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<math>E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|f(t)|^2 dt</math> | <math>E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|f(t)|^2 dt</math> | ||
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| + | <math>E = {1\over(2\pi-0)}\int_{0}^{2\pi}\!|2cos(t)|^2 dt</math> | ||
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| + | <math>E = {1\over(2\pi-0)}{1\over2}(4)\int_{0}^{2\pi}\!(1+cos(2t)) dt</math> | ||
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| + | <math>E = {1\over\pi}(2\pi+{1\over2}sin(2*2\pi)) dt</math> | ||
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| + | <math>E = {2}</math> | ||
</font> | </font> | ||
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<math>P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math> | <math>P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math> | ||
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| + | <math>P = \int_{0}^{2\pi}\!|2cos(t)|^2\ dt</math> | ||
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| + | <math>P = (4){1\over2}\int_{0}^{2\pi}\!1+cos(2t) dt</math> | ||
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| + | <math>P = (4){1\over2}(2\pi+{1\over2}sin(2*2\pi)</math> | ||
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| + | <math>P = 4\pi</math> | ||
</font> | </font> | ||
Latest revision as of 15:05, 4 September 2008
Energy
$ f(t)=2cos(t) $
$ 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 $
