(New page: <font size="3">Let <math>x(t)=cos(t)</math>, then the steps to calculate its Fourier series coefficients are as follows:</font>)
 
 
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<font size="3">Let <math>x(t)=cos(t)</math>, then the steps to calculate its Fourier series coefficients are as follows:</font>
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[[Category:problem solving]]
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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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<font size="3">Let <math>x(t)=cos(4 \pi t) + sin(6 \pi t)</math>, then its Fourier series coefficients are as follows:
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<math>x(t)=\frac{e^{4 \pi jt} + e^{-4 \pi jt}}{2} + \frac{e^{6 \pi jt} + e^{-6 \pi jt}}{2j}</math> and <math>\omega_{0} = \pi</math>
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<math>a_{4} = a_{-4} = \frac{1}{2}</math>
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<math>a_{6} = -a_{-6} = \frac{1}{2j}</math>
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All other <math>a_{k}</math> values are 0
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</font>
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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 09:57, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Let $ x(t)=cos(4 \pi t) + sin(6 \pi t) $, then its Fourier series coefficients are as follows:

$ x(t)=\frac{e^{4 \pi jt} + e^{-4 \pi jt}}{2} + \frac{e^{6 \pi jt} + e^{-6 \pi jt}}{2j} $ and $ \omega_{0} = \pi $

$ a_{4} = a_{-4} = \frac{1}{2} $

$ a_{6} = -a_{-6} = \frac{1}{2j} $

All other $ a_{k} $ values are 0


Back to Practice Problems on Signals and Systems

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