Latest revision as of 10:04, 17 June 2009

Proof

$P_\infty\equiv\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T|x(t)|^2dt$

$P_\infty\equiv(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt)$

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

@@ Line 1: / Line 1: @@
-==<math>Insert formula here</math>==
+=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