CHAPTER III BOOLEAN ALGEBRA

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1 CHAPTER III- CHAPTER III CHAPTER III R.M. Dansereau; v..

2 CHAPTER III-2 BOOLEAN VALUES INTRODUCTION BOOLEAN VALUES Boolean algebra is a form of algebra that deals with single digit binary values and variables. Values and variables can indicate some of the following binary pairs of values: ON / OFF TRUE / FALSE HIGH / LOW CLOSED / OPEN / R.M. Dansereau; v..

3 CHAPTER III-3 BOOL. OPERATIONS FUNDAMENTAL OPERATORS BOOLEAN VALUES -INTRODUCTION Three fundamental operators in Boolean algebra NOT: unary operator that complements represented as A, A, or A AND: binary operator which performs logical multiplication i.e. A ANDed with B would be represented as AB or A B OR: binary operator which performs logical addition i.e. A ORed with B would be represented as A+ B NOT AND OR A A A B AB A B A + B R.M. Dansereau; v..

4 CHAPTER III-4 BOOL. OPERATIONS BINARY BOOLEAN OPERATORS BOOLEAN OPERATIONS -FUNDAMENTAL OPER. Below is a table showing all possible Boolean functions given the twoinputs A and B. F N A B F F F 2 F 3 F 4 F 5 F 6 F 7 F 8 F 9 F F F 2 F 3 F 4 F 5 AB A B A + B B A AB Null A B A B Identity Inhibition A + B Implication R.M. Dansereau; v..

5 CHAPTER III-5 PRECEDENCE OF OPERATORS BOOLEAN OPERATIONS -FUNDAMENTAL OPER. -BINARY BOOLEAN OPER. Boolean expressions must be evaluated with the following order of operator precedence parentheses NOT AND OR Example: F = ( A( C+ BD) + BC)E F = A C + BD { + BC { E { R.M. Dansereau; v..

6 CHAPTER III-6 FUNCTION EVALUATION BOOLEAN OPERATIONS -PRECEDENCE OF OPER. Example : Evaluate the following expression when A =, B =, C = Solution Example 2: Evaluate the following expression when A =, B =, C =, D = Solution F = C+ CB + BA F = + + = + + = F = D( BCA + ( AB+ C) + C) F = ( + ( + ) + ) = ( + + ) = = R.M. Dansereau; v..

7 CHAPTER III-7 BASIC IDENTITIES BOOLEAN OPERATIONS -PRECEDENCE OF OPER. -FUNCTION EVALUATION X + = X X + = X + X = X X + X = ( X ) = X X + Y = Y + X X = X X = X X = XY = YX X + ( Y + Z) = ( X + Y) + Z XYZ ( ) = ( XY)Z XY ( + Z) = XY + XZ X + YZ = ( X + Y) ( X + Z) X X + XY = X + X Y = X + Y ( X + Y) = X Y X X = X XX ( + Y) = X XX ( + Y) = XY ( XY) = X + Y XY + X Z + YZ ( X + Y) ( X + Z) ( Y + Z) = XY + X Z = ( X + Y) ( X + Z) Identity Idempotent Law Complement Involution Law Commutativity Associativity Distributivity Absorption Law Simplification DeMorgan s Law Consensus Theorem R.M. Dansereau; v..

8 CHAPTER III-8 DUALITY PRINCIPLE -PRECEDENCE OF OPER. -FUNCTION EVALUATION -BASIC IDENTITIES Duality principle: States that a Boolean equation remains valid if we take the dual of the expressions on both sides of the equals sign. The dual can be found by interchanging the AND and OR operators along with also interchanging the s and s. This is evident with the duals in the basic identities. For instance: DeMorgan s Law can be expressed in two forms ( X + Y) = X Y as well as ( XY) = X + Y R.M. Dansereau; v..

9 CHAPTER III-9 FUNCTION MANIPULATION () -FUNCTION EVALUATION -BASIC IDENTITIES -DUALITY PRINCIPLE Example: Simplify the following expression F = BC + BC + BA Simplification F = B( C + C) + BA F = B + BA F = B( + A) F = B R.M. Dansereau; v..

10 CHAPTER III- FUNCTION MANIPULATION (2) -BASIC IDENTITIES -DUALITY PRINCIPLE -FUNC. MANIPULATION Example: Simplify the following expression F = A + AB + ABC + ABCD + ABCDE Simplification F = A + A( B + BC + BCD + BCDE) F = A+ B+ BC+ BCD+ BCDE F = A+ B+ B( C+ CD+ CDE) F = A+ B+ C+ CD+ CDE F = A+ B+ C+ C( D+ DE) F = A+ B+ C+ D+ DE F = A+ B+ C+ D+ E R.M. Dansereau; v..

11 CHAPTER III- FUNCTION MANIPULATION (3) -BASIC IDENTITIES -DUALITY PRINCIPLE -FUNC. MANIPULATION Example: Show that the following equality holds ABC ( + BC) = A+ ( B + C) ( B + C) Simplification ABC ( + BC) = A + ( BC+ BC) = A + ( BC) ( BC) = A + ( B + C) ( B+ C) R.M. Dansereau; v..

12 CHAPTER III-2 STANDARD FORMS SOP AND POS -BASIC IDENTITIES -DUALITY PRINCIPLE -FUNC. MANIPULATION Boolean expressions can be manipulated into many forms. Some standardized forms are required for Boolean expressions to simplify communication of the expressions. Sum-of-products (SOP) Example: F( A, B, C, D) = AB + BCD + AD Products-of-sums (POS) Example: FABCD (,,, ) = ( A+ B) ( B+ C + D) ( A + D) R.M. Dansereau; v..

13 R.M. Dansereau; v.. INTRO. TO COMP. ENG. CHAPTER III-3 STANDARD FORMS MINTERMS STANDARD FORMS -SOP AND POS The following table gives the minterms for a three-input system A B C ABC ABC ABC ABC ABC ABC ABC ABC m 2 m 3 m 4 m m m 5 m 6 m 7

14 CHAPTER III-4 STANDARD FORMS SUM OF MINTERMS STANDARD FORMS -SOP AND POS -MINTERMS Sum-of-minterms standard form expresses the Boolean or switching expression in the form of a sum of products using minterms. For instance, the following Boolean expression using minterms FABC (,, ) = ABC + ABC + ABC + ABC could instead be expressed as FABC (,, ) = m + m + m 4 + m 5 or more compactly FABC (,, ) = m(,, 4, 5) = one-set(,, 4, 5) R.M. Dansereau; v..

15 CHAPTER III-5 STANDARD FORMS MAXTERMS STANDARD FORMS -SOP AND POS -MINTERMS -SUM OF MINTERMS The following table gives the maxterms for a three-input system M M M 2 M 3 M 4 M 5 M 6 M 7 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 R.M. Dansereau; v..

16 CHAPTER III-6 STANDARD FORMS PRODUCT OF MAXTERMS STANDARD FORMS -MINTERMS -SUM OF MINTERMS -MAXTERMS Product-of-maxterms standard form expresses the Boolean or switching expression in the form of product of sums using maxterms. For instance, the following Boolean expression using maxterms FABC (,, ) = ( A+ B+ C) ( A+ B+ C) ( A+ B + C) could instead be expressed as FABC (,, ) = M M 4 M 7 or more compactly as FABC (,, ) = M(, 4, 7) = zero-set(, 4, 7) R.M. Dansereau; v..

17 CHAPTER III-7 STANDARD FORMS MINTERM AND MAXTERM EXP. STANDARD FORMS -SUM OF MINTERMS -MAXTERMS -PRODUCT OF MAXTERMS Given an arbitrary Boolean function, such as FABC (,, ) = AB + B( A + C) how do we form the canonical form for: sum-of-minterms Expand the Boolean function into a sum of products. Then take each term with a missing variable X and AND it with X + X. product-of-maxterms Expand the Boolean function into a product of sums. Then take each factor with a missing variable X and OR it with XX. R.M. Dansereau; v..

18 CHAPTER III-8 STANDARD FORMS FORMING SUM OF MINTERMS STANDARD FORMS -MAXTERMS -PRODUCT OF MAXTERMS -MINTERM & MAXTERM Example FABC (,, ) = AB + B( A + C) = AB + AB + BC = = = AB( C + C) + AB( C + C) +( A + A)BC ABC + ABC + ABC + ABC + ABC m(,, 4, 6, 7) A B C F Minterms listed as s in Truth Table R.M. Dansereau; v..

19 CHAPTER III-9 STANDARD FORMS FORMING PROD OF MAXTERMS STANDARD FORMS -PRODUCT OF MAXTERMS -MINTERM & MAXTERM -FORM SUM OF MINTERMS Example F( A, B, C) = AB + B( A + C) = AB + AB + BC = = = = ( A + B) ( A + B + C) ( A + B + C) ( A + B + CC) ( A+ B+ C) ( A+ B+ C) ( A + B + C) ( A + B + C) ( A + B + C) M( 2, 3, 5) A B C F (using distributivity) Maxterms listed as s in Truth Table R.M. Dansereau; v..

20 CHAPTER III-2 STANDARD FORMS CONVERTING MIN AND MAX STANDARD FORMS -MINTERM & MAXTERM -SUM OF MINTERMS -PRODUCT OF MAXTERMS Converting between sum-of-minterms and product-of-maxterms The two are complementary, as seen by the truth tables. To convert interchange the and, then use missing terms. Example: The example from the previous slides FABC (,, ) = m(,, 4, 6, 7) is re-expressed as F( A, B, C) = M( 2, 3, 5) where the numbers 2, 3, and 5 were missing from the minterm representation. R.M. Dansereau; v..

21 CHAPTER III-2 SIMPLIFICATION KARNAUGH MAPS STANDARD FORMS -SUM OF MINTERMS -PRODUCT OF MAXTERMS -CONVERTING MIN & MAX Often it is desired to simplify a Boolean function. A quick graphical approach is to use Karnaugh maps. 2-variable Karnaugh map 3-variable Karnaugh map 4-variable Karnaugh map CD AB A B BC A F = AB F = AB + C F = AB + CD R.M. Dansereau; v..

22 CHAPTER III-22 SIMPLIFICATION KARNAUGH MAP ORDERING STANDARD FORMS SIMPLIFICATION -KARNAUGH MAPS Notice that the ordering of cells in the map are such that moving from one cell to an adjacent cell only changes one variable. 2-variable Karnaugh map 3-variable Karnaugh map 4-variable Karnaugh map CD D D D AB BC A B C C C 3 2 A A A 3 2 A A 2 3 A A 8 9 B B B B C C B B B This ordering allows for grouping of minterms/maxterms for simplification. R.M. Dansereau; v..

23 CHAPTER III-23 SIMPLIFICATION IMPLICANTS STANDARD FORMS SIMPLIFICATION -KARNAUGH MAPS -KARNAUGH MAP ORDER Implicant Bubble covering only s (size of bubble must be a power of 2). Prime implicant Bubble that is expanded as big as possible (but increases in size by powers of 2). Essential prime implicant Bubble that contains a covered only by itself and no other prime implicant bubble. Non-essential prime implicant A that can be bubbled by more then one prime implicant bubble. CD AB R.M. Dansereau; v..

24 CHAPTER III-24 SIMPLIFICATION PROCEDURE FOR SOP SIMPLIFICATION -KARNAUGH MAPS -KARNAUGH MAP ORDER -IMPLICANTS Procedure for finding the SOP from a Karnaugh map Step : Form the 2-, 3-, or 4-variable Karnaugh map as appropriate for the Boolean function. Step 2: Identify all essential prime implicants for s in the Karnaugh map Step 3: Identify non-essential prime implicants for s in the Karnaugh map. Step 4: For each essential and one selected non-essential prime implicant from each set, determine the corresponding product term. Step 5: Form a sum-of-products with all product terms from previous step. R.M. Dansereau; v..

25 CHAPTER III-25 SIMPLIFICATION EXAMPLE FOR SOP () SIMPLIFICATION -KARNAUGH MAP ORDER -IMPLICANTS -PROCEDURE FOR SOP Simplify the following Boolean function F( A, B, C) = m(,, 4, 5) = ABC + ABC + ABC + ABC Solution: BC A zero-set( 2, 3, 6, 7) one-set(,, 4, 5) The essential prime implicants are B. There are no non-essential prime implicants. The sum-of-products solution is F = B. R.M. Dansereau; v..

26 CHAPTER III-26 SIMPLIFICATION EXAMPLE FOR SOP (2) SIMPLIFICATION -IMPLICANTS -PROCEDURE FOR SOP -EXAMPLE FOR SOP Simplify the following Boolean function FABC (,, ) = m(,, 4, 6, 7) = ABC + ABC + ABC + ABC + ABC Solution: BC A zero-set( 235,, ) one-set(,, 4, 6, 7) The essential prime implicants are AB and AB. The non-essential prime implicants are BC or AC. The sum-of-products solution is F = AB+ AB+ BC or F = AB+ AB+ AC. R.M. Dansereau; v..

27 CHAPTER III-27 SIMPLIFICATION PROCEDURE FOR POS SIMPLIFICATION -IMPLICANTS -PROCEDURE FOR SOP -EXAMPLE FOR SOP Procedure for finding the SOP from a Karnaugh map Step : Form the 2-, 3-, or 4-variable Karnaugh map as appropriate for the Boolean function. Step 2: Identify all essential prime implicants for s in the Karnaugh map Step 3: Identify non-essential prime implicants for s in the Karnaugh map. Step 4: For each essential and one selected non-essential prime implicant from each set, determine the corresponding sum term. Step 5: Form a product-of-sums with all sum terms from previous step. R.M. Dansereau; v..

28 CHAPTER III-28 SIMPLIFICATION EXAMPLE FOR POS () SIMPLIFICATION -PROCEDURE FOR SOP -EXAMPLE FOR SOP -PROCEDURE FOR POS Simplify the following Boolean function FABC (,, ) = M( 2, 3, 5) = ( A + B+ C) ( A+ B+ C) ( A+ B+ C) Solution: BC A zero-set( 2, 3, 5) one-set(,, 4, 6, 7) The essential prime implicants are A + B + C and A + B. There are no non-essential prime implicants. The product-of-sums solution is F = ( A+ B) ( A + B+ C). R.M. Dansereau; v..

29 CHAPTER III-29 SIMPLIFICATION EXAMPLE FOR POS (2) SIMPLIFICATION -EXAMPLE FOR SOP -PROCEDURE FOR POS -EXAMPLE FOR POS Simplify the following Boolean function Solution: The essential prime implicants are B + C and B+ C+ D. The non-essential prime implicants can be A + B + D or A+ C+ D. The product-of-sums solution can be either or F( A, B, C) = M(,, 5, 7, 8, 9, 5) F = ( B+ C) ( B+ C+ D) ( A + B + D) F = ( B+ C) ( B+ C+ D) ( A + C + D) zero-set(,, 5, 7, 8, 9, 5) one-set( 2, 3, 4, 6,,, 2, 3, 4) CD AB R.M. Dansereau; v..

30 CHAPTER III-3 SIMPLIFICATION DON T-CARE CONDITION SIMPLIFICATION -EXAMPLE FOR SOP -PROCEDURE FOR POS -EXAMPLE FOR POS Switching expressions are sometimes given as incomplete, or with don tcare conditions. Having don t-care conditions can simplify Boolean expressions and hence simplify the circuit implementation. Along with the zero-set () and one-set (), we will also have dc (). Don t-cares conditions in Karnaugh maps Don t-cares will be expressed as an X or - in Karnaugh maps. Don t-cares can be bubbled along with the s or s depending on what is more convenient and help simplify the resulting expressions. R.M. Dansereau; v..

31 CHAPTER III-3 SIMPLIFICATION DON T-CARE EXAMPLE () SIMPLIFICATION -PROCEDURE FOR POS -EXAMPLE FOR POS -DON T-CARE CONDITION Find the SOP simplification for the following Karnaugh map CD AB zero-set(,, 5, 7, 8, 9, 5) one-set( 2, 3, 4, 6,, 2) dc(, 3, 4) Taken to be Solution: X X X The essential prime implicants are BD and BC. There are no non-essential prime implicants. Taken to be The sum-of-products solution is F = BC+ BD. R.M. Dansereau; v..

32 CHAPTER III-32 SIMPLIFICATION DON T-CARE EXAMPLE (2) SIMPLIFICATION -EXAMPLE FOR POS -DON T-CARE CONDITION -DON T-CARE EXAMPLE Find the POS simplification for the following Karnaugh map CD AB Taken to be Solution: X X X The essential prime implicants are B + C and B + D. There are no non-essential prime implicants. The product-of-sums solution is F = ( B+ C) ( B + D). zero-set(,, 5, 7, 8, 9, 5) one-set( 2, 3, 4, 6,, 2) dc(, 3, 4) Taken to be R.M. Dansereau; v..