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Math Help - The Fourier Transform Shift property

  1. #1
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    The Fourier Transform Shift property

    Hello there,

    I have a question about the Shift property of the Fourier Transform.

    I understood that this is the property:
    $FourierTransform(\ \{\ f(t-t_0)\ \}\ ) = e^{-iwt_0} FourierTransform(\ \{ \ f(t)\ \}\ )$

    However, this doesn't really work for me when I apply it to Sin...

    This is the Transform of Sin(t):
    $FourierTransform(\ \{\ Sin(t)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}[ \delta(w-1)-\delta(w+1)]$

    So according to the shift property, I would have expected to see this as the Transform of Sin(t-t0):
    $FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-iwt_0}[\delta(w-1)-\delta(w+1)]$

    However, this is the actual transform of Sin(t-t0):
    $FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-it_0}[e^{2it_0}\delta(w-1)-\delta(w+1)]$


    So the shift property doesn't work? What am I doing wrong?

    Thank you (:
    Last edited by sapsapz; December 1st 2013 at 07:22 AM.
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  2. #2
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    Re: The Fourier Transform Shift property

    Quote Originally Posted by sapsapz View Post
    Hello there,

    I have a question about the Shift property of the Fourier Transform.

    I understood that this is the property:
    $FourierTransform(\ \{\ f(t-t_0)\ \}\ ) = e^{-iwt_0} FourierTransform(\ \{ \ f(t)\ \}\ )$

    However, this doesn't really work for me when I apply it to Sin...

    This is the Transform of Sin(t):
    $FourierTransform(\ \{\ Sin(t)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}[ \delta(w-1)-\delta(w+1)]$

    So according to the shift property, I would have expected to see this as the Transform of Sin(t-t0):
    $FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-iwt_0}[\delta(w-1)-\delta(w+1)]$

    However, this is the actual transform of Sin(t-t0):
    $FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-it_0}[e^{2it_0}\delta(w-1)-\delta(w+1)]$


    So the shift property doesn't work? What am I doing wrong?

    Thank you (:
    Where did you obtain the form of the transform you have there? It's incorrect.
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  3. #3
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    Re: The Fourier Transform Shift property

    Here is a short proof I encountered:
    The Fourier Transform Shift property-fourpr.png

    Edit:
    Unless you're asking about the transforms of the Sin functions, in which case I calculated them in Wolfram Mathematica
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  4. #4
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    Re: The Fourier Transform Shift property

    I was asking about the final form of the FT of Sin[t-t0]

    That's not what I get from Mathematica but it doesn't matter.

    Multiply your Exp[- i t0] factor through and you obtain the correct answer.
    Thanks from sapsapz
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