| Line 1: | Line 1: | ||
| − | == 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 always ''zero'' == |
---- | ---- | ||
| Line 7: | Line 7: | ||
<math>P_\infty\equiv(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}\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>, | + | 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})*E_\infty</math> | ||
| Line 17: | Line 17: | ||
This limit will always evaluate to zero as long as <math>E_\infty</math> is finite. | 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'' | + | <math>\therefore</math> If <math>E_\infty</math> is ''finite'', then <math>P_\infty</math> is always ''zero'' <math>\square</math> |
| + | |||
| + | -Adam Siembida | ||
Revision as of 10:01, 17 June 2009
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 $
-Adam Siembida
