Fourier transform of the components of bandpass signals

Question

I'm referring to a book called Communication Systems by Simon Haykin, 3rd Ed.

It says that a band-pass signal \$s(t)\$ with mid-band frequency \$f_c\$ and bandwidth \$2W\$ can be represented as follows:
$$ s(t) = s_I(t)\cos(2\pi f_c t) - s_Q\sin(2\pi f_c t) $$ where \$s_I(t)\$ is the in-phase component of \$s(t)\$, and \$s_Q(t)\$ is the quadrature-phase component of \$s(t)\$. Following this, it says that the Fourier transform of \$s_I(t)\$ is related to that of \$s(t)\$ by $$ S_I(f) = \left\{\begin{matrix} S(f-f_c) + S(f+f_c), & -W\leq f\leq W \\ 0, & \text{elsewhere} \end{matrix}\right. $$ Similarly for \$s_Q(t)\$, $$ S_Q(f) = \left\{\begin{matrix} j[S(f-f_c) - S(f+f_c)], & -W\leq f\leq W \\ 0, & \text{elsewhere} \end{matrix}\right. $$

I can't understand how the relations between \$S_I(f)\$ and \$S(f)\$, and \$S_Q(f)\$ and \$S(f)\$ are being derived.

Answer 1

If you multiply \$s(t)\$ with \$2\cos(2\pi f_c t)\$ and with \$-2\sin(2\pi f_c t)\$, respectively (i.e. you are actually demodulating the signal), you get:

$$s(t)\cdot2\cos(2\pi f_c t)=2s_I(t)\cos^2(2\pi f_c t)-2s_Q(t)\sin(2\pi f_c t)\cos(2\pi f_c t)=\\ =s_I(t)+[s_I(t)\cos(2\cdot2\pi f_c t)-s_Q(t)\sin(2\cdot2\pi f_c t)]$$

and

$$-s(t)\cdot2\sin(2\pi f_c t)=-2s_I(t)\cos(2\pi f_c t)\sin(2\pi f_c t)+2s_Q(t)\sin^2(2\pi f_c t)=\\ =s_Q(t)-[s_I(t)\sin(2\cdot2\pi f_c t)+s_Q(t)\cos(2\cdot2\pi f_c t)]$$

Note that the terms in brackets are centered at twice the carrier frequency \$f_c\$. So in the band \$-W<f<W\$ (i.e. by low-pass filtering) we get

$$s(t)\cdot2\cos(2\pi f_c t)=s_I(t)\\ -s(t)\cdot2\sin(2\pi f_c t)=s_Q(t)$$

If you take the Fourier transform of these two equations you obtain the Fourier relations stated in your question.