Revision as of 08:27, 4 September 2008 by Serdbrue (Talk)

Given the Signal $ x(t)=3sin(2*pi*3t) $, Find the energy and power of the signal from 0 to 5 seconds.

Energy

$ \int_1^5 |x(t)|^2 dt $

$ \int_1^5 |3sin(6\pi t)|^2 dt $


$ 9*\int_1^5 sin(6\pi t)^2 dt $


$ 9*\int_1^5 sin(6\pi t)^2 dt $

$ 9*(\dfrac{6 \pi t}{2}-\dfrac{sin(2*(6\pi t))}{4}) $evaluated at 5 and 1

$ 9*(3\pi t-\dfrac{sin(12\pi t)}{4}) $evaluated at 5 and 1

$ 27\pi *t-\dfrac{9sin(12\pi *t)}{4} $evaluated at 5 and 1

$ 27\pi *5-\dfrac{9sin(12\pi *5)}{4}-(27\pi *1-\dfrac{9sin(12\pi *1)}{4} $

$ \int_1^5 |3sin(6\pi t)|^2 dt=108\pi $

Power

$ \dfrac{1}{t2-t1}\int_1^5 |x(t)|^2 dt $

$ \dfrac{1}{5-1}\int_1^5 |3sin(6\pi t)|^2 dt $


$ \dfrac{9}{4}*\int_1^5 sin(6\pi t)^2 dt $


$ \dfrac{9}{4}*\int_1^5 sin(6\pi t)^2 dt $

$ \dfrac{9}{4}*(\dfrac{6 \pi t}{2}-\dfrac{sin(2*(6\pi t))}{4}) $evaluated at 5 and 1

$ \dfrac{9}{4}*(3\pi t-\dfrac{sin(12\pi t)}{4}) $evaluated at 5 and 1

$ \dfrac{27}{4}\pi *t-\dfrac{9sin(12\pi *t)}{16} $evaluated at 5 and 1

$ \dfrac{27}{4}\pi *5-\dfrac{9sin(12\pi *5)}{16}-(\dfrac{27}{4}\pi *1-\dfrac{9sin(12\pi *1)}{16} $

$ \int_1^5 |3sin(6\pi t)|^2 dt=27\pi $

Alumni Liaison

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

Francisco Blanco-Silva