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基于蒙特卡罗 AWGN信道调制信号互信息(信道容量)

news 来源：原创 2025/5/25 10:30:02

假设高斯白噪声信道：

$z_{n}=a_{n}+w_{n}$

$a_{n}$ 定义为实或者复离散信号， $w_{n}$ 表示均值为0，方差为 $\sigma^2$ (实信号)或者 $2\sigma^2$ (复信号)，则SNR可以定义为：

$SNR=\frac{E\{|a_{n}|^2\}}{E\{|w_{n}|^2\}}=\left\{\begin{matrix} \frac{E\{|a_{n}|^2\}}{\sigma^2}---real \ signal\\ \frac{E\{|a_{n}|^2\}}{2\sigma^2}--complex\ signal \end{matrix}\right.$

根据信息论知识：

$I(Z,A)=max\{\iint_{}^{}p(z,a)log(\frac{p(z|a)}{p(z)})dadz \}=max\{H(Z)-H(Z|A)\}$

$H(Z)=-\sum p(z)log p(z),H(Z|A)=-\sum\limits_{a}^{}\sum \limits_{z}p(z,a)logp(z|a)$

将 $z$ 扩展至连续信号:

$I(Z,A)=max \{\sum \limits_{a}\sum \limits_{z}p(z,a)log\ p(z|a)-\sum \limits_{a}\sum \limits_{z}p(z,a)log\ p(z)\}$

$I(Z,A)=max\{\sum \limits_{a} \int p(z,a)log\ p(z|a)dz-\sum \limits_{a} \int p(z,a)log\ p(z)dz\}$

假设 $A$ 为离散信号 $(a_{0},a_{1},a_{2}\cdots ,a_{N-1})$ , $Q(k)$ 是 $a_{k}$ 出现概率则：

$p(z,a)=\sum \limits_{k=0}^{N-1}Q(k)p(z|a_{k})$

带入上式，假设离散信号是等概率出现，上式中的max可以去掉(有限信源等概率时信道容量最大，这个我记得张灏老师的随机过程好像提到过)：

$I(Z,A)\\ =\sum \limits_{a} \int Q(k)p(z|a_{k})log\ p(z|a)dz-\sum \limits_{a} \int A(k)p(z|a_{k})log\ p(z)dz\\ =\sum \limits_{a}Q(k)\{\int p(z|a_{k})log\ p(z|a_{k}))dz-\int p(z|a_{k}p(z)dz) \}\\ =\sum \limits_{a}Q(k)\{ \int p(z|a_{k})(log(p(z|a_{k})-logp(z)))dz \}$

$=\sum \limits_{k}Q(k)\{\int p(z|a_{k})log\frac{p(z|a_{k})}{\sum \limits_{i=0}^{N-1}Q(k)p(z|a_{k})}dz \}\\ =\sum \limits_{k}\frac{1}{N}\{\int p(z|a{k})log\frac{p(z|a_{k})}{\sum \limits_{i=0}^{N-1}\frac{1}{N}p(z|a_{k}) }dz \}\\ =\sum \limits_{k}\frac{1}{N}\{\int p(z|a{k})log\frac{Np(z|a_{k})}{\sum \limits_{i=0}^{N-1}p(z|a_{k}) }dz \}$

$=\sum \limits_{k}\frac{1}{N}\{\int p(z|a_{k})(logN-log\frac{\sum \limits_{i=0}^{N-1}p(z|a_{i})}{p(z|a_{k})} )dz \}\\ =\sum \limits_{k}\frac{1}{N}\{\int p(z|a_{k})logNdz\}-\sum \limits_{k}\frac{1}{N}\{ \int p(z|a_{k})log\frac{\sum \limits_{i=0}^{N-1}p(z|a_{i})}{p(z|a_{k})}dz \}\\ =logN\sum \limits_{k}\int p(z|a_{k})Q(k)dz-\sum \limits_{k}\frac{1}{N}\int p(z|a_{k})log\frac{\sum \limits_{i=0}^{N-1}p(z|a_{i})}{p(z|a_{k})}dz$

$=logN-\sum \limits_{k}\frac{1}{N}\int p(z|a_{k})log\frac{\sum \limits_{i=0}^{N-1}p(z|a_{i})}{p(z|a_{k})}dz \ ( \sum \limits_{k}\int p(z|a_{k}Q(k))dz=1)$

总结：离散等概率信号发生最大互信息可表示为：

$I(Z,A)=logN-\sum \limits_{k}\frac{1}{N}\int p(z|a_{k})log\frac{\sum \limits_{i=0}^{N-1}p(z|a_{i})}{p(z|a_{k})}dz$

当信道为高斯白噪声信道： $z_{n}=a_{n}+w_{n}, w\sim C(0,\sigma^2)$ ,所以 $z_{n}-a_{n}$ 服从 $C(0,\sigma^2)$ ：

$p(z|a)=e^{\frac{-|z-a_{k}|^2}{2\sigma^2}}*\left\{\begin{matrix} (2\pi\sigma^2)^{-1/2}---real \ signal\\ (2\pi\sigma^2)^{-1}--complex \ signal \end{matrix}\right.$

带入上式：

$I(Z,A)=logN-\sum \limits_{k}\frac{1}{N}\int p(z|a_{k})log\frac{\sum \limits_{i=0}^{N-1}p(z|a_{i})}{p(z|a_{k})}dz\\ =logN-\frac{1}{N}\sum \limits_{k}\int p(z|a_{k})log\sum \limits_{i=0}^{N-1}e^{\frac{-(z-a_{i}^2)+(z-a_{k})^2}{2\sigma^2}}dz$

采用蒙特卡洛重要采样仿真令 $F(x)=g(x)f(x)$ 则 $\int F(x)dx=E\left ( \frac{F(x)}{f(x)} \right )=E(g(x))$

如果 $f(x)=p(z|a_{k})=e^{\frac{-|z-a_{k}|^2}{2\sigma^2}}*\left\{\begin{matrix} (2\pi\sigma^2)^{-1/2}\\ (2\pi\sigma^2)^{-1} \end{matrix}\right.$ , $g(x)=log\sum \limits_{i=0}^{N-1}e^{(\frac{-(z-a_{i})^2+(z-a_{k})^2}{2\sigma^2})}$ 则：