(New page: <math>x(t) = u(t)\frac{d}{dt}cos(t-2\pi)</math> <math>X(\omega) = j\omega\int\limits_{-\infty}^{\infty} cos(t-2\pi)u(t)e^{-j\omega t}dt</math> <math>= j\omega\int\limits_{0}^{\inft...)
 
 
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[[Category:problem solving]]
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[[Category:ECE301]]
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[[Category:ECE]]
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[[Category:Fourier transform]]
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[[Category:signals and systems]]
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== Example of Computation of Fourier transform of a CT SIGNAL ==
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A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]]
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<math>x(t) = u(t)\frac{d}{dt}cos(t-2\pi)</math>
 
<math>x(t) = u(t)\frac{d}{dt}cos(t-2\pi)</math>
  
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     <math>=  j\omega e^{-j\omega 2\pi} \int\limits_{0}^{\infty}frac{1}{2}(e^{j\tau} e^{-j\omega \tau}dt</math>
 
     <math>=  j\omega e^{-j\omega 2\pi} \int\limits_{0}^{\infty}frac{1}{2}(e^{j\tau} e^{-j\omega \tau}dt</math>
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----
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[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]

Latest revision as of 12:32, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


$ x(t) = u(t)\frac{d}{dt}cos(t-2\pi) $

$ X(\omega) = j\omega\int\limits_{-\infty}^{\infty} cos(t-2\pi)u(t)e^{-j\omega t}dt $

    $ =  j\omega\int\limits_{0}^{\infty} cos(t-2\pi)e^{-j\omega t}dt $
    $ \tau  = t - 2\pi $
    $ =  j\omega\int\limits_{0}^{\infty} cos(\tau)e^{-j\omega(\tau -2\pi)}dt $
    $ =  j\omega\int\limits_{0}^{\infty} cos(\tau)e^{-j\omega \tau}e^{-j\omega 2\pi}dt $
    $ =  j\omega e^{-j\omega 2\pi} \int\limits_{0}^{\infty} cos(\tau)e^{-j\omega \tau}dt $
    $ =  j\omega e^{-j\omega 2\pi} \int\limits_{0}^{\infty}frac{1}{2}(e^{j\tau} e^{-j\omega \tau}dt $

Back to Practice Problems on CT Fourier transform

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Ryne Rayburn