Реферат: Проектирование устройства, осуществляющего перемножение двух четырехразрядных чисел

Часть 2.

х1х2х3х4 х5х6х7х8

(х1х8)(х2х8)(х3х8)(х4х8)

(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 |

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:

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:

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 |


После упрощения:

Y3=(ab+ab)(p3’p4+p3’p4)+(ab+ab)(p3’p4+p3’p4)

P5=p4b(a+p3’)+b(ap3’+p3’p4+ap4)

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:

Y2=p4’(ap5+ap5)+p4’(ap5+ap5)

P6= p4’p5+a(p4’+p5) a=x1x5

Y1=P6= p4’p5+a(p4’+p5)
Y8=x4x8


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