11. simpl met-algebraicos
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Transcript of 11. simpl met-algebraicos
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Minimización de funciones Booleanas
• Manipulación Algebraica• Mapas de Karnaugh
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Manipulación Algebraica• Factorización• Duplicando un termino ya existente• Teorema del consenso• Propiedad distributiva• Identidades• Teorema de Dmorgan
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Mapas de Karnaugh
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FactorizaciónManipulación Algebraica
Factor común B
m A B F0 0 0 01 0 1 12 1 0 03 1 1 1
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Factorización
m A B F0 0 0 01 0 1 12 1 0 03 1 1 1
F(A,B)=AB+AB = B
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la Factorización se efectúa cuando solo cambia una variable entre dos términos
y esta variable se elimina
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m S P E CS CP CE0 0 0 0 0 0 01 0 0 1 0 0 12 0 1 0 0 1 03 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) =
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m S P E CS CP CE0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’
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m S P E CS CP CE0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E
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m S P E CS CP CE0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E + S P E’
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m S P E CS CP CE0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E + S P E’ + S P E
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CS (S, P, E ) = S P’ E’ + S P’ E + S P E’ + S P E
CS (S, P, E ) = S P’ (E’+E)+ S P (E’+E)
CS (S, P, E ) = S P’ + S P
CS (S, P, E ) = S (P’+P)
CS (S, P, E ) = S
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m S P E CS CP CE0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E + S P E’ + S P E
CS (S,) = S
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m S P E CS CP CE
0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CS (S, P, E ) = S P’ E’ + S P’ E + S P E’ + S P E = S
CS (S,) = S
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m S P E CS CP CE0 0 0 0 0 0 0
1 0 0 1 0 0 1
2 0 1 0 0 1 0
3 0 1 1 0 1 0
4 1 0 0 1 0 0
5 1 0 1 1 0 0
6 1 1 0 1 0 0
7 1 1 1 1 0 0
CP (S, P, E ) = S’ P E’ + S’ P E
CP (S, P, E ) = S’ P (E’+E)
CP (S, P, E ) = S’ P
CP (S, P ) = S’ P
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Duplicando un termino ya existente
F= A + B
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m A B C X0 0 0 0 11 0 0 1 12 0 1 0 13 0 1 1 04 1 0 0 05 1 0 1 06 1 1 0 07 1 1 1 0
FX (A,B,C) = A’ B’ C’ + A’ B’ C + A’ B C’
FX (A,B,C) = A’ B’ C’ + A’ B’ C + A’ B C’ + A’ B’ C’
FX (A,B,C) = A’ B’ + A’ C’
FX (A,B,C) = A’ (B’ + C’)
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S(A,B,C,D)= A’B’C'+ A’B’CS(A,B,C,D)= A’B’CS(A,B,C,D)= A’B’C'+ A’B’C + AB’C' +’BC'D'S(A,B,C,D)= A’B’C'+ A’B’C + AB’C'
M A B C D S
0 0 0 0 0 1
1 0 0 0 1 1
2 0 0 1 0 1
3 0 0 1 1 1
4 0 1 0 0 1
5 0 1 0 1 0
6 0 1 1 0 0
7 0 1 1 1 0
8 1 0 0 0 1
9 1 0 0 1 1
10 1 0 1 0 0
11 1 0 1 1 0
12 1 1 0 0 1
13 1 1 0 1 0
14 1 1 1 0 0
15 1 1 1 1 0
S(A,B,C,D)= A'B' + B'C' + C'D'
S(A,B,C,D)= A’B’C'D'+ A’B’C'D + A’B’CD'+A’B’CD
+A’BC'D' + AB’C'D' + AB’C'D + ABC'D'
0 1 2 3
4 8 9 12
0-1 2-3 8-9 4-12
S(A,B,C,D)= B’C'+ A’B’ + +’BC'D'0-1, 8-9
0-1, 2,3
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Teorema del consenso
m. Acuerdo producido por consentimiento entre todos los miembros de un grupo o entre varios grupos.
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Teorema del consenso
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Teorema del consenso
1
1
1
11
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Teorema del consenso
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Propiedad Distributiva
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Propiedad Distributiva
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F= A’ B + A B’ + A B + A’ C’
F= B + A + A’ C’
F= B + (A+ A’)(A+ C’)
F= B + A+ C’ F= A+B+C’
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Actividad
Usando como recursos • Factorización• Duplicando un termino ya existente• Teorema del consenso• Propiedad distributiva• Identidades• Teorema de Dmorgan Resuelva las siguientes funciones
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1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)X (Y+Z) = XY +XZ4.-Teorema del consenso
AB+A’C+BC = AB+A’C5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=AA 0 =0 A + 0 = AA 1 =A A + 1 =1A A’ =0 A+A’ =1 1 1+ B’+ C
2 DC’(0) 3 A’+B+A4 A+ A’ BC5 A’BC+A’BC’
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F1 (B,C)= 1+B’+C
F1 (B,C)= 1
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)X (Y+Z) = XY +XZ4.-Teorema del consenso
AB+A’C+BC = AB+A’C5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=AA 0 =0 A + 0 = AA 1 =A A + 1 =1A A’ =0 A+A’ =1
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F2 (D,C)= DC’(0)
F2 (D,C)= 0
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)X (Y+Z) = XY +XZ4.-Teorema del consenso
AB+A’C+BC = AB+A’C5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=AA 0 =0 A + 0 = AA 1 =A A + 1 =1A A’ =0 A+A’ =1
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F3 (A, B) = A’+B+A
F3 (A, B) = 1
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)X (Y+Z) = XY +XZ4.-Teorema del consenso
AB+A’C+BC = AB+A’C5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=AA 0 =0 A + 0 = AA 1 =A A + 1 =1A A’ =0 A+A’ =1
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F4 (A,,B,C) = A+A’BC
F4 (A,,B,C)=(A+A’)(A+BC)
F4 (A,,B,C)=A+BC
1.-Identidades
2.- Factorización
AB’ + AB = A(B’+B)= A3.- Propiedad Distributiva
X+YZ = (X+Y) (X+Z)X (Y+Z) = XY +XZ4.-Teorema del consenso
AB+A’C+BC = AB+A’C5.-Teorema de Dmorgan
(AB)’=A’+ B’ (A+B)’=A’ B’
A+B =(A’ B’)’ AB =(A’+B’)’
AND OR
A A=A A + A=AA 0 =0 A + 0 = AA 1 =A A + 1 =1A A’ =0 A+A’ =1