| 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} $
