Can someone explain the theory behind the Fourier Transform?

- mattfeury

Can someone explain the theory behind the Fourier Transform?

- chestercat

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- mattfeury

The math is a little complex, but it hasn't really clicked how it can transform from the time domain to the frequency domain.

- anonymous

jamesj posted the most wonderful lecture let me see if i can find it

- anonymous

well no i can't. when he is on, ask him

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- mattfeury

...He has so many answers. Wow.

- mattfeury

Might be this one, although it's a bit mathy ;)
http://openstudy.com/updates/4ecf7339e4b04e045af3211c

- myininaya

lol what did you expect?
not so much math?
what is your interest in fourier tranform?

- UnkleRhaukus

any function can be expressed as a infinite sum of waves

- mattfeury

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.

- myininaya

neat.

- mattfeury

How does the transform decompose those waves? Perhaps this is where the math kicks in, lol...

- UnkleRhaukus

well if you have an odd function the waves are all sinusoidal , and only vary by frequency and amplitude

- UnkleRhaukus

to get an even function you need cosines,
and many functions will need both

- mattfeury

and transforming into the imaginary domain somehow gives us magical frequency values? sweet.

- UnkleRhaukus

##### 1 Attachment

- UnkleRhaukus

the fact that it has the imaginary component just says the wave is Waving in and out of phase

- mattfeury

I understand complex waveform but I guess I'm just not sure how the transform *decomposes* it.

- UnkleRhaukus

well the transform equation decomposed the original function into even and odd bits

- anonymous

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.

- anonymous

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.

- mattfeury

Interesting. So could you technically do a transform with real numbers only?
Not that I plan on trying or anything.

- mattfeury

Also, is it pretty "brute-force" in terms of finding the coefficient of each sine curve or is there a more clever algorithm? That may be an absurd question.

- anonymous

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 \[f(x)=\frac{a_0}{2}\sum_{n=1}^{\infty}\left( a_n\cos(nx)+b_n\sin(nx)\right)\]

- anonymous

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/(b-a)

- mattfeury

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.

- anonymous

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.

- anonymous

Always better to have your orthogonal base to be also orthonormal, makes things much much easier!

- mattfeury

Wow, man. That's awesome. Exactly what I was looking for. Thank you.

- anonymous

No problem at all :)

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