(Energy)
(Energy)
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<math>E = {1\over8\pi}(4\pi+{1\over2}sin(2\pi)) dt</math>
+
<math>E = {1\over8\pi}(4\pi+{1\over2}sin(8\pi)) dt</math>
  
  

Revision as of 14:07, 5 September 2008

Energy

Consider the signal $ x(t)=cos(t) $ over the interval 0 to $ 4\pi $


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


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


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


$ E = {1\over8\pi}(4\pi+{1\over2}sin(8\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

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

Jeff McNeal