# Manipulating Fourier transforms

## Homework Statement ## The Attempt at a Solution

$\widetilde{f}(p) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{-ipx}f(x) \, \mathrm{d}x$

Then moving on to the transform for $e^{ip_0 x}f(x)$ I get:

$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{-ipx}e^{ip_0 x}f(x) \, \mathrm{d}x$

$= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{-i(p-p_0)x}f(x) \, \mathrm{d}x$

$=\widetilde{f}(p-p_0)$

However I'm really not sure what to do for part b).

Here's my attempt so far:

Starting with:

$F_D(x_D) = \int_{-\infty}^{\infty} \! f(x)\Big[w(x_D - x)\Big] \, \mathrm{d}x$

The $w(x_D - x)$ part is similar to the previous section so I plug that in to get:

$F_D(x_D) = \int_{-\infty}^{\infty} \! f(x) \, \Big[\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{-i(x_D -x)p}\widetilde{w}(p) \, \mathrm{d}p\Big] \mathrm{d}x$

then rearrange a little to get:

$F_D(x_D) = \Big(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ixp}f(x) \, \mathrm{d}x\Big) \int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{w}(p) \, \mathrm{d}p$

The part in brackets is the transform giving $\widetilde{f}(p)$ so,

$F_D(x_D) = \widetilde{f}(p) \int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{w}(p) \, \mathrm{d}p$

which can be neatened a little into:

$F_D(x_D) = \widetilde{f}(p)\widetilde{w}(p) \int_{-\infty}^{\infty} \! e^{-ix_D p} \, \mathrm{d}p$

so now I have $F_D(x_D)$ written in a way that's already starting to look like the desired outcome. However it only gets messy now as I do the fourier transform:

$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{ix_D p}\Big[F_D(x_D)\Big] \, \mathrm{d}x_D = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \! e^{ix_D p}\Big[\widetilde{f}(p)\widetilde{w}(p) \int_{-\infty}^{\infty} \! e^{-ix_D p} \, \mathrm{d}p\,\Big] \mathrm{d}x_D$

The exponents cancel to leave

$\frac{1}{\sqrt{2\pi}} \widetilde{f}(p)\widetilde{w}(p) \int_{-\infty}^{\infty} \! \int_{-\infty}^{\infty} \! \, \mathrm{d}p\, \mathrm{d}x_D$

which I really don't think is a very good result!

Any input would be much appreciated. Thanks.

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which can be neatened a little into:
Can you explain why it is allowed to move $\tilde w(p)$ in front of the integral sign ?

• sa1988
Can you explain why it is allowed to move $\tilde w(p)$ in front of the integral sign ?
Ah, I see now, it can't be shifted over because it's a function of p so it needs to stay within the integral over dp.

Furthermore I've just been looking through some old lecture notes and it turns out I was very much on the right track (woohoo!). The question is essentially proving the convolution theorem.

So I'd got to the point:

$F_D(x_D) = \widetilde{f}(p)\int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{w}(p) \, \mathrm{d}p$

which actually neatens up to

$F_D(x_D) = \int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{f}(p)\widetilde{w}(p) \, \mathrm{d}p$

which can be made into a standard fourier transform by adding factors of $\sqrt{2\pi}$:

$\frac{1}{\sqrt{2\pi}}F_D(x_D) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{-ix_D p}\widetilde{f}(p)\widetilde{w}(p) \, \mathrm{d}p$

then taking the inverse:

$\frac{1}{\sqrt{2\pi}}\widetilde{F}_D(p) = \widetilde{f}(p)\widetilde{w}(p)$

which clearly leads to the desired result:

$\widetilde{F}_D(p) = \sqrt{2\pi}\widetilde{f}(p)\widetilde{w}(p)$