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?
Best Answer
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.