Revision as of 15:04, 4 September 2008 by Jkkietzm (Talk)

Energy of a Signal

$ y = 3e^{t} $ from (0,5)
$ Energy = \int_{t1}^{t2} x(t) $
$ Energy = \int_{0}^{5}3e^{t}dt $
$ Energy = 3e^{5} - 3 $


Power of a Signal

$ y = 3e^{t} $ from (0,5)
$ Average Power = \frac{1}{t2 - t1}\int_{t1}^{t2}x(t)^2 $
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Alumni Liaison

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

Francisco Blanco-Silva