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February 26

Fourier transform

What kind of restriction does the requirement $|{\hat {f}}(\omega )|=|{\hat {f}}(-\omega )|$ place on the time domain function $f(t)\in \mathbb {C}$ .

Is there an easy interpretation?

I can see that both even and odd $f(t)$ satisfy this but I think there is a more general restriction. — Preceding unsigned comment added by 128.40.61.82 (talk) 11:17, 26 February 2014 (UTC)[reply]

I'm not sure very much can be got, even with a real function one could swap the sign at random every time it passed zero. Dmcq (talk) 12:59, 26 February 2014 (UTC)[reply]

Don't take my word for it, but I think that this translates into the autocorrelation being a real function. —Quondum 14:42, 26 February 2014 (UTC)[reply]

For complex functions the absolute values being equal means a value for −ω is the +ω one multiplied by any

e^{i\theta (\omega )}

. Dmcq (talk) 15:57, 26 February 2014 (UTC)[reply]

I'm not sure what point you're trying to make. Sum them for a given

| ω |

, which means that each frequency component is "plane polarized" with arbitrary complex argument (no circular component, graphed as a complex number against time). This has the effect that when multiplied by a time-shifted version of the complex complement, the result is real. Integrate over time. Repeat for every time offset, and you have the definition of autocorrelation of complex functions. Hence my conclusion. —Quondum 16:49, 26 February 2014 (UTC)[reply]

~~Sorry yes I believe you're right. That's quite interesting.~~ Dmcq (talk) 17:42, 26 February 2014 (UTC)[reply]

Um, no the cross-products shouldn't be formed but you get them from the sums being multiplied together. Dmcq (talk) 08:24, 27 February 2014 (UTC)[reply]

I don't follow you. (Aside: I left out mention of how products of distinct frequencies interact, but the time integral of these cross-frequency products is zero and does not affect the result.) I'll check my conclusion as originally stated using an FFT on constrained random data and let you know. —Quondum 15:07, 27 February 2014 (UTC)[reply]

I don't think it works, e.g., with a fundamental solution of the Schrodinger equation. Sławomir Biały (talk) 15:21, 27 February 2014 (UTC)[reply]

I've verified my original conjecture with an FFT (generate random complex values in the frequency domain, then adjust only magnitude to match that of negative frequency, do inverse FFT, do autocorrelation). The autocorrelation has to be done in a cyclic fashion, but the imaginary part of the autocorrelation is zero to the precision of the internal representation. —Quondum 16:55, 27 February 2014 (UTC)[reply]

@Dmcq and Sławomir Biały: C'mon, guys, you can't express doubt and then evaporate. The Wiener–Khinchin theorem (though that article seems a mess, so see Autocorrelation#Properties) says essentially of the autocorrelation that

R_{f}(\tau )=\int _{-\infty }^{+\infty }f(t)f^{*}(t-\tau )\,dt={\frac {1}{2\pi }}\int _{-\infty }^{+\infty }|{\hat {f}}(\omega )|^{2}e^{i\omega \tau }\,d\omega .

Since it is given that

|{\hat {f}}(\omega )|

is a symmetric function, the conclusion that

R_{f}(\tau )

is real follows pretty directly. —Quondum 05:45, 28 February 2014 (UTC)[reply]

Sorry, you're quite right I believe. I had to do something else and it isn't something I'm altogether well informed about. It is still surprising I think. Dmcq (talk) 11:09, 28 February 2014 (UTC)[reply]

Neither am I (undergraduate days, too long ago), and my reaction is the same. And since I derived it afresh, I guess I felt I needed confirmation that I was correct, though as is often the case, this is merely a special case of a more general theorem. In hindsight, this looks like it could have been a homework question for someone who'd just covered the theorem in class. —Quondum 14:53, 28 February 2014 (UTC)[reply]

It wasn't a homework question. I appreciate the help you guys have tried to offer, but unfortunately your answer isn't very useful for my purposes. I might as well calculate the FT and see if the modulus is even, as calculate the auto-correlation and check it is real. I was looking for a property on the original function space if possible, but I suppose it is possible no obvious property exists. — Preceding unsigned comment added by 128.40.61.82 (talk) 10:29, 3 March 2014 (UTC)[reply]

If it's any comfort, my gut says that the time-domain information given by the constraint is exactly what I've stated, nothing more, nothing less. Which in effect is saying that I don't believe other properties exist that are not directly implied by the autocorrelation being real, obvious or otherwise, and to make it worse, the properties of the autocorrelation are not too obviously tied to the time-domain function. I would not be surprised if the constraint did not place any constraints on any chosen finite time segment of the function. —Quondum 07:39, 4 March 2014 (UTC)[reply]

Yes, you're right. Sławomir Biały (talk) 13:13, 2 March 2014 (UTC)[reply]