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Real roots of Fourier transform

  1. May 29, 2006 #1
    If we define the function:

    [tex] F[w]=\int_{-\infty}^{\infty}dxe^{-iwx}g(x) [/tex]

    my question is..what would be the criterion to decide if F[w] has all the roots real (w=w*) and how is derived?..thanks.
  2. jcsd
  3. Jun 7, 2006 #2
    If anybody knows the answer to this question, I would be very interested in learning the answer also!

    At least in the case of a similar transformation, a Laplace like transform,

    Int[e^(-s*sqrt(t))*sin(pi*sqrt(t)),0, infinity] = T{sin(pi*sqrt(t))} = 4*pi*s/(s^2 + pi^2)^2

    Int[e^(-s*sqrt(t+c))*sin(pi*sqrt(t+c)),0, infinity] = T{sin(pi*sqrt(t+c))}

    = q(s)*e^[-sqrt(c)*s]/(s^2 + pi^2)^2, where q(s) is a cubic polynomial in s, c is an arbitrary positive constant, and t is greater than or equal to 0.

    It seems that the information about the zeros of the sine functions above is contained in the residue at the poles of there s-like-transforms, T{f(t)} = F(s), while the kind of function f(t) is, is indicated by the position and order of the poles of F(s). This is only a guess, based on the fact that the transformations above yield functions of s with a common denominator of (s^2 + pi^2)^2, but a different expression in the variable s in the numerator. The two original sine functions in the real variable t have zeros in different locations, but the position and order of the singularities of their s-like-transforms above are identical. As a result, I would presume that all the information about the zeros of the original sine functions in t would have to be tied up in the residues of the singularities of the s-like-transforms

    What are your thoughts?


    Edwin G. Schasteen
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