Line 7: Line 7:
 
<math>x(t)=6sin(2\pi t) + 4cos(4\pi t)</math>
 
<math>x(t)=6sin(2\pi t) + 4cos(4\pi t)</math>
  
<math>=6*\frac{e^{2j\pi t} - e^{-2j\pi t}}{2j}</math>
+
<math>=6*\frac{e^{2j\pi t} - e^{-2j\pi t}}{2j} + 4 *\frac{e^{4j\pi t} + e^{-4j\pi t}}{2} </math>
 +
 
 +
<math>=3*\frac{e^{2j\pi t} - e^{-2j\pi t}}{j}+ 2 * (e^{4j\pi t}+e^{-4j\pi t})</math>
 +
 
 +
<math>a_1 = \frac{3}{j}</math>
 +
 
 +
<math>a_2 = \frac{-3}{j}</math>

Revision as of 17:01, 25 September 2008

Defines the Fourier series of a periodic ct signal as

$ x(t) = \sum_{k=-\infty}^\infty a_k e^{jkw_0t} $

I set a example as

$ x(t)=6sin(2\pi t) + 4cos(4\pi t) $

$ =6*\frac{e^{2j\pi t} - e^{-2j\pi t}}{2j} + 4 *\frac{e^{4j\pi t} + e^{-4j\pi t}}{2} $

$ =3*\frac{e^{2j\pi t} - e^{-2j\pi t}}{j}+ 2 * (e^{4j\pi t}+e^{-4j\pi t}) $

$ a_1 = \frac{3}{j} $

$ a_2 = \frac{-3}{j} $

Alumni Liaison

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

Dr. Paul Garrett