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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:
 
<br>
 
<br>
<math>X(t) = 6\cos(2t) + 8\sin(4t)\,</math><br>
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<math>X(t) = 6\cos(2\pi t) + 8\sin(4\pi t)\,</math><br>
<math>X(t) = 6\frac{e^{j2t}+e^{-j2t}}{2} + 8\frac{e^{j4t}-e^{-j4t}}{2})\,</math><br>
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<math>X(t) = 6\frac{e^{j2\pi t}+e^{-j2\pi t}}{2} + 8\frac{e^{j4\pi t}-e^{-j4\pi t}}{2}\,</math><br>
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<math>X(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\,</math><br>
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With this expression we can conclude:<br>
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<math>a_1 = 3\,</math><br>
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<math>a_{-1} = 3\,</math><br>
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<math>a_2 = 4\,</math><br>
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<math>a_{-2} = -4\,</math><br>
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To calculate <math>a_k\,</math>, we use this equation with <math>\omega_0\,</math> value as 2:
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<br><math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\,</math><br>
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Then you will find the value of <math>a_k \,</math> is zero. -- For any k other than 1, -1, 2, -2
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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) = 6\cos(2\pi t) + 8\sin(4\pi t)\, $
$ X(t) = 6\frac{e^{j2\pi t}+e^{-j2\pi t}}{2} + 8\frac{e^{j4\pi t}-e^{-j4\pi t}}{2}\, $
$ X(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = -4\, $

To calculate $ a_k\, $, we use this equation with $ \omega_0\, $ value as 2:
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\, $
Then you will find the value of $ a_k \, $ is zero. -- For any k other than 1, -1, 2, -2


Back to Practice Problems on Signals and Systems

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