Сложим вместе целую и дробную часть вот так:
10011002 + 0.100110011002 = 1001100.100110011002
Задание № 3
Изобразите логический элемент, обладающий Х=10 входами и одним выходом. Приведите таблицу истинности Вашего логического элемента. Выбор элемента выполните из условий, что N – номер варианта, причём варианты с чётными номерами используют логическую конъюнкцию, а варианты с нечётными номерами используют логическую дизъюнкцию в качестве основы логического элемента.
Решение
X1
| X2
| X3
| X4
| X5
| X6
| X7
| X8
| X9
| X10
| Y
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 1
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 1
| 1
| 1
| 1
| 1
| 1
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| 0
| 1
| 0
| 0
| 0
| 0
| |