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

Two-variable K-MAP y y x x y y y y x x x x

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.

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

Three-Variable K-Map Example 1

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

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

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

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)

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

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

Four-Variable K-Map

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

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

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’

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

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)

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

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

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

NAND IMPLEMENTATION

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

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

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

SIMPLIFICATION WITH TABULATION METHOD DO IT ON BOARD