(Energy)
(Power)
 
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<math>P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math>
+
<math>P = \int_{0}^{2\pi}\!|2cos(t)|^2\ dt</math>
  
  
 +
<math>P = (4){1\over2}\int_{0}^{2\pi}\!1+cos(2t) dt</math>
  
 +
 +
<math>P = (4){1\over2}(2\pi+{1\over2}sin(2*2\pi)</math>
 +
 +
 +
<math>P = 4\pi</math>
 
</font>
 
</font>

Latest revision as of 15:05, 4 September 2008

Energy

$ f(t)=2cos(t) $


$ E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|f(t)|^2 dt $


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


$ E = {1\over(2\pi-0)}{1\over2}(4)\int_{0}^{2\pi}\!(1+cos(2t)) dt $


$ E = {1\over\pi}(2\pi+{1\over2}sin(2*2\pi)) dt $


$ E = {2} $

Power

$ f(t)=2cos(t) $


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


$ P = \int_{0}^{2\pi}\!|2cos(t)|^2\ dt $


$ P = (4){1\over2}\int_{0}^{2\pi}\!1+cos(2t) dt $


$ P = (4){1\over2}(2\pi+{1\over2}sin(2*2\pi) $


$ P = 4\pi $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang