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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
| + | [[Category:signals and systems]] | ||
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| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
== Define a CT Signal and its Fourier Series Coefficients == | == Define a CT Signal and its Fourier Series Coefficients == | ||
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e^{-j3t} </math> | e^{-j3t} </math> | ||
| − | '''<math>a1 = \frac { | + | '''<math>a1 = \frac {5}{2}</math>''' |
'''<math>a-1 = \frac{5}{2}</math>''' | '''<math>a-1 = \frac{5}{2}</math>''' | ||
== a2= a-2 = j == | == a2= a-2 = j == | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:05, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Define a CT Signal and its Fourier Series Coefficients
Periodic CT Signal: x(t) = 5cos(6t) + 2sin(3t)
= $ 5 * \frac {e^{j6t} + e^{-j6t}}{2} + 2 * \frac {e^{j3t} - e^{-j3t}}{2j} $
= $ \frac{5}{2} * e^{j6t} - \frac{5}{2} * e^{-j6t} + j * e^{j3t} -j* e^{-j3t} $
$ a1 = \frac {5}{2} $
$ a-1 = \frac{5}{2} $
