波束成形的物理意义

如图， 是在各种教材中经常看到的天线方向图。 上图表示的是当前天线经过波束成形后在空间中 指向 30度方向 （一般考虑的方向是0-180度）。这里解释一下天线方向：
不失一般性的，一个MISO系统的接收信号（简洁起见，省略噪声）可以表示为：

         y 
        
       
         = 
        
       
         h 
        
       
         v 
        
       
         x 
        
       
      
      
      
      
        (1) 
       
      
     
    
   
     \mathbf{y} = \mathbf{h}\mathbf{v}x \tag{1} 
    
   
 </span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.63888em; vertical-align: -0.19444em;"></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">y</span></span><span class="mspace" style="margin-right: 0.277778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right: 0.277778em;"></span></span><span class="base"><span class="strut" style="height: 0.69444em; vertical-align: 0em;"></span><span class="mord"><span class="mord mathbf">h</span></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span><span class="mord mathdefault">x</span></span><span class="tag"><span class="strut" style="height: 1em; vertical-align: -0.25em;"></span><span class="mord text"><span class="mord">(</span><span class="mord"><span class="mord">1</span></span><span class="mord">)</span></span></span></span></span></span></span><br> 那么， <strong>如何把数学建模里的 代表 波束成形向量 的 <span class="katex--inline"><span class="katex"><span class="katex-mathml"> 
  
   
    
    
      v 
     
    
   
     \mathbf{v} 
    
   
 </span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.44444em; vertical-align: 0em;"></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span></span></span></span></span> 与 我们平常所说的 将 天线阵列对准某个方向发射的物理意义对应呢</strong>？</p> 

正如我们生活中使用手电筒一样， 把手电筒照向某个方向， 其实就对应于最后某个方向是亮的， 而其他方向是暗的。 波束成形也是如此。 让波束成形后的天线阵列所谓对准某一方向， 指的就是在该方向上的信噪比显著地高于其他方向。

由于噪声的功率是一定的， 信噪比其实就是(1)中

     y 
    
   
  
    \mathbf{y} 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.63888em; vertical-align: -0.19444em;"></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">y</span></span></span></span></span></span>的能量。 也就是：<br> <span class="katex--display"><span class="katex-display"><span class="katex"><span class="katex-mathml"> 
  
   
    
    
      P 
     
    
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      ∣ 
     
     
     
       v 
      
     
       H 
      
     
     
     
       h 
      
     
       H 
      
     
    
      h 
     
    
      v 
     
    
   
     P=\mathbf{y}^H\mathbf{y}=|x|^2|\mathbf{v}^H\mathbf{h}^H\mathbf{h}\mathbf{v} 
    
   
 </span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.68333em; vertical-align: 0em;"></span><span class="mord mathdefault" style="margin-right: 0.13889em;">P</span><span class="mspace" style="margin-right: 0.277778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right: 0.277778em;"></span></span><span class="base"><span class="strut" style="height: 1.08577em; vertical-align: -0.19444em;"></span><span class="mord"><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">y</span></span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height: 0.891331em;"><span class="" style="top: -3.113em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight" style="margin-right: 0.08125em;">H</span></span></span></span></span></span></span></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">y</span></span><span class="mspace" style="margin-right: 0.277778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right: 0.277778em;"></span></span><span class="base"><span class="strut" style="height: 1.14133em; vertical-align: -0.25em;"></span><span class="mord">∣</span><span class="mord mathdefault">x</span><span class="mord"><span class="mord">∣</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height: 0.864108em;"><span class="" style="top: -3.113em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mord">∣</span><span class="mord"><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height: 0.891331em;"><span class="" style="top: -3.113em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight" style="margin-right: 0.08125em;">H</span></span></span></span></span></span></span></span><span class="mord"><span class="mord"><span class="mord mathbf">h</span></span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height: 0.891331em;"><span class="" style="top: -3.113em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight" style="margin-right: 0.08125em;">H</span></span></span></span></span></span></span></span><span class="mord"><span class="mord mathbf">h</span></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span></span></span></span></span></span></p> 

由于

     ∣ 
    
   
     x 
    
    
    
      ∣ 
     
    
      2 
     
    
   
  
    |x|^2 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 1.06411em; vertical-align: -0.25em;"></span><span class="mord">∣</span><span class="mord mathdefault">x</span><span class="mord"><span class="mord">∣</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height: 0.814108em;"><span class="" style="top: -3.063em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span></span></span></span></span>也是相同的， 因此对于同一波束成形的天线阵列， 不同方向的信噪比就取决于<span class="katex--inline"><span class="katex"><span class="katex-mathml"> 
 
  
   
   
     ∣ 
    
   
     h 
    
   
     v 
    
    
    
      ∣ 
     
    
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    |\mathbf{h}\mathbf{v}|^2 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 1.06411em; vertical-align: -0.25em;"></span><span class="mord">∣</span><span class="mord"><span class="mord mathbf">h</span></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span><span class="mord"><span class="mord">∣</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height: 0.814108em;"><span class="" style="top: -3.063em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span></span></span></span></span>.<br> 这里先类比一下， 方便大家理解物理意义：</p> 

明白了物理意义， 现在我们来看如何判断当前的某个

     v 
    
   
  
    \mathbf{v} 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.44444em; vertical-align: 0em;"></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span></span></span></span></span>，对应于朝向哪个方向发射。<br> 首先， 根据MISO信道的建模， 我们有， 对于某个方向 <span class="katex--inline"><span class="katex"><span class="katex-mathml"> 
 
  
   
   
     ϕ 
    
   
  
    \phi 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.88888em; vertical-align: -0.19444em;"></span><span class="mord mathdefault">ϕ</span></span></span></span></span>,<br> 其对应的归一化ULA阵列信道为：<br> <span class="katex--display"><span class="katex-display"><span class="katex"><span class="katex-mathml"> 
  
   
    
     
     
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         ⁡ 
        
       
         ( 
        
       
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       ] 
      
     
    
   
     \mathbf{h}_{\mathrm{ULA}}(\phi)=\frac{1}{\sqrt{N}}\left[1, e^{-j \pi \cos(\phi)}, \ldots, e^{-j \pi(N-1) \cos (\phi)}\right] 
    
   
 </span><span class="katex-html"><span class="base"><span class="strut" style="height: 1em; vertical-align: -0.25em;"></span><span class="mord"><span class="mord"><span class="mord mathbf">h</span></span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.328331em;"><span class="" style="top: -2.55em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight"><span class="mord mathrm mtight">U</span><span class="mord mathrm mtight">L</span><span class="mord mathrm mtight">A</span></span></span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height: 0.15em;"><span class=""></span></span></span></span></span></span><span class="mopen">(</span><span class="mord mathdefault">ϕ</span><span class="mclose">)</span><span class="mspace" style="margin-right: 0.277778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right: 0.277778em;"></span></span><span class="base"><span class="strut" style="height: 2.25144em; vertical-align: -0.93em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 1.32144em;"><span class="" style="top: -2.18333em;"><span class="pstrut" style="height: 3em;"></span><span class="mord"><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.926665em;"><span class="svg-align" style="top: -3em;"><span class="pstrut" style="height: 3em;"></span><span class="mord" style="padding-left: 0.833em;"><span class="mord mathdefault" style="margin-right: 0.10903em;">N</span></span></span><span class="" style="top: -2.88666em;"><span class="pstrut" style="height: 3em;"></span><span class="hide-tail" style="min-width: 0.853em; height: 1.08em;"> 
               <svg width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"> 
                <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,

-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,
-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422
s-65,47,-65,47z M834 80H400000v40H845z">
​1​[1,e−jπcos(ϕ),…,e−jπ(N−1)cos(ϕ)]
将不同方向的信道络成一个大矩阵

     H 
    
   
  
    \mathbf{H} 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.68611em; vertical-align: 0em;"></span><span class="mord"><span class="mord mathbf">H</span></span></span></span></span></span>， 其每一行对应于一个方向的<span class="katex--inline"><span class="katex"><span class="katex-mathml"> 
 
  
   
    
    
      h 
     
     
     
       U 
      
     
       L 
      
     
       A 
      
     
    
   
     ( 
    
   
     ϕ 
    
   
     ) 
    
   
  
    \mathbf{h}_{\mathrm{ULA}}(\phi) 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 1em; vertical-align: -0.25em;"></span><span class="mord"><span class="mord"><span class="mord mathbf">h</span></span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.328331em;"><span class="" style="top: -2.55em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight"><span class="mord mathrm mtight">U</span><span class="mord mathrm mtight">L</span><span class="mord mathrm mtight">A</span></span></span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height: 0.15em;"><span class=""></span></span></span></span></span></span><span class="mopen">(</span><span class="mord mathdefault">ϕ</span><span class="mclose">)</span></span></span></span></span>信道， 那么我们只需要计算 <span class="katex--inline"><span class="katex"><span class="katex-mathml"> 
 
  
   
    
    
      v 
     
    
      H 
     
    
    
    
      H 
     
    
      H 
     
    
   
     H 
    
   
     v 
    
   
  
    \mathbf{v}^H\mathbf{H}^H\mathbf{H}\mathbf{v} 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.841331em; vertical-align: 0em;"></span><span class="mord"><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height: 0.841331em;"><span class="" style="top: -3.063em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight" style="margin-right: 0.08125em;">H</span></span></span></span></span></span></span></span><span class="mord"><span class="mord"><span class="mord mathbf">H</span></span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height: 0.841331em;"><span class="" style="top: -3.063em; margin-right: 0.05em;"><span class="pstrut" style="height: 2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mathdefault mtight" style="margin-right: 0.08125em;">H</span></span></span></span></span></span></span></span><span class="mord"><span class="mord mathbf">H</span></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span></span></span></span></span>， 就能得到一个列向量， 其每一项代表了当前波束成形向量<span class="katex--inline"><span class="katex"><span class="katex-mathml"> 
 
  
   
   
     v 
    
   
  
    \mathbf{v} 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.44444em; vertical-align: 0em;"></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span></span></span></span></span>在这个方向上的能量大小。</p> 

通过天线方向图和Matlab实例说明

接下来通过实例来说明：
令

      v 
     
    
      = 
     
     
     
       1 
      
      
      
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       T 
      
     
    
   
     \mathbf{v}=\frac{1}{\sqrt{N}}\left[1, e^{j \pi \cos(\theta)}, \ldots, e^{j \pi(N-1) \cos (\theta)}\right]^T 
    
   
 </span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.44444em; vertical-align: 0em;"></span><span class="mord"><span class="mord mathbf" style="margin-right: 0.01597em;">v</span></span><span class="mspace" style="margin-right: 0.277778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right: 0.277778em;"></span></span><span class="base"><span class="strut" style="height: 2.31123em; vertical-align: -0.93em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 1.32144em;"><span class="" style="top: -2.18333em;"><span class="pstrut" style="height: 3em;"></span><span class="mord"><span class="mord sqrt"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 0.926665em;"><span class="svg-align" style="top: -3em;"><span class="pstrut" style="height: 3em;"></span><span class="mord" style="padding-left: 0.833em;"><span class="mord mathdefault" style="margin-right: 0.10903em;">N</span></span></span><span class="" style="top: -2.88666em;"><span class="pstrut" style="height: 3em;"></span><span class="hide-tail" style="min-width: 0.853em; height: 1.08em;"> 
               <svg width="400em" height="1.08em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"> 
                <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,

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-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,
35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,
-221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467
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​1​[1,ejπcos(θ),…,ejπ(N−1)cos(θ)]T
这其实是将天线阵列对准

     θ 
    
   
  
    \theta 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.69444em; vertical-align: 0em;"></span><span class="mord mathdefault" style="margin-right: 0.02778em;">θ</span></span></span></span></span> 方向发送的波束成形向量。<br> 我们来验证他就是对准了 <span class="katex--inline"><span class="katex"><span class="katex-mathml"> 
 
  
   
   
     θ 
    
   
  
    \theta 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.69444em; vertical-align: 0em;"></span><span class="mord mathdefault" style="margin-right: 0.02778em;">θ</span></span></span></span></span> 方向， 比如我们令其为 空间30度方向。</p> 

N = 16; 
idx = (0 : N-1)';
angle_set =  (1 : 1 : 360) / 180 * pi;
Hset = exp(1j * pi * idx * cos(angle_set));

theta = 30;
x = exp(1j * pi * idx * cos(theta / 180 * pi));

r =  Hset' * x;
polarplot(angle_set, abs(r))
1
2
3
4
5
6
7
8
9
10

运行后， 就能得到文章开头的那张图了。 这是一个极坐标图， 可以清楚的看到， 在30度方向上是能量最大的， 而在其他绝大部分方向上， 能量为0。这也就是我们一开始说的， 通过有效的波束成形， 可以将能量集中， 让天线阵列对准某个方向发送。

也可以根据自己的画图需求改变一下坐标，比如将上面代码第三行替换为

angle_set =  (0 : 0.01 : 180) / 180 * pi;
1

就是画出0~180度的响应， 并且将分辨率提升至0.01度。
效果如图：

我们可以试着改变

     N 
    
   
  
    N 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.68333em; vertical-align: 0em;"></span><span class="mord mathdefault" style="margin-right: 0.10903em;">N</span></span></span></span></span>, 即天线数的大小。<br> 如<code>N = 64;</code><br> <img src="https://img-blog.csdnimg.cn/20201015201954304.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zOTI3NDY1OQ==,size_16,color_FFFFFF,t_70#pic_center" alt="在这里插入图片描述"><br> 可以明显的看到，波束更窄了， 能量更集中了。 这也是为什么天线数越多，通信性能越好的原因， 指向性也越为精准。</p> 

UPA的方向图

UPA的话就涉及两个角度了 仰角

     ϕ 
    
   
  
    \phi 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.88888em; vertical-align: -0.19444em;"></span><span class="mord mathdefault">ϕ</span></span></span></span></span> 和 水平角 <span class="katex--inline"><span class="katex"><span class="katex-mathml"> 
 
  
   
   
     θ 
    
   
  
    \theta 
   
  
</span><span class="katex-html"><span class="base"><span class="strut" style="height: 0.69444em; vertical-align: 0em;"></span><span class="mord mathdefault" style="margin-right: 0.02778em;">θ</span></span></span></span></span>。 因此很难用二维的图进行绘制。<br> 这里给出代码， 细节不再赘述， 和ULA的方法类似：<br> 效果图：<br> <img src="https://img-blog.csdnimg.cn/20201015204318391.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl8zOTI3NDY1OQ==,size_16,color_FFFFFF,t_70#pic_center" alt="在这里插入图片描述"></p> 

主函数main.m

clear
M = UPA_code3(16, 32);
phi =  -1 : 0.01 : 1;
theta = -1 : 0.01 : 1;
[aa, bb] = meshgrid(phi, theta);
Aset = []';
G = zeros(512, 1);

G(256+11) = 1;

s = M * G;
for i = 1 : length(phi)

a = phi(i);

for j = 1 : length(theta)

tmp = acos(a);

b = theta(j) / sin(tmp);

ar = UPA_arrayresponse2(16,32, acos(b), acos(a));

C(j,i) = abs(s' * ar);

end

end
surf(aa, bb, C)
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调用的两个函数文件：

• UPA_code3.m

function [Aset] = UPA_code3(P,Q)
% P, Q: rows and columns of UPA

% theta, phi: AODs

Aset = [];

for n = 0 : 1 / P * pi : (P-1)/Ppi

for m = 0 : 1 / Q * pi : (Q-1)/Qpi

A = zeros(P, Q);

for p = 1 : P

for q = 1 : Q

A(p, q) = exp(1jpi(q-1)cos(m) * sin(n)+1jpi(p-1)cos(n));

end

end

a = A(?;

Aset = [Aset, a];

end

end
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• UPA_arrayresponse2.m

function [A] = UPA_arrayresponse2(P,Q, theta, phi)
% P, Q: rows and columns of UPA

% theta, phi: AODs

A = zeros(P, Q);

for p = 1 : P

for q = 1 : Q

A(p, q) = exp(1jpi(q-1)sin(phi)cos(theta)+1jpi(p-1)*cos(phi));

end

end
A = A(?;
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