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Homework Help: Problem with Fourier Transform

  1. May 18, 2006 #1
    Here's my problem:
    I've got the Fourier transform of f(x) as

    [tex]
    F(K) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(x) e^{-ikx} dx
    [/tex]

    Likewise for g(x) I have

    [tex]
    G(K) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty g(x) e^{-ikx} dx
    [/tex]

    for F(K/lambda) I have

    [tex]
    F(K/\lambda) = \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(\lambda x + y) e^{-ik(\lambda x + y)/\lambda} d(\lambda x + y)
    [/tex]

    My guess is that the answer is relatively simple from here, but I'm not sure how to go about it. Does anyone understand this?
     
    Last edited: May 18, 2006
  2. jcsd
  3. May 18, 2006 #2

    nrqed

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    Gold Member

    You should simply apply the defintion to g(x), so

    [tex] G(k) = {1 \over {\sqrt{ 2 \pi}} } \int dx \,f(\lambda x + y) \, e^{-i kx} [/tex]

    Now, define a new variable x'= lambda x + y so that x=(x'-y)/lambda and dx= dx'/lambda.

    You get

    [tex] G(k) = { 1\over \lambda} {1 \over {\sqrt{ 2 \pi}}} \int \, dx' \, f(x') \, e^{-i k x'/ \lambda} e^{i k y / \lambda} [/tex]

    which directly gives the result you were looking for (I am asusming there was typo in the fomula you gave and that the y was missing in the exponent).

    Patrick
     
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