(54 intermediate revisions by one other user not shown) | |||
Line 1: | Line 1: | ||
− | The function y(t) in this example is | + | [[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"]] | ||
+ | ---- | ||
+ | The function y(t) in this example is the periodic continuous-time signal cos(t) such that | ||
<math> | <math> | ||
y(t) = | y(t) = | ||
− | \ | + | \ cos(t) |
+ | </math> | ||
+ | |||
+ | where cos(t) can be expressed by the Maclaurin series expansion | ||
+ | |||
+ | <math> | ||
+ | \ cos(t) = \sum_{k=0}^\infty \left (-1 \right )^k \frac{t^{2k}}{ \left(2k \right )!} | ||
+ | </math> | ||
+ | |||
+ | and its Fourier series coefficients are described by the equations below. | ||
+ | |||
+ | <math> | ||
+ | \ a_k = \frac{1}{T}\int_{0}^{T} y(t)e^{-jk\omega_0t}\, dt | ||
+ | </math> | ||
+ | |||
+ | <math> | ||
+ | \omega_0 = \frac{2\pi}{T} = 1 | ||
+ | </math> | ||
+ | |||
+ | With this information it is possible to determine the first fiew coefficients of the Fourier series of the signal, as seen below. | ||
+ | |||
+ | Also noting that | ||
+ | |||
+ | <math> | ||
+ | \cos(t) = \frac{e^{jt} - e^{-jt}}{2} | ||
</math> | </math> | ||
− | + | then the solution for solving the coefficients becomes | |
<math> | <math> | ||
− | \ | + | \ a_k = \frac{1}{2\pi}\int_{0}^{2\pi} \frac{e^{jt} - e^{-jt}}{2}e^{-jkt}\, dt |
− | + | ||
− | \ | + | |
</math> | </math> | ||
+ | ---- | ||
+ | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] |
Latest revision as of 11:07, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
The function y(t) in this example is the periodic continuous-time signal cos(t) such that
$ y(t) = \ cos(t) $
where cos(t) can be expressed by the Maclaurin series expansion
$ \ cos(t) = \sum_{k=0}^\infty \left (-1 \right )^k \frac{t^{2k}}{ \left(2k \right )!} $
and its Fourier series coefficients are described by the equations below.
$ \ a_k = \frac{1}{T}\int_{0}^{T} y(t)e^{-jk\omega_0t}\, dt $
$ \omega_0 = \frac{2\pi}{T} = 1 $
With this information it is possible to determine the first fiew coefficients of the Fourier series of the signal, as seen below.
Also noting that
$ \cos(t) = \frac{e^{jt} - e^{-jt}}{2} $
then the solution for solving the coefficients becomes
$ \ a_k = \frac{1}{2\pi}\int_{0}^{2\pi} \frac{e^{jt} - e^{-jt}}{2}e^{-jkt}\, dt $