Logic Design I (17.341) Fall 2011 Lecture Outline Class # 06 October 24, 2011 Dohn Bowden 1
Today s Lecture Administrative Main Logic Topic Homework 2
Course Admin 3
Administrative Admin for tonight Syllabus Review 4
Syllabus Syllabus Moved the Lab Lecture to next week 5
Syllabus Review Week Date Topics Chapter Lab Report Due 1 09/12/11 Introduction to digital systems and number systems 1 2 09/19/11 Binary Codes and Boolean Algebra 2 3 09/26/11 Boolean Algebra (continued) 3 4 10/03/11 Examination 1 X 10/10/11 No Class - Holiday 5 10/17/11 Application of Boolean Algebra 4 6 10/24/11 Karnaugh Maps and Lab lecture 5 7 10/31/11 Multi-Level Gate Circuits. NAND and NOR Gates 7 1 8 11/07/11 Examination 2 9 11/14/11 Combinational Circuit Design and Simulation Using Gates 10 11/23/11 Multiplexers, Decoders. Encoder, and PLD 9 8 2 11 11/28/11 Introduction to VHDL 10 3 12 12/05/11 Examination 3 13 12/12/11 Review 4 14 12/19/11 Final Exam 6
Exam #1 Exam #1 Will be passed back 7
Exam #2 Will be in two weeks November 07, 2011 8
Questions? 9
Chapter 5 KARNAUGH MAPS 10
Introduction Switching functions can generally be simplified by using Algebraic techniques Two problems arise when algebraic procedures are used The procedures are difficult to apply in a systematic way It is difficult to tell when you have arrived at a minimum solution The Karnaugh map aka K-Map method is Generally faster and easier to apply than other simplification methods 11
Minimum Forms of Switching Functions 12
Minimum Forms of Switching Functions Given a minterm expansion the minimum sum-of-products format can often be obtained by the following procedure 1. Combine terms by using XY + XY = X Do this repeatedly to eliminate as many literals as possible A given term may be used more than once because X + X = X 2. Eliminate redundant terms by using the consensus theorem or other theorems 13
Minimum Forms of Switching Functions Unfortunately the result of this procedure may depend on The order in which the terms are combined or Eliminated so that the final expression obtained is not necessarily minimum 14
Minimum Forms of Switching Functions Example Find a minimum sum-of-products expression for None of the terms in the above expression can be eliminated by consensus However 15
Minimum Forms of Switching Functions However Combining terms in a different way leads directly to a minimum sum of products So 16
Minimum Forms of Switching Functions So different solutions can result depending on the order of combining terms 17
Two and Three Variable K-Maps 18
Two Variable K-Maps Like a truth table the K-map of a function specifies The value of the function for every combination of values of the independent variables 19
Two Variable K-Maps The truth table for the function F 20
Two Variable K-Maps Truth table and corresponding K-map for the function F F for A = B = 0 is plotted in the upper left square, Each 1 on the map corresponds to a minterm of F 21
Two Variable K-Maps Read minterms from the map A 1 in square 00 indicates that A B is a minterm of F A 1 in square 01 indicates that A B is a minterm 22
Two Variable K-Maps Minterms in adjacent squares of the map can be combined since they differ in only one variable A B and A B combine to form A Indicate by looping the corresponding 1 s on the map 23
Three Variable K-Maps 24
Three Variable K-Maps Three variable truth table and the corresponding K-map Note row labels 00, 01, 11, and 10 differ by only one variable 25
Three Variable K-Maps Location of Minterms on a Three Variable K-Map 26
Three Variable K-Maps Minterms in adjacent squares of the map differ in only one variable therefore can be combined using the theorem XY + XY = X 27
Three Variable K-Maps K-Map of F (a, b, c) = Σ m(1, 3, 5) = M(0, 2, 4, 6, 7) 28
K-Maps for Product Terms 29
K-Maps for Product Terms 30
K-Maps for Product Terms f(a,b,c) = abc' + b'c + a' 31
Simplification of a Three-Variable Function 32
Simplification of a Three-Variable Function 33
Complement 34
Complement The map for the complement of F is formed by replacing 0 s with 1 s and 1 s with 0 s on the map of F therefore F = c + ab 35
Consensus Theorem 36
Consensus Theorem K-Maps Which Illustrate the Consensus Theorem 37
Function with Two Minimal Forms 38
Function with Two Minimal Forms 39
Four-Variable K-Map 40
Four Variable K-Map Each minterm is located adjacent to the four terms with which it can combine For example m 5 (0101) could combine with m 1 (0001) m 4 (0100) m 7 (0111) or m 13 (1101) 41
Four Variable K-Map Example Plot of acd + a b + d 42
Simplification of Four Variable Functions 43
Simplification of Four Variable Functions 44
Simplification of Four Variable Functions 45
Simplification of an Incompletely Specified Function 46
Simplification of an Incompletely Specified Function 47
Example 48
Example EXAMPLE Use a four variable K-map to find the minimum product of sums for f f = x z + wyz + w y z + x y 49
Example First plot the 1 s on a K-Map f = x z + wyz + w y z + x y 50
Example From the 0 s we get f ' = y'z + wxz' + w'xy 51
Example Complement f to get a minimum product of sums of f f ' = y'z + wxz' + w'xy f = (y + z )(w + x + z)(w + x + y ) 52
Determination of Minimum Expressions Using Essential Prime Implicants 53
Implicants and Prime Implicants Any single 1 or any group of 1 s which Can be combined together on a map of the function F Represents a product term which is called an implicant of F 54
Implicants and Prime Implicants A product term implicant is called A prime implicant if It cannot be combined with another term to eliminate a variable 55
Implicants and Prime Implicants Implicant Prime Implicant Prime Implicant Implicant Implicant Prime Implicant 56
Determination of All Prime Implicants 57
Determination of All Prime Implicants 58
Determination of All Prime Implicants 59
Determination of All Prime Implicants 60
Essential Prime Implicants 61
Essential Prime Implicants If a minterm is covered by only one prime implicant That prime implicant is said to be essential and It must be included in the minimum sum of products Essential Prime Implicants Prime Implicants Implicants 62
Essential Prime Implicants 1 s shaded in blue are covered by only one prime implicant (essential) All other 1 s are covered by at least two prime implicants 63
Flowchart for Determining a Minimum Sum of Products Using a K-Map 64
Flowchart for Determining a Minimum Sum of Products Using a Karnaugh Map 65
Flowchart Procedure Illustration Shaded 1 s are covered by only one prime implicant Essential prime implicants A B AB D Then AC D covers the remaining 1 s 66
Five Variable K-Maps 67
Five Variable K-Maps A five-variable map can be constructed in three dimensions by placing one four-variable map on top of a second one. Terms in the bottom layer are numbered 0 through 15 and corresponding terms in the top layer are numbered 16 through 31, so that the terms in the bottom layer contain A' and those in the top layer contain A. To represent the map in two dimensions, we will divide each square in a four-variable map by a diagonal line and place terms in the bottom layer below the line and terms in the top layer above the line. 68
Note that terms m 0 and m 20 do not combine because they are in different layers and different columns (they differ in two variables). A Five-Variable Karnaugh Map 69
Each term can be adjacent to exactly five other terms: four in the same layer and one in the other layer. 70
When checking for adjacencies, each term should be checked against the five possible adjacent squares. P 1 and P 2 are essential prime implicants. 71
P 1, P 2, P 3, and P 4 are essential prime implicants. F(A, B, C, D, E) = Ʃ m(0, 1, 3, 8, 9, 14, 15, 16, 17, 19, 25, 27, 31) 72
Using a Karnaugh map to facilitate factoring, we see that the two terms in the first column have A B in common; the two terms in the lower right corner have AC in common. 73
F = ABCD + B CDE + A B + BCE We can use a Karnaugh map for guidance in algebraic simplification. From Figure 5-26, we can add the term ACDE by the consensus theorem and then eliminate ABCD and B CDE. 74
75
Other Forms of Karnaugh Maps Instead of labeling the sides of a Karnaugh map with 0 s and 1 s, some people prefer to use Veitch diagrams. In Veitch diagrams, A = 1 for the half of the map labeled A, and A = 0 for the other half. The other variables have a similar interpretation. 76
Veitch Diagrams 77
Two alternative forms of five-variable maps are also used. One form simply consists of two four-variable maps side-by-side. A modification of this uses a mirror image map. In this map, first and eighth columns are adjacent as are second and seventh columns, third and sixth columns, and fourth and fifth columns. 78
Side-by-side Form of Five-Variable K-Maps 79
Mirror Image Form of Five-Variable K-Maps 80
Lab 81
Lab No topics this week 82
Next Week 83
Next Week Topics Chapter 7 Multi-Level Gate Circuits NAND and NOR Gates Lab introduction 84
Home Work 85
Homework 1. Read Chapters 7 2. Solve the following Chapter 5 problems Programmed Exercise 5.1 Programmed Exercise 5.2 86
References 1. None 87