(→Solution) |
(→Solution) |
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| Line 10: | Line 10: | ||
==Solution== | ==Solution== | ||
| − | <math>x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{ | + | <math>x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{2j} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{2}</math> |
<math>\omega_0 = \pi</math> | <math>\omega_0 = \pi</math> | ||
| − | <math>a_1=\frac{1}{ | + | <math>a_1=\frac{1}{2j} </math> |
| − | <math>a_2=\frac{1}{ | + | <math>a_2=\frac{1}{2} </math> |
else <math>a_k</math> equals 0 | else <math>a_k</math> equals 0 | ||
Revision as of 05:40, 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}}{2j} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{2} $
$ \omega_0 = \pi $
$ a_1=\frac{1}{2j} $
$ a_2=\frac{1}{2} $
else $ a_k $ equals 0
