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| − | = | + | =Proof= |
| − | + | == If <math>E_\infty</math> is ''finite'', then <math>P_\infty</math> is always ''zero'' == | |
| − | + | ---- | |
| + | |||
| + | <math>P_\infty\equiv\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T|x(t)|^2dt</math> | ||
| + | |||
| + | <math>P_\infty\equiv(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt)</math> | ||
| + | |||
| + | Because <math>E_\infty\equiv\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt</math>, it follows that by substitution | ||
| + | |||
| + | <math>P_\infty=(\lim_{T\to\infty}\frac{1}{2T})*E_\infty</math> | ||
| + | |||
| + | <math>P_\infty=(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}E_\infty)</math> | ||
| + | |||
| + | <math>P_\infty=\lim_{T\to\infty}\frac{E_\infty}{2T}</math> | ||
| + | |||
| + | This limit will always evaluate to zero as long as <math>E_\infty</math> is finite. | ||
| + | |||
| + | <math>\therefore</math> If <math>E_\infty</math> is ''finite'', then <math>P_\infty</math> is always ''zero'' <math>\square</math> | ||
| + | |||
| + | ---- | ||
| + | |||
| + | --[[User:Asiembid|Asiembid]] 14:04, 17 June 2009 (UTC) - Adam Siembida | ||
Latest revision as of 09:04, 17 June 2009
Proof
If $ E_\infty $ is finite, then $ P_\infty $ is always 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 $, it follows that by substitution
$ 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.
$ \therefore $ If $ E_\infty $ is finite, then $ P_\infty $ is always zero $ \square $
--Asiembid 14:04, 17 June 2009 (UTC) - Adam Siembida
