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Transcript of puertas logicas
![Page 1: puertas logicas](https://reader035.fdocumento.com/reader035/viewer/2022081507/587ec2ef1a28abf37b8b56d9/html5/thumbnails/1.jpg)
Álgebra Booleana
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Operadores Lógicos
•And•Or•Not•Nand•Nor•Exor•Exnor
• Nombre• Característica• Símbolo• Expresión Matemática• Tabla de verdad• Circuito Equivalente• Diagrama de Tiempos
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Nombre AND OR NOT
Característica Condición Alternativa Negar
Símbolo
ExpresiónMatemática S=AB S=A+B S=A
Tabla de Verdad
Circuitoeléctrico
equivalente
Diagramade
Tiempos
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Ejercicio 1 a que operación booleana se refiere el enunciado
La salida es cero cuando cualquier entrada es igual a cero
A B
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Cualquier entrada uno produce una salida uno.
Ejercicio 2 a que operación booleana se refiere el enunciado
A + B
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solamente cuando todas las entradas son cero producen una salida cero.
Ejercicio 3a que operación booleana se refiere el enunciado
A + B
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La salida es uno solamente cuando todas las entradas son uno.
Ejercicio 4 a que operación booleana se refiere el enunciado
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La salida es siempre lo contrario de la entrada.
Ejercicio 5 a que operación booleana se refiere el enunciado
m A S
0 0 1
1 1 0
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NANDLa operación Nand es el negado de
la salida de la operación And.
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La operación Nand es el negado de las entradas de la operación OR.
NAND
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Tabla de verdad
m A B AB0 0 0 11 0 1 12 1 0 13 1 1 0
NAND
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Circuito Eléctrico equivalente
m A B AB0 0 0 11 0 1 12 1 0 13 1 1 0
NAND
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Nand de 3 entradas F(A, B, C) = A B C
m A B C ABC0 0 0 0 11 0 0 1 12 0 1 0 13 0 1 1 14 1 0 0 15 1 0 1 16 1 1 0 17 1 1 1 0
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La operación Nor es el negado de la salida de la operación OR.
NOR
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La operación Nor es el negado de las entradas de la operación AND.
NOR
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Tabla de Verdad
m A B A+B0 0 0 11 0 1 02 1 0 03 1 1 0
NOR
X = A +B
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Circuito eléctrico equivalente
m A B A+B0 0 0 11 0 1 02 1 0 03 1 1 0
NOR
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NOR de tres entradas
m A B C A+B+C
0 0 0 0 11 0 0 1 02 0 1 0 03 0 1 1 04 1 0 0 05 1 0 1 06 1 1 0 07 1 1 1 0
F(A, B, C) = A+B+C
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Alternativa Exclusiva (Opción entre dos cosas, una, otra pero no ambas)
EXOR
La operación Exor produce un resultado 1, cuando un número impar de variables de entrada valen 1.
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AB
EXOR
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AB
EXOR
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Exor , produce un resultado 1, cuando un número impar de Variables de entrada valen 1.
m A B C X0 0 0 01 0 0 12 0 1 03 0 1 14 1 0 05 1 0 16 1 1 07 1 1 1
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m A B C X0 0 0 01 0 0 1 12 0 1 03 0 1 14 1 0 05 1 0 16 1 1 07 1 1 1
Exor , produce un resultado 1, cuando un número impar de Variables de entrada valen 1.
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m A B C X0 0 0 01 0 0 1 12 0 1 0 13 0 1 14 1 0 05 1 0 16 1 1 07 1 1 1
Exor , produce un resultado 1, cuando un número impar de Variables de entrada valen 1.
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m A B C X0 0 0 01 0 0 1 12 0 1 0 13 0 1 14 1 0 0 15 1 0 16 1 1 07 1 1 1
Exor , produce un resultado 1, cuando un número impar de Variables de entrada valen 1.
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m A B C X0 0 0 01 0 0 1 12 0 1 0 13 0 1 14 1 0 0 15 1 0 16 1 1 07 1 1 1 1
Exor , produce un resultado 1, cuando un número impar de Variables de entrada valen 1.
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m A B C X0 0 0 01 0 0 1 12 0 1 0 13 0 1 14 1 0 0 15 1 0 16 1 1 07 1 1 1 1
Exor , produce un resultado 1, cuando un número impar de Variables de entrada valen 1.
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m A B C X0 0 0 0 01 0 0 1 12 0 1 0 13 0 1 1 04 1 0 0 15 1 0 1 06 1 1 0 07 1 1 1 1
Exor , produce un resultado 1, cuando un número impar de Variables de entrada valen 1.
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Exor produce un resultado 1, cuando
un número impar
de variables de entrada valen 1.
m A B C D X0 0 0 0 01 0 0 0 12 0 0 1 03 0 0 1 14 0 1 0 05 0 1 0 16 0 1 1 07 0 1 1 18 1 0 0 09 1 0 0 1
10 1 0 1 011 1 0 1 112 1 1 0 013 1 1 0 114 1 1 1 015 1 1 1 1
X = A B C D
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Exor produce un resultado 1, cuando
un número impar
de variables de entrada valen 1.
m A B C D X0 0 0 0 01 0 0 0 1 12 0 0 1 0 13 0 0 1 14 0 1 0 0 15 0 1 0 16 0 1 1 07 0 1 1 1 18 1 0 0 0 19 1 0 0 1
10 1 0 1 011 1 0 1 1 112 1 1 0 013 1 1 0 1 114 1 1 1 0 115 1 1 1 1
X = A B C D
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Exor produce un resultado 1, cuando
un número impar
de variables de entrada valen 1.
m A B C D X0 0 0 0 0 01 0 0 0 1 12 0 0 1 0 13 0 0 1 1 04 0 1 0 0 15 0 1 0 1 06 0 1 1 0 07 0 1 1 1 18 1 0 0 0 19 1 0 0 1 0
10 1 0 1 0 011 1 0 1 1 112 1 1 0 0 013 1 1 0 1 114 1 1 1 0 115 1 1 1 1 0
X = A B C D
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La operación Exnor es el negado de la salida de la operación Exor.
AB
A
B
EXNOR
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Condición Alternativa Impar Negado de And
Negado de Exor
Negado de Or
m A B C And Or Exor Nand Ex-Nor Nor0 0 0 0 0 0 0 1 1 11 0 0 1 0 1 1 1 0 02 0 1 0 0 1 1 1 0 03 0 1 1 0 1 0 1 1 04 1 0 0 0 1 1 1 0 05 1 0 1 0 1 0 1 1 06 1 1 0 0 1 0 1 1 07 1 1 1 1 1 1 0 0 0
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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a) 1*1= 1
Evaluar las siguiente Operación
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b) 0*0 = 0
Evaluar las siguiente Operación
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c) 1*0*0 = 0
Evaluar las siguiente Operación
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c) 1*A*0 = 0
Evaluar las siguiente Operación
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Leyes y teoremas del álgebra Booleana
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Evaluar las siguiente operación
a) 1+1= 1
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a) 1+0 = 1
Evaluar las siguiente operación
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a)0+0+0 = 0
Evaluar las siguiente operación
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Leyes y teoremas del álgebra Booleana
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And y Nand
1
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A
And y Nand
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A
And y Nand
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1
And y Nand
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Or y Nor
A
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0
Or y Nor
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A
Or y Nor
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0
Or y Nor
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Resuelva las siguientes proposiciones
1.- A 0 =2.- A 1 =3.- A A =4.- A A =
5.- A 0 =6.- A 1 =7.- A A =8.- A A =
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Propiedades
•Conmutativa
•Asociativa
•Distributiva
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Conmutativa
AND
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Conmutativa
Or
A+B = B+A
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Conmutativa
Exor
AB = BA
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Conmutativa
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Asociativa
And A(B C) = (A B) C = A B C
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Asociativa
(A B) C = A B C
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Asociativa
Or A+(B+C) = (A+B)+C = A+B+C
Exor A(BC) = (AB)C = ABC
And A(B C) = (A B) C = A B C
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Asociativa
Or A+B+C+D
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Asociativa
Or (A+B)+C+D = (A+B)+(C+D)
Or A+B+C+D
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Asociativa
Nand [A(B C)’]’ ≠ [(A B)’ C]’ ≠ (A B C)’
Nor [A+(B+C)’]’ ≠ [(A+B)’+C]’≠ (A+B+C)’
Enxor [A(BC)’]’ ≠ [(A B)’C]’≠ (ABC)’
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Asociativa
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Distributiva
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Distributiva
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A + AC + AB + BC
Distributiva
AA + AC + AB + BC=A
A + AC + AB + BC
A (1+C+B)+ BC=1A*1+ BC
A+ BC = A+ BC
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Distributiva
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Resuelva las siguientes proposiciones
1.- A 0 =2.- A 1 =3.- A A =4.- A A =