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== If <math>E_\infty</math> is ''finite'', then <math>P_\infty</math> is ''zero'' == | == If <math>E_\infty</math> is ''finite'', then <math>P_\infty</math> is ''zero'' == | ||
| − | <math>P_\infty\equiv\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T|x(t)| | + | ---- |
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| + | <math>P_\infty\equiv\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T|x(t)|^2dt</math> | ||
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| + | <math>P_\infty\equiv(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt)</math> | ||
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| + | Because <math>E_\infty\equiv\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt</math>, because of substitution it follows that | ||
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| + | <math>P_\infty=(\lim_{T\to\infty}\frac{1}{2T})*E_\infty</math> | ||
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| + | <math>P_\infty=(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}E_\infty)</math> | ||
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| + | <math>P_\infty=\lim_{T\to\infty}\frac{E_\infty}{2T}</math> | ||
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| + | This limit will always evaluate to zero as long as <math>E_\infty</math> is finite. | ||
| + | |||
| + | If <math>E_\infty</math> is ''finite'', then <math>P_\infty</math> is ''zero'' | ||
Revision as of 09:53, 17 June 2009
If $ E_\infty $ is finite, then $ P_\infty $ is zero
$ P_\infty\equiv\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T|x(t)|^2dt $
$ P_\infty\equiv(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt) $
Because $ E_\infty\equiv\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt $, because of substitution it follows that
$ P_\infty=(\lim_{T\to\infty}\frac{1}{2T})*E_\infty $
$ P_\infty=(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}E_\infty) $
$ P_\infty=\lim_{T\to\infty}\frac{E_\infty}{2T} $
This limit will always evaluate to zero as long as $ E_\infty $ is finite.
If $ E_\infty $ is finite, then $ P_\infty $ is zero
