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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}}{2}</math> | <math>x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{2} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{2}</math> | ||
| + | |||
| + | <math>\omega_0 = \pi</math> | ||
| + | |||
| + | <math>a_1= </math> | ||
| + | |||
| + | <math>a_2= </math> | ||
Revision as of 06:16, 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}}{2} $
$ \omega_0 = \pi $
$ a_1= $
$ a_2= $
