Реферат Проектирование устройства, осуществляющего перемножение двух четырехразрядных чисел
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Подписываем
договор
Часть 2.
х1х2х3х4
х5х6х7х8
(х1х8)(х2х8)(х3х8)(х4х8)
(х1х7)(х2х7)(х3х7)(х4х7)
(х1х6)(х2х6)(х3х6)(х4х6)
х3х8 | Х4х7 | Y7 | Р1 |
0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
Y7=(а+b)+(a+b); a=x3x8; b=x4x7 P1=ab
a | b | c | P1 | P2 | Y6 | P2’ |
0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 1 | 0 | 1 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 1 | 0 | 0 | 1 |
P2’=abcp1
a | b | c | d | P2 | P3 | Y5 | P3’ |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
После упрощения:
A | b | c | P2’ | P3 | P4 | Y4 | P4’ |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 | 1 | 0 |
1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
После упрощения:
P4’=p2’p3(bc+ac+ab)+abc(p2’+p3) a=x1x7;b=x2x6;c=x3x5
A | b | P3’ | P4 | P5 | Y3 |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 | 1 | 0 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 |
После упрощения:
a=x1x6;b=x2x5
A | P4’ | P5 | P6 | Y2 |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 1 | 1 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 1 | 1 |
После упрощений:
P6= p4’p5+a(p4’+p5) a=x1x5
Y1=P6= p4’p5+a(p4’+p5)
Y8=x4x8