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fourier transform

Discussion in 'Electronic Design' started by [email protected], Jan 16, 2009.

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  1. Guest

    I have been trying to solve fourier transform of f(t)^n...where f(t)
    is any random function.I require an answer in terms of fourier
    transform of f(t)... Thanks a lot.
  2. Hello
    Thats nice, hope the answer finds you :eek:)

  3. oopere

    oopere Guest

    If f(t) is a random process then you will be having trouble computing
    it's Fourier Transform.

    If you mean that f(t) could be any function, then you should pay more
    attention in class:

    The Fourier transform of f(t)·g(t) is F(f)*G(f), where * stands for
    convolution. Next, work out f(t)·f(t), f(t)·f(t)·f(t) and so on...

  4. Jamie

    Jamie Guest

    Maybe this will help in your journey.
  5. Guest

    F[f^n] = F[f*f^(n-1)]

    the fourier transform of a product is a convolution

    F[f] * F[f^(n-1)]

    Hence you end up with something like F[f] * .... * F[f]

    which is in terms of f only(where f is not the original f but the
    transform of f).


    I[n] = int(f^n*e^(iwx))

    can be reduced by parts

    u = f^n
    du = n*f^(n-1)*f'dx

    dv = e^(iwx)dx
    v = 1/iw*e^(iwx)

    f^n*e^(iwx)/iw - n/iw*int(f^(n-1)*f'*e^(iwx))

    the last integral, by parts again

    u = f'
    du = f''dx

    dv = f^(n-1)*e^(iwx)dx
    v = I[n-1]

    I[n-1]*f' - int(I[n-1]*f')


    I[n] = f^n*e^(iwx)/iw - n/iw*int(f^(n-1)*f'*e^(iwx)) = f^n*e^(iwx)/iw
    - n/iw*(I[n-1]*f' - int(I[n-1]*f''))

    or something similar.

    The point here is that you can find a recursive formula for computing
    it. You can then potentially reduce it even farther to a close
    formula. At least for polynomials you can easily do this since the
    formula will definitely terminate after a finite number of steps. Also
    with sinusoids the formula should simplify to a closed form after a
    few steps.

    I'll leave you to do the rest of the work since it's difficult to show
    mathematical derivations in ascii.
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