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CT Periodic Signal : <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math> | CT Periodic Signal : <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math> | ||
| − | <math>x(t) = 1+\frac {1}{2j} (e^{j\omega_0 t}-e^{-j\omega_0 t})+\frac{1}{2}(e^{j(2\omega_0 t+ | + | <math>x(t) = 1+\frac {1}{2j} (e^{j\omega_0 t}-e^{-j\omega_0 t})+\frac{1}{2}(e^{j(2\omega_0 t+\frac {\pi}{4})}+e^{-j(2\omega_0 t+\frac {\pi}{4})})</math> |
<math>= \frac{e^{3j\pi t}}{2} + \frac{e^{-3j\pi t}}{2} + \frac{e^{4j\pi t}}{2j} - \frac{e^{-4j\pi t}}{2j} \,</math> | <math>= \frac{e^{3j\pi t}}{2} + \frac{e^{-3j\pi t}}{2} + \frac{e^{4j\pi t}}{2j} - \frac{e^{-4j\pi t}}{2j} \,</math> | ||
Revision as of 03:55, 25 September 2008
CT Periodic Signal : $ x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}) $
$ x(t) = 1+\frac {1}{2j} (e^{j\omega_0 t}-e^{-j\omega_0 t})+\frac{1}{2}(e^{j(2\omega_0 t+\frac {\pi}{4})}+e^{-j(2\omega_0 t+\frac {\pi}{4})}) $
$ = \frac{e^{3j\pi t}}{2} + \frac{e^{-3j\pi t}}{2} + \frac{e^{4j\pi t}}{2j} - \frac{e^{-4j\pi t}}{2j} \, $
I take $ \omega_o \, $ as $ \pi \, $ since both functions have a period based on it.
The following is the coefficient of the signal:
$ a_3 = \frac{1}{2}\, $
$ a_{-3} = \frac{1}{2}\, $
$ a_{4} = \frac{1}{2j}\, $
$ a_{-4} = -\frac{1}{2j}\, $
We can write the function in the following illiterations:
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $ where
$ a_3 = a_{-3} = \frac{1}{2}\, $
$ a_{4} = \frac{1}{2j} = -a_{-4}\, $
$ a_k = 0 , k \neq 3,-3,4,-4\, $
