Simplification Using Map Method

Transcription

1 Philadelphia University Faculty of Information Technology Department of Computer Science Computer Logic Design By Dareen Hamoudeh Dareen Hamoudeh 1 Simplification Using Map Method Dareen Hamoudeh 2 1

2 Why map method? Complex algebraic expression Complex Logic gates. Several algebraic expressions for same function. Function minimization using algebraic expression is awkward no specific rules to predict each step in the manipulative process. Map Method: Provides simple, straightforward procedure in minimizing functions. Dareen Hamoudeh 3 Map method (K-map) Also known as: Veitch diagram. Karnaugh map. The Diagram made up of squares, each square represents one minterm. Represents visual diagram of all possible ways a function may expressed in standard form. We will assume: the simplest algebraic expression is any one in(sop) or (POS) that has minimum numbers of literals. Dareen Hamoudeh 4 2

3 Two Variables Map Dareen Hamoudeh 5 Two Variables Map There are 4 minterms for two variables, so the map consists of 4 squares one for each minterm. We mark 0 and 1 for each row & column designate x and y: X: primed in row 0. Unprimed in row 1. y: primed in col. 0. Unprimed in col. 1. Dareen Hamoudeh 6 3

4 Two Variables Map We only mark the squares whose minterm belong to the given function. If we have F=x.y, it is equal to m 3,because it is = 1 when x=1 and y=1. so, we place 1 inside the square that belong to m 3 : Dareen Hamoudeh 7 Two Variables Map If we have F=x+y, then its minterms are: X+y=X.y+x.y +x.y=m 1 +m 2 +m 3 Dareen Hamoudeh 8 4

5 Three Variables Map Dareen Hamoudeh 9 Three Variables Map There are 8 minterms. Map consists of 8 squares. Minterms are arranged in a sequence similar to reflected code. Only one bit changes from 1 to 0 or from 0 to 1 in the sequence. There are 4 squares where each variable =1, and 4 squares where each variable =0. We write the variable with its letter symbol under the four squares where it is unprimed. Dareen Hamoudeh 10 5

6 Map in simplification Basic property for adjacent squares in the map: Any two adjacent squares differ by only one variable: primed in square & unprimed in the other. EX: In m5 & m7 : y is primed in m5 and unprimed in m7, from postulates m5+m7= xy z+ xyz = xy(y +y) = xy. Sum of minterms in adjacent squares can simplified to a single AND term with 2 literals. Dareen Hamoudeh 11 Example 1 Simplify the Boolean function using K-map F=x yz + x yz + xy z + xy z Solution: Dareen Hamoudeh 12 6

7 Represents x.y Represents x.y Dareen Hamoudeh 13 (m0 + m2) and (m4 + m6) Dareen Hamoudeh 14 7

8 Solution: Dareen Hamoudeh 15 Four adjacent squares Dareen Hamoudeh 16 8

9 Example 2 Dareen Hamoudeh 17 Solution: Dareen Hamoudeh 18 9

10 Example 3 Dareen Hamoudeh 19 Solution: Dareen Hamoudeh 20 10

11 Most minimization example F= Z Dareen Hamoudeh 21 Most minimization example F= x +y Dareen Hamoudeh 22 11

12 Four Variables Map Dareen Hamoudeh 23 Four Variables Map Dareen Hamoudeh 24 12

13 Four Variables Map Like three-variable map: we minimize function using Adjacent squares property. In addition the map is considered to lie on surface with the top and bottom edges as well as the right and the left, for Example: m 0 and m 2 form adjacent squares. m 3 and m 11 form adjacent squares. Dareen Hamoudeh 25 Four Variables Map Combination of adjacent squares is easily determined: Dareen Hamoudeh 26 13

14 Example 1 Solution: Dareen Hamoudeh 27 Dareen Hamoudeh 28 14

15 Example 2 Dareen Hamoudeh 29 Solution: Dareen Hamoudeh 30 15

16 Self Study & Practice Five Variables Map Dareen Hamoudeh 31 Five Variables Map Number of squares = number of minterms: 2 5 =32 Rows & columns are numbered in reflected code sequence. There are 16 squares where each variable =1, and 16 squares where each variable =0. As it consists of 2 four-variable maps. Each four-variable maps is recognized from the double line in the center: Each retains the previously defined adjacency, individually. In addition, the center lines considered as the center of a book, with each half of the map being a page Dareen Hamoudeh 32 16

17 Dareen Hamoudeh 33 Represented as 2 four-variable map Dareen Hamoudeh 34 17

18 Five Variables Map When the book is closed, two adjacent squares will fall one in each other, beside its four neighboring squares. Example: m 31 is adjacent to m 30,m 15,m 29,m 23 and m 27 Dareen Hamoudeh 35 Five Variables Map Dareen Hamoudeh 36 18

19 Solution Dareen Hamoudeh 37 NAND and NOR Implementation Dareen Hamoudeh 38 19

20 NAND and NOR Implementation Digital circuits are frequently constructed with only NAND or NOR gates. because these gates are easier to fabricate with electronic components. Because of the importance of NAND and NOR in the design of digital circuits. rules and procedures have been developed for the conversion from Boolean functions in terms of AND, OR and NOT into equivalent NAND or NOR logic diagrams. NAND and NOR are called universal gates. because any digital system or Boolean function can be implemented with only these gates. Dareen Hamoudeh 39 NAND and NOR Implementation Two-level implementation is presented here. There are two other graphic symbols for these gates, to facilitate conversions. Dareen Hamoudeh 40 20

21 NAND and NOR Implementation NAND equivalent symbols: Dareen Hamoudeh 41 NAND equivalent symbols Consists of an AND symbol followed by small circle. Dareen Hamoudeh 42 21

22 NAND equivalent symbols: OR symbol preceded by small circles in all the inputs. It follows DeMorgan s theorem where small circles denote complementation. Dareen Hamoudeh 43 NAND and NOR Implementation NOR equivalent symbols: Dareen Hamoudeh 44 22

23 NOR equivalent symbols Consists of an OR symbol followed by small circle. Dareen Hamoudeh 45 NOR equivalent symbols: AND symbol preceded by small circles in all the inputs. It follows DeMorgan s theorem where small circles denote complementation. Dareen Hamoudeh 46 23

24 NAND and NOR Implementation One-input NAND or NOR gate: Inverter. Three different graphic symbols for inverter: Dareen Hamoudeh 47 NAND and NOR Implementation NAND Simple Examples: Dareen Hamoudeh 48 24

25 NAND and NOR Implementation NOR Simple Examples: Dareen Hamoudeh 49 Implementing Functions using NAND First Way: Boolean function can be implemented with 2- levels of NAND gates by following the rules: Dareen Hamoudeh 50 25

26 Implementing Functions using NAND Second Way: Boolean function can be implemented with 3- levels of NAND gates by following the rules: 1) combine the 0 s in the map(obtain complement in SOP). 2) Implement the complement with 2-way NAND. 3) To obtain the normal output: insert one-input NAND (or inverter)to generate the true value of the output. Dareen Hamoudeh 51 Example Implement the following function with NAND gates: F(x,y,z)= (0,6) Dareen Hamoudeh 52 26

27 1)simplify function: Solution First Way F=x y z +xyz Dareen Hamoudeh 53 2) Solution Dareen Hamoudeh 54 27

28 Solution Second Way 1) simplify Complement (0 s): F = x y + xy + z Dareen Hamoudeh 55 2) Dareen Hamoudeh 56 28

29 Implementing Functions using NOR NOR function is the dual of the NAND. Require that the function simplified in POS. First Way: Boolean function can be implemented with 2- levels of NOR gates by following the rules: 1) combine the 0 s in the map(obtain complement). 2) Complement the function. 2) Implement the complement with 2-way NOR. Dareen Hamoudeh 57 Implementing Functions using NOR Second Way: Boolean function can be implemented with 3- levels of NOR gates by following the rules: 1) combine the 1 s in the map. 2) Obtain the function in SOP. 3) Complement the function to obtain the complement in POS (DeMorgan s). 4) Implement the complement with 2-way NOR. 5) To obtain the normal output: insert one-input NOR(or inverter)to generate the true value of the output. Dareen Hamoudeh 58 29

30 Example Implement the following function with NOR gates: F(x,y,z)= (0,6) Dareen Hamoudeh 59 1)combine the 0 s: Solution First Way F =x y+xy +z Dareen Hamoudeh 60 30

31 Solution 2) complement the function to obtain POS. F= (x+y )(x +y)z Dareen Hamoudeh 61 1) Combine (1 s): Solution Second Way F = x y z + xyz Dareen Hamoudeh 62 31

32 2) complement the function F =(x+y+z)(x +y +z) Dareen Hamoudeh 63 Rules for NAND & NOR Dareen Hamoudeh 64 32

33 Don t Care Conditions Dareen Hamoudeh 65 Don t Care Conditions Dareen Hamoudeh 66 33

34 Don t Care Conditions Dareen Hamoudeh 67 Example Simplify the Boolean function F(w,x,y,z)= (1,3,7,11,15) That has the do not-care conditions: d(w,x,y,z)= (0,2,5) Dareen Hamoudeh 68 34

35 Solution 1 Dareen Hamoudeh 69 Solution 2 Dareen Hamoudeh 70 35

36 We notice that don t care conditions are treated differently Dareen Hamoudeh 71 We can obtain simplified POS for the function: We combine the 0 s. We see that the only way to combine them is to include don t-care minterms 0 & 2: F = z +wy Take the complement of F F(w,x,y,z)=z(w +y)= (1,3,5,7,11,15) We include : minterms 0&2 with the 0 s. Minterm 5 with the 1 s. Dareen Hamoudeh 72 36

37 Example Dareen Hamoudeh 73 37