(New page: ==Energy== <math>E = int_0^{2pi}{cos(2t)}dt</math>)
 
(Power)
 
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==Energy==
 
==Energy==
 +
Energy of cos(2t) from t= 0 to <math>2\pi</math>
  
<math>E = int_0^{2pi}{cos(2t)}dt</math>
+
<math>E = \int_{t1}^{t2}{|(f(t)|^2}dt</math>
 +
 
 +
<math>E = \int_{0}^{2\pi}{|cost(2t)|^2}dt</math>
 +
 
 +
<math>E = \frac{1}{2} \int_{0}^{2\pi}{|cost(4t)|^2}dt</math>
 +
 
 +
<math>E = \frac{1}{2} (t + \frac{1}{4}(sin(4t))| t= 0 to 2\pi</math>
 +
 
 +
<math>E = \frac{1}{2} (2\pi + 0)</math>
 +
 
 +
<math>E = \pi</math>
 +
 
 +
==Power==
 +
Power of cos(2t)
 +
 
 +
<math>P = \frac{1}{t2-t1}\int_{t1}^{t2}{|f(t)|^2}dt</math>
 +
 
 +
<math>P = \frac{1}{2\pi-0}\int_{0}{2\pi}{|cos(2t)|^2}dt</math>
 +
 
 +
<math>P = \frac{1}{2\pi} * E</math>
 +
 
 +
<math>P = \frac{1}{2\pi} * \pi</math>
 +
 
 +
<math>P = \frac{1}{2}</math>

Latest revision as of 10:53, 5 September 2008

Energy

Energy of cos(2t) from t= 0 to $ 2\pi $

$ E = \int_{t1}^{t2}{|(f(t)|^2}dt $

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

$ E = \frac{1}{2} \int_{0}^{2\pi}{|cost(4t)|^2}dt $

$ E = \frac{1}{2} (t + \frac{1}{4}(sin(4t))| t= 0 to 2\pi $

$ E = \frac{1}{2} (2\pi + 0) $

$ E = \pi $

Power

Power of cos(2t)

$ P = \frac{1}{t2-t1}\int_{t1}^{t2}{|f(t)|^2}dt $

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

$ P = \frac{1}{2\pi} * E $

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

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

Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal