(Lu Zhang-show the property)
 
(if E_{\infty} is finite, then P_\infty equals to zero- Ali Alyoussef: new section)
 
(One intermediate revision by one other user not shown)
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Proof:
 
Proof:
<math>E_\infty</math> = <math>\int^{+\infty}_{-\infty}</math><math>|x(t)|^2</math> <math>dt</math>
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<math>E_\infty = \int^{+\infty}_{-\infty}|x(t)|^2 dt</math>
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<math>P_\infty = \displaystyle\lim_{T\to\infty} \dfrac{1}{2T} \int^{+T}_{-T}{|x(t)|^2}{dt}</math>
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 +
We see from the equations above that,
 +
 
 +
<math>P_\infty = \displaystyle\lim_{T\to\infty} \dfrac{E_\infty}{2T}</math>
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For <math>E_{\infty} < {\infty}</math>, we got that,
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<math>P_\infty = \displaystyle\lim_{T\to\infty} \dfrac{E_\infty}{2T} = 0 </math>
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== if E_{\infty} is finite, then P_\infty equals to zero- Ali Alyoussef ==
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from the formula, it can be seen that
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P = the limit of  (E/2T) when T goes to infinity.
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 +
and if E is a fixed value  < infinity
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 +
 
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=> P = E/infinity which will guarantees that we will have a result of zero for P.

Latest revision as of 17:28, 21 June 2009

Property: if $ E_{\infty} $ is finite, then $ P_\infty $ equals to zero.


Proof:

$ E_\infty = \int^{+\infty}_{-\infty}|x(t)|^2 dt $

$ P_\infty = \displaystyle\lim_{T\to\infty} \dfrac{1}{2T} \int^{+T}_{-T}{|x(t)|^2}{dt} $

We see from the equations above that,

$ P_\infty = \displaystyle\lim_{T\to\infty} \dfrac{E_\infty}{2T} $

For $ E_{\infty} < {\infty} $, we got that,

$ P_\infty = \displaystyle\lim_{T\to\infty} \dfrac{E_\infty}{2T} = 0 $

if E_{\infty} is finite, then P_\infty equals to zero- Ali Alyoussef

from the formula, it can be seen that

P = the limit of  (E/2T) when T goes to infinity.

and if E is a fixed value < infinity


=> P = E/infinity which will guarantees that we will have a result of zero for P.

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