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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"]] | ||
| + | ---- | ||
==Define a Periodic CT Signal and Compute its Fourier Series Coefficients== | ==Define a Periodic CT Signal and Compute its Fourier Series Coefficients== | ||
Let's start this process by defining our signal. For simplicities sake lets use the the signal | Let's start this process by defining our signal. For simplicities sake lets use the the signal | ||
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all other <math> \ a_k = 0 </math> | all other <math> \ a_k = 0 </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"
Define a Periodic CT Signal and Compute its Fourier Series Coefficients
Let's start this process by defining our signal. For simplicities sake lets use the the signal
$ x(t) = 4sin(3t) + 8cos(7t) $
The Fourier Series can be easily found by treating
$ Asin(\omega_0t) = \frac{A*(e^{j\omega_0t} - e^{-j\omega_0t})}{2j} $
and
$ Acos(\omega_0t) = \frac{A*(e^{j\omega_0t} + e^{-j\omega_0t})}{2} $
This alows us to to put x(t) in the form of
$ x(t) = \sum_{k=- \infty }^ \infty a_ke^{jk\omega_0t} $
which gives us
$ x(t) = \frac{4*(e^{j3t} - e^{-j3t})}{2j} + \frac{8*(e^{j7t} + e^{-j7t})}{2} $
Simplifying and distributing
$ x(t) = \frac{2*e^{j3t} }{j}- \frac{2*e^{j3t} }{j} + 4e^{j7t} + 4e^{-j7t} $
$ \ a_{-3} = \frac{2}{j} $
$ \ a_{3}= \frac{-2}{j} $
$ \ a_{-7} = 4 $
$ \ a{_7} = 4 $
all other $ \ a_k = 0 $
