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Homework Help: Translations, Modulations, & Dilations of Fourier Transform

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Express the Fourier Transform of the following function

    [tex]ae^{2\pi iabx}f(ax-c)[/tex]

    terms of the Fourier Transform of f . (Here a, b, c are positive constants.)

    2. Relevant equations
    Define the following operators acting on function f(x):

    [tex]M_{b}(f)(x)=e^{-2\pi ibx}f(x)[/tex]

    We have the following properties of these operators and the Fourier transform:


    3. The attempt at a solution
    My first attempt was to reformat the formula in terms of the operators given above:

    [tex]ae^{2\pi iabx}f(ax-c)=\sqrt{a}M_{-ab}(D_{a}(T_{\frac{c}{a}}(f)(x)))[/tex]

    Simply replacing the operators with the counter parts for the Fourier transform would give


    Carrying out these operators I have

    [tex]\sqrt{a}T_{-ab}(D_{\frac{1}{a}}(M_{\frac{c}{a}}(\hat{f})(s)))=e^{2\pi i\frac{c}{a^{2}}(s-ab)}\hat{f}(\frac{s}{a}-ab)[/tex]

    I was skeptical that this was the right answer, since I thought maybe my description of the given expression in terms of operators was incorrect. Therefore I tried again with a different description:

    [tex]ae^{2\pi iabx}f(ax-c)=\sqrt{a}D_{a}(M_{b}(T_{\frac{c}{a}}(f)(x)))[/tex]


    [tex]\sqrt{a}D_{\frac{1}{a}}(T_{b}(M_{\frac{c}{a}}(\hat{f})(s)))=e^{2\pi i\frac{c}{a^{2}}(s+b)}f(\frac{s+b}{a})[/tex]

    These two answers aren't consistent with each other (although they are pretty close, differing only by a factor of 1/a). At this point I decided to try to solve the problem explicitly.

    [tex]\int_{-\infty}^{\infty}ae^{2\pi iabx}f(ax-c)e^{-2\pi isx}dx[/tex]

    let y=ax-c and dy=adx

    [tex]a\int_{-\infty}^{\infty}e^{2\pi ib(y+c)}f(y)e^{-2\pi is(\frac{y+c}{a})}dy[/tex]

    [tex]\int^{\infty}_{-\infty}e^{2\pi ibc}f(y)e^{-2\pi is(\frac{y+c}{a}-by)}dy[/tex]

    [tex]e^{2\pi ibc}\int_{-\infty}^{\infty}f(y)e^{-2\pi is[\frac{1}{a}(y(1-ab)+c)]}dy[/tex]


    This answer is also clearly inconsistent with the last two, which I'm guessing means I made some kind of gross error in the explicit calculation.

    Any guidance would be greatly appreciated.

    (Note: Hats should denote Fourier transform, but the hats didn't expand to cover long expressions like I hoped they would. Therefore, anytime you see a hat over f(x) on the left hand side of the equation, it should be understood that I wanted to take the Fourier transform of the whole side, not just the function, since then the problem would be trivial.)
    Last edited: Sep 29, 2010
  2. jcsd
  3. Sep 30, 2010 #2


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    I can't really help with the operators, since it would take some time to really remember how they work, but there's an algebra error in the equation (***) above. It should be

    [tex]\int^{\infty}_{-\infty}e^{2\pi ibc}f(y)e^{-2\pi i(\frac{y+c}{a}s-by)}dy[/tex]

    Without using the operators, I think you find something like

    [tex] e^{-2\pi i c (s/a-b)} \hat{f}(s/a -b).[/tex]
  4. Sep 30, 2010 #3
    Just to clarify, is that [itex]\frac{s}{a-b}[/itex] or [itex]\frac{s}{a}-b[/itex]?

    Thanks for your help by the way. I corrected the algebra mistake but have not had time to carry out evaluating the integral yet.
  5. Sep 30, 2010 #4


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    It's [itex]\frac{s}{a}-b[/itex]. I thought it would be harder to read typeset that way, but I guess you have to be careful since so many posters forget to put parentheses around denominators.
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