APPLİCATION OF BOOLEAN ALGEABRA

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1 APPLİCATION OF BOOLEAN ALGEABRA CHAPTER 4 Prof. Dr. Mehmet Akbaba Prof. Mehmet Akbaba Digital Logic 1

2 Multiplying Out and Factoring Expressions Given an expression in product-of-sums form, the corresponding sum-of-products expression can be obtained by multiplying out, using the two distributive laws: X(Y + Z) = XY + XZ (X + Y)(X + Z) = X + YZ (4-1) (4-2) In addition, the following theorem is very useful for factoring and multiplying out (X + Y)(X' +Z) = XZ + X'Y (4.3) Note that the variable that is paired with X on one side of the equation is paired with X' on the other side, and vice versa. Prof. Mehmet Akbaba Digital Logic 2

3 Proof: If X = 0, (3-3) reduces to Y(1 + Z) = Y or Y = Y If X = 1, (3-3) reduces to (1 + Y)Z = Z + 0. Y or Z = Z. Because the equation is valid for both X = 0 and X = 1, it is always valid. The following example illustrates the use of Theorem (3-3) for factoring: AB+A'C = (A + C)(A' + B) Note that the theorem can be applied when we have two terms, one which contains a variable and another which contains its complement. Prof. Mehmet Akbaba Digital Logic 3

4 Theorem (3-3) is very useful for multiplying out expressions. In the following example, we can apply (3-3) because one factor contains the variable Q, and the other factor contains Q'. (Q + AB')(CD+ Q') = QCD + Q'AB' If we simply multiplied out by using the distributive law, we would get four terms instead of two: (Q + AB')(CD + Q') = QCD + QQ' + AB'CD + AB'Q' Because the term A B' C D is difficult to eliminate, it is much better to use (3-3)instead of the distributive law. Prof. Mehmet Akbaba Digital Logic 4

5 In general, when we multiply out an expression, we should use (4-3) along with (4-1) and (4-2). To avoid generating unnecessary terms when multiplying out, (4-2) and (4-3) should generally be applied before (4-1), and terms should be grouped to expedite their application. [(X+A)(X+B)=X+AB, (X+A)(X +A)=XA+X A] What theorem was used to eliminate ABC? (Hint: let X = AC.) Prof. Mehmet Akbaba Digital Logic 5

6 In this example, if the ordinary distributive law (4-1) had been used to multiply out the expression by brute force, 162 terms would have been generated, and 158 of these terms would then have to be eliminated. Prof. Mehmet Akbaba Digital Logic 6

7 Example of Factoring The same theorems that are useful for multiplying out expressions are useful for factoring. By repeatedly applying (4-1), (4-2), and (4-3), any expression can be converted to a product-of-sums (POS) form. Prof. Mehmet Akbaba Digital Logic 7

8 = (A + B + C DE)(A + C DE + D' + E)(A' + C) =(A+B+C )(A+B+DE)(A+D +E)(A +C) = (A + B + C )(A + B + D)(A + B + E)(A + D' + E)(A' + C) (4-5; This is the same expression we started with in (4-4). Prof. Mehmet Akbaba Digital Logic 8

9 Exclusive-OR (EX-OR) Operations or EX-OR GATE The exclusive-or operation ( ) is defined as follows: 0 0=0 0 1=1 1 1=0 1 0=1 The truth table for X Y is Prof. Mehmet Akbaba Digital Logic 9

10 The truth table for X Y is X Y X Y EX-OR GATE SYMBOL Prof. Mehmet Akbaba Digital Logic 10

11 Prof. Mehmet Akbaba Digital Logic 11

12 From this table, we can see that X Y = 1 iff X = 1 anad Y=0 or X=0 and Y = 1, but not both =1. The ordinary OR operation, which we have previously defined, is sometimes called inclusive OR because X + Y = 1 iff X = 1 or Y = 1, or both. Exclusive OR can be expressed in terms of AND and OR. Because X Y = 1 iff X is 0 and Y is 1 or X is 1 and Y is 0, we can write, X Y = X' Y + XY' (4.6) Prof. Mehmet Akbaba Digital Logic 12

13 X Y = (X + Y)(XY)' = (X + Y)(X' + Y') = X' Y + XY' (4.7) In (4-7), note that (X Y)' = 1 if X and Y are not both 1. We will use the following symbol for an exclusive-or gate: Prof. Mehmet Akbaba Digital Logic 13

14 The following theorems apply to exclusive OR X 0=X X X=0 X Y=Y X 1=X X X =1 X (commutative law) (X Y) Z = X (Y Z)=X Y Z (Associative law) X(Y Z)= XY XZ (Distributive law) (X Y) =X Y =X Y=X Y +XY Prof. Mehmet Akbaba Digital Logic 14

15 Any of these theorems can be proved by using a truth table or by replacing X Y with one of the equivalent expressions from Equation (4-7). Proof of the distributive law follows: XY XZ=XY(XZ) +(XY) XZ=XY(X +Z )+(X +Y )XZ =XYZ +XY Z =X(YZ +Y Z)=X(Y Z) Prof. Mehmet Akbaba Digital Logic 15

16 EQUIVALENT OR EX-NOR OPERATION OR EX-NOR GATE The equivalence operation ( ) or EX-NOR operation (or EX-NOR GATE) is defined by truth table: EX-NOR=(EX-OR) (4.8) Therefore EX-NOR is defined As complement (NOT) of EX-OR Prof. Mehmet Akbaba Digital Logic 16

17 (4.9) Prof. Mehmet Akbaba Digital Logic 17

18 Prof. Mehmet Akbaba Digital Logic 18

19 Just as for exclusive-or, the equivalence (EX-NOR ) operation is commutative and associative. We will use the following symbol for an equivalence gate: Or another notation is EX-NOR GATE shown below A B Prof. Mehmet Akbaba Digital Logic 19

20 Because equivalence is the complement of exclusive-or, an alternate symbol for the equivalence gate is an exclusive-or gate with a complemented output: (A B) =(A B) F=(A B C)+(B AC ) Prof. Mehmet Akbaba Digital Logic 20

21 EX-NOR=(EX-OR) A B= (A B) =(AB +A B) = (A +B)(A+B ) =A A+A B +AB+BB = A B +AB Prof. Mehmet Akbaba Digital Logic 21

22 Example 1 (Örnek 1) Open the following expression (Aşağıdaki ifadeyi açınız) F=(A B C)+(B AC ) = Prof. Mehmet Akbaba Digital Logic 22

23 Example 2 (Örnek 2) Open the following expression (Aşağıdaki ifadeyi açınız) A B C=[A B +AB] C =(A B +AB)C +(A B +AB) C =(A B +AB)C +((A+B).(A +B ))C =A B C +ABC +A BC+AB C =A(BC +B C)+A (B C +BC) =A(B C)+A (B C) =B(A C)+B (A C) =C(A B)+C (A B) Prof. Mehmet Akbaba Digital Logic 23

24 CONSENSUS THEOREM The consensus theorem is very useful in simplifying Boolean expressions. Given an expression of the form XY + X' Z + YZ, the term YZ is redundant and can be eliminated to form the equivalent expression XY + X' Z. The term that was eliminated is referred to as the consensus term. Given a pair of terms for which a variable appears in one term and the complement of that variable in another, the consensus term is formed by multiplying the two original terms together, leaving out the selected variable and its complement. Prof. Mehmet Akbaba Digital Logic 24

25 For example, the consensus of ab and a' c is bc. Also the consensus of abd and b'de' is (ad)(de') = ade'. The consensus of terms ab'd and a' bd' is 0. The consensus theorem can be stated as follows: XY + X'Z + YZ = XY + X'Z Proof: XY + X'Z + YZ = XY + X'Z + (X + X')YZ = (XY + XYZ) + (X'Z + X'YZ) = XY(1 + Z) + X'Z(1 + Y) = XY + X'Z Prof. Mehmet Akbaba Digital Logic 25

26 The consensus theorem can be used to eliminate redundant terms from Boolean expressions. For example, in the following expression, b' c is the consensus of a' b' and ac, and ab is the consensus of ac and bc', so both consensus terms can be eliminated: (a b +ac+b c=a b +ac and ac+bc +ab=ac+bc ) Prof. Mehmet Akbaba Digital Logic 26

27 The brackets indicate how the consensus terms are formed. The dual form of the consensus theorem is: (X+Y)(X +Z)(Y+Z)=(X+Y)(X +Z) (4.9) Note again that the key to recognizing the consensus term is to first find a pair of terms, one of which contains a variable and the other its complement. In this case, the consensus is formed by adding this pair of terms together leaving out the selected variable and its complement. In the following expression, (a + b + d') is a consensus term and can be eliminated by using the dual consensus theorem: Prof. Mehmet Akbaba Digital Logic 27

28 The final result obtained by application of the consensus theorem may depend on the order in which terms are eliminated. Example: First, we eliminate BCD as shown. (Why can it be eliminated?) Now that BCD has been eliminated, it is no longer there, and it cannot be used to eliminate another term. Checking all pairs of terms shows that no additional terms can be eliminated by the consensus theorem. (4.10) Prof. Mehmet Akbaba Digital Logic 28

29 This time, we do not eliminate BCD; instead we eliminate two other terms by the consensus theorem. After doing this, observe that BCD can no longer be eliminated. Note that the expression reduces to four terms if BCD is eliminated first, but that it can be reduced to three terms if BCD is not eliminated. (C and C also D and D is considered) (4.11) Prof. Mehmet Akbaba Digital Logic 29

30 Sometimes it is impossible to directly reduce an expression to a minimum number of terms by simply eliminating terms. It may be necessary to first add a term using the consensus theorem and then use the added term to eliminate other terms. For example, consider the expression F=ABCD+B CDE+A B +BCE Consensus of ABCD+B CDE (choosing variable B) is ACDE. Adding this term gives: F=ABCD+B CDE+A B +BCE +ACDE Prof. Mehmet Akbaba Digital Logic 30

31 Now we can see that ABCD and B CDE are consensus terms they can be eliminated. Then, we can eliminate ABCD and B'CDE using the consensus theorem, and F reduces to: F=A B +BCE +ACDE The term ACDE is no longer redundant and cannot be eliminated from the final expression. Prof. Mehmet Akbaba Digital Logic 31

32 Algebraic Simplification of Switching Expressions In this section we review and summarize methods for simplifying switching expressions, using the laws and theorems of Boolean algebra. This is important because simplifying an expression reduces the cost of realizing the expression using gates. Later, we will learn graphical methods for simplifying switching functions, but we will learn algebraic methods first. In addition to multiplying out and factoring, three basic ways of simplifying switching functions are combining terms, eliminating terms, and eliminating literals. Prof. Mehmet Akbaba Digital Logic 32

33 1. Combining terms. Use the theorem XY + XY' = X to combine two terms. For example, abc d +abcd =abd (X=abd, Y=c) (4.12) When combining terms by this theorem, the two terms to be combined should contain exactly the same variables, and exactly one of the variables should appear complemented in one term and not in the other. Because X + X = X, a given term may be duplicated and combined with two or more other terms. For example, ab c+abc+a bc=ab c+abc+abc+a bc=ac+bc (=ac( b +b )+bc( a+a )=ac+bc) Prof. Mehmet Akbaba Digital Logic 33

34 The theorem still can be used, of course, when X and Y are replaced with more complicated expressions. For example, (Notice that (a+bc) =a (b +c ) X=a+bc X =a (b +c ) X+X =1 has been used) (a+bc)(d+e )+a (b +c )(d+e )=d+e =(d+e )[(a+bc)+a (b +c )]=d+e 2. Eliminating terms. Use the theorem X + XY = X to eliminate redundant terms if possible; then try to apply the consensus theorem (XY + X' Z + YZ = XY + X' Z) to eliminate any consensus terms. For example, Prof. Mehmet Akbaba Digital Logic 34

35 a b+a bc=a b (X=a b) a bc +bcd+a bd=a bc +bcd (X=c, Y=bd, Z=a b) (consensus theorem) [=a b c +b c d + a bbd = a bc +bcd)] (b=bb) Prof. Mehmet Akbaba Digital Logic 35

36 Eliminating literals. Use the theorem X + X' Y = X + Y to eliminate redundant literals. Simple factoring may be necessary before the theorem is applied. Example: (B+B C D =(B+B )(B+C D )=B+C D) A B+A B C +ABCD =A (B+B C )+ABCD =A (B+C D )+ABCD =B(A +ACD )+A C D =B(A +CD )+A C D =A B+BCD +A C D Prof. Mehmet Akbaba Digital Logic 36

37 The expression obtained after applying steps 1, 2, and 3 will not necessarily have a minimum number of terms or a minimum number of literals. If it does not and no further simplification can be made using steps 1,2, and 3, the deliberate introduction of redundant terms may be necessary before further simplification can be made. 4. Adding redundant terms. Redundant terms can be introduced in several ways such as adding xx', multiplying by (x + x'), adding yz to xy + x' Z, or adding xy to x. When possible, the added terms should be chosen so that they will combine with or eliminate other terms. Prof. Mehmet Akbaba Digital Logic 37

38 Example (4.13) The following comprehensive example illustrates the use of all four methods: Prof. Mehmet Akbaba Digital Logic 38

39 ACD (4.14) What theorems were used in steps 1,2,3, and 4 Prof. Mehmet Akbaba Digital Logic 39

40 If the simplified expression is to be left in a product-ofsums form instead of a sum-of-products form, the duals of the preceding theorems should be applied. (4.15) What theorems were used in steps 1,2, and 3? In general, there is no easy way of determining when a Boolean expression has a minimum number of terms or a minimum number of literals. Systematic methods for finding minimum sum-of-products and minimum productof-sums expressions will be discussed in coming chapters. Prof. Mehmet Akbaba Digital Logic 40

41 REFEENCES» 1. Prof. M. Akbaba Mantık Devreleri Notları 2. Hüseyin EKİZ, Mantık Devreleri, Değişim Yayınları, 4. Baskı, Thomas L. Floyd, Digital Fundamentals, Prentice-Hall Inc. New Jersey, M. Morris Mano, Michael D. Ciletti, Digital Design, Prentice-Hall, Inc.,New Jersey, /25/2015

UNIT 3 BOOLEAN ALGEBRA (CONT D)

UNIT 3 BOOLEAN ALGEBRA (CONT D) Spring 2011 Boolean Algebra (cont d) 2 Contents Multiplying out and factoring expressions Exclusive-OR and Exclusive-NOR operations The consensus theorem Summary of algebraic