(Average Power)
(Energy)
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== Energy ==
 
== Energy ==
 
<font size="4">
 
<font size="4">
 
<math>f(t)=cos(t)</math>
 
 
  
 
<math>P = \int_{t_1}^{t_2}\!|x(t)|^2\ dt</math>
 
<math>P = \int_{t_1}^{t_2}\!|x(t)|^2\ dt</math>

Revision as of 14:19, 5 September 2008

Consider the signal $ x(t)=cos(t) $ over the interval 0 to $ 4\pi $

Average Power

$ Avg. Power = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt $


$ Avg. Power = {1\over(4\pi-0)}\int_{0}^{4\pi}\!|cos(t)|^2 dt $


$ Avg. Power = {1\over(4\pi)}{1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt $


$ Avg. Power = {1\over8\pi}(4\pi+{1\over2}sin(8\pi)) $


$ Avg. Power = {1\over2} $

Energy

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


$ P = \int_{0}^{4\pi}\!|cos(t)|^2\ dt $


$ P = {1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt $


$ P = {1\over2}(4\pi+{1\over2}sin(8\pi)) $


$ P = 2\pi $

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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