(New page: <math>\ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) </math> <math>x(t)=({ e^{j 4t\pi/6} + e^{-j4t\pi/6} \over 2}) + ({ e^{j3t\pi/6} - e^{-j3t\pi/6} \over 2j })</math> <math>a_k=\frac{1}{T}...)
 
Line 9: Line 9:
  
 
<math>T = 2\pi</math>
 
<math>T = 2\pi</math>
 +
  
 
<math>x(t)=({ e^{2\pi jt/3} + e^{-2\pi jt/3} \over 2}) + ({ e^{2\pi j3t/12} - e^{-2\pi j3t/12} \over 2j })</math>
 
<math>x(t)=({ e^{2\pi jt/3} + e^{-2\pi jt/3} \over 2}) + ({ e^{2\pi j3t/12} - e^{-2\pi j3t/12} \over 2j })</math>
 +
 +
 +
<math>\ a_{1}= \frac{1}{2}</math>
 +
 +
<math>a_{-1}= \frac{-1}{2}</math>
 +
 +
<math>a_{3}= \frac{1}{2j} = \frac{j}{2}</math>
 +
 +
<math>a_{-3}= \frac{-1}{2j} = \frac{-j}{2}</math>

Revision as of 16:41, 26 September 2008

$ \ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) $


$ x(t)=({ e^{j 4t\pi/6} + e^{-j4t\pi/6} \over 2}) + ({ e^{j3t\pi/6} - e^{-j3t\pi/6} \over 2j }) $


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


$ T = 2\pi $


$ x(t)=({ e^{2\pi jt/3} + e^{-2\pi jt/3} \over 2}) + ({ e^{2\pi j3t/12} - e^{-2\pi j3t/12} \over 2j }) $


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

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

$ a_{3}= \frac{1}{2j} = \frac{j}{2} $

$ a_{-3}= \frac{-1}{2j} = \frac{-j}{2} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett