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

2. De Morgan’s Theorem

3. 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

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

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

6. 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

7. 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

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

9. 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.

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

11. 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

12. 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.

13. 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.

14. 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

15. 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

16. 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:

17. The Karnaugh-MAP (K-Map)

18. 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.

19. 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

20. K-Map SOP Minimization

21. 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 if X= A B then put an X in that cell A 1
This show the expression true when A = 0 and B = 0

22. 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

23. 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

24. 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
C C 1 1
Each 3 variable term is one cell on a 4 X 2 Karnaugh map 1 1

25. 3 Variables K-Map (3) X = A B C + A B C + A B C + A B C Gray Code
C C
One simplification could be
X = A B + A B 1 1 1 1

26. 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

27. 3 Variables K-Map (4) X = A B C + A B C + A B C + A B C Gray Code
C C
The Best simplification would be
X = B 1 1 1 1

28. 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

29. 4 Variables K-Map Gray Code 00 A B 01 A B 11 A B 10 A B
C D C D C D C D

30. 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
C D C D C D C D
Now try it with Boolean reductions 1 1 1 1 1 1
X = ABD + ABC + CD

31. 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

32. 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
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

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

34. 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

35. Simplify using K-Map (3)
Gray Code
00 A B
01 A B
11 A B
10 A B
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

36. K-Map POS Minimization

37. 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

38. 3 Variables K-Map (2)

39. 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

40. 4 Variables K-Map (2)

41. 4 Variables K-Map (3)

42. 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:

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

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

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

46. Minimum SOP expression is
Solution : CD AB X
X X X X 00 01 11 10 AD BC CD
Minimum SOP expression is

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

48. 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

49. 5 variable K-map (1)
5 variables -> 32 minterms, hence 32 squares required

50. 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.

51. 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’

52. 6 variable K-map
6 variables -> 64 minterms, hence 64 squares required

53. Thank you!