\section{确定信号 DoA 估计的 CRB 推导}
\label{sec:det_doa_crb}

本附录对应第~\ref{sec:det_doa_est} 节的单阵列确定信号 DoA 模型。根据 \gls{fim} 的定义 $\eqref{eq:fisher_element}$，需要先求出对数似然函数 $\eqref{eq:det_ml_log_likelihood}$ 对各类未知量的一阶偏导，再构造 Fisher 信息矩阵并通过 Schur 补消去确定信号快拍等干扰参数。

\subsection{一阶偏导求取}

\subsubsection{对噪声方差的一阶偏导}

\begin{equation}
\frac{\partial \ln \mathcal{L}}{\partial \sigma} = -\frac{MT}{\sigma}+\frac{1}{\sigma^2} \sum_{t=1}^T \mathbf{n}^\Herm _t\mathbf{n}_t.
\end{equation}

\subsubsection{对信源的一阶偏导}

\begin{mdquote}
复向量的偏导数（Wirtinger 微分）：令复数 $z=x+jy$，则其 Wirtinger 微分定义为
%
\begin{equation}
    \label{eq:wirtinger_derivative}
    \begin{cases}
    \dfrac{\partial}{\partial z} = \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} - j\dfrac{\partial}{\partial y} \right)
    \\
    \dfrac{\partial}{\partial z^*} = \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} + j\dfrac{\partial}{\partial y} \right)
    \end{cases}
    \quad\Rightarrow\quad
    \begin{cases}
    \dfrac{\partial}{\partial x} = \dfrac{\partial}{\partial z} + \dfrac{\partial}{\partial z^*}
    \\
    \dfrac{\partial}{\partial y} = j\left(\dfrac{\partial}{\partial z} - \dfrac{\partial}{\partial z^*} \right)
    \end{cases},
\end{equation}
%
且有
\begin{subequations}
  \begin{equation}
  \frac{\partial z^*}{\partial z}
  = \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} - j\dfrac{\partial}{\partial y} \right) (x - jy)
  = \frac{1}{2}\left(\frac{\partial x}{\partial x} - j\frac{\partial (-jy)}{\partial y} \right) = 0,
  \end{equation}
  %
  \begin{equation}
  \frac{\partial z}{\partial z^*}
  = \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} + j\dfrac{\partial}{\partial y} \right) (x + jy)
  = \frac{1}{2}\left(\frac{\partial x}{\partial x} + j\frac{\partial (jy)}{\partial y} \right) = 0,
  \end{equation}
\end{subequations}
%
也即在 Wirtinger 微分定义下，复变量 $z$ 及其复共轭 $z^*$ 可以看做看作两个不相关的变量，对其中一个求导时，将另一项看作常量即可。
\end{mdquote}

首先按 Wirtinger 微分求取信源 $\mathbf{x}_t$ 的偏导
%
\begin{align*}
\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t} &= - \frac{\partial }{\partial \mathbf{x}_t} \left[\frac{1}{\sigma} \left( \mathbf{y}_t-\mathbf{A}\mathbf{x}_t \right)^\Herm \left( \mathbf{y}_t-\mathbf{A}\mathbf{x}_t \right) \right] \nonumber
= - \frac{\partial }{\partial \mathbf{x}_t} \left[\frac{1}{\sigma} \left( \mathbf{y}_t^\Herm  - \mathbf{x}_t^\Herm  \mathbf{A}^\Herm  \right)\left( \mathbf{y}_t-\mathbf{A}\mathbf{x}_t \right) \right] \nonumber
\\
&= \frac{1}{\sigma} \left( \mathbf{y}_t^\Herm  \mathbf{A} - \mathbf{x}_t^\Herm  \mathbf{A}^\Herm \mathbf{A} \right)
= \frac{1}{\sigma} \left( \mathbf{y}_t-\mathbf{A}\mathbf{x}_t \right)^\Herm \mathbf{A} = \frac{1}{\sigma} \mathbf{n}_t^\Herm \mathbf{A}.
\end{align*}
%
对其进行维度修正（取转置）\footnote{%
由于按 Jacobian 定义对列向量 $\mathbf{x}_t$ 求导得到的是 $1\times K$ 的行向量，而后续推导采用列向量形式的梯度表示，因此对结果取转置以统一维度。该转置仅源于导数布局的不同约定，不影响最终结果。}，得到列向量
%
\begin{equation}
\label{eq:first_order_derivative_source}
\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t} = \frac{1}{\sigma} \left(\mathbf{n}_t^\Herm \mathbf{A} \right)^\Tran = \frac{1}{\sigma} \mathbf{A}^\Tran\mathbf{n}_t^*.
\end{equation}
%
然后类似求取信源的共轭转置  $\mathbf{x}_t^\Herm $ 的偏导，得
%
\begin{equation}
\label{eq:first_order_derivative_source_hermit}
\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t^\Herm } = \frac{1}{\sigma} \mathbf{A}^\Herm \mathbf{n}_t.
\end{equation}
%
然后将式 $\eqref{eq:first_order_derivative_source}$ 和 $\eqref{eq:first_order_derivative_source_hermit}$ 代入 Writinger 微分 $\eqref{eq:wirtinger_derivative}$，可得信源实部 $\mathbf{x}_t^{(r)}$ 和虚部 $\mathbf{x}_t^{(i)}$ 的一阶偏导
%
\begin{equation}
\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t^{(r)}}
= \frac{1}{\sigma} \left( \mathbf{A}^\Tran\mathbf{n}_t^* + \mathbf{A}^\Herm \mathbf{n}_t \right)
= \frac{2}{\sigma} \Re\{\mathbf{A}^\Herm \mathbf{n}_t\},
\end{equation}%
%
\begin{equation}
\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t^{(i)}}
= \frac{1}{\sigma} j\left( \mathbf{A}^\Tran\mathbf{n}_t^* - \mathbf{A}^\Herm \mathbf{n}_t \right)
= \frac{2}{\sigma} \Im\{\mathbf{A}^\Herm \mathbf{n}_t\}.
\end{equation}

\subsubsection{对角度的一阶偏导}

\begin{align}
\frac{\partial\ln \mathcal{L}}{\partial \theta_k} &= -\frac{1}{\sigma} \sum_{t=1}^T \left[ - \mathbf{x}_t^\Herm  \frac{\mathrm{d}\mathbf{A}^\Herm }{\mathrm{d}\theta_k }   (\mathbf{y}_t - \mathbf{A} \mathbf{x}_t) - (\mathbf{y}_t - \mathbf{A} \mathbf{x}_t)^\Herm  \frac{\mathrm{d}\mathbf{A}}{\mathrm{d}\theta_k } \mathbf{x}_t \right] \nonumber
\\
&= \frac{1}{\sigma} \sum_{t=1}^T \left[ \mathbf{x}_t^\Herm  \frac{\mathrm{d}\mathbf{A}^\Herm }{\mathrm{d}\theta_k } \mathbf{n}_t + \mathbf{n}_t^\Herm  \frac{\mathrm{d}\mathbf{A}}{\mathrm{d}\theta_k } \mathbf{x}_t \right] \nonumber
\\
&= \frac{2}{\sigma} \sum_{t=1}^T \Re\left[ \mathbf{x}^\Herm _t \frac{\mathrm{d}\mathbf{A}^\Herm }{\mathrm{d}\theta_k } \mathbf{n}_t \right].
\end{align}
%
令 $\mathbf{d}(\theta_k) \triangleq \dfrac{\diff}{\diff\theta_k} \mathbf{a} (\theta_k) \in \mathbb{C}^{M\times 1}$，$\mathbf{d}^\Herm(\theta_k) \triangleq \dfrac{\diff}{\diff\theta_k} \mathbf{a}^\Herm (\theta_k)$，$\mathbf{D} = [\mathbf{d}(\theta_1), \ldots, \mathbf{d}(\theta_K)] \in \mathbb{C}^{M\times K}$，$\mathbf{X}_t \triangleq \diag ( x_1(t), \allowbreak \ldots, \allowbreak x_K (t) ) \in \mathbb{C}^{K\times K}$，则~\footnote{对于多维 DoA 参数，需要将 $\mathbf{D}$ 扩展为关于所有角度分量的 Jacobian。此时仍可通过相同的 Schur 补思想得到等效 Fisher 信息矩阵，但最终矩阵维度和 Hadamard 表达需要按角度分量重新展开。}
%
\begin{equation}
\frac{\partial\ln \mathcal{L}}{\partial \bm{\theta}}
= \begin{bmatrix}
  \dfrac{\partial}{\partial \theta_1}
  \\
  \vdots
  \\
  \dfrac{\partial}{\partial \theta_K}
\end{bmatrix} \ln \mathcal{L}
= \frac{2}{\sigma} \sum_{t=1}^T \Re\left[ \mathbf{X}^\Herm _t \mathbf{D}^\Herm  \mathbf{n}_t \right].
\end{equation}


\subsection{二阶偏导及其均值}

\subsubsection{噪声方差的二阶偏导及均值}

\begin{mdquote}
\vspace{-1em}
\begin{equation*}
  \mathbb{E} \left[ \mathbf{n}^\Herm_t \mathbf{n}_t \right] = M \sigma_n^2 = M \sigma.
\end{equation*}
%
\begin{equation*}
  \mathbb{E} \left[ \mathbf{n}^\Herm_t \mathbf{n}_t \mathbf{n}^\Herm_s \mathbf{n}_s \right] = \begin{cases}
  M^2 \sigma^2, & \text{for}\ t\ne s \\
  M(M+1)\sigma^2, & \text{for}\ t = s
  \end{cases}.
\end{equation*}
\end{mdquote}
%
\begin{align}
\mathbb{E}\left\{\left[\frac{\partial \ln \mathcal{L}}{\partial \sigma}\right]^2\right\}
=& \mathbb{E}\left[\left(-\frac{MT}{\sigma}+\frac{1}{\sigma^2} \sum_{t=1}^T \mathbf{n}^\Herm _t\mathbf{n}_t\right)^2 \right]
\nonumber \\
=& \frac{M^2T^2}{\sigma^2} - \frac{2MT}{\sigma^3} \sum_{t=1}^T \mathbb{E}\left[\mathbf{n}^\Herm _t\mathbf{n}_t\right] + \frac{1}{\sigma^4} \sum_{t=1}^T \sum_{s=1}^T \mathbb{E}\left[ \mathbf{n}^\Herm _t\mathbf{n}_t\mathbf{n}^\Herm _s\mathbf{n}_s \right]
\nonumber \\
=& \frac{M^2T^2}{\sigma^2} - \frac{2M^2T^2}{\sigma^2} + \frac{1}{\sigma^4}[(T^2-T)M^2\sigma^2 + TM(M+1)\sigma^2]
\nonumber \\
=& \frac{MT}{\sigma^2}.
\end{align}

\subsubsection{噪声方差和其他参量的二阶混合偏导及均值}

由于不同时刻（$t\ne s$）的噪声不相关，相同时刻（$t=s$）高斯变量的三阶矩为零，因此有
%
\begin{equation}
\mathbb{E}\left[ \mathbf{n}_t^\Herm  \mathbf{n}_t \mathbf{n}_s^\Tran \right] = \mathbf{0},
\end{equation}
%
因此，噪声方差和其他参量的混合二阶偏导均为零。

\subsubsection{信源实虚部的二阶偏导、二阶混合偏导及均值}

\begin{mdquote}
\begin{enumerate}
    \item 关于噪声的结论：噪声是时间平稳的圆对称复高斯随机过程，也即
    %
    \begin{equation*}
    \mathbb{E}\left[\mathbf{n}_t\mathbf{n}_s^\Herm \right]
    = \sigma \mathbf I_M \delta_{ts},
    \end{equation*}
    %
    \begin{equation*}
    \mathbb{E}[\mathbf{n}_t\mathbf{n}_s^\Tran] = 0,\quad \forall t,s.
    \end{equation*}

    \item 关于实虚部乘法运算的结论（将复数按实虚部展开后可得）
    \begin{align*}
    &\Re\{\mathbf{x}\} \Re\{\mathbf{y}^\Tran\} = \frac{1}{2}[\Re\{\mathbf{x} \mathbf{y}^\Tran\} + \Re\{\mathbf{x} \mathbf{y}^\Herm \}],
    \\
    &\Im\{\mathbf{x}\} \Im\{\mathbf{y}^\Tran\} = -\frac{1}{2}[\Re\{\mathbf{x} \mathbf{y}^\Tran\} - \Re\{\mathbf{x} \mathbf{y}^\Herm \}],
    \\
    &\Re\{\mathbf{x}\} \Im\{\mathbf{y}^\Tran\} = \frac{1}{2}[\Im\{\mathbf{x} \mathbf{y}^\Tran\} - \Im\{\mathbf{x} \mathbf{y}^\Herm \}].
    \end{align*}
\end{enumerate}
\end{mdquote}
%
\begin{subequations}
  \begin{align}
  %信源实部的二阶偏导
  \mathbb{E}\left\{\left[\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t^{(r)}}\right]\left[\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_s^{(r)}}\right]^\Tran\right\}
  &= \frac{1}{\sigma^2} \mathbb{E} \left[ \left( \mathbf{A}^\Tran \mathbf{n}_t^* + \mathbf{A}^\Herm \mathbf{n}_t \right)\left( \mathbf{A}^\Tran\mathbf{n}_s^* + \mathbf{A}^\Herm \mathbf{n}_s \right)^\Tran  \right] \nonumber
  \\
  =& \frac{1}{\sigma^2} \mathbb{E}\left[\left( \mathbf{A}^\Tran \mathbf{n}_t^* \mathbf{n}_s^\Herm \mathbf{A} + \mathbf{A}^\Tran \mathbf{n}_t^* \mathbf{n}_s^\Tran \mathbf{A}^\ast + \mathbf{A}^\Herm \mathbf{n}_t \mathbf{n}_s^\Herm \mathbf{A} + \mathbf{A}^\Herm \mathbf{n}_t \mathbf{n}_s^\Tran \mathbf{A}^\ast \right)\right] \nonumber
  \\
  =& \frac{1}{\sigma^2} \mathbb{E} \left[ \left( \mathbf{A}^\Tran \mathbf{n}_t^* \mathbf{n}_s^\Tran \mathbf{A}^\ast + \mathbf{A}^\Herm  \mathbf{n}_t \mathbf{n}_s^\Herm  \mathbf{A} \right) \right]
  = \frac{2}{\sigma^2}\Re\left\{\mathbf{A}^\Herm \mathbb{E}\left[ \mathbf{n}_t \mathbf{n}_s^\Herm  \right] \mathbf{A}\right\} \nonumber
  \\
  =& \frac{2}{\sigma} \Re\left[\mathbf{A}^\Herm \mathbf{A}\right]\delta(t-s).
  \end{align}
  %信源实虚部的混合二阶偏导
  \begin{equation}
  \mathbb{E}\left\{\left[\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t^{(r)}}\right]\left[\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_s^{(i)}}\right]^\Tran\right\}
  = \frac{4}{\sigma^2} \left(-\frac{1}{2}\right)\Im\left\{\mathbb{E}\left[\mathbf{A}^\Herm  \mathbf{n}_t \mathbf{n}_s^\Herm  \mathbf{A} \right]\right\}
  = -\frac{2}{\sigma} \Im\left[\mathbf{A}^\Herm \mathbf{A}\right]\delta(t-s).
  \end{equation}
  %信源虚部的二阶偏导
  \begin{equation}
  \mathbb{E}\left\{\left[\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t^{(i)}}\right]\left[\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_s^{(i)}}\right]^\Tran\right\}
  = \frac{4}{\sigma^2} \frac{1}{2}\Re\left\{\mathbb{E}\left[\mathbf{A}^\Herm  \mathbf{n}_t \mathbf{n}_s^\Herm  \mathbf{A} \right]\right\}
  = \frac{2}{\sigma} \Re\left[\mathbf{A}^\Herm \mathbf{A}\right]\delta(t-s).
  \end{equation}
\end{subequations}


\subsubsection{信源实虚部与角度的混合二阶偏导及均值}

\begin{subequations}
  \begin{equation}
  \mathbb{E}\left\{\left[\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t^{(r)}}\right]\left[\frac{\partial\ln \mathcal{L}}{\partial \bm{\theta}}\right]^\Tran\right\}
  = \frac{4}{\sigma^2} \sum_{t=1}^T\frac{1}{2}\Re\{\mathbf{A}^\Herm  \mathbb{E}\left[ \mathbf{n}_t\mathbf{n}_t^\Herm  \right] \mathbf{D} \mathbf{X}_t\}
  = \frac{2}{\sigma}\Re\left[ \mathbf{A}^\Herm \mathbf{D} \mathbf{X}_t \right].
  \end{equation}
  %
  \begin{equation}
  \mathbb{E}\left\{\left[\frac{\partial \ln \mathcal{L}}{\partial \mathbf{x}_t^{(i)}}\right]\left[\frac{\partial\ln \mathcal{L}}{\partial \bm{\theta}}\right]^\Tran\right\}
  = \frac{4}{\sigma^2} \sum_{t=1}^T\left(-\frac{1}{2}\right)\Im\{\mathbf{A}^\Herm  \mathbb{E}\left[ \mathbf{n}_t\mathbf{n}_s^\Herm \right] \mathbf{D} \mathbf{X}_t\}
  = \frac{2}{\sigma}\Im\left[ \mathbf{A}^\Herm \mathbf{D} \mathbf{X}_t \right].
  \end{equation}
\end{subequations}


\subsubsection{角度的二阶偏导及均值}

\begin{equation}
\mathbb{E}\left\{\left[\frac{\partial\ln \mathcal{L}}{\partial \bm{\theta}}\right]\left[\frac{\partial\ln \mathcal{L}}{\partial \bm{\theta}}\right]^\Tran\right\}
= \frac{4}{\sigma^2} \sum_{t=1}^T \sum_{s=1}^T \frac{1}{2} \Re\{\mathbf{X}_t^\Herm  \mathbf{D}^\Herm  \mathbb{E}\left[ \mathbf{n}_t\mathbf{n}_s^\Herm  \right] \mathbf{D} \mathbf{X}_s\}
= \frac{2}{\sigma} \sum_{t=1}^T \Re\left[ \mathbf{X}_t^\Herm \mathbf{D}^\Herm \mathbf{D} \mathbf{X}_t \right].
\end{equation}

\subsection{Fisher 信息矩阵/CRB 协方差矩阵构造}

\begin{mdquote}
  \begin{enumerate}
    \item 复矩阵求逆法则：令 $\mathbf{H}$ 是非奇异矩阵，且有 $\mathbf{G}\triangleq\mathbf{H}^{-1}$，则
    \begin{equation}
      \left[ \begin{matrix}
        \Re\{\mathbf{H}\} & -\Im\{\mathbf{H}\} \\
        \Im\{\mathbf{H}\} & \Re\{\mathbf{H}\} \\
      \end{matrix}\right] ^{-1} = \left[ \begin{matrix}
        \Re\{\mathbf{G}\} & -\Im\{\mathbf{G}\} \\
        \Im\{\mathbf{G}\} & \Re\{\mathbf{G}\} \\
      \end{matrix}\right].
    \end{equation}

    \item Schur 补在 \gls{fim} 中的应用：当 \gls{fim} 被分块为
    %
    \begin{equation}
      \mathcal{I} = \left[ \begin{matrix}
        \mathcal{I}_{\theta\theta} & \mathcal{I}_{\theta\psi} \\
        \mathcal{I}_{\psi\theta} & \mathcal{I}_{\psi\psi}
      \end{matrix} \right],
    \end{equation}
    %
    其中 $\bm{\theta}$ 为目标参数，$\bm{\psi}$ 为扰动（nuisance）参数，可以通过 Schur 补获得 CRB
    %
    \begin{equation}
      \label{eq:schur_complement}
      \mathrm{CRB}^{-1}(\bm{\theta})
      = \mathcal{I}_{\mathrm e}
      = \mathcal{I}_{\theta\theta} - \mathcal{I}_{\theta\psi} \mathcal{I}_{\psi\psi}^{-1} \mathcal{I}_{\psi\theta}.
    \end{equation}
  \end{enumerate}
\end{mdquote}

\Gls{fim} 可以写为
%
\begin{align}
  &\mathcal{I} \left( \sigma, \mathbf{x}_{1}^{{(r)}}, \mathbf{x}_{1}^{{(i)}}, \ldots, \mathbf{x}_{T}^{{(r)}}, \mathbf{x}_{T}^{{(i)}},\bm{\theta } \right) \nonumber
  \\
  =& \left[ \begin{array}{c:c:c}
    \text{CRB}^{-1}(\sigma) & 0 & 0 \\ \hdashline
    0 & \begin{array}{c:c:c}
      \begin{matrix}
        \Re\{\mathbf{H}\}\quad & -\Im\{\mathbf{H}\}\quad \\
        \Im\{\mathbf{H}\}\quad & \Re\{\mathbf{H}\}\quad \\
      \end{matrix} & & 0 \\
      \hdashline
      & \ddots & \\
      \hdashline
      0 & & \begin{matrix}
        \Re\{\mathbf{H}\}\quad & -\Im\{\mathbf{H}\}\quad\,\, \\
        \Im\{\mathbf{H}\}\quad & \Re\{\mathbf{H}\}\quad \\
    \end{matrix} \\
    \end{array} & \begin{matrix}
      \Re\{\bm{\Delta}_1\} \\ \Im\{\bm{\Delta}_1\} \\ \hdashline \vdots \\ \hdashline \Re\{\bm{\Delta}_T\} \\ \Im\{\bm{\Delta}_T\}
    \end{matrix}  \\
    \hdashline
    0 & \begin{array}{cc:c:cc}
        \Re\{\bm{\Delta}_1\}^\Tran &
        \Im\{\bm{\Delta}_1\}^\Tran &
        \cdots &
        \Re\{\bm{\Delta}_T\}^\Tran &
        \Im\{\bm{\Delta}_T\}^\Tran
      \end{array} &\bm{\Gamma}
  \end{array}
  \right],
\end{align}
%
其中 $\text{CRB}(\sigma) = \dfrac{\sigma^2}{MT}$，$\mathbf{H} = \dfrac{2}{\sigma}\mathbf{A}^\Herm \mathbf{A}$，$\bm{\Delta}_t = \dfrac{2}{\sigma} \mathbf{A}^\Herm \mathbf{D} \mathbf{X}_t$，$\bm{\Gamma} = \dfrac{2}{\sigma}\displaystyle\sum_{t=1}^T \Re\left[ \mathbf{X}_t^\Herm \mathbf{D}^\Herm \mathbf{D} \mathbf{X}_t \right]$。注意到，由于噪声方差和其他各参量的联合二阶偏导数为零，因此噪声方差这个参量不会通过 Schur 补影响 $\bm{\theta}$，也即可以只考虑除噪声方差之外的参量构成的 \gls{fim}。

我们的目标是通过 Schur 补消去确定信号快拍 $\{\mathbf x_t\}_{t=1}^T$，得到关于 $\bm\theta$ 的等效 Fisher 信息矩阵；随后对该等效 Fisher 信息矩阵求逆，得到 $\bm\theta$ 的 CRB 协方差矩阵。首先定义几个辅助变量，令
%
\begin{equation*}
  \mathbf{F}_{\psi\psi}=\left[ \begin{array}{c:c:c}
    \begin{matrix}
      \Re\{\mathbf{H}\} & -\Im\{\mathbf{H}\} \\
      \Im\{\mathbf{H}\} & \Re\{\mathbf{H}\} \\
    \end{matrix}
    & & 0 \\
    \hdashline
    & \ddots & \\
    \hdashline
    0 & & \begin{matrix}
        \Re\{\mathbf{H}\} & -\Im\{\mathbf{H}\} \\
        \Im\{\mathbf{H}\} & \Re\{\mathbf{H}\} \\
      \end{matrix} \\
    \end{array} \right],\quad
  %
  \mathbf{F}_{\psi\theta} = \left[ \begin{array}{c}
    \Re\{\bm{\Delta}_1\} \\
    \Im\{\bm{\Delta}_1\} \\
    \hdashline
    \vdots \\
    \hdashline
    \Re\{\bm{\Delta}_T\} \\
    \Im\{\bm{\Delta}_T\}
  \end{array}\right],
\end{equation*}
%
\begin{equation*}
  \mathbf{F}_{\theta\psi} = \left[
    \begin{array}{c:c:c}
      \begin{matrix}
        \Re\{\bm{\Delta}_1\}^\Tran & \Im\{\bm{\Delta}_1\}^\Tran
      \end{matrix}
      &
      \cdots
      &
      \begin{matrix}
        \Re\{\bm{\Delta}_T\}^\Tran & \Im\{\bm{\Delta}_T\}^\Tran
      \end{matrix}
    \end{array} \right].
\end{equation*}
%
因此，我们有
%
\begin{align}
  \mathrm{CRB}^{-1}(\bm{\theta}) = \bm{\Gamma}-\mathbf{F}_{\theta\psi}\mathbf{F}_{\psi\psi}^{-1}\mathbf{F}_{\psi\theta}
\end{align}
%
观察矩阵结构，通过分块矩阵乘法，令 $\mathbf{G} = \mathbf{H}^{-1}$，可以有~\footnote{将右侧按照复数乘法展开即可得到。}
%
\begin{equation*}
  \left[ \begin{matrix}
    \Re\{\mathbf{G}\} & -\Im\{\mathbf{G}\} \\
    \Im\{\mathbf{G}\} & \Re\{\mathbf{G}\} \\
  \end{matrix} \right] \left[ \begin{matrix}
    \Re\{\bm{\Delta}_t\}  \\
    \Im\{\bm{\Delta}_t\}  \\
  \end{matrix} \right] = \left[ \begin{matrix}
    \Re\{\mathbf{G}\}\Re\{\bm{\Delta}_t\} - \Im\{\mathbf{G}\}\Im\{\bm{\Delta}_t\} \\
    \Im\{\mathbf{G}\}\Re\{\bm{\Delta}_t\} + \Re\{\mathbf{G}\}\Im\{\bm{\Delta}_t\} \\
  \end{matrix} \right] = \left[ \begin{matrix}
    \Re\{\mathbf{G}\bm{\Delta}_t\}  \\
    \Im\{\mathbf{G}\bm{\Delta}_t\}  \\
  \end{matrix} \right],
\end{equation*}
%
以及~\footnote{将右侧展开，$\Re\{\bm{\Delta}_t^\Herm \mathbf{G}\bm{\Delta}_t\} = (\Re\{\bm{\Delta}_t\} - j \Im\{\bm{\Delta}_t\})^\Tran (\Re\{\mathbf{G}\bm{\Delta}_t\} + j \Im\{\mathbf{G}\bm{\Delta}_t\}) $，即可得到。}
%
\begin{equation*}
\left[ \begin{matrix}
    \Re\{\bm{\Delta}_t\}^\Tran & \Im\{\bm{\Delta}_t\}^\Tran  \\
  \end{matrix} \right] \left[ \begin{matrix}
    \Re\{\mathbf{G}\bm{\Delta}_t\}  \\
    \Im\{\mathbf{G}\bm{\Delta}_t\}  \\
  \end{matrix} \right] = \Re\{\bm{\Delta}_t^\Herm \mathbf{G}\bm{\Delta}_t\},
\end{equation*}
%
因此，角度的 \gls{crb} 可以表示为
%
\begin{align}
  \mathrm{CRB}^{-1}(\bm{\theta}) &= \bm{\Gamma} - \sum_{t=1}^T \Re\{\bm{\Delta}_t^\Herm \mathbf{G}\bm{\Delta}_t\} \nonumber
  = \dfrac{2}{\sigma} \sum_{t=1}^T \Re\left\{ \mathbf{X}_t^\Herm \mathbf{D}^\Herm \mathbf{D} \mathbf{X}_t - \mathbf{X}_t^\Herm \mathbf{D}^\Herm  \mathbf{A}(\mathbf{A}^\Herm  \mathbf{A})^{-1}\mathbf{A}^\Herm \mathbf{D} \mathbf{X}_t \right\} \nonumber
  \\
  &= \dfrac{2}{\sigma} \sum_{t=1}^T \Re\left\{ \mathbf{X}_t^\Herm \mathbf{D}^\Herm  \left[ \mathbf{I} - \mathbf{A}(\mathbf{A}^\Herm \mathbf{A})^{-1}\mathbf{A}^\Herm \right] \mathbf{D} \mathbf{X}_t \right\}.
\end{align}

参考 \cite{stoicaMusicMaximumLikelihood1989} 进一步定义阵列导向矩阵的列空间的投影矩阵 $\mathcal{P}_{\mathbf{A}} = \mathbf{A}(\mathbf{A}^\Herm \mathbf{A})^{-1}\mathbf{A}^\Herm$，及其正交补空间的投影矩阵 $\mathcal{P}_{\mathbf{A}}^\perp = \mathbf{I} -\mathcal{P}_{\mathbf{A}}$，投影矩阵下的等效信息矩阵 $\mathcal{H} = \mathbf{D}^\Herm \mathcal{P}_{\mathbf{A}}^\perp \mathbf{D}$，确定信号的经验协方差 $\mathbf R_{x,\mathrm{det}} = \dfrac{1}{T} \displaystyle\sum_{t=1}^T \mathbf{x}_t \mathbf{x}_t^\Herm$，则关于角度的等效 Fisher 信息矩阵可以表示为
%
\begin{align}
  \mathcal I_{\mathrm e}(\bm\theta)
  = \mathrm{CRB}^{-1}(\bm{\theta})
  = \frac{2T}{\sigma} \Re\left\{ \mathcal{H} \odot \mathbf R_{x,\mathrm{det}}^\Tran \right\}.
\end{align}
%
因此，条件（conditional）\gls{crb} 或确定（deterministic）\gls{crb} 为
%
\begin{align}
  \mathcal{B}_c
  := \mathrm{CRB}(\bm{\theta})
  = \frac{\sigma}{2T}
  \left[\Re\left\{ \mathcal{H} \odot \mathbf R_{x,\mathrm{det}}^\Tran \right\}\right]^{-1},
\end{align}
%
其中 $\odot$ 表示 Hadamard 积。

该结果是单阵列确定信号 DoA 的条件 CRB 基线。后续单阵列 DoA--Doppler 和多星 DoA--Doppler 的 CRB 推导会沿用“先构造完整 Fisher 信息，再对干扰参数做 Schur 补”的处理方式。