0 1] [x] [ ω] = [-1 0] [ω] [-x] see also https://en.wikipedia.org/wiki/Linear_canonical_transformation#Fourier_transform https://en.wikipedia.org/wiki/Linear_canonical_transformatio... the rotation matrix [[ 0 1], [-1 0]] is a 90 degree rotation.
    - meindnoch 3y ago
      Ok, now I see it!
- nextaccountic 3y ago
  Ok you're right! I originally wrote down 90 degrees but then I had a conflicting view about being the inverse and then reasoned it must be 180 degrees So the fourier transform of the fourier transform isn't the same as the inverse fourier transform? (ignoring the scaling bits that can be normalized I think), so I've been lied to? Anyway here is a funny pair of questions https://math.stackexchange.com/questions/1472528/why-is-the-fourier-transform-self-inverse https://math.stackexchange.com/questions/1472528/why-is-the-... https://math.stackexchange.com/questions/3922412/why-isnt-the-fourier-transform-self-inverse https://math.stackexchange.com/questions/3922412/why-isnt-th...
meindnoch 3y ago