(A periodic CT signal)
 
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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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== A periodic CT signal ==
 
== A periodic CT signal ==
  
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Input CT signal: <math> x(t) = cos2t+sin2t</math>
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Input CT signal: <math> x(t) = cos4t+sin2t</math>
  
 
<math>\,x(t)=\frac {e^{j4\pi t}+e^{-j4 \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j}</math>
 
<math>\,x(t)=\frac {e^{j4\pi t}+e^{-j4 \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j}</math>
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otherwise <math>\,a_k</math> values are zero.
 
otherwise <math>\,a_k</math> values are zero.
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[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Latest revision as of 11:08, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


A periodic CT signal

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

, where $ a_k $ is
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.


Input CT signal: $ x(t) = cos4t+sin2t $

$ \,x(t)=\frac {e^{j4\pi t}+e^{-j4 \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j} $


$ x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t} $

$ a_4=\frac{1}{2} $

$ a_{-4}=\frac{1}{2} $

$ a_2=\frac{1}{2j} $

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

otherwise $ \,a_k $ values are zero.


Back to Practice Problems on Signals and Systems

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