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mattfeuryBest ResponseYou've already chosen the best response.0
The math is a little complex, but it hasn't really clicked how it can transform from the time domain to the frequency domain.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
jamesj posted the most wonderful lecture let me see if i can find it
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
well no i can't. when he is on, ask him
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
...He has so many answers. Wow.
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
Might be this one, although it's a bit mathy ;) http://openstudy.com/updates/4ecf7339e4b04e045af3211c
 2 years ago

myininayaBest ResponseYou've already chosen the best response.0
lol what did you expect? not so much math? what is your interest in fourier tranform?
 2 years ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
any function can be expressed as a infinite sum of waves
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
Mostly for digital sound. The ability to turn a waveform onto the frequency spectrum is very intriguing. I've made some digital plugins so the next step is diving into Fourier land.
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
How does the transform decompose those waves? Perhaps this is where the math kicks in, lol...
 2 years ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
well if you have an odd function the waves are all sinusoidal , and only vary by frequency and amplitude
 2 years ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
to get an even function you need cosines, and many functions will need both
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
and transforming into the imaginary domain somehow gives us magical frequency values? sweet.
 2 years ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
the fact that it has the imaginary component just says the wave is Waving in and out of phase
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
I understand complex waveform but I guess I'm just not sure how the transform *decomposes* it.
 2 years ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
well the transform equation decomposed the original function into even and odd bits
 2 years ago

imperialistBest ResponseYou've already chosen the best response.4
The way it works is a bit technical, but basically it relies on the same principles as Fourier series combined with a bit of complex analysis. According to the theory of Fourier series, all functions (satisfying particular properties and over a certain range) can be written as sums of sinusoidal functions. What the Fourier transform does is to measure, over an epsilon range, how much each of those sine curves contributes to the integral. The introduction of the complex variable just allows us to not be restricted to the real numbers, since we may need some of the coefficients of the curves to be complex valued.
 2 years ago

imperialistBest ResponseYou've already chosen the best response.4
Since after all, complex numbers are far far more natural than the reals, so there is no reason why we should be restricted to only working over the reals.
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
Interesting. So could you technically do a transform with real numbers only? Not that I plan on trying or anything.
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
Also, is it pretty "bruteforce" in terms of finding the coefficient of each sine curve or is there a more clever algorithm? That may be an absurd question.
 2 years ago

imperialistBest ResponseYou've already chosen the best response.4
Hm, not quite sure what you mean. There is a relatively elegant integral equation to find the coefficients of each sine and cosine curve in the decomposition, if that is what you are asking for. If your function is f(x), then \[a_n=\frac{1}{\pi}\int_{\pi}^{\pi}f(x)\cos(nx)dx\] \[b_n=\frac{1}{\pi}\int_{\pi}^{\pi}f(x)\sin(nx)dx\] With the series being \[f(x)=\frac{a_0}{2}\sum_{n=1}^{\infty}\left( a_n\cos(nx)+b_n\sin(nx)\right)\]
 2 years ago

imperialistBest ResponseYou've already chosen the best response.4
That is for f defined over [pi,pi], otherwise if it is [a,b], replace the limits of integration with a and b and the coefficient of the integral with 2/(ba)
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
Wow that actually makes a lot of sense. What is the importance of the integral's coefficient? It seems to me that doing the integral would cover it.
 2 years ago

imperialistBest ResponseYou've already chosen the best response.4
Unfortunately you do need the coefficient. It is there because the Fourier transform is actually a type of inner product over the orthonormal basis vectors of sin(nx) and cos(nx). The coefficient of the integral acts as a sort of normalization, to ensure that the magnitude of all of this integrals, with respect to the inner product, is one.
 2 years ago

imperialistBest ResponseYou've already chosen the best response.4
Always better to have your orthogonal base to be also orthonormal, makes things much much easier!
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.0
Wow, man. That's awesome. Exactly what I was looking for. Thank you.
 2 years ago

imperialistBest ResponseYou've already chosen the best response.4
No problem at all :)
 2 years ago
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