文章目录
- UPA的阵列响应向量（暂不考虑双极化天线）
- UPA阵列响应：从单极化天线到双极化天线
- UPA双极化天线的协方差矩阵结构
- 参考文献
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 d_H dH，垂直(Vertical)方向的天线间距为 d V d_V dV，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
ϕ
sin
θ
+
d
H
,
r
cos
ϕ
cos
θ
,
r
sin
ϕ
)
(0,0,0),(d_H,0,0),(r \cos \phi \sin \theta + d_H, r \cos \phi \cos \theta, r \sin \phi)
(0,0,0),(dH,0,0),(rcosϕsinθ+dH,rcosϕcosθ,rsinϕ)，可以进一步计算该三角形三条边的长度
O
A
=
d
H
O
B
=
∣
(
r
cos
ϕ
sin
θ
+
d
H
,
r
cos
ϕ
cos
θ
,
r
sin
ϕ
)
∣
A
B
=
r
\begin{aligned} OA &= d_H \\ OB &= \left | (r \cos \phi \sin \theta + d_H, r \cos \phi \cos \theta, r \sin \phi) \right| \\ AB &= r \end{aligned}
OAOBAB=dH=∣(rcosϕsinθ+dH,rcosϕcosθ,rsinϕ)∣=r
根据三角余弦定理，我们可以得到
cos
∠
O
A
B
=
∣
A
B
∣
2
+
∣
O
A
∣
2
−
∣
O
B
∣
2
2
∣
A
B
∣
⋅
∣
O
A
∣
=
r
2
+
d
H
2
−
(
r
2
+
d
H
2
+
2
d
H
r
cos
ϕ
sin
θ
)
2
r
d
H
=
cos
ϕ
sin
θ
\begin{aligned} \cos {\angle {OAB}} &= \frac { |AB|^2 + |OA|^2 - |OB|^2 }{2 |AB| \cdot |OA|} \\ &= \frac{ r^2 + d^2_H - (r^2 + d^2_H+ 2 d_H r \cos \phi \sin \theta) } {2 r d_H} \\ &= \cos \phi \sin \theta \end{aligned}
cos∠OAB=2∣AB∣⋅∣OA∣∣AB∣2+∣OA∣2−∣OB∣2=2rdHr2+dH2−(r2+dH2+2dHrcosϕsinθ)=cosϕsinθ
因此水平方向的路径差为：
Δ
H
=
d
H
cos
∠
O
A
B
=
d
H
cos
ϕ
sin
θ
\Delta_H = d_H \cos {\angle {OAB}} = d_H \cos \phi \sin \theta
ΔH=dHcos∠OAB=dHcosϕsinθ
（2）垂直方向的路径差
不难看出，垂直方向的路径差为
Δ
V
=
d
V
sin
ϕ
\Delta_V= d_V \sin \phi
ΔV=dVsinϕ
因此阵列响应向量对应的延时（相位）部分可以表征为：
Ψ
(
u
−
1
)
N
H
+
v
(
ϕ
,
θ
)
=
2
π
λ
[
(
u
−
1
)
d
V
sin
ϕ
+
(
v
−
1
)
d
H
cos
ϕ
sin
θ
]
\Psi_{(u-1)N_H+v}(\phi, \theta) = \frac{2 \pi}{\lambda} \left [ (u-1) d_V \sin \phi + (v-1)d_H \cos \phi \sin \theta \right]
Ψ(u−1)NH+v(ϕ,θ)=λ2π[(u−1)dVsinϕ+(v−1)dHcosϕsinθ]
其中 1 ≤ u ≤ N V , 1 ≤ v ≤ N H 1 \leq u \leq N_V, 1 \leq v \leq N_H 1≤u≤NV,1≤v≤NH。
为了与论文[1]的符号对齐，这里我们令
cos
θ
1
=
sin
ϕ
sin
θ
2
=
sin
θ
\begin{aligned} \cos \theta_1 &= \sin \phi \\ \sin \theta_2&= \sin \theta \end{aligned}
cosθ1sinθ2=sinϕ=sinθ
令
θ
=
(
θ
1
,
θ
2
)
T
\boldsymbol \theta = (\theta_1, \theta_2)^T
θ=(θ1,θ2)T，这时UPA阵列响应向量中含相位的项为
e
j
Ψ
(
θ
)
:
=
[
e
j
Ψ
1
(
θ
)
,
e
j
Ψ
2
(
θ
)
,
⋯
,
e
j
Ψ
N
V
N
H
(
θ
)
]
T
∈
C
N
V
N
H
×
1
e^{j \boldsymbol \Psi(\boldsymbol \theta)} := \left [ e^{j \Psi_1(\boldsymbol \theta)}, e^{j \Psi_2(\boldsymbol \theta)}, \cdots, e^{j \Psi_{N_V N_H}(\boldsymbol \theta)} \right]^T \in \mathbb C^{N_V N_H \times 1}
ejΨ(θ):=[ejΨ1(θ),ejΨ2(θ),⋯,ejΨNVNH(θ)]T∈CNVNH×1
其中：
Ψ
(
u
−
1
)
N
H
+
v
(
θ
)
=
2
π
λ
[
(
u
−
1
)
d
V
cos
θ
1
+
(
v
−
1
)
d
H
sin
θ
1
sin
θ
2
]
,
1
≤
u
≤
N
V
,
1
≤
v
≤
N
H
\Psi_{(u-1)N_H+v}(\boldsymbol \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 \leq u \leq N_V, 1 \leq v \leq N_H
Ψ(u−1)NH+v(θ)=λ2π[(u−1)dVcosθ1+(v−1)dHsinθ1sinθ2], 1≤u≤NV,1≤v≤NH
更进一步，UPA阵列响应向量为（包含水平方向和垂直方向）：
a
V
(
θ
)
=
a
V
(
θ
)
e
j
Ψ
(
θ
)
∈
C
N
V
N
H
×
1
a
H
(
θ
)
=
a
H
(
θ
)
e
j
Ψ
(
θ
)
∈
C
N
V
N
H
×
1
\begin{aligned} \boldsymbol a_V (\boldsymbol \theta) &= a_V(\boldsymbol \theta) e^{j \boldsymbol \Psi(\boldsymbol \theta)} \in \mathbb C^{N_V N_H \times 1} \\ \boldsymbol a_H (\boldsymbol \theta) &= a_H(\boldsymbol \theta) e^{j \boldsymbol \Psi(\boldsymbol \theta)} \in \mathbb C^{N_V N_H \times 1} \\ \end{aligned}
aV(θ)aH(θ)=aV(θ)ejΨ(θ)∈CNVNH×1=aH(θ)ejΨ(θ)∈CNVNH×1
其中 a V ( θ ) , a H ( θ ) ∈ R a_V(\boldsymbol \theta),a_H(\boldsymbol \theta) \in \mathbb R aV(θ),aH(θ)∈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
a
V
(
θ
)
=
[
a
V
,
1
(
θ
)
a
V
,
2
(
θ
)
]
⊗
e
j
Ψ
(
θ
)
∈
C
2
N
V
N
H
×
1
a
H
(
θ
)
=
[
a
H
,
1
(
θ
)
a
H
,
2
(
θ
)
]
⊗
e
j
Ψ
(
θ
)
∈
C
2
N
V
N
H
×
1
\begin{aligned} \boldsymbol a_V (\boldsymbol \theta) &= \left[ \begin{array}{c} a_{V,1}\left( \boldsymbol{\theta } \right)\\ a_{V,2}\left( \boldsymbol{\theta } \right)\\ \end{array} \right] \otimes e^{j \boldsymbol \Psi(\boldsymbol \theta)} \in \mathbb C^{2 N_V N_H \times 1} \\ \boldsymbol a_H (\boldsymbol \theta) &= \left[ \begin{array}{c} a_{H,1}\left( \boldsymbol{\theta } \right)\\ a_{H,2}\left( \boldsymbol{\theta } \right)\\ \end{array} \right] \otimes e^{j \boldsymbol \Psi(\boldsymbol \theta)} \in \mathbb C^{2 N_V N_H \times 1} \end{aligned}
aV(θ)aH(θ)=[aV,1(θ)aV,2(θ)]⊗ejΨ(θ)∈C2NVNH×1=[aH,1(θ)aH,2(θ)]⊗ejΨ(θ)∈C2NVNH×1
其中 a V , 1 , a V , 2 ∈ R a_{V,1},a_{V,2} \in \mathbb R aV,1,aV,2∈R分别表示垂直方向上， + 45 ° +45 \degree +45°和 − 45 ° -45 \degree −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
R
=
∫
Ω
ρ
V
(
θ
)
a
V
(
θ
)
a
V
H
(
θ
)
d
θ
+
∫
Ω
ρ
H
(
θ
)
a
H
(
θ
)
a
H
H
(
θ
)
d
θ
∈
C
2
N
V
N
H
×
2
N
V
N
H
\boldsymbol R= \int_{\Omega} \rho_V(\boldsymbol \theta) \boldsymbol a_V(\boldsymbol \theta) \boldsymbol a_V^H(\boldsymbol \theta) d \boldsymbol \theta + \int_{\Omega} \rho_H(\boldsymbol \theta) \boldsymbol a_H(\boldsymbol \theta) \boldsymbol a_H^H(\boldsymbol \theta) d \boldsymbol \theta \in \mathbb C^{2 N_V N_H \times 2 N_V N_H}
R=∫ΩρV(θ)aV(θ)aVH(θ)dθ+∫ΩρH(θ)aH(θ)aHH(θ)dθ∈C2NVNH×2NVNH
不失一般性，这里我们只关注
a
V
(
θ
)
a
V
H
(
θ
)
\boldsymbol a_V(\boldsymbol \theta) \boldsymbol a_V^H(\boldsymbol \theta)
aV(θ)aVH(θ)
a
V
(
θ
)
a
V
H
(
θ
)
=
(
[
a
V
,
1
(
θ
)
a
V
,
2
(
θ
)
]
⊗
e
j
Ψ
(
θ
)
)
(
[
a
V
,
1
(
θ
)
a
V
,
2
(
θ
)
]
⊗
e
j
Ψ
(
θ
)
)
H
=
(
[
a
V
,
1
(
θ
)
a
V
,
2
(
θ
)
]
⊗
e
j
Ψ
(
θ
)
)
(
[
a
V
,
1
(
θ
)
a
V
,
2
(
θ
)
]
T
⊗
(
e
j
Ψ
(
θ
)
)
H
)
=
[
a
V
,
1
2
(
θ
)
a
V
,
1
(
θ
)
a
V
,
2
(
θ
)
a
V
,
2
(
θ
)
a
V
,
1
(
θ
)
a
V
,
2
2
(
θ
)
]
⊗
(
e
j
Ψ
(
θ
)
(
e
j
Ψ
(
θ
)
)
H
)
\begin{aligned} \boldsymbol a_V(\boldsymbol \theta) \boldsymbol a_V^H(\boldsymbol \theta) &= \left ( \left[ \begin{array}{c} a_{V,1}\left( \boldsymbol{\theta } \right)\\ a_{V,2}\left( \boldsymbol{\theta } \right)\\ \end{array} \right] \otimes e^{j \boldsymbol \Psi(\boldsymbol \theta)} \right) \left ( \left[ \begin{array}{c} a_{V,1}\left( \boldsymbol{\theta } \right)\\ a_{V,2}\left( \boldsymbol{\theta } \right)\\ \end{array} \right] \otimes e^{j \boldsymbol \Psi(\boldsymbol \theta)} \right)^H \\ &=\left ( \left[ \begin{array}{c} a_{V,1}\left( \boldsymbol{\theta } \right)\\ a_{V,2}\left( \boldsymbol{\theta } \right)\\ \end{array} \right] \otimes e^{j \boldsymbol \Psi(\boldsymbol \theta)} \right) \left ( \left[ \begin{array}{c} a_{V,1}\left( \boldsymbol{\theta } \right)\\ a_{V,2}\left( \boldsymbol{\theta } \right)\\ \end{array} \right]^T \otimes \left( e^{j \boldsymbol \Psi(\boldsymbol \theta)} \right )^H \right) \\ &= \left[ \begin{matrix} a_{V,1}^{2}\left( \boldsymbol{\theta } \right)& a_{V,1}\left( \boldsymbol{\theta } \right) a_{V,2}\left( \boldsymbol{\theta } \right)\\ a_{V,2}\left( \boldsymbol{\theta } \right) a_{V,1}\left( \boldsymbol{\theta } \right)& a_{V,2}^{2}\left( \boldsymbol{\theta } \right)\\ \end{matrix} \right] \otimes \left ( e^{j \boldsymbol \Psi(\boldsymbol \theta)} \left( e^{j \boldsymbol \Psi(\boldsymbol \theta)} \right )^H \right) \end{aligned}
aV(θ)aVH(θ)=([aV,1(θ)aV,2(θ)]⊗ejΨ(θ))([aV,1(θ)aV,2(θ)]⊗ejΨ(θ))H=([aV,1(θ)aV,2(θ)]⊗ejΨ(θ))([aV,1(θ)aV,2(θ)]T⊗(ejΨ(θ))H)=[aV,12(θ)aV,2(θ)aV,1(θ)aV,1(θ)aV,2(θ)aV,22(θ)]⊗(ejΨ(θ)(ejΨ(θ))H)
我们不难看出，协方差矩阵
R
∈
C
2
N
V
N
H
×
2
N
V
N
H
\boldsymbol R \in \mathbb C^{2 N_V N_H \times 2 N_V N_H}
R∈C2NVNH×2NVNH，具有特定的块结构，可写为
R
=
[
B
1
B
2
H
B
2
B
3
]
∈
C
2
N
V
N
H
×
2
N
V
N
H
\boldsymbol R = \left[ \begin{matrix} \boldsymbol{B}_1& \boldsymbol{B}_{2}^{H}\\ \boldsymbol{B}_2& \boldsymbol{B}_3\\ \end{matrix} \right] \in \mathbb C^{2 N_V N_H \times 2 N_V N_H}
R=[B1B2B2HB3]∈C2NVNH×2NVNH
其中
B
1
,
B
2
,
B
3
∈
C
N
V
N
H
×
N
V
N
H
\boldsymbol{B}_1,\boldsymbol{B}_2,\boldsymbol{B}_3 \in \mathbb C^{ N_V N_H \times N_V N_H}
B1,B2,B3∈CNVNH×NVNH具有相同的结构性质，且不难看出，该性质取决于
e
j
Ψ
(
θ
)
(
e
j
Ψ
(
θ
)
)
H
∈
C
N
V
N
H
×
N
V
N
H
e^{j \boldsymbol \Psi(\boldsymbol \theta)} \left( e^{j \boldsymbol \Psi(\boldsymbol \theta)} \right )^H \in \mathbb C^{ N_V N_H \times N_V N_H}
ejΨ(θ)(ejΨ(θ))H∈CNVNH×NVNH。回顾
e
j
Ψ
(
θ
)
e^{j \boldsymbol \Psi(\boldsymbol \theta)}
ejΨ(θ)的表达式：
e
j
Ψ
(
θ
)
=
[
e
j
Ψ
1
(
θ
)
,
e
j
Ψ
2
(
θ
)
,
⋯
,
e
j
Ψ
N
V
N
H
(
θ
)
]
T
Ψ
(
u
−
1
)
N
H
+
v
(
θ
)
=
2
π
λ
[
(
u
−
1
)
d
V
cos
θ
1
+
(
v
−
1
)
d
H
sin
θ
1
sin
θ
2
]
\begin{aligned} e^{j \boldsymbol \Psi(\boldsymbol \theta)} &= \left [ e^{j \Psi_1(\boldsymbol \theta)}, e^{j \Psi_2(\boldsymbol \theta)}, \cdots, e^{j \Psi_{N_V N_H}(\boldsymbol \theta)} \right]^T \\ \Psi_{(u-1)N_H+v}(\boldsymbol \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{aligned}
ejΨ(θ)Ψ(u−1)NH+v(θ)=[ejΨ1(θ),ejΨ2(θ),⋯,ejΨNVNH(θ)]T=λ2π[(u−1)dVcosθ1+(v−1)dHsinθ1sinθ2]
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
(
e
j
Ψ
k
(
θ
)
:
k
=
(
u
0
−
1
)
N
H
+
v
,
v
=
1
,
⋯
,
N
H
)
∈
C
N
H
×
1
\left (e^{j \Psi_k(\boldsymbol \theta)}: k=(u_0-1)N_H+v, v=1,\cdots,N_H \right) \in \mathbb C^{N_H \times 1}
(ejΨk(θ):k=(u0−1)NH+v,v=1,⋯,NH)∈CNH×1，令
∣
u
0
⟩
=
(
e
j
Ψ
k
(
θ
)
:
k
=
(
u
0
−
1
)
N
H
+
v
,
v
=
1
,
⋯
,
N
H
)
∈
C
N
H
×
1
|u_0 \rangle = \left (e^{j \Psi_k(\boldsymbol \theta)}: k=(u_0-1)N_H+v, v=1,\cdots,N_H \right) \in \mathbb C^{N_H \times 1}
∣u0⟩=(ejΨk(θ):k=(u0−1)NH+v,v=1,⋯,NH)∈CNH×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
Ψ
(
θ
)
(
e
j
Ψ
(
θ
)
)
H
∈
C
N
V
N
H
×
N
V
N
H
e^{j \boldsymbol \Psi(\boldsymbol \theta)} \left( e^{j \boldsymbol \Psi(\boldsymbol \theta)} \right )^H \in \mathbb C^{ N_V N_H \times N_V N_H}
ejΨ(θ)(ejΨ(θ))H∈CNVNH×NVNH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
∣
,
∀
d
|i \rangle \langle j| = |i+d \rangle \langle j+d|, \ \ \forall d
∣i⟩⟨j∣=∣i+d⟩⟨j+d∣, ∀d
因此，我们只需要关注 ∣ 1 ⟩ ⟨ 1 ∣ , ∣ 2 ⟩ ⟨ 1 ∣ , ⋯ , ∣ N V ⟩ ⟨ 1 ∣ |1\rangle \langle 1|, |2\rangle \langle 1|, \cdots, |N_V\rangle \langle 1| ∣1⟩⟨1∣,∣2⟩⟨1∣,⋯,∣NV⟩⟨1∣即可
当
u
0
=
1
u_0=1
u0=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∣=
1ejλ2πdHsinθ1sinθ2ejλ2π2dHsinθ1sinθ2⋮ejλ2π(NH−1)dHsinθ1sinθ2
1ejλ2πdHsinθ1sinθ2ejλ2π2dHsinθ1sinθ2⋮ejλ2π(NH−1)dHsinθ1sinθ2
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
u_0 > 1
u0>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)dVcosθ1
1ejλ2πdHsinθ1sinθ2ejλ2π2dHsinθ1sinθ2⋮ejλ2π(NH−1)dHsinθ1sinθ2
1ejλ2πdHsinθ1sinθ2ejλ2π2dHsinθ1sinθ2⋮ejλ2π(NH−1)dHsinθ1sinθ2
H=ejλ2π(u0−1)dVcosθ1
1ejλ2πdHsinθ1sinθ2ejλ2π2dHsinθ1sinθ2⋮ejλ2π(NH−1)dHsinθ1sinθ2
1ejλ2πdHsinθ1sinθ2ejλ2π2dHsinθ1sinθ2⋮ejλ2π(NH−1)dHsinθ1sinθ2
H
注意到，无论是
∣
1
⟩
⟨
1
∣
|1\rangle \langle 1|
∣1⟩⟨1∣还是
∣
u
0
⟩
⟨
1
∣
,
u
0
>
1
|u_0 \rangle \langle 1|,u_0 > 1
∣u0⟩⟨1∣,u0>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}
1ejλ2πdHsinθ1sinθ2ejλ2π2dHsinθ1sinθ2⋮ejλ2π(NH−1)dHsinθ1sinθ2
1ejλ2πdHsinθ1sinθ2ejλ2π2dHsinθ1sinθ2⋮ejλ2π(NH−1)dHsinθ1sinθ2
H=
α1α2⋮αNHα2∗α1⋱⋯⋯⋱⋱α2αNH∗⋮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
R
=
[
B
1
B
2
H
B
2
B
3
]
∈
C
2
N
V
N
H
×
2
N
V
N
H
\boldsymbol R = \left[ \begin{matrix} \boldsymbol{B}_1& \boldsymbol{B}_{2}^{H}\\ \boldsymbol{B}_2& \boldsymbol{B}_3\\ \end{matrix} \right] \in \mathbb C^{2 N_V N_H \times 2 N_V N_H}
R=[B1B2B2HB3]∈C2NVNH×2NVNH
的每一个块矩阵
B
1
,
B
2
,
B
3
∈
C
N
V
N
H
×
N
V
N
H
\boldsymbol{B}_1,\boldsymbol{B}_2,\boldsymbol{B}_3 \in \mathbb C^{ N_V N_H \times N_V N_H}
B1,B2,B3∈CNVNH×NVNH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,1Bl,2Bl,3⋮Bl,NVBl,2HBl,1Bl,2⋱⋯Bl,3HBl,2HBl,1⋱Bl,3⋯⋱⋱⋱Bl,2Bl,NVH⋮Bl,3HBl,2HBl,1
∈CNVNH×NVNH
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⋅ α1α2⋮αNHα2∗α1⋱⋯⋯⋱⋱α2αNH∗⋮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)dVcosθ1⋅ α1α2⋮αNHα2∗α1⋱⋯⋯⋱⋱α2αNH∗⋮α2∗α1 ∈CNH×NH,β∈R,u0>1
- B l , 1 \boldsymbol{B}_{l,1} Bl,1的自由度为 N H N_H NH
- B l , u 0 , u 0 > 1 \boldsymbol{B}_{l,u_0}, u_0 > 1 Bl,u0,u0>1的自由度为 N H + N H − 1 N_H+N_H-1 NH+NH−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.
