Материал: Отчет по лабораторной работе №7 МСвСПД
Рисунок 5. Граф состояний аддитивного скремблера для .
3.
Для заданного неприводимого полинома
построить схему аддитивного
скремблера/дескремблера.
Рисунок 6. Схема аддитивного скремблера для .
Рисунок 7. Схема аддитивного дескремблера для .
4. Передать через скремблер информационную комбинацию длиной 32 разряда. Комбинация задана в табл. 7.1 в двоично-десятичном виде. Процедуру работы скремблера и результат записать в виде табл. 7.2.
IN =
OUT
=
Такт |
IN |
|
|
|
|
|
|
|
|
|
OUT |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
2 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
3 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
4 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
5 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
6 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
7 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
8 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
0 |
9 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
10 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
11 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
12 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
13 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
14 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
15 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
16 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
17 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
18 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
19 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
20 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
21 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
22 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
23 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
24 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
25 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
26 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
27 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
28 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
29 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
30 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
31 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
32 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
5. Получившийся результат передать через схему дескремблера. Процедуру работы дескремблера и результат записать в виде табл. 7.2. Убедиться, что результат соответствует заданной информационной комбинации.
OUT =
IN’ =
Такт |
OUT |
|
|
|
|
|
|
|
|
|
IN’ |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
2 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
3 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
4 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
5 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
6 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
7 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
8 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
0 |
9 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
10 |
0 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
11 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
12 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
13 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
14 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
15 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
16 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
17 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
18 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
19 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
20 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
21 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
22 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
23 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
24 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
25 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
26 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
27 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
28 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
29 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
30 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
31 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
32 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
6. Внести одиночную ошибку в результат скремблирования и передать комбинацию с ошибкой через схему дескремблера. Процедуру работы дескремблера и результат записать в виде табл. 7.2. Сравнить результат с исходной информационной комбинацией. Сделать выводы по полученным результатам.
OUT
=
IN’
=
Такт |
OUT |
|
|
|
|
|
|
|
|
|
IN’ |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
2 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
3 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
4 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
5 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
6 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
7 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
8 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
0 |
9 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
10 |
0 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
11 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
12 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
13 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
14 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
15 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
16 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
17 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
18 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
19 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
20 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
21 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
22 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
23 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
24 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
25 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
26 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
27 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
28 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
29 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
30 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
31 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
32 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
18 Вариант
8.3.1. Самосинхронизирующиеся скремблеры
Исходные данные:
1. Для заданных параметров (длина регистра и разряды регистра и = , охваченные обратной связью) нарисовать схемы самосинхронизирующегося скремблера и соответствующего ему дескремблера.
Рисунок 1. Схема самосинхронизирующегося скремблера.
Рисунок 2. Схема самосинхронизирующегося дескремблера.
2. Передать через скремблер двоичную информационную комбинацию длиной 32 разряда. Комбинация задана в табл. 7.1 в двоично-десятичном виде. Процедуру работы скремблера (состояния ячеек регистра ri) и результат записать в виде таблицы.
IN
=
OUT
=
Такт |
IN |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
OUT |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
2 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
3 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
4 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
5 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
6 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
7 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
8 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
9 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
10 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
11 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
12 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
13 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
14 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
15 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
16 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
17 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
18 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
19 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
20 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
21 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
22 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
23 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
24 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
25 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
26 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
27 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
28 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
29 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
30 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
31 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
32 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
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