mattfeury
  • mattfeury
Can someone explain the theory behind the Fourier Transform?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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mattfeury
  • 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
  • anonymous
jamesj posted the most wonderful lecture let me see if i can find it
anonymous
  • anonymous
well no i can't. when he is on, ask him

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mattfeury
  • mattfeury
...He has so many answers. Wow.
mattfeury
  • mattfeury
Might be this one, although it's a bit mathy ;) http://openstudy.com/updates/4ecf7339e4b04e045af3211c
myininaya
  • myininaya
lol what did you expect? not so much math? what is your interest in fourier tranform?
UnkleRhaukus
  • UnkleRhaukus
any function can be expressed as a infinite sum of waves
mattfeury
  • 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
  • myininaya
neat.
mattfeury
  • mattfeury
How does the transform decompose those waves? Perhaps this is where the math kicks in, lol...
UnkleRhaukus
  • UnkleRhaukus
well if you have an odd function the waves are all sinusoidal , and only vary by frequency and amplitude
UnkleRhaukus
  • UnkleRhaukus
to get an even function you need cosines, and many functions will need both
mattfeury
  • mattfeury
and transforming into the imaginary domain somehow gives us magical frequency values? sweet.
UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
the fact that it has the imaginary component just says the wave is Waving in and out of phase
mattfeury
  • mattfeury
I understand complex waveform but I guess I'm just not sure how the transform *decomposes* it.
UnkleRhaukus
  • UnkleRhaukus
well the transform equation decomposed the original function into even and odd bits
anonymous
  • 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
  • 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
  • mattfeury
Interesting. So could you technically do a transform with real numbers only? Not that I plan on trying or anything.
mattfeury
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
Always better to have your orthogonal base to be also orthonormal, makes things much much easier!
mattfeury
  • mattfeury
Wow, man. That's awesome. Exactly what I was looking for. Thank you.
anonymous
  • anonymous
No problem at all :)

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