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Convolution integral

  1. Mar 5, 2012 #1
    Determine the confolution of f with itself where f is:

    f(t) = 1 for ltl<1 and 0 everywhere else

    Then deduce that:

    -∞ sin2ω/ω2 dω = ∏

    Fouriertransform of f gives:

    f(ω) = 2/√(2∏) sin(ω)/ω

    and using the convolution theorem gives:

    f*f = 4/√(2∏) sin2(ω)/ω2

    But I'm clueless of what to do from this point. Should I evaluate the convolution integral and equate that to the above?

    In that case we have:

    f*f = ∫-∞ f(τ)f(t-τ) dτ

    But how do I evaluate that? I can see that τ must be between -1 and 1. Thus t must be between -2 and 2? But what does that make the integral look like?
     
    Last edited: Mar 5, 2012
  2. jcsd
  3. Mar 5, 2012 #2

    Dick

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    You are integrating a step function times a shifted step function where t is the size of the shift. The product is going to be 1 on some interval and 0 off that interval. Draw a picture and try working out say f*f(0), f*f(1), f*f(2)... until it becomes clear what function of t that is. Though for your given problem you only need the value of f*f(0).

    And the convolution theorem actually tells you that you take the inverse fourier transform of f(ω)^2 to get the convolution f*f.
     
    Last edited: Mar 5, 2012
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