(New page: ===Useful Info=== <math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. :Let :<math> x(t) = 2sin(2\pi t) + cos(\pi t)....)
 
 
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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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===Useful Info===
 
===Useful Info===
  
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:<math> a_1 = \frac{1}{j} </math>
 
:<math> a_1 = \frac{1}{j} </math>
 
:<math> a_2 = \frac{1}{2} </math>
 
:<math> a_2 = \frac{1}{2} </math>
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:<math> \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 </math>
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:else
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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 09:53, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Useful Info

$ 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 $.

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

Solution

$ x(t) = 2\frac{e^{2 \pi jt}+e^{-2 \pi jt}}{2j} + \frac{e^{\pi jt}+e^{-\pi jt}}{2} $
$ x(t) = \frac{1}{j}(e^{2 \pi jt} + e^{-2 \pi jt}) + \frac{1}{2}(e^{\pi jt}+e^{-\pi jt}) $
$ a_1 = \frac{1}{j} $
$ a_2 = \frac{1}{2} $
$ \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 $
else
$ a_k = 0 $

Back to Practice Problems on Signals and Systems

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