(New page: <math>X(\omega) = 2\pi e^{4j\omega}\,</math> We kenw that the Fourier transformation of <math>x(t - t_o)\,</math> is equal to <math>e^{-j\omega t_o} X(\omega)\,</math> Using this propert...) |
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Revision as of 14:14, 6 October 2008
$ X(\omega) = 2\pi e^{4j\omega}\, $
We kenw that the Fourier transformation of $ x(t - t_o)\, $ is equal to $ e^{-j\omega t_o} X(\omega)\, $
Using this property, we can solve the question
$ X(\omega) = 2\pi e^{4j\omega} Y(\omega)\, $ where $ Y(\omega) = 1\, $
$ y(t) = \delta(t)\, $
$ x(t) = \frac{1}{2\pi} \int^{\infty}_{-\infty}2\pi e^{4j\omega} Y(\omega) dw\, $
$ x(t) = \int^{\infty}_{-\infty}e^{4j\omega} Y(\omega) dw\, $
$ x(t) = \hat{f} (e^{4j\omega } Y(\omega))\, $
$ x(t) = \delta(t+4)\, $
