MA251 Computer Organization and Architecture [ ]

Transcription

1 MA251 Computer Organization and Architecture [ ] Lecture 2: Simplification of Boolean Functions/ minimization of boolean functions Spring 2011 Partha Sarathi Mandal

2 Simplification of Boolean Functions

3 Examples of simplification A. B + A. B = A. (B + B) = A.1 = A (A + B).(A + B ) = (A + B.B ) = A + 0 = A In the reverse way, you can expand a expression, Like X.Y = X.Y.1 = X.Y (Z + Z ) = X.Y.Z + X.Y.Z

4 Simplification of Boolean Functions An implementation of a Boolean Function requires the use of logic gates. A smaller number of gates, with each gate (other then Inverter) having less number of inputs, may reduce the cost of the implementation. There are 3 methods for simplification of Boolean functions. 1. Algebraic Method 2. Karnaugh Map (K-Map) 3. Quine-McCluskey 4

5 Methods : Simplification of Boolean Functions The algebraic method by using Identities. The graphical method by using Karnaugh Map (K-Map) method. The tabular method by Quine-McCluskey The K-map method is easy and straightforward. A K-map for a function of n variables. Consists of 2 n cells, and, in every row and column, two adjacent cells should differ in the value of only one of the logic variables. 5

6 Examples of K-Maps Examples: Cell numbers are written in the cells. 2-variable K-Map x' 1 x 1 x x 1 x 2 x 1 x x 2 x 1 x 2 x 1 x 2 x 2 x

7 Examples: Examples of K-Maps Y= x 1 x 2 + x 1 x 2 = (0,2) x 2 x x 2 x

8 3-variable K-map 3-Variable K-Map: x 3 x 1 x x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 3 x 1 x Example: Y= x 1 x 2 x 3 + x 1 x 2 x 3 = (2,6) 8

9 3-variable K-map 3-Variable K-Map: x 3 x 1 x x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 1 x 2 x 3 x 3 x 1 x

10 4-variable K-Map 4-variable K-map x 3 x 4 x 1 x

11 K-map & minterm A variable or its complement is called a literal. For every K-map, each cell has a minterm associated with it. Thus for cell no. 13 in the 4-variable K-map, the minterm is A.B.C.D Or m 13 = x 1 x 2 x 3 x 4 12

12 K-map & maxterms For every K-map, each cell has one Maxterm associated with it. Thus for cell no.13 in the 4-variable K-map, M 13 = x 1 +x 2 +x 3 +x 4 By De Morgan s theorem, m i = M i ADJACENT minterms (Maxterms): Minterm which are identical, except for one variable, are considered to be adjacent to one another. In a K-map, the corresponding cells are said to be adjacent cells. 13

13 Adjacent minterms: Thus in K-4, Cell 0 is adjacent to cells 1, 4, 2 and 8. In a K-map, the corresponding cells in the top and the bottom rows are adjacent to each other. Similarly the corresponding cells in the leftmost column and the rightmost column are adjacent to each other. An Example: A function F, of 4 variables, is defined by the truth table given in the next slide. ( and again given in the next 3 slides): 14

14 Example: Truth Table 15

15 Example: TruthTable Dec number A B C D F

16 Example: TruthTable (continued) Dec number A B C D F

17 Example: TruthTable (continued) Dec Number A B C D F

18 Sum of Products form: The above table can be described by F = m(0, 2, 3, 5, 6, 7, 8, 10, 11, 14, 15) The function can be written as: F = A B C D + A B CD + A B CD + A BC D + A BCD + A BCD + AB C D + AB CD + AB CD + ABCD + ABCD (1) Each term on the RHS is a minterm. The above function can be simplified by using the Identities. 19

19 The graphical method steps: The graphical method steps: Insert 1 in those cells where the function F has a value of 1. Put 0 in the other cells. Examples: AB CD

20 Steps of graphical method (continued): Combine adjacent 1 s into group of 2 n each such that Each group contains only 1 s not 0 s. The group is not completely a part of a larger group. Choose the minimum number of the largest sized groups needed to cover all the 1 s. Each group is represented by an expression which is an intersection of the minterm in the group. The simplified solution is a logical OR of the expressions of all the groups chosen in steps 3 above. 21

21 Product of Sums Form: Using Maxterms For the same example, F= M(1,4,9,12,13) = (A + B + C+ D ).(A + B + C + D).(A + B + C +D ). (A + B + C + D).( A + B + C +D )..(2) The simplification process is a dual of the process for the SOP form. 22

22 Use of KARNAUGH MAP for Simplification of Logic Functions SOL: On reading the three sets of adjacent boxes of 8, 4 and 2 cells respectively, we get: F = C + B.D + A.B.D CD AB 24

23 SIMPLIFICATION using KARNAUGH MAP Exam 2: F= m(0,2,8,9,10,11,14,15) CD AB F= A.B +A.C+B.D 25

24 SIMPLIFICATION using KARNAUGH MAP Exam 3: Full-adder: A B C S Carry Carry= A.C+A.B+B.C S=A.B.C + A.B.C+A.B.C+A.B.C 26

25 Quine-McCluskey Algorithm Given Boolean function in the form of sum of Minterms find min sum of products. by Exploit the adjacency to find prime implicants. Find essential primes among primes via prime implicant chart.

26 Quine-McCluskey Algorithm The method involves two steps: Finding all prime implicants of the function. Use those prime implicants in a prime implicant chart to find the essential prime implicants of the function, as well as other prime implicants that are necessary to cover the function.

27 Some Definitions The definitions: Given a function F of n variables. Implicant: A single 1, or group of 1's combined together on a map, is called an implicant of F. Prime Implicant : An implicant which cannot further combine with other implicants is called a Prime Implicant of F. A prime implicate is a group of 1,2,4,8 that cannot be contained in any larger group. 30

28 Prime implicant Single 1 on a K-map represents a prime implicant if it is not adjacent to any other 1 Two adjacent 1 s on a K-map form a prime implicant if they are not contained in a group of four 1 s Four adjacent 1 s on a K-map form a prime implicant if they are not contained in a group of eight 1 s etc. 31

29 Essential Prime Implicant If minterm (i.e. a 1) can be covered by only one prime implicant, then that prime implicant is called an Essential Prime Implicant All essential prime implicants must be present in the minimum expression for F. Therefore find all essential prime implicants first. examine each 1 on map if that 1 + all adjacent 1's covered by one term only (i.e. encircled by one loop only), that term is essential cover remaining 1's by minimum set of prime implicants 33

30 Groups of the minterms 1. Forms Groups of the minterms is the following ways. Group 1: terms with no 1s. Group 2: terms which have only one 1. Group 3 terms which have only two 1s and so on.. The terms should be written in the ascending order of their values. Example: 0001, 1000, 0100, 0010 should be written as 0001, 0010, 0100, 1000 and forms group 2. Each group has an equal number of ones in the input combination. 2. Each minterm can only be adjacent to the minterms in the next group.

31 Combining Minterms 2. Any two minterms which differ from each other by only one variable can be combined, and the unmatched variable removed (by marking - or x). E.g., 0000 vs yields 0-00 or 0x vs yields -000 or x When used in a combination, mark with a check. If cannot be combined, mark with a star. These are the prime implicants. 4. Continue again. Pair up rows from adjacent regions if they differ by exactly one bit. 5. Repeat until nothing left.

32 Prime implicant chart The table is required to find essential prime implicant Prime-implicant table rows = prime implicants columns = minterms of the function place an "X" if the minterm is covered by the prime implicant. If column has a single X, than the implicant associated with the row is essential. It must appear in minimum cover as well as other prime implicants that are necessary to cover the function.

33 Determine the prime implicants of the function given below F(w,x,y,z)= (1,4,6,7,8,9,10,11,15) Grouping & Combining the minterms X001 * 01X0 * 100X 10X0 8, 10, 9, 11 10XX * 10XX * 011X * 10X1 101X X111 * 1X11 *

34 Prime implicants

35 Prime implicant chart Eliminate all columns covered by essential primes Find minimum set of rows that cover the remaining columns

36 Prime implicant chart the table shows that the selection of essential prime implicants covers all the minterms of the function except 7 and 15. these two minterms must be included by the selection of one or more prime implicants. Here xyz covers both minterms and hence one to be selected. We have thus found the following minimum set of prime implicants whose sum gives the required minimized function.

37 Problem set 1

38 Problem set 1 contd