Chapter3: Gate-Level Minimization Part 1

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Chapter3: Gate-Level Minimization Part 1. Origionally By Reham S. Al-Majed. Outline . Introduction The Map Method Two-Variable Map Three Variable Map Four variable Map Prime implicant Product – of – Sum simplification . 3.1 Introduction.

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### Chapter3: Gate-Level MinimizationPart 1

Origionally By Reham S. Al-Majed

Outline
• Introduction
• The Map Method
• Two-Variable Map
• Three Variable Map
• Four variable Map
• Prime implicant
• Product – of – Sum simplification
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
3.2 The Map Method
• Simple and straightforward
• Pictorial form of a truth table
• Known as the Karnaugh map or k-map.
• 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!
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
Three-variable Map Representation
• 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.
Minimization by K-map

Steps:

• Mark the K-map with 1s in each minterm that represents the function.
• 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.
• Analyze each group to find the term it represents.
• Write the minimized Boolean expression by OR-ing the terms of the groups.
Example 1
• Express the Boolean function from the truth table and then Simplify using Boolean Algebra & k-map

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

By K-map

Both in the same column (y)

But different rows  cancel each other

Both in the same row (x)

But different columns cancel each other

f = x + y

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

y

yz

x

x

z

f = x’y + xy’

Example 3
• Simplify the function : f(x,y,z) = ∑ (0,2,4,5,6)

y

yz

x

x

z

f = z’+xy’

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.
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’

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)
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.
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)
Exercise
• Simplify the boolean function into Sum-of-Product

F(x,y,z) = ∏ ( 0, 2, 5, 7)