Revision as of 09:47, 4 September 2008 by Cjuzeszy (Talk)

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Signal

$ f(t)= 2sin(t) \mbox{ from 0 to 2}\pi $

Energy

$ E=\int_{t_1}^{t_2}|x(t)|^2dt $

$ E=\int_{0}^{2\pi}|2sin(t)|^2dt $

$ E=\int_{0}^{2\pi}|2\Big(2sin^2(t)-1\Big)+2|dt $

$ E=\int_{0}^{2\pi}|-2cos(2t)+2|dt $

$ E=\int_{0}^{2\pi}2cos(2t)+2dt $

$ E=sin(2t)+2t\Big|\begin{matrix}2\pi\\0\end{matrix} $

$ E=\int_{0}^{2\pi}|2sin(t)|^2dt= 4\pi $


Power

$ P=\frac{1}{t_2-t_1}\int_{t_1}^{t_2}|x(t)|^2dt $

$ P=\frac{1}{2\pi-0}\int_{0}^{2\pi}|2sin(t)|^2dt $

$ P=\frac{1}{2\pi}\int_{0}^{2\pi}|2\Big(2sin^2(t)-1\Big)+2|dt $

$ P=\frac{1}{2\pi}\int_{0}^{2\pi}2cos(2t)+2dt $

$ P=\frac{1}{2\pi}\bigg[sin(2t)+2t\Big|\begin{matrix}2\pi\\0\end{matrix}\bigg] $

$ P=\frac{1}{2\pi-0}\int_{0}^{2\pi}|2sin(t)|^2dt=2\pi $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva