1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]T_{a}(f)(x)=f(x+a)[/tex]
    [tex]M_{b}(f)(x)=e^{-2\pi ibx}f(x)[/tex]
    [tex]D_{c}(f)(x)=\sqrt{c}f(cx)[/tex]

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

    [tex]\hat{T_{a}(f)(x)}=M_{a}\hat{(f)}(s)[/tex]
    [tex]\hat{M_{b}(f)(x)}=T_{b}\hat{(f)}(s)[/tex]
    [tex]\hat{D_{c}(f)(x)}=D_{1/c}\hat{(f)}(s)[/tex]


    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

    [tex]\hat{\sqrt{a}M_{-ab}(D_{a}(T_{\frac{c}{a}}(f)(x)))}=\sqrt{a}T_{-ab}(D_{\frac{1}{a}}(M_{\frac{-c}{a}}(\hat{f})(s)))[/tex]

    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]\hat{\sqrt{a}D_{a}(M_{b}(T_{\frac{c}{a}}(f)(x)))}=\sqrt{a}D_{\frac{1}{a}}(T_{b}(M_{\frac{-c}{a}}(\hat{f})(s)))[/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]

    [tex]M_{-bc}(D_{\frac{1}{a}}(D_{1-ab}(T_{\frac{c}{1-ab}}(\hat{f})(s))))[/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

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook