信道容量

128 阅读 0 评论 85 点赞

我是靠谱客的博主矮小招牌，这篇文章主要介绍信道容量，现在分享给大家，希望可以做个参考。

信道容量的迭代算法

实验目的
熟悉信道容量的迭代算法；
学习如何将复杂的公式转化为程序。
实验要求
已知：信源符号个数r、信宿符号个数s、信道转移概率矩阵P。
输入：任意的一个信道转移概率矩阵，r、s、P在运行时从键盘输入。
输出：最佳信源分布P’,信道容量C。
实验代码

复制代码

package com.yrwan.xindao;
import java.util.Scanner;
public class Diedai {
    public static int M;
    public static int N;
    public static double C;
    public static void main(String[] args) {
        int t = 0; // 迭代次数
        Scanner input = new Scanner(System.in);
        System.out.println("输入状态转移矩阵的行列数M，N：");
        N = input.nextInt();
        M = input.nextInt();
        double[][] p = new double[N][M];
        double[] S = new double[N];
        double[] SS = new double[N];    // 存入迭代计算后的S
        System.out.println("输入状态转移矩阵：");
        // 输入转移矩阵
        for (int i = 0; i < N; i++)
            for (int j = 0; j < M; j++) {
                p[i][j] = input.nextDouble();
            }
        input.close();
        // 初始化数据
        for (int i = 0; i < N; i++)
            S[i] = (double) 1 / N;// 赋值
        // 迭代
        boolean flag = true;
        double[][] fa = new double[M][N];
        while (flag) {
            // 计算fa[j][i]
            for (int j = 0; j < M; j++) {
                double sum = 0;
                for (int i = 0; i < N; i++) {
                    sum = sum + S[i] * p[i][j];
                }
                for (int i = 0; i < N; i++)
                    fa[j][i] = (S[i] * p[i][j]) / sum;
            }
            // 迭代计算S
            // 计算分母
            double sum1 = 0;
            for (int i = 0; i < N; i++) {
                boolean flag3 = true;
                double sum2 = 0;
                for (int j = 0; j < M; j++) {
                    if (fa[j][i] != 0)
                        sum2 = sum2+ (p[i][j] * (Math.log(fa[j][i]) / Math.log(Math.E)));
                    else if (fa[j][i] == 0 && p[i][j] != 0) {
                        flag3 = false;
                    }// exp(log0)=0,下同
                    else if (fa[j][i] == 0 && p[i][j] == 0)
                        sum2 = sum2 + 0;// 0log0=0
                }
                if (flag3)
                    sum1 = sum1 + Math.exp(sum2);
                else
                    sum1 = sum1 + 0;
            }
            // 计算SS[i]4
            for (int i = 0; i < N; i++) {
                boolean flag1 = true;// 若有无穷比无穷
                double sum6 = 0;
                for (int j = 0; j < M; j++) {
                    if (fa[j][i] != 0)
                        sum6 = sum6 + p[i][j]* (Math.log(fa[j][i]) / Math.log(Math.E));
                    else if (fa[j][i] == 0 && p[i][j] != 0) {
                        flag1 = false;
                    } else if (fa[j][i] == 0 && p[i][j] == 0)
                        sum6 = sum6 + 0;
                }
                if (flag1)
                    SS[i] = Math.exp(sum6) / sum1;
                else
                    SS[i] = 0;
            }
            double epsilon = 0; //停止迭代的容差限
            for (int i = 0; i < N; i++) {
                epsilon = epsilon + Math.pow(SS[i] - S[i], 2);
            }
            if (epsilon < 0.00001)
                flag = false;
            else {
                for (int i = 0; i < N; i++)
                    S[i] = SS[i];
            }
            C = Math.log(sum1) / Math.log(2);
            t++;
        }
        System.out.println("迭代次数为：" + t);
        System.out.print("最佳信源分布为：");
        for (int i = 0; i < N - 1; i++)
            System.out.print(S[i] + "，");
        System.out.println(S[N - 1]);
        System.out.println("信道容量为：" + C);
    }
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
package com.yrwan.xindao;
import java.util.Scanner;
public class Diedai {
    public static int M;
    public static int N;
    public static double C;
    public static void main(String[] args) {
        int t = 0; // 迭代次数
        Scanner input = new Scanner(System.in);
        System.out.println("输入状态转移矩阵的行列数M，N：");
        N = input.nextInt();
        M = input.nextInt();
        double[][] p = new double[N][M];
        double[] S = new double[N];
        double[] SS = new double[N];    // 存入迭代计算后的S
        System.out.println("输入状态转移矩阵：");
        // 输入转移矩阵
        for (int i = 0; i < N; i++)
            for (int j = 0; j < M; j++) {
                p[i][j] = input.nextDouble();
            }
        input.close();
        // 初始化数据
        for (int i = 0; i < N; i++)
            S[i] = (double) 1 / N;// 赋值
        // 迭代
        boolean flag = true;
        double[][] fa = new double[M][N];
        while (flag) {
            // 计算fa[j][i]
            for (int j = 0; j < M; j++) {
                double sum = 0;
                for (int i = 0; i < N; i++) {
                    sum = sum + S[i] * p[i][j];
                }
                for (int i = 0; i < N; i++)
                    fa[j][i] = (S[i] * p[i][j]) / sum;
            }
            // 迭代计算S
            // 计算分母
            double sum1 = 0;
            for (int i = 0; i < N; i++) {
                boolean flag3 = true;
                double sum2 = 0;
                for (int j = 0; j < M; j++) {
                    if (fa[j][i] != 0)
                        sum2 = sum2+ (p[i][j] * (Math.log(fa[j][i]) / Math.log(Math.E)));
                    else if (fa[j][i] == 0 && p[i][j] != 0) {
                        flag3 = false;
                    }// exp(log0)=0,下同
                    else if (fa[j][i] == 0 && p[i][j] == 0)
                        sum2 = sum2 + 0;// 0log0=0
                }
                if (flag3)
                    sum1 = sum1 + Math.exp(sum2);
                else
                    sum1 = sum1 + 0;
            }
            // 计算SS[i]4
            for (int i = 0; i < N; i++) {
                boolean flag1 = true;// 若有无穷比无穷
                double sum6 = 0;
                for (int j = 0; j < M; j++) {
                    if (fa[j][i] != 0)
                        sum6 = sum6 + p[i][j]* (Math.log(fa[j][i]) / Math.log(Math.E));
                    else if (fa[j][i] == 0 && p[i][j] != 0) {
                        flag1 = false;
                    } else if (fa[j][i] == 0 && p[i][j] == 0)
                        sum6 = sum6 + 0;
                }
                if (flag1)
                    SS[i] = Math.exp(sum6) / sum1;
                else
                    SS[i] = 0;
            }
            double epsilon = 0; //停止迭代的容差限
            for (int i = 0; i < N; i++) {
                epsilon = epsilon + Math.pow(SS[i] - S[i], 2);
            }
            if (epsilon < 0.00001)
                flag = false;
            else {
                for (int i = 0; i < N; i++)
                    S[i] = SS[i];
            }
            C = Math.log(sum1) / Math.log(2);
            t++;
        }
        System.out.println("迭代次数为：" + t);
        System.out.print("最佳信源分布为：");
        for (int i = 0; i < N - 1; i++)
            System.out.print(S[i] + "，");
        System.out.println(S[N - 1]);
        System.out.println("信道容量为：" + C);
    }
}

二元对称信道模拟器

实验目的
加深理解二进制对称信道的工作原理；
掌握通过高级编程语言生成伪随机数的方法。
实验要求
输入：BSC信道的错误概率，任意的二进制序列
输出：经BSC信道传输后的二进制序列
源程序格式整齐清晰，注释简单明了
实验代码

复制代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
package com.yrwan.xindao;
import java.util.Scanner;
public class Eryuanduichen {
    public static void main(String[] args){
        double errPr;
        int[] data;
        Scanner input = new Scanner(System.in);
        System.out.print("输入BSC信道的错误概率(0~1)：");
        errPr=input.nextDouble();
        System.out.print("输入要传输的二进制序列长度：");
        data=new int[input.nextInt()];
        System.out.print("输入要传输的二进制序列：");
        for (int i = 0; i < data.length; i++) {
            data[i]=input.nextInt();
        }
        input.close();
        BSC bsc=new BSC(errPr);
        int[] result=bsc.send(data);
        System.out.printf("传输后的二进制序列为：");
        for (int r:result) {
            System.out.printf(r+" ");
        }
    }
}

复制代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
package com.yrwan.xindao;
public class BSC {//第一种方法
    double errPr;//BSC信道的错误概率
    public BSC(double Pr){
        errPr = Pr;
    }
//经BSC传输信号，返回传输后的值
    public int[] send(int[] data){
        boolean [] x=int2boolean(data);
        for(int i = 0; i<x.length;i++)
            if(Math.random()<errPr){
                x[i] = !x[i];
            }
        return boolean2int(x);
    }
 //将int型数组转为boolean型
    private boolean[] int2boolean(int[] data){
        boolean [] x=new boolean[data.length];
        for (int i = 0; i < data.length; i++) {
            x[i] = data[i] == 1;
        }
        return x;
    }
 //将boolean型数组转为int型
    private int[] boolean2int(boolean[] data){
        int[] x=new int[data.length];
        for (int i = 0; i < data.length; i++) {
            x[i]=data[i]?1:0;
        }
        return x;
    }
}

复制代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
package com.yrwan.xindao;
public class BSC {//第二种方法
    double errPr;//BSC信道的错误概率
    public BSC(double Pr){
        errPr = Pr;
    }
//经BSC传输信号，返回传输后的值
    public int[] send(int[] data){
        int[] x=new int[data.length];
        x = data;
        for(int i = 0; i<data.length;i++)
            if(Math.random()<errPr){
                x[i] = 1 - x[i];
            }
        return x;
    }
}

复制代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
//C语言版
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
void BSC(int *p, int n, float errPr){
  srand(time(NULL));
  int i;
  for(i = 0; i < n; i++)
    if((double)rand()/RAND_MAX < errPr)
      *(p+i) = 1 - *(p+i);
}
int main(void){
  float errPr;
  printf("请输入信道的错误概率：");
  scanf("%f",&errPr);
  printf("请输入二进制序列长度：");
  int n;
  scanf("%d",&n);
  int data[n];
  printf("请输入二进制序列：");
  int i;
  for(i=0; i < n; i++)
    scanf("%d",&data[i]);
  BSC(data, n, errPr);
  printf("经信道传输后的二进制序列为：");
  for(i=0; i < n; i++)
    printf("%d ",data[i]);
  return 0;
}