Probelm getting started with Fouriertransforms

  • Context: Graduate 
  • Thread starter Thread starter toofle
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the challenges of understanding Fourier transforms, particularly in relation to discrepancies in results obtained from Mathematica and manual calculations. Participants explore normalization constants and the implications of different conventions in Fourier transform definitions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about their Fourier transform calculations and the differences in results when using Mathematica.
  • Another participant explains that normalization constants can vary and that Mathematica allows for optional parameters to adjust these constants.
  • A participant questions the equivalence of two complex expressions involving exponential functions and Heaviside theta functions, suggesting that linearity might lead to a different form without normalization.
  • Another participant points out that the two expressions in question are likely not equivalent due to differences in their factors, specifically noting the presence of a square root of pi in one expression and a factor of pi in the other.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the equivalence of the two mathematical expressions discussed. There is acknowledgment of differing views regarding normalization constants and their impact on results.

Contextual Notes

There are unresolved issues regarding the normalization constants used in Fourier transforms and how they affect the results. The discussion highlights the importance of clarity in mathematical expressions and the potential for confusion arising from different conventions.

toofle
Messages
20
Reaction score
0
I have trouble getting started with Fouriertransforms. I have access to Mathematica but don't obtain the same results(see below).
Isn't my reasoning correct?

Example:
Fouriertransform 1/(t^2+1):
F19: e^(-a|t|),a>0 <-> 2a/(a^2+w^2)

1/(t^2+1) = 1/2 * 2/(t^2+1)
Linearity => f(t) = 1/2*g(t), g(t) = 2/(t^2+1)

Symmetry(F10)+ f19 => gtop(w) = e^-|w| * 2*Pi

=> ftop(w) = 1/2*2*Pi*e^(-|w|)
<=> ftop(w) = Pi*e^(-|w|)

Answer: ftop(w) = Pi*e^(-|w|)

But Mathematica:
FourierTransform[1/(1 + t^2), t, \[Omega]]
E^-Abs[\[Omega]] Sqrt[\[Pi]/2]

Where does the root come from?
 
Physics news on Phys.org
Welcome to PF, toofle! :smile:

The Fourier transform and the inverse Fourier transform have normalization constants ([itex]1 \over \sqrt{2\pi}[/itex]).
There are different conventions where to put these constants.

If you look in Mathematica you'll see that the FourierTransform function has optional FourierParameters specifying what to do with these normalization constants.
 
Thanks that clears some things up.

But is:
-(1/6) E^(-3 \[Omega]) Sqrt[\[Pi]/2] (E^(
5 \[Omega]) (-3 + 2 E^\[Omega]) HeavisideTheta[-\[Omega]] + (2 -
3 E^\[Omega]) HeavisideTheta[\[Omega]])

Equivalent to:
1/2 E^(-2 Abs[\[Omega]]) \[Pi] + -(1/3) E^(-3 Abs[\[Omega]]) \[Pi]


?
Because if I split into partial fractions and use linearity I get the second one without norming.
So it seems so but it is kind of difficult to see.
 
Let's first see if I can make some sense of your formulas.
As they are, they are rather off-putting to analyze.
Let me (just this once) set them in LaTeX first.


toofle said:
Thanks that clears some things up.

But is:
[tex]-{1 \over 6} e^{-3 \omega} \sqrt{\pi \over 2} \left(e^{<br /> 5 \omega} (-3 + 2 e^\omega) \Theta(-\omega) + (2 - <br /> 3 e^\omega) \Theta(\omega) \right)[/tex]

Equivalent to:
[tex]{1 \over 2} e^{-2 |\omega|} \pi + -{1 \over 3} e^{-3 |\omega|} \pi[/tex]


?
Because if I split into partial fractions and use linearity I get the second one without norming.
So it seems so but it is kind of difficult to see.

Let me know if I have your formulas right...?
 
Looking at the formulas I can see immediately that they will not be equivalent.

I can't see straight away if the e-powers match up, but I can see that the factor of pi cannot be equivalent.
The first one has a square root of pi, while the second has a factor of pi.
So it seems likely there is a factor of about [itex]\sqrt{2\pi}[/itex] between them, which is a normalization constant.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K