• 3 Pages

ee261_lecture_24

Montana, EE 261

Excerpt: ... ion to Logic Circuits Fall 2008 Lecture #24 Page 3 Karnaugh Map Minimization Logic Minimization Prime Implicant - a Normal Product Term of F (i.e., a P that implies F) where if any variable is removed from P, the resulting product does NOT imply F - K-maps: a circled set of 1's that cannot be larger without circling 1 or more 0's Prime Implicant Theorem - a Minimal Sum is a sum of Prime Implicants BUT Does not need to include ALL prime Implicants EE 261 Introduction to Logic Circuits Fall 2008 Lecture #24 Page 4 2 Karnaugh Map Minimization Logic Minimization Complete Sum - the product of all Prime Implicants (not minimized) Distinguished 1-Cell - an "input combination" that is covered by only ONE Prime Implicant Essential Prime Implicant - a Prime Implicant that covers one or more "Distinguished 1-Cells" NOTE: - the sum of Essential Prime Implicants is the Minimal Sum - this means we're done minimizing EE 261 Introduction to Logic Circuits Fall 2008 Lecture #24 Page 5 Karnaugh Map Minimi ...

• 3 Pages

ece-301-10

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 ...

• 4 Pages

Lecture7Note

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 ...

• 4 Pages

Lecture7Note

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 ...

• 4 Pages

ee261_lecture_26

Montana, EE 261

Excerpt: ... EE 261 Introduction to Logic Circuits Lecture #26 Agenda 1. Combinational Logic Design Flow 2. Exam #2 Announcements : Wednesday (11/5) 1. Exam #2 Friday (11/7) EE 261 Introduction to Logic Circuits Fall 2008 Lecture #26 Page 1 Combinational ...

• 2 Pages

ee261_lecture_26

Montana, EE 261

Excerpt: ... EE 261 Introduction to Logic Circuits Lecture #26 Agenda 1. Combinational Logic Design Flow 2. Exam #2 Announcements : Wednesday (11/5) 1. Exam #2 Friday (11/7) EE 261 Introduction to Logic Circuits Fall 2008 Lecture #26 Page 1 Combinational ...

• 1 Pages

exam1.obj

Valparaiso, ECE 221

Excerpt: ... xpressions in either SOP or POS form (recognize prime implicants , essential prime implicants and secondary prime implicants ). create a minimized logic expression from a specification (word problem). Test #1 Information: Open Book, Closed Note 4-5 Problems ...

• 11 Pages

lecture_15

Penn State, CSE 271

Excerpt: ... CSE271 Section 1 Introduction to Digital Systems Instructor: Sunho Lim Lecture 15 : Review of Chapter 3 http:/www.cse.psu.edu/~slim/CSE271_001 Implicant A product term that can be used in a sum of products for that function A rectangle of 1, 2, 4, 8, (any power of 2) adjacent 1s in the map. Lecture 15 CSE271 Section 1 Fall 04 2 1 Implicant (contd) a cd ab 1 Implicants v Minterms v Rectangles b 1 1 1 1 1 1 of 2 1s q d c acd, bcd, acd, bcd, abd, abc v Rectangles q of 4 1s cd Lecture 15 CSE271 Section 1 Fall 04 3 Prime implicant An implicant that is not fully contained in any one other implicant. Intuitively, the largest implicant that can include all the minterms in the implicant Lecture 15 CSE271 Section 1 Fall 04 4 2 Prime implicant (contd) a cd ab 1 Prime Implicants b 1 1 1 1 1 1 Minterm abcd Rectangles of 2 1s acd, bcd, acd, bcd, abc, abd d Rectangles of 4 1s ...

• 45 Pages

exact2lev

Georgia Tech, ECE 3060

Excerpt: ... a 111 110 100 000 F = abc + abc + abc + abc + abc F = ab + bc + ac + ab ECE 3060 Lecture 107 Prime Denitions Prime implicant implicant not contained by any other implicant Prime cover cover of prime implicants Essential Prime Implicant (EPI): there is at least one minterm covered by EPI and not covered by any other prime implicant ECE 3060 Lecture 108 Two level logic optimization Assumptions: primary goal is to reduce the number of implicants all implicants have the same cost secondary goal is to reduce the number of literals Minimum cover cover of the function with the minimum number of implicants global optimum c b a 011 101 111 001 010 f= 110 100 Lecture 109 000 ECE 3060 Minimal or Irredundant Cover Cover of the function that is not a proper superset of another cover no implicant can be dropped local optimum ECE 3060 Lecture 1010 ECE 3060 Lecture 1011 Minimal Cover with respect ...

• 30 Pages

lec10

Georgia Tech, ECE 3060

Excerpt: ... 11 110 100 000 F = abc + abc + abc + abc + abc F = ab + bc + ac + ab ECE 3060 Lecture 107 Prime Denitions Prime implicant implicant not contained by any other implicant Prime cover cover of prime implicants Essential Prime Implicant (EPI): there is at least one minterm covered by EPI and not covered by any other prime implicant ECE 3060 Lecture 108 Two level logic optimization Assumptions: primary goal is to reduce the number of implicants all implicants have the same cost secondary goal is to reduce the number of literals Minimum cover cover of the function with the minimum number of implicants global optimum c b a 011 101 111 001 010 f = 110 100 Lecture 109 000 ECE 3060 Minimal or Irredundant Cover Cover of the function that is not a proper superset of another cover no implicant can be dropped local optimum 011 101 001 010 c b a ECE 3060 111 110 100 Lecture 1010 000 Minimal Cover with res ...

• 6 Pages

algorithms-1

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 ...

• 16 Pages

lect6

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) * ...

• 4 Pages

L15

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 ...

• 14 Pages

Q_M_method_supp

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 ...

• 4 Pages

hw2_2009

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 ...

• 6 Pages

lecture_12

Penn State, CSE 271

Excerpt: ... CSE271 Section 1 Introduction to Digital Systems Instructor: Sunho Lim Lecture 12 http:/www.cse.psu.edu/~slim/CSE271_001 Terminology Implicant v an implicant of a function is a product term that can be used in an SOP expression for that function. On the K-map, it's a rectangle in which the number of squares having "1" is a power of 2. cover of function F is a set of implicants that each of the 1's of F are included in at least one of these implicants. CSE271 Section 1 Fall 04 2 Cover vA Lecture 12 1 Terminology (cont'd) Prime v an implicant implicant that is not fully contained in any other implicant. Essential vA prime implicant prime implicant that includes at least one 1 that is not included in any other prime implicant. Essential prime implicants must be used in any minimum sum of products expression. CSE271 Section 1 Fall 04 3 Lecture 12 Use K-map to Find the Minimum SOP Map method 1 v Find all essential prime implicants , starting with the most isolated 1's. Mark the m ...

• 12 Pages

lec7_2

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 ...

• 6 Pages

lec7_4

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) ...

• 15 Pages

quine-mccluskey-handout

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 ...

• 2 Pages

quine-mccluskey-handout

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 ...

• 2 Pages

practice4

Arizona, ECE 474

Excerpt: ... tion D {area = 8, delay = 9 } Option E {area = 1, delay = 8} Option F {area = 5, delay = 4 } Option G {area = 1, delay = 5} Option H {area = 9, delay = 6} 6. Use DeMorgan's Law to find the inverse of the following equations. Provide you answer in sum-of-products form. (a) (b) F(a, b, c, d) = a' + b + c'd F = a'bc' + ab 7. Given F(a, b, c) = ab + ac + a'b'c (a) (b) (c) (d) List the variables in F List the literals in F List the product terms in F List the minterms in F 8. Consider the equation F(a,b,c, d) = m(4, 5, 7, 12, 14, 15). Which of the following product terms are prime implicants of the equation (Hint: draw a Kmap). (a) (b) (c) (d) (e) a'bc'd' ab'c ad' bc'd' There are no prime implicants in this equation. 9. Using the K-map provided in Fig2, identify (a) (b) (c) (d) minterms implicants prime implicants essential prime implicants Fig 2: K-map used in Problem 9. bc a 0 1 00 01 11 10 1 1 1 1 1 1 10. What is the difference between an exact algorithm and a heuristic? 11. Perform two-level log ...

• 15 Pages

quine-mccluskey-handout

Columbia, CS 4823

Excerpt: ... CSEE W4823x Prof. Steven Nowick The Quine-McCluskey Method Handout 5 September 4, 2008 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. 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 are the prime implicants of the function. The rows are minterms of where the function is 1, called ON-set minterms. The goal of the method is to ...

• 2 Pages

hw6

Columbia, CS 4823

Excerpt: ... p, indicate if the transition is function-hazard-free. If the transition has a function hazard, explain why. (a) ABCD = 0000 1101 (b) ABCD = 0010 0110 (c) ABCD = 0001 0111 (d) ABCD = 1111 0010 (e) ABCD = 0101 1111 3. Exact Hazard-Free Two-Level Logic Minimization. Assume the following specied input transitions for the given Karnaugh map: (i) ABCD = 0001 0100 (ii) ABCD = 1100 1101 (iii) ABCD = 1001 1010 (iv) ABCD = 0111 1010 You are to nd an exactly minimal two-level implementation for the above function which is hazardfree for each of these 4 specied input transitions. Follow the steps in Handout #39, and as covered in class lectures. In particular, do the following: (a) Step 0: Identify Required Cubes and Privileged Cubes. In the K-map, add the four input transitions specied above (as arrows). Then draw the: (a) required cubes; (b) privileged cubes (and highlight their start points). (b) Step 1: Generate DHF- Prime Implicants . Recopy the Karnaugh map ...

• 1 Pages

final.obj

Valparaiso, ECE 221

Excerpt: ... ECE 221 Digital System Design Sectional Objectives Final - Dewey Chapters 1-5 and 8-10 After reading and studying Chapters 1-5 and 8-10 you should be able to: convert values between decimal, binary, hexadecimal and octal number systems. perform addition, subtraction and multiplication on binary, hexadecimal and octal numbers (unsigned and twos complement representations). represent a combinational expression in the following forms: Sum-of-Products (SOP) Product-ofSums (POS), minterm list, maxterm list. create a truth table for components that contain a combination of active-1 and active-0 inputs and outputs. use DeMorgans theorem to complement a function or convert a function to a new representation. use Karnaugh maps to minimize logic expressions in either SOP or POS form (recognize prime implicants , essential prime implicants and secondary prime implicants ). create a minimized implementation of a combinational boolean expression using simple gates (AND, OR, INV), complex ga ...