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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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----
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==Fourier Transform==
 
==Fourier Transform==
  
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==Solution==
 
==Solution==
<math>x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{2} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{2}</math>
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<math>x(t)= \frac{e^{\pi jt}+e^{-\pi jt}}{2j} + \frac{e^{2\pi jt}+e^{-2\pi jt}}{2}</math>
  
 
<math>\omega_0 = \pi</math>
 
<math>\omega_0 = \pi</math>
  
<math>a_1= </math>
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<math>a_1=\frac{1}{2j} </math>
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<math>a_2=\frac{1}{2} </math>
  
<math>a_2= </math>
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else <math>a_k</math> equals 0
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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 10:54, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Fourier Transform

Let $ x(t)=sin(\pi t) + cos(2\pi t) $

Remember that the formula for CT Fourier Series are:

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

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

Solution

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

$ \omega_0 = \pi $

$ a_1=\frac{1}{2j} $

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

else $ a_k $ equals 0


Back to Practice Problems on Signals and Systems

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Correspondence Chess Grandmaster and Purdue Alumni

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