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<math>\ x(t) = \sin(4\pi t) + \sin(6\pi t)</math>
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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) = \sin(4\pi t) \sin(6\pi t)</math>
  
  
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<math>\ x(t) = \frac{-1}{4}(e^{5(j2\pi t)} - e^{-1(j2\pi t)} - e^{1(j2\pi t)} + e^{-5(j2\pi t)}</math>
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<math>\ x(t) = \frac{-1}{4}(e^{5(j2\pi t)} - e^{-1(j2\pi t)} - e^{1(j2\pi t)} + e^{-5(j2\pi t)})</math>
  
  
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All other values <math>\ a_{n} = 0</math>
 
All other values <math>\ a_{n} = 0</math>
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[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Latest revision as of 11:01, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


$ \ x(t) = \sin(4\pi t) \sin(6\pi t) $


$ \ x(t) = (\frac{e^{j4\pi t} - e^{-j4\pi t}}{2}) (\frac{e^{j6\pi t} - e^{-j6\pi t}}{2j}) $


$ \ x(t) = \frac{-1}{4}(e^{j10\pi t} - e{-j2\pi t} - e^{j2\pi t} + e^{-j10\pi t}) $


$ \ x(t) = \frac{-1}{4}(e^{5(j2\pi t)} - e^{-1(j2\pi t)} - e^{1(j2\pi t)} + e^{-5(j2\pi t)}) $


$ a_{5} = \frac{-1}{4}, a_{-1} = \frac{1}{4}, a_{1} = \frac{1}{4}, a_{-5} = \frac{-1}{4} $


All other values $ \ a_{n} = 0 $


Back to Practice Problems on Signals and Systems

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