| 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> = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}{4} + </math> |
Revision as of 16:32, 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} + \frac{e^{j*2*\pi*t}{4} + $
