Chapter3: Gate-Level MinimizationPart 1

Origionally By Reham S. Al-Majed

Imam Muhammad Bin Saud University

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Outline

Introduction

The Map Method Two-Variable Map Three Variable Map

Four variable Map Prime implicant

Product – of – Sum simplification

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3.1 Introduction

The complexity of digital logic gates depends on the complexity

of the corresponding Boolean function.

Gate-level minimization: finding an optimal gate-level

implementation of the Boolean functions that describing circuits.

Boolean expression may be simplified by: Boolean Algebra

lack specific rules for each succeeding step. Map method ( See next slides) Synthesis tools

Efficient and quick

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3.2 The Map Method

Simple and straightforward

Pictorial form of a truth table

Known as the Karnaugh map or k-map.

Made up of squares An n-variable K-map has 2n squares/cells.

Each square represents one minterm.

Square’s value corresponds to one value in truth table.

Mark with 1 the squares at which the function minterms produce 1.

Any two adjacent squares in the map differ by only one variable Gray Code

Uppermost cells are adjacent to the lowermost cells!

The leftmost cells are adjacent to the rightmost cells!

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Two-variable Map Representation Four rows in truth table

Four minterms four squares in the map

Example: consider the boolean function f = x.y

Mark square correspond to m3 with 1

Row 1 x unprimed , column 1 y unprimed

Minterm f y x

x’ y’m 0

0 0 0

x’ ym 1

0 1 0

x y’m 2

0 0 1

x ym 3

1 1 1

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Eight minterms eight squares

Minterms are arranged in Gray code not in a binary sequence !

Check adjacency of leftmost and rightmost cells (m0-m2 , m4-m6)

Try to use (+) between any two, four, eight squares ?

For convenience, variable is written under squares in which it is

unprimed.

Three-variable Map Representation

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Minimization by K-mapSteps:

1. Mark the K-map with 1s in each minterm that represents the function.

2. Group the adjacent marked squares.

• Groups must contain power of 2 (e.g. 2,4,8) ones.

• Grouping can be side to side or top to bottom but not diagonally.

• It is desirable to use the same minterm with other groups (if adjacent).

• The goal is to find the fewest number of groups combine the maximum

number of adjacent squares.

3. Analyze each group to find the term it represents.

4. Write the minimized Boolean expression by OR-ing the terms of the groups.

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Example 1 Express the Boolean function from the truth table and then

Simplify using Boolean Algebra & k-map Minterm f y x

x’ y’m 0

0 0 0

x’ ym 1

1 1 0

x y’m 2

1 0 1

x ym 3

1 1 1

f = ∑(1,2,3) = m1+m2+m3

= x’y +xy’+xy

By BA

f = x’y +xy’+xy

= x’y + x(y’+y)

= x’y + x . 1

= (x’+x) (y+x)

= 1. y+x

= x + y

Both in the same column (y)But different rows cancel each otherBoth in the same row (x)But different columns cancel each other

f = x + y

By K-map

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Example 2

Simplify the function : f(x,y,z) = ∑ (2,3,4,5)

10 11 01 00

1 1 0

1 1 1

f = x’y + xy’

x

yz

x

z

y

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10 11 01 00

11 0

1 1 1 1

Example 3

Simplify the function : f(x,y,z) = ∑ (0,2,4,5,6)

f = z’+xy’

x

yz

x

z

y

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3.3 Four-Variable Map 16 minterms 16 squares arranged in Gray code sequence

Concatenate row number with column to obtain minterm.

Remember: the larger combined squares the smaller number of

literals in term.

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Examples

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

F = A’B’C’+ B’CD’ + A’BCD’ + AB’C’

F = y’ + w’z’ + xz’

F = B’D’ + B’C’ + A’CD’

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Prime Implicants

Prime Implicant: Product term obtained by combining the maximum numbers of adjacent squares. Single 1 is PI if it is not adjacent to any other 1’s. Two adjacent 1’s form PI if they aren’t within a group of four. Four ?

Essential Prime Implicant: If a minterm in a square is covered by only one prime implicant. Look at each 1 and check the number of PIs that cover it

If only one PI it is EPI

Example: F(A, B, C, D) = ∑ ( 0, 2, 3, 5, 7, 8, 9, 10, 11, 13,

15)

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3.5 Product – of – Sums Simplification

The 1’s represent minterms of the function.

The minterms not included in SoP represent complement of the

function.

Mark empty squares by 0’s and group them obtain the

simplified expression of the complement of the function.

Apply DeMorgan’s theorem The function will be in product of

sums.

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Example

Simplify the following Boolean function into SoP and PoS :

F(A,B,C,D)= ∑ (0, 1, 2, 5, 8, 9, 10) SoP

Mark 1’s Combine squares marked with 1’s F= B’D’ + B’C’ + A’C’D

PoS Mark 0’s Combine squares marked with 0’s obtain simplified complement of F F’ = AB + CD + BD’ Apply DeMorgan’s theorem F= (A’+B)(C’+D’)(B’+D)

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Exercise

Simplify the boolean function into Sum-of-Product

F(x,y,z) = ∏ ( 0, 2, 5, 7)