Boolean Algebra: Introduction, Laws, Functions, and Simplification Techniques, Exams of Algebra

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BOOLEAN ALGEBRA

Compiled By: Afaq Alam Khan

Index

 Introduction

 Boolean Algebra Laws

 Boolean functions

 Operation Precedence

 Boolean Algebra Function

 Canonical Forms

 SOP

 POS

 Simplification of Boolean Functions

 Algebric simplification

 K-Map

 Quine –McCluskey Method (Tabular Method)

Introduction

 Variable used in Boolean algebra can have only two

values. Binary 1 for HIGH and Binary 0 for LOW.

 Complement of a variable is represented by an

overbar (-). Thus, complement of variable B is

represented as B’. Thus if B = 0 then B’= 1 and if B =

1 then B’= 0.

 ORing of the variables is represented by a plus (+) sign

between them. For example ORing of A, B, C is

represented as A + B + C.

 Logical ANDing of the two or more variable is

represented by writing a dot between them such as

A.B.C. Sometime the dot may be omitted like ABC.

Boolean Operations

A B A.B

A B A+B

A A’

AND

OR

Not

Operator Presedence

 The operator Precedence for evaluating Boolean

expression is:

 1. Parentheses

 2. NOT

 3. AND

 4. OR

Example

 Using the Theorems and Laws of Boolean algebra,

Prove the following.

(A+B) .(A+A’B’).C + (A’.(B+C’))’ + A’.B + A.B.C = A+B+C

Boolean Algebric Function

 Consider the following Boolean

function:

F1= x’y’z+xy’z’+xy’z+xyz’+xyz

After Simplification

F1 = x + y’z

 A Boolean function can be

represented in a truth table.

x y z F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1 Truth Table

y

z

x F

Realization of Boolean Function using Gates Canonical Form Non Canonical Form

 The purpose of Boolean algebra is to facilitate the analysis and design of digital circuits. It provides a convenient tool to:

 Express in algebraic form a truth table relationship between

binary variables.

 Express in algebraic form the input-output relationship of logic

diagrams.

 Find simpler circuits for the same function.

 A Boolean function specified by a truth table can be expressed algebraically in many different ways. Two ways of forming Boolean expressions are Canonical and Non- Canonical forms.

Canonical Forms For Boolean Function  SoP form – Example x y z F1 Minterms 0 0 0 0 x’y’z’ m 0 0 1 0 x’y’z m 0 1 0 1 x’yz’ m 0 1 1 0 x’yz m 1 0 0 0 xy’z’ m 1 0 1 1 xy’z m 1 1 0 1 xyz’ m 1 1 1 1 xyz m

F1= x’yz’ + xy’z + xyz’ + xyz

F1 = (m2+m5+m6+m7)

F1 =∑(m2,m5,m6,m7)

F1 = ∑ (2, 5,6,7)

Decimal numbers in the above

expression indicate the subscript of

the minterm notation

Canonical Forms For Boolean Function  PoS Form: The canonical PoS form for Boolean function of truth table are obtained by ANDing the ORed terms corresponding to the 0’s in the output column of the truth table  The product terms also known as Maxterms are formed by ORing the complemented and un- complemented variables in such a way that the 1 in the truth table is represented by a complement of variable 0 in the truth table is represented by a variable itself.

Canonical Forms For Boolean Function  Example: Express the following in SoP form F1 = x + y’z  Solution: =(y+y’)x + y’z(x+x’) [because x+x’=1] =xy + xy’ + xy’z + x’y’z =xy(z+z’) + xy’(z+z’) + xy’z + x’y’z =xyz + xyz’ + xy’z + xy’z’ + xy’z + x’y’z =xyz + xyz’ + (xy’z + xy’z) + xy’z’ + x’y’z = xyz + xyz’ + xy’z + xy’z’ + x’y’z [because x+x =x] = m7 + m6 + m5 + m4 + m = ∑(m7, m6, m5, m4, m1) = ∑(1,4,5,6,7)

Canonical Forms - Exercises

 Exercise 1: Express G(A,B,C)=A.B.C + A’.B + B’.C in

SoP form.

 Exercise 2: Express F(A,B,C)=A.B’ + B’.C in PoS form

Algebraic Simplification

 Using Boolean algebra techniques, simplify this

expression: AB + A(B + C) + B(B + C)

 Solution

=AB + AB + AC + BB + BC (Distributive law)

=AB + AB + AC + B + BC (B.B=B)

= AB + AC + B + BC (AB+AB=AB)

= AB + AC + B (B+BC =B)

=B+AC (AB+B =B)

Algebric Simplification

 Minimize the following Boolean expression using Algebric

Simplification

F(A,B,C)=A′B+BC′+BC+AB′C′

 Solution

=A′B+(BC′+BC′)+BC+AB′C′ [indeponent law]

= A′B+(BC′+BC)+(BC’+AB′C′)

= A′B+B(C′+C)+C’(B+AB′)

=A’B + B.1+ c’ (B+A)

= B(A′+1)+C′(B+A)

=B + C′(B+A) [A’+1=1]

= B+BC′+AC′

= B(1+C′)+AC′

= B+AC′ [1+C’ = 1]