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<math>      E_\infty = \int_{0}^{3} [1]^2 </math>
 
<math>      E_\infty = \int_{0}^{3} [1]^2 </math>
  
<math> 1 = 1+1+1+1 = 4 </math>
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<math> E_\infty = 1+1+1+1 = 4 </math>
  
  
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<math>P_\infty lim N-> - \infty = \frac{1}{2*N+1}\int_{-N}^{N}[x(t)]^2 dt</math>
 
<math>P_\infty lim N-> - \infty = \frac{1}{2*N+1}\int_{-N}^{N}[x(t)]^2 dt</math>
  
<math> P_\infty 1\2*N+1 * lim N-> -\infty \int_{-N}^{N} [x(N)]^2 </math>
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<math> P_\infty \frac{1}{2*N+1} * lim N-> -\infty \int_{-N}^{N} [x(N)]^2 </math>
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<math> P_\infty = \frac{1}{\infty} * 4 = 0 * 4 = 0 </math>

Latest revision as of 09:41, 7 September 2008

Energy

$ E_\infty = \frac{1}{t_2-t_1}\int_{t_1}^{t_2}[x(t)]^2 dt $

ex: $ E_\infty = \int_{-\infty}^{\infty} [x(t)]^2 dt $

$ E_\infty = \int_{0}^{3} [1]^2 $

$ E_\infty = 1+1+1+1 = 4 $


Power

$ P_\infty lim N-> - \infty = \frac{1}{2*N+1}\int_{-N}^{N}[x(t)]^2 dt $

$ P_\infty = \frac{1}{2*N+1} * lim N-> -\infty \int_{-N}^{N} [x(N)]^2 $


$ P_\infty = \frac{1}{\infty} * 4 = 0 * 4 = 0 $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin