George Mason, ECE 301
Excerpt: ... Plan for 1 March Midterm #1: add five points Policy: if you dispute grading of a quiz or homework, you must see me during office hours; not at end of lecture Reminder: midterm #2 on Wed, 8 March Review previous lecture (3-5; 3-6; table 3-3) Discuss other implementations (NOR-OR; NAND-AND; OR-NAND; AND-NOR) Wired OR and Wired AND Explain Table 3-4 04/26/09 ece-301 1 Plan for 1 March (contd) Dont cares Tabulation Method Prime Implicants and Essential Prime Implicants Work some problems Reading for next Monday: 4-1 to 4-3 (Adders) NOTE: lecture on Monday will spend 50% of time on review for midterm 04/26/09 ece-301 2 Tabulation Method Tedious for humans; easy for computers Works for higher dimensions Two step process: (1) determine prime implicants and (2) select the necessary prime implicants Two methods: (1) binary, or (2) decimal 04/26/09 ece-301 3 ...
Wisconsin, ECE 352
Excerpt: ... Lecture 7 (09/18/2002): Supplementary Note on Karnaugh-Map Instructor: Yong Kim (Section 1) During the lecture, we haven't really finished up the example on less than & equivalent relations of 4 variables K-map. Here is a example we partly covered in class with detailed explanation. Problem: Given (0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15) find all prime implicants , essential prime implicants and find a sum of product expression with minimum literals. Solution: First we find all prime implicants (rectangles containing largest number of 1's in power of 2, say one 1, two 1s, four 1s, eight 1s, .). We have total of six prime implicants : B'D', CD, B'C, BD, AB', AD) as shown below. BD 1 BD A 1 1 CD C 1 1 1 1 D 1 1 BC B AB 1 1 AD An essential prime implicant is the only prime implicant that covers a minterm or minterms. In our problem, Minterm m0 (=A'B'C'D') is covered by only one prime implicants B'D'. Thus B'D' is one of the essential prime implicants . Also Minterm m5 (=AB'C'D) is only covered by the prime ...
Wisconsin, ECE 352
Excerpt: ... Lecture 7 (09/18/2002): Supplementary Note on Karnaugh-Map Instructor: Yong Kim (Section 1) During the lecture, we havent really finished up the example on less than & equivalent relations of 4 variables K-map. Here is a example we partly covered in class with detailed explanation. Problem: Given ? (0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15) find all prime implicants , essential prime implicants and find a sum of product expression with minimum literals. Solution: First we find all prime implicants (rectangles containing largest number of 1s in power of 2, say one 1, two 1s, four 1s, eight 1s, ). We have total of six prime implicants : BD, CD, BC, BD, AB, AD) as shown below. BD 1 BD A 1 1 CD C 1 1 1 BC B 1 1 D AD An essential prime implicant is the only prime implicant that covers a minterm or minterms. In our problem, Minterm m0 (=ABCD) is covered by only one prime implicants BD. Thus BD is one of the essential prime implicants . Also Minterm m5 (=ABCD) is only c ...
University of Illinois, Urbana Champaign, ECE 462
Excerpt: ... new table. The function's prime implicants correspond to the codes left unmarked. Given the code, the associated product is easily constructed. This description is mathematically precise, albeit a bit terse. As you unpack you discover some implications : when you separate rows of a table into sets of codes that have the same number of 1's set, you may have some empty sets. The only way two codes can differ in exactly one variable position is if one of the codes uses '1' in that position, and the other uses '0'. A code c might result from combining code pairs a and b, and also from combining code pairs e and f (etc.). The set notation used above deals with the duplication issue-an element is in the set just once. In the tabular application of the algorithm, if you create a new code and find that it exists already in the table, then mark the codes that combined to form it, but do not replicate it in the table. With each successive table the largest possible number of 1's decreases, so some of the sets ...
FAU, CDA 3201
Excerpt: ... Lecture 6 1 Simplification Techinques *So far, approach has been heuristic depends completely on experience. *Not sure if it is minimum or not. *Cumbersome to do algebraic simplification with 4 or + variables. 2 ways 1. k-maps are a graphical approach 2. (iterated consensus not in this course) k-maps find product terms for use in sum of minimum products (These are called prime implicants ) It's useful to 6 variables. 2 Implicants: of a function can be a product term that can be used in a sum of products expression for it. Function is 1 if implicant is 1. Consider: CD/AB 00 01 11 10 00 01 11 10 CD\AB 00 01 11 10 00 1 1 00 1 1 00 1 1 01 1 01 1 01 1 11 1 1 1 1 11 1 1 1 1 11 1 1 1 1 10 10 10 Implicants of F 3 minterms A'B'C'D' A'B'CD A'BCD ABC'D' ABC'D ABCD AB'CD Groups of 2 A'CD BCD ACD B'CD ABC' ABD Groups of 4 CD Notes: *Any sum of products for a func, must be a sum of implicants. Must choose enough implicants so that each "1" of a function is included in at least 1 implicant. ("cover" of func) * ...
Idaho, ECE 240
Excerpt: ... ECE 240: Session 15; Page 1/4 ECE 240: Lecture 15 Prime Implicants Any single 1 or group of 1's that can be combined together on a Karnaugh map of the function F represents a product term which is called an IMPLICANT. A PRIME IMPLICANT is a product term that cannot be combined with another term to eliminate a variable. If a minterm is only covered by one prime implicant, that prime implicant becomes an ESSENTIAL PRIME IMPLICANT Two adjacent 1's form a prime implicant if they are not contained in a group of four adjacent 1's. Four adjacent 1's form a prime implicant if they are not contained in a group of eight adjacent 1's. The minimum sum-of-products expression for a function consists of some (BUT NOT NECESSARILY ALL) of the prime implicants of a function. If a minterm is only covered by one prime implicant, that prime implicant becomes an ESSENTIAL PRIME IMPLICANT \ AB CD\ 00 01 11 10 \_ | | | | | 00 | | 1 | 1 | | |_|_|_|_| | | | | | 01 | 1 | 1 | 1 | | |_|_|_|_| | | | | | 11 ...
Georgia Tech, ECE 2030
Excerpt: ... Supplement to Logic and Computer Design Fundamentals 3rd Edition 1 MORE OPTIMIZATION elected topics not covered in the third edition of Logic and Computer Design Fundamentals are provided here for optional coverage and for self-study if desired. This material fits well with the desired coverage in some programs but not may not fit within others due to time constraints or local preferences. This supplement provides two optimization algorithms for finding a minimum cost two-level circuit. The first algorithm selects prime implicants for a minimum cost implementation of a two-level, sum of products circuit. The second algorithm replaces K-maps with tabular representations that permit the handling of more that the six variables possible using K-maps. Unfortunately, the latter algorithm is difficult to execute manually and the prime implicant generation approach is not the best for program implementation. The prime implicant selection step is more useful for both manual computation for simple problems and comput ...
University of Illinois, Urbana Champaign, ECE 462
Excerpt: ... ECE 462 Logic Design Homework 2 This homework (Questions 1-5) have an assignment of 60 points. If you do the extra credit question at the end, you will get an equivalent of 15 additional homework points towards your total. Partially correct answers will also be graded. You are highly encouraged to attempt Question 6. Prime implicants , Consensus theorem 1) a) Is it possible to have a 4-variable function f(w,x,y,z) that has more primeimplicants than non- prime implicants ? If yes, justify with an example. If no, prove that this can never happen. [4 points] b) We studied that the consensus theorem states the following. xy + x'z + yz = xy + x'z If xy and x'z are two prime implicants of a given function, where x is a variable and y and z are two product terms, is their consensus yz (if yz 0) also an implicant? Is it a prime implicant? If yes, show proof. If no, give a counterexample. [5 points] Quine-McCluskey method 2) A two-level circuit C with 4 inputs and 4 outputs is to be designed to convert a decimal digit ...
Utah, EE 5740
Excerpt: ... CS/EE 5740/6740: Computer Aided Design of Digital Circuits Chris J. Myers Lecture 7: Prime Implicants Reading: Chapter 4.1-6 Sums of Products A SOP (sum of products) is a join of meets, henceforth called OR of ANDs: f = xy z + x y + wxyz Conjunctions (ANDs) of literals are functions. They are also called cubes, terms, and implicants An SOP is optimum, if no other SOP representing the same function has fewer cubes or literals Boolean Algebra F2 ({0,1}) Boolean 2-cube 11 y 01 x 00 10 2-variable K-map x y0 1 0 1 K-maps and Boolean 4-cubes y 00 01 11 10 z Note these minterms are Distance-2 00 01 11 10 wxy z wxyz w x B4 = {0,1}4 = {0,1} {0,1} {0,1} {0,1} Note: Each vertex is a minterm of F4 ({0,1}) f = wx ( y z + yz ) = wxy z + wxyz Representing k-cubes on the n-cube A k -cube is a product (AND) of k literals: bc A k -cube appears as a dense set of adjacent minterms. On an n -cube. For k = 2 < n = 4, Implicants with Dont Ca ...
Utah, EE 5740
Excerpt: ... Sums of Products A SOP (sum of products) is a join of meets, henceforth called OR of ANDs: CS/EE 5740/6740: Computer Aided Design of Digital Circuits f = xy z + x y + wxyz Chris J. Myers Lecture 7: Prime Implicants Reading: Chapter 4.1-6 Conjunctions (ANDs) of literals are functions. They are also called cubes, terms, and implicants An SOP is optimum, if no other SOP representing the same function has fewer cubes or literals K-maps and Boolean 4-cubes Note these minterms are "Distance-2" Boolean Algebra F2 ({0,1}) z wxy z y Boolean 2-cube 11 y 01 00 01 11 10 00 01 x 00 10 w wxyz x 11 10 B4 = {0,1}4 = {0,1} {0,1} {0,1} {0,1} Note: Each vertex is a minterm of F4 ({0,1}) f = wx ( y z + yz ) = wxy z + wxyz 2-variable K-map x y0 1 0 1 Implicants with Don't Cares Representing k-cubes on the n-cube x y wyz Discriminant 1 (ON-set) Discriminant don't care Discriminant 0 (OFF-set) ...
Columbia, CS 6861
Excerpt: ... CSEE E6861y Prof. Steven Nowick The Quine-McCluskey Method Handout 6 January 22, 2009 Introduction The Quine-McCluskey method is an exact algorithm which nds a minimum-cost sum-of-products implementation of a Boolean function. This handout introduces the method and applies it to several examples. There are 4 main steps in the Quine-McCluskey algorithm: 1. Generate Prime Implicants 2. Construct Prime Implicant Table 3. Reduce Prime Implicant Table (a) Remove Essential Prime Implicants (b) Row Dominance (c) Column Dominance 4. Solve Prime Implicant Table Note: For this course, you are not responsible for Step #1 on this handout: the method for generating all prime implicants of a Boolean function. You can look over this method, but you will be learning, and be responsible for, a more powerful modern prime-generation technique in a few weeks. In Step #1, the prime implicants of a function are generated using an iterative procedure. In Step #2, a prime implicant table is constructed. The columns of the tabl ...
Columbia, CS 4861
Excerpt: ... CSEE W4861y Prof. Steven Nowick The Quine-McCluskey Method Handout 6 January 18, 2007 Introduction The Quine-McCluskey method is an exact algorithm which finds a minimum-cost sum-of-products implementation of a Boolean function. This handout introduces the method and applies it to several examples. There are 4 main steps in the Quine-McCluskey algorithm: 1. Generate Prime Implicants 2. Construct Prime Implicant Table 3. Reduce Prime Implicant Table (a) Remove Essential Prime Implicants (b) Row Dominance (c) Column Dominance 4. Solve Prime Implicant Table Note: For this course, you are not responsible for Step #1 on this handout: the method for generating all prime implicants of a Boolean function. You can look over this method, but you will be learning, and be responsible for, a more powerful modern prime-generation technique in a few weeks. In Step #1, the prime implicants of a function are generated using an iterative procedure. In Step #2, a prime implicant table is constructed. The columns of the table ...