(Energy)
(Power)
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<math>P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math>
 
<math>P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math>
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<math>P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math>
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</font>
 
</font>

Revision as of 14:56, 4 September 2008

Energy

$ f(t)=2cos(t) $


$ E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|f(t)|^2 dt $


$ E = {1\over(2\pi-0)}\int_{0}^{2\pi}\!|2cos(t)|^2 dt $


$ E = {1\over(2\pi-0)}{1\over2}\int_{0}^{2\pi}\!(1+cos(2t)) dt $


$ E = {1\over(4\pi)}(2\pi+{1\over2}sin(2*2\pi)) dt $


$ E = {1\over2} $

Power

$ f(t)=2cos(t) $


$ P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt $


$ P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt $


Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett