Digital Principles and System Design

1. CS8351
DIGITAL PRINCIPLES AND
SYSTEM DESIGN
BY
E.VINOTH, AP/ECE
MOOKAMBIGAI COLLEGE OF ENGINEERING
1

2. UNIT 1
BOOLEAN ALGEBRA AND
LOGIC GATES
2

3. INTRODUCTION
What is Digital System?
 Digital describes electronic technology that generates, stores and processes
data in terms of two states: positive and non-positive.
 Here, Positive state is ‘1’ and Non positive state is ‘0’.
 Digital systems are used in communication, business transactions, traffic
control, spacecraft guidance, medical treatment, weather monitoring, the
Internet and many other commercial, industrial and scientific enterprises.
3

4. NUMBER SYSTEMS
 Number system is a basis for counting various items.
 The decimal number system has 10 digits:0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, but
the modern computers communicate and operate with binary numbers
which use only the digits 0 and 1.
 When decimal quantities are represented in the binary form, they take
more digits.
 For large decimal numbers people have to deal with very large binary
strings and therefore, they do not like working with binary numbers. This
fact gave rise to three new number systems : Octal, Hexadecimal and
Binary Coded Decimal (BCD).
4

5. NUMBER SYSTEMS (CONT.,)
 Binary Number system (0 and 1).
 Octal Number system (0 to 7).
 Decimal Number system ( 0 to 9).
 Hexadecimal Number system (0 to 15) which means 10 as A, 11 as B, 12
as C, 13 as D, 14 as E and 15 as F).
5

6. NUMBER SYSTEMS (CONT.,)
 Single binary digit is called as a bit.
 4 binary digit is called as a nibble.
 8 binary digit is called as a byte.
 16 binary digit is called as a word.
6

7. BINARY NUMBER SYSTEM
 It is base two or radix two system.
 The two binary digits (bits) are 1 and 0.
 The weight is expressed as a power of 2.
 Eg. (1101.1011)2
1 1 0 1 1 0 1 1
7

8. OCTAL NUMBER SYSTEM
 It is base eight or radix eight system.
 The digits are 0, 1, 2, 3, 4, 5, 6 and 7.
 The weight is expressed as a power of 8.
 Eg. (5632.471)8
8
Octal Point

9. DECIMAL NUMBER SYSTEM
 It is base ten or radix ten system.
 The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
 The weight is expressed as a power of 10.
 Eg. (5678.9)10
9
Decimal Point

10. HEXADECIMAL NUMBER
SYSTEM
 It is base 16 or radix 16 system.
 The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F.
 The weight is expressed as a power of 16.
 Eg. (3FD.84)16
10
Hexadecimal Point

11. RELATION BETWEEN 2, 8, 10
AND 16
11

12. CONVERSION
 We have 4 number systems, so that there are 12 conversions such as,
 Binary to Octal, Binary to Decimal and Binary to Hexadecimal .
 Decimal to Binary, Decimal to Octal and Decimal to Hexadecimal.
 Octal to Binary, Octal to Decimal and Octal to Hexadecimal.
 Hexadecimal to Binary, Hexadecimal to Octal and Hexadecimal to
Decimal.
12

13. BINARY TO OCTAL
CONVERSION
Eg. Binary Input : (010100111.100011)2
13
4

14. BINARY TO DECIMAL
CONVERSION
1 0 1 1 . 1 0 1
1 x 2-3 = 0.125
0 x 2-2 = 0
1 x 2-1 = 0.5
1 x 20 = 1
1 x 21 = 2
0 x 22 = 0
1 x 23 = 8
11.625
Eg. Binary Input : (1011.101)2
14

15. BINARY TO HEXADECIMAL
CONVERSION
Eg. Binary Input : (1111001010010100001.010110110011011)2
15

16. OCTAL TO BINARY
CONVERSION
 Eg. Octal Input : (35.346)8
16
011

17. OCTAL TO DECIMAL
CONVERSION
1 2 0 7 . 1 2 5
5 x 8-3 = 0.0097
2 x 8-2 = 0.0313
1 x 8-1 = 0.125
7 x 80 = 7
0 x 81 = 0
2 x 82 = 128
1 x 83 = 512
647.1660
Eg. Octal Input : (1207.125)8
17

18. OCTAL TO HEXADECIMAL
CONVERSION
18

19. DECIMAL TO BINARY
CONVERSION
Eg. Decimal Input : (39.47)10
Binary Output : (100111.01111)2
19
0
1
1
1
1

20. DECIMAL TO OCTAL
CONVERSION
Eg. Decimal Input : (225.225)10
Octal Output : (341.16314)8
20

21. DECIMAL TO HEXADECIMAL
CONVERSION
Eg. Decimal Input : (374.37)10
Hexadecimal Output : (176.5EB8)16
21

22. HEXADECIMAL TO BINARY
CONVERSION
Binary Output : (1111001010010100001.010110110011011)2
Hexadecimal Input : (794A1.5B36)16
7 9 4 A 1 . 5 B 3 6
0111 1001 0100 1010 0001 . 0101 1011 0011 0110
22

23. HEXADECIMAL TO OCTAL
CONVERSION
 There is no direct conversion of hexadecimal to octal form.
 1. convert the hexadecimal number to binary.
 2. convert the binary number to octal number.
Eg. Hexadecimal Input (39. A)16
3 9 . A
0011 1001 . 1010
000 111 001 . 101 000
0 7 1 . 5 0
Octal Output
(71.5)8
Group the converted binary number
into three to convert it into octal number
23

24. HEXADECIMAL TO DECIMAL
CONVERSION
A 2 1 E . 1 0 D
D x 16-3 = 0.0032
0 x 16-2 = 0
1 x 16-1 = 0.0625
E x 160 = 14
1 x 161 = 16
2 x 162 = 512
A x 163 = 40960
41502.0657
24

25. ARITHMETIC OPERATIONS
 Binary Addition.
 Binary Subtraction.
 Binary Multiplication.
 Binary Division.
25

26. BINARY ADDITION
For 2 Inputs For 3 Inputs
26

27. BINARY ADDITION (Cont.,)
27

28. BINARY ADDITION (Cont.,)
28

29. BINARY SUBTRACTION
For 2 Inputs For 3 Inputs
29

30. BINARY SUBTRACTION
(Cont.,)
30

31. BINARY SUBTRACTION
(Cont.,)
Subtract (26)10 and (12)10 by converting them into
binary
31

32. BINARY SUBTRACTION USING
1’s COMPLEMENT
32

33. BINARY SUBTRACTION USING
1’s COMPLEMENT (Cont.,)
33

34. BINARY SUBTRACTION USING
1’s COMPLEMENT (Cont.,)
34

35. BINARY SUBTRACTION USING
2’s COMPLEMENT
35

36. BINARY SUBTRACTION USING
2’s COMPLEMENT
36

37. BINARY SUBTRACTION USING
2’s COMPLEMENT
37

38. BINARY MULTIPLICATION
38

39. BINARY MULTIPLICATION
(Cont.,)
39

40. BINARY DIVISION
40

41. BINARY DIVISION (Cont.,)
41

42. BINARY DIVISION (Cont.,)
Divide (15)10 and (3)10 by converting them into binary
42

43. BINARY CODES
43

44. CLASSIFICATION OF BINARY
CODES
44

45. BCD CODES
45

46. BCD CODES
Note: 0 to 9 only
46

47. EXCESS 3 CODE
47

48. GRAY CODES
48

49. BINARY TO GRAY CODES
Eg. Convert the given binary code into gray code.
(11101)2
49

50. BINARY TO GRAY CODES
50

51. GRAY TO BINARY CODES
Eg. Convert the given gray code into binary code.
(10011)2
51

52. GRAY TO BINARY CODES
52

53. BCD TO EXCESS 3 CODES
Note: Add 3 to every BCD Code i.e., Add 0011 to BCD code we can obtain Excess
53

54. EXCESS 3 TO BCD CODES
54

55. LOGIC GATES
 Logic gates are the basic elements that make up a digital
system.
 The electronic gate is a circuit that is able to operate on a
number of binary inputs in order to perform a particular
logical function.
 The types of gates available are the NOT, AND, OR,
NAND, NOR, exclusive-OR, and the exclusive-NOR.
55

56. LOGIC GATES (Cont.,)
56

57. AND GATE
57

58. OR GATE
58

59. NOT GATE
59

60. NAND GATE
60

61. NOR GATE
61

62. EX-OR GATE
62

63. EX-NOR GATE
63

64. COMPARISON
NAME SYMBOL OPERATION IC NUMBER
AND GATE IC 7408
OR GATE IC 7432
NOT GATE IC 7404
NAND GATE IC 7400
NOR GATE IC 7402
EX-OR GATE IC 7486
EX-NOR GATE IC 74266
B
A
X 

B
A
X 

A
X 
B
A
X 

B
A
X 

B
A
X 

B
A
X 

64

65. BOOLEAN ALGEBRA
 The Boolean algebra is used to express the output of any
combinational network.
 Such a network can be implemented using logic gates.
 Boolean Algebra Terminology
 Variable.
 Constant.
 Complement.
 Literal.
65

66. BOOLEAN FUNCTION
66

67. PROPERTIES AND THEOREMS OF BOOLEAN
ALGEBRA
(Closure)
67

68. PROPERTIES AND THEOREMS OF BOOLEAN
ALGEBRA
68

69. EXAMPLES
69

70. EXAMPLES
70

71. DEMORGAN’S THEOREM
 DeMorgan suggested two theorems that form an important
part of Boolean algebra. In the equation form, they are
71

72. CONSENSUS THEOREM
 In simplification of Boolean expression, an expression of the
form AB+ AC+ BC the term BC is redundant and can be
eliminated to form the equivalent expression AB+ AC. The
theorem used for this simplification is known as consensus
theorem and it is stated as
72

73. HOW TO IDENTIFY
CONSENSUS
 Find a pair of terms, one of which contains a variable and the
other contains its complement.
 Find the third term which should contain the remaining
variables from pair of terms eliminating selected variable and
its complement.
73

74. CONSENSUS THEOREM
EXAMPLES
74

75. EXAMPLES
75

76. EXAMPLES
76

77. EXAMPLES
77

78. EXAMPLES
78

79. EXAMPLES
79

80. BOOLEAN EXPRESSIONS
 Boolean expressions are constructed by connecting the
Boolean constants and variables with the Boolean operations.
These Boolean expressions are also known as Boolean
formulas.
 For example, if the Boolean expression (A + B) C is used to
describe the function f, then Boolean function is written as
f(A, B, C) = (A + B) C or f = (A + B)C
80

81. BOOLEAN EXPRESSIONS
(Cont.,)
 In this Boolean function the variables are appeared either in a
complemented or an uncomplemented form.
 Each occurrence of a variable in either a complemented or an
uncomplemented form is called a literal.
81

82. BOOLEAN EXPRESSIONS
(Cont.,)
 These literals and terms are arranged in one of the two forms
:
 Sum of product form (SOP)
 A product term is any group of literals that are ANDed
together.
 Product of sum form (POS).
 A product of sums is any groups of sum terms ANDed
together.
82

83. BOOLEAN EXPRESSIONS
(Cont.,)
 Sum of product form (SOP)
 Product of sum form (POS).
83

84. STANDARD (CANONICAL)
SOP
 If each term in SOP form contains all the literals then the
SOP form is known as standard or canonical SOP form.
 Each individual term in the standard SOP form is called
minterm.
84

85. STANDARD (CANONICAL) POS FORM
(Cont.,)
 If each term in POS form contains all the literals then the
POS form is known as standard or canonical POS form.
 Each individual term in the standard POS form is called
maxterm.
85

86. CONVERTING EXPRESSIONS IN STANDARD
SOP
 Step 1 : Find the missing literal in each product term if any.
 Step 2 : AND each product term having missing literal/s with
term/s form by Oring the literal and its complement.
 Step 3 : Expand the terms by applying distributive law and
reorder the literals in the product terms.
 Step 4 : Reduce the expression by omitting repeated product
terms if any. Because A+A=A.
86

87. CONVERTING EXPRESSIONS IN STANDARD SOP
(Cont.,)
87

88. CONVERTING EXPRESSIONS IN STANDARD SOP
(Cont.,)
88

89. CONVERTING EXPRESSIONS IN STANDARD SOP
(Cont.,)
89

90. CONVERTING EXPRESSIONS IN STANDARD
POS
 Step 1 : Find the missing literals in each sum term if any.
 Step 2 : OR each sum term having missing literal/s with
term/s form by ANDing the literal and its complement.
 Step 3 : Expand the terms by applying distributive law and
reorder the literals in the sum terms.
 Step 4 : Reduce the expression by omitting repeated sum
terms if any. Because A·A=A
90

91. CONVERTING EXPRESSIONS IN STANDARD POS
(Cont.,)
91

92. CONVERTING EXPRESSIONS IN STANDARD POS
(Cont.,)
92

93. CONVERTING EXPRESSIONS IN STANDARD POS
(Cont.,)
93

94. CONVERTING EXPRESSIONS IN STANDARD POS
(Cont.,)
94

95. MINTERMS AND MAXTERMS
 Each individual term in standard SOP form is called minterm
and each individual term in standard POS form is called
maxterm.
95

96. MINTERMS AND MAXTERMS
(Cont.,)
96

97. COMPLEMENTS OF STANDARD
FORMS
 The POS and SOP functions derived from the same truth
table are logically equivalent.
 In terms of minterms and maxterms we can then write
 Using this complementary relationship we can find logical
function in terms of maxterms if function in minterms is known
or vice-versa.
97

98. EXAMPLES
98

99. EXAMPLES (Cont.,)
99

100. KARNAUGH MAP (K MAP)
 During the process of simplification of Boolean expression we
have to predict each successive step.
 On the other hand, the map method gives us a systematic
approach for simplifying a Boolean expression.
 The map method, first proposed by Veitch and modified by
Karnaugh, hence it is known as the Veitch diagram or the
Karnaugh map.
100

101. K MAP
 The basis of this method is a graphical chart known as
Karnaugh map (K-map).
 It contains boxes called cells.
 Each of the cell represents one of the 2n possible products
that can be formed from n variables.
 Thus, a 2-variable map contains 22= 4 cells, a 3-variable map
contains 23= 8 cells and so forth.
101

102. K MAP (Cont.,)
102

103. ONE VARIABLE
MINTER
M
MAXTER
M
103

104. TWO VARIABLE
MINTER
M
MAXTER
M
104

105. THREE VARIABLE
MINTER
M
MAXTER
M
105

106. FOUR VARIABLE
MINTER
M
MAXTER
M
106

107. PLOTTING A K MAP
107

108. PLOTTING A K MAP (Cont.,)
108

109. PLOTTING A K MAP (Cont.,)
109

110. PLOTTING A K MAP (Cont.,)
110

111. PLOTTING A K MAP (Cont.,)
111

112. PLOTTING A K MAP (Cont.,)
112

113. GROUPING
113

114. GROUPING (Cont.,)
114

115. GROUPING (Cont.,)
115

116. GROUPING (Cont.,)
116

117. ILLEGAL GROUPING
117

118. SIMPLIFICATION OF SOP
EXPRESSION
1. Plot the K-map and place 1s. in those cells corresponding to
the 1s in the truth table or sum of product expression. Place
0s in other cells.
2. Check the K-map for adjacent 1s and encircle those 1s
which are not adjacent to any other 1s. These are called
isolated 1s.
3. Check for those 1s which are adjacent to only one other 1
and encircle such pairs.
118

119. SIMPLIFICATION OF SOP
EXPRESSION
4. Check for quads and octets of adjacent 1s even if it contains
some 1s that have already been encircled. While doing this
make sure that there are minimum number of groups.
5. Combine any pairs necessary to include any 1s that have
not yet been grouped.
6. Form the simplified expression by summing product terms of
all the groups.
119

120. EXAMPLES
120

121. EXAMPLES (Cont.,)
121

122. DON'T CARE TERMS
 In some logic circuits, certain input conditions never occur,
therefore the corresponding output never appears.
 In such cases the output level is not defined, it can be either
HIGH or LOW.
 These output levels are indicated by `X' or `d' in the truth
tables and are called don't care outputs or don't care
conditions or incompletely specified functions.
122

123. DON'T CARE TERMS (Cont.,)
 A circuit designer is free to make the output for any "don't
care" condition either a ‘0’ or a ‘1’ in order to produce the
simplest output expression.
123

124. DON'T CARE TERMS (Cont.,)
124

125. EXAMPLES
125

126. EXAMPLES (Cont.,)
126

127. EXAMPLES (Cont.,)
127

128. EXAMPLES (Cont.,)
128

129. EXAMPLES (Cont.,)
129

130. SIMPLICATION OF POS
EXPRESSION
1. Plot the K-map and place 0s in those cells corresponding to
the 0s in the truth table or maxterms in the product of sums
expression.
2. Check the K-map for adjacent 0s and encircle those 0s
which are not adjacent to any other 0s. These are called
isolated 0s.
3. Check for those 0s which are adjacent to only one other 0
and encircle such pairs.
130

131. SIMPLICATION OF POS
EXPRESSION
4. Check for quads and octets of adjacent 0s even if it contains
some 0s that have already been encircled. While doing this
make sure that there are minimum number of groups.
5. Combine any pairs necessary to include any 0s that have
not yet been grouped.
6. Form the simplified POS expression for F by taking product
of sum terms of all the groups.
131

132. EXAMPLES
132

133. EXAMPLES (Cont.,)
133

134. EXAMPLES (Cont.,)
134

135. EXAMPLES (Cont.,)
135

136. IMPLEMENTATION OF BOOLEAN
FUNCTION
 The Boolean algebra is used to express the output of any
combinational network.
 Such a network can be implemented using logic gates.
 Basic Gates – AND, OR, NOT
 Universal Gate – NAND, NOR
 Other Gate – EX-OR, EX-NOR
136

137. IMPLEMENTATION OF SOP
137

138. IMPLEMENTATION OF POS
138

139. EXAMPLES
139

140. EXAMPLES (Cont.,)
FULL ADDER
140

141. NOT GATE USING NAND
GATE
141

142. AND GATE USING NAND
GATE
142

143. OR GATE USING NAND GATE
143

144. NOT GATE USING NOR GATE
144

145. AND GATE USING NOR GATE
145

146. OR GATE USING NOR GATE
146

147. EXAMPLES
147

148. EXAMPLES (Cont.,)
148

149. THANK YOU
149

150. QUERIES ?
150