# Translations, Modulations, & Dilations of Fourier Transform

1. Sep 29, 2010

### McCoy13

1. The problem statement, all variables and given/known data
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.)

2. Relevant 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:

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

3. 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.

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

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

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

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

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

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.)

Last edited: Sep 29, 2010
2. Sep 30, 2010

### fzero

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

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

3. Sep 30, 2010

### 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.

4. Sep 30, 2010

### fzero

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.