3 ms·
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
by 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 agoFT 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 ago90 degree rotation? That would imply that the fourier transform is orthogonal to the original function.
- nicwilson 3y agoIt 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 agoOk, now I see it!
- nextaccountic 3y agoOk 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>the fourier transform is the inverse of itself Not exactly. Almost.
- ska 3y ago> the best analogy is a basis change in a vector space: It's closer than an analogy. FT is essentially a map of basis into a dual space, but this dual is non uniquely isomorphic with the original so given an arbitrary choice of isomorphism you can turn it into "just" a change of basis.
- dustingetz 3y agofractional FT is the uncertainty principle then? you can know position locally, or momentum locally, or you can know them both blurrily with a rotated basis in between
- nextaccountic 3y agoYes, but that's because the momentum is the fourier transform of the position and vice versa So there's an uncertainty principle for signal processing, entirely because of the fourier transform and not because of quantum mechanics See this video https://www.youtube.com/watch?v=MBnnXbOM5S4 https://www.youtube.com/watch?v=MBnnXbOM5S4