(CT Signal)
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[[Category:ECE301]]
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[[Category:ECE]]
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[[Category:Fourier series]]
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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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Thus, the Fourier series coefficients for this are:<br><br>
 
Thus, the Fourier series coefficients for this are:<br><br>
 
<math> a_0 = 1 </math><br><br><math> a_1 = {1 \over 2j}</math><br><br><math> a_{-1} = {-1 \over 2j}</math><br><br><math> a_2 = {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}})</math><br><br><math> a_{-2} = {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) </math><br><br><math>
 
<math> a_0 = 1 </math><br><br><math> a_1 = {1 \over 2j}</math><br><br><math> a_{-1} = {-1 \over 2j}</math><br><br><math> a_2 = {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}})</math><br><br><math> a_{-2} = {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) </math><br><br><math>
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[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Revision as of 10:34, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


CT Signal

$ x(t) = 1 + sin(w_0 t) + 3cos(2w_0 t + {\pi \over 4}) $

This is a signal with period $ T = {2\pi \over w_0} $

$ x(t) = 1 + {1 \over 2j}[e^{j w_0 t} - e^{-j w_0 t}] + {3 \over 2}[e^{(j 2w_0 t + {\pi \over 4})}+e^{-(j 2w_0 t + {\pi \over 4})}] $

$ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\pi \over 4}}] $

$ e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} $
$ e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} $



$ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\pi \over 4}}] $

$ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) [e^{j 2w_0 t}]+ {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) [e^{-j 2w_0 t}] $


Thus, the Fourier series coefficients for this are:

$ a_0 = 1 $

$ a_1 = {1 \over 2j} $

$ a_{-1} = {-1 \over 2j} $

$ a_2 = {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) $

$ a_{-2} = {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) $

$ ---- [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] $

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Mu Qiao