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.

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    $\begingroup$ The minus sign on your $r_Q$ is in the author’s definition of the whole signal. $\endgroup$
    – Peter K.
    Commented Feb 7, 2022 at 11:50
  • $\begingroup$ @Peter K. Thanks. Just one more question why the author says that $s(t)=cos(2\pi f_c t+\phi_0)$ is narrow band for any any delay spread? $\endgroup$ Commented Feb 8, 2022 at 7:02
  • $\begingroup$ @Peter K. Also, I am unable to get your comment that "minus sign on your $r_Q$ is in the author’s definition of the whole signal" as the final result for me turns out to be "$r_I(t)cos(2\pi f_ct)+r_Q(t)sin(2\pi f_ct)$" $\endgroup$ Commented Feb 8, 2022 at 10:02
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    $\begingroup$ @UserHuffmann for the first question, yes, the $B$ of $s(t)$ is infinitesimal and, therefore, $B < T_m^{-1}$ for any $T_m$. $\endgroup$
    – AlexTP
    Commented Feb 8, 2022 at 12:57
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    $\begingroup$ @UserHuffmann for the second question, just define $\theta_n(t) = -\phi_n(t)$ you will get the final form of the author. As $\phi_n(t)$ and $\theta_n(t)$ have the same stochastic properties, substituing $\theta_n(t)$ for $\phi_n(t)$ does not change the model. IMO, the author wanted to come up with their definition $r_Q(t)$ as this is a common choice for the so-called quadrature component. $\endgroup$
    – AlexTP
    Commented Feb 8, 2022 at 13:24


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