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| − | The function y(t) in this example is the periodic continuous-time signal cos( | + | [[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( | + | \ cos(t) |
</math> | </math> | ||
| − | where cos( | + | 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> | <math> | ||
| − | \ cos( | + | \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 $
