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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0jamesj posted the most wonderful lecture let me see if i can find it

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0well no i can't. when he is on, ask him

mattfeury
 3 years ago
Best ResponseYou've already chosen the best response.0...He has so many answers. Wow.

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

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

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

mattfeury
 3 years ago
Best ResponseYou've already chosen the best response.0Mostly 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.

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

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

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

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

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

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

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

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.4The 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.

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.4Since 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.

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

mattfeury
 3 years ago
Best ResponseYou've already chosen the best response.0Also, 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.

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.4Hm, 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 \[f(x)=\frac{a_0}{2}\sum_{n=1}^{\infty}\left( a_n\cos(nx)+b_n\sin(nx)\right)\]

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.4That 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)

mattfeury
 3 years ago
Best ResponseYou've already chosen the best response.0Wow 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.

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.4Unfortunately 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.

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

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

imperialist
 3 years ago
Best ResponseYou've already chosen the best response.4No problem at all :)
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