(New page: == The Chosen Signal == <math> x(t) = e^{3t} u(t+2) + e^{4t} u(t-2) \!</math> == The Fourier Transform == <math> X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \!</math> <m...)
 
 
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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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== The Chosen Signal ==
 
== The Chosen Signal ==
  
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<math> X(\omega) = \frac{e^{-2*(3-j\omega)}}{(j\omega - 3 )} + \frac{e^{( 4- j\omega )2}}{(j\omega -4 )} </math>
 
<math> X(\omega) = \frac{e^{-2*(3-j\omega)}}{(j\omega - 3 )} + \frac{e^{( 4- j\omega )2}}{(j\omega -4 )} </math>
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[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]

Latest revision as of 12:24, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


The Chosen Signal

$ x(t) = e^{3t} u(t+2) + e^{4t} u(t-2) \! $

The Fourier Transform

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

$ X(\omega) = \int_{-2}^{\infty} e^{3t} e^{-j\omega t} dt + \int_{2}^{\infty} e^{4t} e^{-j\omega t} dt \! $

$ X(\omega) = \int_{-2}^{\infty} e^{(3-j\omega )t} dt + \int_{2}^{\infty} e^{(4-j\omega )t} dt \! $

$ X(\omega) = {\left. \frac{e^{(j\omega -3)t}}{(j\omega - 3 )} \right]^{\infty}_{-2} } + {\left. \frac{e^{( j\omega -4)t}}{(j\omega -4 )} \right]^{\infty}_2 }\, $


$ X(\omega) = \frac{e^{-2*(3-j\omega)}}{(j\omega - 3 )} + \frac{e^{( 4- j\omega )2}}{(j\omega -4 )} $


Back to Practice Problems on CT Fourier transform

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