(New page: CT Periodic Signal : <math>x(t) = \cos(2\pi t) + \sin(3\pi t)\,</math> <math>= \frac{e^{2j\pi t}}{2} + \frac{e^{-2j\pi t}}{2} + \frac{e^{3j\pi t}}{2j} - \frac{e^{-3j\pi t}}{2j} \,</math...)
 
 
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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 Periodic Signal :  
 
CT Periodic Signal :  
  
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<math>a_k = 0 , k \neq 2,-2,3,-3\,</math>
 
<math>a_k = 0 , k \neq 2,-2,3,-3\,</math>
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Reference -* [[HW4.1 Wei Jian Chan_ECE301Fall2008mboutin]]
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[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Latest revision as of 10:09, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


CT Periodic Signal :

$ x(t) = \cos(2\pi t) + \sin(3\pi t)\, $

$ = \frac{e^{2j\pi t}}{2} + \frac{e^{-2j\pi t}}{2} + \frac{e^{3j\pi t}}{2j} - \frac{e^{-3j\pi t}}{2j} \, $

$ \omega_o \, $ = $ \pi \, $

Coefficients of signal:

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

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

$ a_{3} = \frac{1}{2j}\, $

$ a_{-3} = -\frac{1}{2j}\, $

Since

$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $ where

$ a_2 = a_{-2} = \frac{1}{2}\, $

$ a_{3} = -a_{-3}\, $

$ a_k = 0 , k \neq 2,-2,3,-3\, $


Reference -* HW4.1 Wei Jian Chan_ECE301Fall2008mboutin


Back to Practice Problems on Signals and Systems

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