Manipulating Fourier transforms

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In summary, the conversation discusses the process of solving a problem involving the Fourier transform and its implications for the convolution theorem. The speaker presents their attempt at solving the problem, starting with the transform for a given function and then moving on to the transform for a modified function. They then discuss their attempt at solving a second part of the problem, which involves manipulating the transform to reach the desired outcome. After reviewing their notes, the speaker realizes they were on the right track and the solution involves using the convolution theorem. The conversation ends with the speaker presenting the final result, which is that the Fourier transform of the modified function is equal to the product of the Fourier transforms of the original function and the modifying function.
  • #1
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Homework Statement


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Homework Equations

The Attempt at a Solution



I think I'm ok with the first part. I start with:

##\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 into 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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  • #2
sa1988 said:
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 ?
 
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  • #3
BvU said:
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)##
 

1. What is a Fourier transform and how does it work?

A Fourier transform is a mathematical tool used to decompose a function or signal into its constituent frequencies. It works by representing a function as a combination of sine and cosine waves of different frequencies and amplitudes.

2. Why is manipulating Fourier transforms useful?

Manipulating Fourier transforms allows us to analyze and understand the frequency content of a signal or function. This can be useful in various applications such as signal processing, image and sound compression, and data analysis.

3. What are some common manipulations of Fourier transforms?

Some common manipulations of Fourier transforms include scaling, shifting, convolution, and filtering. These operations can be used to modify the frequency content of a signal or function.

4. How do I interpret the results of a manipulated Fourier transform?

The results of a manipulated Fourier transform can be interpreted as the new frequency components of the original signal or function. This can help in understanding how the signal has been modified and what changes have been made to its frequency content.

5. Are there any limitations to manipulating Fourier transforms?

Yes, there are some limitations to manipulating Fourier transforms. For example, if the original signal is not continuous or contains sharp discontinuities, the transformed signal may not accurately represent the original signal. Additionally, some manipulations may introduce artifacts or distortions in the transformed signal.

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