Download presentation

Presentation is loading. Please wait.

Published byLeila Beck Modified over 2 years ago

1
CS 121 Digital Logic Design Gate-Level Minimization Chapter 3

2
Outline 3.1 Introduction 3.2 The Map Method 3.3 Four-Variable Map 3.4 Product of sums simplification 3.5 Don‘t Care Conditions 3.7 NAND and NOR Implementaion 3.8 Other Two-Level Implementaion 3.9 Exclusive-OR function

3
3.1 Introduction (1-1) Gate-Level Minimization refers to the design task of finding an optimal gate-level implementation of the Boolean functions describing a digital circuit. Notes about simplification of Boolean expression: Minimum number of terms and literals in each term (minimum number of gates and inputs in the digram). Reduce the complexity of the digital gates. The simplest expression is not unique. Simplification Methods: Algebraic minimization lack on specific rules. (section 2.4). Karnaugh map or K-map.

4
A Karnaugh map is a graphical tool for assisting in the general simplification procedure. Combination of 2, 4, … adjacent squares The relation is: Logic circuit ↔ Boolean function ↔ Truth table ↔ K- map ↔ conical form ↔ satndrad form. Conical form: ( sum of minterms, product of maxterms. Standrad form: ( simplifier : sum of product, product of sum 3.2 The Map Method (1-12)

5
3.2 The Map Method (2-12) Two-variable maps: Y’Y XX’

6
3.2 The Map Method (3-12) Rules for K-map: o We can reduce functions by circling 1’s in the K- map o Each circle represents a minterm reduction o Following circling, we can deduce minimized and- or form. Rules to consider o Every cell containing a 1 must be included at least once. o The largest possible “power of 2 rectangle” must be enclosed. o The 1’s must be enclosed in the smallest possible number of rectangles.

7
Example 1: F(X,Y) = XY’ + XY Two-Variable maps (cont.) 3.2 The Map Method (4-12) From the map, we see that F (X,Y) = X. Note: There are implied 0s in other boxes. This can be justified using algebraic manipulations: F(X,Y) = XY’ + XY = X(Y’ +Y) = X.1 = X 1X

8
Example 2: G(x,y) = m1 + m2 + m3 Two-Variable maps (cont.) 3.2 The Map Method (5-12) G(x,y) = m1 + m2 + m3 = X’Y + XY’ + XY From the map, we can see that : G = X + Y 1 11 X Y

9
Example 3: F = Σ(0, 1) Two-Variable maps (cont.) 3.2 The Map Method (6-12) Using algebraic manipulations: F = Σ(0,1) = x’y + x’y’ = x’ (y+y’) = x’ 1 1 X’ xyF001011100110xyF001011100110

10
3 variables 8 squares ( minterms). On a 3-variable K-Map: ◦ One square represents a minterm with three variables ◦ Two adjacent squares represent a product term with two variables ◦ Four “adjacent” terms represent a product term with one variables ◦ Eight “adjacent” terms is the function of all ones (logic 1). 3.2 The Map Method (7-12) Three-variable maps:

11
using algebraic manipulations: F = X’Y’Z’ + X’YZ’ + XY’Z’ + XYZ’ = Z’ (X’Y’ + X’Y + XY’ + XY) = Z’ (X’ (Y’+Y) + X (Y’+Y)) = Z’ (X’+ X) = Z’ 3.2 The Map Method (8-12) Three-variable maps (cont.): Example 1: F(X,Y) = X’Y’Z’ + X’YZ’ + XY’Z’ + XYZ’ Y Z x 1 1 1 1

12
Example 2: F=AB’C’ +A BC + ABC +A BC + A’B’C + A’BC’ three-Variable maps (cont.) 3.2 The Map Method (9-12) From the map, we see that F=A + BC + BC B C A 1 1 1 1 1 1 00 01 11 10 0 1

13
Example 4 : F (x, y, z)= Σm (2, 3, 6, 7) three-Variable maps (cont.) 3.2 The Map Method (10-12) using algebraic manipulations: F(x, y, z) = x’yz + xyz + x’yz’ + xyz’ = yz (x’ + x) + yz’ (x’ + x) = yz + yz’ = y (z + z’) = y y z x 11 11 00 01 11 10 0 1 Y

14
Example (3-1), (3-2) : three-Variable maps (cont.) 3.2 The Map Method (11-12)

15
Example (3-3), (3-4) : three-Variable maps (cont.) 3.2 The Map Method (12-12)

16
3.3 Four-Variables Map (1-9) 4 variables 16 squares ( minterms). On a 4-variable K-Map: Two adjacent squares represent a term of three literals. Four adjacent squares represent a term of two literals. Eight adjacent squares represent a term of one literal. Note: The larger the number of squares combined, the smaller the number of literals in the term.

17
3.3 Four-Variables Map (2-9) Flat Map Vs. Torus

18
3.3 Four-Variables Map (3-9) Example 1 (3-5) : F(w,x,y,z) = ∑ ( 0,1,2,4,5,6,8,9,12,13,14) y z w x 1 00 01 11 10 00 01 11 10 1 11 11 11 1 1 1 Y’ W’YZ’ XYZ’ F = y‘ + w‘yz‘ + xyz‘

19
3.3 Four-Variables Map (4-9) Example 2 (3-6) : F = A’B’C’ + B’CD’ + A’BCD’ + AB’C’ C D A B 1 00 01 11 10 1 F = B‘D‘ + B‘C‘ + A‘CD‘ 00 01 11 10 1 1 1 1 1 B’D’ B’C’ A’CD’

20
3.3 Four-Variables Map (5-9) Simplification using Prime Implicants o A Prime Implicant is a product term obtained by combining the maximum possible number of adjacent squares in the map. o If a minterm in a square is covered by only one prime implicant, that implicant is called Essential. o Prime Implicants and Essential Prime Implicants can be determined by inspection of a K-Map. o Notes: Two adjacent 1’s form prime implicant, if they are not within a group of four adjacent squares. Four adjacent 1’s form prime implicant, if they are not within a group of eight adjacent squares and so on.

21
3.3 Four-Variables Map (6-9) Simplification using Prime Implicants Example 1: F(A,B,C,D) = ∑ (0,2,3,5,7,8,9,10,11,13,15) Minterms covered by single prime implicant DB CB 11 1 1 11 B D A 11 11 1 ESSENTIAL Prime Implicants C BD CD BD DB 1 1 1 1 1 1 B C D A 11 11 1 AD BA

22
3.3 Four-Variables Map (7-9) Simplification using Prime Implicants Example 1: F(A,B,C,D) = ∑ (0,2,3,5,7,8,9,10,11,13,15) o Essential prim implicants: BD, B’D’ o Prime implicant: CD, B’C, AD, AB’. o The minterms that not cover by essential implicants are: m3, m9, m11. The simplified expression is optained from the sum of the essential implicants and other prime implicants that may be needed to cover any remaining minterms. o So this function can be written with these ways: F = BD + B’D’ + CD + AD F = BD + B’D’ + CD + AB’ F = BD + B’D’ + B’C + AD F = BD + B’D’ + B’C + AB’

23
3.3 Four-Variables Map (8-9) Simplification using Prime Implicants Example 2: F(W,X,Y,Z) = ∑ (0,2,3,8,9,10,11,12,13,14,15) X Y Z W 111 1 111 111 1 W X’Y X’Z’ Note: that all of these prime implicants are essential.

24
3.3 Four-Variables Map (9-9) Simplification using Prime Implicants Example 3: F(W,X,Y,Z) = ∑ (0,2,3,4,7,12,13,14,15) X Y Z W 1 1 1 11 1111 WX W’Y’Z’ W’X’Y W’YZ XYZ XY’Z’ Essential: WX Prime: XYZ, XY’Z’, W’Y’Z’, W’YZ, W’X’Y, W’X’Z’ W’X’Z’

25
3.5 Producut-of-Sum simplification (1-9) 1. Mark with 1’s the minterms of F. 2. Mark with 0’s the minterms of F’. 3. Circle 0’s to express F’. 4. Complement the result in step 3 to obtain a simplified F in product-of-sums form.

26
3.5 Producut-of-Sum simplification (1-9) Example 1: Simplify : F= ∑(0,1,2,5,8,9,10) in Product-of-Sums Form B C D A 1 1 1 1 11 1 0 0 0 0 0 00 0 0 CD AB BD’ F’ = AB + CD + BD’ F = (F’)’ = (A’+B’) + (C’+D’) + (B’+D)

27
3.5 Producut-of-Sum simplification (1-9) Example 2: Simplify : F(x, y, z) = (0, 2, 5,7)in Product-of-Sums Form y z x 0 00 01 11 10 0 1 0 00 XZ X’Z’ F’ = XZ + X’Z’ F = (F’)’ = (X’+Z’) + (X+Z)

28
3.6 Don't Cares Condition (1-4) Sometimes a function table or map contains entries for which it is known: The input values for the minterm will never occur, or The output value for the minterm is not used. Functions that have unspecified outputs for some input combinations are called incompletely specified functions. In these cases, the output value is defined as a “don't care” ( an “x” entry) assumed to be either 0 or 1. The choice between 0 and 1 is depending on the way the incompletely specified function is simplied. By placing “don't cares” in the function table or map, the cost of the logic circuit may be lowered.

29
3.6 Don't Cares Condition (2-4) Example : A logic function having the binary codes for the BCD digits as its inputs. Only the codes for 0 through 9 are used. The six codes, 1010 through 1111 never occur, so the output values for these codes are “x” to represent “don’t cares.”

30
3.6 Don't Cares Condition (3-4) Example (3.9) : F(W,X,Y,Z) = ∑ (1,3,7,11,15) d(W,X,Y,Z) = ∑ (0,2,5) X Y Z W 1 1 x x 1 0 x 1 1 0 0 0 0 0 0 0 F = YZ + W’Z X Y Z W 1 1 x x 1 0 x 1 1 0 0 0 0 0 0 0 F = YZ + W’X’

31
3.6 Don't Cares Condition (4-4) Example (3.9) : F(W,X,Y,Z) = ∑ (1,3,7,11,15) d(W,X,Y,Z) = ∑ (0,2,5) X Y Z W 1 1 x x 1 0 x 1 1 0 0 0 0 0 0 0 F’ = Z’ + WY’ F = Z ( W’ + Y)

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google