Given the periodic CT signal
$ \,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\, $
compute its Fourier series coefficients.
Answer
We can rewrite the signal $ \,x(t)\, $ as
$ \,x(t)=\frac{3\pi}{2}(\frac{e^{j(\frac{3\pi}{2}t+\pi)}+e^{-j(\frac{3\pi}{2}t+\pi)}}{2})(\frac{e^{j(\frac{3\pi}{4}t+\frac{\pi}{2})}-e^{-j(\frac{3\pi}{4}t+\frac{\pi}{2})}}{2j})\, $
$ \,x(t)=\frac{3\pi}{8j}(e^{j\frac{3\pi}{2}t}e^{j\pi}+e^{-j\frac{3\pi}{2}t}e^{-j\pi})(e^{j\frac{3\pi}{4}t}e^{j\frac{\pi}{2}}-e^{-j\frac{3\pi}{4}t}e^{-j\frac{\pi}{2}})\, $
$ \,x(t)=\frac{3\pi}{8j}(-e^{j\frac{3\pi}{2}t}-e^{-j\frac{3\pi}{2}t})(je^{j\frac{3\pi}{4}t}+je^{-j\frac{3\pi}{4}t})\, $
$ \,x(t)=\frac{-3\pi}{8}(e^{j\frac{3\pi}{2}t}+e^{-j\frac{3\pi}{2}t})(e^{j\frac{3\pi}{4}t}+e^{-j\frac{3\pi}{4}t})\, $
$ \,x(t)=-\frac{3\pi}{8}(e^{j(\frac{3\pi}{2}+\frac{3\pi}{4})t}+e^{j(\frac{3\pi}{2}-\frac{3\pi}{4})t}+e^{j(-\frac{3\pi}{2}+\frac{3\pi}{4})t}+e^{j(-\frac{3\pi}{2}-\frac{3\pi}{4})t})\, $
$ \,x(t)=-\frac{3\pi}{8}(e^{j\frac{9\pi}{4}t}+e^{j\frac{3\pi}{4}t}+e^{-j\frac{3\pi}{4}t}+e^{-j\frac{9\pi}{4}t})\, $
$ \,x(t)=-\frac{3\pi}{8}e^{j3\frac{3\pi}{4}t}-\frac{3\pi}{8}e^{j\frac{3\pi}{4}t}-\frac{3\pi}{8}e^{-j\frac{3\pi}{4}t}-\frac{3\pi}{8}e^{-j3\frac{3\pi}{4}t}\, $
This form of $ \,x(t)\, $ is in the format of a Fourier series, so we can directly get the fundamental frequency $ \,\omega_o\, $ and the coefficients $ \,a_k\, $. Therefore, the answer is:
$ \,\omega_o=\frac{3\pi}{4}\, $
and the coefficients are
$ \,a_{-3}=-\frac{3\pi}{8}\, $
$ \,a_{-1}=-\frac{3\pi}{8}\, $
$ \,a_1=-\frac{3\pi}{8}\, $
$ \,a_3=-\frac{3\pi}{8}\, $
$ \,a_k=0\, $ for all other $ \,k\in\mathbb{Z}\, $
