CT Signal and its Fourier Series Coefficients
Let the signal be $ \ x(t) = \cos(3t) \sin(9t) $ which has a fundamental frequency of Now computing its coefficients:
$ \ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{2j}) $
$ \ x(t)= \frac{e^{j12t} + e^{-j6t} + e^{j6t} - e^{-j12t}}{4j} $
Now rearranging the terms:
$ \ x(t)= \frac{e^{j12t} + e^{-j12t} + e^{j6t} - e^{-j6t}}{4j} $
