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| − | == | + | == Average Power == |
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| − | <math> | + | <math>Avg. Power = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt</math> |
| − | <math> | + | <math>Avg. Power = {1\over(4\pi-0)}\int_{0}^{4\pi}\!|cos(t)|^2 dt</math> |
| − | <math> | + | <math>Avg. Power = {1\over(4\pi)}{1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt</math> |
| − | <math> | + | <math>Avg. Power = {1\over8\pi}(4\pi+{1\over2}sin(8\pi)) dt</math> |
| − | <math> | + | <math>Avg. Power = {1\over2}</math> |
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Revision as of 14:12, 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)) dt $
$ Avg. Power = {1\over2} $
Power
$ f(t)=2cos(t) $
$ P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt $
$ P = \int_{0}^{2\pi}\!|2cos(t)|^2\ dt $
$ P = (4){1\over2}\int_{0}^{2\pi}\!1+cos(2t) dt $
$ P = (4){1\over2}(2\pi+{1\over2}sin(2*2\pi) $
$ P = 4\pi $
