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CHAPTER 1 : INTRODUCTION Lecturer : FAIRUL AFZAL Office : KKF 8 (04-9854150)

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Presentation on theme: "CHAPTER 1 : INTRODUCTION Lecturer : FAIRUL AFZAL Office : KKF 8 (04-9854150)"— Presentation transcript:

1 CHAPTER 1 : INTRODUCTION Lecturer : FAIRUL AFZAL Office : KKF 8 ( )

2  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  Signed numbers  Arithmetic operations with signed numbers  Binary-Coded-Decimal (BCD)  ASCII codes  Gray codes  Digital codes & parity

4  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  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  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  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 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 Continuous values

10

11  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  0 ~ 9  0 ~ 1  0 ~ 7  0 ~ F Decimal Binary Octal Hexadecimal

13 ABCDEF ABCDEF BinaryOctalHexDec NUMBER SYSTEMSNUMBER SYSTEMS

14 Binary: Most significant digit Least significant digit Hexadecimal: 1D63A7A Most significant digit Least significant digit

15  Base 10 - (0,1,2,3,4,5,6,7,8,9)  Example : X X X = = Weights for whole numbers are positive power of ten that increase from right to left, beginning with 10 0

16 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 : X X 2 1 1X 2 2 Weights in a binary number are based on power of two, that increase from right to right to left,beginning with = =5 10

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

18 Base 16 system  Uses digits 0 ~ 9 & letters A,B,C,D,E,F  Groups of four bits represent each base 16 digit HEXADECIMAL DECIMAL BINARY A B C D E F151111

19 Base 16 system –  multiplication and division algorithms for conversion to and from base 10  Example : A9F 16 convert to decimal: A 9 F 15 X X X 16 2 Weights in a hexadecimal number are based on power of sixteen that increase from right to right to left,beginning with = =  Readily converts to binary Groups of four (binary) digits can be used to represent each hexadecimal number Example : A9F 8 convert to binary : A 9 F

20  Any Radix (base) to Decimal Conversion

21  Binary to Decimal Conversion

22 Convert ( ) 2 to its decimal equivalent: Binary Positional Values x x x xx x x x Products

23 Convert to its decimal equivalent: xx x Positional Values Products Octal Digits

24 Convert 3B4F 16 to its decimal equivalent: Hex Digits 3 B 4 F xx x , Positional Values Products x

25  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: to Binary

26 2 5 = = = = = MSBLSB = Remainder

27 Carry x 2 = x 2 = x 2 = x 2 = The Answer: MSBLSB

28 Convert 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 =

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

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

31 Each octal number converts to 3 binary digits To convert to binary, just substitute code:

32  Example:  Convert the following binary numbers to their octal equivalent (vice versa). a) b) c)  Answer: a) b) c)

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

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

35  Example:  Convert the following binary numbers to their hexadecimal equivalent (vice versa). a) b) 1F.C 16  Answer: a) b)

36 Convert to hex using the 4-bit substitution code : A E 6 A 56AE6A 16

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

38 0 + 0 = 0Sum of 0 with a carry of = 1Sum of 1 with a carry of = 1 Sum of 1 with a carry of = 10Sum of 0 with a carry of 1 Example: ???

39  Addition  Example:  Substraction  Example:  Example: C 16

40 0 - 0 = = = = with a borrow of 1 Example: ???

41 0 X 0 = 0 0 X 1 = 0Example: 1 X 0 = X 1 = 1 X

42 Use the same procedure as decimal division

43  Changing all the 1s to 0s and all the 0s to 1s Example: Binary number ’s complement

44  2’s complement  Step 1: Find 1’s complement of the number Binary # ’s complement  Step 2: Add 1 to the 1’s complement

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

46  Left most is the sign bit  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  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

48  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 BCD  Convert 10 to binary and BCD.

49  ASCII (American Standard Code for Information Interchange) Code 1. 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  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 ????? DecimalBinaryGray Code

51 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 b) Truth Table (Jadual Kebenaran) c) Timing Diagram (Rajah Pemasaan)

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

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

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

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

60  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  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 B X A B X

64

65 Commutative law of addition 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) = AB + AC (A+B)(C+D) = AC + AD + BC + BD

70  In math if you add 0 you have changed nothing  In Boolean Algebra ORing with 0 changes nothing

71  ORing with 1 must give a 1 since if any input is 1 an OR gate will give a 1

72  In math if 0 is multiplied with anything you get 0. If you AND anything with 0 you get 0

73  ANDing anything with 1 will yield the anything

74  ORing with itself will give the same result

75  Either A or A must be 1 so A + A =1

76  ANDing with itself will give the same result

77  In digital Logic 1 =0 and 0 =1, so AA=0 since one of the inputs must be 0.

78  If you not something twice you are back to the beginning

79 Proof: A + AB = A(1 +B)DISTRIBUTIVE LAW = A∙1 RULE 2: (1+B)=1 = ARULE 4: A∙1 = A

80  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 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 DE MORGAN’S THEOREM

83 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 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 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 :

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

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

88 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

89 ABC + ABC (A + B +C)D = (A+B+C).(A+B+C) = (A.B.C)+D

90 Y 1. Y=??? 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  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.

93 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 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 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 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 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 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 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 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)

102

103 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 = Variables Karnaugh Map

104 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 Variables Karnaugh Map

105 Gray Code 00A B 01A B 11 A B 10A B 0 1 C C

106 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 Variables Karnaugh Map (cont’d)

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

108 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 Variables Karnaugh Map (cont’d)

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

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

111 Gray Code 00A B 01A B 11A B 10A B C D C D

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

113 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 Gray Code 00A B 01A B 11A B 10A B C D C D Z = C + A B + B D

115 First, we need to change the circuit to an SOP expression

116 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

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

118

119 Gray Code AB C

120

121 A B C D

122

123

124 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 :

125 Input Output Example : 3 variables with output “don’t care (X)”

126 4 variables with output “don’t care (X)”

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

128 AB CD X 1 0 X X Minimum SOP expression is CD AD BC

129  Minimize this expression with a Karnaugh map ABCD + ACD + BCD + ABCD

130  5 variables -> 32 minterms, hence 32 squares required

131 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’)

132  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 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 variables -> 64 minterms, hence 64 squares required

135 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 ICS217-Digital Electronics - Part 1.5 Combinational Logic

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