(Power)
(Power)
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<math>9*\int_1^5 sin(6\pi t)^2 dt</math>
 
<math>9*\int_1^5 sin(6\pi t)^2 dt</math>
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<math>9*\int_1^5 sin(6\pi t)^2 dt</math>
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<math>9*(6\pi *t/2-sin(2*(6\pi *t)/4)</math>evaluated at 5 and 1
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<math>9*(3\pi *t-sin(12\pi *t)/4)</math>evaluated at 5 and 1
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<math>27\pi *t-9/4*sin(12\pi *t)</math>evaluated at 5 and 1
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<math>27\pi *5-9/4*sin(12\pi *5)-(27\pi *1-9/4*sin(12\pi *1)</math>
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<math>\int_1^5 |3sin(6\pi t)|^2 dt=108\pi

Revision as of 08:06, 4 September 2008

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

Power

$ \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*(6\pi *t/2-sin(2*(6\pi *t)/4) $evaluated at 5 and 1

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

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

$ 27\pi *5-9/4*sin(12\pi *5)-(27\pi *1-9/4*sin(12\pi *1) $

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

Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal