Problemas y soluciones de circuitos usando mapas de karnaugh
Entrada | SalidaCódigo Gray |
A B C D | W X Y Z |
0 0 0 0 | 0 0 0 0 |
0 0 0 1 | 0 0 0 1 |
0 0 1 0 | 0 0 1 1 |
0 0 1 1 | 0 0 1 0 |
0 10 0 | 0 1 1 0 |
0 1 0 1 | 0 1 1 1 |
0 1 1 0 | 0 1 0 1 |
0 1 1 1 | 0 1 0 0 |
1 0 0 0 | 1 1 0 0 |
1 0 0 1 | 1 1 0 1 |
1 0 1 0 | X X X X |
1 0 1 1 | X X X X |
1 1 0 0 | X X X X |
1 1 0 1 | X X X X |
1 1 1 0 | X X X X |
1 1 1 1 | X X X X |
W
AB CD | 00 | 01 | 11 | 10 |
00 | | | | |
01| | | | |
11 | x | x | x | x |
10 | 1 | 1 | x | x |
W=A
X
AB CD | 00 | 01 | 11 | 10 |
00 | | | | |
01 | 1 | 1 | 1 | 1 |
11 | x | x | x | x |
10 | 1 | 1 | x | x |
X=A+B
Y
AB CD | 00 | 01 | 11 | 10 |
00 | | | 1 | 1 |
01 | 1 | 1 | | |
11 | x | x | x | x |
10 | | | x | x |
Y=BC+BC=B⊕C
Z
AB CD | 00 | 01 | 11 | 10 |
00 | | 1 | | 1 |
01 | | 1 || 1 |
11 | x | x | x | x |
10 | | 1 | x | x |
Z=CD+DC=C⨁D
Circuito BCD a GRAY
TALLER II: a) Completar la tabla de verdad del Decodificador de BCD a 7 segmentos. b) Realizar la simplificación usando mapas de Karnaugh.
Entrada | Salidas |
W X Y Z | a | b | c | d | e | f | g |
0 0 0 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 |
0 0 0 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
0 01 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 |
0 0 1 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 |
0 1 0 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
0 1 0 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
0 1 1 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
0 1 1 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
1 0 0 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
1 0 0 1 |1 | 1 | 1 | 1 | 0 | 1 | 1 |
1 0 1 0 | X | X | X | X | X | X | X |
1 0 1 1 | X | X | X | X | X | X | X |
1 1 0 0 | X | X | X | X | X | X | X |
1 1 0 1 | X | X | X | X | X | X | X |
1 1 1 0 | X | X | X | X | X | X | X |
1 1 1 1 | X | X | X | X | X | X | X |
A
WX YZ | 00 | 01 | 11 | 10 |
00 | 1 | | 1 | 1 |
01 | | 1 | 1 | 1 |
11 | x | x | x | x |10 | 1 | 1 | x | x |
A=W+Y+XZ+X Z=W+Y+(X⊕Z)
B
WX YZ | 00 | 01 | 11 | 10 |
00 | 1 | 1 | 1 | 1 |
01 | 1 | | 1 | |
11 | x | x | x | x |
10 | 1 | 1 | x | x |
B=YZ+Y Z+X=X+Y⨁Z
C
WX YZ | 00 | 01 | 11 | 10 |
00 | 1 | 1 | 1 | |
01 | 1 | 1 | 1 | 1 |
11 | x | x | x | x |
10 | 1 | 1 | x | x |
C=Y+Z+X
D
WX YZ | 00 | 01 | 11 | 10 |
00 | 1 | | 1 | 1 |
01 | | 1 | |1 |
11 | x | x | x | x |
10 | 1 | 1 | x | x |
D=W+YZ+XY+X Z+XYZ=W+YZ+X(Y+ Z)+XYZ
E
WX YZ | 00 | 01 | 11 | 10 |
00 | 1 | | | 1 |
01 | | | | 1 |
11 | x | x | x | x |
10 | 1 | | x | x |
E=YZ+X Z=Z(Y+X)
F
WX YZ | 00 | 01 | 11 | 10 |
00 | 1 | | | |
01 | 1 | 1 | | 1 |
11 | x | x | x | x |
10 | 1 | 1 | x | x |
F=W+XY+Y Z+XZ=W+YX+Z+XZ
G
WX YZ | 00 | 01| 11 | 10 |
00 | | | 1 | 1 |
01 | 1 | 1 | | 1 |
11 | x | x | x | x |
10 | 1 | 1 | x | x |
G=W+YZ+XY+XY=W+YZ+X+XY=W+YXZ+XY
TALLER 3: a) Deducir las 10 funciones del Mapa de Karnaugh del decodificador de BCD a decimal y b) dibujar el circuito con compuertas lógicas
Tabla de Verdad
Entrada BCD | Salidas (Dígitos decimales del 0 al 9) |
A B C D | D0 | D1 | D2 | D3 | D4 | D5 |D6 | D7 | D8 | D9 |
0 0 0 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 0 0 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 0 1 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 0 1 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
0 1 0 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
0 1 0 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
0 1 1 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0...
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