(New page: <math>\ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) </math> <math>x(t)=({ e^{j 4t\pi/6} + e^{-j4t\pi/6} \over 2}) + ({ e^{j3t\pi/6} - e^{-j3t\pi/6} \over 2j })</math> <math>a_k=\frac{1}{T}...) |
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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
| + | [[Category:signals and systems]] | ||
| + | |||
| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
| + | |||
<math>\ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) </math> | <math>\ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) </math> | ||
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<math>T = 2\pi</math> | <math>T = 2\pi</math> | ||
| + | |||
<math>x(t)=({ e^{2\pi jt/3} + e^{-2\pi jt/3} \over 2}) + ({ e^{2\pi j3t/12} - e^{-2\pi j3t/12} \over 2j })</math> | <math>x(t)=({ e^{2\pi jt/3} + e^{-2\pi jt/3} \over 2}) + ({ e^{2\pi j3t/12} - e^{-2\pi j3t/12} \over 2j })</math> | ||
| + | |||
| + | |||
| + | <math>\ a_{1}= \frac{1}{2}</math> | ||
| + | |||
| + | <math>a_{-1}= \frac{-1}{2}</math> | ||
| + | |||
| + | <math>a_{3}= \frac{1}{2j} = \frac{j}{2}</math> | ||
| + | |||
| + | <math>a_{-3}= \frac{-1}{2j} = \frac{-j}{2}</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:07, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
$ \ x(t) = \cos(4t\pi /6) + \sin(3t \pi /6) $
$ x(t)=({ e^{j 4t\pi/6} + e^{-j4t\pi/6} \over 2}) + ({ e^{j3t\pi/6} - e^{-j3t\pi/6} \over 2j }) $
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $
$ T = 2\pi $
$ x(t)=({ e^{2\pi jt/3} + e^{-2\pi jt/3} \over 2}) + ({ e^{2\pi j3t/12} - e^{-2\pi j3t/12} \over 2j }) $
$ \ a_{1}= \frac{1}{2} $
$ a_{-1}= \frac{-1}{2} $
$ a_{3}= \frac{1}{2j} = \frac{j}{2} $
$ a_{-3}= \frac{-1}{2j} = \frac{-j}{2} $
