对数似然比(LLR)

LLR 基本原理

假设$a[i]=a_I[i]+j{a}_Q[i]$表示调制发送QAM复信号，数据{bI,1,...,bI,k,...,bI,m,bQ,1,...,bQ,k,...,bQ,m}\{b_{I,1},...,b_{I,k},...,b_{I,m},b_{Q,1},...,b_{Q,k},...,b_{Q,m}\}{bI,1​,...,bI,k​,...,bI,m​,bQ,1​,...,bQ,k​,...,bQ,m​}分为$I,Q$两路。假设信道的冲击响应为$H(i)$,则接收信号可表示为：
$$r[i]=H(i).a[i]+w[i]$$
$w[i]$为复高斯白噪声方差${\sigma }^2=N_0$,对于QAM调制信号，分IQ两路分别已$b_{I,k},b_{Q,k}$表示IQ两路第$k$比特取值$(0,1)$,QAM 星座图中我们已第$k$比特取值为0或者1可将星座图分为两分区，其中已$S_k^{(0)}$表示第$k$比特为0的星座点集合，$S_k^{(1)}$表示第$k$比特为1星座点集合,由于I和Q路是一样的我们下来呢仅以I路为例进行说明。
对数似然比$（LLR,Log-LikelihoodRatio）$定义为：
$$\begin{array}{rl}LLR(b_{I,k})& =\log \frac{P[b_{I,k=0}|r[i]]}{P[b_{I,k=1}|r[i]]}\\ & =\log \frac{{\sum }_{\alpha \in S_k^0}P[\alpha [i]=\alpha |r[i]]}{{\sum }_{\alpha \in S_k^1}P[\alpha [i]=\alpha |r[i]]}\end{array}$$
应用贝叶斯公式且发送符号等概率分布可得：
$$\begin{array}{rl}LLR(b_{I,k})& =\log \frac{{\sum }_{\alpha \in S_k^0}P[\alpha [i]=\alpha |r[i]]}{{\sum }_{\alpha \in S_k^1}P[\alpha [i]=\alpha |r[i]]}\\ & =\log \frac{{\sum }_{\alpha \in S_k^0}P[r[i]|\alpha [i]=\alpha ]}{{\sum }_{\alpha \in S_k^1}P[r[i]|\alpha [i]=\alpha ]}\end{array}$$
高斯白噪声信道$P[r[i]|\alpha [i]=\alpha ]$概率密度可表示为：
$$\begin{array}{rl}p(r[i]|\alpha [i]& =\alpha )=\frac{1}{\sqrt{2\pi \sigma }}{e}^{\left\{-\frac{1}{2}\frac{|r[i]-H(i)\alpha {|}^2}{{\sigma }^2}\right\}}\end{array}$$

$$\begin{array}{rl}LLR(b_{I,k})& =\log \frac{{\sum }_{\alpha \in S_k^0}{e}^{\left\{-\frac{1}{2}\frac{|r[i]-H(i)\alpha {|}^2}{{\sigma }^2}\right\}}}{{\sum }_{\alpha \in S_k^1}{e}^{\left\{-\frac{1}{2}\frac{|r[i]-H(i)\alpha {|}^2}{{\sigma }^2}\right\}}}\end{array}$$
又$\log\sum_{j}z_{j}\thickapprox max_{j}\log z_{j} $

LLR(bI,k)=log⁡∑α∈Sk0e{−12∣r[i]−H(i)α∣2σ2}∑α∈Sk1e{−12∣r[i]−H(i)α∣2σ2}≈log⁡maxα∈Sk0e{−12∣r[i]−H(i)α∣2σ2}maxα∈Sk1e{−12∣r[i]−H(i)α∣2σ2}=log⁡maxα∈Sk0e{−12∣r[i]−H(i)α∣2σ2}−log⁡maxα∈Sk1e{−12∣r[i]−H(i)α∣2σ2}=minα∈Sk0{−∣r[i]−H(i)α∣22σ2}−minα∈Sk1{−∣r[i]−H(i)α∣22σ2}=12σ2{minα∈Sk1∣r[i]−H(i)α∣2−minα∈Sk0∣r[i]−H(i)α∣2} \begin{align*} LLR(b_{I,k})&=\log\frac{\sum_{\alpha \in S_{k}^{0}}e^{\left\{ -\frac{1}{2}\frac{|r[i]-H(i)\alpha|^2}{\sigma^2} \right\}}}{\sum_{\alpha \in S_{k}^{1}}e^{\left\{ -\frac{1}{2}\frac{|r[i]-H(i)\alpha|^2}{\sigma^2} \right\}}}\\ &\thickapprox\log\frac{max_{\alpha \in S_{k}^{0}}e^{\left\{ -\frac{1}{2}\frac{|r[i]-H(i)\alpha|^2}{\sigma^2} \right\}}}{max_{\alpha \in S_{k}^{1}}e^{\left\{ -\frac{1}{2}\frac{|r[i]-H(i)\alpha|^2}{\sigma^2} \right\}}}\\ &=\log max_{\alpha \in S_{k}^{0}}e^{\left\{ -\frac{1}{2}\frac{|r[i]-H(i)\alpha|^2}{\sigma^2} \right\}}-\log max_{\alpha \in S_{k}^{1}}e^{\left\{ -\frac{1}{2}\frac{|r[i]-H(i)\alpha|^2}{\sigma^2} \right\}}\\ &=min_{\alpha\in S_{k}^{0}} \{-\frac{|r[i]-H(i)\alpha|^2}{2\sigma^2}\}-min_{\alpha\in S_{k}^{1}} \{-\frac{|r[i]-H(i)\alpha|^2}{2\sigma^2}\}\\ &=\frac{1}{2\sigma^2}\{ min_{\alpha \in S_{k}^{1}}|r[i]-H(i)\alpha|^2-min_{\alpha\in S_{k}^{0}} |r[i]-H(i)\alpha|^2 \} \end{align*} LLR(bI,k​)​=log∑α∈Sk1​​e{−21​σ2∣r[i]−H(i)α∣2​}∑α∈Sk0​​e{−21​σ2∣r[i]−H(i)α∣2​}​≈logmaxα∈Sk1​​e{−21​σ2∣r[i]−H(i)α∣2​}maxα∈Sk0​​e{−21​σ2∣r[i]−H(i)α∣2​}​=logmaxα∈Sk0​​e{−21​σ2∣r[i]−H(i)α∣2​}−logmaxα∈Sk1​​e{−21​σ2∣r[i]−H(i)α∣2​}=minα∈Sk0​​{−2σ2∣r[i]−H(i)α∣2​}−minα∈Sk1​​{−2σ2∣r[i]−H(i)α∣2​}=2σ21​{minα∈Sk1​​∣r[i]−H(i)α∣2−minα∈Sk0​​∣r[i]−H(i)α∣2}​
MMSE均衡矩阵如下：
$$\mathbf{W}=({\mathbf{H}}^{\mathbf{H}}+{\mathbf{N}}_0\mathbf{I}{)}^{-1}{\mathbf{H}}^{\mathbf{H}}$$
则$I$为单位矩阵大小为$N_r$,均衡后的输出可表示为：
$$\begin{array}{rl}\mathbf{z}& ={\mathbf{w}}^{\mathbf{H}}({\mathbf{h}}_{\mathbf{i}}{x}_i+\sum _{k\ne n}{\mathbf{h}}_{\mathbf{k}}{x}_k+\mathbf{n})\\ & =\underset{H(i)}{\underbrace{{\mathbf{w}}^{\mathbf{H}}{\mathbf{h}}_{\mathbf{i}}}}{x}_i+\underset{{\sigma }^2}{\underbrace{{\mathbf{w}}^{\mathbf{H}}(\sum _{k\ne n}{\mathbf{h}}_{\mathbf{k}}{x}_k+\mathbf{n})}}\end{array}$$
LLR(bI,k)=∣H(i)∣22σ2{minα∈Sk1∣x[i]−α∣2−minα∈Sk0∣x[i]−α∣2} \begin{align*} LLR(b_{I,k})&=\frac{|H(i)|^2}{2\sigma^2}\{\mathop {min}\limits_{\alpha \in S_{k}^{1}}|x[i]-\alpha|^2-\mathop {min}\limits_{\alpha\in S_{k}^{0}} |x[i]-\alpha|^2 \}\\ \end{align*} LLR(bI,k​)​=2σ2∣H(i)∣2​{α∈Sk1​min​∣x[i]−α∣2−α∈Sk0​min​∣x[i]−α∣2}​
注意很多参考文献将$LLR$定义为比特1概率比比特0，这个实际上没啥影响因为在后续译码只需要
$$\begin{gathered} LLR(b_{I,k})>0，判定为1 \\ LLR(b_{I,k})<0，判定为0 \end{gathered}$$
$$LLR(b_{I,k})=\log \frac{P[b_{I,k=1}|r[i]]}{P[b_{I,k=0}|r[i]]}$$
LLR(bI,k)=∣H(i)∣22σ2{minα∈Sk0∣x[i]−α∣2−minα∈Sk1∣x[i]−α∣2}≜∣H(i)∣22σ2.DI,kDI,k=minα∈Sk0∣x[i]−α∣2−minα∈Sk1∣x[i]−α∣2 \begin{align*} LLR(b_{I,k})&=\frac{|H(i)|^2}{2\sigma^2}\{\mathop {min}\limits_{\alpha \in S_{k}^{0}}|x[i]-\alpha|^2-\mathop {min}\limits_{\alpha\in S_{k}^{1}} |x[i]-\alpha|^2 \}\\ &\triangleq \frac{|H(i)|^2}{2\sigma^2}.D_{I,k}\\ D_{I,k}&=\mathop {min}\limits_{\alpha \in S_{k}^{0}}|x[i]-\alpha|^2-\mathop {min}\limits_{\alpha\in S_{k}^{1}} |x[i]-\alpha|^2 \end{align*} LLR(bI,k​)DI,k​​=2σ2∣H(i)∣2​{α∈Sk0​min​∣x[i]−α∣2−α∈Sk1​min​∣x[i]−α∣2}≜2σ2∣H(i)∣2​.DI,k​=α∈Sk0​min​∣x[i]−α∣2−α∈Sk1​min​∣x[i]−α∣2​
使用$\frac{2}{{\sigma }^2}$ 对$LLR$做下归一化,则上式可定义为：
LLR(bI,k)=∣H(i)∣24{minα∈Sk0∣x[i]−α∣2−minα∈Sk1∣x[i]−α∣2}≜∣H(i)∣24.DI,kDI,k=14minα∈Sk0∣x[i]−α∣2−minα∈Sk1∣x[i]−α∣2 \begin{align*} LLR(b_{I,k})&=\frac{|H(i)|^2}{4}\{\mathop {min}\limits_{\alpha \in S_{k}^{0}}|x[i]-\alpha|^2-\mathop {min}\limits_{\alpha\in S_{k}^{1}} |x[i]-\alpha|^2 \}\\ &\triangleq \frac{|H(i)|^2}{4}.D_{I,k}\\ D_{I,k}&=\frac{1}{4}\mathop {min}\limits_{\alpha \in S_{k}^{0}}|x[i]-\alpha|^2-\mathop {min}\limits_{\alpha\in S_{k}^{1}} |x[i]-\alpha|^2 \end{align*} LLR(bI,k​)DI,k​​=4∣H(i)∣2​{α∈Sk0​min​∣x[i]−α∣2−α∈Sk1​min​∣x[i]−α∣2}≜4∣H(i)∣2​.DI,k​=41​α∈Sk0​min​∣x[i]−α∣2−α∈Sk1​min​∣x[i]−α∣2​

后续为了方便和参考文献仿真对比我们也已比特1和比特0定义的$LLR$进行推到。
至此$LLR$的基本原理已经介绍完了，但是针对每一个比特计算$D_{I,k}$运算量比较大所以很多的文章又针对其做了很多简化处理已降低运算量。下面我们针对参考文献1中16QAM调制信号简化策略进行说明,其中bit位置顺序$(b_1,b_2,b_3,b_4)$：

第一幅图片中$b_1$为0的星座点为左半轴所有星座点的集合即$S_1^0$,$b_1$为1的星座点集合为右边半轴星座点即$S_1^1$,如图1所示星座点我们可知:
$$\begin{array}{r}{x}_I[i]>2\\ x_Q[i]>2\end{array}$$
我们把bit为1称为正比特用$+$标记，0称为反比特用$-$标记，所以计算$b_1的llr$时，与$x[i]$距离最近的$+$比特星座点为{3+3j}\{3+3j\}{3+3j}，反比特的星座点为{−1+3j}\{-1+3j\}{−1+3j},带入$D_{I,1}$可得：
DI,1=14minα∈Sk0∣x[i]−α∣2−minα∈Sk1∣x[i]−α∣2=14{∣xI[i]+xQ[i]j−(−1+3j)∣2−∣xI[i]+xQ[i]j−(3+3j)∣2}=14{(xI[i]+1)2+(xQ[i]−3)2−(xI[i]−3)2−(xQ[i]−3)2}=2(xI[i]−1)−−−xI[i]>2 \begin{align*} D_{I,1}&=\frac{1}{4}\mathop {min}\limits_{\alpha \in S_{k}^{0}}|x[i]-\alpha|^2-\mathop {min}\limits_{\alpha\in S_{k}^{1}} |x[i]-\alpha|^2 \\ &=\frac{1}{4}\{|x_{I}[i]+x_{Q}[i]j-(-1+3j)|^2-|x_{I}[i]+x_{Q}[i]j-(3+3j)|^2\}\\ &=\frac{1}{4}\{(x_{I}[i]+1)^2+(x_{Q}[i]-3)^2-(x_{I}[i]-3)^2-(x_{Q}[i]-3)^2\}\\ &=2(x_{I}[i]-1) --- x_{I}[i]>2 \end{align*} DI,1​​=41​α∈Sk0​min​∣x[i]−α∣2−α∈Sk1​min​∣x[i]−α∣2=41​{∣xI​[i]+xQ​[i]j−(−1+3j)∣2−∣xI​[i]+xQ​[i]j−(3+3j)∣2}=41​{(xI​[i]+1)2+(xQ​[i]−3)2−(xI​[i]−3)2−(xQ​[i]−3)2}=2(xI​[i]−1)−−−xI​[i]>2​
同理针对$x[i]$可能的星座点位置可得：

$$\begin{array}{rl}{D}_{I,1}& =\{\begin{array}{ll}{x}_I[i],& |x_I[i]|\le 2\\ 2(x_I[i]-1),& x_I[i]>2\\ 2(x_I[i]+1),& x_I[i]<-2\end{array}\\ D_{I,2}& =-|x_I[i]|+2\end{array}$$
同样的原理针对64QAM信号有如下公式(I路举例)：

$$\begin{array}{rl}{D}_{I,1}& =\{\begin{array}{ll}{x}_I[i],& |x_I[i]|\le 2\\ 2(x_I[i]-1),& 2<x_I[i]\le 4\\ 3(x_I[i]-2),& 4<x_I[i]\le 6\\ 4(x_I[i]-3),& x_I[i]>6\\ 2(x_I[i]+1),& -4\le x_I[i]<-2\\ 3(x_I[i]+2),& -6\le x_I[i]<-4\\ 4(x_I[i]+3),& x_I[i]<-6>\end{array}\\ D_{I,2}& =\{\begin{array}{ll}2(-|x_I[i]|+3),& |x_I[i]|\le 2\\ 4-|x_I[i]|,& 2<|x_I[i]|\le 6\\ 2(-|x_I[i]|+5),& |x_I[i]|>6\end{array}\\ D_{I,3}& =\{\begin{array}{ll}|x_I[i]|-2,& |x_I[i]|\le 4\\ -|x_I[i]|+6,& x_I[i]>4\end{array}\end{array}$$
同理256QAM信号有如下公式：
$b_0$星座点

$b_1$星座点

$b_2$星座点

$b_3$星座点

$$\begin{array}{rl}{D}_{I,1}& =\{\begin{array}{ll}8\ast (x_I[i]+7),& x_I[i]<-14\\ 7\ast (x_I[i]+6),& -14\le x_I[i]<-12\\ 6\ast (x_I[i]+5),& -12\le x_I[i]<-10\\ 5\ast (x_I[i]+4),& -10\le x_I[i]<-8\\ 4\ast (x_I[i]+3),& -8\le x_I[i]<-6\\ 3\ast (x_I[i]+2),& -6\le x_I[i]<-4\\ 2\ast (x_I[i]+2),& -4\le x_I[i]<-2\\ x_I[i],& -2\le x_I[i]<2\\ 2\ast (x_I[i]-1),& 2\le x_I[i]<4\\ 3\ast (x_I[i]-2),& 4\le x_I[i]<6\\ 4\ast (x_I[i]-3),& 6\le x_I[i]<8\\ 5\ast (x_I[i]-4),& 8\le x_I[i]<10\\ 6\ast (x_I[i]-5),& 10\le x_I[i]<12\\ 7\ast (x_I[i]-6),& 12\le x_I[i]<14\\ 8\ast (x_I[i]-7),& 14\le x_I[i]\end{array}\\ & \\ D_{I,2}& =\{\begin{array}{ll}4\ast (-|x_I[i]|+11),& |x_I[i]|\ge 14\\ 3\ast (-|x_I[i]|+10),& 12\le |x_I[i]|<14\\ 2\ast (-|x_I[i]|+9),& 10\le |x_I[i]|<12\\ 1\ast (-|x_I[i]|+8),& 6\le |x_I[i]|<10\\ 2\ast (-|x_I[i]|+7),& 4\le |x_I[i]|<6\\ 3\ast (-|x_I[i]|+6),& 2\le |x_I[i]|<4\\ 4\ast (-|x_I[i]|+5),& |x_I[i]|<2\end{array}\\ & \\ D_{I,3}& =\{\begin{array}{ll}2\ast (-|x_I[i]|+13),& 14\le |x_I[i]|\\ 1(-|x_I[i]|+12),& 10\le |x_I[i]|<14\\ 2(-|x_I[i]|+11),& 8\le |x_I[i]|<10\\ 2(|x_I[i]|-5),& 6\le |x_I[i]|<8\\ |x_I[i]|-4,& 2\le |x_I[i]|<6\\ 2(|x_I[i]|-3),& |x_I[i]|<2\end{array}\\ & \\ D_{I,4}& =\{\begin{array}{ll}-|x_I[i]|+14,& 12\le |x_I[i]|\\ |x_I[i]|-10,& 8\le |x_I[i]|<12\\ -|x_I[i]|+6,& 4\le |x_I[i]|<8\\ |x_I[i]|-2,& |x_I[i]|<4\end{array}\end{array}$$
同理针对Q路是同样的原理只需要，将$x_I[i]$替换成$x_Q[i]$。

更进一步的简化：
16QAM

$$\begin{array}{rl}{D}_{I,1}& =x_I[i],\\ D_{I,2}& =-|D_{I,1}|+2\end{array}$$

64QAM

$$\begin{array}{rl}{D}_{I,1}& \backsimeq x_I[i],\\ D_{I,2}& \backsimeq -|D_{I,1}|+4\\ D_{I,3}& \backsimeq -|D_{I,2}||+2\end{array}$$
256QAM

$$\begin{array}{rl}{D}_{I,1}& \backsimeq x_I[i],\\ D_{I,2}& \backsimeq -|D_{I,1}|+8\\ D_{I,3}& \backsimeq -|D_{I,2}|+4\\ D_{I,4}& \backsimeq -|D_{I,3}|+2\end{array}$$

假设QAM信号比特数目为$m$，令$d=2^{\frac{m}{2}-1},m=4,6,8$
$$\begin{array}{rl}{D}_{I,1}& \backsimeq x_I[i],\\ D_{I,i}& \backsimeq -|D_{I,1}|+d>>(i-1),i=0:1:(2,3,4)\end{array}$$

【1】Simplified Soft-Output Demapper for Binary Interleaved COFDM with Application to HIPERLAN/2
【2】A Low Complexity 256QAM Soft Demapper for 5G Mobile System