(New page: For the signal: <math>x(t) = e^{-8(t + 1)} u(t + 1)</math> <math>X(\omega) = \int_{-\infty}^\infty e^{-8(t + 1)} u(t + 1) e^{-j\omega t} \mathrm{d}t</math> when <math>t + 1 < 0 </math> ...)
 
 
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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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----
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For the signal:
 
For the signal:
  
<math>x(t) = e^{-8(t + 1)} u(t + 1)</math>
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<math>x(t) = e^{-8(t + 1)} u(t + 1) \ </math>
  
 
<math>X(\omega) = \int_{-\infty}^\infty e^{-8(t + 1)} u(t + 1) e^{-j\omega t} \mathrm{d}t</math>
 
<math>X(\omega) = \int_{-\infty}^\infty e^{-8(t + 1)} u(t + 1) e^{-j\omega t} \mathrm{d}t</math>
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<math> = e^{-8}  \frac{ -e^{-(8+j\omega)t}}{(8+j\omega)t}\bigg|_{t=-1}^\infty</math>
 
<math> = e^{-8}  \frac{ -e^{-(8+j\omega)t}}{(8+j\omega)t}\bigg|_{t=-1}^\infty</math>
 
<math> = e^{-8} \frac{ e^{(8+j\omega)}}{(8+j\omega)}</math>
 
<math> = e^{-8} \frac{ e^{(8+j\omega)}}{(8+j\omega)}</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 12:34, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform



For the signal:

$ x(t) = e^{-8(t + 1)} u(t + 1) \ $

$ X(\omega) = \int_{-\infty}^\infty e^{-8(t + 1)} u(t + 1) e^{-j\omega t} \mathrm{d}t $

when $ t + 1 < 0 $ or $ t < -1 $, $ u(t + 1) = 0 $ so,

$ X(\omega) = \int_{-1}^\infty e^{-8(t + 1)} e^{-j\omega t}\mathrm{d}t $

$ = e^{-8} \int_{-1}^\infty e^{-8t} e^{-j \omega t}\mathrm{d}t $ $ = e^{-8} \int_{-1}^\infty e^{-(8+j\omega)t} \mathrm{d}t $

$ = e^{-8} \frac{ -e^{-(8+j\omega)t}}{(8+j\omega)t}\bigg|_{t=-1}^\infty $ $ = e^{-8} \frac{ e^{(8+j\omega)}}{(8+j\omega)} $


Back to Practice Problems on CT Fourier transform

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