EE207: Digital Systems I, Semester I 2003/2004. CHAPTER 2-i: Combinational Logic Circuits (Sections )

Transcription

1 EE207: Digital Systems I, Semester I 2003/2004 CHAPTER 2-i: Combinational Logic Circuits (Sections )

2 Overview Binary logic and Gates Boolean Algebra Basic Properties Algebraic Manipulation Standard and Canonical Forms Minterms and Maxterms (Canonical forms) SOP and POS (Standard forms) Karnaugh Maps (K-Maps) 2, 3, 4, and 5 variable maps Simplification using K-Maps K-Map Manipulation Implicants: Prime, Essential Don t Cares 0 Circuits ( ) 2

3 Binary Logic Deals with binary variables that take 2 discrete values (0 and ), and with logic operations Three basic logic operations: AND, OR, NOT Binary/logic variables are typically represented as letters: A,B,C,,X,Y,Z 0 Circuits ( ) 3

4 Binary Logic Function F(vars) = expression set of binary variables Example: F(a,b) = a b + b G(x,y,z) = x (y+z ) Operators ( +,, ) Variables Constants ( 0, ) Groupings (parenthesis) 0 Circuits ( ) 4

5 Basic Logic Operators -bit logic AND resembles binary multiplication: 0 0 = 0, 0 = 0, 0 = 0, = -bit logic OR resembles binary addition, except for one operation: = 0, 0 + =, + 0 =, + = ( 0 2 ) 0 Circuits ( ) 5

6 Truth Tables for logic operators Truth table: tabular form that uniquely represents the relationship between the input variables of a function and its output 2-Input AND A B F=A B Input OR A B F=A+B A NOT F=A Circuits ( ) 6

7 Truth Tables (cont.) Q: Let a function F() depend on n variables. How many rows are there in the truth table of F()? A: 2 n rows, since there are 2 n possible binary patterns/combinations for the n variables 0 Circuits ( ) 7

8 Logic Gates Logic gates are abstractions of electronic circuit components that operate on one or more input signals to produce an output signal. 2-Input AND 2-Input OR NOT (Inverter) A B F = A B F A G A H B G = A+B H = A 0 Circuits ( ) 8

9 Timing Diagram t 0 t t 2 t 3 t 4 t 5 t 6 Input signals A B 0 0 Transitions Gate Output Signals F=A B G=A+B H=A Basic Assumption: Zero time for signals to propagate Through gates 0 Circuits ( ) 9

10 Combinational Logic Circuit from Logic Function Consider function F = A + B C + A B A combinational logic circuit can be constructed to implement F, by appropriately connecting input signals and logic gates: Circuit input signals from function variables (A, B, C) Circuit output signal function output (F) Logic gates from logic operations C A F B 0 Circuits ( ) 0

11 Combinational Logic Circuit from Logic Function (cont.) In order to design a cost-effective and efficient circuit, we must minimize the circuit s size (area) and propagation delay (time required for an input signal change to be observed at the output line) Observe the truth table of F=A + B C + A B and G=A + B C Truth tables for F and G are identical same function Use G to implement the logic circuit (less components) A B C F G Circuits ( )

12 Circuit from Logic Function (cont.) C A F B C B A G 0 Circuits ( ) 2

13 Boolean Algebra VERY nice machinery used to manipulate (simplify) Boolean functions George Boole (85-864): An investigation of the laws of thought Terminology: Literal: A variable or its complement Product term: literals connected by Sum term: literals connected by + 0 Circuits ( ) 3

14 Boolean Algebra Properties Let X: boolean variable, 0,: constants. X + 0 = X -- Zero Axiom 2. X = X -- Unit Axiom 3. X + = -- Unit Property 4. X 0 = 0 -- Zero Property 0 Circuits ( ) 4

15 Boolean Algebra Properties (cont.) Let X: boolean variable, 0,: constants 5. X + X = X -- Idepotence 6. X X = X -- Idepotence 7. X + X = -- Complement 8. X X = 0 -- Complement 9. (X ) = X -- Involution Unchanged in value following multiplication by itself 0 Circuits ( ) 5

16 The Duality Principle The dual of an expression is obtained by exchanging ( and +), and ( and 0) in it, provided that the precedence of operations is not changed. Cannot exchange x with x Example: Find H(x,y,z), the dual of F(x,y,z) = x yz + x y z H = (x +y+z ) (x +y + z) Dual does not always equal the original expression If a Boolean equation/equality is valid, its dual is also valid 0 Circuits ( ) 6

17 The Duality Principle (cont.) With respect to duality, Identities 8 have the following relationship:. X + 0 = X 2. X = X (dual of ) 3. X + = 4. X 0 = 0 (dual of 3) 5. X + X = X 6. X X = X (dual of 5) 7. X + X = 8. X X = 0 (dual of 8) 0 Circuits ( ) 7

18 More Boolean Algebra Properties Let X,Y, and Z: boolean variables 0. X + Y = Y + X. X Y = Y X -- Commutative 2. X + (Y+Z) = (X+Y) + Z 3. X (Y Z) = (X Y) Z -- Associative 4. X (Y+Z) = X Y + X Z 5. X+(Y Z) = (X+Y) (X+Z) -- Distributive 6. (X + Y) = X Y 7. (X Y) = X + Y -- DeMorgan s In general, ( X + X X n ) = X X 2 X n, and ( X X 2 X n ) = X + X X n 0 Circuits ( ) 8

19 . x + x y = x Absorption Property (Covering) 2. x (x+y) = x (dual) Proof: x + x y = x + x y = x (+y) = x = x QED (2 true by duality) 0 Circuits ( ) 9

20 Consensus Theorem. xy + x z + yz = xy + x z 2. (x+y) (x +z) (y+z) = (x+y) (x +z) -- (dual) Proof: xy + x z + yz = xy + x z + (x+x )yz = xy + x z + xyz + x yz = (xy + xyz) + (x z + x zy) = xy + x z QED (2 true by duality). 0 Circuits ( ) 20

21 Truth Tables (revisited) Enumerates all possible combinations of variable values and the corresponding function value Truth tables for some arbitrary functions F (x,y,z), F 2 (x,y,z), and F 3 (x,y,z) are shown to the right. x y z F F 2 F Circuits ( ) 2

22 Truth Tables (cont.) Truth table: a unique representation of a Boolean function If two functions have identical truth tables, the functions are equivalent (and vice-versa). Truth tables can be used to prove equality theorems. However, the size of a truth table grows exponentially with the number of variables involved, hence unwieldy. This motivates the use of Boolean Algebra. 0 Circuits ( ) 22

23 Boolean expressions-not unique 0 Unlike truth tables, expressions representing a Boolean function are NOT unique. Example: F(x,y,z) = x y z + x y z + x y z G(x,y,z) = x y z + y z The corresponding truth tables for F() and G() are to the right. They are identical! Thus, F() = G() Circuits ( ) xy z F G

24 Algebraic Manipulation Boolean algebra is a useful tool for simplifying digital circuits. Why do it? Simpler can mean cheaper, smaller, faster. Example: Simplify F = x yz + x yz + xz. F = x yz + x yz + xz = x y(z+z ) + xz = x y + xz = x y + xz 0 Circuits ( ) 24

25 Algebraic Manipulation (cont.) Example: Prove x y z + x yz + xyz = x z + yz Proof: x y z + x yz + xyz = x y z + x yz + x yz + xyz = x z (y +y) + yz (x +x) = x z + yz = x z + yz QED. 0 Circuits ( ) 25

26 Complement of a Function The complement of a function is derived by interchanging ( and +), and ( and 0), and complementing each variable. Otherwise, interchange s to 0s in the truth table column showing F. The complement of a function IS NOT THE SAME as the dual of a function. 0 Circuits ( ) 26

27 Complementation: Example Find G(x,y,z), the complement of F(x,y,z) = xy z + x yz G = F = (xy z + x yz) = (xy z ) (x yz) DeMorgan = (x +y+z) (x+y +z ) DeMorgan again Note: The complement of a function can also be derived by finding the function s dual, and then complementing all of the literals 0 Circuits ( ) 27

28 Canonical and Standard Forms We need to consider formal techniques for the simplification of Boolean functions. Minterms and Maxterms Sum-of-Minterms and Product-of-Maxterms Product and Sum terms Sum-of-Products (SOP) and Product-of-Sums (POS) 0 Circuits ( ) 28

29 Definitions Literal: A variable or its complement Product term: literals connected by Sum term: literals connected by + Minterm: a product term in which all the variables appear exactly once, either complemented or uncomplemented Maxterm: a sum term in which all the variables appear exactly once, either complemented or uncomplemented 0 Circuits ( ) 29

30 Minterm Represents exactly one combination in the truth table. Denoted by m j, where j is the decimal equivalent of the minterm s corresponding binary combination (b j ). A variable in m j is complemented if its value in b j is 0, otherwise is uncomplemented. Example: Assume 3 variables (A,B,C), and j=3. Then, b j = 0 and its corresponding minterm is denoted by m j = A BC 0 Circuits ( ) 30

31 Maxterm Represents exactly one combination in the truth table. Denoted by M j, where j is the decimal equivalent of the maxterm s corresponding binary combination (b j ). A variable in M j is complemented if its value in b j is, otherwise is uncomplemented. Example: Assume 3 variables (A,B,C), and j=3. Then, b j = 0 and its corresponding maxterm is denoted by M j = A+B +C 0 Circuits ( ) 3

32 Truth Table notation for Minterms and Maxterms Minterms and Maxterms are easy to denote using a truth table. Example: Assume 3 variables x,y,z (order is fixed) x y z Minterm x y z = m x y z = m 0 0 x yz = m 2 0 x yz = m xy z = m 4 0 xy z = m 5 0 xyz = m 6 xyz = m 7 Maxterm x+y+z = M 0 x+y+z = M x+y +z = M 2 x+y +z = M 3 x +y+z = M 4 x +y+z = M 5 x +y +z = M 6 x +y +z = M 7 0 Circuits ( ) 32

33 Canonical Forms (Unique) Any Boolean function F( ) can be expressed as a unique sum of minterms and a unique product of maxterms (under a fixed variable ordering). In other words, every function F() has two canonical forms: Canonical Sum-Of-Products (sum of minterms) Canonical Product-Of-Sums maxterms) (product of 0 Circuits ( ) 33

34 Canonical Forms (cont.) Canonical Sum-Of-Products: The minterms included are those m j such that F( ) = in row j of the truth table for F( ). Canonical Product-Of-Sums: The maxterms included are those M j such that F( ) = 0 in row j of the truth table for F( ). 0 Circuits ( ) 34

35 Example Truth table for f (a,b,c) at right The canonical sum-of-products form for f is f (a,b,c) = m + m 2 + m 4 + m 6 = a b c + a bc + ab c + abc The canonical product-of-sums form for f is f (a,b,c) = M 0 M 3 M 5 M 7 = (a+b+c) (a+b +c ) (a +b+c ) (a +b +c ). Observe that: m j = M j 0 Circuits ( ) a b c f

36 Shorthand: and f (a,b,c) = m(,2,4,6), where indicates that this is a sum-of-products form, and m(,2,4,6) indicates that the minterms to be included are m, m 2, m 4, and m 6. f (a,b,c) = M(0,3,5,7), where indicates that this is a product-of-sums form, and M(0,3,5,7) indicates that the maxterms to be included are M 0, M 3, M 5, and M 7. Since m j = M j for any j, m(,2,4,6) = M(0,3,5,7) = f (a,b,c) 0 Circuits ( ) 36

37 Conversion Between Canonical Forms Replace with (or vice versa) and replace those j s that appeared in the original form with those that do not. Example: f (a,b,c) = a b c + a bc + ab c + abc = m + m 2 + m 4 + m 6 = (,2,4,6) = (0,3,5,7) = (a+b+c) (a+b +c ) (a +b+c ) (a +b +c ) 0 Circuits ( ) 37

38 Standard Forms (NOT Unique) Standard forms are like canonical forms, except that not all variables need appear in the individual product (SOP) or sum (POS) terms. Example: f (a,b,c) = a b c + bc + ac is a standard sum-of-products form f (a,b,c) = (a+b+c) (b +c ) (a +c ) is a standard product-of-sums form. 0 Circuits ( ) 38

39 Conversion of SOP from standard to canonical form Expand non-canonical terms by inserting equivalent of in each missing variable x: (x + x ) = Remove duplicate minterms f (a,b,c) = a b c + bc + ac = a b c + (a+a )bc + a(b+b )c = a b c + abc + a bc + abc + ab c = a b c + abc + a bc + ab c 0 Circuits ( ) 39

40 Conversion of POS from standard to canonical form Expand noncanonical terms by adding 0 in terms of missing variables (e.g., xx = 0) and using the distributive law Remove duplicate maxterms f (a,b,c) = (a+b+c) (b +c ) (a +c ) = (a+b+c) (aa +b +c ) (a +bb +c ) = (a+b+c) (a+b +c ) (a +b +c ) (a +b+c ) (a +b +c ) = (a+b+c) (a+b +c ) (a +b +c ) (a +b+c ) 0 Circuits ( ) 40

41 Karnaugh Maps Karnaugh maps (K-maps) are graphical representations of boolean functions. One map cell corresponds to a row in the truth table. Also, one map cell corresponds to a minterm or a maxterm in the boolean expression Multiple-cell areas of the map correspond to standard terms. 0 Circuits ( ) 4

42 Two-Variable Map x x 2 0 x 2 x m m 2 m m 3 OR m 0 3 m m 2 m 3 NOTE: ordering of variables is IMPORTANT for f(x,x 2 ), x is the row, x 2 is the column. 0 Cell 0 represents x x 2 ; Cell represents x x 2 ; etc. If a minterm is present in the function, then a is placed in the corresponding cell. Circuits ( ) 42

43 Two-Variable Map (cont.) Any two adjacent cells in the map differ by ONLY one variable, which appears complemented in one cell and uncomplemented in the other. Example: m 0 (=x x 2 ) is adjacent to m (=x x 2 ) and m 2 (=x x 2 ) but NOT m 3 (=x x 2 ) 0 Circuits ( ) 43

44 2-Variable Map -- Example f(x,x 2 ) = x x 2 + x x 2 + x x 2 = m 0 + m + m 2 = x + x 2 s placed in K-map for specified minterms m 0, m, m 2 Grouping (ORing) of s allows simplification What (simpler) function is represented by each dashed rectangle? a = m 0 + m a 2 = m 0 + m 2 Note m 0 covered twice x 2 x Circuits ( ) 44

45 Minimization as SOP using K-map Enter s in the K-map for each product term in the function Group adjacent K-map cells containing s to obtain a product with fewer variables. Groups must be in power of 2 (2, 4, 8, ) Handle boundary wrap for K-maps of 3 or more variables. Realize that answer may not be unique 0 Circuits ( ) 45

46 Three-Variable Map yz x m 0 m m m 4 m 5 m 7 0 m 2 m 6 -Note: variable ordering is (x,y,z); yz specifies column, x specifies row. -Each cell is adjacent to three other cells (left or right or top or bottom or edge wrap) 0 Circuits ( ) 46

47 Three-Variable Map (cont.) The types of structures that are either minterms or are generated by repeated application of the minimization theorem on a three variable map are shown at right. Groups of, 2, 4, 8 are possible. group of 2 terms minterm 0 group of 4 terms Circuits ( ) 47

48 Simplification Enter minterms of the Boolean function into the map, then group terms Example: f(a,b,c) = ac + abc + bc Result: f(a,b,c) = ac + b 0 a bc Circuits ( ) 48

49 More Examples f (x, y, z) = m(2,3,5,7) f (x, y, z) = x y + xz yz X f 2 (x, y, z) = m (0,,2,3,6) f 2 (x, y, z) = x +yz 0 Circuits ( ) 49

50 Four-Variable Maps YZ WX m 0 m m 3 m 2 0 m 4 m 5 m 7 m 6 m 2 m 3 m 5 m 4 0 m 8 m 9 m 0 m Top cells are adjacent to bottom cells. Left-edge cells are adjacent to right-edge cells. Note variable ordering (WXYZ). 0 Circuits ( ) 50

51 Four-variable Map Simplification One square represents a minterm of 4 literals. A rectangle of 2 adjacent squares represents a product term of 3 literals. A rectangle of 4 squares represents a product term of 2 literals. A rectangle of 8 squares represents a product term of literal. A rectangle of 6 squares produces a function that is equal to logic. 0 Circuits ( ) 5

52 Example Simplify the following Boolean function (A,B,C,D) = m(0,,2,4,5,7,8,9,0,2,3). First put the function g( ) into the map, and then group as many s as possible. ab cd g(a,b,c,d) = c +b d +a bd 0 Circuits ( ) 52

53 5-Variable K-Map BC DE A= BC DE ABCDE A=0 A BCDE 0 Circuits ( ) 53

54 Implicants and Prime Implicants (PIs) An Implicant (P) of a function F is a product term which implies F, i.e., F(P) =. An implicant (PI) of F is called a Prime Implicant of F if any product term obtained by deleting a literal of PI is NOT an implicant of F Thus, a prime implicant is not contained in any larger implicant. 0 Circuits ( ) 54

55 Example Consider function f(a,b,c,d) whose K- map is shown at right. a b is not a prime implicant because it is contained in b. cd ab acd is not a prime implicant because it is contained in ad. b, ad, and a cd are prime implicants. a b ad b a cd acd 0 Circuits ( ) 55

56 Essential Prime Implicants (EPIs) If a minterm of a function F is included in ONLY one prime implicant p, then p is an essential prime implicant of F. An essential prime implicant MUST appear in all possible SOP expressions of a function To find essential prime implicants: Generate all prime implicants of a function Select those prime implicants that contain at least one that is not covered by any other prime implicant. For the previous example, the PIs are b, ad, and a cd ; all of these are essential. a cd ad b 0 Circuits ( ) 56

57 Another Example Consider f 2 (a,b,c,d), whose K-map is shown below. The only essential PI is b d. cd ab 0 Circuits ( ) 57

58 Systematic Procedure for Simplifying Boolean Functions. Generate all PIs of the function. 2. Include all essential PIs. 3. For remaining minterms not included in the essential PIs, select a set of other PIs to cover them, with minimal overlap in the set. 4. The resulting simplified function is the logical OR of the product terms selected above. 0 Circuits ( ) 58

59 Example f(a,b,c,d) = m(0,,2,3,4,5,7,4,5). ab cd Five grouped terms, not all needed. 3 shaded cells covered by only one term 3 EPIs, since each shaded cell is covered by a different term. F(a,b,c,d) = a b + a c + a d + abc 0 Circuits ( ) 59

60 Product of Sums Simplification Use sum-of-products simplification on the zeros of the function in the K-map to get F. Find the complement of F, i.e. (F ) = F Recall that the complement of a boolean function can be obtained by () taking the dual and (2) complementing each literal. OR, using DeMorgan s Theorem. 0 Circuits ( ) 60

61 POS Example ab cd F (a,b,c,d) = ab + ac + a bcd Find dual of F, dual(f ) = (a+b )(a+c )(a +b+c+d ) Complement of literals in dual(f ) to get F F = (a +b)(a +c)(a+b +c +d) (verify that this is the same as in slide 60) Circuits ( ) 6

62 Don't Care Conditions There may be a combination of input values which will never occur if they do occur, the output is of no concern. The function value for such combinations is called a don't care. They are usually denoted with x. Each x may be arbitrarily assigned the value 0 or in an implementation. Don t cares can be used to further simplify a function 0 Circuits ( ) 62

63 Minimization using Don t Cares Treat don't cares as if they are s to generate PIs. Delete PI's that cover only don't care minterms. Treat the covering of remaining don't care minterms as optional in the selection process (i.e. they may be, but need not be, covered). 0 Circuits ( ) 63

64 0 Example Simplify the function f(a,b,c,d) whose K-map is shown at the right. f = a c d+ab +cd +a bc or f = a c d+ab +cd +a bd The middle two terms are EPIs, while the first and last terms are selected to cover the minterms m, m 4, and m 5. (There s a third solution!) Circuits ( ) ab cd x x x x x x 0 0 x x x x x x 64

65 0 Another Example Simplify the function g(a,b,c,d) whose K-map is shown at right. g = a c + ab or g = a c +b d Circuits ( ) ab cd x 0 0 x 0 x x x 0 x x 0 x 0 0 x 0 x x x 0 x x 0 x 0 0 x 0 x x x 0 x x 0 65

66 Algorithmic minimization What do we do for functions with more than 4-5 variables? You can code up a minimiser (Computer- Aided Design, CAD) Quine-McCluskey algorithm Iterated consensus We won t discuss these techniques here 0 Circuits ( ) 66