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

Latest revision as of 18:38, 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 $

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

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

Alumni Liaison

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

Buyue Zhang