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Question about fourier transform

  • Thread starter fanxilong
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  • #1
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hi every one!

i want to know the fourier transform of x(t)
x(t)=exp(-t/a)*sin(a*t), where a ,b is constant

and can it be work out by matlab?

another question is :
how to proof the fourier transform of x(t) who follows normal distribution n(u, sigm^2 ) is also normal distribution ?

thanks !
 

Answers and Replies

  • #2
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Do you have your powers correct in the exponential? x(t) doesn't look like it would have a converging FT. Note, that as t -> -infinity, x(t) is unbounded.

BTW: sin(a*t) can be written in terms of complex exponentials.
 
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  • #3
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Do you have your powers correct in the exponential? x(t) doesn't look like it would have a converging FT. Note, that as t -> -infinity, x(t) is unbounded.

BTW: sin(a*t) can be writting in terms of complex exponentials.
yes, you are right , it should be like this:
x(t)=0 for t<0
x(t)=exp(-t/a)*sin(b*t) for t>=0

and then ?
 
Last edited:
  • #4
cepheid
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and can it be work out by matlab?
Hi! MATLAB calculates things numerically. If you plug in numbers corresponding values for that function, then I believe the FFT ("fast Fourier transform") algorithm will spit out numbers for the Fourier transform (can't recall if you have to work out the frequencies yourself).

This is NOT what you want though. You want an analytical expression for the Fourier transform of that function, right? So my question is, what is the hang up here? In other words, where are you having a problem. Start with the *definition* of the Fourier transform, and go ahead:

[tex] X(\omega) = \mathcal{F}\{x(t)\} = \int_{-\infty}^{\infty} x(t)e^{-i \omega t}\,dt = \int_{0}^{\infty} (e^{-t/a}\sin(bt))e^{-i \omega t}\,dt [/tex]

I think I got that right, but I'd double check it if I were you.

another question is :
how to proof the fourier transform of x(t) who follows normal distribution n(u, sigm^2 ) is also normal distribution ?
Again, what is the issue? You know the definition of a Fourier transform, right?
 

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