new post, josh-oap-1-2
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@@ -217,6 +217,8 @@ $$\begin{equation}
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定义
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<a id="DefinitionOfArrayManifoldVector"></a>
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$$\begin{equation} \label{DefinitionOfArrayManifoldVector}
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\vec{v}_{\vec{k}}\left(\vec{k}\right) = \left[ \begin{array}{c}
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e^{-j \vec{k}^\mathrm{T}\vec{p}_0} \\
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@@ -0,0 +1,66 @@
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---
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title: 'Josh''s Notes: 最优阵列处理(Part 1.2 —— 均匀线阵模型)'
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mathjax: true
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comments: true
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copyright: true
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toc:
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enable: false
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number: true
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wrap: true
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expand_all: true
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max_depth: 4
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tags:
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- 阵列信号处理
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- 均匀线阵
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- ULA
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categories:
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- - Josh 的学习笔记
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- 最优阵列处理
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- 阵列和空域滤波器
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abbrlink: de20fd09
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date: 2023-04-20 00:14:31
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---
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我们感兴趣的线阵如[图 1-2-1](#fig.1-2-1) 所示,其中有 $N$ 个位于 $z$ 轴上的阵元,阵元间距均匀,记为 $d$。我们把这种阵列称为**{% label primary @均匀线性阵列(uniform linear array, ULA) %}**。我们已经把阵列的中心放在了坐标系的原点上,这种阵列中心放置方法将在计算量上具有优点。
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<a id="fig.1-2-1"></a>
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{width=600px}
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<!-- more -->
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  阵元的位置为
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$$\begin{equation} \label{ElementLocation}
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\begin{cases}
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p_{x_n} = 0 \\
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p_{y_n} = 0 \\
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p_{z_n} = \left( n - \displaystyle\frac{N-1}{2} \right) d
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\end{cases}
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, \, n = 0,1,\cdots,N-1
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\end{equation}$$
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为了确定阵列流形矢量 $\vec{v}_{\vec{k}}\left(\vec{k}\right)$, 把式 $\eqref{ElementLocation}$ 代入到[阵列流形矢量的定义][], 得到
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$$\begin{equation}
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\vec{v}_{\vec{k}}\left(k_z\right) = \left[\begin{array}{c}
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e^{j \left( \frac{N-1}{2} \right) k_z d} \\
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e^{j \left( \frac{N-1}{2} - 1\right) k_z d} \\
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\vdots \\
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e^{-j \left( \frac{N-1}{2} \right) k_z d} \\
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\end{array}\right]
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\end{equation}$$
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这里
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$$\begin{equation}
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k_z = -\frac{2\pi}{\lambda}\cos\theta = -k_0 \cos\theta
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\end{equation}$$
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其中波数的幅度
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$$\begin{equation}
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k_0 \triangleq \left| \vec{k} \right| = \frac{2\pi}{\lambda}
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\end{equation}$$
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[阵列流形矢量的定义]: https://josh-gao.top/posts/8b61f5a7.html#DefinitionOfArrayManifoldVector
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