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. Narrowband noise and spectrum of its quadrature components

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?

  • \$\begingroup\$ @Andyaka That's how it's written in the book. It makes sense to me that \$n(t)\$ should be real. \$\endgroup\$ – user21760 Jul 30 '18 at 11:11
  • \$\begingroup\$ Yeah I wasn't thinking! \$\endgroup\$ – Andy aka Jul 30 '18 at 11:25
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    \$\begingroup\$ That is the voltage or current equation,so the power spectral density is 2x \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 30 '18 at 14:23
  • \$\begingroup\$ @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. \$\endgroup\$ – 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.

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