CRC校验码多项式问题

Question

CRC校验算法，这是一段从设备上采集的一段MSB数据：
1 2 3 4 5 6 7 8 9 10 11
41 C0 0 0 0 0 0 0 0 0 48
41 C0 80 0 0 0 0 0 0 0 38
41 C0 80 0 0 0 0 0 0 0 38
41 C0 40 0 0 0 0 0 0 0 70
41 C0 40 0 0 0 0 0 0 0 70
41 C0 C0 0 0 0 0 0 0 0 0
41 C0 C0 0 0 0 0 0 0 0 0
41 C0 20 0 0 0 0 0 0 0 F8
41 C0 20 0 0 0 0 0 0 0 F8
41 C0 A0 0 0 0 0 0 0 0 88
41 C0 A0 0 0 0 0 0 0 0 88
41 C0 60 0 0 0 0 0 0 0 C0
41 C0 60 0 0 0 0 0 0 0 C0
41 C0 E0 0 0 0 0 0 0 0 B0
41 C0 E0 0 0 0 0 0 0 0 B0
41 C0 10 0 0 0 0 0 0 0 10
41 C0 10 0 0 0 0 0 0 0 10
41 C0 90 0 0 0 0 0 0 0 60
41 C0 90 0 0 0 0 0 0 0 60
41 C0 50 0 0 0 0 0 0 0 28
41 C0 50 0 0 0 0 0 0 0 28
41 C0 D0 0 0 0 0 0 0 0 58
41 C0 D0 0 0 0 0 0 0 0 58
41 C0 30 0 0 0 0 0 0 0 A0
41 C0 30 0 0 0 0 0 0 0 A0
41 C0 B0 0 0 0 0 0 0 0 D0
41 C0 B0 0 0 0 0 0 0 0 D0
41 C0 70 0 0 0 0 0 0 0 98
41 C0 70 0 0 0 0 0 0 0 98
41 C0 F0 0 0 0 0 0 0 0 E8
41 C0 F0 0 0 0 0 0 0 0 E8

下面是一段LSB采集的数据
82 3 0 0 0 0 0 0 0 0 12
82 3 1 0 0 0 0 0 0 0 1C
82 3 1 0 0 0 0 0 0 0 1C
82 3 2 0 0 0 0 0 0 0 0E
82 3 2 0 0 0 0 0 0 0 0E
82 3 3 0 0 0 0 0 0 0 0
82 3 3 0 0 0 0 0 0 0 0
82 3 4 0 0 0 0 0 0 0 1F
82 3 4 0 0 0 0 0 0 0 1F
82 3 5 0 0 0 0 0 0 0 11
82 3 5 0 0 0 0 0 0 0 11
82 3 6 0 0 0 0 0 0 0 3
82 3 6 0 0 0 0 0 0 0 3
82 3 7 0 0 0 0 0 0 0 0D
82 3 7 0 0 0 0 0 0 0 0D
82 3 8 0 0 0 0 0 0 0 8
82 3 8 0 0 0 0 0 0 0 8
82 3 9 0 0 0 0 0 0 0 6
82 3 9 0 0 0 0 0 0 0 6
82 3 0A 0 0 0 0 0 0 0 14
82 3 0A 0 0 0 0 0 0 0 14
82 3 0B 0 0 0 0 0 0 0 1A
82 3 0B 0 0 0 0 0 0 0 1A
82 3 0C 0 0 0 0 0 0 0 5
82 3 0C 0 0 0 0 0 0 0 5
82 3 0D 0 0 0 0 0 0 0 0B
82 3 0D 0 0 0 0 0 0 0 0B
82 3 0E 0 0 0 0 0 0 0 19
82 3 0E 0 0 0 0 0 0 0 19
82 3 0F 0 0 0 0 0 0 0 17
82 3 0F 0 0 0 0 0 0 0 17
82 3 10 0 0 0 0 0 0 0 13
82 3 10 0 0 0 0 0 0 0 13
82 3 11 0 0 0 0 0 0 0 1D
82 3 11 0 0 0 0 0 0 0 1D
82 3 12 0 0 0 0 0 0 0 0F
82 3 12 0 0 0 0 0 0 0 0F
82 3 13 0 0 0 0 0 0 0 1
82 3 13 0 0 0 0 0 0 0 1
82 3 14 0 0 0 0 0 0 0 1E
82 3 14 0 0 0 0 0 0 0 1E
82 3 15 0 0 0 0 0 0 0 10
82 3 15 0 0 0 0 0 0 0 10
82 3 16 0 0 0 0 0 0 0 2
82 3 16 0 0 0 0 0 0 0 2
82 3 17 0 0 0 0 0 0 0 0C
82 3 17 0 0 0 0 0 0 0 0C
82 3 18 0 0 0 0 0 0 0 9
82 3 18 0 0 0 0 0 0 0 9
82 3 19 0 0 0 0 0 0 0 7

最后一个字节为校验码,可能是帧校验序列码FCS，谁能帮忙分析一下这个校验码的规律

Answer 1

似乎找到了。是5bit的CRC，多项式对应的值是0x0B，计算结果后面用3个0补足为8bit，所有数据都通过这个在线CRC计算网页验证过

CRC（循环冗余校验）在线计算_ip33.com CRC（循环冗余校验）在线计算 http://www.ip33.com/crc.html

#include <stdio.h>
#include <stdint.h>

uint8_t crc5(const uint8_t *data, size_t length) {
    uint8_t crc = 0x00;
    for (size_t i = 0; i < length; i++) {
        crc ^= data[i];
        for (size_t j = 0; j < 8; j++) {
            if (crc & 0x80) {
                crc = (crc << 1) ^ 0x58; //// 0x58 = 0x0B<<(8-5)
            } else {
                crc <<= 1;
            }
        }
    }
    return crc ;
}
 

int main() {
    uint8_t data1[] = {0x41, 0xC0, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00};
    uint8_t data2[] = {0x41, 0xC0, 0x80, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00};
    uint8_t crc1;
    uint8_t crc2;
 
    crc1 = crc5(data1, sizeof(data1));
    crc2 = crc5(data2, sizeof(data2));
     printf("CRC5: 0x%02X - 0x%02X\n", crc1, crc2);
    return 0;
}