Line 4: Line 4:
 
      
 
      
 
  =<math>3*\frac{e^jt+e^-jt}{2}+\frac{e^j2t-e^-j2t}{2}\,</math>
 
  =<math>3*\frac{e^jt+e^-jt}{2}+\frac{e^j2t-e^-j2t}{2}\,</math>
  =<math>3\2e^jt}-3\2e^{-jt}+1\2e^{j2t}-1\2e^{-j2t}</math>
+
  =<math>\frac{3}{2}e^jt}-\frac{3}[2}e^{-jt}+\frac{1}{2}e^{j2t}-\frac{1}{2}e^{-j2t}</math>

Revision as of 10:36, 24 September 2008

Fourier series:

Input CT signal: x(t) = 3cost+sin2t
.
=$ 3*\frac{e^jt+e^-jt}{2}+\frac{e^j2t-e^-j2t}{2}\, $
=$ \frac{3}{2}e^jt}-\frac{3}[2}e^{-jt}+\frac{1}{2}e^{j2t}-\frac{1}{2}e^{-j2t} $

Alumni Liaison

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

Buyue Zhang