(New page: ==Energy of a CT signal== <math>E = \int_{t1}^{t2} |x(t)|^2\ dt \!</math> ==Power of a CT signal== <math>P = \frac{1}{t2-t1} \int_{t2}^{t1} |f(t)|^2\ dt \!</math> ==Energy of a DT sign...)
 
(Removing all content from page)
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
==Energy of a CT signal==
 
  
<math>E = \int_{t1}^{t2} |x(t)|^2\ dt \!</math>
 
 
==Power of a CT signal==
 
 
<math>P = \frac{1}{t2-t1} \int_{t2}^{t1} |f(t)|^2\ dt \!</math>
 
 
==Energy of a DT signal==
 
 
<math>E = \sum_{n=n1}^{n2} |x[n]|^2</math>
 
 
==Power of a DT signal==
 
 
This one may not be right, I took an educated guess
 
<math>P = \frac{1}{n2-n1+1} \sum_{n=n1}^{n2} |x[n]|^2</math>
 

Latest revision as of 07:38, 5 September 2008

Alumni Liaison

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

Buyue Zhang