1.-Diseñ un sumador binario de tres numeros de 2 bits cada uno, y que de una suma de 4 bits:
A continuacion de se presenta la tabla de verdad de la funcion a realizar:
Reduccion de salidas [a, b, c, d]
salida c mapa de karnaugh
salida c (simplificado)
A'B'C'D'E+A'B'C'EF'+A'B'CD'E'+A'B'CE'F'+A'BC'E'F+A'C'D'EF'+
A'BC'DE'+A'C'DE'F+A'CD'E'F'+A'BCEF+A'BCDE+A'CDEF+AB'C'D'E'F+
AB'C'DE'F'+AB'CD'E+AB'CEF'+ABC'D'E'F'+ABC'EF+ABC'DE+AC'DEF+
ABCE'F+ACD'EF'+ABCDE'+ACDE'F
salida d mapa de karnaugh
salida d (simplificado)
A'BDF+AB'C'D'E'+B'D'F+B'DF'+BC'DF+BD'F'+BDE'F
Circuito Simulado
2.Diseñe un sistema combinacional operable para multiplicar dos numeros binarios de tres bits cada uno.
A continuacion se presenta la tabla de verdad con su respecta funcion a resolver:A | B | C | D | E | F | miniterminos | |||||||
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 0 | 1 | 0 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 0 | 1 | 1 | 3 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 1 | 0 | 0 | 4 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 1 | 0 | 1 | 5 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 1 | 1 | 0 | 6 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 0 | 1 | 1 | 1 | 7 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 1 | 0 | 0 | 0 | 8 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 0 | 1 | 0 | 0 | 1 | 9 | 0 | 0 | 0 | 0 | 0 | 1 | |
0 | 0 | 1 | 0 | 1 | 0 | 10 | 0 | 0 | 0 | 0 | 1 | 0 | |
0 | 0 | 1 | 0 | 1 | 1 | 11 | 0 | 0 | 0 | 0 | 1 | 1 | |
0 | 0 | 1 | 1 | 0 | 0 | 12 | 0 | 0 | 0 | 1 | 0 | 0 | |
0 | 0 | 1 | 1 | 0 | 1 | 13 | 0 | 0 | 0 | 1 | 0 | 1 | |
0 | 0 | 1 | 1 | 1 | 0 | 14 | 0 | 0 | 0 | 1 | 1 | 0 | |
0 | 0 | 1 | 1 | 1 | 1 | 15 | 0 | 0 | 0 | 1 | 1 | 1 | |
0 | 1 | 0 | 0 | 0 | 0 | 16 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 1 | 0 | 0 | 0 | 1 | 17 | 0 | 0 | 0 | 0 | 1 | 0 | |
0 | 1 | 0 | 0 | 1 | 0 | 18 | 0 | 0 | 0 | 1 | 0 | 0 | |
0 | 1 | 0 | 0 | 1 | 1 | 19 | 0 | 0 | 0 | 1 | 1 | 0 | |
0 | 1 | 0 | 1 | 0 | 0 | 20 | 0 | 0 | 1 | 0 | 0 | 0 | |
0 | 1 | 0 | 1 | 0 | 1 | 21 | 0 | 0 | 1 | 0 | 1 | 0 | |
0 | 1 | 0 | 1 | 1 | 0 | 22 | 0 | 0 | 1 | 1 | 0 | 0 | |
0 | 1 | 0 | 1 | 1 | 1 | 23 | 0 | 1 | 0 | 0 | 0 | 0 | |
0 | 1 | 1 | 0 | 0 | 0 | 24 | 0 | 0 | 0 | 0 | 0 | 0 | |
0 | 1 | 1 | 0 | 0 | 1 | 25 | 0 | 0 | 0 | 0 | 1 | 1 | |
0 | 1 | 1 | 0 | 1 | 0 | 26 | 0 | 0 | 0 | 1 | 1 | 0 | |
0 | 1 | 1 | 0 | 1 | 1 | 27 | 0 | 0 | 1 | 0 | 0 | 1 | |
0 | 1 | 1 | 1 | 0 | 0 | 28 | 0 | 0 | 1 | 1 | 0 | 0 | |
0 | 1 | 1 | 1 | 0 | 1 | 29 | 0 | 0 | 1 | 1 | 1 | 1 | |
0 | 1 | 1 | 1 | 1 | 0 | 30 | 0 | 1 | 0 | 0 | 1 | 0 | |
0 | 1 | 1 | 1 | 1 | 1 | 31 | 0 | 1 | 0 | 1 | 0 | 1 | |
1 | 0 | 0 | 0 | 0 | 0 | 32 | 0 | 0 | 0 | 0 | 0 | 0 | |
1 | 0 | 0 | 0 | 0 | 1 | 33 | 0 | 0 | 0 | 1 | 0 | 0 | |
1 | 0 | 0 | 0 | 1 | 0 | 34 | 0 | 0 | 1 | 0 | 0 | 0 | |
1 | 0 | 0 | 0 | 1 | 1 | 35 | 0 | 0 | 1 | 1 | 0 | 0 | |
1 | 0 | 0 | 1 | 0 | 0 | 36 | 0 | 1 | 0 | 0 | 0 | 0 | |
1 | 0 | 0 | 1 | 0 | 1 | 37 | 0 | 1 | 0 | 1 | 0 | 0 | |
1 | 0 | 0 | 1 | 1 | 0 | 38 | 0 | 1 | 1 | 0 | 0 | 0 | |
1 | 0 | 0 | 1 | 1 | 1 | 39 | 0 | 1 | 1 | 1 | 0 | 0 | |
1 | 0 | 1 | 0 | 0 | 0 | 40 | 0 | 0 | 0 | 0 | 0 | 0 | |
1 | 0 | 1 | 0 | 0 | 1 | 41 | 0 | 0 | 0 | 1 | 0 | 1 | |
1 | 0 | 1 | 0 | 1 | 0 | 42 | 0 | 0 | 1 | 0 | 1 | 0 | |
1 | 0 | 1 | 0 | 1 | 1 | 43 | 0 | 0 | 1 | 1 | 1 | 1 | |
1 | 0 | 1 | 1 | 0 | 0 | 44 | 0 | 1 | 0 | 1 | 0 | 0 | |
1 | 0 | 1 | 1 | 0 | 1 | 45 | 0 | 1 | 1 | 0 | 0 | 1 | |
1 | 0 | 1 | 1 | 1 | 0 | 46 | 0 | 1 | 1 | 1 | 1 | 0 | |
1 | 0 | 1 | 1 | 1 | 1 | 47 | 1 | 0 | 0 | 0 | 1 | 1 | |
1 | 1 | 0 | 0 | 0 | 0 | 48 | 0 | 0 | 0 | 0 | 0 | 0 | |
1 | 1 | 0 | 0 | 0 | 1 | 49 | 0 | 0 | 0 | 1 | 1 | 0 | |
1 | 1 | 0 | 0 | 1 | 0 | 50 | 0 | 0 | 1 | 1 | 0 | 0 | |
1 | 1 | 0 | 0 | 1 | 1 | 51 | 0 | 1 | 0 | 0 | 1 | 0 | |
1 | 1 | 0 | 1 | 0 | 0 | 52 | 0 | 1 | 1 | 0 | 0 | 0 | |
1 | 1 | 0 | 1 | 0 | 1 | 53 | 0 | 1 | 1 | 1 | 1 | 0 | |
1 | 1 | 0 | 1 | 1 | 0 | 54 | 1 | 0 | 0 | 1 | 0 | 0 | |
1 | 1 | 0 | 1 | 1 | 1 | 55 | 1 | 0 | 1 | 0 | 1 | 0 | |
1 | 1 | 1 | 0 | 0 | 0 | 56 | 0 | 0 | 0 | 0 | 0 | 0 | |
1 | 1 | 1 | 0 | 0 | 1 | 57 | 0 | 0 | 0 | 1 | 1 | 1 | |
1 | 1 | 1 | 0 | 1 | 0 | 58 | 0 | 0 | 1 | 1 | 0 | 0 | |
1 | 1 | 1 | 0 | 1 | 1 | 59 | 0 | 1 | 0 | 1 | 0 | 1 | |
1 | 1 | 1 | 1 | 0 | 0 | 60 | 0 | 1 | 1 | 1 | 0 | 0 | |
1 | 1 | 1 | 1 | 0 | 1 | 61 | 1 | 0 | 0 | 0 | 1 | 1 | |
1 | 1 | 1 | 1 | 1 | 0 | 62 | 1 | 0 | 0 | 0 | 0 | 0 | |
1 | 1 | 1 | 1 | 1 | 1 | 63 | 1 | 1 | 0 | 0 | 0 | 1 |
Funcion A
Metodo Tabular
Columna 0 | Columna 1 | Columna 2 | |||||||||
Group 4 | Group 4 | Group 4 | |||||||||
54 | 110110 v | 55,54 | 11011- v | 54, 55, 62, 63 | 11-11- | ||||||
Group 5 | 62,54 | 11-110 v | |||||||||
47 | 101111 v | Group 5 | |||||||||
55 | 110111 v | 47,63 | 1-1111 | ||||||||
61 | 111101 v | 55,63 | 11-111 v | ||||||||
62 | 111110 v | 61,63 | 1111-1 | ||||||||
Group 6 | 62,63 | 11111- v | |||||||||
63 | 111111 v |
Indicaciones de color:
Naranja: Termino escencial.
Azul: Renglon en el cual el termino escencial se encuentra y elimina todas las "x" .
Amarillo: Renglon asginado para eliminar "x" restantes en otros renglones.
Verde:"x" eliminadas por renglones escenciales por el cual se encuentran localizadas en la misma columna.
Rojo:Renglon eliminado
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Mapa de Karnaught
fa=(A*C*D*E*F)+(A*B*D*E)+(A*B*C*D*F) |
Circuito:
Funcion B:
Columna 0 | Columna 1 | Columna 2 | ||||
Group 2 | Group 2 | Group 2 | ||||
36 | 100100 v | 36,37 | 10010- v | 1001--[36, 37, 38, 39] | ||
Group 3 | 36,38 | 1001-0 v | 10-10-[36, 37, 44, 45] | |||
37 | 100101 v | 36,44 | 10-100 v | 1-010-[36, 37, 52, 53] | ||
38 | 100110 v | 36,52 | 1-0100 v | 10-1-0[36, 38, 44, 46] | ||
44 | 101100 v | Group 3 | 1--100[36, 44, 52, 60] | |||
52 | 110100 v | 37,39 | 1001-1 v | |||
Group 4 | 37,45 | 10-101 v | ||||
23 | 010111 v | 37,53 | 1-0101 v | |||
30 | 011110 v | 38,39 | 10011- v | |||
39 | 100111 v | 38,46 | 10-110 v | |||
45 | 101101 v | 44,45 | 10110- v | |||
46 | 101110 v | 44,46 | 1011-0 v | |||
51 | 110011 v | 44,60 | 1-1100 v | |||
53 | 110101 v | 52,53 | 11010- v | |||
60 | 111100 v | 52,60 | 11-100 v | |||
Group 5 | Group 4 | |||||
31 | 011111 v | 23,31 | 01-111 | |||
59 | 111011 v | 30,31 | 01111- | |||
Group 6 | 51,59 | 11-011 | ||||
63 | 111111 v | Group 5 | ||||
31,63 | -11111 | |||||
59,63 | 111-11 |
Naranja: Termino escencial.
Azul: Renglon en el cual el termino escencial se encuentra y elimina todas las "x" .
Amarillo: Renglon asginado para eliminar "x" restantes en otros renglones.
Verde:"x" eliminadas por renglones escenciales por el cual se encuentran localizadas en la misma columna.
Rojo:Renglon eliminado.
fb=(!A*B*D*E*F)+(!A*B*C*D*E)+(A*!B*!C*D)+(A*!B*D*!E)+(A*!B*D*!F)+(A*B*!D*E*F)+(A*!C*D*!E)+(A*D*!E*!F)+(B*C*D*E*F)
Metodo de Mapa de Karnaught
Circuito:
Funcion C
Metodo Tabular:
Column 0 | Column 1 | Column 2 | ||||
Group 2 | Group 2 | Group 2 | ||||
20 | 010100 v | 20,21 | 01010- v | 01-10-[20, 21, 28, 29] | ||
34 | 100010 v | 20,22 | 0101-0 | -1010-[20, 21, 52, 53] | ||
Group 3 | 20,28 | 01-100 v | -1-100[20, 28, 52, 60] | |||
21 | 010101 v | 34,35 | -10100 v | 100-1-[34, 35, 38, 39] | ||
22 | 010110 v | 34,38 | 10001- v | 10-01-[34, 35, 42, 43] | ||
28 | 011100 v | 34,42 | 100-10 v | 10--10[34, 38, 42, 46] | ||
35 | 100011 v | 34,50 | 10-010 v | 1--010[34, 42, 50, 58] | ||
38 | 100110 v | 34,52 | 1-0010 v | |||
42 | 101010 v | Group 3 | ||||
50 | 110010 v | 21,29 | 01-101 v | |||
52 | 110100 v | 22,53 | -10101 v | |||
Group 4 | 28,29 | 01110- v | ||||
27 | 11011 | 28,46 | -11100 v | |||
29 | 011101 v | 35,39 | 100-11 v | |||
39 | 100111 v | 35,43 | 10-011 v | |||
43 | 101011 v | 38,39 | 10011- v | |||
45 | 101101 | 38,46 | 10-110 v | |||
46 | 101110 v | 42,43 | 10101- v | |||
53 | 110101 v | 42,46 | 101-10 v | |||
58 | 111010 v | 42,53 | 1-1010 v | |||
60 | 111100 v | 50,53 | 11-010 v | |||
Group 5 | 52,53 | 11010- v | ||||
55 | 110111 v | 52,60 | 11-100 v | |||
Group 4 | ||||||
39,55 | 1-0111 | |||||
53,55 | 1101-1 |
Indicaciones de color:
Naranja: Termino escencial.
Azul: Renglon en el cual el termino escencial se encuentra y elimina todas las "x" .
Amarillo: Renglon asginado para eliminar "x" restantes en otros renglones.
Verde:"x" eliminadas por renglones escenciales por el cual se encuentran localizadas en la misma columna.
Rojo:Renglon eliminado.
fc=(!A*B*!C*D*!F)+(!A*B*C*!D*E*F)+(!A*B*D*!E)+(A*!B*!D*E)+(A*!B*C*D*!E*F)+(A*!B*E*!F)+(A*!D*E*!F)+(B*D*!E*!F)+(A*!C*D*E*F)+(B*!C*D*!E)
Metodo Mapa de Karnaught
Circuito
Salmon Angel Adrian Gerardo
Mapa de karnaugh
simplificacion
A'B'+A'B'CD'+AB+ABC
implementacion
5.-
Por medio del método tabular se simplifico la expresión
F=(A’B’C’EF’)+(A’BC’DE’)+(A’BCDF’)+(A’BDEF)+(A’D’E’F)+(A’B’D’)
Se simula el circuito de la expresion simplificada.