Вариант 18
14.3.1. Систематический декодер Меггитта
1. Построить схему систематического декодера Меггитта для кода Хэмминга (15,11), образованного полиномом g(x) = 1+x+x4.
2. Для полученного в предыдущей работе кодового слова и вектора ошибки веса 1, вычислить комбинацию с ошибкой, провести ее декодирование в построенном декодере Меггитта, заполнив таблицу процедуры декодирования, и выписать кодовое слово.
Закодированный
вектор с однократной ошибкой –
Декодированная
последовательность –
Такт |
IN |
b0 |
b1 |
b2 |
b3 |
b4 |
b5 |
b6 |
b7 |
b8 |
b9 |
b10 |
b11 |
b12 |
b13 |
b14 |
s0 |
s1 |
s2 |
s3 |
OUT |
0 |
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
1 |
1 |
1 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
0 |
0 |
0 |
|
2 |
0 |
0 |
1 |
|
|
|
|
|
|
|
|
|
|
|
|
|
0 |
1 |
0 |
0 |
|
3 |
0 |
0 |
0 |
1 |
|
|
|
|
|
|
|
|
|
|
|
|
0 |
0 |
1 |
0 |
|
4 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
|
|
|
|
|
|
|
0 |
0 |
0 |
1 |
|
5 |
0 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
|
|
|
|
|
|
1 |
1 |
0 |
0 |
|
6 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
|
|
|
|
|
1 |
1 |
1 |
0 |
|
7 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
|
|
|
|
0 |
1 |
1 |
1 |
|
8 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
|
|
|
1 |
1 |
1 |
1 |
|
9 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
|
|
1 |
0 |
1 |
1 |
|
10 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
|
0 |
0 |
0 |
1 |
|
11 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
1 |
1 |
0 |
0 |
|
12 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
|
|
0 |
1 |
1 |
0 |
|
13 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
|
0 |
0 |
1 |
1 |
|
14 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
1 |
1 |
0 |
1 |
|
15 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
|
16 |
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
17 |
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
18 |
|
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
19 |
|
|
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
20 |
|
|
|
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
21 |
|
|
|
|
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
22 |
|
|
|
|
|
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
23 |
|
|
|
|
|
|
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
24 |
|
|
|
|
|
|
|
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
25 |
|
|
|
|
|
|
|
|
|
|
|
1 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
26 |
|
|
|
|
|
|
|
|
|
|
|
|
1 |
0 |
0 |
0 |
1 |
0 |
1 |
1 |
0 |
27 |
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
28 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
0 |
1 |
0 |
0 |
0 |
1 |
29 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
0 |
1 |
0 |
0 |
0 |
30 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
0 |
0 |
1 |
0 |
1 |