Line 1: | Line 1: | ||
+ | [[Category:problem solving]] | ||
+ | [[Category:ECE301]] | ||
+ | [[Category:ECE]] | ||
+ | [[Category:Fourier series]] | ||
+ | |||
+ | == Example of Computation of Fourier series of a CT SIGNAL == | ||
+ | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
+ | ---- | ||
+ | |||
CT Periodic Signal : <math>x(t) = \cos(3\pi t) + \sin(4\pi t)\,</math> | CT Periodic Signal : <math>x(t) = \cos(3\pi t) + \sin(4\pi t)\,</math> | ||
Line 26: | Line 35: | ||
<math>a_k = 0 , k \neq 3,-3,4,-4\,</math> | <math>a_k = 0 , k \neq 3,-3,4,-4\,</math> | ||
+ | ---- | ||
+ | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] |
Revision as of 09:23, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
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\, $