(Power)
(Energy)
Line 1: Line 1:
 
== x(t) = sin t ==
 
== x(t) = sin t ==
 
== Energy ==
 
== Energy ==
<math>E=\int_0^{2\pi}{|sin(t)|^2dt}</math><br>
+
<math>E=\int_0^{2\pi}{|sin(t)|^2dt}</math><br><br>
<math>=\frac{\int_0^{2\pi}(1-cos(2t))dt}{2}</math><br>
+
<math>=\frac{\int_0^{2\pi}(1-cos(2t))dt}{2}</math><br><br>
<math>=\frac{t-\frac{1}{2}sin(2t)}{2}|_{t=0}^{t=2\pi}</math><br>
+
<math>=\frac{t-\frac{1}{2}sin(2t)}{2}|_{t=0}^{t=2\pi}</math><br><br>
<math>=\frac{1}{2}(2\pi-0-0+0)</math><br>
+
<math>=\frac{1}{2}(2\pi-0-0+0)</math><br><br>
 
'''<math>=\pi</math>'''
 
'''<math>=\pi</math>'''
  

Revision as of 19:37, 2 September 2008

x(t) = sin t

Energy

$ E=\int_0^{2\pi}{|sin(t)|^2dt} $

$ =\frac{\int_0^{2\pi}(1-cos(2t))dt}{2} $

$ =\frac{t-\frac{1}{2}sin(2t)}{2}|_{t=0}^{t=2\pi} $

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

$ =\pi $

Power

$ P=\frac{1}{2\pi-0}\int_0^{2\pi}{|sin(t)|^2dt} $
$ P=\frac{1}{2\pi}E $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood