(A periodic CT signal)
(A periodic CT signal)
Line 22: Line 22:
 
<math>a_{-4}=\frac{1}{2}</math>
 
<math>a_{-4}=\frac{1}{2}</math>
  
<math>a_2=\frac{1+j}{2j}</math>
+
<math>a_2=\frac{1}{2j}</math>
  
<math>a_{-2}=\frac{-1-j}{2j}</math>
+
<math>a_{-2}=\frac{-1}{2j}</math>
  
 
All other <math>\,a_k</math> values are zero.
 
All other <math>\,a_k</math> values are zero.

Revision as of 16:23, 26 September 2008

A periodic CT signal

Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $

, where $ a_k $ is
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.


Input CT signal: $ x(t) = cos2t+sin2t $

$ \,x(t)=\frac {e^{j4\pi t}+e^{-j4 \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j} $


$ x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t} $

$ a_4=\frac{1}{2} $

$ a_{-4}=\frac{1}{2} $

$ a_2=\frac{1}{2j} $

$ a_{-2}=\frac{-1}{2j} $

All other $ \,a_k $ values are zero.

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009