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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"]] | ||
+ | ---- | ||
== CT Fourier Series == | == CT Fourier Series == | ||
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<math>a_{-16} = \frac{\sqrt{2}}{4}(1 + j)</math> | <math>a_{-16} = \frac{\sqrt{2}}{4}(1 + j)</math> | ||
+ | |||
+ | ---- | ||
+ | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] |
Latest revision as of 10:03, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
CT Fourier Series
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_o t} $
Example
$ x(t) = 1 + sin(8\pi t) + 2cos(8\pi t) + cos(16\pi t + \frac{\pi}{4}) $
The fundamental frequency is $ 8\pi $.
$ x(t) = 1 + (\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + (e^{j8 \pi t} + e^{-j8 \pi t}) + (\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2}) $
$ x(t) = 1 + (1 + \frac{1}{2j})e^{j8\pi t} + (1 - \frac{1}{2j})e^{-j8\pi t} + (\frac{1}{2} e^{j(\frac{\pi}{4})}) e^{j16\pi t} + (\frac{1}{2} e^{-j(\frac{\pi}{4})}) e^{-j16\pi t} $
The Fourier series coefficients are:
$ a_0 = 1 $
$ a_8 = 1 - \frac{1}{2}j $
$ a_{-8} = 1 + \frac{1}{2}j $
$ a_{16} = \frac{\sqrt{2}}{4}(1 + j) $
$ a_{-16} = \frac{\sqrt{2}}{4}(1 + j) $