(Average Power)
(Energy)
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<math>P = {1\over2}(4\pi+{1\over2}sin(8\pi)</math>
+
<math>P = {1\over2}(4\pi+{1\over2}sin(8\pi))</math>
  
  
 
<math>P = 2\pi</math>
 
<math>P = 2\pi</math>
 
</font>
 
</font>

Revision as of 14:16, 5 September 2008

Average Power

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


$ 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

$ f(t)=cos(t) $


$ 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 $

Alumni Liaison

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

Dr. Paul Garrett