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Need help with a Fourier Transform

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    f(t) = (sin(2t))/t




    2. Relevant equations



    3. The attempt at a solution
    I know that sin(t)/t has the fourier transform pi(w). I'm just not sure how to apply that fact to this problem. Knowing that sin(t)/t --> pi(w), I reasoned that sin(2t)/t --> 2pi(2w). I'm almost certain this is incorrect. I arrived at this conclusion by noticing that if i multiplied the top and bottom of sin(2t)/t by 2, I would have a sinc function 2sinc(2t). Then I could use the property sin(t)/t --> pi(w). Am I anywhere close?
     
  2. jcsd
  3. Nov 1, 2009 #2

    jbunniii

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    You have the right idea. Let's introduce the following notation:

    [tex]h(t) = \frac{\sin(t)}{t}[/tex]

    and [itex]H(\omega)[/itex] is the Fourier transform of [itex]h(t)[/itex].

    Notice that

    [tex]h(2t) = \frac{\sin(2t)}{2t}[/tex]

    and so

    [tex]f(t) = 2 h(2t) = \frac{\sin(2t)}{t}[/tex]

    Therefore you need to find the Fourier transform of [itex]2 h(2t)[/tex].

    Now suppose I write

    [tex]g(t) = h(2t)[/itex]

    and let [itex]G(\omega)[/itex] denote the Fourier transform of [itex]g(t)[/itex]. Do you know (in general, not necessarily for a specific function) how [itex]G(\omega)[/itex] can be expressed in terms of [itex]H(\omega)[/itex]? If so, then you are essentially done, because

    [tex]\mathcal{F}(f(t)) = \mathcal{F}(2 h(2t)) = 2 \mathcal{F}(h(2t)) = 2 \mathcal{F}(g(t)) = 2 G(\omega)[/tex]
     
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