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Convolution of two delta functions in frequency domain

  1. Sep 8, 2011 #1
    Apparently, when convolving, for example:

    [δ(ω-π) - δ(ω+π)] * (δ(ω+50π)-δ(ω-50π))

    the result is
    δ(ω+49π)-δ(ω-51π)-δ(ω+51π)+δ(ω-49π)

    where δ() is the Dirac delta function, * the convolution operator and ω the frequency variable

    How do we get to this? Can you help me on the intuition in this example and/or general in convolution in the frequency domain?

    thank you.


    edit: i think i understand that we use the distributivity property to expand it.When we have , i.e. δ(ω+π) * δ(ω-50π) how do we continue?
     
  2. jcsd
  3. Sep 8, 2011 #2

    chiro

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    Science Advisor

    In terms of actually computing the convolution, you can use Laplace transforms to get the "frequency information" of both terms, and then use the inverse Laplace Transform to get the actual result. With convolutions you just work out the Laplace transforms of T1, and T2 (in T1 * T2) and then use it in the way that was said above.

    You could alternatively do it just with the actual definition though, but that would require a different approach. The Laplace way is a lot easier.

    Do you know about Laplace transforms?

    What have you tried so far? Maybe you should show some working or any ideas you had or tried that failed.
     
  4. Sep 8, 2011 #3
    The course i'm on doesn't include Laplace transforms, so i guess it's not the expected way.

    The equation i've written is part of a solved problem set, and i'm trying to actually understand why this result is correct because i'm preparing for my exams.
    I know the rule about homework.I have the solution, i'm only asking for tips to understand it.

    In the time domain, it's ok. f(x) convolved with δ(x-k) yields f(k) , as the delta function is 0 everywhere but k.
    In the frequency domain, it seems that the frequency offset of the delta function, is simply added as an offset to the convolving pair.
     
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