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<math>
 
<math>
\ a_k = \frac{1}{T}\int_{T}^{\infty} y(t)e^-jk\omega_0t\, 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 $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett