CHAPTER 1 INTRODUCTION TO DIGITAL LOGIC EKT 124 / 3 ELEKTRONIK DIGIT 1 CHAPTER 1 INTRODUCTION TO DIGITAL LOGIC

De Morgan’s Theorem

De Morgan’s Theorems 16) (X+Y) = X . Y 17) (X.Y) = X + Y Two most important theorems of Boolean Algebra were contributed by De Morgan. Extremely useful in simplifying expression in which product or sum of variables is inverted. The TWO theorems are : 16) (X+Y) = X . Y 17) (X.Y) = X + Y

Implications of De Morgan’s Theorem Input Output X Y X+Y XY 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 (b) (c) (a) Equivalent circuit implied by theorem (16) (b) Negative- AND (c) Truth table that illustrates DeMorgan’s Theorem

Implications of De Morgan’s Theorem Input Output X Y XY X+Y 0 0 1 1 0 1 1 1 0 1 1 1 1 0 0 (b) (c) (a) Equivalent circuit implied by theorem (17) (b) Negative-OR (c) Truth table that illustrates DeMorgan’s Theorem

De Morgan’s Theorem Conversion (1) Step 1: Change all ORs to ANDs and all ANDs to Ors Step 2: Complement each individual variable (short overbar) Step 3: Complement the entire function (long overbars) Step 4: Eliminate all groups of double overbars Example : A . B A .B. C = A + B = A + B + C = A + B = A + B + C = A + B = A + B + C = A + B

De Morgan’s Theorem Conversion (2) ABC + ABC (A + B +C)D = (A+B+C).(A+B+C) = (A.B.C)+D = (A+B+C).(A+B+C) = (A.B.C)+D

Example: Analyze the circuit below 2. Simplify the Boolean expression found in 1

Example: Analyze the circuit below (CONT.) Follow the steps list below (constructing truth table) List all the input variable combinations of 1 and 0 in binary sequentially Place the output logic for each combination of input Base on the result found write out the boolean expression.

Exercises: Simplify the following Boolean expressions (AB(C + BD) + AB)C ABC + ABC + ABC + ABC + ABC Write the Boolean expression of the following circuit.

Standard Forms of Boolean Expressions (1) Sum of Products (SOP) Products of Sum (POS) Notes: SOP and POS expression cannot have more than one variable combined in a term with an inversion bar There’s no parentheses in the expression

Standard Forms of Boolean Expressions (2) Converting SOP to Truth Table Examine each of the products to determine where the product is equal to a 1. Set the remaining row outputs to 0.

Standard Forms of Boolean Expressions (3) Converting POS to Truth Table Opposite process from the SOP expressions. Each sum term results in a 0. Set the remaining row outputs to 1.

Standard Forms of Boolean Expressions (4) The standard SOP Expression All variables appear in each product term. Each of the product term in the expression is called as minterm. Example: In compact form, f(A,B,C) may be written as

Standard Forms of Boolean Expressions (5) The standard POS Expression All variables appear in each product term. Each of the product term in the expression is called as maxterm. Example: In compact form, f(A,B,C) may be written as

Standard Forms of Boolean Expressions (6) Example: Convert the following SOP expression to an equivalent POS expression: Example: Develop a truth table for the expression:

The Karnaugh-MAP (K-Map)

K-Map (1) Karnaugh Mapping is used to minimize the number of logic gates that are required in a digital circuit. This will replace Boolean reduction when the circuit is large. Write the Boolean equation in a SOP form first and then place each term on a map.

K-Map (2) The map is made up of a table of every possible SOP using the number of variables that are being used. If 2 variables are used then a 2X2 map is used If 3 variables are used then a 4X2 map is used If 4 variables are used then a 4X4 map is used If 5 Variables are used then a 8X4 map is used

K-Map SOP Minimization

2 Variables K-Map (1) B B A Notice that the map is going false to true, left to right and top to bottom 0 1 2 3 B B The upper right hand cell is A B if X= A B then put an X in that cell A 1 This show the expression true when A = 0 and B = 0

2 Variables K-Map (2) B B If X=AB + AB then put an X in both of these cells A 1 1 From Boolean reduction we know that A B + A B = B B B From the Karnaugh map we can circle adjacent cell and find that X = B A 1 1

3 Variables K-Map (1) Gray Code 00 A B 01 A B 11 A B 10 A B 0 1 C C 0 1 C C 0 1 2 3 6 7 4 5

3 Variables K-Map (2) X = A B C + A B C + A B C + A B C Gray Code 00 A B 01 A B 11 A B 10 A B 0 1 C C 1 1 Each 3 variable term is one cell on a 4 X 2 Karnaugh map 1 1

3 Variables K-Map (3) X = A B C + A B C + A B C + A B C Gray Code 0 1 C C One simplification could be X = A B + A B 1 1 1 1

3 Variables K-Map (4) X = A B C + A B C + A B C + A B C Gray Code 0 1 C C Another simplification could be X = B C + B C A Karnaugh Map does wrap around 1 1 1 1

3 Variables K-Map (4) X = A B C + A B C + A B C + A B C Gray Code 0 1 C C The Best simplification would be X = B 1 1 1 1

On a 3 Variables K-Map One cell requires 3 Variables Two adjacent cells require 2 variables Four adjacent cells require 1 variable Eight adjacent cells is a 1

4 Variables K-Map Gray Code 00 A B 01 A B 11 A B 10 A B 0 0 0 1 1 1 1 0 C D C D C D C D 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10

Simplify: X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D Gray Code 00 A B 01 A B 11 A B 10 A B 0 0 0 1 1 1 1 0 C D C D C D C D Now try it with Boolean reductions 1 1 1 1 1 1 X = ABD + ABC + CD

On a 4 Variables K-Map One Cell requires 4 variables Two adjacent cells require 3 variables Four adjacent cells require 2 variables Eight adjacent cells require 1 variable Sixteen adjacent cells give a 1 or true

Simplify : Z = B C D + B C D + C D + B C D + A B C Gray Code 00 A B 01 A B 11 A B 10 A B 00 01 11 10 C D C D C D C D 1 1 1 1 1 1 1 1 1 1 1 1 Z = C + A B + B D

Simplify using K-Map (1) Firstly, change the circuit to an SOP expression

Simplify using K-Map (2) Y= A + B + B C + ( A + B ) ( C + D) Y = A B + B C + A B ( C + D ) Y = A B + B C + A B C + A B D Y = A B + B C + A B C A B D Y = A B + B C + (A + B + C ) ( A + B + D) Y = A B + B C + A + A B + A D + B + B D + AC + C D SOP expression

Simplify using K-Map (3) Gray Code 00 A B 01 A B 11 A B 10 A B 00 01 11 10 C D C D C D C D Y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

K-Map POS Minimization

3 Variables K-Map (1) Gray Code C 0 0 0 1 AB 0 1 1 1 1 0 0 1 2 3 6 7 0 1 AB 0 1 2 3 6 7 4 5

3 Variables K-Map (2)

4 Variables K-Map (1) C D 0 0 0 1 0 0 0 1 1 1 1 0 A B 1 1 1 0 0 1 3 2 0 0 0 1 1 1 1 0 A B 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10

4 Variables K-Map (2)

4 Variables K-Map (3)

K-Map - Examples Mapping a Standard SOP expression Answer: Mapping a Standard POS expression Using K-Map, convert the following standard POS expression into a minimum SOP expression Y = AB + AC or standard SOP:

K-Map with “Don’t Care” Conditions (1) Example : Input Output 3 variables with output “don’t care (X)”

K-Map with “Don’t Care” Conditions (2) 4 variables with output “don’t care (X)”

K-Map with “Don’t Care” Conditions (3) Example: Determine the minimal SOP using K-Map: Answer:

Minimum SOP expression is Solution : CD 00 01 11 10 AB 0 1 1 0 1 X 1 0 X X X X 0 0 1 0 00 01 11 10 0 1 3 2 5 7 6 13 15 14 8 9 11 10 AD BC CD Minimum SOP expression is

Exercise Minimize this expression with a K-Map ABCD + ACD + BCD + ABCD

K-map POS & SOP Simplification Example: Simplify the Boolean function F(ABCD)=(0,1,2,5,8,9,10) in (a) S-of-P (b) P-of-S Using the maxterms (0’s) and complimenting F Grouping as if they were minterms, then using De- Morgen’s theorem to get F. F’(ABCD)= BD’+CD+AB F(ABCD)= (B’+D)(C’+D’)(A’+B’) Using the minterms (1’s) F(ABCD)= B’D’+B’C’+A’C’D

5 variable K-map (1) 5 variables -> 32 minterms, hence 32 squares required

5 variable K-map (2) Adjacent squares. E.g. square 15 is adjacent to 7,14,13,31 and its mirror square 11. The centre line must be considered as the centre of a book, each half of the K-map being a page The centre line is like a mirror with each square being adjacent not only to its 4 immediate neighbouring squares, but also to its mirror image.

5 variable K-Map (3) Example: Simplify the Boolean function F(ABCDE) = (0,2,4,6,11,13,15,17,21,25,27,29,31) Soln: F(ABCDE) = BE+AD’E+A’B’E’

6 variable K-map 6 variables -> 64 minterms, hence 64 squares required

Thank you!