- #1

McCoy13

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## Homework Statement

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.)

## Homework 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]

## 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.)

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