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| − | <math>\mathcal{X}(\omega) = \ | + | [[Category:problem solving]] |
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:inverse Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of inverse Fourier transform (CT signals) == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
| + | <math>\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.</math> | ||
| + | |||
<math>\ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega </math> | <math>\ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega </math> | ||
| − | <math>= \frac{1}{2\pi}\int_{- | + | <math>\ 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 </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> | ||
| + | |||
| + | ---- | ||
| + | [[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}) $
