m (Coefficients)
 
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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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== CT signal ==
 
== CT signal ==
<math>x(t) = 2 + cos({\frac{2\pi t}{3}})+ 4sin({\frac{5\pi t}{3}})\,</math>
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<math>x(t) = cos({\frac{2\pi t}{3}})+ 4sin({\frac{5\pi t}{3}})\,</math>
  
 
== Coefficients ==
 
== Coefficients ==
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<math>4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math><br>
 
<math>4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math><br>
  
<math>x(t) = 2 + \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math>
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<math>x(t) = \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math>
  
 
<br>
 
<br>
<math>x(t) = 2 + \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{2j5\pi t}{6}} + 2je^{\frac{-2j5\pi t}{6}}</math>
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<math>x(t) = \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{5j2\pi t}{6}} + 2je^{\frac{-5j2\pi t}{6}}</math>
 
<br>
 
<br>
Then we can know the fundamental frequency is <math>\frac{\pi}{3}</math>. <br>
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Then we can know the fundamental frequency is <math>\frac{\pi}{3}</math>. <br><br>
Also, we can get coefficients <math>a_0</math>,<math>a_2</math>,<math>a_2</math>,<math>a_5</math>,<math>a_-5</math>,.<br>
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Also, we can get coefficients <math>a_2</math>,<math>a_{-2}</math>,<math>a_5</math>,
<math>a_0 = 2, a_2 = a_-2 = \frac{1}{2}, a_5 = -2j, a_-5 = 2j</math>
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<math>a_{-5}</math>.<br><br>
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<math>a_2 = a_{-2} = \frac{1}{2}, a_5 = -2j, a_{-5} = 2j, a_k = 0,</math>where k  is not 2,-2,5,-5
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[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Latest revision as of 10:04, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


CT signal

$ x(t) = cos({\frac{2\pi t}{3}})+ 4sin({\frac{5\pi t}{3}})\, $

Coefficients

$ cos({\frac{2\pi t}{3}}) = \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} $

$ 4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}} $

$ x(t) = \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}} $


$ x(t) = \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{5j2\pi t}{6}} + 2je^{\frac{-5j2\pi t}{6}} $
Then we can know the fundamental frequency is $ \frac{\pi}{3} $.

Also, we can get coefficients $ a_2 $,$ a_{-2} $,$ a_5 $, $ a_{-5} $.

$ a_2 = a_{-2} = \frac{1}{2}, a_5 = -2j, a_{-5} = 2j, a_k = 0, $where k is not 2,-2,5,-5


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