(New page: CT Periodic Signal : <math>x(t) = \cos(3\pi t) + \sin(4\pi t)\,</math> <math>x(t) = \cos(3\pi t) + \sin(4\pi t)\,</math> <math>= \frac{e^{3j\pi t}}{2} + \frac{e^{-3j\pi t}}{2} + \frac{e^...) |
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<math>a_{4} = \frac{1}{2j} = -a_{-4}\,</math> | <math>a_{4} = \frac{1}{2j} = -a_{-4}\,</math> | ||
| − | <math>a_k = 0\,</math> | + | <math>a_k = 0 , k \neq 3,-3,4,-4\,</math> |
Revision as of 18:39, 21 September 2008
CT Periodic Signal : $ x(t) = \cos(3\pi t) + \sin(4\pi t)\, $
$ x(t) = \cos(3\pi t) + \sin(4\pi t)\, $
$ = \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\, $
