CHAPTER 1 INTRODUCTION TO DIGITAL LOGIC
EKT 124 / 3 ELEKTRONIK DIGIT 1 CHAPTER 1 INTRODUCTION TO DIGITAL LOGIC De Morgans Theorem De Morgans 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 Morgans Theorem
InputOutput X YX+Y XY (b) (c) (a) Equivalent circuit implied by theorem (16) (b) Negative- AND (c) Truth table that illustrates DeMorgans Theorem Implications of De Morgans Theorem
InputOutput X YXY X+Y (b) (c) (a) Equivalent circuit implied by theorem (17) (b) Negative-OR (c) Truth table that illustrates DeMorgans Theorem De Morgans 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 Morgans 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 Theres 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 thena 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 B B The upper right hand cell is A B ifX= 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 BB 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
C C 3 Variables K-Map (2) X = A B C + A B C + A B C + A B C
Gray Code 00A B 01A B 11 A B 10A B 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
CC 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
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
CC 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
C DC DC D C D 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 00A B 01A B 11A B 10A B C DC DC 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 00A B 01A B 11A B 10A B C DC DC 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+ ABD 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 00A B 01A B 11A B 10A B C DC DC 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
AB 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
A B 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 orstandard SOP: K-Map with Dont Care Conditions (1)
Example : Input Output 3 variables with output dont care (X) K-Map with Dont Care Conditions (2)
4 variables with output dont care (X) K-Map with Dont Care Conditions (3)
Example: Determine the minimal SOP using K-Map: Answer: Minimum SOP expression is
Solution : CD AB X X XXX 00 01 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 (0s) and complimenting F Grouping as if they were minterms, then using De- Morgens theorem to get F. F(ABCD)= BD+CD+AB F(ABCD)= (B+D)(C+D)(A+B) Using the minterms (1s) F(ABCD)= BD+BC+ACD 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+ADE+ABE 6 variable K-map 6 variables -> 64 minterms, hence 64 squares required Thank you!