Часть 2.
х1х2х3х4
х5х6х7х8
(х1х8)(х2х8)(х3х8)(х4х8)
Возможно вы искали - Реферат: Разработка МПС на базе КР580
(х1х7)(х2х7)(х3х7)(х4х7)
(х1х6)(х2х6)(х3х6)(х4х6)
(х1х5)(х2х5)(х3х5)(х4х5)
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 |
Похожий материал - Реферат: ТОЭ контрольная №5
После упрощения Y6=(cp1+cp1)(ab+ab)+(cp1+cp1)(ab+ab)
P2=a(bp1+bp1)+p1(bc+bc)+abc a=x2x8;b=x3x7;c=x4x6
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:
Очень интересно - Реферат: Усилительные каскады в области высоких частот
Y5=(dp2+dp2)(a(bc+bc)+a(bc+bc))+(dp2+dp2)c(ab+ab)
P3=bd(ac+ac)+cp2(ac+ab)+ab(cp2+cp2)+dp2(ab+ab)+abcp2
P3=bd(ac+ac)+cp2(ad+ad)+(ab+ab)(dp2+cp2)+abcp2 a=x1x8;b=x2x7;c=x3x6;d=x4x5
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:
Вам будет интересно - Реферат: ЭТПиМЭ
Y4=(p2’p3+p2’p3)(abc+abc+abc)+ap2’p3(bc+bc)+abcp2’+abcp2’p3
P4=(p2’+p3)b(ac+ac)+abc(p2’+p3)+abp(p2’+c)+abc(p2’+p3)+abcp2’p3
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 |
Похожий материал - Реферат: Отчет по практике
После упрощения:
Y3=(ab+ab)(p3’p4+p3’p4)+(ab+ab)(p3’p4+p3’p4)
P5=p4b(a+p3’)+b(ap3’+p3’p4+ap4)