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<math>x(t) =\sqrt x <\math>
+
<math>x(t) =\sqrt x </math>
  
<math>P_\infty</math>
+
<math>E_\infty = \int_{-\infty}^\infty x(t) dt</math>
 +
 
 +
<math>E_\infty = \int_{-\infty}^\infty \sqrt x dt
 +
 
 +
E_\infty = \int_{-\infty}^0 j \sqrt -x  dt + \int_0^\infty \sqrt x  dt</math>
 +
 
 +
Solving for the two parts of <math>E_\inf</math>:
 +
 
 +
<math>\int_{-\infty}^0 j \sqrt -x dt = \dfrac {0 + \infty}{2}</math> and  <math>\int_0^\infty \sqrt t dt = \dfrac{\infty + 0}{2} </math>
 +
 
 +
Therefore:
 +
<math>E_\infty = \infty</math>
 +
 
 +
 
 +
 
 +
 
 +
Solving for <math>P_\infty</math>
 +
 
 +
<math>P_\infty = \lim_{t\to\infty} \int_{-t}^t x(t)d\tau</math>
 +
 
 +
<math>P_\infty = \lim_{t\to\infty} \int_{-t}^t \sqrt t  d\tau</math>

Revision as of 07:48, 22 June 2009

$ x(t) =\sqrt x $

$ E_\infty = \int_{-\infty}^\infty x(t) dt $

$ E_\infty = \int_{-\infty}^\infty \sqrt x dt E_\infty = \int_{-\infty}^0 j \sqrt -x dt + \int_0^\infty \sqrt x dt $

Solving for the two parts of $ E_\inf $:

$ \int_{-\infty}^0 j \sqrt -x dt = \dfrac {0 + \infty}{2} $ and $ \int_0^\infty \sqrt t dt = \dfrac{\infty + 0}{2} $

Therefore: $ E_\infty = \infty $



Solving for $ P_\infty $

$ P_\infty = \lim_{t\to\infty} \int_{-t}^t x(t)d\tau $

$ P_\infty = \lim_{t\to\infty} \int_{-t}^t \sqrt t d\tau $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett