Let $ x(t)=cos(4 \pi t) + sin(6 \pi t) $, then its Fourier series coefficients are as follows:
$ x(t)=\frac{e^{4 \pi jt} + e^{-4 \pi jt}}{2} + \frac{e^{6 \pi jt} + e^{-6 \pi jt}}{2j} $ and $ \omega_{0} = \pi $
$ a_{4} = a_{-4} = \frac{1}{2} $
$ a_{6} = -a_{-6} = \frac{1}{2j} $
All other $ a_{k} $ values are 0
