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Algebra Booleana 2

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Published on February 21, 2014

Author: IVN_Galileo

Source: slideshare.net

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Algebra Booleana 2
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ALGEBRA BOOLEANA SIMPLIFICACIÓN DE CIRCUITOS

Tabla de teoremas del Algebra Booleana 1 2 3 4 5 6 7 8 9 10 11 12 13 TEOREMA 0A = 0 1A = A AA = A AA’ = 0 AB = BA ABC = A(BC) (ABC)’ = A’+B’+C’ AB+AC = A(B+C) AB+AB’ = A A+AB = A A+A’B = A+B CA+CA’B = CA+CB AB+A’C+BC=AB+A’C DUAL 1+A=1 0+A=A A+A=A A + A’ = 1 A+ B=B+A A+B+C = A+(B+C) (A+B+C)’ = A’B’C’ (A+B)(A+C) = A+BC (A+B)(A+B’) = A A(A+B) = A A(A’+B) = AB (C+A)(C+A’+B) = (C+A)(C+B) (A+B)(A’+C)(B+C)=(A+B)(A’+C)

Ejemplo  Con la siguiente tabla construya la expresión booleana X 1 1 1 1 0 0 0 0 Y 1 1 0 0 1 1 0 0 Z 1 0 1 0 1 0 1 0 f(X,Y,Z) 1 0 1 0 1 0 0 0 XYZ + XY´Z+X´YZ

Ejemplo XYZ + XY´Z+X´YZ XZ(Y+Y´) +X´YZ Y+Y´= 1 XZ+X´YZ Z(X+X´Y) DONDE (X+X´Y) = X+Y Z(X+Y) = XZ+YZ

Ejemplo Z(X+Y) = XZ+YZ

Ejercicio  Construya la expresión booleana y dibuje el circuito de la siguiente tabla. X 1 1 1 1 0 0 0 0 Y 1 1 0 0 1 1 0 0 Z 1 0 1 0 1 0 1 0 f(X,Y,Z) 1 0 1 0 0 0 1 1

Ejercicio  Construya la expresión booleana y dibuje el circuito de la siguiente tabla. X 1 1 1 1 0 0 0 0 Y 1 1 0 0 1 1 0 0 Z 1 0 1 0 1 0 1 0 f(X,Y,Z) 1 0 1 0 0 0 1 1 XYZ + XY´Z + X´Y´Z + X´Y´Z´

Ejercicio  XYZ + XY´Z + X´Y´Z + X´Y´Z´ XZ(Y+Y´) + X´Y´Z + X´Y´Z´ del teorema 4 Y + Y´ =1 XZ + X´Y´(Z+Z´) del teorema 4 Z + Z´ = 1 XZ + X´Y´

Ejercicio XZ + X´Y´

Ejercicio Construya la expresión booleana y dibuje el circuito de la siguiente tabla. X 1 1 1 1 0 0 0 0 Y 1 1 0 0 1 1 0 0 Z 1 0 1 0 1 0 1 0 f(X,Y,Z) 0 0 1 1 1 0 1 0

Ejercicio Construya la expresión booleana y dibuje el circuito de la siguiente tabla. X 1 1 1 1 0 0 0 0 Y 1 1 0 0 1 1 0 0 Z 1 0 1 0 1 0 1 0 f(X,Y,Z) 0 0 1 1 1 0 1 0 XY´Z + XY´Z´ + X´YZ + X´Y´Z

Ejercicio La expresión booleana resultante es f(X,Y,Z)= XY´Z + XY´Z´ + X´YZ + X´Y´Z XY´(Z+Z´) + X´Z(Y+Y´) XY´ + X´Z

Ejercicio XY´ + X´Z

Ejercicio Simplifique el siguiente circuito de ser posible

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