(New page: For the signal: <math>X(\omega)= </math>) |
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For the signal: | For the signal: | ||
| − | <math>X(\omega)= </math> | + | <math>X(\omega)= 2\pi \delta(\omega) + 3\pi \delta(\omega - 3\pi) - 4\pi \delta(\omega + 5\pi)</math> |
| + | |||
| + | <math>x(t) = \frac{1}{2\pi} \int_{-\infty}^\infty (2\pi \delta(\omega) + 3\pi \delta(\omega - 3\pi) - 4\pi \delta(\omega + 5\pi)) e^{j\omega t} \mathrm{d}\omega</math> | ||
| + | |||
| + | <math> = \int_{-\infty}^\infty ( \delta(\omega) + \frac{3}{2} \delta(\omega - 3\pi) - 2 \delta(\omega + 5\pi)) e^{j\omega t} \mathrm{d}\omega</math> | ||
| + | |||
| + | <math> x(t) = 1 + \frac{3}{2}e^{j3\pi t} - 2e^{-5\pi t}</math> | ||
Revision as of 05:11, 8 October 2008
For the signal:
$ X(\omega)= 2\pi \delta(\omega) + 3\pi \delta(\omega - 3\pi) - 4\pi \delta(\omega + 5\pi) $
$ x(t) = \frac{1}{2\pi} \int_{-\infty}^\infty (2\pi \delta(\omega) + 3\pi \delta(\omega - 3\pi) - 4\pi \delta(\omega + 5\pi)) e^{j\omega t} \mathrm{d}\omega $
$ = \int_{-\infty}^\infty ( \delta(\omega) + \frac{3}{2} \delta(\omega - 3\pi) - 2 \delta(\omega + 5\pi)) e^{j\omega t} \mathrm{d}\omega $
$ x(t) = 1 + \frac{3}{2}e^{j3\pi t} - 2e^{-5\pi t} $
