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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
| + | [[Category:signals and systems]] | ||
| + | |||
| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
==Problem== | ==Problem== | ||
Let the signal <math>x(t) = 5cos(3\pi t) + sin(\pi t)\,</math><br> | Let the signal <math>x(t) = 5cos(3\pi t) + sin(\pi t)\,</math><br> | ||
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:<math>a_{3} = \frac{5}{2}</math> | :<math>a_{3} = \frac{5}{2}</math> | ||
:<math>a_{k} = 0\,</math>, for all other k | :<math>a_{k} = 0\,</math>, for all other k | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:06, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Problem
Let the signal $ x(t) = 5cos(3\pi t) + sin(\pi t)\, $
Find the Fourier Series coefficients
Analysis
First rewrite the signal as a sum of complex exponentials:
- $ x(t) = 5(\frac{e^{3\pi jt} + e^{-3\pi jt}}{2}) + \frac{e^{\pi jt} - e^{-\pi jt}}{2j} $
Simplifying gives:
- $ x(t) = \frac{5}{2} e^{3\pi jt} + \frac{5}{2} e^{-3\pi jt} + \frac{1}{2j} e^{\pi jt} - \frac{1}{2j} e^{-\pi jt} $
The fundamental period is:
- $ \omega_o = \frac{2\pi}{T} = \pi $, with $ T = 2\, $ being the period of the original signal.
From the fundamental period, it is easily seen that the fourier series coefficients are:
- $ a_{-3} = \frac{5}{2} $
- $ a_{-1} = -\frac{1}{2j} $
- $ a_{1} = \frac{1}{2j} $
- $ a_{3} = \frac{5}{2} $
- $ a_{k} = 0\, $, for all other k
