(Power)
(Energy)
Line 2: Line 2:
  
 
== Energy ==
 
== Energy ==
<math>u = (t-\pi)</math>
+
<math>u = (t-2)</math>
  
 
<math>E=\int_0^{2\pi}{|cos(u)|^2du}</math>
 
<math>E=\int_0^{2\pi}{|cos(u)|^2du}</math>
Line 10: Line 10:
  
  
<math>E=\frac{1}{2}((u+\frac{1}{2}sin(2(u)))|_{u=-\pi}^{u=2\pi-\pi}</math>
+
<math>E=\frac{1}{2}((u+\frac{1}{2}sin(2(u)))|_{u=-2}^{u=2\pi-2}</math>
  
  
<math>E=\frac{1}{2}(\pi + 0 -(-\pi - 0))</math>
+
<math>E=\frac{1}{2}(2\pi-2 + .378 -(-2 - .378))</math>
  
  

Revision as of 05:44, 5 September 2008

cos(t-2)

Energy

$ u = (t-2) $

$ E=\int_0^{2\pi}{|cos(u)|^2du} $


$ E=\frac{1}{2}\int_0^{2\pi}(1+cos(2(u)))du $


$ E=\frac{1}{2}((u+\frac{1}{2}sin(2(u)))|_{u=-2}^{u=2\pi-2} $


$ E=\frac{1}{2}(2\pi-2 + .378 -(-2 - .378)) $


$ E=\pi $

Power

$ E=\frac{1}{2\pi-0}\int_0^{2\pi}{|cos(u)|^2du} $


$ =\frac{1}{2\pi-0} *{\frac{1}{2}}\int_0^{2\pi}(1+cos(2u))du $


$ =\frac{1}{4\pi}((u)+\frac{1}{2}sin(2u))|_{u=-\pi}^{u=2\pi-\pi} $


$ =\frac{1}{4\pi}(\pi+0-(0-\pi)) $


$ =\frac{1}{2} $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010