(CT SIGNAL & ITS FOURIER COEFFICIENTS)
 
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== CT SIGNAL & ITS FOURIER 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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----
 
Let the input signal be
 
Let the input signal be
  
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=    <math>\frac{e^{jt}-e^{-jt}}{2i}</math>+<math>2*\frac{e^{2jt}+{e^{-2jt}}}{2}</math>
 
=    <math>\frac{e^{jt}-e^{-jt}}{2i}</math>+<math>2*\frac{e^{2jt}+{e^{-2jt}}}{2}</math>
  
=    <math>\frac{-j {e^{jt}}{2}</math>
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=    <math>\frac{-j{e^{jt}}}{2}+\frac{j{e^{-jt}}}{2}+{e^{2jt}}+{e^{-2jt}}</math>
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The coefficients are as follows:
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<math>a_1 = \frac{-j}{2}</math>
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<math>a_-1 = \frac{j}{2}</math>
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<math>a_2  = 1 </math>
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<math>a_-2  = 1 </math>
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<math>a_k = 0</math>
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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 09:59, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Let the input signal be

$ x(t)=sint +2cos2t $

= $ \frac{e^{jt}-e^{-jt}}{2i} $+$ 2*\frac{e^{2jt}+{e^{-2jt}}}{2} $

= $ \frac{-j{e^{jt}}}{2}+\frac{j{e^{-jt}}}{2}+{e^{2jt}}+{e^{-2jt}} $

The coefficients are as follows:

$ a_1 = \frac{-j}{2} $

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

$ a_2 = 1 $

$ a_-2 = 1 $

$ a_k = 0 $


Back to Practice Problems on Signals and Systems

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