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\ a_k = \frac{1}{2\pi}\int_{0}^{2\pi} \frac{e^{jt} - e^{-jt}}{2}e^{-jkt}\, dt | \ a_k = \frac{1}{2\pi}\int_{0}^{2\pi} \frac{e^{jt} - e^{-jt}}{2}e^{-jkt}\, dt | ||
</math> | </math> | ||
| + | |||
| + | which provides us with the solutions | ||
| + | |||
| + | <math> | ||
| + | \ a_1 = | ||
Revision as of 14:43, 26 September 2008
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 $
which provides us with the solutions
$ \ a_1 = $
