Line 10: Line 10:
 
<math> = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} </math>
 
<math> = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} </math>
  
<math> = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}{4} + </math>
+
<math> = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + </math>

Revision as of 17:32, 23 September 2008

Define a Periodic CT signal and compute its Fourier series coefficients

Consider the following CT signal:

x(t) such that

$ x(t) = cos(2* \pi * t) * cos(4* \pi * t) $

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

$ = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + $

Alumni Liaison

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

Dr. Paul Garrett