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| − | + | == ENERGY == | |
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| + | The energy of a signal can by computed by the following Energy formula: | ||
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| + | <math>E = \int_{t_1}^{t_2} y(t)\, dt</math> | ||
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| + | on the other hand, power of a signal can be calculated by: | ||
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| + | <math>P = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} y(t)\, dt</math> | ||
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| + | Let's now calculate the energy and power of the following signal: <math>y(t) = x^{2}</math> for <math>t_1 = 0</math> and <math>t_2 = 5</math> | ||
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| + | <math>E = \int_{0}^{5} x^{2}\, dt = \frac{1}{3} \left [ x^{3} \right ] _0^5 = \frac{125}{3}</math> | ||
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| + | <math>P = \frac{1}{5 - 0} \int_{0}^{5} x^{2}\, dt = \frac{25}{3}</math> | ||
Latest revision as of 17:32, 5 September 2008
ENERGY
The energy of a signal can by computed by the following Energy formula:
$ E = \int_{t_1}^{t_2} y(t)\, dt $
on the other hand, power of a signal can be calculated by:
$ P = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} y(t)\, dt $
Let's now calculate the energy and power of the following signal: $ y(t) = x^{2} $ for $ t_1 = 0 $ and $ t_2 = 5 $
$ E = \int_{0}^{5} x^{2}\, dt = \frac{1}{3} \left [ x^{3} \right ] _0^5 = \frac{125}{3} $
$ P = \frac{1}{5 - 0} \int_{0}^{5} x^{2}\, dt = \frac{25}{3} $
