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Topic: fourier transform of distributions (Read 1431 times) |
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trusure
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fourier transform of distributions
« on: Mar 16th, 2010, 1:57pm » |
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the Fourier transform of a derivative is:
(d/dx(f))^ = i s (f^)(s)
Now I have the function
f(t)={ sin(t) , if -pi/2 <=t<=pi/2
0 , else
}
how to prove that
(d/dx(f))^ = i s (f^)(s)
in distribution sense; I mean applying the notation
<f^, phi> = <f, (phi)^>
(or any other rules like : <f ', phi> = - <f, (phi)'> )
for any test function phi
thanks
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| « Last Edit: Mar 16th, 2010, 1:57pm by trusure » |
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MonicaMath
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Re: fourier transform of distributions
« Reply #1 on: Mar 17th, 2010, 8:40am » |
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Did you find the FT for your function? I will give it to you
f^(s)=sqrt(2/pi) i s cos(s.pi/2)/(s^2 - 1)
use this now.
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Obob
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Re: fourier transform of distributions
« Reply #2 on: Mar 17th, 2010, 10:50am » |
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Rather than work with this particular function, you can do it for any (tempered) distribution f as follows (writing D for d/ds and p = phi for the test function to save space):
<i s f^, p> = <f^, i s p> = <f, (i s p)^> = - <f, D(p^)> = <Df, p^> = <(Df)^, p>,
which says that (Df)^ = i s f^. The crucial step here is <f, (i s p)^> = - <f, D(p^)>, which follows from (i s p)^ = - D(p^). This is just a statement about test functions; you can prove it by writing out the definition of the Fourier transform and differentiating under the integral.
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