(Equations)
 
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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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----
 
== Equations ==
 
== Equations ==
  
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From Phil Cannon
 
From Phil Cannon
  
==Input Signal
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==Input Signal==
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<math>x(t)=cos(3*pi*t)cos(6*pi*t)\!</math>
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<br>
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<br>
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<math>x(t)=[1/2*e^{j*2*pi*t}+1/2*e^{-j*2*pi*t}]*[1/2*e^{j*4*pi*t}+1/2*e^{-j*4*pi*t}]</math>
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<br>
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<math>    =1/4*e^{j6pit}+1/4*e^{-j2pit}+1/4*e^{j2pit}+1/4*e^{-j6pit}</math>
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<br>
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<br>
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The fundamental frequency is 2*pi
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<br>
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a3=1/4
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a-1=1/4
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a1=1/4
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a3=1/4
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all other ak=0
  
<math>x(t) = (7+2*i)*cos(4*pi*t) + 14*sin(6*pi*t)
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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 11:02, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Equations

Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $

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

From Phil Cannon

Input Signal

$ x(t)=cos(3*pi*t)cos(6*pi*t)\! $

$ x(t)=[1/2*e^{j*2*pi*t}+1/2*e^{-j*2*pi*t}]*[1/2*e^{j*4*pi*t}+1/2*e^{-j*4*pi*t}] $
$ =1/4*e^{j6pit}+1/4*e^{-j2pit}+1/4*e^{j2pit}+1/4*e^{-j6pit} $

The fundamental frequency is 2*pi
a3=1/4 a-1=1/4 a1=1/4 a3=1/4 all other ak=0


Back to Practice Problems on Signals and Systems

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