(New page: == Signal Energy == <math>E = \int_{t_1}^{t_2}\!x(t)^2dt</math> == Power ==)
 
 
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== Signal Energy ==
 
== Signal Energy ==
  
<math>E = \int_{t_1}^{t_2}\!x(t)^2dt</math>
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<math>E = \int_{t_1}^{t_2}\!|x(t)|^2dt</math>
  
 +
 +
----
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==Signal Energy Example==
 +
<math>E = \int_{0}^{4\pi}\!|sin(t)|^2dt</math>
 +
 +
<math>E = \int_{0}^{4\pi}\!(\frac{1-cos(2t)}{2})dt</math>
 +
 +
<math>E = 2 \pi - \frac{1}{4}\sin(8\pi)</math>
 +
 +
<math>E = 2\pi        </math>
  
  
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== Power ==
 
== Power ==
 +
<math>P={1\over(t_2-t_1)}\int_{t_1}^{t_2}\!|x(t)|^2dt</math>
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 +
 +
----
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==Power Example==
 +
<math>P={1\over(4\pi-0)}\int_{0}^{4\pi}\!|sin(t)|^2dt</math>
 +
 +
<math>P={1\over(4\pi-0)}\int_{0}^{4\pi}\!(\frac{1-cos(2t)}{2})dt</math>
 +
 +
<math>P=\frac{1}{2}-\frac{1}{16\pi}sin(8\pi)</math>
 +
 +
<math>P=\frac{1}{2}</math>

Latest revision as of 17:35, 5 September 2008

Signal Energy

$ E = \int_{t_1}^{t_2}\!|x(t)|^2dt $



Signal Energy Example

$ E = \int_{0}^{4\pi}\!|sin(t)|^2dt $

$ E = \int_{0}^{4\pi}\!(\frac{1-cos(2t)}{2})dt $

$ E = 2 \pi - \frac{1}{4}\sin(8\pi) $

$ E = 2\pi $



Power

$ P={1\over(t_2-t_1)}\int_{t_1}^{t_2}\!|x(t)|^2dt $



Power Example

$ P={1\over(4\pi-0)}\int_{0}^{4\pi}\!|sin(t)|^2dt $

$ P={1\over(4\pi-0)}\int_{0}^{4\pi}\!(\frac{1-cos(2t)}{2})dt $

$ P=\frac{1}{2}-\frac{1}{16\pi}sin(8\pi) $

$ P=\frac{1}{2} $

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Ruth Enoch, PhD Mathematics