1. Gate-level Minimization
Although truth tables representation of a function is unique, it can be expressed algebraically in different forms
The procedure of simplifying Boolean expressions (in 2-4) is
difficult since it lacks specific rules to predict the successive steps in the simplification process.
Alternative: Karnaugh Map (K-map) Method.
Straight forward procedure for minimizing Boolean Function
Fact: Any function can be expressed as sum of minterms
K-map method can be seen as a pictorial form of the truth table. y y x m0 m1 m2 m3 x
Two-variable map

2. Two-variable K-MAP y y x x y y y y x x x x

3. Two-variable K-MAP y y y y x x x x
The three squares can be determined from the intersection
of variable x in the second row and variable y in the second
column.

4. Three-Variable K-Map
Any two adjacent squares differ by only one variable.
M5 is row 1 column 01.  101= xy’z=m5
Since adjacent squares differ by one variable (1 primed, 1 unprimed)
From the postulates of Boolean algebra, the sum of two minterms in adjacent squares can be simplified to a simple AND
For example m5+m7=xy’z+xyz=xz(y’+y)=xz

5. Three-Variable K-Map
Example 1

6. Three-Variable K-Map
Example 2
Simplify: m0 m1 m3 m2 m4 m5 m7 m6

7. Three-Variable K-Map
Example 3
Simplify: m0 m1 m3 m2 m4 m5 m7 m6

8. Three-Variable K-Map
Example 3
Simplify: m0 m1 m3 m2 m4 m5 m7 m6

9. Three-Variable K-Map Example 4 Given:
(a) Express F in sum of minterms.
(b) Find the minimal sum of products using K-Map
(a)

10. Three-Variable K-Map
Example 4 (continued) m0 m1 m3 m2 m4 m5 m7 m6

11. Three-variable K-Map: Observations
One square represents one minterm  a term of 3 literals
Two adjacent squares  a term of 2 literals
Four adjacent squares  a term of 1 literal
Eight adjacent squares  the function equals to 1

12. Four-Variable K-Map

13. Four-Variable K-Map Example 5
Simplify F(w,x,y,z) = S(0,1,2,4,5,6,8,9,12,13,14) 1

14. Four-Variable K-Map Example 6 Simplify F(A,B,C,D) =
Represented by 0001 or 0000

16. Prime Implicants
Need to ensure that all Minterms of function are covered
But avoid any redundant terms whose minterms are already covered
Prime Implicant is product Term obtained by combining maximum possible number of adjacent squares
If a minterm in a square is covered by only prime implicant then ESSENTIAL PRIME IMPLICANT
Essential prime implicant BD and B’D’
Non Essential prime implicant CD, B’C, AD and AB’

17. Four-variable K-Map: Observations
One square represents one minterm  a term of 4 literals
Two adjacent squares  a term of 3 literals
Four adjacent squares  a term of 2 literal
Eight adjacent squares  a term of 1 literal
sixteen adjacent squares  the function equals to 1

19. SUM of PRODUCT and PRODUCT OF SUM
Simplify the following Boolean function in:
(a) sum of products (b) product of sums
Combining the one’s:
(a)
Combining the zero’s:
Taking the the complement:
(b)

20. SOP and POS gate implementation
SUM OF PRODUCT (SOP)
PRODUCT OF SUM (POS)

21. Implementation of Boolean Functions
Draw the logic diagram for the following function: F = (a.b)+(b.c) a b F c

22. Implement a circuit using OR and Inverter Gates only
2 Level
More than two level
SOP
POS
Implement a circuit using OR and Inverter Gates only
Implement a circuit using AND and Inverter Gates only
Implement a circuit using NAND Gates only
Implement a circuit using NOR Gates only

23. NAND IMPLEMENTATION

25. F=AB+CD TWO LEVEL IMPLEMENT-ATION F=[(AB)’.(CD)’]’=AB+CD

26. F(X,Y,Z)=(1,2,3,4,5,7) SUM OF PRODUCT

27. CHAPTER 4 COVERT AND TO NAND WITH AND INVER.
CONVERT OR TO NAND WITH INVERT OR. SINGLE BUBBLE WITH INVERTER

42. SIMPLIFICATION WITH TABULATION METHOD DO IT ON BOARD