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Probelm getting started with Fouriertransforms!

by toofle
Tags: fouriertransforms, probelm, started
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toofle
#1
Sep18-11, 07:45 AM
P: 20
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?
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I like Serena
#2
Sep18-11, 09:30 AM
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P: 6,187
Welcome to PF, toofle!

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.
toofle
#3
Sep18-11, 12:44 PM
P: 20
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.

I like Serena
#4
Sep18-11, 12:57 PM
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P: 6,187
Probelm getting started with Fouriertransforms!

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.


Quote Quote by toofle View Post
Thanks that clears some things up.

But is:
[tex]-{1 \over 6} e^{-3 \omega} \sqrt{\pi \over 2} \left(e^{
5 \omega} (-3 + 2 e^\omega) \Theta(-\omega) + (2 -
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...?
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#5
Sep19-11, 11:43 AM
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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.


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