(→Average Power of a Signal) |
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==Energy of a signal== | ==Energy of a signal== | ||
Consider the signal <math>\ y = \sin(t)</math> | Consider the signal <math>\ y = \sin(t)</math> | ||
| − | Lets find the energy over | + | Lets find the energy over two cycles: |
<math>Energy = \int_{t_1}^{t_2}\!|x(t)|^2 dt</math> | <math>Energy = \int_{t_1}^{t_2}\!|x(t)|^2 dt</math> | ||
| − | <math>Energy = \int_{0}^{ | + | <math>Energy = \int_{0}^{4 \pi}\!|sin(t)|^2 dt</math> |
| − | <math>Energy = \int_{0}^{ | + | <math>Energy = \int_{0}^{4 \pi}\!(\frac{1-cos(2t)}{2}) dt</math> |
| − | <math>Energy = \pi - \frac{1}{4} \sin( | + | <math>Energy = 2 \pi - \frac{1}{4} \sin(8 \pi)</math> |
| − | <math>\ Energy = \pi </math> | + | <math>\ Energy = 2 \pi </math> |
==Average Power of a Signal== | ==Average Power of a Signal== | ||
| + | Here we compute the average power of the same signal above over two cycles: | ||
| + | |||
<math>Average Power = {1\over(t_2-t_1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt</math> | <math>Average Power = {1\over(t_2-t_1)}\int_{t_1}^{t_2}\!|x(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}\!|sin(t)|^2 dt</math> | ||
| − | <math>Average Power = {1\over(4 \pi -0)}\int_{0}^{4 \pi}\!\frac{(1-cos(2t)}{2} | + | <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{1}{2} </math> | ||
Latest revision as of 15:58, 5 September 2008
Energy of a signal
Consider the signal $ \ y = \sin(t) $ Lets find the energy over two cycles:
$ Energy = \int_{t_1}^{t_2}\!|x(t)|^2 dt $
$ Energy = \int_{0}^{4 \pi}\!|sin(t)|^2 dt $
$ Energy = \int_{0}^{4 \pi}\!(\frac{1-cos(2t)}{2}) dt $
$ Energy = 2 \pi - \frac{1}{4} \sin(8 \pi) $
$ \ Energy = 2 \pi $
Average Power of a Signal
Here we compute the average power of the same signal above over two cycles:
$ 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} $
