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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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----
 
CT signal:
 
CT signal:
  
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<math>x(t) = 4 * \frac{e^{j5\pi t} - e^{-j5\pi t}}{2} - (2+j)*\frac{e^{j3\pi t} + e^{-j3\pi t}}{2}\,</math>
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<math>x(t) = 4 * \frac{e^{j5\pi t} - e^{-j5\pi t}}{2j} - (2+j)*\frac{e^{j3\pi t} + e^{-j3\pi t}}{2}\,</math>
  
  
  
<math>x(t) = 2e^{j5\pi t} - 2e^{-j5\pi t} - \frac{2+j}{2}e^{j3\pi t} - \frac{2+j}{2}e^{j3\pi t}\,</math>
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<math>x(t) = \frac{2}{j}e^{j5\pi t} - \frac{2}{j}e^{-j5\pi t} - \frac{2+j}{2}e^{j3\pi t} - \frac{2+j}{2}e^{j3\pi t}\,</math>
  
  
  
<math>x(t) = 2e^{5*j\pi t} - 2e^{-5*j\pi t} - \frac{2+j}{2}e^{3*j\pi t} - \frac{2+j}{2}e^{-3*j\pi t}\,</math>
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<math>x(t) = \frac{2}{j}e^{5*j\pi t} - \frac{2}{j}e^{-5*j\pi t} - \frac{2+j}{2}e^{3*j\pi t} - \frac{2+j}{2}e^{-3*j\pi t}\,</math>
  
  
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<math>a_3 = \frac{-2-j}{2}\,</math>
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<math>a_3 = \frac{-2-j}{2}\,</math>  
  
<math>a_(-3) = \frac{-2-j}{2}\,</math>
 
  
<math>a_5 = 2\,</math>
 
  
<math>a_(-5) = -2\,</math>
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<math>a_{-3} = \frac{-2-j}{2}\,</math>
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<math>a_5 = \frac{2}{j}\,</math>
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<math>a_{-5} = \frac{-2}{j}\,</math>
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For every other k:
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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 10:58, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


CT signal:

$ x(t) = 4\sin(5 \pi t) - (2 + j)\cos(3 \pi t)\, $


$ x(t) = 4 * \frac{e^{j5\pi t} - e^{-j5\pi t}}{2j} - (2+j)*\frac{e^{j3\pi t} + e^{-j3\pi t}}{2}\, $


$ x(t) = \frac{2}{j}e^{j5\pi t} - \frac{2}{j}e^{-j5\pi t} - \frac{2+j}{2}e^{j3\pi t} - \frac{2+j}{2}e^{j3\pi t}\, $


$ x(t) = \frac{2}{j}e^{5*j\pi t} - \frac{2}{j}e^{-5*j\pi t} - \frac{2+j}{2}e^{3*j\pi t} - \frac{2+j}{2}e^{-3*j\pi t}\, $


$ \omega_0\, $ ends up being $ \pi\, $ for this signal

So because of that, in my last step, i was able to determine what value of K's i had. (k = 3,-3,5,-5)

So then you just take the coefficients of those terms to get the $ a_k\, $


Therefore:


$ a_3 = \frac{-2-j}{2}\, $


$ a_{-3} = \frac{-2-j}{2}\, $


$ a_5 = \frac{2}{j}\, $


$ a_{-5} = \frac{-2}{j}\, $


For every other k:

$ a_k\, = 0 $


Back to Practice Problems on Signals and Systems

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