(New page: ===Useful Info=== <math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. :Let :<math> x(t) = 2sin(2\pi t) + cos(\pi t)....) |
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:<math> a_1 = \frac{1}{j} </math> | :<math> a_1 = \frac{1}{j} </math> | ||
:<math> a_2 = \frac{1}{2} </math> | :<math> a_2 = \frac{1}{2} </math> | ||
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| + | :<math> \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 | ||
| + | : | ||
Revision as of 07:07, 25 September 2008
Useful Info
$ 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 $.
- Let
- $ x(t) = 2sin(2\pi t) + cos(\pi t). $
Solution
- $ x(t) = 2\frac{e^{2 \pi jt}+e^{-2 \pi jt}}{2j} + \frac{e^{\pi jt}+e^{-\pi jt}}{2} $
- $ x(t) = \frac{1}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{1}{2}(e^{\pi jt}+e^{-\pi jt}) $
- $ a_1 = \frac{1}{j} $
- $ a_2 = \frac{1}{2} $
- $ \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 : $
