chapter3 gate level minimization part 1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter3: Gate-Level Minimization Part 1 PowerPoint Presentation
Download Presentation
Chapter3: Gate-Level Minimization Part 1

Loading in 2 Seconds...

play fullscreen
1 / 16

Chapter3: Gate-Level Minimization Part 1 - PowerPoint PPT Presentation


  • 81 Views
  • Uploaded on

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.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chapter3: Gate-Level Minimization Part 1' - osanna


Download Now An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chapter3 gate level minimization part 1

Chapter3: Gate-Level MinimizationPart 1

Origionally By Reham S. Al-Majed

Imam Muhammad Bin Saud University

outline
Outline
  • Introduction
  • The Map Method
    • Two-Variable Map
    • Three Variable Map
  • Four variable Map
    • Prime implicant
  • Product – of – Sum simplification
3 1 introduction
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
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!
two variable map representation
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
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
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
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
Example 2
  • Simplify the function : f(x,y,z) = ∑ (2,3,4,5)

y

yz

x

x

z

f = x’y + xy’

example 3
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
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
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 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
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
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
Exercise
  • Simplify the boolean function into Sum-of-Product

F(x,y,z) = ∏ ( 0, 2, 5, 7)