Revision as of 05:39, 3 September 2008 by Anders89 (Talk)

The signal is: x(t) = 2cos(2t)


Energy

$ \int_0^{2\pi}{|2cos(2t)|^2dt} $

$ = 4 \times \frac{1}{2}\int_0^{2\pi}(1+cos(4t))dt $


$ =2 \times (t+\frac{1}{4}sin(4t))|_{t=0}^{t=2\pi} $


$ =2 \times (2\pi+\frac{1}{4}-0-0) $


$ =4*\pi + \frac{1}{2} $

Average Power

$ \frac{1}{2\pi - 0}\int_0^{2\pi}{|2cos(2t)|^2dt} $


$ =\frac{1}{2\pi} \times 4 \times \frac{1}{2}\int_0^{2\pi}(1+cos(4t))dt $


$ =\frac{1}{\pi} \times (t+\frac{1}{4}sin(4t))|_{t=0}^{t=2\pi} $


$ =\frac{1}{\pi} \times (2\pi+\frac{1}{4}-0-0) $


$ = 2 + \frac{2}{4\pi} $

Alumni Liaison

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

Francisco Blanco-Silva