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Emily Thompson  erthomps@purdue.edu
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<math>x(t)=\sqrt(5t)</math>
 
<math>x(t)=\sqrt(5t)</math>
  
 
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----
 
<math>E\infty</math>
 
  
 
<math>E\infty=\int_{-\infty}^\infty |x(t)|^2dt</math>
 
<math>E\infty=\int_{-\infty}^\infty |x(t)|^2dt</math>
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<math>P\infty</math>
 
  
 
<math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |x(t)|^2dt</math>
 
<math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |x(t)|^2dt</math>
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<math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T5tdt</math>
 
<math>P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T5tdt</math>
  
<math>P\infty=\lim_{T \to \infty}\frac{|\frac{5}{2}t^2|_{0}^{T}}{2*T}</math>
+
<math>P\infty=\lim_{T \to \infty}\frac{\frac{5}{2}|t^2|_{0}^{T}}{2*T}</math>
  
 
<math>P\infty=\lim_{T \to \infty}\frac{5}{4}\frac{T^2-0}{T}</math>
 
<math>P\infty=\lim_{T \to \infty}\frac{5}{4}\frac{T^2-0}{T}</math>

Latest revision as of 18:59, 21 June 2009

Emily Thompson erthomps@purdue.edu

$ x(t)=\sqrt(5t) $


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

$ E\infty=\int_{-\infty}^\infty |\sqrt(5t)|^2dt $

$ E\infty=\int_{-\infty}^\infty |5t|dt $

$ E\infty=\int_{0}^\infty 5tdt $

$ E\infty=\frac{5}{2}(t^2|_{0}^{\infty}) $

$ E\infty= \infty-0 $

$ E\infty=\infty $


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

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |\sqrt(5t)|^2dt $

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|5t|dt $

$ P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T5tdt $

$ P\infty=\lim_{T \to \infty}\frac{\frac{5}{2}|t^2|_{0}^{T}}{2*T} $

$ P\infty=\lim_{T \to \infty}\frac{5}{4}\frac{T^2-0}{T} $

$ P\infty=\lim_{T \to \infty}\frac{5}{4}T $

$ P\infty=\infty $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

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