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Lets find the energy over one cycle: | Lets find the energy over one cycle: | ||
| − | <math>\ int_ | + | <math>Energy = \int_{t_1}^{t_2}\!|x(t)|^2 dt</math> |
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| + | <math>Energy = \int_{0}^{2 \pi}\!|sin(t)|^2 dt</math> | ||
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| + | <math>Energy = \int_{0}^{2 \pi}\!|frac{1-cos(2t)| dt</math> | ||
| + | |||
| + | <math>Energy = \int_{0}^{2 \pi}\!|sin(t)|^2 dt</math> | ||
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| + | ==Average Power of a Signal== | ||
| + | <math>Avg. Power = {1\over(t_2-t_1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt</math> | ||
Revision as of 15:35, 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)| dt $
$ Energy = \int_{0}^{2 \pi}\!|sin(t)|^2 dt $
Average Power of a Signal
$ Avg. Power = {1\over(t_2-t_1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt $
