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Homework Help: Help in performing convolution

  1. Oct 30, 2011 #1
    Hello. I have a problem convolving two functions. I have attached a file with the problem in details, and will be very grateful if someone can provide me with a proper explanation.
    Thanks! :shy:

    Attached Files:

  2. jcsd
  3. Oct 31, 2011 #2
    It's super easy (though a little abstact) to prove using the convolution theorem.
    [tex] F\{a \ast b \} = F\{a\} \cdot F\{b\}[/tex]
    The Fourier transform of cosine gives you the sum of two delta peaks at [itex]-\omega_0[/itex] and at [itex]\omega_0[/itex]. The Fourier transform of 1/x gives you a sgn-function. You multiply those two together and you get the difference of the delta peaks. You inverse Fourier transform that and you get the sine.

    You might as well compute the convolution directly, but I'm still thinking about how to carry out the integration.

    If you want a rational argument, keep in mind that the convolution measures the two signals correlation where the argument of the convoluted function is the phase difference of your two input functions.
  4. Oct 31, 2011 #3
    Thanks, but I'm still unable to really figure it out

    I made the computation (file attached), and still have some differences in the results I get, so I probably miss something here... but what exactly?

    Attached Files:

    • 1.pdf
      File size:
      140.1 KB
  5. Oct 31, 2011 #4
    Okay, the sign is pretty simple. You know that the delta function is zero except for one certain point, right? So there is actually a very simple relation.
    [tex] \delta(x-x_0) sgn(x) = \delta(x-x_0), x_0>0[/tex]
    [tex] \delta(x-x_0) sgn(x) = -\delta(x-x_0), x_0<0[/tex]
    I think you can figure that out by yourself.
    About the factors... I'll look into it and come back to you, okay?
    Btw: I think I figured out a way to do the convolution integral directly through contour integration. Do you wanna know that or are you happy with the convolution theorem?
  6. Oct 31, 2011 #5

    I'm quite satisfied with knowing the same result could also be recieved using contour integration. The real idea here is to use the Fourier transform's properties (since that's the subject i'm focused on now, really), so no need to deviate from that right now (thanks a lot though :shy:).
    The coefficients issue is, in fact, more critical for me to figure out thoroughly at the moment, and therefore I'll be grateful if you could share your conclusions with me once you reach them.
    Thanks! :smile:
  7. Nov 1, 2011 #6
    It looks like they missed a factor in the exercise sheet.
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