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Fourier Transformation - Convolution quick question

  1. Feb 22, 2014 #1
    Okay the question is to find the fourier transform of:

    rect([itex]\frac{x}{5})[/itex][itex]\otimes[/itex]([itex]\delta[/itex](x+3)-[itex]\delta[/itex](x-3))

    =F[itex]^{\infty}_{\infty}[/itex] [itex]\int[/itex]rect([itex]\frac{x'}{5}[/itex])([itex]\delta[/itex](x+3-x')-[itex]\delta[/itex](x-3-x')) dx' [1]

    - where F represents a fourier transform.
    My Issue
    Okay I am fine doing this using the convolution theorem, that the fourier transform of a convultion is given by the product of the two individual fourier transforms, but I am having trouble doing it explicitly

    So from [1] integrating over each delta function, I deduce that the first term collapses everywhere except x'=x+3, and the second everywhere except x'=x-3, . So I get:

    F(rect[itex]\frac{x+3}{5}[/itex]-rect[itex]\frac{x-3}{5})[/itex]
    = (5sinc[itex]\frac{5k}{2}[/itex]exp[itex]^{\frac{3ik}{5}}[/itex]exp[itex]^{\frac{-3ik}{5}}[/itex])
    using the properties that F(rect([itex]\frac{x}{1}[/itex]))=asinc([itex]\frac{ka}{2}[/itex]) and that F(f(x+a))=F(f(x))exp[itex]^{ika}[/itex]

    Which does not agree with the convultion theorem were I get :

    5sinc[itex]\frac{5k}{2}[/itex]exp[itex]^{3ik}[/itex]exp[itex]^{-3ik}[/itex]

    Thanks alot in advance for any assistance !
     
  2. jcsd
  3. Feb 22, 2014 #2

    vela

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    Why are the exponentials being multiplied in both cases? In fact, if they were supposed to be multiplied, they'd cancel in both cases, leaving you with identical but incorrect answers.

    Your mistake is in determining the shift. You're shifting rect(x/5) by 3, not 3/5, because to go from rect(x/5) to rect((x±3)/5), you replace x by x±3, not x±3/5.
     
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