(New page: == Energy == <font size="5"> Consider the signal <math>x(t)=cos(t)</math> over the interval 0 to <math>pi</math> <math>E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|f(t)|^2 dt</math> <math>...) |
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| − | <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>\pi</math> |
Revision as of 13:48, 5 September 2008
Energy
Consider the signal $ x(t)=cos(t) $ over the interval 0 to $ \pi $
$ E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|f(t)|^2 dt $
$ E = {1\over(2\pi-0)}\int_{0}^{2\pi}\!|2cos(t)|^2 dt $
$ E = {1\over(2\pi-0)}{1\over2}(4)\int_{0}^{2\pi}\!(1+cos(2t)) dt $
$ E = {1\over\pi}(2\pi+{1\over2}sin(2*2\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 $
