(→Solution) |
(→Solution) |
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==Solution== | ==Solution== | ||
| − | <math>x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{2} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{ | + | <math>x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{2} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{2j}</math> |
<math>\omega_0 = \pi</math> | <math>\omega_0 = \pi</math> | ||
| − | <math>a_1= </math> | + | <math>a_1=1/2 </math> |
| − | <math>a_2= </math> | + | <math>a_2=1/2j </math> |
Revision as of 06:18, 25 September 2008
Fourier Transform
Let $ x(t)=sin(\pi t) + cos(2\pi t) $
Remember that the formula for CT Fourier Series are:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Solution
$ x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{2} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{2j} $
$ \omega_0 = \pi $
$ a_1=1/2 $
$ a_2=1/2j $
