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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) = \sin(4\pi t) \sin(6\pi t)</math> | <math>\ x(t) = \sin(4\pi t) \sin(6\pi t)</math> | ||
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All other values <math>\ a_{n} = 0</math> | All other values <math>\ a_{n} = 0</math> | ||
+ | ---- | ||
+ | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] |
Latest revision as of 10:01, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
$ \ x(t) = \sin(4\pi t) \sin(6\pi t) $
$ \ x(t) = (\frac{e^{j4\pi t} - e^{-j4\pi t}}{2}) (\frac{e^{j6\pi t} - e^{-j6\pi t}}{2j}) $
$ \ x(t) = \frac{-1}{4}(e^{j10\pi t} - e{-j2\pi t} - e^{j2\pi t} + e^{-j10\pi t}) $
$ \ x(t) = \frac{-1}{4}(e^{5(j2\pi t)} - e^{-1(j2\pi t)} - e^{1(j2\pi t)} + e^{-5(j2\pi t)}) $
$ a_{5} = \frac{-1}{4}, a_{-1} = \frac{1}{4}, a_{1} = \frac{1}{4}, a_{-5} = \frac{-1}{4} $
All other values $ \ a_{n} = 0 $