(New page: <math>x(t) = e^{-2t} , t>2 \,</math> <math>x(t)= e^{-4t} , 0 \le t \le 2</math> <math>x(t)= 0 , t < 0 \,</math> We can summarize it as <math>x(t)= e^{-2t} u(t-2) + e^{-4t}( u(t-2)-u(t...) |
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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier transform of a CT SIGNAL == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
<math>x(t) = e^{-2t} , t>2 \,</math> | <math>x(t) = e^{-2t} , t>2 \,</math> | ||
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<math>X(\omega) = \frac{e^{-(2j\omega + 4)t}}{j\omega +2} + \frac{1 - e^{-(2j\omega + 8)t}}{-j\omega +4} \,</math> | <math>X(\omega) = \frac{e^{-(2j\omega + 4)t}}{j\omega +2} + \frac{1 - e^{-(2j\omega + 8)t}}{-j\omega +4} \,</math> | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:26, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
$ x(t) = e^{-2t} , t>2 \, $
$ x(t)= e^{-4t} , 0 \le t \le 2 $
$ x(t)= 0 , t < 0 \, $
We can summarize it as
$ x(t)= e^{-2t} u(t-2) + e^{-4t}( u(t-2)-u(t))\, $
$ X(\omega) = \int^\infty_\infty e^{-2t}e^{-j\omega t} dt + \int^2_0 e^{-4t}e^{-j\omega t} dt\, $
$ X(\omega) = \int^\infty_\infty e^{-(2+j\omega)t} dt + \int^2_0 e^{-(4+j\omega) t} dt\, $
$ X(\omega) = {\left. \frac{e^{-(j\omega + 2)t}}{-(j\omega +2)} \right]^{\infty}_0 } + {\left. \frac{e^{-(j\omega + 4)t}}{-(j\omega +4)} \right]^2_0 }\, $
$ X(\omega) = \frac{e^{-(2j\omega + 4)t}}{j\omega +2} - \frac{e^{-(2j\omega + 8)t}}{-j\omega +4} + \frac{1}{4+j\omega} \, $
$ X(\omega) = \frac{e^{-(2j\omega + 4)t}}{j\omega +2} + \frac{1 - e^{-(2j\omega + 8)t}}{-j\omega +4} \, $
