1. CHAPTER 1 : INTRODUCTION
EKT 121 / 4 ELEKTRONIK DIGIT 1
CHAPTER 1 : INTRODUCTION
Lecturer : FAIRUL AFZAL
Office : KKF 8 ( )

2. 1.0 Number & codes Digital vs. analog quantities
Decimal numbering system (Base 10)
Binary numbering system (Base 2)
Hexadecimal numbering system (Base 16)
Octal numbering system (Base 8)
Number conversion
Binary arithmetic
1’s and 2’s complements of binary numbers

3. Number & codes Signed numbers
Arithmetic operations with signed numbers
Binary-Coded-Decimal (BCD)
ASCII codes
Gray codes
Digital codes & parity

4. Digital vs. Analogue
Digital signals are represented by only two possible -
1 (binary 1) 0r 0 (binary 0)
Sometimes call these values “high” and “low” or “ true” and “false””
Example : light switch , it can be in just two position – “ on ” or “ off ”
More complicated signals can be constructed from 1s and 0s by stringing them end-to end. Example : 3 binary digits, have 8 possible combinations :
000,001,010,011,100,101,110 and 111.
The diagram : example a typical digital signal, represented as a series of voltage levels that change as time goes on.

5. Digital vs. Analogue
Analogue electronics can be any value within limits.
Example : Voltage change simultaneously from one value to the next, like gradually turning a light dimmer switch up or down.
The diagram below shows an analoq signal that changes with time.

6. Digital vs. Analogue Why digital ? Problem with all signals – noise
Noise isn't just something that you can hear - the fuzz that appears on old video recordings also qualifies as noise. In general, noise is any unwanted change to a signal that tends to corrupt it.
Digital and analogue signals with added noise:
Analog : never get back a perfect copy of the original signal
Digital : easily be recognized even
among all that noise : either 0 or 1

7. Digital and analog quantities
Two ways of representing the numerical values of quantities :
i) Analog (continuous)
ii) Digital (discrete)
Analog : a quantity represented by voltage, current or meter movement that is proportional to the value that quantity
Digital : the quantities are represented not by proportional quantities but by symbols called digits

8. Digital and analog systems
Digital system:
combination of devices designed to manipulate logical information or physical quantities that are represented in digital forms
Example : digital computers and calculators, digital audio/video equipments, telephone system.
Analog system:
contains devices manipulate physical quantities that are represented in analog form
Example : audio amplifiers, magnetic tape recording and playback equipment, and simple light dimmer switch

9. Analog Quantities
Continuous values

10. Digital Waveform

11. Introduction to Numbering Systems
We are all familiar with decimal number systems - use everyday :calculator, calendar, phone or any common devices use this numbering system :
Decimal = Base 10
Some other number systems:
Binary = Base 2
Octal = Base 8
Hexadecimal = Base 16

12. Numbering Systems Decimal Binary Octal Hexadecimal 0 ~ 9 0 ~ 1 0 ~ 7
0 ~ F

13.
A B C D E F
Binary
Octal
Hex
Dec N U M B E R S Y T

14. Most significant digit Least significant digit
Significant Digits
Binary:
Most significant digit Least significant digit
Hexadecimal: 1D63A7A

15. Decimal numbering system (Base 10)
Example : 39710
Weights for whole numbers are positive power of ten that increase from right to left , beginning with 100
3 X 102 +
9 X 101 +
7 X 100 = =
39710

16. Binary Number System 1X 22 + 0 X 21 + 1 X 20 = 4 + 0 + 1 = 510
Base 2 system – (0 , 1)
used to model the series of electrical signals computers use to represent information
0 represents the no voltage or an off state
1 represents the presence of voltage or an on state
Example :
Weights in a binary number are based on power of two, that increase from right to right to left,beginning with 20
1X 22 +
0 X 21 +
1 X 20 = =
510

17. Octal Number System + + = = Base 8 system – (0,1,2,3,4,5,6,7) 7X 82
multiplication and division algorithms for conversion to and from base 10
Example : convert to decimal:
Weights in a binary number are based on power of eight that increase from right to right to left,beginning with 80 +
7X 82
5 X 81 +
6 X 80 =
49410 =
Readily converts to binary
Groups of three (binary) digits can be used to represent each octal number
Example : convert to binary :

18. Hexadecimal Number System
BINARY
0000 1
0001 2
0010 3
0011 4
0100 5
0101 6
0110 7
0111 8
1000 9
1001 A 10
1010 B 11
1011 C 12
1100 D 13
1101 E 14
1110 F 15
1111
Base 16 system
Uses digits 0 ~ 9 &
letters A,B,C,D,E,F
Groups of four bits represent each base 16 digit

19. Hexadecimal Number System
Base 16 system –
multiplication and division algorithms for conversion to and from base 10
Example : A9F16 convert to decimal:
A F
Weights in a hexadecimal number are based on power of sixteen that increase from right to right to left,beginning with 160
10X 162 +
9 X 161 +
15 X 160 =
271910 =
Readily converts to binary
Groups of four (binary) digits can be used to represent each hexadecimal number
Example : A9F8 convert to binary :
A F16

20. Number Conversion
Any Radix (base) to Decimal Conversion

21. Number Conversion (BASE 2 –BASE 10)
Binary to Decimal Conversion

22. Binary to Decimal Conversion
Convert ( )2 to its decimal equivalent: Binary Positional Values x x x x x x x x 27 26 25 24 23 22 21 20
Products
17310

23. Octal to Decimal Conversion
Convert 6538 to its decimal equivalent:
Octal Digits x x x
Positional Values
Products
42710

24. Hexadecimal to Decimal Conversion
Convert 3B4F16 to its decimal equivalent:
Hex Digits
3 B F x x x x
Positional Values
Products
15,18310

25. Number Conversion INTEGER DIGIT:
Decimal to Any Radix (Base) Conversion
INTEGER DIGIT:
Repeated division by the radix & record the remainder
FRACTIONAL DECIMAL:
Multiply the number by the radix until the answer is in integer
Example:
to Binary

26. Decimal to Binary Conversion
Remainder
2 5 = 2
1 2 =
6 =
3 =
1 =
MSB LSB
2510 =

27. Decimal to Binary Conversion
MSB
LSB
Carry
x 2 =
x 2 =
x 2 =
0.5 x 2 =
The Answer:

28. Decimal to Octal Conversion
Convert to its octal equivalent:
427 / 8 = 53 R3 Divide by 8; R is LSD
53 / 8 = 6 R5 Divide Q by 8; R is next digit
6 / 8 = 0 R6 Repeat until Q = 0
6538

29. Decimal to Hexadecimal Conversion
Convert to its hexadecimal equivalent: 830 / 16 = 51 R14 51 / 16 = 3 R3 3 / 16 = 0 R3
= E in Hex
33E16

30. Number Conversion
Binary to Octal Conversion (vice versa)
Grouping the binary position in groups of three starting at the least significant position.

31. Octal to Binary Conversion
Each octal number converts to 3 binary digits
To convert 6538 to binary, just substitute code:

32. Number Conversion
Example:
Convert the following binary numbers to their octal equivalent (vice versa).
b) 47.38
Answer:
11.748

33. Number Conversion
Binary to Hexadecimal Conversion (vice versa)
Grouping the binary position in 4-bit groups, starting from the least significant position.

34. Binary to Hexadecimal Conversion
The easiest method for converting binary to hexadecimal is to use a substitution code
Each hex number converts to 4 binary digits

35. Number Conversion Example:
Convert the following binary numbers to their hexadecimal equivalent (vice versa).
1F.C16
Answer:
10.816

36. Substitution Code 56AE6A16 0101 0110 1010 1110 0110 1010 5 6 A E 6 A
Convert to hex using the 4-bit substitution code :
A E A
56AE6A16

37. Substitution Code
Substitution code can also be used to convert binary to octal by using 3-bit groupings:

38. Binary Addition 0 + 0 = 0 Sum of 0 with a carry of 0
Example:
???

39. Simple Arithmetic Addition Example: Example: 100011002 5816 + 1011102
Substraction
101102
Example:
5816
7C16

40. Binary Subtraction
0 - 0 = 0
1 - 1 = 0
1 - 0 = 1
10 -1 = with a borrow of 1
Example:
???

41. Binary Multiplication
0 X 0 = 0
0 X 1 = 0 Example:
1 X 0 =
1 X 1 = X
100110
000000

42. Binary Division
Use the same procedure as decimal division

43. 1’s complements of binary numbers
Changing all the 1s to 0s and all the 0s to 1s
Example:
Binary number
’s complement

44. 2’s complements of binary numbers
Step 1: Find 1’s complement of the number
Binary #
1’s complement
Step 2: Add 1 to the 1’s complement

45. Signed Magnitude Numbers …
Sign bit
31 bits for magnitude
0 = positive
1 = negative
This is your basic
Integer format

46. Sign numbers Left most is the sign bit Sign-magnitude 1’s complement
0 is for positive, and 1 is for negative
Sign-magnitude
= +25
sign bit magnitude bits
1’s complement
The negative number is the 1’s complement of the corresponding positive number
Example:
+25 is is

47. Sign numbers 2’s complement Example Express +19 and -19 in
The positive number – same as sign magnitude and 1’s complement
The negative number is the 2’s complement of the corresponding positive number.
Example
Express +19 and -19 in
i. sign magnitude
ii. 1’s complement
iii. 2’s complement

48. Digital Codes
BCD (Binary Coded Decimal) Code
Represent each of the 10 decimal digits (0~9) as a 4-bit binary code.
Example:
Convert 15 to BCD.
BCD
Convert 10 to binary and BCD.

49. Digital Codes
ASCII (American Standard Code for Information Interchange) Code
Used to translate from the keyboard characters to computer language
Can convert from ASCII code to binary / hexadecimal/ or any numbering systems and vice versa
How to convert ?????

50. Digital Codes Decimal Binary Gray Code 0000 1 0001 2 0010 0011 3 4
0000 1
0001 2
0010
0011 3 4
0100
0110 5
0101
0111 6
The Gray Code
Only 1 bit changes
Can’t be used in arithmetic circuits
Can convert from Binary to Gray Code and vice versa.
How to convert ?????

51. 3.0 LOGIC GATES
Inverter (Gate Not)
AND Gate
OR Gate
NAND Gate
NOR Gate
Exclusive-OR and Exclusive-NOR
Fixed-function logic: IC Gates

52. Introduction Three basic logic gates AND Gate – expressed by “ . “
OR Gates – expressed by “ + “ sign (not an ordinary addition)
NOT Gate – expressed by “ ‘ “ or “¯”

53. NOT Gate (Inverter) a) Gate Symbol & Boolean Equation
b) Truth Table (Jadual Kebenaran)
c) Timing Diagram (Rajah Pemasaan)

54. OR Gate a) Gate Symbol & Boolean Equation
b) Truth Table (Jadual Kebenaran)
c) Timing Diagram (Rajah Pemasaan)

55. AND Gate a) Gate Symbol & Boolean Equation
c) Timing Diagram (Rajah Pemasaan)
b) Truth Table (Jadual Kebenaran)

56. a) Gate Symbol, Boolean Equation
NAND Gate
a) Gate Symbol, Boolean Equation
& Truth Table
b) Timing Diagram

57. a) Gate Symbol, Boolean Equation
NOR Gate
a) Gate Symbol, Boolean Equation
& Truth Table
b) Timing Diagram

58. a) Gate Symbol, Boolean Equation
Exclusive-OR Gate
a) Gate Symbol, Boolean Equation
& Truth Table
b) Timing Diagram

59. AND gate NOT gate Examples : Logic Gates IC
Note : x is referring to family/technology (eg : AS/ALS/HCT/AC etc.)

60. 4.0 BOOLEAN ALGEBRA Boolean Operations & expression
Laws & rules of Boolean algebra
DeMorgan’s Theorems
Boolean analysis of logic circuits
Simplification using Boolean Algebra
Standard forms of Boolean Expressions
Boolean Expressions & truth tables
The Karnaugh Map

61. Karnaugh Map SOP minimization
Karnaugh Map POS minimization
5 Variable K-Map
Programmable Logic

62. Boolean Operations & expression
Variable – a symbol used to represent logical quantities (1 or 0)
ex : A, B,..used as variable
Complement – inverse of variable and is indicated by bar over variable
ex : Ā

63. Operation : Boolean Addition – equivalent to the OR operation
X = A + B
Boolean Multiplication – equivalent to the AND operation
X = A∙B A X B A X B

64. Laws & rules of Boolean algebra

65. Commutative law of addition
A+B = B+A
the order of ORing does not matter.

66. Commutative law of Multiplication
AB = BA
the order of ANDing does not matter.

67. Associative law of addition
A + (B + C) = (A + B) + C
The grouping of ORed variables does not matter

68. Associative law of multiplication
A(BC) = (AB)C
The grouping of ANDed variables does not matter

69. (A+B)(C+D) = AC + AD + BC + BD
Distributive Law
A(B + C) = AB + AC
(A+B)(C+D) = AC + AD + BC + BD

70. Boolean Rules 1) A + 0 = A
In math if you add 0 you have changed nothing
In Boolean Algebra ORing with 0 changes nothing

71. Boolean Rules 2) A + 1 = 1
ORing with 1 must give a 1 since if any input
is 1 an OR gate will give a 1

72. Boolean Rules 3) A • 0 = 0
In math if 0 is multiplied with anything you
get 0. If you AND anything with 0 you get 0

73. Boolean Rules 4) A • 1 = A
ANDing anything with 1 will yield the anything

74. Boolean Rules 5) A + A = A
ORing with itself will give the same result

75. Boolean Rules 6) A + A = 1
Either A or A must be 1 so A + A =1

76. Boolean Rules 7) A • A = A
ANDing with itself will give the same result

77. Boolean Rules 8) A • A = 0
In digital Logic 1 =0 and 0 =1, so AA=0 since one of the inputs must be 0.

78. Boolean Rules 9) A = A
If you not something twice you are back to the beginning

79. Boolean Rules 10) A + AB = A
Proof:
A + AB = A(1 +B) DISTRIBUTIVE LAW
= A∙ RULE 2: (1+B)=1
= A RULE 4: A∙1 = A

80. Boolean Rules 11) A + AB = A + B
If A is 1 the output is 1 , If A is 0 the output is B
Proof:
A + AB = (A + AB) + AB RULE 10
= (AA +AB) + AB RULE 7
= AA + AB + AA +AB RULE 8
= (A + A)(A + B) FACTORING
= 1∙(A + B) RULE 6
= A + B RULE 4

81. Boolean Rules 12) (A + B)(A + C) = A + BC
PROOF
(A + B)(A +C) = AA + AC +AB +BC DISTRIBUTIVE LAW
= A + AC + AB + BC RULE 7
= A(1 + C) +AB + BC FACTORING
= A.1 + AB + BC RULE 2
= A(1 + B) + BC FACTORING
= A.1 + BC RULE 2
= A + BC RULE 4

82. De Morgan’s Theorem

83. Theorems of Boolean Algebra
1) A + 0 = A 2) A + 1 = 1 3) A • 0 = 0 4) A • 1 = A 5) A + A = A 6) A + A = 1 7) A • A = A 8) A • A = 0

84. Theorems of Boolean Algebra
9) A = A 10) A + AB = A 11) A + AB = A + B 12) (A + B)(A + C) = A + BC 13) Commutative : A + B = B + A AB = BA 14) Associative : A+(B+C) =(A+B) + C A(BC) = (AB)C 15) Distributive : A(B+C) = AB +AC (A+B)(C+D)=AC + AD + BC + BD

85. De Morgan’s Theorems
Two most important theorems of Boolean Algebra were contributed by De Morgan.
Extremely useful in simplifying expression in which product or sum of variables is inverted.
The TWO theorems are :
16) (X+Y) = X . Y 17) (X.Y) = X + Y

86. Implications of De Morgan’s Theorem
Input Output
X Y X+Y XY
(b)
(c)
(a) Equivalent circuit implied by theorem (16)
(b) Negative- AND
(c) Truth table that illustrates DeMorgan’s Theorem

87. Implications of De Morgan’s Theorem
Input Output
X Y XY X+Y
(b)
(c)
(a) Equivalent circuit implied by theorem (17)
(b) Negative-OR
(c) Truth table that illustrates DeMorgan’s Theorem

88. De Morgan’s Theorem Conversion
Step 1: Change all ORs to ANDs and all ANDs to Ors Step 2: Complement each individual variable (short overbar) Step 3: Complement the entire function (long overbars) Step 4: Eliminate all groups of double overbars Example : A . B A .B. C = A + B = A + B + C = A + B = A + B + C = A + B

89. De Morgan’s Theorem Conversion
ABC + ABC (A + B +C)D = (A+B+C).(A+B+C) = (A.B.C)+D = (A+B+C).(A+B+C) = (A.B.C)+D

90. Examples: Analyze the circuit below
2. Simplify the Boolean expression found in 1

91. Follow the steps list below (constructing truth table)
List all the input variable combinations of 1 and 0 in binary sequentially
Place the output logic for each combination of input
Base on the result found write out the boolean expression.

92. Exercises: Simplify the following Boolean expressions
(AB(C + BD) + AB)C
ABC + ABC + ABC + ABC + ABC
Write the Boolean expression of the following circuit.

93. Standard Forms of Boolean Expressions
Sum of Products (SOP)
Products of Sum (POS)
Notes:
SOP and POS expression cannot have more than
one variable combined in a term with an inversion bar
There’s no parentheses in the expression

94. Standard Forms of Boolean Expressions
Converting SOP to Truth Table
Examine each of the products to determine where
the product is equal to a 1.
Set the remaining row outputs to 0.

95. Standard Forms of Boolean Expressions
Converting POS to Truth Table
Opposite process from the SOP expressions.
Each sum term results in a 0.
Set the remaining row outputs to 1.

96. Standard Forms of Boolean Expressions
The standard SOP Expression
All variables appear in each product term.
Each of the product term in the expression is called
as minterm.
Example:
In compact form, f(A,B,C) may be written as

97. Standard Forms of Boolean Expressions
The standard POS Expression
All variables appear in each product term.
Each of the product term in the expression is called as
maxterm.
Example:
In compact form, f(A,B,C) may be written as

98. Standard Forms of Boolean Expressions
Example:
Convert the following SOP expression to an equivalent POS expression:
Example:
Develop a truth table for the expression:

99. THE K-MAP

100. Karnaugh Map (K-Map)
Karnaugh Mapping is used to minimize the number of logic gates that are required in a digital circuit.
This will replace Boolean reduction when the circuit is large.
Write the Boolean equation in a SOP form first and then place each term on a map.

101. Karnaugh Map (K-Map)
The map is made up of a table of every possible SOP using the number of variables that are being used.
If 2 variables are used then a 2X2 map is used
If 3 variables are used then a 4X2 map is used
If 4 variables are used then a 4X4 map is used
If 5 Variables are used then a 8X4 map is used

102. K-Map SOP Minimization

103. 2 Variables Karnaugh Map
B B A
Notice that the map is going false to true, left to right and top to bottom
B B
The upper right hand cell is A B if X= A B then put an X in that cell A 1
This show the expression true when A = 0 and B = 0

104. 2 Variables Karnaugh Map
B B
If X=AB + AB then put an X in both of these cells A 1 1
From Boolean reduction we know that A B + A B = B
B B
From the Karnaugh map we can circle adjacent cell and find that X = B A 1 1

105. 3 Variables Karnaugh Map
C C
Gray Code
00 A B
01 A B
11 A B
10 A B

106. 3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
C C 1 1
Each 3 variable term is one cell on a 4 X 2 Karnaugh map 1 1

107. 3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
C C
One simplification could be
X = A B + A B 1 1 1 1

108. 3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
C C
Another simplification could be
X = B C + B C
A Karnaugh Map does wrap around 1 1 1 1

109. 3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
C C
The Best simplification would be
X = B 1 1 1 1

110. On a 3 Variables Karnaugh Map
One cell requires 3 Variables
Two adjacent cells require 2 variables
Four adjacent cells require 1 variable
Eight adjacent cells is a 1

111. 4 Variables Karnaugh Map
Gray Code
00 A B
01 A B
11 A B
10 A B
C D C D C D C D

112. Simplify : X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D
Gray Code
00 A B
01 A B
11 A B
10 A B
C D C D C D C D
Now try it with Boolean reductions 1 1 1 1 1 1
X = ABD + ABC + CD

113. On a 4 Variables Karnaugh map
One Cell requires 4 variables
Two adjacent cells require 3 variables
Four adjacent cells require 2 variables
Eight adjacent cells require 1 variable
Sixteen adjacent cells give a 1 or true

114. Simplify : Z = B C D + B C D + C D + B C D + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
C D C D C D C D 1 1 1 1 1 1 1 1 1 1 1 1
Z = C + A B + B D

115. Simplify using Karnaugh map
First, we need to change the circuit to an SOP expression

116. Simplify using Karnaugh map (cont’d)
Y= A + B + B C + ( A + B ) ( C + D)
Y = A B + B C + A B ( C + D )
Y = A B + B C + A B C + A B D
Y = A B + B C + A B C A B D
Y = A B + B C + (A + B + C ) ( A + B + D)
Y = A B + B C + A + A B + A D + B + B D + AC + C D
SOP expression

117. Simplify using Karnaugh map (cont’d)
Gray Code
00 A B
01 A B
11 A B
10 A B
C D C D C D C D
Y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

118. K-Map POS Minimization

119. 3 Variables Karnaugh Map
Gray Code
0 0
0 1
1 1
1 0 C AB

120. 3 Variables Karnaugh Map (cont’d)

121. 4 Variables Karnaugh Map
0 0
0 1
1 1
1 0
C D
A B

122. 4 Variables Karnaugh Map (cont’d)

123. 4 Variables Karnaugh Map (cont’d)

124. Karnaugh Map - Example Mapping a Standard SOP expression Example:
Answer:
Mapping a Standard POS expression
Using K-Map, convert the following standard POS
expression into a minimum SOP expression
Y = AB + AC or standard SOP :

125. K-Map with “Don’t Care” Conditions
Example :
Input Output
3 variables with output “don’t care (X)”

126. K-Map with “Don’t Care” Conditions (cont’d)
4 variables with output “don’t care (X)”

127. K-Map with “Don’t Care” Conditions (cont’d)
Example:
Determine the minimal SOP using K-Map:
Answer:

128. Minimum SOP expression is
Solution : CD AB X
X X X X 00 01 11 10 AD BC CD
Minimum SOP expression is

129. Extra Exercise Minimize this expression with a Karnaugh map
ABCD + ACD + BCD + ABCD

130. 5 variable K-map
5 variables -> 32 minterms, hence 32 squares required

131. K-map Product of Sums simplification
Example: Simplify the Boolean function F(ABCD)=(0,1,2,5,8,9,10) in (a) S-of-p (b) P-of-s
Using the maxterms (0’s) and complimenting F Grouping as if they were minterms, then using De Morgen’s theorem to get F.
F’(ABCD)= BD’+CD+AB
F(ABCD)= (B’+D)(C’+D’)(A’+B’)
Using the minterms (1’s)
F(ABCD)= B’D’+B’C’+A’C’D

132. 5 variable K-map
Adjacent squares. E.g. square 15 is adjacent to 7,14,13,31 and its mirror square 11.
The centre line must be considered as the centre of a book, each half of the K-map being a page
The centre line is like a mirror with each square being adjacent not only to its 4 immediate neighbouring squares, but also to its mirror image.

133. 5 variable K-map
Example: Simplify the Boolean function F(ABCDE) = (0,2,4,6,11,13,15,17,21,25,27,29,31) Soln: F(ABCDE) = BE+AD’E+A’B’E’

134. 6 variable K-map
6 variables -> 64 minterms, hence 64 squares required

135. Tutorial 1.5
ICS217-Digital Electronics - Part 1.5 Combinational Logic
1. Simplify the Boolean function F(ABCDE) = (0,1,4,5,16,17,21,25,29) Soln: F(ABCDE) = A’B’D’+AD’E+B’C’D’ 2. Simplify the following Boolean expressions using K-maps. (a) BDE+B’C’D+CDE+A’B’CE+A’B’C+B’C’D’E’ Soln: DE+A’B’C’+B’C’E’ (b) A’B’CE’+A’B’C’D’+B’D’E’+B’CD’+CDE’+BDE’ Soln: BDE’+B’CD’+B’D’E’+A’B’D’+CDE’ (c) F(ABCDEF) = (6,9,13,18,19,27,29,41,45,57,61) Soln: F(ABCDEF) = A’B’C’DEF’+A’BC’DE+CE’F+A’BD’EF

136. END OF Chapter 1