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我已经计算了奇偶校验位,它们都是不同的。所以我添加不同奇偶校验位的点来找出错误的位,我得到 7(而总共有 6 位)。

如何确定哪个位有错误。每个数据位都覆盖有 2 个奇偶校验位。因此,如果我更改 1 位,则 2 个奇偶校验位变为正确,而 1 仍然是错误的。

我该怎么做呢? 计算奇偶校验位

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1 回答 1

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if your final code is 101010

 
Decimal Number  Binary Number
    0                   000
    1                   001
    2                   010
    3                   011
    4                   100
    5                   101
    6                   110
    
now calculating the error position as follows with odd parity 

we have taken position as no of one in the binary no like for e1 1,3,5 no has 1 in last place likewise for e2 

2,3,6 has one in last second place and so on 

        position   bits on position 
E1 ->   1,3,5   ->      111             -> 0 
E2 ->   2,3,6   ->      010             -> 0
E3 ->   4,5,6   ->      010             -> 0

so by using odd parity there is no error 


reference : https://www.youtube.com/watch?v=UY0VpqyJ3U4
于 2021-12-02T18:15:40.470 回答