UPA/URA双极化天线的协方差矩阵结构

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 u 0 ∈ { 1 , ⋯   , N V } 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 Ψ ( θ ) ( e j Ψ ( θ ) ) H = [ ∣ 1 ⟩ ⟨ 1 ∣ ∣ 1 ⟩ ⟨ 2 ∣ ∣ 1 ⟩ ⟨ 3 ∣ ⋯ ∣ 1 ⟩ ⟨ N V ∣ ∣ 2 ⟩ ⟨ 1 ∣ ∣ 2 ⟩ ⟨ 2 ∣ ∣ 2 ⟩ ⟨ 3 ∣ ⋯ ⋮ ∣ 3 ⟩ ⟨ 1 ∣ ∣ 3 ⟩ ⟨ 2 ∣ ∣ 3 ⟩ ⟨ 3 ∣ ⋯ ∣ N V − 1 ⟩ ⟨ N V ∣ ⋮ ⋮ ⋮ ⋱ ∣ N V − 1 ⟩ ⟨ N V ∣ ∣ N V ⟩ ⟨ 1 ∣ ⋯ ∣ N V ⟩ ⟨ N V − 2 ∣ ∣ N V ⟩ ⟨ N V − 1 ∣ ∣ N V ⟩ ⟨ N V ∣ ] e^{j \boldsymbol \Psi(\boldsymbol \theta)} \left( e^{j \boldsymbol \Psi(\boldsymbol \theta)} \right )^H=\left[ \begin{matrix}{} |1\rangle \langle 1|& |1\rangle \langle 2|& |1\rangle \langle 3|& \cdots& |1\rangle \langle N_V|\\ |2\rangle \langle 1|& |2\rangle \langle 2|& |2\rangle \langle 3|& \cdots& \vdots\\ |3\rangle \langle 1|& |3\rangle \langle 2|& |3\rangle \langle 3|& \cdots& |N_V-1\rangle \langle N_V|\\ \vdots& \vdots& \vdots& \ddots& |N_V-1\rangle \langle N_V|\\ |N_V\rangle \langle 1|& \cdots& |N_V\rangle \langle N_V-2|& |N_V\rangle \langle N_V-1|& |N_V\rangle \langle N_V|\\ \end{matrix} \right] ejΨ(θ)(ejΨ(θ))H= ​∣1⟩⟨1∣∣2⟩⟨1∣∣3⟩⟨1∣⋮∣NV​⟩⟨1∣​∣1⟩⟨2∣∣2⟩⟨2∣∣3⟩⟨2∣⋮⋯​∣1⟩⟨3∣∣2⟩⟨3∣∣3⟩⟨3∣⋮∣NV​⟩⟨NV​−2∣​⋯⋯⋯⋱∣NV​⟩⟨NV​−1∣​∣1⟩⟨NV​∣⋮∣NV​−1⟩⟨NV​∣∣NV​−1⟩⟨NV​∣∣NV​⟩⟨NV​∣​ ​

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 ∣ = [ 1 e j 2 π λ d H sin ⁡ θ 1 sin ⁡ θ 2 e j 2 π λ 2 d H sin ⁡ θ 1 sin ⁡ θ 2 ⋮ e j 2 π λ ( N H − 1 ) d H sin ⁡ θ 1 sin ⁡ θ 2 ] [ 1 e j 2 π λ d H sin ⁡ θ 1 sin ⁡ θ 2 e j 2 π λ 2 d H sin ⁡ θ 1 sin ⁡ θ 2 ⋮ e j 2 π λ ( N H − 1 ) d H sin ⁡ θ 1 sin ⁡ θ 2 ] H |1\rangle \langle 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} \\ \vdots \\ 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} \\ \vdots \\ e^{j\frac{2\pi}{\lambda}(N_H-1)d_H \sin \theta_1 \sin \theta_2}\\ \end{array} \right]^H ∣1⟩⟨1∣= ​1ejλ2π​dH​sinθ1​sinθ2​ejλ2π​2dH​sinθ1​sinθ2​⋮ejλ2π​(NH​−1)dH​sinθ1​sinθ2​​ ​ ​1ejλ2π​dH​sinθ1​sinθ2​ejλ2π​2dH​sinθ1​sinθ2​⋮ejλ2π​(NH​−1)dH​sinθ1​sinθ2​​ ​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
∣ u 0 ⟩ ⟨ 1 ∣ = ( e j 2 π λ ( u 0 − 1 ) d V cos ⁡ θ 1 [ 1 e j 2 π λ d H sin ⁡ θ 1 sin ⁡ θ 2 e j 2 π λ 2 d H sin ⁡ θ 1 sin ⁡ θ 2 ⋮ e j 2 π λ ( N H − 1 ) d H sin ⁡ θ 1 sin ⁡ θ 2 ] ) [ 1 e j 2 π λ d H sin ⁡ θ 1 sin ⁡ θ 2 e j 2 π λ 2 d H sin ⁡ θ 1 sin ⁡ θ 2 ⋮ e j 2 π λ ( N H − 1 ) d H sin ⁡ θ 1 sin ⁡ θ 2 ] H = e j 2 π λ ( u 0 − 1 ) d V cos ⁡ θ 1 [ 1 e j 2 π λ d H sin ⁡ θ 1 sin ⁡ θ 2 e j 2 π λ 2 d H sin ⁡ θ 1 sin ⁡ θ 2 ⋮ e j 2 π λ ( N H − 1 ) d H sin ⁡ θ 1 sin ⁡ θ 2 ] [ 1 e j 2 π λ d H sin ⁡ θ 1 sin ⁡ θ 2 e j 2 π λ 2 d H sin ⁡ θ 1 sin ⁡ θ 2 ⋮ e j 2 π λ ( N H − 1 ) d H sin ⁡ θ 1 sin ⁡ θ 2 ] H \begin{aligned} |u_0 \rangle \langle 1| &=\left ( e^{j \frac{2\pi}{\lambda}(u_0-1)d_V \cos \theta_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} \\ \vdots \\ 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} \\ \vdots \\ 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 \cos \theta_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} \\ \vdots \\ 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} \\ \vdots \\ e^{j\frac{2\pi}{\lambda}(N_H-1)d_H \sin \theta_1 \sin \theta_2}\\ \end{array} \right]^H \end{aligned} ∣u0​⟩⟨1∣​= ​ejλ2π​(u0​−1)dV​cosθ1​ ​1ejλ2π​dH​sinθ1​sinθ2​ejλ2π​2dH​sinθ1​sinθ2​⋮ejλ2π​(NH​−1)dH​sinθ1​sinθ2​​ ​ ​ ​1ejλ2π​dH​sinθ1​sinθ2​ejλ2π​2dH​sinθ1​sinθ2​⋮ejλ2π​(NH​−1)dH​sinθ1​sinθ2​​ ​H=ejλ2π​(u0​−1)dV​cosθ1​ ​1ejλ2π​dH​sinθ1​sinθ2​ejλ2π​2dH​sinθ1​sinθ2​⋮ejλ2π​(NH​−1)dH​sinθ1​sinθ2​​ ​ ​1ejλ2π​dH​sinθ1​sinθ2​ejλ2π​2dH​sinθ1​sinθ2​⋮ejλ2π​(NH​−1)dH​sinθ1​sinθ2​​ ​H​

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
[ 1 e j 2 π λ d H sin ⁡ θ 1 sin ⁡ θ 2 e j 2 π λ 2 d H sin ⁡ θ 1 sin ⁡ θ 2 ⋮ e j 2 π λ ( N H − 1 ) d H sin ⁡ θ 1 sin ⁡ θ 2 ] [ 1 e j 2 π λ d H sin ⁡ θ 1 sin ⁡ θ 2 e j 2 π λ 2 d H sin ⁡ θ 1 sin ⁡ θ 2 ⋮ e j 2 π λ ( N H − 1 ) d H sin ⁡ θ 1 sin ⁡ θ 2 ] H = [ α 1 α 2 ∗ ⋯ α N H ∗ α 2 α 1 ⋱ ⋮ ⋮ ⋱ ⋱ α 2 ∗ α N H ⋯ α 2 α 1 ] ∈ C N H × N H \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} \\ \vdots \\ 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} \\ \vdots \\ e^{j\frac{2\pi}{\lambda}(N_H-1)d_H \sin \theta_1 \sin \theta_2}\\ \end{array} \right]^H = \left[ \begin{matrix} \alpha _1& \alpha _{2}^{*}& \cdots& \alpha _{N_H}^{*}\\ \alpha _2& \alpha _1& \ddots& \vdots\\ \vdots& \ddots& \ddots& \alpha _{2}^{*}\\ \alpha _{N_H}& \cdots& \alpha _2& \alpha _1\\ \end{matrix} \right] \in \mathbb C^{N_H \times N_H} ​1ejλ2π​dH​sinθ1​sinθ2​ejλ2π​2dH​sinθ1​sinθ2​⋮ejλ2π​(NH​−1)dH​sinθ1​sinθ2​​ ​ ​1ejλ2π​dH​sinθ1​sinθ2​ejλ2π​2dH​sinθ1​sinθ2​⋮ejλ2π​(NH​−1)dH​sinθ1​sinθ2​​ ​H= ​α1​α2​⋮αNH​​​α2∗​α1​⋱⋯​⋯⋱⋱α2​​αNH​∗​⋮α2∗​α1​​ ​∈CNH​×NH​

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：
考虑协方差矩阵
$$\mathbf{R}=\left[\begin{array}{cc}{\mathbf{B}}_1& {\mathbf{B}}_2^H\\ {\mathbf{B}}_2& {\mathbf{B}}_3\end{array}\right]\in {\mathbb{C}}^{2N_V{N}_H\times 2N_V{N}_H}$$

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
B l = [ B l , 1 B l , 2 H B l , 3 H ⋯ B l , N V H B l , 2 B l , 1 B l , 2 H ⋱ ⋮ B l , 3 B l , 2 B l , 1 ⋱ B l , 3 H ⋮ ⋱ ⋱ ⋱ B l , 2 H B l , N V ⋯ B l , 3 B l , 2 B l , 1 ] ∈ C N V N H × N V N H \boldsymbol{B}_l=\left[ \begin{matrix}{} \boldsymbol{B}_{l,1}& \boldsymbol{B}_{l,2}^{H}& \boldsymbol{B}_{l,3}^{H}& \cdots& \boldsymbol{B}_{l,N_V}^{H}\\ \boldsymbol{B}_{l,2}& \boldsymbol{B}_{l,1}& \boldsymbol{B}_{l,2}^{H}& \ddots& \vdots\\ \boldsymbol{B}_{l,3}& \boldsymbol{B}_{l,2}& \boldsymbol{B}_{l,1}& \ddots& \boldsymbol{B}_{l,3}^{H}\\ \vdots& \ddots& \ddots& \ddots& \boldsymbol{B}_{l,2}^{H}\\ \boldsymbol{B}_{l,N_V}& \cdots& \boldsymbol{B}_{l,3}& \boldsymbol{B}_{l,2}& \boldsymbol{B}_{l,1}\\ \end{matrix} \right] \in \mathbb{C}^{ N_V N_H \times N_V N_H } Bl​= ​Bl,1​Bl,2​Bl,3​⋮Bl,NV​​​Bl,2H​Bl,1​Bl,2​⋱⋯​Bl,3H​Bl,2H​Bl,1​⋱Bl,3​​⋯⋱⋱⋱Bl,2​​Bl,NV​H​⋮Bl,3H​Bl,2H​Bl,1​​ ​∈CNV​NH​×NV​NH​

其中

B l , 1 = β ⋅ [ α 1 α 2 ∗ ⋯ α N H ∗ α 2 α 1 ⋱ ⋮ ⋮ ⋱ ⋱ α 2 ∗ α N H ⋯ α 2 α 1 ] ∈ C N H × N H , β ∈ R \boldsymbol{B}_{l,1} = \beta \cdot \left[ \begin{matrix} \alpha _1& \alpha _{2}^{*}& \cdots& \alpha _{N_H}^{*}\\ \alpha _2& \alpha _1& \ddots& \vdots\\ \vdots& \ddots& \ddots& \alpha _{2}^{*}\\ \alpha _{N_H}& \cdots& \alpha _2& \alpha _1\\ \end{matrix} \right] \in \mathbb C^{N_H \times N_H}, \beta \in \mathbb R Bl,1​=β⋅ ​α1​α2​⋮αNH​​​α2∗​α1​⋱⋯​⋯⋱⋱α2​​αNH​∗​⋮α2∗​α1​​ ​∈CNH​×NH​,β∈R

B l , u 0 = β ⋅ e j 2 π λ ( u 0 − 1 ) d V cos ⁡ θ 1 ⋅ [ α 1 α 2 ∗ ⋯ α N H ∗ α 2 α 1 ⋱ ⋮ ⋮ ⋱ ⋱ α 2 ∗ α N H ⋯ α 2 α 1 ] ∈ C N H × N H , β ∈ R , u 0 > 1 \boldsymbol{B}_{l,u_0} = \beta \cdot e^{j \frac{2\pi}{\lambda}(u_0-1)d_V \cos \theta_1} \cdot \left[ \begin{matrix} \alpha _1& \alpha _{2}^{*}& \cdots& \alpha _{N_H}^{*}\\ \alpha _2& \alpha _1& \ddots& \vdots\\ \vdots& \ddots& \ddots& \alpha _{2}^{*}\\ \alpha _{N_H}& \cdots& \alpha _2& \alpha _1\\ \end{matrix} \right] \in \mathbb C^{N_H \times N_H}, \beta \in \mathbb R, u_0 > 1 Bl,u0​​=β⋅ejλ2π​(u0​−1)dV​cosθ1​⋅ ​α1​α2​⋮αNH​​​α2∗​α1​⋱⋯​⋯⋱⋱α2​​αNH​∗​⋮α2∗​α1​​ ​∈CNH​×NH​,β∈R,u0​>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$N_H+N_H-1$

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.文章来源地址https://www.toymoban.com/news/detail-685486.html

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