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<math>P=\frac{1}{2T}\int_{-T}^{T}\!|f(t)|^2\ dt</math> | <math>P=\frac{1}{2T}\int_{-T}^{T}\!|f(t)|^2\ dt</math> | ||
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<math>=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{|2cos(t)|^2dt}</math> | <math>=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{|2cos(t)|^2dt}</math> |
Revision as of 16:00, 3 September 2008
Signal
$ y(t)=2cos(t) $
Energy
According to formula of Energy of a singal,we can get:
$ E = \int_{t_1}^{t_2} \! |f(t)|^2\ dt $
$ =\int_0^{2\pi}{|2cos(t)|^2dt} $
$ =\int_0^{2\pi}{(2(2cos(t)^2-1)+2)dt} $
$ =\int_0^{2\pi}{2+cos(2t))dt} $
$ =(2t+sin(2t))|_{t=0}^{t=2\pi} $
$ =(4\pi+0-0-0) $
$ =(4\pi) $
Power
According to formula of Power of a singal,we can get: $ P=\frac{1}{2T}\int_{-T}^{T}\!|f(t)|^2\ dt $
$ =\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{|2cos(t)|^2dt} $
$ =\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{(2(2cos(t)^2-1)+2)dt} $
$ =\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{(2+cos(2t))dt} $
$ =\frac{1}{4\pi}(2t+sin(2t))|_{t={-2\pi}}^{t=2\pi} $
$ =\frac{1}{4\pi}(4\pi+0-(-4\pi)-0) $
$ =\frac{1}{4\pi}(8\pi) $
$ =2 $