CH41 Chapter 4 Boolean Algebra and Logic Simplification By Taweesak Reungpeerakul

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

CH Boolean Operations & Expressions Boolean Addition is equivalent to the OR operation. 0+0 = = = = 1 Boolean multiplication is equivalent to the AND operation. 0·0 = 0 0·1 = 0 1·0 = 0 1·1 = 1

CH 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)

CH 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

CH 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

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

CH 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

CH Boolean Analysis of Logic Circuits (cont.) Truth Table A B C D (AB+C)D 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

CH 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

CH 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

CH 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 !!!!

CH 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

CH 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 !!

CH 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.

CH 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

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)

CH418 Std. SOP to std. POS Example: ABC+ABC+ABC+ABC+ABC variables 2 3 = 8 possible combinations Remained terms: 111, 110, 010 Std. POS = (A+B+C)(A+B+C)(A+B+C)

CH Boolean Expressions and Truth Tables EX: ABC+ABC+ABC+ABC  out=1 ABCOut 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

CH Boolean Expressions and Truth Tables (cont.) EX: (A+B+C)(A+B+C)(A+B+C)  out=0 ABCOut 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

CH 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

CH 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

CH 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 ?

CH 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 !! ABC+ABC+ABC+ABC

CH 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 Observe that A is related to all cells related to the binary “1” of A

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 ?

CH 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.

CH 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

CH429 A BA B C Some examples of grouping BC ABC ACBC AB ABC D 4.9 Karnaugh Map SOP Minimization (cont.)

CH 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 Answer: A B+AC AB AC

CH431 Ex2: Map and minimize the following SOP expression on a Karnaugh map: A B +ABC+A B C Answer: B B 4.9 Karnaugh Map SOP Minimization (cont.)

CH432 ABCOut x x Out = AB+BC Mapping Directly from a Truth Table 4.9 Karnaugh Map SOP Minimization (cont.)

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) Answer: (A+B)(A+C)(A+B+C) A+BA+C

CH434 Ex2: Map and minimize the following POS expression on a Karnaugh map: (A+B)(A+B+C)(A+B+C) Answer: A 0000 A 4.10 Karnaugh Map POS Minimization (cont.)

CH (B+C+D)(A+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D) (A+B) (A+C+D) Answer: (B+D)(A+B)(A+C+D) (B+D) 4.10 Karnaugh Map POS Minimization (cont.)

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) (A+B) (A+C+D) Min POS: (B+D)(A+B)(A+C+D) (B+D) Min SOP: AB+BC+AD AD BC AB

CH437 7-segment decoding Logic DigitD C B A a b c d e f g x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

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