Sistemas digitales - Universidad de Concepción

Capítulo 4: Funciones Booleanas 25
Solución
a)
b)
c)
ABCD bits en 0 W XY Z
0000 4 0100
0001 3 0101
0010 3 0101
0011 2 0111
0100 3 0101
0101 2 0111
0110 2 0111
0111 1 0110
1000 3 0101
1001 2 0111
1010 2 0111
1011 1 0110
1100 2 0111
1101 1 0110
1110 1 0110
1111 0 0010
W (A,B,C,D) = 0
∑
X(A,B,C,D) = m(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14)
∑
Y (A,B,C,D) = m(3,5,6,7,9,10,11,12,13,14,15)
∑
Z(A,B,C,D) = m(1,2,3,4,5,6,8,9,10,12)
W (A,B,C,D) = 0
X(A,B,C,D) = (A ′ + B ′ + C ′ + D ′ )
Y (A,B,C,D) = (B + C + D)(A + C + D)(A + B + C)(A + B + D)
Z(A,B,C,D) = (A + B + C + D)(A ′ + B ′ + D ′ )(A ′ + B ′ + C ′ )
(A ′ + C ′ + D ′ )(B ′ + C ′ + D ′ )