UPA的阵列响应向量（暂不考虑双极化天线）

下图形象描述了UPA阵列的接收信号

UPA阵列的水平(Horizontal)方向的天线间距为$d_H$，垂直(Vertical)方向的天线间距为$d_V$，图中BA是点A处的阵元接收到的信号方向，我们需要衡量水平、垂直两个方向的路径差。

（1）水平方向的路径差
考虑三角形OAB，我们从图中可以看出三个点的坐标分别为：$(0,0,0),(d_H,0,0),(r\cos \varphi \sin \theta +d_H,r\cos \varphi \cos \theta ,r\sin \varphi )$，可以进一步计算该三角形三条边的长度
$$\begin{array}{rl}OA& =d_H\\ OB& =\left|(r\cos \varphi \sin \theta +d_H,r\cos \varphi \cos \theta ,r\sin \varphi )\right|\\ AB& =r\end{array}$$

根据三角余弦定理，我们可以得到
$$\begin{array}{rl}\cos \angle OAB& =\frac{|AB{|}^2+|OA{|}^2-|OB{|}^2}{2|AB|\cdot |OA|}\\ & =\frac{r^2+d_H^2-(r^2+d_H^2+2d_{H}r\cos \varphi \sin \theta )}{2r{d}_H}\\ & =\cos \varphi \sin \theta \end{array}$$

因此水平方向的路径差为：
$${\Delta }_H=d_H\cos \angle OAB=d_H\cos \varphi \sin \theta$$

（2）垂直方向的路径差
不难看出，垂直方向的路径差为
$${\Delta }_V=d_V\sin \varphi$$

因此阵列响应向量对应的延时（相位）部分可以表征为：
$${\Psi }_{(u-1)N_H+v}(\varphi ,\theta )=\frac{2\pi }{\lambda }\left[(u-1)d_V\sin \varphi +(v-1)d_H\cos \varphi \sin \theta \right]$$

其中$1\le u\le N_V,1\le v\le N_H$。

为了与论文[1]的符号对齐，这里我们令
$$\begin{array}{rl}\cos {\theta }_1& =\sin \varphi \\ \sin {\theta }_2& =\sin \theta \end{array}$$

令$\mathbf{\theta }=({\theta }_1,{\theta }_2{)}^T$，这时UPA阵列响应向量中含相位的项为
$$e^{j\mathbf{\Psi }(\mathbf{\theta })}:={\left[e^{j{\Psi }_1(\mathbf{\theta })},e^{j{\Psi }_2(\mathbf{\theta })},\cdots ,e^{j{\Psi }_{N_V{N}_H}(\mathbf{\theta })}\right]}^T\in {\mathbb{C}}^{N_V{N}_H\times 1}$$

其中：
$${\Psi }_{(u-1)N_H+v}(\mathbf{\theta })=\frac{2\pi }{\lambda }\left[(u-1)d_V\cos {\theta }_1+(v-1)d_H\sin {\theta }_1\sin {\theta }_2\right],1\le u\le N_V,1\le v\le N_H$$

更进一步，UPA阵列响应向量为（包含水平方向和垂直方向）：
$$\begin{array}{rl}{\mathbf{a}}_V(\mathbf{\theta })& =a_V(\mathbf{\theta })e^{j\mathbf{\Psi }(\mathbf{\theta })}\in {\mathbb{C}}^{N_V{N}_H\times 1}\\ {\mathbf{a}}_H(\mathbf{\theta })& =a_H(\mathbf{\theta })e^{j\mathbf{\Psi }(\mathbf{\theta })}\in {\mathbb{C}}^{N_V{N}_H\times 1}\end{array}$$

其中$a_V(\mathbf{\theta }),a_H(\mathbf{\theta })\in \mathbb{R}$表示天线本身的field pattern（对应幅度的概念）。

UPA阵列响应：从单极化天线到双极化天线

注意到，上一章节所推演的阵列响应响应为单极化UPA阵列。对于双极化UPA，其阵列响应向量定义为
$$\begin{array}{rl}{\mathbf{a}}_V(\mathbf{\theta })& =\left[\begin{array}{c}{a}_{V,1}\left(\mathbf{\theta }\right)\\ a_{V,2}\left(\mathbf{\theta }\right)\end{array}\right]\otimes e^{j\mathbf{\Psi }(\mathbf{\theta })}\in {\mathbb{C}}^{2N_V{N}_H\times 1}\\ {\mathbf{a}}_H(\mathbf{\theta })& =\left[\begin{array}{c}{a}_{H,1}\left(\mathbf{\theta }\right)\\ a_{H,2}\left(\mathbf{\theta }\right)\end{array}\right]\otimes e^{j\mathbf{\Psi }(\mathbf{\theta })}\in {\mathbb{C}}^{2N_V{N}_H\times 1}\end{array}$$

其中$a_{V,1},a_{V,2}\in \mathbb{R}$分别表示垂直方向上，$+45°$和$-45°$极化天线的field pattern，为天线固有的值，不受环境影响。

UPA双极化天线的协方差矩阵结构

双极化UPA阵列的协方差矩阵为
$$\mathbf{R}={\int }_{\Omega }{\rho }_V(\mathbf{\theta }){\mathbf{a}}_V(\mathbf{\theta }){\mathbf{a}}_V^H(\mathbf{\theta })d\mathbf{\theta }+{\int }_{\Omega }{\rho }_H(\mathbf{\theta }){\mathbf{a}}_H(\mathbf{\theta }){\mathbf{a}}_H^H(\mathbf{\theta })d\mathbf{\theta }\in {\mathbb{C}}^{2N_V{N}_H\times 2N_V{N}_H}$$

不失一般性，这里我们只关注${\mathbf{a}}_V(\mathbf{\theta }){\mathbf{a}}_V^H(\mathbf{\theta })$
$$\begin{array}{rl}{\mathbf{a}}_V(\mathbf{\theta }){\mathbf{a}}_V^H(\mathbf{\theta })& =\left(\left[\begin{array}{c}{a}_{V,1}\left(\mathbf{\theta }\right)\\ a_{V,2}\left(\mathbf{\theta }\right)\end{array}\right]\otimes e^{j\mathbf{\Psi }(\mathbf{\theta })}\right){\left(\left[\begin{array}{c}{a}_{V,1}\left(\mathbf{\theta }\right)\\ a_{V,2}\left(\mathbf{\theta }\right)\end{array}\right]\otimes e^{j\mathbf{\Psi }(\mathbf{\theta })}\right)}^H\\ & =\left(\left[\begin{array}{c}{a}_{V,1}\left(\mathbf{\theta }\right)\\ a_{V,2}\left(\mathbf{\theta }\right)\end{array}\right]\otimes e^{j\mathbf{\Psi }(\mathbf{\theta })}\right)\left({\left[\begin{array}{c}{a}_{V,1}\left(\mathbf{\theta }\right)\\ a_{V,2}\left(\mathbf{\theta }\right)\end{array}\right]}^T\otimes {\left(e^{j\mathbf{\Psi }(\mathbf{\theta })}\right)}^H\right)\\ & =\left[\begin{array}{cc}{a}_{V,1}^2\left(\mathbf{\theta }\right)& a_{V,1}\left(\mathbf{\theta }\right)a_{V,2}\left(\mathbf{\theta }\right)\\ a_{V,2}\left(\mathbf{\theta }\right)a_{V,1}\left(\mathbf{\theta }\right)& a_{V,2}^2\left(\mathbf{\theta }\right)\end{array}\right]\otimes \left(e^{j\mathbf{\Psi }(\mathbf{\theta })}{\left(e^{j\mathbf{\Psi }(\mathbf{\theta })}\right)}^H\right)\end{array}$$

我们不难看出，协方差矩阵$\mathbf{R}\in {\mathbb{C}}^{2N_V{N}_H\times 2N_V{N}_H}$，具有特定的块结构，可写为
$$\mathbf{R}=\left[\begin{array}{cc}{\mathbf{B}}_1& {\mathbf{B}}_2^H\\ {\mathbf{B}}_2& {\mathbf{B}}_3\end{array}\right]\in {\mathbb{C}}^{2N_V{N}_H\times 2N_V{N}_H}$$

其中${\mathbf{B}}_1,{\mathbf{B}}_2,{\mathbf{B}}_3\in {\mathbb{C}}^{N_V{N}_H\times N_V{N}_H}$具有相同的结构性质，且不难看出，该性质取决于$e^{j\mathbf{\Psi }(\mathbf{\theta })}{\left(e^{j\mathbf{\Psi }(\mathbf{\theta })}\right)}^H\in {\mathbb{C}}^{N_V{N}_H\times N_V{N}_H}$。回顾$e^{j\mathbf{\Psi }(\mathbf{\theta })}$的表达式：
$$\begin{array}{rl}{e}^{j\mathbf{\Psi }(\mathbf{\theta })}& ={\left[e^{j{\Psi }_1(\mathbf{\theta })},e^{j{\Psi }_2(\mathbf{\theta })},\cdots ,e^{j{\Psi }_{N_V{N}_H}(\mathbf{\theta })}\right]}^T\\ {\Psi }_{(u-1)N_H+v}(\mathbf{\theta })& =\frac{2\pi }{\lambda }\left[(u-1)d_V\cos {\theta }_1+(v-1)d_H\sin {\theta }_1\sin {\theta }_2\right]\end{array}$$

我们不妨先固定u0∈{1,⋯,NV}u_0 \in \{1,\cdots,N_V\}u0​∈{1,⋯,NV​}，取子向量$\left(e^{j{\Psi }_k(\mathbf{\theta })}:k=(u_0-1)N_H+v,v=1,\cdots ,N_H\right)\in {\mathbb{C}}^{N_H\times 1}$，令
$$|u_0⟩=\left(e^{j{\Psi }_k(\mathbf{\theta })}:k=(u_0-1)N_H+v,v=1,\cdots ,N_H\right)\in {\mathbb{C}}^{N_H\times 1}$$

（这里存在一定程度的符号滥用，但为了方便叙述，我们选择采用量子力学中常用的狄拉克符号），则$e^{j\mathbf{\Psi }(\mathbf{\theta })}{\left(e^{j\mathbf{\Psi }(\mathbf{\theta })}\right)}^H\in {\mathbb{C}}^{N_V{N}_H\times N_V{N}_H}$可以写为
$$e^{j\mathbf{\Psi }(\mathbf{\theta })}{\left(e^{j\mathbf{\Psi }(\mathbf{\theta })}\right)}^H=\left[\begin{array}{ccccc}|1⟩⟨1|& |1⟩⟨2|& |1⟩⟨3|& \cdots & |1⟩⟨N_V|\\ |2⟩⟨1|& |2⟩⟨2|& |2⟩⟨3|& \cdots & ⋮\\ |3⟩⟨1|& |3⟩⟨2|& |3⟩⟨3|& \cdots & |N_V-1⟩⟨N_V|\\ ⋮& ⋮& ⋮& \ddots & |N_V-1⟩⟨N_V|\\ |N_V⟩⟨1|& \cdots & |N_V⟩⟨N_V-2|& |N_V⟩⟨N_V-1|& |N_V⟩⟨N_V|\end{array}\right]$$

首先，我们不难发现
$$|i⟩⟨j|=|i+d⟩⟨j+d|,\forall d$$

因此，我们只需要关注$|1⟩⟨1|,|2⟩⟨1|,\cdots ,|N_V⟩⟨1|$即可

当$u_0=1$时
$$|1⟩⟨1|=\left[\begin{array}{l}1\\ e^{j\frac{2\pi }{\lambda }{d}_H\sin {\theta }_1\sin {\theta }_2}\\ e^{j\frac{2\pi }{\lambda }2d_H\sin {\theta }_1\sin {\theta }_2}\\ ⋮\\ e^{j\frac{2\pi }{\lambda }(N_H-1)d_H\sin {\theta }_1\sin {\theta }_2}\end{array}\right]{\left[\begin{array}{l}1\\ e^{j\frac{2\pi }{\lambda }{d}_H\sin {\theta }_1\sin {\theta }_2}\\ e^{j\frac{2\pi }{\lambda }2d_H\sin {\theta }_1\sin {\theta }_2}\\ ⋮\\ e^{j\frac{2\pi }{\lambda }(N_H-1)d_H\sin {\theta }_1\sin {\theta }_2}\end{array}\right]}^H$$

不难发现，上述写法直接对应到ULA阵，因此对应部分的块矩阵满足Toplitz性。

当$u_0>1$时
$$\begin{array}{rl}|u_0⟩⟨1|& =\left(e^{j\frac{2\pi }{\lambda }(u_0-1)d_V}\left[\begin{array}{l}1\\ e^{j\frac{2\pi }{\lambda }{d}_H\sin {\theta }_1\sin {\theta }_2}\\ e^{j\frac{2\pi }{\lambda }2d_H\sin {\theta }_1\sin {\theta }_2}\\ ⋮\\ e^{j\frac{2\pi }{\lambda }(N_H-1)d_H\sin {\theta }_1\sin {\theta }_2}\end{array}\right]\right){\left[\begin{array}{l}1\\ e^{j\frac{2\pi }{\lambda }{d}_H\sin {\theta }_1\sin {\theta }_2}\\ e^{j\frac{2\pi }{\lambda }2d_H\sin {\theta }_1\sin {\theta }_2}\\ ⋮\\ e^{j\frac{2\pi }{\lambda }(N_H-1)d_H\sin {\theta }_1\sin {\theta }_2}\end{array}\right]}^H\\ & =e^{j\frac{2\pi }{\lambda }(u_0-1)d_V}\left[\begin{array}{l}1\\ e^{j\frac{2\pi }{\lambda }{d}_H\sin {\theta }_1\sin {\theta }_2}\\ e^{j\frac{2\pi }{\lambda }2d_H\sin {\theta }_1\sin {\theta }_2}\\ ⋮\\ e^{j\frac{2\pi }{\lambda }(N_H-1)d_H\sin {\theta }_1\sin {\theta }_2}\end{array}\right]{\left[\begin{array}{l}1\\ e^{j\frac{2\pi }{\lambda }{d}_H\sin {\theta }_1\sin {\theta }_2}\\ e^{j\frac{2\pi }{\lambda }2d_H\sin {\theta }_1\sin {\theta }_2}\\ ⋮\\ e^{j\frac{2\pi }{\lambda }(N_H-1)d_H\sin {\theta }_1\sin {\theta }_2}\end{array}\right]}^H\end{array}$$

注意到，无论是$|1⟩⟨1|$还是$|u_0⟩⟨1|,u_0>1$，大家都有一个公共的Toeplitz结构，即
$$\left[\begin{array}{l}1\\ e^{j\frac{2\pi }{\lambda }{d}_H\sin {\theta }_1\sin {\theta }_2}\\ e^{j\frac{2\pi }{\lambda }2d_H\sin {\theta }_1\sin {\theta }_2}\\ ⋮\\ e^{j\frac{2\pi }{\lambda }(N_H-1)d_H\sin {\theta }_1\sin {\theta }_2}\end{array}\right]{\left[\begin{array}{l}1\\ e^{j\frac{2\pi }{\lambda }{d}_H\sin {\theta }_1\sin {\theta }_2}\\ e^{j\frac{2\pi }{\lambda }2d_H\sin {\theta }_1\sin {\theta }_2}\\ ⋮\\ e^{j\frac{2\pi }{\lambda }(N_H-1)d_H\sin {\theta }_1\sin {\theta }_2}\end{array}\right]}^H=\left[\begin{array}{cccc}{\alpha }_1& {\alpha }_2^{\ast }& \cdots & {\alpha }_{N_H}^{\ast }\\ {\alpha }_2& {\alpha }_1& \ddots & ⋮\\ ⋮& \ddots & \ddots & {\alpha }_2^{\ast }\\ {\alpha }_{N_H}& \cdots & {\alpha }_2& {\alpha }_1\end{array}\right]\in {\mathbb{C}}^{N_H\times N_H}$$

因为积分无非就是加权求和，并不改变上述积分内部矩阵的结构，因此我们得出如下结论：
（1）若波长$\lambda$和$d_V$未知，即系数$e^{j\frac{2\pi }{\lambda }(u_0-1)d_V}$未知，则协方差矩阵
$$\mathbf{R}=\left[\begin{array}{cc}{\mathbf{B}}_1& {\mathbf{B}}_2^H\\ {\mathbf{B}}_2& {\mathbf{B}}_3\end{array}\right]\in {\mathbb{C}}^{2N_V{N}_H\times 2N_V{N}_H}$$

的每一个块矩阵${\mathbf{B}}_1,{\mathbf{B}}_2,{\mathbf{B}}_3\in {\mathbb{C}}^{N_V{N}_H\times N_V{N}_H}$具有相同的结构，即
$${\mathbf{B}}_l=\left[\begin{array}{ccccc}{\mathbf{B}}_{l,1}& {\mathbf{B}}_{l,2}^H& {\mathbf{B}}_{l,3}^H& \cdots & {\mathbf{B}}_{l,N_V}^H\\ {\mathbf{B}}_{l,2}& {\mathbf{B}}_{l,1}& {\mathbf{B}}_{l,2}^H& \ddots & ⋮\\ {\mathbf{B}}_{l,3}& {\mathbf{B}}_{l,2}& {\mathbf{B}}_{l,1}& \ddots & {\mathbf{B}}_{l,3}^H\\ ⋮& \ddots & \ddots & \ddots & {\mathbf{B}}_{l,2}^H\\ {\mathbf{B}}_{l,N_V}& \cdots & {\mathbf{B}}_{l,3}& {\mathbf{B}}_{l,2}& {\mathbf{B}}_{l,1}\end{array}\right]\in {\mathbb{C}}^{N_V{N}_H\times N_V{N}_H}$$

其中

$${\mathbf{B}}_{l,1}=\beta \cdot \left[\begin{array}{cccc}{\alpha }_1& {\alpha }_2^{\ast }& \cdots & {\alpha }_{N_H}^{\ast }\\ {\alpha }_2& {\alpha }_1& \ddots & ⋮\\ ⋮& \ddots & \ddots & {\alpha }_2^{\ast }\\ {\alpha }_{N_H}& \cdots & {\alpha }_2& {\alpha }_1\end{array}\right]\in {\mathbb{C}}^{N_H\times N_H},\beta \in \mathbb{R}$$

$${\mathbf{B}}_{l,u_0}=\beta \cdot e^{j\frac{2\pi }{\lambda }(u_0-1)d_V}\cdot \left[\begin{array}{cccc}{\alpha }_1& {\alpha }_2^{\ast }& \cdots & {\alpha }_{N_H}^{\ast }\\ {\alpha }_2& {\alpha }_1& \ddots & ⋮\\ ⋮& \ddots & \ddots & {\alpha }_2^{\ast }\\ {\alpha }_{N_H}& \cdots & {\alpha }_2& {\alpha }_1\end{array}\right]\in {\mathbb{C}}^{N_H\times N_H},\beta \in \mathbb{R},u_0>1$$

• ${\mathbf{B}}_{l,1}$的自由度为$N_H$
• ${\mathbf{B}}_{l,u_0},u_0>1$的自由度为$N_H+N_H-1$（甚至还可以再压缩）

（2）若系数$e^{j\frac{2\pi }{\lambda }(u_0-1)d_V}$已知，则

• ${\mathbf{B}}_{l,1}$的自由度为$N_H$
• ${\mathbf{B}}_{l,u_0},u_0>1$的自由度为$N_H$
  本质上，此时协方差矩阵的自由度可以降为$N_H$。

参考文献

[1] L. Miretti, R. L. G. Cavalcante and S. Stańczak, “Channel Covariance Conversion and Modelling Using Infinite Dimensional Hilbert Spaces,” in IEEE Transactions on Signal Processing, vol. 69, pp. 3145-3159, 2021, doi: 10.1109/TSP.2021.3082461.