Boolean Algebra

Understand about Boolean Algebra and its functions

Understand and apply the Logic gates Use the Karnaugh Maps for

simplification of sum-of-product expansions

Introduction Boolean Expressions and Boolean

Functions Minterm and Maxterm Logic Gates Karnaugh Map

Invented by and named after George Boole. Commonly, and especially in computer science and digital

electronics, this term is used to mean two-value logic. Provide the operations and rules for working with the set

{0,1}. Map logical propositions to symbols. Permit manipulation of logic statements using Mathematics. A Boolean algebra B consists of a set S containing distinct

elements 0 and 1, binary operators + and . on S, and a unary operator on S, we write B = (S, +, . , , 0, 1).

Note: • 0 and 1 in the foregoing definition are merely symbolic names.

In general, they have nothing to do with the numbers 0 and 1.• + and are merely binary operators. In general, they have nothing to do with

the ordinary addition and multiplication.

Introduction

Boolean algebra deals with

- set of 2 elements {0, 1}- binary operators: OR ( / +), AND( / .) - unary operator NOT { / ‘ }

Rules:

/ +0 0 0

0 1 1

1 0 1

1 1 1

OR / .

0 0 0

0 1 0

1 0 0

1 1 1

AND

/ ‘0 0

1 1

NOT

Boolean Algebra

Boolean functions can be represented using expressions made up from variables and Boolean operations.

Example:

rprqprqpF ),,(

zxyzxzyxF ),,(

Example F(x, y, z) = xy’ + z is a Boolean function where xy’ + z is a Boolean expression.

The truth table for F(x, y, z) = xy’ + z is

x y z y’ xy’ xy’+z0 0 0 1 0 00 0 1 1 0 10 1 0 0 0 00 1 1 0 0 11 0 0 1 1 11 0 1 1 1 11 1 0 0 0 01 1 1 0 0 1

Any boolean expression may be expressed in terms of either minterms or maxterms.

To do this we must first define the concept of a literal.

A literal is a single variable within a term which may or may not be complemented. For an expression with N variables, minterms and maxterms are defined as follows : A minterm is the product of N distinct literals where each

literal occurs exactly once. A maxterm is the sum of N distinct literals where each

literal occurs exactly once.

For a two-variable expression, the minterms and maxterms are as follows

For a three-variable expression, the minterms and maxterms are as follows

Also known as sum-of-products expansion. The Sum of Products form represents an

expression as a sum of minterms. Sum of minterms corresponding to input

combinations for which the function produces a 1 as output.

To derive the Sum of Products form from a truth table, OR together all of the minterms which give a value of 1.

Consider the truth table;

F(X,Y) = X.Y’ + X.Y

Also known as product-of-sums expansion. The Product of Sums form represents an

expression as a product of maxterms. Product of maxterms corresponding to input

combinations for which the function produces a 0 as output.

Any Boolean Expression has a DNF and a CNF. Given a Boolean function, we can obtain the DNF

and CNF either by Truth table or Boolean identity laws.

To derive the Product of Sums form from a truth To derive the Product of Sums form from a truth table, AND together all of the maxterms which table, AND together all of the maxterms which give a value of 0. give a value of 0.

Consider the truth table; Consider the truth table;

F(X,Y) = X + Y’

Recall that a boolean equation can be represented by a Truth Table

A B C F 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1

A truth table for a boolean function of N variables has 2N entries.

The ‘1’s represent F(A,B,C).

The ‘0’s represent F’(A,B,C)

Can write SOP form of equation directly from truth table.

A B C F 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1

A’BCAB’C’AB’CABC’ABC

F(A,B,C) = A’BC + AB’C’ + AB’C + ABC’ + ABC

Note that each term in has ALL variables present. If a product term has ALL variables present, it is a MINTERM.

To get POS form of F, write SOP form of F’, then use DeMorgan’s Law.

A B C F 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1

A’B’C’A’B’CA’BC’

F’(A,B,C) = A’B’C’ + A’B’C + A’BC’ Take complement of both sides:

(F’(A,B,C))’ = (A’B’C’ + A’B’C + A’BC’)’ Apply DeMorgan’s Law to right side. Left side is (F’)’ = F.

F(A,B,C) = (A’B’C’)’ (A’B’C)’ (A’BC’)’apply DeMorgan’s Law to each term

F(A,B,C) = (A+B+C) (A+B+C’)(A+B’+C)

POS Form!!

We saw that:

F(A,B,C) = (A+B+C) (A+B+C’)(A+B’+C)

F(A,B,C) = A’BC + AB’C’ + AB’C + ABC’ + ABC’ + ABC

SOP form. If a product term has all variables present, it is a MINTERM.

POS form. If a sum term has all variables present, it is a MAXTERM.

All Boolean functions can be written in terms of either Minterms or Maxterms.

Give the expression represented by the following truth table in both Sum of Products and Product of Sums forms.

The basic element in a digital electronic system Performs a logical operation on one or more logic

inputs and produces a single logic output Because the output is also a logic-level value, an

output of one logic gate can connect to the input of one or more other logic gates.

The logic normally performed is Boolean logic and is most commonly found in digital circuits.

Logic gates are primarily implemented electronically using diodes or transistors,

Logic Gates

OR AND NOT NOR NAND XOR

Not Gate

Inputx

Outputx

Two inputs AND Gate

Outputxy

xx00 1111 00

x

xx yy00 00 0000 11 0011 00 0011 11 11

xy

Outputx + y

xx yy x + yx + y00 00 0000 11 1111 00 1111 11 11

Two Inputs OR Gate

NOR gate

NAND gate

)( BAINPUT XORA B X=

0 0 0

0 1 1

1 0 1

1 1 0

XOR Gate

B)(A

Construct circuits that produce the following outputs:

Find the output of the given circuit shown below and write the logic table:

x y z0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1

29

(xy)’+(x’z)

Karnaugh Maps (K-Maps) are a graphical method of visualizing the 0’s and 1’s of a boolean function K-Maps are very useful for performing Boolean

minimization. Will work on 2, 3, and 4 variable K-Maps in

this class. Karnaugh maps can be easier to use than

boolean equation minimization once you get used to it.

Karnaugh Map

A K-map has a square for each ‘1’ or ‘0’ of a boolean function.

Two variable K-map has 22 =4 squaresThree variable K-map has 23 = 8 squaresFour variable K-map has 24 = 16 squares

2 variable 3 variable 4 variable

In order to use a Karnaugh map, it is best if the Boolean expression is in Disjunctive Normal Form (DNF).

1. This means that every variable should be represented in every term (for a variable X, either X or ~X should be in every term) Ex: ABC v ~ABC v ~A~B~C is in DNF, but AC v ~ABC v

~A~B~C is not [no B in the first term]

2. The variables within a term are ANDed together, and the terms are ORed together. Ex: (AvBvC)(~AvBvC)(~Av~Bv~C) [variables ORed and terms ANDed instead of vice-versa]

3. There should be no parentheses Ex: ABC v ~A(BCv~B~C) [use of parenthesis] are not in DNF.

Two-variable Two-variable mapmap

Ex:Ex: A’B’ v A’B v AB A’B’ v A’B v AB

Two-variable Two-variable mapmap

Ex:Ex: A’B’ v A’B v AB A’B’ v A’B v ABA A’

B

B’ 1

11

0

Two-variable map

Ex: ~A~B v ~AB v AB A A’

B

B’ 1

11

0

On a two-variable map, look for a pair of 1's that are either in the same row or column. If we find such a pair, we record the common variable.

= B+ A’

Three-variable map Ex:

XY XY’ X’Y’ X’Y

Z

Z’

X’Y’Z’ v X’YZ v XYZ’ v XY’Z v XYZ v X’YZ’

Three-variable map Ex:

XY XY’ X’Y’ X’Y

Z

Z’ 1

X’Y’Z’ v X’YZ v XYZ’ v XY’Z v XYZ v X’YZ’

1

1

11

1

On a three-variable map we begin by looking for groups of four 1's.

The four squares must form a rectangle (i.e. four in a row or a 2x2 square) - 'L' shapes and zigzags aren't a valid group of 4.

A group of four will have one common variable. A pair must have two common variables.

In the examples below, the common variable for the left map is ~C and the common variable for the right group is B.

In the examples below, the left pair has the common variables BC, while the right pair has the common variables A~B.

Note that the common variables within a group of 1's are ANDed. A solo 1 must have all 3 variables.

The map "wraps around" from the left edge to the right, like the center of a paper towel roll.

The common variables for the pair on the left are ~B~C, while the group of four on the right has the common variable ~B.

Three-variable map Ex:

XY XY’ X’Y’ X’Y

Z

Z’ 1

X’Y’Z’ v X’YZ v XYZ’ v XY’Z v XYZ v X’YZ’

1

1

11

1

= Y + XZ + X’Z’

Four-variable mapFour-variable map Ex: Ex: abcd’ v ab’cd v ab’cd’ v abcd v abcd’ v ab’cd v ab’cd’ v abcd v

a’b’d’c v a’b’d’c’ v a’b’dc v a’b’cd’ a’b’d’c v a’b’d’c’ v a’b’dc v a’b’cd’ v a’bcd v a’bc’d’ v a’bd’cv a’bcd v a’bc’d’ v a’bd’c

Four-variable mapFour-variable map Ex:Ex: abcd’ v ab’cd v ab’cd’ v abcd v a’b’d’c v abcd’ v ab’cd v ab’cd’ v abcd v a’b’d’c v

a’b’d’c’ v a’b’dc v a’bcd v a’bc’d’ v a’bd’ca’b’d’c’ v a’b’dc v a’bcd v a’bc’d’ v a’bd’c

AB AB’ A’B’ A’B

CD

C’D

C’D’

CD’ 1

1

1

1

1

1

1 1

1

1

On a four-variable map, begin by looking for groups of eight 1's. The eight squares must form a rectangle - odd shapes aren't a valid group of eight.

Next look for groups of four, then pairs, then solos.

A group of eight will have one common variable, a group of four will have two common variables, a pair will have three common variables, and a solo will have all four variables.

The map "wraps around" from the left edge to the right, AND from top to bottom, kind of like a map of the world. This means groups can overlap the left and right, or top and bottom edges.

The four corners form a "double wraparound" group of four, with the common variables ~B~D:

Four-variable mapFour-variable map Ex: abcd’ v ab’cd v ab’cd’ v abcd v a’b’d’c v

a’b’d’c’ v a’b’dc v a’bcd v a’bc’d’ v a’bd’c

AB AB’ A’B’ A’B

CD

C’D

C’D’

CD’ 1

1

1

1

1

1

1 1

1

1

= A’D’ + C

QUESTIONSQUESTIONS

Write the minimize expression for

cda v c’d’a v da’c’ v cd’a’ v ad’c v adc’

Construct a circuit to produce the following

output:

( ) ( ' ) 'ab c b d