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Homework Help: Convolution in Frequency Domain

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the Fourier transform of the following signal JUST by using the FT table and the FT properties

    x(t) = sin(t) -pi<=t<=pi
    0 otherwise

    NOTE: I am using CONVOLVED WITH as a substitute for * (the real convolution operator) because I cannot express multiplication in any other way that I know of).

    2. Relevant equations

    Frequency Convolution : x1(t)*x2(t) ==> (1/2pi)X1([itex]\omega[/itex]) CONVOLVED WITH X2([itex]\omega[/itex])

    Fourier Transforms:
    sin(ω0t) ===> (pi/j)[δ(ω-ω0)-δ(ω+ω0)]

    rect(t/[itex]\tau[/itex]) ===> [itex]\tau[/itex]sinc([itex]\omega[/itex][itex]\tau[/itex]/2pi)

    3. The attempt at a solution

    I know that x(t) can be expressed as a sine multiplied with a rectangle function

    x(t) = sin(t) * rect(t/2pi)

    so by the frequency convolution property, X(ω)= (1/2pi)[[itex]\frac{pi}{j}[/itex][δ(ω-1)+δ(ω+1)] CONVOLVED WITH 2pi*sinc(ω)]

    My problem is how to do the convolution. Is it easier than it looks? I don't know how the imaginary (j) factors into the convolution. Can someone help explain how to simplify X(ω)? Thank you
  2. jcsd
  3. Dec 11, 2012 #2

    rude man

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    I get from the problem statement that you're not supposed to doany convolving.

    How about using u(t-π) and u(t+π) along with sin(t) and forgetting about your rectangle function?
  4. Dec 11, 2012 #3
    My teacher actually said that using the rectangle function was the correct method of doing it, but didn't elaborate. All he said was set x(t) = sin(t)*rect(t/2pi)
  5. Dec 11, 2012 #4

    rude man

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    OK, then, how about the fact that

    f(ω)**δ(ω-ω0) = f(ω-ω0) where I use ** to denote convolution. You have delta functions in your transform of sin(t); these are easily eliminated by using this theorem in convolving with the transform of the rectangular time function. The rest is very messy algebra.
    Last edited: Dec 11, 2012
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