1. 241-208 CH41 Chapter 4 Boolean Algebra and Logic Simplification By Taweesak Reungpeerakul

2. 241-208 CH42 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

3. 241-208 CH43 4.1 Boolean Operations & Expressions 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. 241-208 CH44 4.2 Laws&Rules of Boolean Algebra 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)

5. 241-208 CH45 4.2 Laws&Rules of Boolean Algebra (cont.) 1) A+0=A10) A·A=A 2) A+1=111) A·A=0 3) A·0=012) 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 Rules of Boolean Algebra

6. 241-208 CH46 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

7. 241-208 CH47 Examples of DeMorgan’s Theorems Ex#1: (AB+C)(BC) = (AB+C) +(BC) = (AB)C +(B+C) = (A+B)C + B+C Question: (A+B)C D Ans: (A ·B)+C+D Ex# 2: AB + CDE = (AB) · (CDE) = (A+B) · (CD+E) Question: A+B+C+ DE Ans: A B C+D+E

8. 241-208 CH48 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

9. 241-208 CH49 4.4 Boolean Analysis of Logic Circuits (cont.) Truth Table A B C D (AB+C)D 0 0 0 00 0 0 0 10 0 0 1 00 0 0 1 11 0 1 0 00 0 1 0 10 0 1 1 00 0 1 1 11 1 0 0 00 1 0 0 10 1 0 1 00 1 0 1 11 1 1 0 00 1 1 0 11 1 1 1 00 1 1 1 11 A D C B AB AB+C (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

10. 241-208 CH410 4.5 Simplification using Boolean Algebra EX#1: AB+A(B+C)+B(B+C) = AB+AB+AC+BB+BC = AB + AC + B + BC = AB+AC+B(1+C) 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

11. 241-208 CH411 4.5 Simplification using Boolean Algebra (cont.) = AB+AC+B = AB+B+AC = B+BA+AC = B(1+A)+AC = B+AC Answer Use commutative law A+B = B+A Use distributive law A(B+C) = AB+AC Use commutative law A+B = B+A, AB = BA Use rule #2 (1+A) = 1

12. 241-208 CH412 4.5 Simplification using Boolean Algebra (cont.) EX#2: A B C+A B C+A B C+A B C+A B C = B C+A B C+A B C+A B C = B C+ B C+A B C = BC+B(C+AC) = BC+B(C+A) = BC+B C+AB Answer 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 !!!!

13. 241-208 CH413 4.6 Standard Forms of Boolean Expressions 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

14. 241-208 CH414 4.6 Standard Forms of Boolean Expressions (cont.) Ex# 2: (A + B) + C = (A+B)C DeMorgan’s = (A+B)C Distributive = AC+BC Domain is the set of literals (or variables) contained in the Boolean expression !!

15. 241-208 CH415 4.6 Standard Forms of Boolean Expressions (cont.) 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.

16. 241-208 CH416 4.6 Standard Forms of Boolean Expressions (cont.) 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

17. 241-208 CH417 Standard Forms (cont.) 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 Ans: (A+B+C) (A+B+C) (A+B+C)(A+B+C)

18. 241-208 CH418 Std. SOP to std. POS Example: ABC+ABC+ABC+ABC+ABC 101 011 100 001 000 3 variables 2 3 = 8 possible combinations Remained terms: 111, 110, 010 Std. POS = (A+B+C)(A+B+C)(A+B+C)

19. 241-208 CH419 4.7 Boolean Expressions and Truth Tables EX: ABC+ABC+ABC+ABC 000 010 101 110  out=1 ABCOut 0001 0010 0101 0110 1000 1011 1101 1110 What u should know: Be able to convert SOP and POS expressions to truth tables and vice versa !! 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 Convert SOP to truth table

20. 241-208 CH420 4.7 Boolean Expressions and Truth Tables (cont.) EX: (A+B+C)(A+B+C)(A+B+C) 100 010 011  out=0 ABCOut 0001 0011 0100 0110 1000 1011 1101 1111 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 Convert POS to truth table

21. 241-208 CH421 4.7 Boolean Expressions and Truth Tables (cont.) Convert truth table to SSOP Convert truth table to SPOS 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 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

22. 241-208 CH422 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 2 n, n is number of variables Each cell differs from an adjacent cell by only one variable 3 variables so 8 cells The numbers are entered in gray code, to force adjacent cells to be different by only one variable. Gray code

23. 241-208 CH423 4.8 The Karnaugh Map (cont.) AB C How to read !! ABC Full representation for 3 variables. (in fact it is n-Dimension truth table) Question : What about the map for 4 variables ?

24. 241-208 CH424 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 !! 11 1 1 ABC+ABC+ABC+ABC

25. 241-208 CH425 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 11 1 Observe that A is related to all cells related to the binary “1” of A

26. 241-208 CH426 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. 4.9 Karnaugh Map SOP Minimization (cont.) How to minimize SOP expression ?

27. 241-208 CH427 4.9 Karnaugh Map SOP Minimization (cont.) B changes across this boundary C changes across this boundary X = AC +AB 1. Group the 1’s into two overlapping groups as indicated. 2.Read each group by eliminating any variable that changes across a boundary. 3.The vertical group is read AC. 4.The horizontal group is read AB.

28. 241-208 CH428 4.9 Karnaugh Map SOP Minimization (cont.) X 1. Group the 1’s into two separate groups as indicated. 2.Read each group by eliminating any variable that changes across a boundary. 3.The upper (yellow) group is read as AD. 4.The lower (green) group is read as AD. X = AD +AD B changes C changes B changes C changes across outer boundary

29. 241-208 CH429 A BA B C Some examples of grouping 1 1 1 1 1 1 1 11 1 1 1 1 1 11 1 1 1 11 BC ABC ACBC AB ABC D 4.9 Karnaugh Map SOP Minimization (cont.)

30. 241-208 CH430 1 4.9 Karnaugh Map SOP Minimization (cont.) Ex1: Map and minimize the following std. SOP expression on a Karnaugh map: A B C+ABC+ABC+A B C 000 001 110 100 1 1 1 1 1 1 1 Answer: A B+AC AB AC

31. 241-208 CH431 Ex2: Map and minimize the following SOP expression on a Karnaugh map: A B +ABC+A B C 110 111 010 011 11 11 11 11 Answer: B B 4.9 Karnaugh Map SOP Minimization (cont.)

32. 241-208 CH432 ABCOut 000 1 001 0 010 0 011 0 100 1 101 0 110 1 111 x 1 1 1 x Out = AB+BC Mapping Directly from a Truth Table 4.9 Karnaugh Map SOP Minimization (cont.)

33. 241-208 CH433 A+B+C 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 0 0 0 0 0 0 0 0 Answer: (A+B)(A+C)(A+B+C) A+BA+C

34. 241-208 CH434 Ex2: Map and minimize the following POS expression on a Karnaugh map: (A+B)(A+B+C)(A+B+C) 000 001 010 011 0000 Answer: A 0000 A 4.10 Karnaugh Map POS Minimization (cont.)

35. 241-208 CH435 00 0 0 00 (B+C+D)(A+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D) 0000 1000 0010 01101011 1001 1010 0 (A+B) (A+C+D) Answer: (B+D)(A+B)(A+C+D) (B+D) 4.10 Karnaugh Map POS Minimization (cont.)

36. 241-208 CH436 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 01101011 1001 1010 0 0 00 0 00 (A+B) (A+C+D) Min POS: (B+D)(A+B)(A+C+D) (B+D) Min SOP: AB+BC+AD 0 0 0 0 0 0 0 1 1 111 11 1 1 AD BC AB

37. 241-208 CH437 7-segment decoding Logic DigitD C B A a b c d e f g 00 0 0 0 1 1 1 1 1 1 0 10 0 0 1 0 1 1 0 0 0 0 20 0 1 0 1 1 0 1 1 0 1 30 0 1 1 1 1 1 1 0 0 1 40 1 0 0 0 1 1 0 0 1 1 50 1 0 1 1 0 1 1 0 1 1 60 1 1 0 1 0 1 1 1 1 1 70 1 1 1 1 1 1 0 0 0 0 81 0 0 0 1 1 1 1 1 1 1 91 0 0 1 1 1 1 1 0 1 1 101 0 1 0 x x x x x x x 111 0 1 1 x x x x x x x 121 1 0 0 x x x x x x x 131 1 0 1 x x x x x x x 141 1 1 0 x x x x x x x 151 1 1 1 x x x x x x x

38. 241-208 CH438 Karnaugh Map Minimization of the Segment Logic SOP for segment a: DC BA+DCBA+DCBA+ DCBA+DCBA+DCBA+DC BA+DCBA 1 1 11 1 x x D 11 1 xxxx B CA Minimum SOP expression: D+B+CA+CA