CT Signal and its Fourier Series Coefficients
Let the signal be $ \ x(t) = \cos(3t) \sin(9t) $
Now computing its coefficients:
$ \ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{2}) $
Let the signal be $ \ x(t) = \cos(3t) \sin(9t) $
Now computing its coefficients:
$ \ x(t)= (\frac{e^{j3t} + e^{-j3t}}{2}) (\frac{e^{j9t} - e^{-j9t}}{2}) $