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Properties of the Fourier Transform - Time Differentitation

  1. Aug 8, 2012 #1
    1. The problem statement, all variables and given/known data
    This is copied from a book:
    $$\eqalign{
    & {\rm{Time Differentitation}} \cr
    & {\rm{Given that: }}F(\omega ) = F\left[ {f(t)} \right] \cr
    & F\left[ {f'(t)} \right] = jwF(\omega ) \cr
    & {\rm{Proof:}} \cr
    & f(t) = {F^{ - 1}}\left[ {F\left( \omega \right)} \right] = {1 \over {2\pi }}\int_{ - \infty }^\infty {F\left( \omega \right){e^{j\omega t}}d\omega } \cr
    & {\rm{Taking the derivative of both sides with respect to }}t{\rm{ gives:}} \cr
    & {d \over {dt}}f(t) = {{j\omega } \over {2\pi }}\int_{ - \infty }^\infty {F\left( \omega \right){e^{j\omega t}}d\omega } = j\omega {F^{ - 1}}\left[ {F(\omega )} \right]{\rm{ or }}F\left[ {f'(t)} \right] = jwF(\omega ) \cr} $$

    Can somebody explain why the jw is outside the integral? I can't see how that happens using Leibniz's integral rule - http://en.wikipedia.org/wiki/Leibniz_integral_rule

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 8, 2012 #2

    marcusl

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    Science Advisor
    Gold Member

    The factor of jw should be inside the integral, as you surmise. Then take the forward FT of both sides to get the answer.
     
  4. Aug 9, 2012 #3
    Thanks marcusl. I see now.
     
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