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Fourier Transform Time Shifting Property

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data

    I tried to work out the FT of a sin function with a time delay using first mathematical manipulation, and then using the time shifting property.

    However I get two very similar, but for some reason not identical answers.

    2. Relevant equations

    Please open the .jpg to see my working out.
    The answers are identical except for the 'f' and 'f0' which are circled in red pen.

    3. The attempt at a solution

    Thanks for your help!
     

    Attached Files:

  2. jcsd
  3. Sep 12, 2011 #2

    vela

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    It follows from the fact that [itex]f(x)\delta(x-x_0) = f(x_0)\delta(x-x_0)[/itex].
     
  4. Sep 12, 2011 #3
    could you please explain further?
     
  5. Sep 12, 2011 #4

    Hootenanny

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    Are you familiar with the Dirac Delta function and its properties?
     
  6. Sep 12, 2011 #5
    Yeh I am. And I understand the equation given. But where is that happening in my working out?

    The exponential with f0 and t0 given is simply a complex number, and can't be considered a dirac delta function in the frequency domain right?
     
  7. Sep 12, 2011 #6

    vela

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    What do you get if you apply that property to, say, [itex]e^{j2\pi f t_0} \delta(f-f_0)[/itex] (where [itex]e^{j2\pi f t_0}[/itex] is the function multiplying the delta function)?
     
  8. Sep 12, 2011 #7
    I fully get it now!

    Basically it doesn't matter if you use f or f0, the result is the same since you are multiplying with an impulse at f=f0!

    Thanks so much for your help!
     
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