Number System conversions

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1 Number System conversions

2 Number Systems The system used to count discrete units is called number system. There are four systems of arithmetic which are often used in digital electronics. Decimal Number System Binary Number System Octal Number System Hexa Decimal System

3 Decimal Number system Decimal number system contains 0 digits: 0,,2,3,4,5,6,7,8,9; and that is why its base or radix is 0. Here radix means total number of digits used in any system. To indicate digits greater than 9 (nine) we shall use the sum of these digits times power of 0.

4 Fractional DECIMAL SYSTEM Decimal System The decimal system is composed of 0 numerals or symbols. These 0 symbols are 0,, 2, 3, 4, 5, 6, 7, 8, 9; using these symbols as digits of a number, we can express any quantity. The decimal system, also called the base- 0 system because it has 0 digits

5 Decimal Number System The decimal number system is a positional number system. Example: X 0 0 = X 0 = 20 6 X 0 2 = X 0 3 = 5000

6 Binary Number System The binary number system is also a positional numbering system. Instead of using ten digits, 0-9, the binary system uses only two digits, 0 and.

7 Binary Number System The binary number system is also known as base 2. The values of the positions are calculated by taking 2 to some power. Why is the base 2 for binary numbers? o Because we use 2 digits, the digits 0 and.

8 Fractional BINARY SYSTEM In the binary system, there are only two symbols or possible digit values, 0 and. This base-2 system can be used to represent any quantity that can be represented in decimal or other number system In digital systems the information that is being processed is usually presented in binary form. Binary quantities can be represented by any device that has only two operating states or possible conditions. Eg. a switch has only open or closed. We arbitrarily (as we define them) let an open switch represent binary 0 and a closed switch represent binary. Thus we can represent any binary number by using series of switches.

9 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 0

10 Fractional OCTAL NUMBER SYSTEM The octal number system has a base of eight, meaning that it has eight possible digits: 0,,2,3,4,5,6, Use to represent long binary numbers in computers and microprocessors.

11 Hexadecimal Number Base 6 system Uses digits 0-9 & letters A,B,C,D,E,F Groups of four bits represent each base 6 digit System

12 Fractional HEXADECIMAL NUMBER SYSTEM The hexadecimal system uses base 6. Thus, it has 6 possible digit symbols. It uses the digits 0 through 9 plus the letters A, B, C, D, E, and F as the 6 digit symbols Use to represent long binary numbers in computers and microprocessors. These digits can use to program machine language.

13 Binary Numbering Scale Base 2 Number Base 0 Equivalent Power Positional Value

14 Significant Digits Binary: 00 Most significant digit Least significant digit Hexadecimal: D63A7A Most significant digit Least significant digit

15 Converting From Decimal to Binary Make a list of the binary place values up to the number being converted. Perform successive divisions by 2, placing the remainder of 0 or in each of the positions from right to left. Continue until the quotient is zero. Example:

16 DECIMAL TO BINARY CONVERTION There are two methods to convert it:- i. Revese of Binary-To-Digital Method Decimal number write as the sum of square 0 & is write on the byte Example : Convert 450 to the binary value Solve = 45 = = = 0 0 2

17 ii. Repeat division method The numbers is divide by 2. Balance for the question is written until the last answer. Example : convert 250 to binary Solve = 250 =?2 = balance LSB = balance 0 = balance 0 = 3 balance 2 = 2 0 balance MSB... Answer = 002

18 Decimal to Binary Conversion The easiest way to convert a decimal number to its binary equivalent is to use the Division Algorithm This method repeatedly divides a decimal number by 2 and records the quotient and remainder The remainder digits (a sequence of zeros and ones) form the binary equivalent in least significant to most significant digit sequence

19 Division Algorithm Convert 67 to its binary equivalent: 67 0 = x 2 Step : 67 / 2 = 33 R Step 2: 33 / 2 = 6 R Step 3: 6 / 2 = 8 R 0 Step 4: 8 / 2 = 4 R 0 Step 5: 4 / 2 = 2 R 0 Step 6: 2 / 2 = R 0 Divide 67 by 2. Record quotient in next row Again divide by 2; record quotient in next row Repeat again Repeat again Repeat again Repeat again Step 7: / 2 = 0 R STOP when quotient equals

20 Decimal to binary Examples: (fractional) Convert the decimal number (2.0625) 0 into binary number. Solution: Fractional part: x 2 = x 2 = x 2 = x 2 =.000 (2.0625) 0 = (00.000) 2

21 Decimal to Octal Examples: Conversion (35) 0 = (473) LSD MSD

22 Convert from decimal to octal by using the repeated division method used for decimal to binary conversion. Divide the decimal number by 8 The first remainder is the LSB and the last is the MSB. Example : convert 3590 to Decimal Value Solve = DECIMAL TO OCTAL CONVERTION 3590 =?8 = balance 7 LSB 44 = 8 5 balance 4 = balance 5... Answer = 5478 MSB

23 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 =

24 Decimal to Octal Examples: (fractional) Convert the decimal number ( ) 0 into octal number. Solution: Fractional part: x 8 = x 8 = x 8 = x 8 = x 8 = ( ) 0 = (34.634) 8

25 Decimal to Hexadecimal Conversion Examples (35) 0 = (3B) B 3 LSD MSD

26 Decimal to Hexadecimal Conversion Convert to its hexadecimal equivalent: 830 / 6 = 5 R4 5 / 6 = 3 R3 3 / 6 = 0 R3 = E in Hex 33E 6

27 Decimal to Hexadecimal (fractional) Examples: Convert the decimal number ( ) 0 into hexadecimal number. Solution: Fractional part: x 6 = x 6 = ( ) 0 = (E.39) 6

28 Converting from Binary to Decimal Example of a binary number and the values of the positions:

29 Converting from Binary to Decimal X 2 0 = X 2 = = 2 4 = 6 X 2 3 = 8 2 = = 32 0 X 2 4 = 0 X 2 2 = = = 64 0 X 2 5 = = 8 X 2 6 =

30 Binary to Decimal Conversion The easiest method for converting a binary number to its decimal equivalent is to use the Multiplication Algorithm Multiply the binary digits by increasing powers of two, starting from the right Then, to find the decimal number equivalent, sum those products

31 Multiplication Algorithm Convert (000) 2 to its decimal equivalent: Binary Positional Values x x x x x x x x Products

32 Converting from Binary to Decimal Practice conversions: Binary Decimal

33 Converting From Decimal to Binary Practice conversions: Decimal Binary

34 BINARY TO OCTAL CONVERSION Can be converted by grouping the binary bit in group of three starting from LSB Octal is a base-8 system and equal to two the power of three, so a digit in Octal is equal to three digit in binary system.

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36 binary to hexadecimal system Group the digits of the binary number by four starting from the right. 2 Replace each group of four digits by an equivalent hexadecimal digit. Convert into a hexadecimal number = B5 6 B 5

37 Exercise Convert into a hexadecimal number. 2 Convert 0 2 into a hexadecimal number.

38 Octal to Decimal Conversion Convert to its decimal equivalent: Octal Digits Positional Values Products x x x

39 OCTAL TO BINARY CONVERTION Convert from octal to binary by converting each octal digit to a three bit binary equivalent Octal digit Binary Equivalent Convert from binary to octal by grouping bits in threes starting with the LSB. Each group is then converted to the octal equivalent Leading zeros can be added to the left of the MSB to fill out the last group.

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

42 Hexadecimal to Decimal Conversion Convert 3B4F6 to its decimal equivalent: Hex Digits Positional Values Products 3 B 4 F x x x x ,83 0

43 HEXADECIMAL TO binary To convert a hexadecimal to binary number, convert each hexadecimal digit to its 4 bit equivalent using the hexa number. Example: (23.AB) 6 = () 2 Solution: (23.AB) 6 = 2 3. A B (23.AB) 6 = ( ) 2

44 HEXADECIMAL TO OCTAL CONVERTION There is two ways to convert it:- i. Hexadecimal Decimal Octal ii. Hexadecimal Binary Octal i. Hexadecimal Decimal Octal

45 i. Hexadecimal Binary Octal

46 Hexadecimal Number System Binary Decimal Hexadecimal Binary Decimal Hexadecimal A 0 B C D E F

47 Binary Logic and Gates Binary variables take on one of two values. Logical operators operate on binary values and binary variables. Basic logical operators are the logic functions AND, OR and NOT. Logic gates implement logic functions. Boolean Algebra: a useful mathematical system for specifying and transforming logic functions. We study Boolean algebra as a foundation for designing and analyzing digital Boolean Algebra and Logic Gates 47

48 Binary Variables Recall that the two binary values have different names: True/False On/Off Yes/No /0 We use and 0 to denote the two values. Variable identifier examples: A, B, y, z, or X for now RESET, START_IT, or ADD later Boolean Algebra and Logic Gates 48

49 Logical Operations The three basic logical operations are: AND OR NOT AND is denoted by a dot ( ). OR is denoted by a plus (+). NOT is denoted by an overbar ( ), a single quote mark (') after, or (~) before the variable. Boolean Algebra and Logic Gates 49

50 Examples: Notation Examples Y A B is read Y is equal to A AND B. z x y X A is read z is equal to x OR y. is read X is equal to NOT A. Note: The statement: + = 2 (read one plus one equals two ) is not the same as + = (read or equals ). Boolean Algebra and Logic Gates 50

51 Operator Definitions Operations are defined on the values "0" and "" for each operator: AND OR 0 0 = 0 0 = 0 0 = 0 = = = + 0 = + = NOT 0 0 Boolean Algebra and Logic Gates 5

52 Boolean Algebra and Logic Gates 52 Truth Tables Tabular listing of the values of a function for all possible combinations of values on its arguments Example: Truth tables for the basic logic operations: AND OR NOT X 0 0 Y Z = X Y X Y Z = X+Y X 0 Z 0 X

53 Boolean Algebra and Logic Gates 53 Truth Tables Cont d Used to evaluate any logic function Consider F(X, Y, Z) = X Y + Y Z X Y Z X Y Y Y Z F = X Y + Y Z

54 Using Switches Inputs: Boolean Algebra and Logic Gates 54 logic is switch closed logic 0 is switch open Outputs: logic is light on logic 0 is light off. NOT input: Logic Function Implementation Normally-closed switch => NOT logic is switch open C logic 0 is switch closed Switches in parallel => OR Switches in series => AND

55 Logic Function Implementation cont d Example: Logic Using Switches A B C Light is on (L = ) for L(A, B, C, D) = A (B C + D) = A B C + A D and off (L = 0), otherwise. Useful model for relay and CMOS gate circuits, the foundation of current digital logic circuits Boolean Algebra and Logic Gates 55 D

56 Boolean Algebra and Logic Gates 56 Logic Gates In the earliest computers, switches were opened and closed by magnetic fields produced by energizing coils in relays. The switches in turn opened and closed the current paths. Later, vacuum tubes that open and close current paths electronically replaced relays. Today, transistors are used as electronic switches that open and close current

57 X Y Logic Gate Symbols and Behavior Logic gates have special symbols: AND gat e Z = X Y X Y OR gat e Z = X + Y And waveform behavior in time as follows: X NOT gat e or i nver t er Z = X X 0 0 Y 0 0 ( AND) X Y ( OR) X + Y 0 Boolean Algebra and Logic Gates 57 ( NOT) X 0 0

58 Logic Diagrams and Expressions X Y Z Boolean equations, truth tables and logic diagrams describe the same function! Truth tables are unique, but expressions and logic diagrams are not. This gives flexibility in implementing functions. Boolean Algebra and Logic Gates 58 Truth Table F X Y Z X Y Z Logic Equation F X Y Z Logic Diagram F

59 Boolean Algebra and Logic Gates 59 Gate Delay In actual physical gates, if an input changes that causes the output to change, the output change does not occur instantaneously. The delay between an input change and the output change is the gate delay denoted by t G : Input Output 0 0 tg tg tg = 0.3 ns Time (ns)

60 Boolean Algebra Invented by George Boole in 854 An algebraic structure defined by a set B = {0, }, together with two binary operators (+ and ) and a unary operator ( ). X + 0 = X 2. X. = X Identity element 3. X + = 4. X. 0 0 = 5. X + X = X 6. X. X = X Idempotence 7. X + X = 8. X. X = 0 Complement 9. X = X Involution 0. X + Y = Y + X. XY = YX Commutative 2. (X + Y) + Z = X + (Y + Z) 3. (XY) Z X(Y Z) = Associative 4. X(Y + Z) = XY XZ + 5. X + YZ = (X + Y) (X + Z) Distributive 6. X + Y = X. Y 7. X. Y = X + Y DeMorgan s Boolean Algebra and Logic Gates 60

61 Some Properties of Boolean Algebra Boolean Algebra is defined in general by a set B that can have more than two values A two-valued Boolean algebra is also know as Switching Algebra. The Boolean set B is restricted to 0 and. Switching circuits can be represented by this algebra. The dual of an algebraic expression is obtained by interchanging + and and interchanging 0 s and s. The identities appear in dual pairs. When there is only one identity on a line the identity is self-dual, i. e., the dual expression = the original expression. Sometimes, the dot symbol (AND operator) is not written when the meaning is clear Boolean Algebra and Logic Gates 6

62 Dual of a Boolean Expression Example: F = (A + C) B + 0 Boolean Algebra and Logic Gates 62 dual F = (A C + B) = A C + B Example: G = X Y + (W + Z) dual G = (X+Y) (W Z) = (X+Y) (W+Z) Example: H = A B + A C + B C dual H = (A+B) (A+C) Unless it happens (B+C) to be self-dual, the dual of an expression does not equal the expression itself H is self- Are any of these functions self-dual? (A+B)(A+C)(B+C)=(A+BC)(B+C)=AB+AC+BC

63 Boolean Algebra and Logic Gates 63 Boolean Operator Precedence The order of evaluation is:. Parentheses 2. NOT 3. AND 4. OR Consequence: Parentheses appear around OR expressions Example: F = A(B + C)(C + D)

64 Boolean Algebraic Proof Example A + A B = A (Absorption Theorem) Proof Steps Justification A + A B = A + A B Identity element: A = A = A ( + B) Distributive = A + B = = A Identity element Our primary reason for doing proofs is to learn: Careful and efficient use of the identities and theorems of Boolean algebra, and How to choose the appropriate identity or theorem to apply to make forward progress, irrespective of the Boolean Algebra and Logic Gates 64 application.

65 Boolean Algebraic Proof Example 2 AB + AC + BC = AB + AC (Consensus Theorem) Proof Steps Justification = AB + AC + BC = AB + AC + BC Identity element = AB + AC + (A + A) BC Complement = AB + AC + ABC + ABC Distributive = AB + ABC + AC + ACB Commutative = AB + ABC + AC + ACB Identity element = AB (+C) + AC ( + B) Distributive = AB. + AC. +X = = AB + AC Identity element Boolean Algebra and Logic Gates 65

66 Useful Theorems Minimization X Y + X Y = Y Absorption X + X Y = X Simplification X + X Y = X + Y DeMorgan s X + Y = X Y Boolean Algebra and Logic Gates 66 Minimization (dual) (X+Y)(X+Y) = Y Absorption (dual) X (X + Y) = X Simplification (dual) X (X + Y) = X Y DeMorgan s (dual) X Y = X + Y

67 Truth Table to Verify DeMorgan s X + Y = X Y X Y = X + Y X Y X Y X+Y X Y X+Y X Y X Y X+Y Generalized DeMorgan s Theorem: X + X X n = X X 2 X n X X 2 X n = X + X X n Boolean Algebra and Logic Gates 67

68 Complementing Functions Use DeMorgan's Theorem:. Interchange AND and OR operators 2. Complement each constant and literal Example: Complement F = F = (x + y + z)(x + y + z) Example: Complement G = (a + bc)d + e Boolean Algebra and Logic Gates 68 x y z x y z

69 Expression Simplification An application of Boolean algebra Simplify to contain the smallest number of literals (variables that may or may not be complemented) A B = AB + ABCD + A C D + A C D + A B D = AB + AB(CD) + A C (D + D) + A B D = AB + A C + A B D = B(A + AD) +AC = B (A + D) + A C (has only 5 literals) Boolean Algebra and Logic Gates 69 A C D A B D A C D A B C D

70 Next Canonical Forms Minterms and Maxterms Sum-of-Minterm (SOM) Canonical Form Product-of-Maxterm (POM) Canonical Form Representation of Complements of Functions Conversions between Representations Boolean Algebra and Logic Gates 70

71 Boolean Algebra and Logic Gates 7 Minterms Minterms are AND terms with every variable present in either true or complemented form. Given that each binary variable may x appear normal (e.g., x) or complemented (e.g., ), there are 2 n minterms for n variables. Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: XY X Y X Y X Y (both normal) (X normal, Y complemented) (X complemented, Y normal) (both complemented) Thus there are four minterms of two

72 Maxterms Maxterms are OR terms with every variable in true or complemented form. Given that each binary variable may appear normal (e.g., x) or complemented (e.g., x), there are 2 n maxterms for n variables. Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: X Y (both normal) X Y (x normal, y complemented) X Y (x complemented, y normal) X Boolean Algebra and Logic Gates 72 Y (both complemented)

73 Two variable minterms and maxterms. 3 m 3 = x y M 3 = x + y The minterm m i should evaluate to for each combination of x and y. The maxterm is the complement of the minterm Boolean Algebra and Logic Gates 73 Minterms & Maxterms for 2 variables x y Index Minterm Maxterm m 0 = x y M 0 = x + y 0 m = x y M = x + y 0 2 m 2 = x y M 2 = x + y

74 x Boolean Algebra and Logic Gates 74 Minterms & Maxterms for 3 y z Index variables Minterm m0 = x y z m = x y z m2 = x y z m3 = x y z m4 = x y z m5 = x y z m6 = x y z m7 = x y z Maxterm M0 = x + y + z M = x + y + z M2 = x + y + z M3 = x + y + z M4 = x + y + z M5 = x + y + z M6 = x + y + z M7 = x + y + z Maxterm Mi is the complement of minterm mi

75 Purpose of the Index Minterms and Maxterms are designated with an index The index number corresponds to a binary pattern The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true or complemented form For Minterms: means the variable is Not Complemented and 0 means the variable is Complemented. For Maxterms: 0 means the variable is Not Complemented and Boolean Algebra and Logic Gates 75 means the variable is Complemented.

76 Standard Order All variables should be present in a minterm or maxterm and should be listed in the same order (usually alphabetically) Example: For variables a, b, c: Maxterms (a + b + c), (a + b + c) are in standard order However, (b + a + c) is NOT in standard order Boolean Algebra and Logic Gates 76 (a + c) does NOT contain all variables Minterms (a b c) and (a b c) are in standard order However, (b a c) is not in standard order

77 Sum-Of-Minterm (SOM) Sum-Of-Minterm (SOM) canonical form: Sum of minterms of entries that evaluate to x y z F Minterm m = x y z m 6 = x y z m 7 = x y z F = m + m6 + m7 = (, 6, 7) = x y z + x y z + x y z Boolean Algebra and Logic Gates 77 Focus on the entries

78 Sum-Of-Minterm Examples F(a, b, c, d) = (2, 3, 6, 0, ) F(a, b, c, d) = m 2 + m 3 + m 6 + m 0 + m a b c d + a b c d + a b c d + a b c d + a b c d G(a, b, c, d) = (0,, 2, 5) G(a, b, c, d) = m 0 + m + m a b 2 + m a b c d + a b c d + a b c d c d 5 Boolean Algebra and Logic Gates 78

79 Product-Of-Maxterm (POM) Product-Of-Maxterm (POM) canonical form: Product of maxterms of entries that evaluate to 0 x y z F Maxterm M 2 = (x + y + z) M 4 = (x + y + z) M 6 = (x + y + z) Focus on the 0 entries F = M2 M4 M6 = (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z) Boolean Algebra and Logic Gates 79

80 F(a, b, c, d) = (, 3, 6, ) F(a, b, c, d) = M M 3 M 6 M (a+b+c+d) (a+b+c+d (a+b+c+d (a+b+c+d G(a, b, c, d) = (0, 4, 2, 5) ) ) ) G(a, b, c, d) = M 0 M 4 M 2 M 5 Boolean Algebra and Logic Gates 80 Product-Of-Maxterm Examples (a+b+c+d) (a+b+c+d ) (a+b+c+d ) (a+b+c+d )

81 Boolean Algebra and Logic Gates 8 Observations We can implement any function by "ORing" the minterms corresponding to the entries in the function table. A minterm evaluates to for its corresponding entry. We can implement any function by "ANDing" the maxterms corresponding to 0 entries in the function table. A maxterm evaluates to 0 for its corresponding entry. The same Boolean function can be expressed in two canonical ways: Sum-of-Minterms (SOM) and Productof-Maxterms (POM). If a Boolean function has fewer entries then the SOM canonical form will contain fewer literals than POM. However, if it has fewer 0 entries then the POM

82 Converting to Sum-of-Minterms Boolean Algebra and Logic Gates 82 Form A function that is not in the Sum-of-Minterms form can be converted to that form by means of a truth table Consider F = y + x z x y z F Minterm m 0 = x y z 0 0 m = x y z 0 0 m 2 = x y z m 4 = x y z 0 m 5 = x y z F = (0,, 2, 4, 5) = m0 + m + m2 + m4 + m5 = x y z + x y z + x y z + x y z + x y z

83 Converting to Product-of-Maxterms Boolean Algebra and Logic Gates 83 Form A function that is not in the Product-of-Minterms form can be converted to that form by means of a truth table Consider x y z again: F Minterm F = y + x z F = (3, 6, 7) = M 3 = (x+y+z) M 6 = (x+y+z) 0 M 7 = (x+y+z) M3 M6 M7 = (x+y+z) (x+y+z) (x+y+z)

84 Conversions Between Canonical Forms x y z F Minterm Maxterm M 0 = (x + y + z) 0 0 m = x y z 0 0 m 2 = x y z 0 m 3 = x y z M 4 = (x + y + z) 0 m 5 = x y z 0 0 M 6 = (x + y + z) m 7 = x y z F = m +m 2 +m 3 +m 5 +m 7 = (, 2, 3, 5, 7) = x y z + x y z + x y z + x y z + x y z F = M 0 M 4 M 6 = (0, 4, 6) = (x+y+z)(x+y+z)(x+y+z) Boolean Algebra and Logic Gates 84

85 Algebraic Conversion to Sum-of- Minterms Expand all terms first to explicitly list all minterms AND any term missing a variable v with (v + v) Example : f = x + x y f = x (y + y) + x y f = x y + x y + x y f = m 3 + m 2 + m 0 = (0, 2, 3) Boolean Algebra and Logic Gates 85 (2 variables) Example 2: g = a + b c (3 variables) g = a (b + b)(c + c) + (a + a) b c g = a b c + a b c + a b c + a b c + a b c + a b c g = a b c + a b c + a b c + a b c + a b c g = m + m 4 + m 5 + m 6 + m 7 = (, 4, 5, 6, 7)

86 Algebraic Conversion to Product-of- Maxterms Expand all terms first to explicitly list all maxterms OR any term missing a variable v with v v Example : f = x + x y (2 variables) Apply 2 nd distributive law: f = (x + x) (x + y) = (x + y) = (x + y) = M Example 2: g = a c + b c + a b (3 variables) g = (a c + b c + a) (a c + b c + b) (distributive) g = (c + b c + a) (a c + c + b) (x + x y = x + y) Boolean Algebra and Logic Gates 86

87 Function Complements The complement of a function expressed as a sum of minterms is constructed by selecting the minterms missing in the sum-of-minterms canonical form Alternatively, the complement of a function expressed by a Sum of Minterms form is simply the Product of Maxterms with the same indices Example: Given F(x, y, z) = (, 3, 5, 7) F(x, y, z) = (0, 2, 4, 6) F(x, y, z) = (, 3, 5, 7) Boolean Algebra and Logic Gates 87

88 Boolean Algebra and Logic Gates 88 Summary of Minterms and Maxterms There are 2 n minterms and maxterms for Boolean functions with n variables. Minterms and maxterms are indexed from 0 to 2 n Any Boolean function can be expressed as a logical sum of minterms and as a logical product of maxterms The complement of a function contains those minterms not included in the original function The complement of a sum-of-minterms is a product-of-maxterms with the same indices

89 Standard Forms Standard Sum-of-Products (SOP) form: equations are written as an OR of AND terms Standard Product-of-Sums (POS) form: equations are written as an AND of OR terms Examples: A B C A B C B SOP: (A B) (A B C ) C POS: These (A B mixed C) (A forms C) are neither SOP nor POS A B C A C (A B) Boolean Algebra and Logic Gates 89

90 Standard Sum-of-Products Boolean Algebra and Logic Gates 90 (SOP) A sum of minterms form for n variables can be written down directly from a truth table. Implementation of this form is a two-level network of gates such that: The first level consists of n-input AND gates The second level is a single OR gate This form often can be simplified so that the corresponding circuit is simpler.

91 Standard Sum-of-Products A Simplification Example: F ( A, B, C ) Writing the minterm expression: F = A B C + A B C + A B C + ABC + ABC Simplifying: F = A B C + A (B C + B C + B C + B C) F = A B C + A (B (C + C) + B (C + C)) F = A B C + A (B + B) F = A B C + A F = B C + A Simplified F contains 3 literals compared to 5 Boolean Algebra and Logic Gates 9 (SOP) (, 4, 5, 6, 7 )

92 Boolean Algebra and Logic Gates 92 AND/OR Two-Level Implementation The two implementations for F are shown below A B C A B C A B C A B C A B C F A B C It is quite apparent which is simpler! F

93 SOP and POS Observations The previous examples show that: Canonical Forms (Sum-of-minterms, Product-of- Maxterms), or other standard forms (SOP, POS) differ in complexity Boolean algebra can be used to manipulate equations into simpler forms Simpler equations lead to simpler implementations Questions: How can we attain a simplest expression? Is there only one minimum cost circuit? The next part will deal with these issues Boolean Algebra and Logic Gates 93

94 Terms of Use All (or portions) of this material 2008 by Pearson Education, Inc. Permission is given to incorporate this material or adaptations thereof into classroom presentations and handouts to instructors in courses adopting the latest edition of Logic and Computer Design Fundamentals as the course textbook. These materials or adaptations thereof are not to be sold or otherwise offered for consideration. This Terms of Use slide or page is to be included within the original materials or any adaptations thereof. Boolean Algebra and Logic Gates 94

95 Introduction to Karnaugh Maps If implemented correctly, they almost always produce a minimum solution. They are more straightforward that algebraic manipulations They generally produce SOPs, but POS can be generated from their complements if required.

96 Two-variable Karnaugh maps A B 0 0 B A A B AB A B A B AB A B 0 0 m 0 m 2 m m 3

97 Three-variable Karnaugh maps Product terms corresponding to groups of two.

98 Four-variable Karnaugh maps

99 Implicants An implicant of a function is a product term. From the point of view of the map, an implicant is a rectangle of, 2, 4, 8,... (any power of 2) s. That rectangle may not include any 0 s. Example: f = A B C D +A B CD+A BCD+AB CD+ABC D +ABC D+ABCD AB CD

100 Implicants The implicants of f are: Minterms ( implicant) A B C D A B CD A BCD ABC D ABC D ABCD AB CD Groups of 2 A CD BCD ACD B CD ABC ABD Groups of 4 CD

101 Prime and Essential Prime Implicants Prime Implicant: o an implicant that (from the point of view of the map) is not fully contained in any one other implicant. Essential Prime Implicant: o a prime implicant that includes at least one that is not included in any other prime implicant. CD AB

102 Minimum SOP Expressions From Karnaugh Maps

103 Minimum SOP Expressions From Karnaugh Maps. Find all essential prime implicants. Circle them on the map and mark the minterm(s) that make them essential with an asterisk (*). 2. Find enough other prime implicants to cover the function. Do this using two criteria: a. Choose a prime implicant that covers as many new s (that is, those not already covered by a chosen prime implicant). b. Avoid leaving isolated uncovered s. The main idea is. To Have all ones covered 2. To Have as few terms as possible 3. To have several rectangles with more s and few rectangles with less s

104 Example f = w x y z +w xy z + w xy z+ w xyz+ wx y z + w xyz+ wxy z + wxyz AB CD * * * 0 unnecessary * 0 f = y z +wyz+w xz

105 f = b + a' c In Groups

106 Prime implicant Don t Cares A rectangle of, 2, 4, 8,... s or X s not included in any one larger rectangle. From the point of view of finding prime implicants, X s (don t cares) are treated as s. Essential prime implicant A prime implicant that covers at least one not covered by any other prime implicant (as always). Don t cares (X s) do not make a prime implicant essential.

107 Example f = Σm(,7,0,,3) + Σd(5,8,5) CD 00 AB x 0 x x 0 Use don t cares to get as many minterms in each tem as possible F = BD + A C D + AB C

108 For the following problem, find the minimum SOP expression within the options given h(a,b,c) = Σm(0,,5) + d(3,4,6,7) In groups a) h = a'b' + c + a b) h = a + c + b c) h = c + b d) h = b e) h = c

109 Implementation of Two Functions

110 F = A B C +A BC +ABC +ABC; G = A B C+A BC+ABC +ABC AB C 0 AB C 0 Example F = A C +AB G = A C+AB

111 Example F = A B + ABC G = A B + BC F = A B + ABC G = A B + ABC

112 Example f = a b + bc g = ab + ac f = a b + abc g = ab + abc

113 Example F = A C + A C D + AB C G = AC + A C D + AB C

114 Try

115 Many electronic systems automatically invert gates Easier to fabricate with electronic components Basic gates used in integrated circuits (IC) digital logic families. NAND gate universal gate Could be used to construct any logic gate NAND, NOR

116 NAND gates. Alternate symbol for NAND. Symbols for NOR gate.

117 NAND Gate Implementation When we have a circuit consisting of AND and OR gates such that. the output of the circuit comes from an OR, 2. the inputs to all OR gates come either from a system input or from the output of an AND, and 3. the inputs to all AND gates come either from a system input or from the output of an OR. All gates are replaced by NAND gates, and any input coming directly into an OR is complemented.

118 Try: g = wx(y+z)+x y Example

119 NOR Gate Implementation When we have a circuit consisting of AND and OR gates such that. the output of the circuit comes from an AND, 2. the inputs to all OR gates come either from a system input or from the output of an AND, and 3. the inputs to all AND gates come either from a system input or from the output of an OR. All gates are replaced by NOR gates, and any input coming directly into an AND is complemented.

120 Example Try: g = (x+y )(x +y)(x +z)

121 XOR and XNOR A xor B is if a = or b is and 0 if both are or 0; Develop a truth table for XOR

122 Homework

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