Проектиране на пълен суматор

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Проектиране на пълен суматор / субтрактор XOR – инвертиране на битове от B i при Add/Sub=0 изходът на XOR повтаря B Cin – 1) определя типа на операцията – 2) добавяне на 0 или 1 XOR Y = A + B A B Y 0 0 0 0 1 1 1 0 1 1 1 0 A B Y допълнителен код Add/Sub
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доц. д-р Ясен Горбунов, 2017 14 Суматор Проектиране на пълен суматор / субтрактор - пример 1 2 – 4 = -2 A = 2 DEC → 0010 B = 4 DEC → 0100 -4 → (1011 + 1 = 1100 Д.К. ) A 0010 + ~B + 1011 Cin 0001 -------------- 1110 Д.К. 0 1 0 0 0 0 1 0 1 1 0 1 0 1 1 1 Д.К. 0
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доц. д-р Ясен Горбунов, 2017 15 Суматор Проектиране на пълен суматор / субтрактор - пример 1 2 – 4 = -2 A = 2 DEC → 0010 B = 4 DEC → 0100 -4 → (1011 + 1 = 1100 Д.К. ) A 0010 + ~B + 1011 Cin 0001 -------------- 1110 Д.К. 0 1 0 0 0 0 1 0 1 1 0 1 0 1 1 1 Д.К. 1 4 – 4 = 0 A = 4 DEC → 0100 B = 4 DEC → 0100 -4 → (1011 + 1 = 1100 Д.К. ) A 0100 + ~B + 1011 Cin 0001 -------------- Cout 1 0000 Д.К. 0 0 1 0 0 0 1 0 1 1 0 1 0 0 0 0 Д.К. 0 1
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доц. д-р Ясен Горбунов, 2017 16 Суматор Проектиране на пълен суматор със знак ако и двата операнда притежават знак → допълнителен код Кодов преобразувател подаване на 0 към eдиния вход на пълен суматор → опростяване на уравненията A B 0 0 0 1 1 0 1 1 0 1 1 0 S C out 0 0 0 1 S = A B C in C out = AB + AC in + BC in C in 0 0 0 1 1 0 1 1 0 0 0 0 1 1 1 1 1 0 0 1 0 1 1 1 A B 0 0 0 1 0 0 0 1 0 1 0 1 S C out 0 0 0 0 C in 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 1 1 0 1 0 0 1 0 1 S = B Cin 0 Si gn
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доц. д-р Ясен Горбунов, 2017 17 Умножител Проектиране на 2-битово устройство за умножение 2 x 2 Multiplier X 0 X 1 Y 0 Y 1 Z 0 Z 1 Z 2 Z 3 Вход X Вход Y Изход Z
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доц. д-р Ясен Горбунов, 2017 18 Умножител Проектиране на 2-битово устройство за умножение таблица на истинност Номер на набора X x Y = Z Входове Изход X Y Z X 1 X 0 Y 1 Y 0 Z 3 Z 2 Z 1 Z 0 0 0 x 0 = 0 0 0 0 0 0 0 0 0 1 0 x 1 = 0 0 0 0 1 0 0 0 0 2 0 x 2 = 0 0 0 1 0 0 0 0 0 3 0 x 3 = 0 0 0 1 1 0 0 0 0 4 1 x 0 = 0 0 1 0 0 0 0 0 0 5 1 x 1 = 1 0 1 0 1 0 0 0 1 6 1 x 2 = 2 0 1 1 0 0 0 1 0 7 1 x 3 = 3 0 1 1 1 0 0 1 1 8 2 x 0 = 0 1 0 0 0 0 0 0 0 9 2 x 1 = 2 1 0 0 1 0 0 1 0 10 2 x 2 = 4 1 0 1 0 0 1 0 0 11 2 x 3 = 6 1 0 1 1 0 1 1 0 12 3 x 0 = 0 1 1 0 0 0 0 0 0 13 3 x 1 = 3 1 1 0 1 0 0 1 1 14 3 x 2 = 6 1 1 1 0 0 1 1 0 15 3 x 3 = 9 1 1 1 1 1 0 0 1
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доц. д-р Ясен Горбунов, 2017 19 Умножител Проектиране на 2-битово устройство за умножение Z 0 = 0 15 ( 5,7,13,15 ) Z 1 = 0 15 ( 6,7,9,11, 13,14 ) Z 2 = 0 15 ( 10,11, 14 ) Z 3 = 0 15 ( 15 ) Z X 1 X X Y Y 1 0 0 X 0 Y 0 X 0 Y Y X 1 0 1 1 X 0 Y 1 X 0 Y 1 0 Z Z Z 1 2 3 0 X 1 X 0 Y 0 Y X 1 Y 1 Y 0 X 1 X 0 Y 1 X 1 Y 1 Y 0 X 1 X 0 Y 1 Y 0 Z 0 = X 0 Y 0 Z 1 = X 1 X 0 Y 1 + X 0 Y 1 Y 0 + X 1 X 0 Y 0 + X 1 Y 1 Y 0 Z 2 = X 1 X 0 Y 1 + X 1 Y 1 Y 0 Z 3 = X 1 X 0 Y 1 Y 0
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доц. д-р Ясен Горбунов, 2017 20 Умножител Проектиране на 4-битово устройство за умножение x B 3 B 2 B 1 B 0 A 3 B 0 A 2 B 0 A 1 B 0 A 0 B 0 A 3 A 2 A 1 A 0 A 3 B 1 A 2 B 1 A 1 B 1 A 0 B 1 A 3 B 2 A 2 B 2 A 1 B 2 A 0 B 2 A 3 B 3 A 2 B 3 A 1 B 3 A 0 B 3 + P 7 P 6 P 5 P 4 P 3 P 2 P 1 P 0 0 P 2 0 0 0 P 1 P 0 P 5 P 4 P 3 P 7 P 6 A 3 A 2 A 1 A 0 B 0 B 1 B 2 B 3 x A B P 4 4 8
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доц. д-р Ясен Горбунов, 2017 21 Умножител Други начини за проектиране и използване на умноженители 1. Комбинационни логически схеми – бързи, но обемисти 2. Последователно изместване и събиране (shift & add) –
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