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Question about Fourier Transforms

  1. Jan 6, 2010 #1
    Hi!

    I have a question about F. transforms. In my exams (for mathematical methods in physics) I'm usually given a function such as:

    [tex]f(t)=\begin{cases} 1,\ t\in(-T,T)\\
    0,\ else\\
    \end{cases}[/tex]

    Now I can find easily [tex]F(\omega)[/tex], but then I need to find the original function again (to confirm the validity of the found transform). And when I integrate I'm converting it into complex integral. And there are the troubles! I end up with exponential functions, but since t is arbitrary I can choose two contours of integration. And the question is: based on what do I choose those contours? Do I have to make it such that I have decaying exponential? My assistant wasn't quite clear on that and I didn't find it anywhere else...

    So if anyone could help me I'd appreciate it! Thanx!
     
  2. jcsd
  3. Jan 6, 2010 #2
  4. Jan 7, 2010 #3
    Yeah, the contour part (and complex integration part) is clear, but say I have two poles on the negative Im part. Choosing the upper half plane as a place to put my contour I will have [tex]\oint f(z)=0[/tex] as Cauchy-Goursat theorem states, and if I choose the bottom half my integral will be [tex]\oint f(z)=2\pi i\sum Res(f)[/tex]. Those two differ, and I am choosing the contours based on a parameter t.

    For my original function it's:

    [tex]f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \underbrace{\sqrt{\frac{2}{\pi}}\frac{\sin(\omega T)}{\omega}e^{-i \omega t}}_{F(\omega)}d\omega[/tex]
    After a slight manipulation I have:
    [tex]\frac{1}{2\pi i}\int_{-\infty}^\infty\frac{1}{\omega}\left[e^{i\omega (T-t)}-e^{-i\omega(T+t)}\right]d\omega[/tex]

    I can split those two into separate integrals. We don't know how T and t are behaving respectably to each other (T<t,T>t ?) so if we put different relations we get different values of integral. But based on what do I choose which contour goes for what?
     
  5. Jan 7, 2010 #4
    I only see one pole at the origin. Why did you say there're two ?
     
  6. Jan 7, 2010 #5
    Oh my bed, anywho, I get that I have one pole, but still I have [tex]e^{i\omega(T-t)}[/tex] and [tex]e^{-i\omega(T+t)}[/tex], and based on T and t i have different contour. Is it because I'm dealing with the transform which, by definition must exponentially die out? When dealing with something like damped oscillator, I know that I must get exponentially decaying function, but in my case I have rectangle puls...
     
  7. Jan 7, 2010 #6
    The exponential is imaginary, so it won't "die out" or decay, but the path integral, on the other hand, can "die out", if you can choose an appropriate contour, that's why you need Jordan's lemma.
     
  8. Jan 8, 2010 #7
    Oh, I see... Thanx!
     
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