Pune Vidyarthi Griha s COLLEGE OF ENGINEERING, NSIK - 4 Minimization techniques By Prof. nand N. Gharu ssistant Professor Computer Department
Combinational Logic Circuits Introduction Standard representation of canonical forms (SOP & POS), Maxterm and Minterm, Conversion between SOP and POS forms K-map reduction techniques upto 4 variables (SOP & POS form), Design of Half dder, Full dder, Half Subtractor & Full Subtractor using k-map Code Converter using K-map: Gray to Binary Binary to Gray Code Converter (upto 4 bit) IC 7447 as BCD to 7- Segment decoder driver IC 7483 as dder & Subtractor, 1 Digit BCD dder Block Schematic of LU IC 74181 IC 74381 8/29/2017 2
Standard Representation ny logical expression can be expressed in the following two forms: Sum of Product (SOP) Form Product of Sum (POS) Form 8/29/2017 3
SOP Form For Example, logical expression given is; Sum Y. B B. C C. Product 8/29/2017 4
POS Form For Example, logical expression given is; Product Y ( B).( B C).( C) Sum 8/29/2017 5
Standard or Canonical SOP & POS Forms We can say that a logic expression is said to be in the standard (or canonical) SOP or POS form if each product term (for SOP) and sum term (for POS) consists of all the literals in their complemented or uncomplemented form. 8/29/2017 6
Standard SOP Y C C C Each product term consists all the literals 8/29/2017 7
Standard POS Y ( B C).( B C).( B C) Each sum term consists all the literals 8/29/2017 8
Examples Sr. No. Expression Type 1 Non Standard SOP Y C C 2 Standard SOP Y 3 Standard POS Y ( B).( B).( B) 4 Non Standard POS Y ( B).( B C) 8/29/2017 9
Conversion of SOP form to Standard SOP Procedure: 1. Write down all the terms. 2. If one or more variables are missing in any product term, expand the term by multiplying it with the sum of each one of the missing variable and its complement. 3. Drop out the redundant terms 8/29/2017 10
Example 1 Convert given expression into its standard SOP form Y C BC Y C BC Missing literal is C Missing literal is B Missing literal is Y.( C C) C.( B B) BC.( ) Term formed by ORing of missing literal & its complement 8/29/2017 11
Example 1 Continue. Y.( C C) C.( B B) BC.( ) Y C C C C C C Y C C C C C C Y C C C C Standard SOP form Each product term consists all the literals 8/29/2017 12
Conversion of POS form to Standard POS Procedure: 1. Write down all the terms. 2. If one or more variables are missing in any sum term, expand the term by adding the products of each one of the missing variable and its complement. 3. Drop out the redundant terms 8/29/2017 13
Example 2 Convert given expression into its standard SOP form Y ( B).( C) ( B C) Y ( B).( C) ( B C) Missing literal is C Missing literal is B Missing literal is Y ( B CC).( C BB).( B C ) Term formed by NDing of missing literal & its complement 8/29/2017 14
Example 2 Continue. Y ( B CC).( C BB).( B C ) Y ( B C)( B C).( B C)( B C).( B C)( B C) Y ( B C)( B C)( B C)( B C) Y ( B C)( B C)( B C)( B C) Standard POS form Each sum term consists all the literals 8/29/2017 15
Concept of Minterm and Maxterm Minterm: Each individual term in the standard SOP form is called as Minterm. Maxterm: Each individual term in the standard POS form is called as Maxterm. 8/29/2017 16
The concept of minterm and max term allows us to introduce a very convenient shorthand notation to express logic functions 8/29/2017 17
Minterms & Maxterms for 3 variable/literal logic function Variables Minterms Maxterms B C mi Mi 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 C m0 C m1 C m2 C m3 C m4 C m5 C m6 C m7 B C M 0 B C M 1 B C M 2 B C M 3 B C M 4 B C M 5 B C M 6 B C M 7 8/29/2017 mit Nevase 18
Minterms and maxterms Each minterm is represented by m i where i=0,1,2,3,.,2 n-1 Each maxterm is represented by M i where i=0,1,2,3,.,2 n-1 If n number of variables forms the function, then number of minterms or maxterms will be 2 n i.e. for 3 variables function f(,b,c), the number of minterms or maxterms are 2 3 =8 8/29/2017 19
Minterms & Maxterms for 2 variable/literal logic function Variables Minterms Maxterms B mi Mi 0 0 m0 B M 0 0 1 1 0 1 1 m1 m2 B M 1 B M 2 B M m3 3 8/29/2017 20
Representation of Logical expression using minterm Y C C C C m 7 m 3 m 4 m 5 Y m7 m3 m4 m5 Logical Expression Corresponding minterms Y m(3,4,5,7) OR Y f (, B, C) m(3,4,5,7) where denotes sum of products 8/29/2017 21
Representation of Logical expression using maxterm Y ( B C).( B C).( B C) M 2 M 0 M 6 Logical Expression Corresponding maxterms Y M. M. M 2 0 6 Y M(0,2,6) OR Y f (, B, C) M (0,2,6) where denotes product of sum 8/29/2017 22
Conversion from SOP to POS & Vice versa The relationship between the expressions using minters and maxterms is complementary. We can exploit this complementary relationship to write the expressions in terms of maxterms if the expression in terms of minterms is known and vice versa 8/29/2017 23
Conversion from SOP to POS & Vice versa For example, if a SOP expression for 4 variable is given by, Y m(0,1, 3, 5, 6, 7,11,12,15) Then we can get the equivalent POS expression using the complementary relationship as follows, Y M(2, 4,8, 9,10,13,14) 8/29/2017 24
Examples 1. Convert the given expression into standard form Y BC C 2. Convert the given expression into standard form Y ( B).( C) 8/29/2017 25
Combinational Logic Circuits Introduction Standard representation of canonical forms (SOP & POS), Maxterm and Minterm, Conversion between SOP and POS forms K-map reduction techniques upto 4 variables (SOP & POS form), Design of Half dder, Full dder, Half Subtractor & Full Subtractor using k-map Code Converter using K-map: Gray to Binary, Binary to Gray Code Converter (upto 4 bit) IC 7447 as BCD to 7- Segment decoder driver IC 7483 as dder & Subtractor, 1 Digit BCD dder Block Schematic of LU IC 74181 IC 74381 8/29/2017 26
Karnaugh Map (K-map) In the algebraic method of simplification, we need to write lengthy equations, find the common terms, manipulate the expressions etc., so it is time consuming work. Thus K-map is another simplification technique to reduce the Boolean equation. 8/29/2017 27
Karnaugh Map (K-map) It overcomes all the disadvantages of algebraic simplification techniques. The information contained in a truth table or available in the SOP or POS form is represented on K-map. 8/29/2017 28
Karnaugh Map (K-map) K-map Structure - 2 Variable & B are variables or inputs 0 & 1 are values of & B 2 variable k-map consists of 4 boxes i.e. 2 2 =4 B 0 1 0 1 8/29/2017 29
Karnaugh Map (K-map) K-map Structure - 2 Variable Inside 4 boxes we have enter values of Y i.e. output B B B 0 1 0 1 B 0 1 0 1 m0 m1 m2 m3 K-map & its associated minterms 8/29/2017 30
Karnaugh Map (K-map) Relationship between Truth Table & K-map B Y 0 0 0 0 1 1 B B B 0 1 0 1 0 0 1 1 1 0 0 1 1 1 B B B 0 1 0 1 0 1 0 1 8/29/2017 31
C Karnaugh Map (K-map) K-map Structure - 3 Variable, B & C are variables or inputs 0 1 3 variable k-map consists of 8 boxes i.e. 2 3 =8 BC 0 1 00 01 BC 00 0 1 01 11 10 11 10 8/29/2017 32
C Karnaugh Map (K-map) 3 Variable K-map & its associated product terms 0 1 BC 00 0 00 01 11 10 C C C C C C C C C 01 11 10 C C C BC 0 1 00 01 11 10 C C C C C C C C 1 C C C C 8/29/2017 33
Karnaugh Map (K-map) 3 Variable K-map & its associated minterms C 0 1 BC 00 00 01 11 10 m0 m2 m1 m3 m6 m7 m4 m5 01 11 10 BC 0 1 00 01 11 m0 m4 m1 m5 m3 m7 0 m0 m1 m3 m2 10 m2 m6 1 m4 m5 m7 m6 8/29/2017 34
Karnaugh Map (K-map) K-map Structure - 4 Variable, B, C & D are variables or inputs 4 variable k-map consists of 16 boxes i.e. 2 4 =16 CD 00 01 11 10 CD 00 01 11 10 00 00 01 01 11 11 10 10 8/29/2017 35
Karnaugh Map (K-map) 4 Variable K-map and its associated product terms CD 00 01 11 10 CD 00 01 11 10 00 CD CD CD CD 00 CD CD CD CD 01 CD CD CD CD 01 CD CD CD CD 11 CD CD CD CD 11 CD CD CD CD 10 CD CD CD CD 10 CD CD CD CD 8/29/2017 36
Karnaugh Map (K-map) 4 Variable K-map and its associated minterms CD 00 01 11 10 CD 00 01 11 10 00 m 0 m 4 m 12 m 8 00 m 0 m 1 m 3 m 2 01 m 1 m 5 m 13 m 9 01 m 4 m 5 m 7 m 6 11 m 3 m 7 m 15 m 11 11 m 12 m 13 m 15 m 14 10 m 2 m 6 m 11 m 10 10 m 8 m 9 m 11 m 10 8/29/2017 37
Representation of Standard SOP form expression on K-map For example, SOP equation is given as Y C C C C C The given expression is in the standard SOP form. Each term represents a minterm. We have to enter 1 in the boxes corresponding to each minterm as below C C BC BC BC BC BC 00 01 11 10 0 1 1 1 0 0 1 0 1 1 C C C 8/29/2017 38
Simplification of K-map Once we plot the logic function or truth table on K-map, we have to use the grouping technique for simplifying the logic function. Grouping means the combining the terms in adjacent cells. The grouping of either 1 s or 0 s results in the simplification of boolean expression. 8/29/2017 39
Simplification of K-map If we group the adjacent 1 s then the result of simplification is SOP form If we group the adjacent 0 s then the result of simplification is POS form 8/29/2017 40
Grouping While grouping, we should group most number of 1 s. The grouping follows the binary rule i.e we can group 1,2,4,8,16,32,.. number of 1 s. We cannot group 3,5,7, number of 1 s Pair: group of two adjacent 1 s is called as Pair Quad: group of four adjacent 1 s is called as Quad Octet: group of eight adjacent 1 s is called as Octet 8/29/2017 41
Grouping of Two djacent 1 s : Pair pair eliminates 1 variable C C BC BC BC BC BC 00 01 11 10 0 1 0 0 1 1 0 0 0 0 Y C C Y ( C C) Y ( C C 1) 8/29/2017 42
Grouping of Two djacent 1 s : Pair BC BC BC BC BC 00 01 11 10 0 1 0 0 0 0 1 0 0 1 BC BC BC BC BC 00 01 11 10 0 1 0 1 1 1 0 0 1 0 BC BC BC BC BC 00 01 11 10 0 1 0 1 0 0 0 1 0 0 B B 0 B 1 0 1 1 1 1 0 8/29/2017 43
Grouping of Two djacent 1 s : Pair CD CD CD CD CD 00 01 11 10 00 0 1 0 0 01 11 10 0 0 0 0 0 0 0 0 0 1 0 0 8/29/2017 44
Possible Grouping of Four djacent 1 s : Quad Quad eliminates 2 variable CD CD CD CD CD 00 01 11 10 00 0 0 0 0 CD CD CD CD CD 00 01 11 10 00 0 1 0 0 01 0 0 0 0 01 0 1 0 0 11 0 0 0 0 11 0 1 0 0 10 1 1 1 1 10 0 1 0 0 8/29/2017 45
Possible Grouping of Four djacent 1 s : Quad Quad eliminates 2 variable CD CD CD CD CD 00 01 11 10 00 0 0 0 0 CD CD CD CD CD 00 01 11 10 00 0 1 1 0 01 1 1 0 0 01 0 0 0 0 11 1 1 0 0 11 0 0 0 0 10 0 0 0 0 10 0 1 1 0 8/29/2017 46
Possible Grouping of Four djacent 1 s : Quad Quad eliminates 2 variable CD CD CD CD CD 00 01 11 10 00 1 0 0 1 CD CD CD CD CD 00 01 11 10 00 0 0 0 0 01 0 0 0 0 01 1 0 0 1 11 0 0 0 0 11 1 0 0 1 10 1 0 0 1 10 0 0 0 0 8/29/2017 47
Possible Grouping of Four djacent 1 s : Quad Quad eliminates 2 variable CD CD CD CD CD 00 01 11 10 00 0 0 0 0 CD CD CD CD CD 00 01 11 10 00 0 0 0 0 01 0 1 1 1 01 0 1 1 0 11 0 1 1 1 11 0 1 1 0 10 0 0 0 0 10 0 1 1 0 8/29/2017 48
Possible Grouping of Eight djacent 1 s : Octet Octet eliminates 3 variable CD CD CD CD CD 00 01 11 10 00 0 0 0 0 CD CD CD CD CD 00 01 11 10 00 0 1 1 0 01 0 0 0 0 01 0 1 1 0 11 1 1 1 1 11 0 1 1 0 10 1 1 1 1 10 0 1 1 0 8/29/2017 49
Possible Grouping of Eight djacent 1 s : Octet Octet eliminates 3 variable CD CD CD CD CD 00 01 11 10 00 1 1 1 1 CD CD CD CD CD 00 01 11 10 00 1 0 0 1 01 0 0 0 0 01 1 0 0 1 11 0 0 0 0 11 1 0 0 1 10 1 1 1 1 10 1 0 0 1 8/29/2017 50
Rules for K-map simplification 1. Groups may not include any cell containing a zero. B B B 0 1 0 1 0 1 B B B 0 1 0 1 0 1 1 Not ccepted ccepted 8/29/2017 51
Rules for K-map simplification 2. Groups may be horizontal or vertical, but may not be diagonal B B B 0 1 0 1 0 1 1 0 B B B 0 1 0 1 0 1 1 1 Not ccepted ccepted 8/29/2017 52
Rules for K-map simplification 3. Groups must contain 1,2,4,8 or in general 2 n cells BC BC BC BC BC BC BC BC BC 00 0 1 01 11 10 0 1 1 1 0 0 0 0 BC 00 0 1 01 11 10 0 1 1 1 0 0 0 0 B B B 0 1 0 1 1 1 0 1 B B B 0 1 0 1 1 1 0 1 Not ccepted ccepted 8/29/2017 53
Rules for K-map simplification 4. Each group should be as large as possible BC BC 00 BC 01 BC 11 BC 10 0 1 1 1 1 1 0 0 1 1 BC BC 00 BC 01 BC 11 BC 10 0 1 1 1 1 1 0 0 1 1 Not ccepted ccepted 8/29/2017 54
Rules for K-map simplification 5. Each cell containing a one must be in at least one group BC BC BC BC BC 00 01 11 10 0 0 0 0 1 1 0 0 1 0 8/29/2017 55
Rules for K-map simplification 6. Groups may be overlap BC BC BC BC BC 00 01 11 10 0 1 1 1 1 1 0 0 1 1 8/29/2017 56
Rules for K-map simplification 7. Groups may wrap around the table. The leftmost cell in a row may be grouped with rightmost cell and the top cell in a column may be grouped with bottom cell CD CD CD CD CD 00 01 11 10 00 01 11 1 1 1 1 0 0 0 0 0 0 0 0 BC BC BC BC BC 00 01 11 10 0 1 1 0 0 1 1 0 0 1 10 1 1 1 1 8/29/2017 57
Rules for K-map simplification 8. There should be as few groups as possible, as long as this does not contradict any of the previous rules. BC BC 00 BC 01 BC 11 BC 10 0 1 1 1 1 1 0 0 1 1 BC BC 00 BC 01 BC 11 BC 10 0 1 1 1 1 1 0 0 1 1 Not ccepted ccepted 8/29/2017 58
Rules for K-map simplification 9. pair eliminates one variable. 10. Quad eliminates two variables. 11. octet eliminates three variables 8/29/2017 59
Example 1 For the given K-map write simplified Boolean expression C C C 00 01 11 10 0 0 1 1 1 1 0 0 1 0 8/29/2017 60
Example 1 continue.. C C C 00 01 11 10 0 0 1 1 1 1 0 0 1 0 C BC Simplified Boolean expression Y BC C 8/29/2017 61
Example 2 For the given K-map write simplified Boolean expression C C C 00 01 11 10 0 1 1 0 1 1 1 0 0 1 8/29/2017 62
Example 2 continue.. C C C 00 01 11 10 0 1 1 0 1 1 1 0 0 1 C Simplified Boolean expression Y B C B 8/29/2017 63
Example 3 logical expression in the standard SOP form is as follows; Y C C C C Minimize it with using the K-map technique 8/29/2017 64
Example 3 continue Y C C C C BC BC 00 BC 01 BC 11 BC 10 0 1 0 1 1 1 0 1 0 0 C Simplified Boolean expression C Y C C 8/29/2017 65
Example 4 logical expression representing a logic circuit is; Y m(0,1, 2, 5,13,15) Draw the K-map and find the minimized logical expression 8/29/2017 66
Example 4 continue.. Y m(0,1, 2, 5,13,15) CD CD CD CD CD 00 01 11 10 00 01 11 10 1 0 1 1 0 3 1 2 0 4 1 5 0 7 0 6 12 13 15 14 Simplified Boolean expression 0 1 1 0 0 8 0 9 0 11 0 10 D Y D CD D CD D 8/29/2017 67
Example 5 Minimize the following Boolean expression using K-map ; f (, B, C, D) m(1,3,5,9,11,13) 8/29/2017 68
Example 5 continue.. f (, B, C, D) m(1,3,5,9,11,13) CD CD CD CD CD 00 01 11 10 00 01 11 10 0 0 1 1 1 3 0 2 0 4 1 5 0 7 0 6 12 13 15 14 Simplified Boolean expression 0 1 0 0 0 8 1 9 1 11 0 10 f BD CD BD CD f D( B C) 8/29/2017 69
Example 6 Minimize the following Boolean expression using K-map ; f (, B, C, D) m(4,5,8,9,11,12,13,15) 8/29/2017 70
Example 6 continue.. f (, B, C, D) m(4,5,8,9,11,12,13,15) CD CD CD CD CD 00 01 11 10 00 01 11 10 0 0 0 1 0 3 0 2 1 4 1 5 0 7 0 6 12 13 15 14 Simplified Boolean expression 1 1 1 0 1 8 1 9 1 11 0 10 BC f BC C D C D 8/29/2017 71
Example 7 Minimize the following Boolean expression using K-map ; f 2(, B, C, D) m(0,1,2,3,11,12,14,15) 8/29/2017 72
Example 7 continue.. f 2(, B, C, D) m(0,1,2,3,11,12,14,15) CD CD CD CD CD 00 01 11 10 00 01 11 10 1 0 1 1 1 3 1 2 0 4 0 5 0 7 0 6 12 13 15 14 Simplified Boolean expression 1 0 1 1 0 8 0 9 1 11 0 10 D CD f 2 D CD 8/29/2017 73
Example 8 Solve the following expression with K-maps; 1. 2. f 1(, B, C) m(0,1,3,4,5) f 2(, B, C) m(0,1,2,3,6,7) 8/29/2017 74
Example 8 continue f 1(, B, C) m(0,1,3,4,5) f 2(, B, C) m(0,1,2,3,6,7) C BC BC 00 BC 01 BC 11 BC 10 1 0 1 1 1 3 0 2 0 4 1 1 1 5 0 7 0 6 B Simplified Boolean expression f 1 C B BC BC 00 BC 01 BC 11 BC 10 1 0 1 1 1 3 1 2 0 4 0 1 0 5 1 7 1 6 Simplified Boolean expression f 2 B B 8/29/2017 75
Example 9 Simplify ; f (, B, C, D) m(0,1,4,5,7,8,9,12,13,15) 8/29/2017 76
Example 9 continue.. CD CD CD CD CD 00 01 11 10 00 01 11 10 f (, B, C, D) m(0,1,4,5,7,8,9,12,13,15) 1 0 1 1 0 3 0 2 1 4 1 5 1 7 0 6 12 13 15 14 Simplified Boolean expression f C BD 1 1 1 0 1 8 1 9 0 11 0 10 C BD 8/29/2017 77
Example 10 Solve the following expression with K-maps; 1. 2. f 1(, B, C, D) m(0,1,3,4,5,7) f 2(, B, C) m(0,1,3,4,5,7) 8/29/2017 78
Example 10 continue f 1(, B, C, D) m(0,1,3,4,5,7) f 2(, B, C ) m (0,1,3,4,5,7) CD CD CD CD CD 00 01 11 10 00 01 11 10 C 1 0 1 1 0 3 0 2 1 4 1 5 1 7 0 6 0 12 0 13 0 15 0 14 0 8 0 9 0 11 0 10 Simplified Boolean expression D f C D BC BC 00 BC 01 BC 11 BC 10 1 0 1 1 1 3 0 2 0 4 1 1 1 5 1 7 0 6 Simplified Boolean expression 1 8/29/2017 79 B f 2 B C C
K-map and don t care conditions For SOP form we enter 1 s corresponding to the combinations of input variables which produce a high output and we enter 0 s in the remaining cells of the K-map. For POS form we enter 0 s corresponding to the combinations of input variables which produce a high output and we enter 1 s in the remaining cells of the K-map. 8/29/2017 80
K-map and don t care conditions But it is not always true that the cells not containing 1 s (in SOP) will contain 0 s, because some combinations of input variable do not occur. lso for some functions the outputs corresponding to certain combinations of input variables do not matter. 8/29/2017 81
K-map and don t care conditions In such situations we have a freedom to assume a 0 or 1 as output for each of these combinations. These conditions are known as the Don t Care Conditions and in the K-map it is represented as X, in the corresponding cell. The don t care conditions may be assumed to be 0 or 1 as per the need for simplification 8/29/2017 82
K-map and don t care conditions - Example Simplify ; f (, B, C, D) m(1,3,7,11,15) d(0,2,5) 8/29/2017 83
K-map and don t care conditions - Example f (, B, C, D) m(1,3,7,11,15) d(0,2,5) CD CD CD CD CD 00 01 11 10 00 01 11 10 X 0 1 1 1 3 X 2 0 4 X 5 1 7 0 6 12 13 15 14 Simplified Boolean expression 0 0 1 0 0 8 0 9 1 11 0 10 f CD D D CD 8/29/2017 mit Nevase 84