Line 11: Line 11:
 
\left ( \frac{1}{jk\omega_0} \right )
 
\left ( \frac{1}{jk\omega_0} \right )
 
=
 
=
\left ( \frac{1}{jk \left (2pi/T \right)} \right )
+
\left ( \frac{1}{jk \left (2\pi/T \right)} \right )
 
</math>
 
</math>

Revision as of 11:59, 26 September 2008

The function y(t) in this example is the signal equal to the periodic continuous-time integral of cos(x) such that

$ y(t) = \int_{-\infty}^{t} cos(x)\, dx $

where its Fourier series coefficients are described by the equation

$ \left ( \frac{1}{jk\omega_0} \right ) = \left ( \frac{1}{jk \left (2\pi/T \right)} \right ) $

Alumni Liaison

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

Dr. Paul Garrett