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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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----
 
Defines the Fourier series of a periodic ct signal as
 
Defines the Fourier series of a periodic ct signal as
  
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I set a example as
 
I set a example as
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<math>x(t)=6sin(2\pi t) + 4cos(4\pi t)</math>
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<math>=6*\frac{e^{2j\pi t} - e^{-2j\pi t}}{2j} + 4 *\frac{e^{4j\pi t} + e^{-4j\pi t}}{2} </math>
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<math>=3*\frac{e^{2j\pi t} - e^{-2j\pi t}}{j}+ 2 * (e^{4j\pi t}+e^{-4j\pi t})</math>
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<math>a_1 = \frac{3}{j}</math>
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<math>a_2 = \frac{-3}{j}</math>
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<math>a_3 = 2</math>
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<math>a_4=2</math>
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else
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<math>a_k = 0</math>
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<math>w_0 = \pi</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 11:03, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Defines the Fourier series of a periodic ct signal as

$ x(t) = \sum_{k=-\infty}^\infty a_k e^{jkw_0t} $

I set a example as

$ x(t)=6sin(2\pi t) + 4cos(4\pi t) $

$ =6*\frac{e^{2j\pi t} - e^{-2j\pi t}}{2j} + 4 *\frac{e^{4j\pi t} + e^{-4j\pi t}}{2} $

$ =3*\frac{e^{2j\pi t} - e^{-2j\pi t}}{j}+ 2 * (e^{4j\pi t}+e^{-4j\pi t}) $

$ a_1 = \frac{3}{j} $

$ a_2 = \frac{-3}{j} $

$ a_3 = 2 $

$ a_4=2 $

else

$ a_k = 0 $

$ w_0 = \pi $


Back to Practice Problems on Signals and Systems

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