(→Energy) |
(→Energy) |
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Consider the signal | Consider the signal | ||
| − | <math>x(t)=cos(t)</math> over the interval 0 to <math>\pi</math> | + | <math>x(t)=cos(t)</math> over the interval 0 to <math>4\pi</math> |
| − | <math>E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!| | + | <math>E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt</math> |
| − | <math>E = {1\over( | + | <math>E = {1\over(4\pi-0)}\int_{0}^{4\pi}\!|cos(t)|^2 dt</math> |
| − | <math>E = {1\over( | + | <math>E = {1\over(4\pi-0)}{1\over2}(4)\int_{0}^{4\pi}\!(1+cos(t)) dt</math> |
| − | <math>E = {1\over\pi}( | + | <math>E = {1\over\pi}(4\pi+{1\over2}sin(4\pi)) dt</math> |
Revision as of 13:49, 5 September 2008
Energy
Consider the signal $ x(t)=cos(t) $ over the interval 0 to $ 4\pi $
$ E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt $
$ E = {1\over(4\pi-0)}\int_{0}^{4\pi}\!|cos(t)|^2 dt $
$ E = {1\over(4\pi-0)}{1\over2}(4)\int_{0}^{4\pi}\!(1+cos(t)) dt $
$ E = {1\over\pi}(4\pi+{1\over2}sin(4\pi)) dt $
$ E = {2} $
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 $
