Talk:Fractional Fourier transform
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fractional fourier transfrom - a significant revolution??
- Yes, it is. Namias introduced it to quantum mechanics, but I'm not sure what they use it for in that field. In optics though, it's quite important. The relevant paper is, A.W. Lohmann, “Image rotation, Wigner rotation and the fractional Fourier transform,” J. Opt. Soc. Am. A 10, 2181–2186 (1993). That introduced it to the optics community, which uses it to model the effect of quadratic phase systems. - jhealy 22:13, Aug 24, 2006 (GMT)
Figures and meaning ?
How does one actually understand the Figure "Time/Frequency Distribution of Fractional Fourier Transform."? I find this quite confusing, are these actually spectrograms? —Preceding unsigned comment added by Dai mingjie (talk • contribs) 19:55, 27 July 2009 (UTC)
- I believe they are spectrograms, or at least something similar, like a Cohen class distribution. The behaviour is consistent with what I'd expect of those, though it's all clean enough that it may be illustrative rather than really generated from data. If you have a signal processing background, you might be confused by the origin being at the centre - a convention from Fourier optics.
- The first image is a sinusoid, and the last one is a pair of delta functions, the Fourier transform of the first. The others are intermediate rotations, and would appear to be chirps if viewed as a time- or space-varying signal. Jhealy (talk) 18:57, 14 August 2009 (UTC)
It would be nice to have a plot of what the basis functions look like, that the FrFT is analysing the signal into linear combinations of.
FRFT as an operator
My original research shows that FRFT by angle alpha applied to f(w) is equivalent to applying the operator exp(i*alpha*T) where T = 0.5 * (1-w^2+d^2/dw^2) With this representation you can do some things easily, like use it to show that exp(-w^2/2) is unchanged by FRFT regardless of alpha, and you can also show that exp(w^2/2) simply changes to exp(i*alpha)*exp(w^2/2) under FRFT. If there's a trusted reference somewhere it would be nice to mention this operator representation in the article. Doubledork (talk) 21:03, 6 September 2011 (UTC)
- That doesn't look right. If α=π/2 the FRFT is just an FT, and applied to the function f(ω)=1 it should give a delta function, but your operator in that case gives exp(iπ(1-ω2)/4). Note also that Fourier transformation of exp(+ω2/2) makes no sense at all because that is not a tempered distribution (it is not in the domain where the Fourier transform is even defined). — Steven G. Johnson (talk) 22:56, 6 September 2011 (UTC)
- "your operator in that case gives exp(iπ(1-ω2)/4" - Not at all - note that for the operator "T" I defined above, T*1 = (1-w^2)/2, but T*T*1 does not equal (1-w^2)^2/4. The concatenation of operator T's does not yield a simple power of a number that would result in the expression you gave.
- "Fourier transformation of exp(+ω2/2) makes no sense at all" - like it or not, it does make sense when using the operator I've defined.
- Even if I got the operator wrong somehow (which I don't think I did), surely you'll agree with the hunch that the FRFT would be expressible as exp(i*alpha*T) for some operator T, given its cyclic pattern?
- BTW the T I showed above can also be written as Tf = 0.5 * (d/dw-w)*(d/dw+w)f = 0.5 * (exp(w^2/2)*d/dw*exp(-w^2)*d/dw*exp(w^2/2))f
- Edit #1: I think you can see basically the same result at the bottom of page 2 if you google for "An introduction to the Fractional Fourier Transform and friends"
- but they wrote theirs a little differently.
- Edit #2: but watch out, that source says on page 7 that the FRFT is like LCT (linear canonical transformation) in that there is some ambiguity due to having to choose a "branch of the square root". I think that statement is incorrect. The operator has no ambiguity. In my experience it's only the LCT that has such problems.
- Doubledork (talk) 00:19, 7 September 2011 (UTC)
- Whoops, you're right, I wasn't thinking clearly when I acted it on 1. Your T is the equivalent to the operator for the 1d quantum harmonic oscillator, which means that its eigenfunctions are Hermite functions, which are the same as the eigenfunctions of the Fourier transform, and the eigenvalues are right too (up to a scale factor, which I haven't looked closely enough to check). So yes, it does indeed look like this might give the FRFT. This does seem like a nice way of doing it.
- If it works like it appears to, I would be surprised if it weren't known, since the FRFT is often defined precisely by looking at Hermite functions, and the connection of Hermite functions to the harmonic-oscillator eigenproblem is well known. However, we would have to dig up a clear reference before we could put it into the article. Edit: Ah, good, it looks like you found a reference. — Steven G. Johnson (talk) 00:38, 7 September 2011 (UTC)
- Regarding the branch cut of the square root, I think that the point is that if you define the FRFT by its action on the Hermite eigenfunctions, there is still a choice of branch cuts when you take a fractional power of an eigenvalue like -1. At first glance, this seems like it corresponds to replacing your T with T+4k for integers k (since for α=π/2, an FT, the 4k term vanishes). — Steven G. Johnson (talk) 00:45, 7 September 2011 (UTC)
Elaborate on the meaning of bosonic and fermionic in this context
Could you elaborate on the meaning of bosonic and fermionic in the context of harmonic decompositions? The links of the adjectives go to rather general articles about the particles in physics. — Preceding unsigned comment added by 2001:4CA0:4E01:0:221:28FF:FE3C:B197 (talk) 09:22, 4 April 2013 (UTC)