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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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== Define a Periodic CT signal and compute its Fourier series coefficients ==
 
== Define a Periodic CT signal and compute its Fourier series coefficients ==
  
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K = 3
 
K = 3
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K= -1
 
K= -1
 +
 
K= 1
 
K= 1
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K= -3
 
K= -3
  
  
<math> a^{3} = \frac{1}{4} </math>   
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<math> a^{3} = \frac{1}{4} </math>  
 +
   
 
<math> a^{-1} = \frac{1}{4} </math>
 
<math> a^{-1} = \frac{1}{4} </math>
 +
 
<math> a^{1} = \frac{1}{4} </math>
 
<math> a^{1} = \frac{1}{4} </math>
 +
 
<math> a^{-3} = \frac{1}{4} </math>
 
<math> a^{-3} = \frac{1}{4} </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 10:58, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Define a Periodic CT signal and compute its Fourier series coefficients

Consider the following CT signal:


x(t) such that

$ ak = \frac{1}{T} \int_{0}^{T} x(t) * e^{-j*k} * \frac{2*\pi}{T} *dt $


$ x(t) = cos(2* \pi * t) * cos(4* \pi * t) $


$ = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} $


$ = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}}{4} + \frac{e^{-j*6*\pi*t}}{4} $


K = 3

K= -1

K= 1

K= -3


$ a^{3} = \frac{1}{4} $

$ a^{-1} = \frac{1}{4} $

$ a^{1} = \frac{1}{4} $

$ a^{-3} = \frac{1}{4} $


Back to Practice Problems on Signals and Systems

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