Compute Fourier Transform of x/(x^2+1)^2 using e^|x|

In summary, the task was to compute the Fourier transform of \frac{x}{(x^2+1)^2}. After trying a few different approaches, it was recognized that the problem could be rewritten as \frac{d}{dx} \frac{-1}{x^2+1} and transformed using formula 106 and 107 from the wiki page on Fourier transforms. The final result is i \xi e^{|x|}.
  • #1
saxen
44
0

Homework Statement


Using that the Fourier transform of e[itex]^{|x|}[/itex] is [itex]\frac{2}{\xi^2+1}[/itex]. Compute the Fourier transform of [itex]\frac{x}{(x^2+1)^2}[/itex]

Homework Equations


The Attempt at a Solution



My first thought was to try and rewrite the problem in a form I recognized, tried a couple of things but what I though was best was to write:

[itex]\frac{d}{dx}[/itex] [itex]\frac{-1}{x^2+1}[/itex]

And transform that to e[itex]^{i*\xi}[/itex]*f([itex]\xi[/itex]). This was wrong. Very wrong actually.

Anyone have any hints for me?

thanks!

edit: Missed some things but should be right this time.
 
Last edited:
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  • #2
http://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships Is has nothing to do with convolution! Fourier transform turns DERIVATIVE into MULTIPLICATION with i times argument, so [itex]i \xi[/itex] in this case, so the result should be something like [itex]i \xi e^{|x|}[/itex] (i assume you are supposed to perform the inverse FT, right?
 
  • #3
susskind_leon said:
http://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships Is has nothing to do with convolution! Fourier transform turns DERIVATIVE into MULTIPLICATION with i times argument, so [itex]i \xi[/itex] in this case, so the result should be something like [itex]i \xi e^{|x|}[/itex] (i assume you are supposed to perform the inverse FT, right?

Hello sussking_leon,

My bad, the * means multiplication and not convolution. How do you get the [itex]i \xi [/itex] infront of the exponent?
 
  • #4
Check the wiki page, forumla 106 and 107 do the trick. So if FT{e^|x|} = f(v), then FT{x e^|x|} = i d/dv f(v). (Fourier transform non-unitary, angular frequency)
 
  • #5
susskind_leon said:
Check the wiki page, forumla 106 and 107 do the trick. So if FT{e^|x|} = f(v), then FT{x e^|x|} = i d/dv f(v). (Fourier transform non-unitary, angular frequency)

Ah, I was thinking right at least. Its so stupid, we don't get any of these rules on our exam so without wiki, I would have never solved this exercise. Thanks for the help.
 

What is a Hard Fourier Transform?

A Hard Fourier Transform (HFT) is a mathematical technique used to decompose a function into its constituent frequencies. It is an extension of the traditional Fourier Transform, which is used to analyze periodic functions. HFT is used to analyze non-periodic functions and can provide more accurate results than the traditional Fourier Transform.

How does a Hard Fourier Transform work?

The HFT works by breaking down a function into a series of sine and cosine waves of different frequencies. These waves are then combined to reconstruct the original function. The HFT also takes into account the amplitude and phase of each wave, resulting in a more accurate representation of the original function.

What are the applications of a Hard Fourier Transform?

The HFT has many applications in various fields, including signal processing, image and audio compression, and data analysis. It is commonly used in fields such as physics, engineering, and mathematics to analyze and model complex systems.

What are the advantages of using a Hard Fourier Transform?

The HFT offers several advantages over the traditional Fourier Transform. It can accurately analyze non-periodic functions, provides more accurate results, and can handle a wider range of functions. Additionally, it can be used to analyze functions with discontinuities and sharp changes, which the traditional Fourier Transform may struggle with.

Are there any limitations to using a Hard Fourier Transform?

While the HFT is a powerful tool, there are some limitations to its use. It requires a large amount of computational power and may be more time-consuming compared to the traditional Fourier Transform. Additionally, the accuracy of the results can be affected by noise and other errors in the data. Therefore, it is important to carefully consider the limitations and potential sources of error before using the HFT.

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