Noise covariance matrix

Question

I've been trying to understand this paper.

The paper seems to be about analyzing noise properties of a superconducting coplanar waveguide microwave resonator.

They use an IQ mixer to perform measurements and it returns in phase and out of phase quadrature amplitudes. $\xi=[I,Q]^T $ and $\delta\xi=[\delta I,\delta Q]^T $, the latter of which is shown here:

$$\langle \delta \xi(\nu)\delta \xi ^\dagger (\nu') \rangle = S(\nu)\delta(\nu - \nu'), \, S(\nu) = \left( \begin{array}{cc} S_{II}(\nu) & S_{IQ}(\nu) \\ S_{IQ}^*(\nu) & S_{QQ}(\nu) \end{array} \right) \tag{1} $$ where $\delta \xi (\nu)$ is the Fourier transform of the time-domain data, the dagger represents Hermitian conjugate, $S_{II}$ and $S_{QQ}$ are the auto-power spectra, and $S_{IQ}$ is the cross-power spectrum. The matrix $S$ is Hermitian and may be diagonalized using a unitary transformation; however, we find that the imaginary part of $S_{IQ}$ is negligible and that an ordinary rotation applied to the real part $\text{Re} S$ gives almost identical results. The eigenvectors are calculated at every frequency: $$O^T(\nu)\text{Re} S(\nu) O(\nu)= \left( \begin{array}{cc} S_{aa}(\nu) & 0 \\ 0 & S_{bb}(\nu) \end{array} \right) \tag{2}$$

First off, I noticed $\langle \delta \xi \delta\xi^\dagger \rangle$, which is just the standard covariant matrix. Aside from that, I wasn't able to get much out of it though.

I would like to understand where they got the first two equations shown below. I think that would be a decent starting point for me.

I was hoping someone could point me in the direction of references that could help me understand the paper.

Answer 1

In general, a wide-sense stationary (wss) process, say $\textbf{x}(t)$, does not have a Fourier transform because it is not square integrable (it has finite power but not finite energy). But let us ignore this mathematical detail, instead proceed formally and write fearlessly:

$$\textbf{X}(u)=\int_{-\infty}^{\infty}\textbf{x}(t)e^{-\iota 2 \pi ut}dt \\ \textbf{x}(t)=\int_{-\infty}^{\infty}\textbf{X}(t)e^{\iota 2 \pi ut}du$$

Now calculate, again formally, the correlation of the spectral amplitudes at different frequencies by fearlessly interchanging the integration and the statistical expectation operator $\mathbf{E}$, etc.:

$$\mathbf{E}[\textbf{X}(u)\textbf{X}^{*}(u')] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mathbf{E}[\textbf{x}(t)\textbf{x}^{*}(t')]e^{-\iota 2 \pi (ut-u't')}dtdt'$$

Here $K_x(t-t')=\mathbf{E}[\textbf{x}(t)\textbf{x}^{*}(t')]$ is the (time) auto-correlation of the process itself, so with this you can write now $$\mathbf{E}[\textbf{X}(u)\textbf{X}^{*}(u')] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}K_x(t-t')e^{-\iota 2 \pi (ut-u't')}dtdt'$$ Next, change the integration variables to $\tau=t-t'$ followed by integration with respect to $\tau$ and $t'$ after which you arrive to

$$\mathbf{E}[\textbf{X}(u)\textbf{X}^{*}(u')] = S_x(u)\delta(u-u')$$ where the spectral density $S_x(u)$ is defined as the Fourier transform of the auto-correlation function $S_x(u)=\int_{-\infty}^{\infty}K_x(\tau)e^{-\iota 2 \pi u\tau}d\tau$.

Although here it was defined and not derived the relationship is usually referred to as the Wiener-Khinchin theorem (see comment of @IamAStudent above).

The same can be "derived" for a matrix process as was used in the paper you quoted.