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mattfeury

Can someone explain the theory behind the Fourier Transform?

  • 2 years ago
  • 2 years ago

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  1. mattfeury
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    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
  2. satellite73
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    jamesj posted the most wonderful lecture let me see if i can find it

    • 2 years ago
  3. satellite73
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    well no i can't. when he is on, ask him

    • 2 years ago
  4. mattfeury
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    ...He has so many answers. Wow.

    • 2 years ago
  5. mattfeury
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    Might be this one, although it's a bit mathy ;) http://openstudy.com/updates/4ecf7339e4b04e045af3211c

    • 2 years ago
  6. myininaya
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    lol what did you expect? not so much math? what is your interest in fourier tranform?

    • 2 years ago
  7. UnkleRhaukus
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    any function can be expressed as a infinite sum of waves

    • 2 years ago
  8. mattfeury
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    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
  9. myininaya
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    neat.

    • 2 years ago
  10. mattfeury
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    How does the transform decompose those waves? Perhaps this is where the math kicks in, lol...

    • 2 years ago
  11. UnkleRhaukus
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    well if you have an odd function the waves are all sinusoidal , and only vary by frequency and amplitude

    • 2 years ago
  12. UnkleRhaukus
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    to get an even function you need cosines, and many functions will need both

    • 2 years ago
  13. mattfeury
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    and transforming into the imaginary domain somehow gives us magical frequency values? sweet.

    • 2 years ago
  14. UnkleRhaukus
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    • 2 years ago
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  15. UnkleRhaukus
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    the fact that it has the imaginary component just says the wave is Waving in and out of phase

    • 2 years ago
  16. mattfeury
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    I understand complex waveform but I guess I'm just not sure how the transform *decomposes* it.

    • 2 years ago
  17. UnkleRhaukus
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    well the transform equation decomposed the original function into even and odd bits

    • 2 years ago
  18. imperialist
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    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
  19. imperialist
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    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
  20. mattfeury
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    Interesting. So could you technically do a transform with real numbers only? Not that I plan on trying or anything.

    • 2 years ago
  21. mattfeury
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    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.

    • 2 years ago
  22. imperialist
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    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
  23. imperialist
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    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)

    • 2 years ago
  24. mattfeury
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    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
  25. imperialist
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    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
  26. imperialist
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    Always better to have your orthogonal base to be also orthonormal, makes things much much easier!

    • 2 years ago
  27. mattfeury
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    Wow, man. That's awesome. Exactly what I was looking for. Thank you.

    • 2 years ago
  28. imperialist
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    No problem at all :)

    • 2 years ago
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