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Real and imaginary of F(ω)

  1. May 17, 2014 #1
    I never see the following hypothesis but I believe that they are true...

    ##\text{Re}(\hat f (\omega)) = a(\omega)##

    ##\text{Im}(\hat f (\omega)) = b(\omega)##


    where:

    ##f(t) = \int_{-\infty}^{+\infty}\hat f(\omega) \exp(i \omega t) d\omega = \int_{0}^{\infty} a(\omega) \cos(\omega t) + b(\omega) \sin(\omega t) d\omega##


    ##\hat f (\omega) = \int_{-\infty}^{+\infty}f(t) \exp(-i \omega t) dt##


    ##a(\omega) = \frac{1}{\pi}\int_{-\infty}^{+\infty} f(t) \cos(\omega t) dt##

    ##b(\omega) = \frac{1}{\pi}\int_{-\infty}^{+\infty} f(t) \sin(\omega t) dt##


    So, the two first equations are true?
     
    Last edited: May 17, 2014
  2. jcsd
  3. May 17, 2014 #2

    Simon Bridge

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    You are missing the definition of ##\hat f(\omega)##.
    From there you should be able to prove (or disprove) the relationships yourself.
     
  4. May 17, 2014 #3
    Is unmistakable that ##\hat f## represents the Fourier transform of ##f## !!!
     
  5. May 17, 2014 #4

    HallsofIvy

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    Use the fact that [itex]e^{-i\omega t}= cos(\omega t)- i sin(\omega t)[/itex].

     
  6. May 17, 2014 #5
    An phasor in the complex form is ##A \exp(i(\omega t + \varphi)) = A \exp(i \varphi) \exp(i \omega t)##, the summation of phasors wrt angular frequency is ##\sum A(\omega) \exp(i \varphi(\omega)) \exp(i \omega t) \Delta \omega = \sum \hat f(\omega) \exp(i \omega t) \Delta \omega##. So, becomes clear that the ##\text{Abs}(\hat f(\omega)) = A(\omega)## and ##\text{Arg}(\hat f(\omega)) = \varphi(\omega)##.

    What I want mean is that I don't understand the relation that a(ω) and b(ω) has with f(ω).
     
  7. May 17, 2014 #6

    Simon Bridge

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    ... but you edited post #1 to include that anyway - thank you ;)
    So your next step was to relate the sine and cosine form to the exponential in the fourier transform re post #4.

    As in post #4. ##e^{-i\omega t}=\cos\omega t - i\sin\omega t##
    Make the substitution in the Fourier transformation definition... which you gave as:
    $$\hat f(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}\;dt$$ ... and follow your nose.

    Did you try that?
     
  8. May 17, 2014 #7
    Yeah, but I don't see in none place a direct connection between a(ω) and b(ω) with Re(f(ω)) and Im(f(ω)).
     
  9. May 17, 2014 #8

    Simon Bridge

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    Please show your working.
     
  10. May 18, 2014 #9
    $$

    \\f(x) = \int_{0}^{\infty} A(\omega) \cos(x \omega) + B(\omega) \sin(x \omega) d\omega

    \\ = \frac{1}{\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)(\cos(\omega t) \cos(\omega x) + \sin(\omega t) \sin(\omega x))dt d\omega

    \\ = \frac{1}{\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)\cos(\omega(x-t))dt d\omega

    \\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t)(\exp(i \omega (x-t)) + \exp(-i \omega(x-t)))dt d\omega

    \\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega (x-t))dt d\omega + \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(-i \omega(x-t)) dt d\omega

    \\ = \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega (x-t))dt d\omega + \frac{1}{2 \pi} \int_{-infty}^{0} \int_{-\infty}^{\infty}f(t) \exp(i \omega(x-t)) dt d\omega

    \\ = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(t) \exp(i \omega(x-t))dt d\omega

    \\ = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \left( \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(t) \exp(-i \omega t) dt \right) \exp(i \omega x) d \omega

    \\ = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \left( \hat f(\omega) \right) \exp(i \omega x) d \omega

    $$
     
  11. May 18, 2014 #10
    So...?
     
  12. May 18, 2014 #11

    Simon Bridge

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    You started at the wrong place.

    Start from your stated definition for ##\hat f## ... the one with the exponential in it. That is your first line.

    Your second line should use the substitution for the exponential in terms of sine and cosine.
     
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