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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:inverse Fourier transform]]
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[[Category:signals and systems]]
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== Example of Computation of inverse Fourier transform (CT signals) ==
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A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]]
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<math>\mathcal{X}(\omega) = \left\{ {\begin{array}{*{20}c}
 
<math>\mathcal{X}(\omega) = \left\{ {\begin{array}{*{20}c}
 
   {1,} & {-2 \le  \omega \le 0}  \\
 
   {1,} & {-2 \le  \omega \le 0}  \\
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<math>x(t)=\frac{1}{2\pi}(\frac{1}{jt} - \frac{e^{-j 2 t}}{jt}) + \frac{1}{2\pi}(-\frac{e^{j 2 t}}{jt} + \frac{1}{jt}) </math>
 
<math>x(t)=\frac{1}{2\pi}(\frac{1}{jt} - \frac{e^{-j 2 t}}{jt}) + \frac{1}{2\pi}(-\frac{e^{j 2 t}}{jt} + \frac{1}{jt}) </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 11:47, 16 September 2013

Example of Computation of inverse Fourier transform (CT signals)

A practice problem on CT Fourier transform


$ \mathcal{X}(\omega) = \left\{ {\begin{array}{*{20}c} {1,} & {-2 \le \omega \le 0} \\ { -1,} & {0 \le g(x) \ge 2} \\ {0,} & {| \omega| > 2} \\ \end{array}} \right. $


$ \ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega $

$ \ x(t)=\frac{1}{2\pi}\int_{-2}^{0}e^{j\omega t}\,d\omega + \frac{1}{2\pi}\int_{0}^{2}-e^{j\omega t}\,d\omega $

$ x(t)=\frac{1}{2\pi}(\frac{1}{jt} - \frac{e^{-j 2 t}}{jt}) + \frac{1}{2\pi}(-\frac{e^{j 2 t}}{jt} + \frac{1}{jt}) $


Back to Practice Problems on CT Fourier transform

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Questions/answers with a recent ECE grad

Ryne Rayburn