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Chapter 4

Chapter 4. Boolean Algebra and Logic Simplification By Taweesak Reungpeerakul. Contents. Boolean Operations & Expressions Rules of Boolean Algebra DeMorgan’s Theorems Simplification Using Boolean Algebra Standard Forms of Boolean Algebra Karnaugh Map Five-variable Karnaugh Map.

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Chapter 4

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  1. Chapter 4 Boolean Algebra and Logic Simplification By Taweesak Reungpeerakul 241-208 CH4

  2. Contents • Boolean Operations & Expressions • Rules of Boolean Algebra • DeMorgan’s Theorems • Simplification Using Boolean Algebra • Standard Forms of Boolean Algebra • Karnaugh Map • Five-variable Karnaugh Map 241-208 CH4

  3. Boolean Addition is equivalent to the OR operation. 0+0 = 0 0+1 = 1 1+0 = 1 1+1 = 1 Boolean multiplication is equivalent to the AND operation. 0·0 = 0 0·1 = 0 1·0 = 0 1·1 = 1 4.1 Boolean Operations & Expressions 241-208 CH4

  4. Laws (CAD) Commutative, Associative, and Distributive A+B = B+A (C for addition) AB = BA (C for multiplication) A+ (B+C) = (A+B)+C (A for addition) A(BC) = (AB)C (A for multiplication) A(B+C) = AB+AC (distributive) 4.2 Laws&Rules of Boolean Algebra 241-208 CH4

  5. 1) A+0=A 10) A·A=A 2) A+1=1 11) A·A=0 3) A·0=0 12) A=A 4) A·1=1 5) A+A=A 6) A+A=1 7) A+AB=A 8) A+AB=A+B 9) (A+B)(A+C)=A+BC 4.2 Laws&Rules of Boolean Algebra(cont.) Rules of Boolean Algebra 241-208 CH4

  6. 4.3 DeMorgan’s Theorems • The complement of a product of variables is equal to the sum of the complements of the variables. XY = X + Y • The complement of a sum of variables is equal to the product of complements of the variables. X + Y = X·Y 241-208 CH4

  7. Ex#1: (AB+C)(BC) = (AB+C) +(BC) = (AB)C +(B+C) = (A+B)C + B+C Question: (A+B)CD Ans: (A ·B)+C+D Ex# 2: AB + CDE = (AB) · (CDE) = (A+B) · (CD+E) = (A+B) · (CD+E) Question: A+B+C+DE Ans: ABC+D+E Examples of DeMorgan’s Theorems 241-208 CH4

  8. 4.4 Boolean Analysis of Logic Circuits • You should be able to: • Determine the Boolean expression for a • combination logic gates. • Evaluate the logic operation of a circuit from the Boolean expression • Construct a truth table 241-208 CH4

  9. 4.4 Boolean Analysis of Logic Circuits(cont.) Truth Table A B C D (AB+C)D 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 A AB B AB+C C D (AB+C)D Cause 1 when D and (AB+C) = 1 If AB = 0, C = 1 If AB = 1, C = 0 or 1 AB = 1 when both A = B = 1 241-208 CH4

  10. EX#1: AB+A(B+C)+B(B+C) = AB+AB+AC+BB+BC = AB + AC + B + BC = AB+AC+B(1+C) 4.5 Simplification using Boolean Algebra Use distributive law A(B+C) = AB+AC Use rule #5, A+A = A then A+A = A Use distributive law A(B+C) = AB+AC Use rule #2 (1+A) = 1 241-208 CH4

  11. = AB+AC+B = AB+B+AC = B+BA+AC = B(1+A)+AC = B+AC Answer 4.5 Simplification using Boolean Algebra (cont.) Use commutative law A+B = B+A Use commutative law A+B = B+A, AB = BA Use distributive law A(B+C) = AB+AC Use rule #2 (1+A) = 1 241-208 CH4

  12. EX#2: ABC+ABC+ABC+ABC+ABC = BC+ABC+ABC+ABC = BC+BC+ABC = BC+B(C+AC) = BC+B(C+A) = BC+BC+AB Answer 4.5 Simplification using Boolean Algebra (cont.) Can u follow up this example by yourself ? Use rule #11 A+AB = A+B How about trying more questions, for example : Some questions in pp. 179 !!!! 241-208 CH4

  13. Sum-of-Products (SOP): 2 or more product terms are summed by Boolean addition such as AB+ABC+AC Watch out ! each bar if any must denote on only a single literal (variable) (in brief watch out the NAND), for example AB+ABC+AC is not SOP Ex# 1: convert (A+B)(C+D) into SOP form apply distributive law, hence = AC+AD+BC+BD 4.6 Standard Forms of Boolean Expressions 241-208 CH4

  14. Ex# 2: (A + B) + C = (A+B)C DeMorgan’s = (A+B)C Distributive = AC+BC 4.6 Standard Forms of Boolean Expressions (cont.) Domain is the set of literals (or variables) contained in the Boolean expression !! 241-208 CH4

  15. Standard SOP Form: All variables in the domain appear in each product term such as ABC+ABC+ABC Convert SOP to SSOP Step 1: Consider domain of SOP Step 2: Multiply each nonstandard term by (L+L) Step3 : Repeat step 2 until no nonstandard term left. 4.6 Standard Forms of Boolean Expressions (cont.) 241-208 CH4

  16. Ex# 1: AB+ABC  standard SOP = AB(C+C)+ABC = ABC+ABC+ABC Ex# 2: B+ABC = B(A+A)+ABC = AB+AB+ABC = AB(C+C)+AB(C+C)+ABC = ABC+ABC+ABC+ABC+ABC 4.6 Standard Forms of Boolean Expressions (cont.) 241-208 CH4

  17. Product-of-Sum (POS): 2 or more sum terms are multiplied such as (A+B)(A+B+C) Watch out the NOR term !! Standard POS: all variables in the domain appear in each sum term such as (A+B+C)(A+B+C) Ex# 1: (A+C)(A+B+C)  standard POS = (A+C+BB)(A+B+C) =(A+B+C) (A+B+C) (A+B+C) Question: (A+C)(A+B)  std. POS Standard Forms (cont.) Ans: (A+B+C) (A+B+C) (A+B+C)(A+B+C) 241-208 CH4

  18. Std. SOP to std. POS Example: ABC+ABC+ABC+ABC+ABC 101 011 100 001 000 3 variables 23 = 8 possible combinations Remained terms: 111, 110, 010 Std. POS = (A+B+C)(A+B+C)(A+B+C) 241-208 CH4

  19. EX: ABC+ABC+ABC+ABC 000 010 101 110 out=1 A B C Out 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 4.7 Boolean Expressions and Truth Tables What u should know: Be able to convert SOP and POS expressions to truth tables and vice versa !! Convert SOP to truth table FACT SSOP is equal to 1 if at least one of the product term is 1. Step I : construct truth table for all possible inputs. Step II: convert SOP to SSOP Step III: Place “1” in the output column that makes the SSOP expression a “1” Step IV: Place “0” for all the remaining apart from Step III 241-208 CH4

  20. EX: (A+B+C)(A+B+C)(A+B+C) 100 010 011 out=0 A B C Out 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 4.7 Boolean Expressions and Truth Tables (cont.) Convert POS to truth table FACT SPOS is equal to 0 if at least one of the sum term is 0. Step I : construct truth table for all possible inputs. Step II: convert POS to SPOS Step III: Place “0” in the output column that makes the SPOS expression a “0” Step IV: Place “1” for all the remaining apart from Step III 241-208 CH4

  21. 4.7 Boolean Expressions and Truth Tables (cont.) Convert truth table to SSOP Step 1: Consider only output “1” Step 2: Convert each binary value to the corresponding product term Step 3: Repeat step 1&2 to get other product terms Step 4: Write all product terms in a summation expression Convert truth table to SPOS Step 1: Consider only output “0” Step 2: Convert each binary value to the corresponding sum term (1-> complement literal and 0-> for literal) Step 3: Repeat step 1&2 to get other sum terms Step 4: Write all product terms in a product expression Check this out: Example 4-20, pp. 187 241-208 CH4

  22. Gray code 4.8 The Karnaugh Map • The Karnaugh map is an array of cells in which each cell represents a binary value of the input variables. • Can facilitate to produce the simplest SOP or POS expression • The number of cells is 2n, n is number of variables 3 variables so 8 cells The numbers are entered in gray code, to force adjacent cells to be different by only one variable. Each cell differs from an adjacent cell by only one variable 241-208 CH4

  23. 4.8 The Karnaugh Map (cont.) C C AB AB AB AB ABC ABC ABC ABC ABC ABC ABC ABC Full representation for 3 variables. (in fact it is n-Dimension truth table) How to read !! Question : What about the map for 4 variables ? 241-208 CH4

  24. 4.9 Karnaugh Map SOP Minimization Aim: You should be able to utilise K-Map to simplify Boolean expression to their minimum form. Convert SSOP to K-Map What we know is each product term in SSOP relates to “1” in corresponding truth table, but K-Map is, in deed, a form of n-dims truth table. Hence the way to convert SSOP to K-map is similar to the way to convert SSOP to truth table !! 1 1 ABC+ABC+ABC+ABC 1 1 241-208 CH4

  25. 4.9 Karnaugh Map SOP Minimization (cont.) Convert non-standard SOP to K-Map Assume that the domain of Boolean is {A,B,C} and the expression we consider is A convert A (non-SOP) to SSOP as follows:- A = A(B+B)(C+C) = (AB+AB)(C+C) = ABC+ABC+ABC+ABC 1 1 Observe that A is related to all cells related to the binary “1” of A 1 1 241-208 CH4

  26. 4.9 Karnaugh Map SOP Minimization (cont.) How to minimize SOP expression ? • Grouping 1s - Each group must contain 1,2,4,8,or 16 cells - Each cell in a group must be adjacent to one or more cells in that same group, but all cells in the group do not have to be adjacent to each other. - Always include the largest possible number of 1s in a group - Each 1 on the map must be included in at least one group. The 1s already in a group can be included in another group as long as the overlapping groups include non-common 1s. 241-208 CH4

  27. B changes across this boundary C changes across this boundary • The vertical group is read AC. • The horizontal group is read AB. X = AC +AB 4.9 Karnaugh Map SOP Minimization (cont.) • 1.Group the 1’s into two overlapping groups as indicated. • Read each group by eliminating any variable that changes across a boundary. 241-208 CH4

  28. C changes across outer boundary B changes • The upper (yellow) group is read as AD. B changes C changes X = AD +AD 4.9 Karnaugh Map SOP Minimization (cont.) • 1.Group the 1’s into two separate groups as indicated. • Read each group by eliminating any variable that changes across a boundary. • The lower (green) group is read as AD. X 241-208 CH4

  29. 4.9 Karnaugh Map SOP Minimization (cont.) Some examples of grouping ABC AC BC D BC ABC 1 1 1 1 1 1 1 1 1 AB C AB 1 1 1 1 1 1 1 1 1 1 1 1 241-208 CH4

  30. 4.9 Karnaugh Map SOP Minimization (cont.) Ex1: Map and minimize the following std. SOP expression on a Karnaugh map: ABC+ABC+ABC+ABC 000 001 110 100 AB 1 1 1 1 1 1 1 1 AC Answer: AB+AC 241-208 CH4

  31. 4.9 Karnaugh Map SOP Minimization (cont.) Ex2: Map and minimize the following SOP expression on a Karnaugh map: AB+ABC+ABC 110 111 010 011 B 1 1 1 1 1 1 1 1 Answer: B 241-208 CH4

  32. 4.9 Karnaugh Map SOP Minimization (cont.) Mapping Directly from a Truth Table A B C Out 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 x 1 1 x 1 Out = AB+BC 241-208 CH4

  33. 4.10 Karnaugh Map POS Minimization Ex1: Map and minimize the following std. POS expression on a Karnaugh map: (A+B+C)(A+B+C)(A+B+C)(A+B+C) 000 001 111 110 A+B A+C 0 0 0 0 0 0 0 0 A+B+C Answer: (A+B)(A+C)(A+B+C) 241-208 CH4

  34. 4.10 Karnaugh Map POS Minimization (cont.) Ex2: Map and minimize the following POS expression on a Karnaugh map: (A+B)(A+B+C)(A+B+C) 000 001 010 011 A 0 0 0 0 0 0 0 0 Answer: A 241-208 CH4

  35. 4.10 Karnaugh Map POS Minimization (cont.) (B+C+D)(A+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D) 0000 1000 0010 0110 1011 1001 1010 (B+D) (A+C+D) 0 0 0 (A+B) 0 0 0 0 Answer: (B+D)(A+B)(A+C+D) 241-208 CH4

  36. Converting Between POS and SOP Using Karnaugh Map (B+C+D)(A+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D) 0000 1000 0010 0110 1011 1001 1010 (B+D) AD BC (A+C+D) 0 0 0 1 1 0 0 0 1 1 1 AB (A+B) 1 1 1 1 0 0 0 0 0 0 0 0 Min POS: (B+D)(A+B)(A+C+D) Min SOP: AB+BC+AD 241-208 CH4

  37. 7-segment decoding Logic Digit D C B A a b c d e f g 0 0 0 0 0 1 1 1 1 1 1 0 1 0 0 0 1 0 1 1 0 0 0 0 2 0 0 1 0 1 1 0 1 1 0 1 3 0 0 1 1 1 1 1 1 0 0 1 4 0 1 0 0 0 1 1 0 0 1 1 5 0 1 0 1 1 0 1 1 0 1 1 6 0 1 1 0 1 0 1 1 1 1 1 7 0 1 1 1 1 1 1 0 0 0 0 8 1 0 0 0 1 1 1 1 1 1 1 9 1 0 0 1 1 1 1 1 0 1 1 10 1 0 1 0 x x x x x x x 11 1 0 1 1 x x x x x x x 12 1 1 0 0 x x x x x x x 13 1 1 0 1 x x x x x x x 14 1 1 1 0 x x x x x x x 15 1 1 1 1 x x x x x x x 241-208 CH4

  38. Karnaugh Map Minimization of the Segment Logic SOP for segment a: DC BA+DCBA+DCBA+ DCBA+DCBA+DCBA+DC BA+DCBA CA 1 1 1 CA B 1 1 1 D x x x x 1 x 1 x Minimum SOP expression: D+B+CA+CA 241-208 CH4

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