Translations, Modulations, & Dilations of Fourier Transform

[McCoy13]

Homework Statement

Express the Fourier Transform of the following function

ae^{2\pi iabx}f(ax-c)

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):

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

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

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

The Attempt at a Solution

My first attempt was to reformat the formula in terms of the operators given above:

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

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

\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)))

Carrying out these operators I have

\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)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:

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

\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)))

\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})

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.

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

let y=ax-c and dy=adx

a\int_{-\infty}^{\infty}e^{2\pi ib(y+c)}f(y)e^{-2\pi is(\frac{y+c}{a})}dy\int^{\infty}_{-\infty}e^{2\pi ibc}f(y)e^{-2\pi is(\frac{y+c}{a}-by)}dye^{2\pi ibc}\int_{-\infty}^{\infty}f(y)e^{-2\pi is[\frac{1}{a}(y(1-ab)+c)]}dy

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

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.)

 

[fzero]

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

let y=ax-c and dy=adx

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


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


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

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

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

Without using the operators, I think you find something like

e^{-2\pi i c (s/a-b)} \hat{f}(s/a -b).

 

[McCoy13]

Just to clarify, is that \frac{s}{a-b} or \frac{s}{a}-b?

Thanks for your help by the way. I corrected the algebra mistake but have not had time to carry out evaluating the integral yet.

 

[fzero]

Just to clarify, is that \frac{s}{a-b} or \frac{s}{a}-b?

Thanks for your help by the way. I corrected the algebra mistake but have not had time to carry out evaluating the integral yet.

It's \frac{s}{a}-b. 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.