(New page: <math>x(t) = e^{-3t} , t>3 \,</math> <math>x(t)= e^{-6t} , 0 \le t \le 3</math> <math>x(t)= 0 , t < 0 \,</math> <math>x(t)= e^{-3t} u(t-3) + e^{-6t}( u(t-3)-u(t))\,</math> <math>X(\omega)...)
 
 
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<math>x(t) = e^{-3t} , t>3 \,</math>
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[[Category:problem solving]]
<math>x(t)= e^{-6t} , 0 \le t \le 3</math>
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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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<math>x(t) = e^{-3t} , t>3 \,</math>,
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<math>x(t)= e^{-6t} , 0 \le t \le 3</math>,
 
<math>x(t)= 0 , t < 0 \,</math>
 
<math>x(t)= 0 , t < 0 \,</math>
  
 
<math>x(t)= e^{-3t} u(t-3) + e^{-6t}( u(t-3)-u(t))\,</math>
 
<math>x(t)= e^{-3t} u(t-3) + e^{-6t}( u(t-3)-u(t))\,</math>
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<math>X(\omega) = \int^\infty_\infty e^{-3t}e^{-j\omega t} dt + \int^2_0 e^{-6t}e^{-j\omega t} dt\,</math>
 
<math>X(\omega) = \int^\infty_\infty e^{-3t}e^{-j\omega t} dt + \int^2_0 e^{-6t}e^{-j\omega t} dt\,</math>
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<math>X(\omega) = \int^\infty_\infty e^{-(3+j\omega)t} dt + \int^3_0 e^{-(6+j\omega) t} dt\,</math>
 
<math>X(\omega) = \int^\infty_\infty e^{-(3+j\omega)t} dt + \int^3_0 e^{-(6+j\omega) t} dt\,</math>
<math>X(\omega) = {\left. \frac{e^{-(j\omega + 3)t}}{-(j\omega +3)} \right]^{\infty}_0 } + {\left. \frac{e^{-(j\omega + 6)t}}{-(j\omega +6)} \right]^3_0 }\,</math>
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<math>X(\omega) = {\left. \frac{e^{-(j\omega + 3)t}}{-(j\omega +3)} \right]^{\infty}_3 } + {\left. \frac{e^{-(j\omega + 6)t}}{-(j\omega +6)} \right]^3_0 }\,</math>
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<math>X(\omega) =  \frac{e^{-(3j\omega + 9)}}{j\omega +3}  -  \frac{e^{-(3j\omega + 18)t}}{-j\omega +6} + \frac{1}{6+j\omega} \,</math>
 
<math>X(\omega) =  \frac{e^{-(3j\omega + 9)}}{j\omega +3}  -  \frac{e^{-(3j\omega + 18)t}}{-j\omega +6} + \frac{1}{6+j\omega} \,</math>
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<math>X(\omega) =  \frac{e^{-(3j\omega + 9)}}{j\omega +3}  +  \frac{1 - e^{-(3j\omega + 18)t}}{-j\omega +6} \,</math>
 
<math>X(\omega) =  \frac{e^{-(3j\omega + 9)}}{j\omega +3}  +  \frac{1 - e^{-(3j\omega + 18)t}}{-j\omega +6} \,</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 11:36, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform


$ x(t) = e^{-3t} , t>3 \, $, $ x(t)= e^{-6t} , 0 \le t \le 3 $, $ x(t)= 0 , t < 0 \, $

$ x(t)= e^{-3t} u(t-3) + e^{-6t}( u(t-3)-u(t))\, $

$ X(\omega) = \int^\infty_\infty e^{-3t}e^{-j\omega t} dt + \int^2_0 e^{-6t}e^{-j\omega t} dt\, $

$ X(\omega) = \int^\infty_\infty e^{-(3+j\omega)t} dt + \int^3_0 e^{-(6+j\omega) t} dt\, $

$ X(\omega) = {\left. \frac{e^{-(j\omega + 3)t}}{-(j\omega +3)} \right]^{\infty}_3 } + {\left. \frac{e^{-(j\omega + 6)t}}{-(j\omega +6)} \right]^3_0 }\, $

$ X(\omega) = \frac{e^{-(3j\omega + 9)}}{j\omega +3} - \frac{e^{-(3j\omega + 18)t}}{-j\omega +6} + \frac{1}{6+j\omega} \, $

$ X(\omega) = \frac{e^{-(3j\omega + 9)}}{j\omega +3} + \frac{1 - e^{-(3j\omega + 18)t}}{-j\omega +6} \, $


Back to Practice Problems on CT Fourier transform

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