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Chapter3: Gate-Level Minimization Part 1

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 Minimization Part 1

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  1. Chapter3: Gate-Level MinimizationPart 1 Origionally By Reham S. Al-Majed Imam Muhammad Bin Saud University

  2. Outline • Introduction • The Map Method • Two-Variable Map • Three Variable Map • Four variable Map • Prime implicant • Product – of – Sum simplification

  3. 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

  4. 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!

  5. 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

  6. 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.

  7. 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.

  8. 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

  9. Example 2 • Simplify the function : f(x,y,z) = ∑ (2,3,4,5) y yz x x z f = x’y + xy’

  10. Example 3 • Simplify the function : f(x,y,z) = ∑ (0,2,4,5,6) y yz x x z f = z’+xy’

  11. 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.

  12. 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’

  13. 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)

  14. 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.

  15. 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)

  16. Exercise • Simplify the boolean function into Sum-of-Product F(x,y,z) = ∏ ( 0, 2, 5, 7)

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