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| − | :<math>E = \frac{1}{2}\left(t + \frac{1}{2}sin(2t)\right)\,</math> | + | :<math>E = \frac{1}{2}\left(t + \frac{1}{2}sin(2t)\right)|_{0}^{2\pi}\,</math> |
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| + | |||
| + | :<math>E = \frac{1}{2}\left(2\pi + \frac{1}{2}sin(4\pi)\right) - 0\,</math> | ||
| + | |||
| + | |||
| + | :<math>E = \frac{1}{2}\left(2\pi\right) = \pi\,</math> | ||
Revision as of 18:59, 4 September 2008
Suppose a signal is defined by $ cos(t) $
The energy can be computed using the formula:
- $ E = \int_{b}^{a}{|x(t)|^2}dt\, $
Suppose we want to compute the energy of the signal $ cos(t) $ in the interval $ 0 $ to $ 2\pi $.
The formula then becomes:
- $ E = \int_{0}^{2\pi}{|cos(t)|^2}dt\, $
Using trigonometric identity, $ cos^2(t) = \frac{1}{2} + \frac{1}{2}cos(2t)\, $
This implies:
- $ E = \frac{1}{2}\int_{0}^{2\pi}1 + cos(2t)dt\, $
Integrating yields
- $ E = \frac{1}{2}\left(t + \frac{1}{2}sin(2t)\right)|_{0}^{2\pi}\, $
- $ E = \frac{1}{2}\left(2\pi + \frac{1}{2}sin(4\pi)\right) - 0\, $
- $ E = \frac{1}{2}\left(2\pi\right) = \pi\, $
