1. EKT 121 / 4 ELEKTRONIK DIGIT 1 CHAPTER 1 : INTRODUCTION

2. 1.0Number & Codes Digital and 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. Signed numbers Arithmetic operations with signed numbers Binary-Coded-Decimal (BCD) ASCII codes Gray codes Digital codes & parity

4. 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

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

6. Analog Quantities Continuous values

7. Digital Waveform

8. Introduction to Numbering Systems We are all familiar with the decimal number system (Base 10). Some other number systems that we will work with are: Binary  Base 2 Octal  Base 8 Hexadecimal  Base 16

9. Number Systems  0 ~ 9  0 ~ 1  0 ~ 7  0 ~ F Decimal Binary Octal Hexadecimal

10. Characteristics of Numbering Systems 1) The digits are consecutive. 2) The number of digits is equal to the size of the base. 3) Zero is always the first digit. 4) When 1 is added to the largest digit, a sum of zero and a carry of one results. 5) Numeric values determined by the implicit positional values of the digits.

11. 00000000 00000001 00000010 00000011 00000100 00000101 00000110 00000111 00001000 00001001 00001010 00001011 00001100 00001101 00001110 00001111 000 001 002 003 004 005 006 007 010 011 012 013 014 015 016 017 0123456789ABCDEF0123456789ABCDEF 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 BinaryOctalHexDec NUMBER SYSTEMSNUMBER SYSTEMS

12. Significant Digits Binary: 11101101 Most significant digit Least significant digit Hexadecimal: 1D63A7A Most significant digit Least significant digit

13. Binary Number System Also called the “Base 2 system” The binary number system is 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

14. Binary Numbering Scale Base 2 NumberBase 10 EquivalentPowerPositional Value 00002020 1 00112121 2 01022 4 01132323 8 10042424 16 10152525 32 11062626 64 11172727 128

15. Octal Number System Also known as the Base 8 System Uses digits 0 - 7 Readily converts to binary Groups of three (binary) digits can be used to represent each octal digit Also uses multiplication and division algorithms for conversion to and from base 10

16. Hexadecimal Number System Base 16 system Uses digits 0-9 & letters A,B,C,D,E,F Groups of four bits represent each base 16 digit

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

18. Number Conversion Binary to Decimal Conversion

19. Convert (10101101) 2 to its decimal equivalent: Binary 1 0 1 0 1 1 0 1 Positional Values x x x xx x x x 2020 21212 2323 2424 2525 2626 2727 128 + 32 + 8 + 4 + 1 Products 173 10

20. Octal to Decimal Conversion Convert 653 8 to its decimal equivalent: 6 5 3 xx x 8 2 8 1 8 0 384 + 40 + 3 427 10 Positional Values Products Octal Digits

21. Hexadecimal to Decimal Conversion Convert 3B4F 16 to its decimal equivalent: Hex Digits 3 B 4 F xx x 16 3 16 2 16 1 16 0 12288 +2816 + 64 +15 15,183 10 Positional Values Products x

22. Number Conversion Decimal to Any Radix (Base) Conversion 1. INTEGER DIGIT: Repeated division by the radix & record the remainder 2. FRACTIONAL DECIMAL: Multiply the number by the radix until the answer is in integer Example: 25.3125 to Binary

23. Decimal to Binary Conversion 2 5 = 12 + 1 2 1 2 = 6 + 0 2 6 = 3 + 0 2 3 = 1 + 1 2 1 = 0 + 1 2 MSBLSB 25 10 = 1 1 0 0 1 2 Remainder

24. Decimal to Binary Conversion Carry. 0 1 0 1 0.3125 x 2 = 0.625 0 0.625 x 2 = 1.25 1 0.25 x 2 = 0.50 0 0.5 x 2 = 1.00 1 The Answer: 1 1 0 0 1.0 1 0 1 MSBLSB

25. Decimal to Octal Conversion Convert 427 10 to its octal equivalent: 427 / 8 = 53 R3Divide by 8; R is LSD 53 / 8 = 6 R5Divide Q by 8; R is next digit 6 / 8 = 0 R6Repeat until Q = 0 653 8

26. Decimal to Hexadecimal Conversion Convert 830 10 to its hexadecimal equivalent: 830 / 16 = 51 R14 51 / 16 = 3 R3 3 / 16 = 0 R3 33E 16 = E in Hex

27. Number Conversion Binary to Octal Conversion (vice versa) 1. Grouping the binary position in groups of three starting at the least significant position.

28. Octal to Binary Conversion Each octal number converts to 3 binary digits To convert 653 8 to binary, just substitute code: 6 5 3 110 101 011

29. Number Conversion  Example: Convert the following binary numbers to their octal equivalent (vice versa). a) 1001.1111 2 b) 47.3 8 c) 1010011.11011 2 Answer: a) 11.74 8 b) 100111.011 2 c) 123.66 8

30. Number Conversion Binary to Hexadecimal Conversion (vice versa) 1. Grouping the binary position in 4-bit groups, starting from the least significant position.

31. 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

32. Number Conversion  Example: Convert the following binary numbers to their hexadecimal equivalent (vice versa). a) 10000.1 2 b) 1F.C 16 Answer: a) 10.8 16 b) 00011111.1100 2

33. Convert 010101101010111001101010 2 to hex using the 4-bit substitution code : 0101 0110 1010 1110 0110 1010 Substitution Code 5 6 A E 6 A 56AE6A 16

34. Substitution code can also be used to convert binary to octal by using 3-bit groupings: 010 101 101 010 111 001 101 010 Substitution Code 2 5 5 2 7 1 5 2 25527152 8

35. Binary Addition 0 + 0 = 0Sum of 0 with a carry of 0 0 + 1 = 1Sum of 1 with a carry of 0 1 + 0 = 1 Sum of 1 with a carry of 0 1 + 1 = 10Sum of 0 with a carry of 1 Example: 11001 111 + 1101+ 11 100110 ???

36. Simple Arithmetic Addition  Example: 10001100 2 + 101110 2 10111010 2 Substraction  Example: 1000100 2 - 101110 2 10110 2  Example: 58 16 + 24 16 7C 16

37. Binary Subtraction 0 - 0 = 0 1 - 1 = 0 1 - 0 = 1 10 -1 = 10 -1 with a borrow of 1 Example: 1011 101 - 111- 11 100 ???

38. Binary Multiplication 0 X 0 = 0 0 X 1 = 0Example: 1 X 0 = 0100110 1 X 1 = 1 X 101 100110 000000 + 100110 10111110

39. Binary Division Use the same procedure as decimal division

40. 1’s complements of binary numbers Changing all the 1s to 0s and all the 0s to 1s Example: 1 1 0 1 0 0 1 0 1Binary number 0 0 1 0 1 1 0 1 01’s complement

41. 2’s complements of binary numbers 2’s complement Step 1: Find 1’s complement of the number Binary #11000110 1’s complement00111001 Step 2: Add 1 to the 1’s complement 00111001 + 00000001 00111010

42. Signed Magnitude Numbers Sign bit 0 = positive 1 = negative 31 bits for magnitude This is your basic Integer format 110010..…00101110010101

43. Sign numbers Left most is the sign bit 0 is for positive, and 1 is for negative Sign-magnitude 0 0 0 1 1 0 0 1 = +25 sign bit magnitude bits 1’s complement The negative number is the 1’s complement of the corresponding positive number Example: +25 is 00011001 -25 is 11100110

44. Sign numbers 2’s complement 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

45. Digital Codes BCD (Binary Coded Decimal) Code 1. Represent each of the 10 decimal digits (0~9) as a 4-bit binary code.  Example: Convert 15 to BCD. 1 5 0001 0101 BCD Convert 10 to binary and BCD.

46. Digital Codes ASCII (American Standard Code for Information Interchange) Code 1. Used to translate from the keyboard characters to computer language

47. Digital Codes The Gray Code Only 1 bit changes Can’t be used in arithmetic circuits Binary to Gray Code and vice versa. DecimalBinaryGray Code 00000 10001 200100011 3 0010 401000110 501010111 601100101

48. 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

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

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

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

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

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

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

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

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

57. 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

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

59. Boolean Operations & expression 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

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

61. Laws & rules of Boolean algebra

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

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

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

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

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

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

68. 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

69. 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

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

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

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

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

74. 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.

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

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

77. 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

78. 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

79. De Morgan’s Theorem,

80. 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

81. 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

82. De Morgan’s Theorems 16) (X+Y) = X. Y 17) (X.Y) = X + Y 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 :

83. Implications of De Morgan’s Theorem (a) Equivalent circuit implied by theorem (16) (b) Alternative symbol for the NOR function (c) Truth table that illustrates DeMorgan’s Theorem (a) (b) Input Output X Y X+Y XY 0 0 1 1 0 1 0 0 10 0 0 1 1 0 0 (c)

84. Implications of De Morgan’s Theorem (a) Equivalent circuit implied by theorem (17) (b) Alternative symbol for the NAND function (c) Truth table that illustrates DeMorgan’s Theorem (a) (b) Input Output X Y XY X+Y 0 0 1 1 0 1 1 1 10 1 1 1 1 0 0 (c)

85. 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

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

87. Examples: Analyze the circuit below Y 1. Y=??? 2. Simplify the Boolean expression found in 1

88. 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.

89. Exercises: Simplify the following Boolean expressions 1. (AB(C + BD) + AB)C 2. ABC + ABC + ABC + ABC + ABC Write the Boolean expression of the following circuit.

90. 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

91. 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.

92. 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.

93. 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

94. 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

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

96. THE K-MAP

97. 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.

98. 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 Karnaugh Map (K-Map)

99. K-Map SOP Minimization

100. AAAA B Notice that the map is going false to true, left to right and top to bottom The upper right hand cell is A B if X= A B then put an X in that cell AAAA B 1 This show the expression true when A = 0 and B = 0 0 1 2 3 2 Variables Karnaugh Map

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

102. 3 Variables Karnaugh Map Gray Code 00A B 01A B 11 A B 10A B 0 1 C C 0 1 0 1 2 3 2 3 6 7 6 7 4 5 4 5

103. X = A B C + A B C + A B C + A B C Gray Code 00A B 01A B 11 A B 10A B 0 1 C C Each 3 variable term is one cell on a 4 X 2 Karnaugh map 11 11 3 Variables Karnaugh Map (cont’d)

104. X = A B C + A B C + A B C + A B C Gray Code 00A B 01A B 11 A B 10A B 0 1 C C One simplification could be X = A B + A B 11 1 1 3 Variables Karnaugh Map (cont’d)

105. X = A B C + A B C + A B C + A B C Gray Code 00A B 01A B 11 A B 10A B 0 1 C C Another simplification could be X = B C + B C A Karnaugh Map does wrap around 11 11 3 Variables Karnaugh Map (cont’d)

106. X = A B C + A B C + A B C + A B C Gray Code 00A B 01A B 11 A B 10A B 0 1 C C The Best simplification would be X = B 11 11 3 Variables Karnaugh Map (cont’d)

107. 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

108. 4 Variables Karnaugh Map Gray Code 00A B 01A B 11A B 10A B 0 0 0 1 1 1 1 0 C D C D 0 1 3 2 0 1 3 2 4 5 7 6 4 5 7 6 12 13 15 14 12 13 15 14 8 9 11 10 8 9 11 10

109. Gray Code 00A B 01A B 11A B 10A B 0 0 0 1 1 1 1 0 C D C D 1 1 1 1 1 1 X = ABD + ABC + CD Now try it with Boolean reductions 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

110. 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

111. Simplify : Z = B C D + B C D + C D + B C D + A B C Gray Code 00A B 01A B 11A B 10A B 00 01 11 10 C D C D 1111 11 11 11 1 1 Z = C + A B + B D

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

113. 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 Simplify using Karnaugh map (cont’d) SOP expression

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

115. K-Map POS Minimization

116. 3 Variables Karnaugh Map Gray Code 0 0 1 1 1 0 0 1 AB C 0 1 0 1 2 3 2 3 6 7 6 7 4 5 4 5

117. 3 Variables Karnaugh Map (cont’d)

118. 4 Variables Karnaugh Map 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 0 1 3 2 4 5 7 6 4 5 7 6 12 13 15 14 12 13 15 14 8 9 11 10 8 9 11 10

119. 4 Variables Karnaugh Map (cont’d)

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

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

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

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

125. Solution : AB CD 00 01 11 10 00 01 11 10 0 1 1 0 1 X 1 0 X X 0 0 1 0 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 Minimum SOP expression is CD AD BC

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

127. 5 variable K-map 5 variables -> 32 minterms, hence 32 squares required

128. 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 minterms (1’s) F (ABCD) = B’D’+B’C’+A’C’D 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’)

129. 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.

130. 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’

132. E = 0 How many k-map is needed? If you have 5 variables, you’ll need 2 k-map… Let’s say the variables are A, B, C, D and E. 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 0 1 3 2 4 5 7 6 4 5 7 6 12 13 15 14 12 13 15 14 8 9 11 10 8 9 11 10

133. How many k-map is needed? E = 1 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 0 1 3 2 4 5 7 6 4 5 7 6 12 13 15 14 12 13 15 14 8 9 11 10 8 9 11 10

134. Try this out… Simplify the Boolean function F(A,B,C,D,E) =  (0,1,4,5,16,17,21,25,29) Soln: F(A,B,C,D,E) = A’B’D’+AD’E+B’C’D’

135. 1 st step – convert the minterm into Boolean equation F(A,B,C,D,E) =  (0,1,4,5,16,17,21,25,29) How to convert … ?? 0 = 00000  1 = 00001 

136. 1 st step – convert the minterm into Boolean equation F(A,B,C,D,E) =  (0,1,4,5,16,17,21,25,29)

137. 2 nd – prepare 2 k-map 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0E = 1

138. 3 rd - plug in the Boolean term/minterm into k-map 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0E = 1 1 1 1 1 1 1 1 1 1

139. 4 th - look for similar grouping that can be done in both k-map 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0E = 1 1 1 1 1 1 1 1 1 1 That will give us

140. 4 th - look for similar grouping that can be done in both k-map 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0E = 1 1 1 1 1 1 1 1 1 1 That will give us

141. The combination will look like this… 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0E = 1 1 1 1 1 1 1 1 1 1 That will give us

142. 5 th- look for the remaining 1’s that has not been included yet 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0E = 1 1 1 1 1 1 1 1 1 1

143. 6 th - group the remaining 1’s 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0E = 1 1 1 1 1 1 1 1 1 That new grouping will give us Note that this time, E need to be included in the term, since the grouping is in the E = 1 k-map only. 1

144. 6 th - group the remaining 1’s 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0E = 1 1 1 1 1 1 1 1 1 Full expression will be, 1

145. 6 variable K-map 6 variables -> 64 minterms, hence 64 squares required

146. ICS217-Digital Electronics - Part 1.5 Combinational Logic Tutorial 1.5 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

147. END OF Chapter 1