Question

我使用C ++中非常基本的CRC子例程创建了CCSDS BCH（64,56）编码器。我打算在GNU Radio应用程序中使用它。 BCH（64,56）码块的格式如下所示。。可以将一组代码块组合起来，以形成一个数据单元，称为数据链路传输单元（CLTU），如下所示。

据我了解，BCH和CRC都使用相同的计算形式在数据末尾附加了“余数/奇偶校验”，如this线程中所述。

我要做的就是修改标准的CRC C ++ crcFast() subroutine 。该子例程通过迭代由给定多项式（crcInit（））预初始化的数组（表）来计算CRC。下面的代码显示了两个子例程crcInit（）和crcFast（）。

typedef unsigned char uint8_t;
typedef unsigned short crc;
#define WIDTH (8 * sizeof(crc))
#define TOPBIT (1 << (WIDTH - 1))
crc crcTable[256];
#define POLYNOMIAL 0xD8  /* 11011 followed by 0's */
void
crcInit(void)
{
    crc remainder;


    /*
* Compute the remainder of each possible dividend.
*/
    for (int dividend = 0; dividend < 256; ++dividend)
    {
        /*
* Start with the dividend followed by zeros.
*/
        remainder = dividend << (WIDTH - 8);

        /*
* Perform modulo-2 division, a bit at a time.
*/
        for (uint8_t bit = 8; bit > 0; --bit)
        {
            /*
* Try to divide the current data bit.
*/
            if (remainder & TOPBIT)
            {
                remainder = (remainder << 1) ^ POLYNOMIAL;
            }
            else
            {
                remainder = (remainder << 1);
            }
        }

        /*
* Store the result into the table.
*/
        crcTable[dividend] = remainder;
    }

} /* crcInit() */


crc
crcFast(uint8_t const message[], int nBytes)
{
    uint8_t data;
    crc remainder = 0;


    /*
     * Divide the message by the polynomial, a byte at a time.
     */
    for (int byte = 0; byte < nBytes; ++byte)
    {
        data = message[byte] ^ (remainder >> (WIDTH - 8));
        remainder = crcTable[data] ^ (remainder << 8);
    }

    /*
     * The final remainder is the CRC.
     */
    return (remainder);

}   /* crcFast() */

修改后的代码如下所示。表生成函数crcInit（）不变。对crcFast算法进行了稍微修改，以将格式所指定的奇偶校验字节（补码和填充位）的更改并入其中。 CRC类型已从short更改为unsigned char（1个字节）。排除汉明码中的BCH（64,56），其生成多项式为g（x）= x ^ 7 + x ^ 6 + x ^ 2 +1，在我看来相当于0xC5。

#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
#include <vector>
#include <iostream>
#include "debug.h"
typedef unsigned char uint8_t;
typedef unsigned char crc;
#define WIDTH  (8 * sizeof(crc))
#define TOPBIT (1 << (WIDTH - 1))
crc  crcTable[256];
#define POLYNOMIAL 0xC5  // x^7 + x^6 + x^2 + 1
#define INITIAL_REMAINDER 0x00
#define BCH_INFORMATION_BLOCK 7
void
crcInit(void)
{
    crc  remainder;


    /*
     * Compute the remainder of each possible dividend.
     */
    for (int dividend = 0; dividend < 256; ++dividend)
    {
        /*
         * Start with the dividend followed by zeros.
         */
        remainder = dividend << (WIDTH - 8);

        /*
         * Perform modulo-2 division, a bit at a time.
         */
        for (uint8_t bit = 8; bit > 0; --bit)
        {
            /*
             * Try to divide the current data bit.
             */         
            if (remainder & TOPBIT)
            {
                remainder = (remainder << 1) ^ POLYNOMIAL;
            }
            else
            {
                remainder = (remainder << 1);
            }
        }

        /*
         * Store the result into the table.
         */
        crcTable[dividend] = remainder;
    //std::cout << "Remainder from table : " << int (remainder&0xffff) << std::endl;
    }

}   /* crcInit() */

void
crcEncoder(std::vector<unsigned char> &message, const crc initial_remainder)
{
    uint8_t data;
    crc remainder = initial_remainder;

    /*
     * Divide the message by the polynomial, a byte at a time.
     */
    for (int byte = 0; byte < message.size(); ++byte)
    {
        data = message.at(byte) ^ (remainder >> (WIDTH - 8));
        remainder = crcTable[data] ^ (remainder << 8);
    }

    //Flip the remainder and move by 1 bit
    remainder ^= 0xFF;
    remainder <<= 1;

    //Set filler bit to 0 (anding with 1111 1110)
    remainder &= 0xFE;

    /*
     * The final remainder is the CRC.
     */
    message.push_back(remainder);
    //return message;
}


void bchEncoder(std::vector<unsigned char> &message)
{
    std::vector<unsigned char> information; // 7 bytes
    std::vector<unsigned char> codewords; // Encoded message

    //Ensure integral information symbols
    while(!(message.size() % BCH_INFORMATION_BLOCK) == 0)
      {
        message.push_back(0x55);
      }

    for(int i = 0; i < message.size(); i += BCH_INFORMATION_BLOCK)
    {
        //Copy 7 information bytes
        std::copy(message.begin() + i, message.begin() + i + BCH_INFORMATION_BLOCK,
                      std::back_inserter(information));
        //BCH encoding
        crcEncoder(information,INITIAL_REMAINDER);

        //Copy encoded information bits
        codewords.insert(codewords.end(), information.begin(), information.end());

        //Clear information bytes
        information.clear();
    }
    message = codewords;
}


int main()
{
  crcInit();
  //hexdump(crcTable,256);
  unsigned char message[] = {0xaa, 0xbb, 0xcd, 0xdd, 0xee, 0xff, 0x11,0x00};
  //unsigned char tail[] = {0xC5,0xC5,0xC5,0xC5,0xC5,0xC5,0xC5,0x79};
  std::vector<unsigned char> info(message, message + sizeof(message)/sizeof(unsigned char));

  bchEncoder(info);
  hexdump(info.data(),info.size());

  //Vector hex dump

  return 0;
}

我某种程度上觉得我的方法太幼稚了。我想知道它是否准确。

此致

Answer 1

如果BCH码的距离为3，仅用于纠正单个位错误或检测（但不纠正）所有两个位错误，则BCH多项式将与字段多项式相同。如果需要更高级别的校正或检测，则BCH多项式会变得复杂。 Wiki文章对此进行了解释：

https://en.wikipedia.org/wiki/BCH_code

由于消息长度（包括填充位）为64位，因此使用的是7位字段（最多127位有效），但是所示的表生成是针对多项式0x1C5的。要解决此问题，请将POLYNOMIAL更改为0x8A，即（（0xC5 << 1）＆0xFE））。这将导致7位奇偶校验最终以字节的高7位结尾。

编码循环的内部应该是：

  Application.Wait Now + TimeValue("00:00:05")
  SendMessage h, WM_CLOSE, 0, 0

使用基本的CRC C ++子例程来实现CCSDS BCH（64,56）FEC代码

1 个答案: