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Fourier transform of a shifted and time-reversed sign

  1. May 27, 2015 #1
    1. The problem statement, all variables and given/known data
    given a continuous-time signal g(t) . Its fourier transform is G(f) ( see definition in picture / "i" is the imaginary number) . It is required to find the fourier transform of the shifted-time-reversed signal g(a-t) where a is a real constant .
    That is , find the fourier transform of g(a-t) based on the knowledge of the fourier transform G(f) of g(t)

    2. Relevant equations
    The defition of the fourier transform is shown in the attached picture

    3. The attempt at a solution
    There are 2 properties of the fourier transform : shift property + time scaling.
    But i'm not sure how to use them both . I prefer to use the definition of the fourier transform to find the relationship between the fourier transform of g(a-t) and the fourier transform of g(t)
     

    Attached Files:

  2. jcsd
  3. May 27, 2015 #2

    Zondrina

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    So you know ##\mathcal{F} \{g(t) \} = G(f)##.

    Try writing this:

    $$g(a - t) = g((-1) (t - a))$$

    Now you can use both the time scaling and time shifting properties.
     
  4. May 27, 2015 #3
    Can you check the workout i did in the picture below ?
     

    Attached Files:

  5. May 27, 2015 #4

    Zondrina

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    The time scaling property states:

    $$\mathcal{F} \{g(\alpha x) \} = \frac{1}{|\alpha|} \hat G \left( \frac{f}{\alpha} \right)$$

    You know ##\mathcal{F} \{g(t) \} = \hat G(f)## and ##g(a - t) = g((-1) (t - a))##.

    Let ## u = t - a##. Then ##g((-1) (t - a)) = g((-1)u) = g(\alpha u)## with ##\alpha = -1##.

    Then you know ##\mathcal{F} \{g(\alpha u) \} = \frac{1}{|\alpha|} \hat G (\frac{f}{\alpha}) = \hat G (-f)##.

    This amounts to saying: Since you know ##\mathcal{F} \{g(t) \} = \hat G(f)##, then you can simply apply a negative sign to the argument and get ##\mathcal{F} \{ g(a - t) \}##.
     
  6. May 27, 2015 #5
    There has to be an exponential factor . Did you check my solution ?
     
  7. May 27, 2015 #6

    Zondrina

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    The solution is fine, I thought it would be insightful to elaborate on the time-reversal property where ##\alpha = -1##.

     
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