(Power)
(Energy)
Line 11: Line 11:
  
  
<math>E = {1\over(2\pi-0)}{1\over2}\int_{0}^{2\pi}\!(1+cos(2t)) dt</math>
+
<math>E = {1\over(2\pi-0)}{1\over2}(4)\int_{0}^{2\pi}\!(1+cos(2t)) dt</math>
  
  
<math>E = {1\over(4\pi)}(2\pi+{1\over2}sin(2*2\pi)) dt</math>
+
<math>E = {1\over\pi}(2\pi+{1\over2}sin(2*2\pi)) dt</math>
  
  
<math>E = {1\over2}</math>
+
<math>E = {2}</math>
  
 
</font>
 
</font>

Revision as of 14:58, 4 September 2008

Energy

$ f(t)=2cos(t) $


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


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


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


$ E = {1\over\pi}(2\pi+{1\over2}sin(2*2\pi)) dt $


$ E = {2} $

Power

$ f(t)=2cos(t) $


$ P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt $


$ P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt $


Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman