Gate-level Minimization

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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

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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 2

Simplify:

Three-Variable K-Map

Example 3

Simplify:

Three-Variable K-Map

Example 3

Simplify:

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)

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

Simplify F(w,x,y,z) = S(0,1,2,4,5,6,8,9,12,13,14)

Example 5

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

Non Essential prime implicant CD, B’C, AD and AB’

Essential prime implicant BD and B’D’

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

PRODUCT OF SUM (POS)

SUM OF PRODUCT (SOP)

Implementation of Boolean Functions

• Draw the logic diagram for the following function: F = (a.b)+(b.c)

a

b

F

c

Implement a circuit
• 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

F=AB+CD

F=(A’B’)’+(C’D’)’

F=[(AB)’.(CD)’]’=AB+CD

TWO LEVEL

IMPLEMENT-ATION

CHAPTER 4

COVERT AND TO NAND WITH AND INVER.

CONVERT OR TO NAND WITH INVERT OR. SINGLE BUBBLE WITH INVERTER