Algebra de Boole y Simplificacion

8/13/2019 Algebra de Boole y Simplificacion

1/45

Seminario: lgebra de Boole y

simplificacin lgicaAutomtica Industrial

Grado en Ingeniera QumicaUniversidad de Santiago de Compostela


2/45

Funciones bsicas

Boolean functions : NOT, AND, OR,exclusive OR(XOR) : odd function

exclusive NOR(XNOR) : even function(equivalence)


3/45

Funciones bsicas

AND Z=X Y or Z=XY

Z=1 if and only if X=1 and Y=1, otherwise Z=0

OR Z=X + Y

Z=1 if X=1 or if Y=1, or both X=1and Y=1. Z=0 if and only ifX=0 and Y=0

NOT Z=Xor

Z=1 if X=0, Z=0 if X=1


4/45

Funciones bsicas


5/45

Operaciones booleanas

Boolean AdditionLogical OR operation

Ex 4-1) Determine the values of A, B, C, and D that make the sum termA+B+C+D

Sol) all literals must be 0 for the sum term to be 0A+B+C+D=0+1+0+1=0A=0, B=1, C=0, and D=1

Boolean Multiplication

Logical AND operation

Ex 4-2) Determine the values of A, B, C, and D for ABCD=1

Sol) all literals must be 1 for the product term to be 1

ABCD=1010=1A=1, B=0, C=1, and D=0


6/45

Identidades bsicas


7/45

Propiedades del lgebra de Boole

Propiedad commutativaEl orden de los literales no importa

A + B = B + A

A B = B A


8/45


Propiedad asociativa:

A + (B + C) = (A + B) + C (=A+B+C)

A(BC) = (AB)C (=ABC)


9/45


Propiedad distributiva : A(B + C) = AB + AC


10/45


(A+B)(C+D) = AC + AD + BC + BD


11/45

Teorema de DeMorgan

Teorema de DeMorgan

F(A,A, , + , 1,0) = F(A, A, + , ,0,1)

(A B) = A + B and (A + B) = A B


12/45

Teorema de DeMorgan


13/45

Anlisis de circuitos lgicos

Expresin Booleana de un circuito lgico:


14/45

Tabla de verdad de un circuito lgico

Input OutputA B C D A(B+CD)

0 0 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 1 1 0

0 1 0 0 00 1 0 1 0

0 1 1 0 0

0 1 1 1 0

1 0 0 0 0

1 0 0 1 0

1 0 1 0 0

1 0 1 1 1

1 1 0 0 1

1 1 0 1 1

1 1 1 0 1

1 1 1 1 1

A(B+CD)=AB(C+C) (D+D) +A(B+B)CD=ABC(D+D) +ABC(D+D) +ABCD+ABCD=ABCD+ABCD+ABCD+ABCD+ABCD+ABCD

=ABCD+ABCD+ABCD+ABCD +ABCD

A(B+CD)=m11+m12+m13+m14+m15=(11,12,13,14,15)


15/45

Simplificacin usando lgebraBooleana

Usando lgebra Booleana , simplificar la expresin:

AB+A(B+C)+B(B+C)

Sol) AB+AB+AC+BB+BC =B(1+A+A+C)+AC=B+AC


16/45

Mapas de Karnaugh (*)This method coverts the truth table information into a

two-dimensional map. It then converts areas of 1s onthe map into groups. These groups are then identifiedand this gives us the simplest expression.

(*) Fuente: Logic Equation Simplification.University of Wales, Newport


17/45


18/45

STEP ONE

of the simplification process would be to fill in theKarnaugh Map

(Note: we normally only transfer 1s onto the map.)



19/45

STEP TWO

is to group the 1s. Groups are formed using the

following rules:1. Group sizes must be powers of 2 1, 2, 4, 8, 16, etc

no other size groups are allowed.

2. Groups must be square of rectangles (1 x 4 or 2 x 2

etc)3. Groups must be as large as possible (never group 2

groups of 2 if a group of 4 can be made.)

4. All 1s mustbe grouped.

5. A 1 may be grouped more than once.

6. Do not include redundant groups a redundant groupis a group that contains 1s which have all beenpreviously grouped.



20/45

STEP THREEidentifies the expression for each group. The groups

are examined one at a time. For a group the followingquestion is asked for each input one at a time:

For the group:Is the input logic state for every square in the group:

Always 1 if it is then the input appears in theexpression Always 0 - if it is then the not input appears in

the expression Both 1 and 0 - if it is then the input does not

appears in the expressionAfter each input has been checked, the expression is

the AND of the inputs states identified.



21/45

STEP FOUR

identifies the complete expression for the function.The individual group expressions are OR-ed togetherto give the simplified expression.



22/45

Example

The Truth Table and Karnaugh Map are shown below:

BABABAY

A B Y B A 0 1 Y0 0 0

0 1

1 0 1

1 1



23/45

111

1101

010

0100

Y10B AYBA



24/45

111

11101

01011

0100

Y10B AYBA

STEP 1



25/45

A B Y B A 0 1 Y

0 0 1 01 1

0 1 0

1 0 1 11

1 1 1

STEP 2



26/45

A B Y B A 0 1 Y

0 0 1 01 1

0 1 0

1 0 1 11

1 1 1

STEP 3

A always 1 so AB 1 and 0 so no BExpression

A 1 and 0 so no A

B always 0 so not BExpression

AB



27/45


28/45


29/45

Note there is one additional rule for grouping 1s on this

map and larger maps:

Rule: 1s may be grouped between the left hand columnand the right hand column.

(*) Fuente: Logic Equation Simplification.

University of Wales, Newport


30/45

ExampleA function F has the truth table shown below. Determine the

simplest Boolean Expression for the function.

A B C F A 0 0 1 1

0 0 0 1 C B 0 1 1 0 F

0 0 1 0 00 1 0 0

0 1 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1




31/45

1111

1011

11011001

11111110

0010111

00100

F0110C B1000

1100AFCBA




32/45

1111

1011

11011001

11111110

0010111

00100

F0110C B1000

1100AFCBA




33/45

1111

1011

11011001

11111110

0010111

00100

F0110C B1000

1100AFCBA

A always 1 so AB 1 and 0 so no B

C 1 and 0 so no CExpression

A 1 and 0 so no AB always 1 so BC always 1 so CExpression

A

CB

A 1 and 0 so no AB always 0 so not BC always 0 so not C

Expression CB


34/45

1111

1011

11011001

11111110

0010111

00100

F0110C B1000

1100AFCBA

Complete expression CBCBAF




35/45

ExampleThree judges A, B and C vote: 1 guilty and 0 not guilty. Design

a logic circuit using NAND only which will allow a majority

decision (F) to be found. e.g. A = 1, B = 0, C = 0 gives anoutput of 0 (not guilty)

A B C F A 0 0 1 1

0 0 0 C B 0 1 1 0 F0 0 1 0

0 1 0

0 1 1 1

1 0 01 0 1

1 1 0

1 1 1(*) Fuente: Logic Equation Simplification.



36/45

A B C D Y A 0 0 1 1

0 0 0 0 C D B 0 1 1 0 Y

0 0 0 1

0 00 0 1 0

0 0 1 1 0 1

0 1 0 0

0 1 0 1 1 1

0 1 1 0

0 1 1 1 1 0

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 11 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

4-input

Karnaugh Map

This has 16 entrieson the Truth Table

and so theKarnaugh Map has16 squares




37/45

1111

0111

1011

00111101

0101

1001

0001

1 01110

0110

1 11010

0010

0 11100

0100 0 0

1000

Y0110C D B0000

1100AYDCBA

Note there is oneadditional rulefor grouping 1s

on this map and

larger maps:Rule: 1s maybe groupedbetween the top

row and thebottom row.




38/45

Example

Four judges A, B, C and D vote: 1 guilty and 0 not guilty.Obtain a Boolean Expression that will allow a majoritydecision to be found. In the case of a split decision

the vote of A determines the outcome Y.




39/45

11111

10111

11011

1001111101

10101

11001

0000111

1 01111000110

1111 101010

0001011

0 101100

00100 10 001000

Y0110C D B00000

1100AYDCBA




40/45

11111

10111

11011

1001111101

10101

11001

0000111

1 01111000110

1111 101010

0001011

0 101100

00100 10 001000

Y0110C D B00000

1100AYDCBA

Now formgroups



CA


41/45

11111

10111

11011

1001111101

10101

11001

0000111

1 01111000110

1111 101010

0001011

0 101100

00100 10 001000

Y0110C D B00000

1100AYDCBA

Identify groups

A always 1 so AB 1 and 0 so no BC always 1 so CD 1 and 0 so no DExpression

CA

A always 1 so AB 1 and 0 so no B

C 1 and 0 so no CD always 1 so DExpression

DA

A always 1 so AB always 1 so BC 1 and 0 so no CD 1 and 0 so no DExpression

BA

A 1 and 0 so no A

B always 1 so BC always 1 so CD always 1 so DExpression

DCB

CA


42/45

11111

10111

11011

1001111101

10101

11001

0000111

1 01111000110

1111 101010

0001011

0 101100

00100 10 001000

Y0110C D B00000

1100AYDCBA

Boolean Expression

DCBDACABAY




43/45

ExampleTwo 2-bit numbers (A,B) and (C,D) are to be compared.

If (A,B) > (C,D)then the G (greater than) output is to equal 1

If (A,B) < (C,D)then the L (less than) output is to equal 1

If (A,B) = (C,D)

then the E (Equal to) output is to equal 1e.g. A = 1, B = 0, C = 1, D = 1

10 (2) is less than 11 (3) so L = 1




44/45

1111

0111

1011

00111101

0101

1001

0001

1 01110

0110

1 11010

0010

0 11100

0100 0 0

1000

G0110C D B0000

1100AELGDCBA




45/45

A 0 0 1 1

C D B 0 1 1 0 L

0 0

0 1

1 1

1 0

A 0 0 1 1

C D B 0 1 1 0 E

0 0

0 1

1 1

1 0

Expression GExpression L

Expression E(*) Fuente: Logic Equation Simplification.

Algebra de Boole y Simplificacion

Documents

Transcript of Algebra de Boole y Simplificacion