Boolean algebra and logic gates

1. BOOLEAN ALGEBRA
AND
LOGIC GATE
BY
Prof. K Adisesha

2. OUTLINE OF BOOLEAN ALGEBRA
2
 2.1 Introduction to Boolean algebra
 2.2 History of Boolean algebra
 2.3 Logical Operators
 2.4 Basic theorems and properties of Boolean algebra
 2.5 Boolean functions
 2.6 Canonical and standard forms
 2.7 Digital logic gates
Prof. K Adisesha

3. INTRODUCTION
 An algebra that deals with binary number
system is called “Boolean Algebra”.
 It is very power in designing logic circuits used by
the processor of computer system.
 The logic gates are the building blocks of all the
circuit in a computer.
 Boolean algebra deals with truth table TRUE
and FALSE.
 If result of any logical statement or expression is
always TRUE or 1, it is called Tautology and if
the result is always FALSE or 0, it is called
Fallacy
 It is also called as “Switching Algebra”. 3
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4. GEORGE BOOLE
 Father of Boolean algebra
 Boolean algebra derives its name from the
mathematician George Boole (1815-1864) who
is considered the “Father of symbolic logic”.
 He came up with a type of boolean algebra, the
three most basic operations of which were (and
still are) AND, OR and NOT.
 It was these three functions that formed the
basis of his premise, and were the only
operations necessary to perform comparisons
or basic mathematical functions.
George Boole (1815 - 1864)
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5. BOOLEAN ALGEBRA
 A variable used in Boolean algebra or Boolean
equation can have only one of two variables. The two
values are FALSE (0) and TRUE (1)
 A Sentence which can be determined to be TRUE or
FALSE are called logical statements or truth
functions and the results TRUE or FALSE is called
Truth values.
 Boolean Expression consists of
 Literal: A variable or its complement
 Product term: literals connected by •
 Sum term: literals connected by +
• A truth table is a mathematical table used in logic to
computer functional values of logical expressions.
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6. TRUTH TABLE
 A truth table is a mathematical table used in logic to
computer functional values of logical expressions.
 A truth table is a table whose columns are statements
and whose rows are possible scenarios.
 Example: Consider the logical expression
Logical Statement: Sports = “Sunny can Play
Cricket OR Football”
Y = A OR B (Logical Variables: Y, A, B,
Logical Operator OR)
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A=Cricket B=Football Y=A OR B
0 0 0
0 1 1
1 0 1
1 1 1

7. LOGICAL OPERATORS
 There are three logical operator, AND, OR and NOT.
 These operators are now used in computer
construction known as switching circuits.
 B = {0, 1} and two binary operators, ‘+’ and ‘.’
 The rules of operations: AND, OR and NOT.
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8. AND OPERATOR
 The AND operator is a binary operator. This operator
operates on two variables.
 The operation performed by AND operator is called
logical multiplication.
 The symbol we use for it is ‘.’
 Example: X . Y can be read as X AND Y
 The Truth table and the Venn diagram for the NOT
operator is:
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10. OR OPERATOR
 The OR operator is a binary operator. This operator
operates on two variables.
 The operation performed by OR operator is called
logical addition.
 The symbol we use for it is ‘+’.
 Example: X + Y can be read as X OR Y
 The Truth table and the Venn diagram for the NOT
operator is:
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12. NOT OPERATOR
 The Not operator is a unary operator. This operator
operates on single variable.
 The operation performed by Not operator is called
complementation.
 The symbol we use for it is bar.
 𝐗 means complementation of X
 If X=1, X =0 If X=0, X =1
 The Truth table and the Venn diagram for the NOT
operator is:
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14. EVALUATION OF BOOLEAN EXPRESSION
USING TRUTH TABLE
To create a truth table, follow the steps given below.
 Step 1: Determine the number of variables, for n
variables create a table with 2n rows.
 For two variables i.e. X, Y then truth table will need 22 or
4 rows.
 For three variables i.e. X, Y, Z, then truth table will need
23 or 8 rows.
 Step 2: List the variables and every combination of 1
(TRUE) and 0 (FALSE) for the given variables
 Step 3: Create a new column for each term of the
statement or argument.
 Step 4: If two statements have the same truth values,
then they are equivalent. 14
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15. CONSIDER THE FOLLOWING BOOLEAN
EXPRESSION F=X+Y
 Step 1: This expression as two variables X and Y,
then 22 or 4 rows.
 Step 2: List the variables and every combination
of X and Y.
 Step 3: The final column contain the values of
F=X+ Y.
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CONSIDER THE FOLLOWING BOOLEAN
EXPRESSION BOOLEAN ALGEBRA
 The truth table for the
Boolean function:
is shown at the right.
 To make evaluation of the
Boolean function easier, the
truth table contains extra
(shaded) columns to hold
evaluations of subparts of
the function.
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17. OPERATOR PRECEDENCE
 The operator precedence for evaluating Boolean
Expression is
 Parentheses
 NOT
 AND
 OR
 Examples
 x y' + z
 (x y + z)'
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18. BOOLEAN FUNCTIONS
 A Boolean function
 Binary variables
 Binary operators OR and AND
 Unary operator NOT
 Parentheses
 Examples
 F1= x y z'
 F2 = x + y'z
 F3 = x' y' z + x' y z + x y'
 F4 = x y' + x' z
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19. BASIC THEOREMS
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21. DUALITY
 The principle of duality is an important concept. It states
that every algebraic expression deducible from the
postulates of Boolean algebra, remains valid if the
operators identity elements are interchanged.
 To form the dual of an expression, replace all + operators
with . operators, all . operators with + operators, all ones
with zeros, and all zeros with ones.
 Form the dual of the expression
a + (bc) = (a + b)(a + c)
 Following the replacement rules…
a(b + c) = ab + ac
 Take care not to alter the location of the parentheses if
they are present. 21
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Indempotence Law:
“This law states that when a variable is combines with itself
using OR or AND operator, the output is the same variable”.

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24. ABSORPTION LAW:
“THIS LAW ENABLES A REDUCTION OF COMPLICATED
EXPRESSION TO A SIMPLER ONE BY ABSORBING COMMON TERMS”.
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25. DEMORGAN’S THEOREM
 Theorem 5(a):Statement: “When the OR sum of two variables is inverted, this is same
as inverting each variable individually and then AND ing these inverted variables”
 (x + y)’ = x’y’
 Theorem 5(b): “When the AND product of two variables is inverted, this is same as
inverting each variable individually and then OR ing these inverted variables”
 (xy)’ = x’ + y’
 By means of truth table
x y x’ y’ x+y (x+y)’ x’y’ xy x’+y' (xy)’
0 0 1 1 0 1 1 0 1 1
0 1 1 0 1 0 0 0 1 1
1 0 0 1 1 0 0 0 1 1
1 1 0 0 1 0 0 1 0 0
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26. DEMORGAN’S FIRST THEOREM
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FIRST LAW: “The complement of a logical sum equals the logical
product of the complements.

27. DEMORGAN’S SECOND THEOREM
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SECOND LAW: “The complement of a logical product equals the
logical sum of the complements.

28. SIMPLIFICATION OF BOOLEAN EXPRESSION:
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Simplification of Boolean expression can
be achieved by two popular methods:
o Algebraic Manipulation
o Karnaugh Maps (K Map)

29. ALGEBRAIC MANIPULATION
 To minimize Boolean expressions
 Literal: single variable in a term (complemented or uncomplemented )
(an input to a gate)
 Term: an implementation with a gate
 The minimization of the number of literals and the number of terms → a
circuit with less equipment
 It is a hard problem (no specific rules to follow)
 Example 2.1
1. x(x'+y) = xx' + xy = 0+xy = xy
2. x+x'y = (x+x')(x+y) = 1 (x+y) = x+y
3. (x+y)(x+y') = x+xy+xy'+yy' = x(1+y+y') = x or x+yy’ = x+ 0 = x
4. xy + x'z + yz = xy + x'z + yz(x+x') = xy + x'z + yzx + yzx' = xy(1+z)
+ x'z(1+y) = xy +x'z
5. (x+y)(x'+z)(y+z) = (x+y)(x'+z), by duality from function 4.
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Examples

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EXAMPLES
 Example 2.2
 F1' = (x'yz' + x'y'z)' = (x'yz')' (x'y'z)' = (x+y'+z) (x+y+z')
 F2' = [x(y'z'+yz)]' = x' + (y'z'+yz)' = x' + (y'z')' (yz)’
= x' + (y+z) (y'+z')
 Example 2.3: a simpler procedure
 Take the dual of the function and complement each
literal
1. F1 = x'yz' + x'y'z.
The dual of F1 is (x'+y+z') (x'+y'+z).
Complement each literal: (x+y'+z)(x+y+z') = F1'
2. F2 = x(y' z' + yz).
The dual of F2 is x+(y'+z') (y+z).
Complement each literal: x'+(y+z)(y' +z') = F2'
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34. 34
CANONICAL AND STANDARD FORMS
Minterms and Maxterms
 Boolean expression expressed as sum of Minterms or
product of Maxterms are called canonical forms.
 For example, the following expressions are the
Minterm canonical form and Maxterm canonical form
of two variables X and Y.
 Minterm Canonical = f(X, Y) = X’Y’ + X’ Y +X Y’+ X Y
Maxterm Canonical = f(X, Y) = (X+Y).(X +Y’).(X’+Y’)
 The Minterm canonical expression is the sum of all
Minterms.
 The maxterm canonical expression is the product of all
Maxterms.
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MINTERMS AND MAXTERMS
 Each maxterm is the complement of its corresponding
minterm, and vice versa.
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MINTERMS AND MAXTERMS
 Any Boolean function can be expressed by
 A truth table
 Sum of minterms
 f1 = x'y'z + xy'z' + xyz = m1 + m4 +m7 = S(1, 4, 7) (Minterms)
 f2 = x'yz+ xy'z + xyz'+xyz = m3 + m5 +m6 + m7 (Minterms)
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MINTERMS AND MAXTERMS
 The complement of a Boolean function
 The minterms that produce a (0)
 f1' = m0 + m2 +m3 + m5 + m6
= x'y'z'+x'yz'+x'yz+xy'z+xyz'
 f1 = (f1')' = m’0 . m’2 . m’3 . m’5 . m’6
= (x+y+z)(x+y'+z) (x+y'+z') (x'+y+z')(x'+y'+z)
= M0 M2 M3 M5 M6
 f2 = (x+y+z)(x+y+z')(x+y'+z)(x'+y+z)=M0M1M2M4
 Any Boolean function can be expressed as
 A sum of minterms (“sum” meaning the ORing of terms).
 A product of maxterms (“product” meaning the ANDing of
terms).
 Both Boolean functions are said to be in Canonical form.
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SUM OF MINTERMS (SOP)
 Sum of minterms: there are 2n minterms
 Example 4: Express F = A+BC' as a sum of minterms.
 F(A, B, C) = S(1, 4, 5, 6, 7)
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PRODUCT OF MAXTERMS
 Product of maxterms: there are 2n maxterms
 Example 5: express F = xy + x'z as a product of
maxterms.
F(x, y, z) = P(0, 2, 4, 5)
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40. MINTERMS AND MAXTERMS
(WITH THREE VARIABLES)
[ Figure 2.22 from the textbook

41. MINTERMS AND MAXTERMS
(WITH THREE VARIABLES)
The function is
1 for these rows

42. MINTERMS AND MAXTERMS
(WITH THREE VARIABLES)
The function is
1 for these rows
The function is
0 for these rows

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CONVERSION BETWEEN CANONICAL
FORMS
 The complement of a function expressed as the
sum of minterms equals the sum of minterms
missing from the original function.
 F(A, B, C) = S(1, 4, 5, 6, 7)
 Thus, F'(A, B, C) = S(0, 2, 3) = m0 + m2 +m3
 By DeMorgan's theorem
F(A, B, C) = (m0 + m2 +m3)' = M0 M2 M3 = P(0, 2, 3)
F'(A, B, C) =P (1, 4, 5, 6, 7)
 mj' = Mj
 Interchange the symbols S and P and list those
numbers missing from the original form
 S of 1's
 P of 0's
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44. TWO DIFFERENT WAYS TO SPECIFY THE SAME
FUNCTION F OF THREE VARIABLES
SOP Form
f(x1, x2, x3) = Σ m(0, 2, 4, 5, 6, 7)
POS Form
f(x1, x2, x3) = Π M(1, 3)

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 Example
 F = xy + xz
 F(x, y, z) = S(1, 3, 6, 7)
 F(x, y, z) = P (0, 2, 4, 5)
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STANDARD FORMS
 The two canonical forms of Boolean algebra are
basic forms that one obtains from reading a given
function from the truth table.
 We do not use it, because each minterm or maxterm
must contain, by definition, all the variables, either
complemented or uncomplemented.
 Standard forms: the terms that form the function
may obtain one, two, or any number of literals.
 Sum of products: F1 = y' + xy+ x'yz'
 Product of sums: F2 = x(y'+z)(x'+y+z')
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CONVERSION OF SOP INTO CANONICAL FORM
 Convert the Boolean function f(X, Y) = X + X Y
into canonical form.
 Solution:
 The given Boolean function f(X, Y) = X + X Y ------ (i)
 It has two variables and sum of two Minterms. The first term
X is missing one variable.
 So to make it of two variables it can be multiplied by (Y+Y’)=1.
 Therefore, X = X (Y+Y’) =XY+XY’
 Substitute the value of X in (i) we get f(X, Y) = X Y+X Y’ + X Y
 Here, the term X Y appear twice, it is possible to remove one of
them. f(X, Y) = X Y+X Y’
 Therefore: SOP Expression is f(X, Y) = Σ (2, 3)
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48. 48
CONVERSION OF POS INTO CANONICAL FORM
 Convert the Boolean function F(X, Y, Z) = X+Y
(Y+Z) into canonical form.
 Solution:
 The given Boolean function F(X,Y,Z) = (X+Y).(Y+Z) ------(i)
 It has three variables and product of two Maxterms. Each
Maxterm is missing one variable.
 The first term can be written as X+Y = (X+Y+Z. Z ) Since Z. Z =0
 Using distributive law (X + YZ) = (X + Y) (X + Z), we can write
 X+Y = (X+Y+Z) (X+Y+ Z ) ------(ii)
 The Second term can be written as
 Y + Z = (Y+Z+X. X ) Y+Z = (Y+Z+X) (Y+Z+X ) ------(iii)
 Substitute (ii) and (iii) in (i) we get
 F(X, Y, Z) = (X+Y+Z) (X+Y+ Z ) (Y+Z+X) (Y+Z+X )
 Therefore: POS Expression is F(X, Y, Z) = π (0, 1, 4)
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49. 49
KARNAUGH MAP:
o A graphical display of the fundamental products in a
truth table.
o Fundamental Product: The logical product of variables
and complements that produces a high output for a
given input condition.
o The map method provides simple procedure for
minimizing the Boolean function.
o The map method was first proposed by E.W. Veitch in
1952 known as “Veitch Diagram”.
o In 1953, Maurice Karnaugh proposed “Karnaugh Map”
also known as “K-Map”.
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CONSTRUCTION OF K-MAP
• The K-Map is a pictorial representation of a truth table
made up of squares.
• Each square represents a Minterm or Maxterm.
• A K-Map for n variables is made up of 2n squares.
 Single Variable K-Map:
 The map consists of 2 squares (i.e. 2n square, 21 = 2 square)
 Two Variable K-Map
• The map consists of 4 squares (i.e. 2n square, 22 = 4 square)
 Three Variable K-Map:
 The map consists of 8 squares (i.e. 2n square, 23 = 8 square)
 Four Variable K-Map
• The map consists of 16 squares (i.e. 2n square, 24 = 16 square)
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51. KARNAUGH MAPS
 Karnaugh maps, or K-maps, are often used to simplify logic problems
with 2, 3 or 4 variables.
BA
For the case of 2 variables, we form a map consisting of 22=4 cells
as shown in Figure
A
B
0 1
0
1
Cell = 2n ,where n is a number of variables
00 10
01 11
A
B
0 1
0
1
A
B
0 1
0
1
BA
BA AB
BA BA
BA BA
Maxterm Minterm
0 2
1 3

52. KARNAUGH MAPS
 3 variables Karnaugh map
AB
C 00 01 11 10
0
1
CBA CBA CAB CBA
CBA BCA ABC CBA
0 2 6 4
531 7
Cell = 23=8

53. KARNAUGH MAPS
 4 variables Karnaugh map
AB
CD 00 01 11 10
00
01
11
10
5
3
1
7
62
0 4
9
15
13
11
1014
12 8

54.  The Karnaugh map is completed by entering a
'1‘(or ‘0’) in each of the appropriate cells.
 Within the map, adjacent cells containing 1's
(or 0’s) are grouped together in twos, fours, or
eights.
KARNAUGH MAPS

55. 55
Figure: Digital logic gates
DIGITAL LOGIC GATES
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56. 56
Figure: Digital logic gates
Summary of Logic Gates
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NAND Gate is a Universal Gate:
To prove that any Boolean function can be implemented using only NAND
gates, we will show that the AND, OR, and NOT operations can be
performed using only these gates

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NAND Gate is a Universal Gate:
To prove that any Boolean function can be implemented using only NOR
gates, we will show that the AND, OR, and NOT operations can be
performed using only these gates

60. BOOLEAN FUNCTIONS
 Implementation with logic gates
F4 = x y' + x' z
F3 = x' y' z + x' y z + x y'
F2 = x + y'z
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IMPLEMENTATION
 Two-level implementation
 Multi-level implementation
nonstandard form standard form
F1 = y' + xy+ x'yz' F2 = x(y'+z)(x'+y+z')
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MULTIPLE INPUTS
 Extension to multiple inputs
 A gate can be extended to more than two inputs.
If its binary operation is commutative and
associative.
 AND and OR are commutative and associative.
OR
 x+y = y+x
 (x+y)+z = x+(y+z) = x+y+z
AND
 xy = yx
 (x y)z = x(y z) = x y z
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MULTIPLE INPUTS
 NAND and NOR are commutative but not associative →
they are not extendable.
Figure: Demonstrating the nonassociativity of the NOR operator; (x ↓
y) ↓ z ≠ x ↓(y ↓ z)
z
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MULTIPLE INPUTS
 Multiple input NOR = a complement of OR gate,
Multiple input NAND = a complement of AND.
 The cascaded NAND operations = sum of products.
 The cascaded NOR operations = product of sums.
Figure: Multiple-input and cascaded NOR and NAND gates
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 The XOR and XNOR gates are commutative and
associative.
 XOR is an odd function: it is equal to 1 if the inputs
variables have an odd number of 1's.
Figure 2.8 3-input XOR gate
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POSITIVE AND NEGATIVE LOGIC
 Positive and Negative Logic
 Two signal values <=> two
logic values
 Positive logic: H=1; L=0
 Negative logic: H=0; L=1
 The positive logic is used in
this book
Figure 2.9 Signal assignment and logic polarity
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Figure 2.10 Demonstration of positive and negative logic
POSITIVE AND NEGATIVE LOGIC
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68. THANK YOU
68
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