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\ a_k = \frac{1}{T}\int_{T}^{\infty} y(t)e^{-jk\omega_0t}\, dt | \ a_k = \frac{1}{T}\int_{T}^{\infty} y(t)e^{-jk\omega_0t}\, dt | ||
</math> | </math> | ||
| + | |||
| + | where | ||
| + | <math> | ||
| + | \omega_0 = \frac{2\pi}{T} | ||
Revision as of 12:34, 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_{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}\, dt $
where $ \omega_0 = \frac{2\pi}{T} $
