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<math>Average Power = {1\over(4 \pi -0)}\int_{0}^{4 \pi}\!|sin(t)|^2 dt</math> | <math>Average Power = {1\over(4 \pi -0)}\int_{0}^{4 \pi}\!|sin(t)|^2 dt</math> | ||
| − | <math>Average Power = {1\over(4 \pi -0)}\int_{0}^{4 \pi}\! | + | <math>Average Power = {1\over(4 \pi -0)}\int_{0}^{4 \pi}\!\frac{(1-cos(2t)}{2} dt</math> |
<math>Average Power= \frac{2 \pi - \frac{1}{4} \sin(8 \pi)}{4 \pi}</math> | <math>Average Power= \frac{2 \pi - \frac{1}{4} \sin(8 \pi)}{4 \pi}</math> | ||
<math>\ Average Power= \frac{1}{2} </math> | <math>\ Average Power= \frac{1}{2} </math> | ||
Revision as of 15:56, 5 September 2008
Energy of a signal
Consider the signal $ \ y = \sin(t) $ Lets find the energy over one cycle:
$ Energy = \int_{t_1}^{t_2}\!|x(t)|^2 dt $
$ Energy = \int_{0}^{2 \pi}\!|sin(t)|^2 dt $
$ Energy = \int_{0}^{2 \pi}\!(\frac{1-cos(2t)}{2}) dt $
$ Energy = \pi - \frac{1}{4} \sin(4 \pi) $
$ \ Energy = \pi $
Average Power of a Signal
$ Average Power = {1\over(t_2-t_1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt $
$ Average Power = {1\over(4 \pi -0)}\int_{0}^{4 \pi}\!|sin(t)|^2 dt $
$ Average Power = {1\over(4 \pi -0)}\int_{0}^{4 \pi}\!\frac{(1-cos(2t)}{2} dt $
$ Average Power= \frac{2 \pi - \frac{1}{4} \sin(8 \pi)}{4 \pi} $
$ \ Average Power= \frac{1}{2} $
