Conversion of Additive Noise to Phase Noise

RF Microelectronics by Razavi contains the following snippet in section 8.7.3 concerning the analysis of phase noise in oscillators:

We write $x(t)=A\cos(\omega_0t)+n(t)$ where $n(t)$ denotes the narrowband additive noise (voltage or current). It can be proved that narrowband noise in the vicinity of $\omega_0$ can be expressed in terms of its quadrature components: $$n(t) = n_I(t)\cos(\omega_0t) - n_Q(t)\sin(\omega_ot)$$ where $n_I(t)$ and $n_Q(t)$ have the same spectrum of $n(t)$ but translated down by $\omega_0$ (Fig. below) and doubled in spectral density.

I don't see how the math adds up though. Taking the Fourier transform of $n(t)$, $$S_n(\omega) = \frac{1}{2}\left[ S_{nI}(\omega-\omega_0) + S_{nI}(\omega+\omega_0)\right] + \frac{j}{2}\left[ S_{nQ}(\omega+\omega_0) - S_{nQ}(\omega-\omega_0)\right]$$ If the quadrature components are the same as mentioned so $S_{nQ}(\omega) = S_{nI}(\omega)$, $$S_n(\omega) = \frac{1-j}{2} S_{nI}(\omega-\omega_0) + \frac{1+j}{2} S_{nI}(\omega+\omega_0)$$ Doesn't this show that the spectral density of $S_{nI}$ and $S_{nQ}$ is $\frac{2}{\sqrt{2}}$ that of $S_n$ rather than double, in order for the magnitude to equal?

• @Andyaka That's how it's written in the book. It makes sense to me that $n(t)$ should be real. – user21760 Jul 30 '18 at 11:11
• Yeah I wasn't thinking! – Andy aka Jul 30 '18 at 11:25
• That is the voltage or current equation,so the power spectral density is 2x – Tony Stewart Sunnyskyguy EE75 Jul 30 '18 at 14:23
• @TonyEErocketscientist I think you’re right in that I wrongly assumed that $S_n$ denotes the voltage/current spectrum. I now see that it is the power spectral density. However, the power spectral density is not twice the voltage/current spectrum; rather, it involves the limit of a squared absolute value. See my answer below. For the case of quadrature signals here, each of the quadrature signals is twice the two-sided spectrum of the RF power spectrum. – user21760 Aug 4 '18 at 21:23

In the following, $S_n$ denotes the voltage/current spectrum, as considered in the original question. Considering the spectrum around $\omega=\omega_0$,
$$S_n(\omega) = \frac{1-j}{2} S_{nI}(\omega-\omega_0) + \frac{1+j}{2} S_{nI}(\omega+\omega_0)$$ $$S_n(\omega_0) = \frac{1-j}{2} S_{nI}(0) + \frac{1+j}{2} S_{nI}(2\omega_0)$$ $$S_n(\omega_0) = \frac{1-j}{2} S_{nI}(0)$$
since $S_{nI}(2\omega_0)\approx0$. The power spectral density is $\lim_{T\rightarrow 0}\frac{|S_T(\omega)|^2}{T}$, where $T$ is the period and $S_T$ is the voltage/current spectrum of a periodic waveform truncated to one period. For energy waveforms (i.e. for the energy spectral density) the limit and division by the period are not required. For ease of notation I use the latter: $$|S_n(\omega_0)|^2 = \left|\frac{1-j}{2} S_{nI}(0)\right|^2$$ $$|S_n(\omega_0)|^2 = \frac{1}2{}\left|S_{nI}(0)\right|^2$$
This derivation assumes that $S_n$ is the voltage or current spectrum. I believe that the figure corresponds to having $S_n$ denote the power spectrum instead. Thus, it agrees with the figure and the power spectral density of the quadrature components is double that of the RF signal.