(New page: E<math>\infty</math> = <math>\int_{-<math>\infty</math>}^{<math>\infty</math>}</math>(<math>\sqrt{t}</math>)^2 dt E<math>\infty</math> = <math>\int</math> t dt E<math>\infty</math> = ...)
 
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E<math>\infty</math> = <math>\int_{-<math>\infty</math>}^{<math>\infty</math>}</math>(<math>\sqrt{t}</math>)^2 dt
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<math>x(t)=\sqrt{t}</math>
 +
----
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<math>E\infty= \int_{-\infty}^{\infty}|x(t)|^2dt</math>
  
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E<math>\infty</math> = <math>\int_{-\infty}^{\infty} tdt</math>
  
E<math>\infty</math> = <math>\int</math> t dt
 
  
  
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E<math>\infty</math> = <math>\frac{1}{2} t^2</math> evaluated from -<math>\infty</math> to +<math>\infty</math> = <math>\infty</math>
  
E<math>\infty</math> = (<math>\frac{1}{2}</math>)t^2 evaluated from -<math>\infty</math> to +<math>\infty</math> = <math>\infty</math>
 
  
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P<math>\infty</math> = lim T<math>\to</math><math>\infty</math> <math>\frac{1}{2T}</math> <math>\int_{-T}^{T}\ tdt</math>
  
P<math>\infty</math> = lim T<math>\to</math><math>\infty</math> <math>\frac{1}{2T}</math> <math>\int_{-T}^{T}</math>
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<math>\frac{1}{2T} (.5t^2)|_{-T}^{T} = \frac{T}{4}</math>
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lim T<math>\to</math><math>\infty</math> = <math>\infty</math> = P<math>\infty</math>

Revision as of 07:48, 17 June 2009

$ x(t)=\sqrt{t} $


$ E\infty= \int_{-\infty}^{\infty}|x(t)|^2dt $

E$ \infty $ = $ \int_{-\infty}^{\infty} tdt $


E$ \infty $ = $ \frac{1}{2} t^2 $ evaluated from -$ \infty $ to +$ \infty $ = $ \infty $


P$ \infty $ = lim T$ \to $$ \infty $ $ \frac{1}{2T} $ $ \int_{-T}^{T}\ tdt $

$ \frac{1}{2T} (.5t^2)|_{-T}^{T} = \frac{T}{4} $

lim T$ \to $$ \infty $ = $ \infty $ = P$ \infty $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva