# Fourier transform of the components of bandpass signals

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.

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.