CT Signal and its Fourier Series Coefficients
Let the signal be $ \ x(t) = \cos(2t\pi /3) \sin(2t \pi /9) $
which has a fundamental frequency of 9 Now computing its coefficients:
$ \ x(t)= (\frac{e^{j2t\pi /3} + e^{-j 2t\pi/ 3}}{2}) (\frac{e^{j 2t\pi /9} - e^{-j2t \pi /9}}{2j}) $
$ \ x(t)= \frac{e^{-j4t\pi /9} - e^{-j8t\pi /9} + e^{j8t\pi /9} - e^{j4t\pi /9}}{4j} $
Now rearranging the terms:
$ \ x(t)= \frac{-e^{j4t\pi /9} + e^{-j4t\pi /9} + e^{j8t\pi /9} - e^{j8t\pi /9}}{4j} $
$ \ x(t)= \frac{-e^{j(2)2t\pi /9} + e^{-j(2)2t\pi /9} + e^{j(4)2t\pi /9} - e^{j(4)2t\pi /9}}{4j} $
Now, the Fourier coefficients can clearly be seen:
$ \ a_{2}= \frac{-1}{4j} = \frac{j}{4} $
$ a_{-2}= \frac{1}{4j} = \frac{-j}{4} $
$ a_{4}= \frac{1}{4j} = \frac{-j}{4} $
$ a_{-4}= \frac{-1}{4j} = \frac{j}{4} $
