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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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Let the signal x(t) be equal to:
 
Let the signal x(t) be equal to:
  
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<math>X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \,</math>
 
<math>X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \,</math>
  
 
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==Comments/questions==
 
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*Note: This could also be done knowing that the Fourier transform of <math>e^{j\omega_0 t} = 2\pi \delta(\omega - \omega_0) \,</math>.
Note: This could also be done knowing that the Fourier transform of <math>e^{j\omega_0 t} = 2\pi \delta(\omega - \omega_0) \,</math>.
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[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]

Latest revision as of 12:29, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


Let the signal x(t) be equal to:

$ x(t) = cos(2\pi t) \, $


The Fourier Transform of a signal in Continuous Time is defined by:

$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \, $


Using this, we obtain:

$ X(\omega) = \int_{-\infty}^{\infty}cos(2\pi t)e^{-j\omega t}dt \, $


Knowing that cos(t) is equal to: $ \frac{e^{jt}+e^{-jt}}{2} $:

$ X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{j2\pi t}+e^{-j2\pi t})e^{-j\omega t}dt \, $


$ X(\omega) = \pi \delta(\omega - 2\pi) + \pi \delta(\omega - 2\pi) \, $


Comments/questions

  • Note: This could also be done knowing that the Fourier transform of $ e^{j\omega_0 t} = 2\pi \delta(\omega - \omega_0) \, $.

Back to Practice Problems on CT Fourier transform

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