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How the heck are you supposed to use FFT?

  1. Jul 16, 2008 #1
    Hey,

    This may seem very stupid but how the heck are you supposed to use FFT? The FFT transformations take a vector as in input but I want to use it to work out FT of a perticular value. I have the charecteristic function of a distribution and wish to numericaly work out values for the distribution function. Hope this makes sense.

    Thanks in advance
     
  2. jcsd
  3. Jul 16, 2008 #2

    marcusl

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    Re: Fft

    The FT gives the complex spectrum of the input signal, the FFT gives the discrete complex spectrum of an input sequence or vector. In other words, it decomposes the input into frequency components. I don't follow how you intend to apply this to distributions.
     
  4. Jul 17, 2008 #3

    HallsofIvy

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    Re: Fft

    The Fourier transform itself is a transform of a function not a number. The FFT (Fast Fourier Transform) thinks of the function as being given by a discrete "vector" of values at different points. I have no idea what you could mean by taking the FFT of a single value.
     
  5. Jul 17, 2008 #4
    Re: Fft

    Sorry maybe I was not so clear on this. I have a function
    [tex] \psi(\lambda)=\int_{-\infty}^{\infty}e^{i \lambda t}f(t) dt [/tex]

    I was wondering if you could tell me how I could use the FFT (i.e. what values I should sample) to obtain [tex] f(\alpha) [/tex] for some [tex] \alpha [/tex] I choose?

    Thanks
     
  6. Jul 17, 2008 #5

    marcusl

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    Re: Fft

    EDIT: I couldn't get equations in this post to render properly, please see the next.
     
    Last edited: Jul 17, 2008
  7. Jul 17, 2008 #6

    marcusl

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    Re: Fft

    So you know [tex]\psi(\lambda)[/tex] and want to find [tex]f(\alpha)[/tex], which is given by the continuous inverse FT or IFT
    [tex]f(\alpha)=\int_{-\infty}^{\infty}e^{-i \lambda \alpha}\psi(\lambda) d\lambda[/tex]

    You want to know how to evaluate this using the FFT, right? This is a classic problem in digital signal processing, and the approach depends on the nature of f and psi.

    A) f(t) is "bandlimited", that is, it is zero for [tex]t > t_c[/tex] where [tex]t_c[/tex] is a "cutoff time." (Because you are doing an IFT, the usual conventions of sampling a time function to find frequency are swapped here--you are sampling a spectrum to find a time function). An example is f(t)=gaussian, which is effectively zero after 5 standard deviations in time. This is the cutoff time t_c. Psi is also limited in frequency, after, say [tex]5\sigma[/tex]. So you sample psi uniformly over the interval
    [tex]|\lambda| \leq 5\sigma[/tex], at a rate [tex]\Delta\lambda[/tex] that satisfies the sampling theorem

    [tex]\Delta\lambda \geq \frac{2}{t_c}[/tex].

    Use a number of points N that is a power of 2 if you want to use an FFT, otherwise compute with the DFT, and take the value closest to [tex]\alpha[/tex].

    B) If you are not bandlimited, you will alias and your results will suffer in accuracy. We won't discuss this unless it's necessary.
     
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