DigitalLogic_BooleanAlgebra_P.pdf

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
 Boolean Algebra is used to analyze
and simplify the digital (logic)
circuits.
 It uses only the binary numbers i.e.
0 and 1. It is also called as Binary
Algebra or logical Algebra.
 It is a convenient way and
systematic way of expressing and
analyzing the operation of logic
circuits
 Boolean algebra was invented
by George Boole in 1854.

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
0 0 0
0 1 0
1 0 0
1 1 1
A B A+B
0 0 0
0 1 1
1 0 1
1 1 1
A A’
0 1
1 0
AND OR Not

Laws in Boolean Algebra
 Commutative Law
A.B = B.A
A+B = B+A
 Associative Law
(A.B).C = A.(B.C)
(A+B) + C = A+ (B+C)
 Distributive Law
A.(B+C)=A.B+A.C
A+(B.C)=(A+B).(A+C)
 Absorption
A+ (A.B)=A
A.(A+B)=A
 AND Law
A.0 = 0
A.1 =A
A.A = A
A.A’ =0
 OR law
A+0 = A
A+1=1
A+A=A
A+A’ = 1
 Inversion
Law(Involution)
A’’ = A
 DeMorgan’s
Theorm
(x.y)’ = x’ + y’
(x+y)’ = x’ . y’
A+AB = A
A+A’B =A+B
(A+B)(A+C) = A+BC
Idempotent Law
Complement Law

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
 A Boolean function can be expressed algebraically with binary variables, the logic
operation symbols, parentheses and equal sign.
 For a given combination of values of the variables, the Boolean function can be
either 1 or 0.
 Consider for example, the Boolean Function:
F1 = x + y’z
The Function F1 is equal to 1 if x is 1 or if both y' and z are equal to 1; F1 is equal to
0 otherwise.
 The relationship between a function and its binary variables can be represented in
a truth table. To represent a function in a truth table we need a list of
the 2n combinations of the n binary variables.
 A Boolean function can be transformed from an algebraic expression into a logic
diagram composed of different Gates

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 F1
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 F1
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: The canonical SoP form for Boolean
function of truth table are obtained by ORing the
ANDed terms corresponding to the 1’s in the output
column of the truth table
 The product terms also known as minterms are
formed by ANDing the complemented and un-
complemented variables in such a way that the 0 in
the truth table is represented by a complement of
variable 1 in the truth table is represented by a
variable itself.

 SoP form – Example x y z F1 Minterms
0 0 0 0 x’y’z’ m0
0 0 1 0 x’y’z m1
0 1 0 1 x’yz’ m2
0 1 1 0 x’yz m3
1 0 0 0 xy’z’ m4
1 0 1 1 xy’z m5
1 1 0 1 xyz’ m6
1 1 1 1 xyz m7
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

 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.

 PoS form –
Example
x y z F2 Maxterms
0 0 0 0 x + y+z M1
0 0 1 0 x+y+z’ M2
0 1 0 1 x+y’ + z M3
0 1 1 0 x+y’+z’ M4
1 0 0 0 x’+y+z M5
1 0 1 1 x’ +y+z’ M6
1 1 0 1 x’+y’+z M7
1 1 1 1 x’+y’+z’ M8
F2=(x+y+z).(x+y+z’).(x+y’+z’).(x’+y+z)
F2 = (M1.M2.M4.M5)
F2 =∏(M1,M2,M4,M5)
F2 = ∏(1, 2,4,5)
Decimal numbers in the above expression
indicate the subscript of the Maxterm
notation

 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 + m1
= ∑(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

Simplification of Boolean functions
 Algebric simplification
 K-Map simplification
 Quine-McLusky Method of simplification

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]

 Simplify: C + (BC)’
=C + (BC)’ Original Expression
=C + (B’ + C’) DeMorgan's Law.
=(C + C’) + B’ Commutative, Associative Laws.
=1 + B’ Complement Law.
=1 Identity Law.

 Exercise 3: Using the theorems and laws of Boolean
Algebra, reduce the following functions
F1(A,B,C,D) = ∑(0,1,2,3,6,7,14,15)
 Solution:
= A’B’C’D’ + A’B’C’D + A’B’CD’ + A’B’CD +A’BCD’ + A’BCD + ABCD’ + ABCD
= ?
 Exercise 4: Using the theorems and laws of Boolean
Algebra, reduce the following functions
F1(X,Y,Z) = ∏(0,1,4,5,7)
 Solution:
=(X+Y+Z) (X+Y+Z’) (X’+Y+Z) (X’+Y+Z’) (X’+Y’+Z’)
= ?

Simplification Using K-Map
 Karnaugh Maps
 The Karnaugh map (K–map), introduced
by Maurice Karnaugh in 1953, is a grid-
like representation of a truth table which
is used to simplify boolean algebra
expressions.
 A Karnaugh map has zero and one
entries at different positions. It provides
grouping together Boolean expressions
with common factors and eliminates
unwanted variables from the expression.
 In a K-map, crossing a vertical or
horizontal cell boundary is always a
change of only one variable.

K-Map Simplification
 A Karnaugh map provides a systematic method for
simplifying Boolean expressions and, if properly used, will
produce the simplest expression possible, known as the
minimum expression.
 Karnaugh maps can be used for expressions with two, three,
four. and five variables. Another method, called the Quine-
McClusky method can be used for higher numbers of
variables.
 The number of cells in a Karnaugh map is equal to the total
number of possible input variable combinations as is the
number of rows in a truth table. For three variables, the
number of cells is 23 = 8. For four variables, the number of
cells is 24 = 16.

 The 4-Variable Karnaugh Map
 The 4-variable Karnaugh map is an array of sixteen
cells,
 Binary values of A and B are along the left side and
the values of C and D are across the top.
 The value of a given cell is the binary values of A and
B at the left in the same row combined with the binary
values of C and D at the top in the same column.
 For example, the cell in the upper right corner has a
binary value of 0010 and the cell in the lower right
corner has a binary value of 1010.

The 4-Variable Karnaugh Map
Figure shows the standard product terms that are represented by each cell
in the 4-variable Karnaugh map.

The 3-Variable Karnaugh Map
 A 3-variable Karnaugh map showing product terms

 Procedure
 After forming the K-Map, enter 1s for the min terms that
correspond to 1 in the truth table (or enter 1s for the min terms of
the given function to be simplified). Enter 0s for the remaining
minterms.
 Encircle octets, quads and pairs taking in use adjecency,
overlapping and rolling. Try to form the groups of maximum
number of 1s
 If any such 1s occur which are not used in any of the encircled
groups, then these isolated 1s are encircled separately.
 Review all the encircled groups and remove the redundant
groups, if any.
 Write the terms for each encircled group.
 The final minimal Boolean expression corresponding to the K-Map
will be obtained by ORing all the terms obtained above

K-Map Simplification – Example 1
 Simplify
F=A’B’C’D’ + A’B’C’D + A’BC’D’ + A’BC’D + A’BCD’ + A’BCD + AB’C’D
+ AB’CD
Solution:
Step 1: Draw the K-Map and label Properly
Step 2: Fill up the cells by 1s as per the given function which you want to
simplify
Step 3: Encircle adjacent 1s making groups of 16, 8, 4 ,2 and single 1’s
starting from big to small
Step 4: write the terms representing the groups
Step 5: The final minimal Boolean expression corresponding to the K-
Map will be obtained bu Oring all the terms obtained above

Simplify
F=A’B’C’D’ + A’B’C’D + A’BC’D’ + A’BC’D + A’BCD’ + A’BCD + AB’C’D
+ AB’CD
Step 4
Step 1 Step 2 Step 3
Step 5:
F = A’C’ + A’B + AB’D

K-Map Example 2
 Simplify F=
 Solution
The given expression is obviously not in standard form because
each product term does not have four variables.
 Map each of the resulting binary values by placing a 1 in the appropriate
cell of the 4- variable Karnaugh map.

Simplify: F=
Step 1,2 Step 3,4
Step 5
F= AB’ + AC’ + B’C’

K-Map
 For a 4-variable map:
 1-cell group yields a 4-variable product term
 8-cell group yields a 1-variable term
 16-cell group yields a value of 1 for the expression
 For a 3-variable map:
 l-cell group yields a 3-variable product term
 4-cell group yields a 1-variable term
 8-cell group yields a value of 1 for the expression

K-Map Example 3
 Simplify the following three variable function
F = A’ + AB’ + ABC’
Solution:
The given function is not in standard SoP form, so the
standard form will be
F= ∑(0,1,2,3,4,5,6)
F = A’ + B’ + C’

K-Map Simplification - Exercise
 Minimize the following function using K-Map
i) P(A,B,C,D) = ∑(0,1,2,5,8,10,11,14,15)
ii) F(x,y,z)=x’y’z’ + x’y’z + xyz’ + xyz
iii) S(a,b,c,d) = a’b’c’ + b’cd’ + a’bc’d +ab’c’d’ + ab’cd + acbd’ + abcd

Quine- McCluskey Method
 K-Map Method is a useful tool for the simplification of
Boolean function up to four variables. Although this
method can be used for 5 or 6 variables but it is not
simple to use.
 Another method developed by Quine and improved by
McCluskey was found to be good for simplification of
Boolean functions of any number of variables.
Self Study

DigitalLogic_BooleanAlgebra_P.pdf

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