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− | \ a_k = \frac{1}{T}\int_{T}^{\infty} y(t)\, dx | + | \ a_k = \frac{1}{T}\int_{T}^{\infty} y(t)e^-jk\omega_0t\, dx |
</math> | </math> |
Revision as of 12:30, 26 September 2008
The function y(t) in this example is the periodic continuous-time signal cos(x) such that
$ y(t) = \ cos(t) $
where cos(x) can be expressed by the Maclaurin series expansion
$ \ cos(t) = \sum_{n=0}^\infty \left (-1 \right )^n \frac{t^{2n}}{ \left(2n \right )!} $
and its Fourier series coefficients are described by the equation
$ \ a_k = \frac{1}{T}\int_{T}^{\infty} y(t)e^-jk\omega_0t\, dx $