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| + | [[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>X(\omega ) = \delta(\omega ) + \delta(\omega - 5) + \delta(\omega - 5)\,</math> | <math>X(\omega ) = \delta(\omega ) + \delta(\omega - 5) + \delta(\omega - 5)\,</math> | ||
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<math> = \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega )e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega - 5)e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega + 5)e^{j\omega t}d\omega\,</math> | <math> = \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega )e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega - 5)e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega + 5)e^{j\omega t}d\omega\,</math> | ||
| + | |||
| + | <math> = \frac{1}{2\pi}*1 + \frac{1}{2\pi}*e^{5jt} + \frac{1}{2\pi}*e^{-5jt}\,</math> | ||
| + | |||
| + | <math> = \frac{1}{2\pi} * (1 + 2cos(5t))\,</math> | ||
| + | |||
| + | |||
| + | I'll add another one when i have time | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:46, 16 September 2013
Example of Computation of inverse Fourier transform (CT signals)
A practice problem on CT Fourier transform
$ X(\omega ) = \delta(\omega ) + \delta(\omega - 5) + \delta(\omega - 5)\, $
$ x(t) = \int_{-\infty}^{\infty}X(\omega )e^{j\omega t}d\omega\, $
$ = \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega )e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega - 5)e^{j\omega t}d\omega + \frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega + 5)e^{j\omega t}d\omega\, $
$ = \frac{1}{2\pi}*1 + \frac{1}{2\pi}*e^{5jt} + \frac{1}{2\pi}*e^{-5jt}\, $
$ = \frac{1}{2\pi} * (1 + 2cos(5t))\, $
I'll add another one when i have time
