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==Signal==
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[[Category:problem solving]]
Let the signal be <math>x(t) = 5cos(3\pi t) + sin(\pi t)\,</math>
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[[Category:ECE301]]
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[[Category:ECE]]
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[[Category:Fourier series]]
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[[Category:signals and systems]]
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== Example of Computation of Fourier series of a CT SIGNAL ==
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A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]]
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----
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==Problem==
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Let the signal <math>x(t) = 5cos(3\pi t) + sin(\pi t)\,</math><br>
 
Find the Fourier Series coefficients
 
Find the Fourier Series coefficients
 
==Analysis==
 
==Analysis==
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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
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----
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[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Latest revision as of 11:06, 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

Back to Practice Problems on Signals and Systems

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