(New page: == Signal == <math> x(t) = e^{-2t} u(t) + e^{-5t} u(t-4) \!</math> == Fourier Transform == <math> X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \!</math> <math> X(\omega) ...)
 
 
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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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== Signal ==
 
== Signal ==
  
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<math> X(\omega) = \frac{1}{2+j\omega } + \frac{e^{-(20+4j\omega )}}{5+j\omega } \!</math>
 
<math> X(\omega) = \frac{1}{2+j\omega } + \frac{e^{-(20+4j\omega )}}{5+j\omega } \!</math>
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[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]]

Latest revision as of 12:28, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


Signal

$ x(t) = e^{-2t} u(t) + e^{-5t} u(t-4) \! $

Fourier Transform

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

$ X(\omega) = \int_{0}^{\infty} e^{-2t} e^{-j\omega t} dt + \int_{4}^{\infty} e^{-5t} e^{-j\omega t} dt \! $

$ X(\omega) = \int_{0}^{\infty} e^{-(2+j\omega )t} dt + \int_{4}^{\infty} e^{-(5+j\omega )t} dt \! $

$ X(\omega) = {\left. \frac{e^{-(2+j\omega )t}}{-(2+j\omega )} \right]^{\infty}_0 } + {\left. \frac{e^{-(5 + j\omega )t}}{-(5 +j\omega )} \right]^{\infty}_4 }\, $


$ X(\omega) = \frac{1}{2+j\omega } + \frac{e^{-(20+4j\omega )}}{5+j\omega } \! $


Back to Practice Problems on CT Fourier transform

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