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Suppose a signal is defined by <math>cos(t)</math> | Suppose a signal is defined by <math>cos(t)</math> | ||
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| + | The energy can be computed using the formula: | ||
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| + | :<math>E = \int_{b}^{a}{|x(t)|^2}dt\,</math> | ||
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| + | Suppose we want to compute the energy of the signal <math>cos(t)</math> in the interval <math>0</math> to <math>2\pi</math>. | ||
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| + | The formula then becomes: | ||
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| + | :<math>E = \int_{0}^{2\pi}{|cos(t)|^2}dt\,</math> | ||
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| + | Using trigonometric identity, <math>cos^2(t) = \frac{1}{2} + \frac{1}{2}cos(2t)\,</math> | ||
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| + | This implies: | ||
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| + | :<math>E = \frac{1}{2}\int_{0}^{2\pi}1 + cos(2t)dt\,</math> | ||
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| + | Integrating yields | ||
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| + | :<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> | ||
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| + | :<math>E = \frac{1}{2}\left(2\pi\right) = \pi\,</math> | ||
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| + | The formula for calculating average power is similar to energy: | ||
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| + | :<math>P = \frac{1}{T}\int_{0}^{T}{|x(t)|^2}dt\,</math> | ||
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| + | Applying the signal and interval, the following formula is obtained: | ||
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| + | :<math>P = \frac{1}{2\pi-0}\int_{0}^{2\pi}|cos(t)|^2dt\,</math> | ||
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| + | :<math>P = \frac{1}{4\pi}\int_{0}^{2\pi}1 + cos(2t)dt\,</math> | ||
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| + | :<math>P = \frac{1}{4\pi}\left(t + \frac{1}{2}sin(2t)\right)|_{0}^{2\pi}\,</math> | ||
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| + | :<math>P = \frac{1}{4\pi}\left(2\pi + \frac{1}{2}sin(4\pi)\right) - 0\,</math> | ||
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| + | :<math>P = \frac{1}{4\pi}\left(2\pi\right) = \frac{1}{2}</math> | ||
Latest revision as of 20:14, 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\, $
The formula for calculating average power is similar to energy:
- $ P = \frac{1}{T}\int_{0}^{T}{|x(t)|^2}dt\, $
Applying the signal and interval, the following formula is obtained:
- $ P = \frac{1}{2\pi-0}\int_{0}^{2\pi}|cos(t)|^2dt\, $
- $ P = \frac{1}{4\pi}\int_{0}^{2\pi}1 + cos(2t)dt\, $
- $ P = \frac{1}{4\pi}\left(t + \frac{1}{2}sin(2t)\right)|_{0}^{2\pi}\, $
- $ P = \frac{1}{4\pi}\left(2\pi + \frac{1}{2}sin(4\pi)\right) - 0\, $
- $ P = \frac{1}{4\pi}\left(2\pi\right) = \frac{1}{2} $
