Álgebra booleana
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2013ORGANIZAÇÃO DE COMPUTADORES
Álgebra Booleana
Aluno: Lucas Santos Camurça Nº Matrícula: 2013.01.70241-2 E-mail: Lucascamurca14@bol.com.br Professor: Péricles
ATENÇÃO o símbolo representa o operador XOR
1. Desenvolva a tabela verdade para as seguintes expressões booleanas.
a) A . B . C + A . B . C
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
C 0 1 0 1 0 1 0 1
A.B.C 0 0 0 0 0 0 0 1
. . 1 1 1 1 1 1 1 0
A . B . C + . . 1 1 1 1 1 1 1 1
b) A . (C + B + D ) A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 C + + 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 A . (C + + ) 0 0 0 0 0 0 0 0 1 1 1 1 1 0 1 1
c) A . B . C + A . B. C + A . B . C
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
C A.B.C A. . . . 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 0 0
A.B.C + A.. + .. 1 0 0 0 1 0 0 1
d) (A + B) . ( A + C) . (A B)
A B C (A+B) ( + ) ( ) 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 1 1 1 0 1 1 1 0 1 1 0 0 1 0 1 1 0 1 1 0 1 1 1 0 1 0 0 1 1 1 1 0 0
e) A . B + A . B
(A + B).( + ).( ) 0 0 1 0 0 0 0 0
A 0 0 1 1
B 0 1 0 1
A.B 0 0 0 1
A . 0 0 1 0
A . B + A . 0 0 1 1
f) A + (B + A . C) D A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 A.C ( + . ) ( + . ) 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 0 0 1 0 1 1 0 0 0 1 0 0 1 1 1 0 1 1 0 0 A + ( + . ) 1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1
2. Considere os seguintes valores binários: A = 1011 B = 1110 C = 0011 D = 1010 a) X= A. (B C) X= 1011 . (1110 OO11) X= 1011 . 1101 X= 1001 b) X= (A + B) . (C (A + )) X= (1011 + 1110) . (0011 ( 1011 + 0101 )) X= 0000 . (0011 1111) X= 0000 . 1100 X= 0000 c) X= B . . A + (C D) X= 1110 . 1100 . 1011 + (1100 1010 ) X= 1110 . 1100 . 1011 + 0001 X= 1000 + 0001 X= 1001 d) X = (( + B D) . (C + A) + B) . A + B) X= (( + 0100) .