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Help finding Fourier Transform!

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the Fourier Transform of:

    f(t)=[itex]\frac{cos(\alpha t)}{t^2+\beta^2}[/itex]

    2. Relevant equations

    F[itex](\omega)[/itex]=[itex]\frac{1}{2\pi}[/itex][itex]\int[/itex][itex]^{∞}_{-∞}[/itex][itex]\frac{cos(\alpha t)exp(i \omega t)}{t^2+\beta^2}[/itex]

    3. The attempt at a solution
    I start with:

    [itex]cos(\alpha t)[/itex]=[itex]\frac{exp(i \alpha t)+exp(-i \alpha t)}{2}[/itex]

    Substituting this in, I get:

    [itex]F(\omega)[/itex]=[itex]\frac{1}{4\pi}\int^{∞}_{-∞}\frac{exp(it(\alpha+\omega))+exp(it(\omega-\alpha))}{t^2+\beta^2}[/itex]

    From here I know I should be able to get this in the form of some delta functions but I can't figure out the manipulation. I'd appreciate any help!
     
  2. jcsd
  3. Nov 12, 2012 #2

    Dick

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    If you want to evaluate integrals like that you need to use contour integration. Find the poles, figure out contours and use the residue theorem. Have you done stuff like that?
     
  4. Nov 12, 2012 #3
    Yes I've used the residue theorem to evaluate contour integrals. We've covered so much in this class I had already forgotten that that technique would work.

    Since this involves a [itex]2\pi[/itex] periodic function, I'd integrate over the unit circle in the complex plane... Use Euler's formula for the cosine substitution to get an integral in Z, and then find the poles and residues, Right?

    Thanks Dick!
     
  5. Nov 12, 2012 #4

    Dick

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    That's the general technique alright, but the details are wrong. Your function isn't periodic and you don't integrate over the unit circle. You integrate over part of the real axis and then close the contour in the upper or lower half plane. But sure, review the technique and see how far you get.
     
  6. Nov 12, 2012 #5
    Right, this acts like a 'sinc' function, so the Jordan contour is the right selection for integration.

    So the function is:

    [itex]f(t)=\frac{cos(\alpha t)}{\beta^2+t^2}[/itex]

    Let:
    [itex]z=e^{i\alpha t}[/itex]
    [itex]dz=i\alpha e^{i\alpha t}dt[/itex]
    [itex]cos(\alpha t)=\frac{z+z^{-1}}{2}[/itex]

    Transforming [itex] f(t)[/itex] to [itex] f(z)[/itex]

    [itex]f(z)=\frac{z+z^{-1}}{2(\beta^2+t^2)} \cdot \frac{dz}{i\alpha z}[/itex]

    So here with this [itex]t[/itex] in the denominator, can I replace it directly with a [itex]z[/itex], or do I need to use [itex]t=\frac{ln(z)}{i\alpha}[/itex] which I got from the second equation above?

    Once I do that, I'll plug into the F.T. integral and find the value from the Residue Thm.
     
  7. Nov 12, 2012 #6

    Dick

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    That is WAY too fast and sloppy. You aren't going to get anywhere like this. Work out a simple example of this kind of problem first. Integrate e^(it)/(1+t^2) over the real line first. Where are the poles? What's a good contour? If you get that you are probably ready to do to think about the full problem.
     
  8. Nov 12, 2012 #7
    Okay, taking your advice.

    For the example problem,

    [itex]\int^{∞}_{-∞} \frac{e^{it}}{1+t^2}[/itex]
    We can directly substitute t=z
    [itex]\oint ^{∞}_{-∞} \frac{e^{iz}}{1+z^2} = \oint ^{\infty}_{-\infty} \frac{e^{iz}}{(1+i)(1-i)}[/itex]

    Now since it's clear to see that f(z)→0 as R→∞, we can integrate over the upper half plane using a semicircular contour.

    Only the pole z=i lies inside this contour.

    So,

    [itex] \oint ^{\infty}_{-\infty} \frac{e^{iz}}{(1+i)(1-i)}=2\pi i Res(z=i) = lim_{z→i} (z-i)\frac{e^{iz}}{(z+i)(z-i)} [/itex]
    [itex] lim_{z→i}=\frac{2\pi i}{2ie} = \frac{\pi}{e}[/itex]
     
  9. Nov 12, 2012 #8
    I figured out what I was doing wrong. I realize now the substitutions I was making from the Euler formula don't work over the real line, those are more suited for integration over the unit circle. Once I chose a better way to transform into an integral in Z, this turned out to be a very simple problem.

    Thanks for helping me and for scolding me for blindly pulling a formula out of the book.
     
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