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<math>    = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\pi t}</math>
 
<math>    = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\pi t}</math>
  
<math>x(t) =  \frac{4\pi}{3} + \frac{1}{j2000}(e^{j1000\pi t}+e^{j-1000\pi t}) - \frac{1}{j1000}(e^{j1000\pi t}-e^{j-1000\pi t})</math>
 
  
 
==Fourier Series Coefficients==
 
==Fourier Series Coefficients==
<math>a_0 = \frac{4\pi}{3}</math>
+
<math>a_3 = -\frac{1}{2}</math>
  
<math>a_1 = \frac{1}{1000}</math>
+
<math>a_-3 = -\frac{1}{2}</math>
 
+
<math>w_0 = 1000\pi\ </math>
+

Revision as of 16:43, 26 September 2008

Periodic CT Signal

$ x(t) = cos(3\pi t+\pi) \! $ with fundamental frequency of $ \pi $


$ x(t) = \frac{e^{j(3\pi t+\pi)}+e^{-j(3\pi t+\pi)}}{2} $

$ = \frac{e^{j3\pi t}e^{\pi}+e^{-j3\pi t}e^{\pi}}{2} $

$ = \frac{-e^{j3\pi t}-e^{-j3\pi t}}{2} $

$ = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\pi t} $


Fourier Series Coefficients

$ a_3 = -\frac{1}{2} $

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

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

Dr. Paul Garrett