| Line 21: | Line 21: | ||
K = 3 | K = 3 | ||
| + | |||
K= -1 | K= -1 | ||
| + | |||
K= 1 | K= 1 | ||
| + | |||
K= -3 | K= -3 | ||
| − | <math> a^{3} = \frac{1}{4} </math> | + | <math> a^{3} = \frac{1}{4} </math> |
| + | |||
<math> a^{-1} = \frac{1}{4} </math> | <math> a^{-1} = \frac{1}{4} </math> | ||
| + | |||
<math> a^{1} = \frac{1}{4} </math> | <math> a^{1} = \frac{1}{4} </math> | ||
| + | |||
<math> a^{-3} = \frac{1}{4} </math> | <math> a^{-3} = \frac{1}{4} </math> | ||
Revision as of 17:40, 23 September 2008
Define a Periodic CT signal and compute its Fourier series coefficients
Consider the following CT signal:
x(t) such that
$ ak = \frac{1}{T} \int_{0}^{T} x(t) * e^{-j*k} * \frac{2*\pi}{T} *dt $
$ 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} + \frac{e^{-j*6*\pi*t}}{4} $
K = 3
K= -1
K= 1
K= -3
$ a^{3} = \frac{1}{4} $
$ a^{-1} = \frac{1}{4} $
$ a^{1} = \frac{1}{4} $
$ a^{-3} = \frac{1}{4} $
