I am reading Wireless Communications by Andrea Goldsmith. There the author mentions that for narrow band channel the received signal can be given as: $$r(t)=\text{Re} \left\{u(t)\text{exp}(j2\pi f_ct)\left(\sum_{n}\alpha_n(t)\text{exp}(-j\phi_n(t))\right) \right\}$$ where $u(t)$ is the low pass equivalent signal for $s(t)$ $\alpha_n(t)$ is the attenuation due to $n\text{th}$ resolvable component, and $\phi_n(t)$ is the phase term for the $n\text{th}$ resolvable component.
She now assumes $s(t)$ to be just an unmodulated carrier with random phase offset $\phi_0$ and writes: $$s(t)=\text{Re}\{\text{exp}(j2\pi f_c t+\phi_0)\}=\cos(2\pi f_c t+\phi_0)$$ and says that this is narrow band for any any delay spread $T_m$ for narrowband $B_u<{T_m}^{-1}$ where $B_u$ is the bandwidth of the signal. Why is it so? Is it because $\cos(2\pi f_c t+\phi_0)$ has just one frequency?
Moving further, the author writes the received signal as:
\begin{align} r(t)&=\text{Re}\{\text{exp}(j2\pi f_ct)(\sum_{n}\alpha_n(t)\text{exp}(-j\phi_n(t)))\}\\ &=r_I(t)\cos(2\pi f_c t)-r_Q(t)\sin(2\pi f_c t) \end{align}
where she says
$$r_I(t)=\sum_n \alpha_n(t) \cos(\phi_nt)$$ and $$r_Q(t)=\sum_n \alpha_n(t) \sin(\phi_nt)$$
But when I tried to solve I got $$r_Q(t)=-\sum_n \alpha_n(t) \sin(\phi_nt)$$ I am totally confused about how the calculations are happening? Am I ignoring the negative sign? Can somebody please help me out.
