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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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<math>\ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) </math>
 
<math>\ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) </math>
  
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<math>a_{-3}= \frac{-1}{2j} = \frac{-j}{2}</math>
 
<math>a_{-3}= \frac{-1}{2j} = \frac{-j}{2}</math>
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[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Latest revision as of 10:07, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


$ \ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) $


$ x(t)=({ e^{j 4t\pi/6} + e^{-j4t\pi/6} \over 2}) + ({ e^{j3t\pi/6} - e^{-j3t\pi/6} \over 2j }) $


$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $


$ T = 2\pi $


$ x(t)=({ e^{2\pi jt/3} + e^{-2\pi jt/3} \over 2}) + ({ e^{2\pi j3t/12} - e^{-2\pi j3t/12} \over 2j }) $


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

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

$ a_{3}= \frac{1}{2j} = \frac{j}{2} $

$ a_{-3}= \frac{-1}{2j} = \frac{-j}{2} $


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