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(→Power) |
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</font> | </font> | ||
| − | == | + | == Energy == |
<font size="4"> | <font size="4"> | ||
| − | <math>f(t)= | + | <math>f(t)=cos(t)</math> |
| − | <math>P = \int_{t_1}^{t_2}\!| | + | <math>P = \int_{t_1}^{t_2}\!|x(t)|^2\ dt</math> |
| − | <math>P = \int_{0}^{ | + | <math>P = \int_{0}^{4\pi}\!|cos(t)|^2\ dt</math> |
| − | <math>P = | + | <math>P = {1\over2}\int_{0}^{4\pi}\!1+cos(2t) dt</math> |
Revision as of 14:14, 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} $
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 = (4){1\over2}(2\pi+{1\over2}sin(2*2\pi) $
$ P = 4\pi $
