BOOLEAN ALGEBRA CLASS XII. Presented By : Dinesh Patel PGT CS KV IIT Powai

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1 BOOLEAN ALGEBRA CLASS II Presented By : Dinesh Patel PGT CS KV IIT Powai

2 Introduction Boolean Algebra is a set of rules and regulation which is suitable for Digital Circuits, whose answer is either True or False Open Close

3 History Mr. Aristotal constructed a compete system of formal logic to organize man s reasoning. Only Mathematician George Boole become able to manipulate these symbols to arrive the solution with mathematical system of Logic and produce new system the Algebra of logic.i.e. Boolean Algebra.

4 Truth Table Present all the possible values and result of logical variable with given combinations of values. No. of combination = 2 n [n is no. of variables / Options]

5 Eg. (1) I want to have tea (2) Tea is available I want to have tea Tea is available Result T T T T F F F T F F F F I want to have tea Tea is available Result OR

6 TAUTOLOG : If result of any logical statement or expression if always TRUE or 1 called Tautology. FALLAC : If result of any logical statement or expression if always FALSE or 0 called Fallacy.

7 Logical Operators NOT OR

8 NOT Operator eg. I want to have tea. NOT [I want to have tea] means I don t want to have tea Truth table for NOT operator I want to have tea Result

9 OR Operator OR operator being denoted as Logical Addition and symbol used for it is + Truth Table for OR operator

10 Operator operator being denoted as Logical Multiplication and symbol used for it is. Truth Table for OR operator

11 NOT, OR, Operator Truth Table for NOT, OR, operator

12 Truth Table for.z,.z and (Z) Truth Table for NOT, OR, operator Z Z.. Z Z (Z)

13 Basic Logic Gates Gate is an electronic circuit being operates on one or more signals to produce output signals. NOT NO T OR +.

14 OR ( + ) (. ) OR OR NOR OR ( )

15 Principal of Duality Using this Principal Dual relation can be obtained by : Changing each OR sign (+) with sign (.) Changing each sign (.) with OR sign (+) Replacing each 0 by 1 and 1 by 0. eg = = 0 [Dual relation]

16 BASIC THEOREMS OF BOOLEAN ALGEBRA

17 Property of 0 and = 0 OR 1 + = 1 1 OR 1 0. = = 1

18 Indempotence Law + = OR. =

19 Involution Law NOT NOT ( ) = Truth Table of Involution Law ( )

20 Complementarity Law + = 1 OR + = 1. = 0. =0

21 + = + Commutative Law OR = OR. =. =

22 + (+Z) = (+) + Z Associative Law Z OR OR = OR Z OR (Z) = () Z Z Z = Z

23 Truth Table of Associative Law + (+Z) = (+) + Z Z + Z + + ( + Z) ( + ) + Z (Z) = () Z can be obtain as above

24 Distributive Law (+Z) = + Z Z OR +Z = Z Z +(Z) = (+) (+Z) Z OR = Z OR OR + +Z

25 +(Z) = (+) (+Z) 3 rd Law OR R = OR R

26 Truth Table of Distributive Law (+Z) = + Z Z + Z Z ( + Z) + Z (Z) = (+) (+Z) can be obtain as above

27 Absorption Law (+) = OR + + = OR

28 (+) = OR De Morgan s Theorems 1 st Theorem R = R + (+) (.) = +. R () 2 nd Theorem = OR R +

29 Prove of 1 st De Morgan s Theorem ( + ) = Suppose : = 1 then ( + ) = 1 so (+) = 0 So ( + ) + = 1 [ = 1] = (( + ) + ). (( + ) + ) [+Z=(+)(+Z)] = ( + + ). ( + + ) = ( 1 + ). ( + 1 ) = 1. 1 = 1 PROVED Suppose : = 0 then ( + ) = 0 so (+) = 1 So ( + ). = 0 [ 0. 1 = 0] =. ( + ) [ (Z)=().Z)] = ( + + ). ( + + ) [+Z=(+)(+Z)] = ( 0. ) + (. 0 ) = = 0 PROVED Distribu tive Law Associat ive Law Distribu tive Law

30 Minterms It is a product of all the literals (with or without bar) within the logic system. Find the minterms of +. Sol. + = =. (+ ) +.(+ ) + =1 = = = = Comple mentry Law Insert Find the minterms of AB + C to make Sol. Write the terms AB + C equal to Insert s where letter is missing AB + C variables Write all the combinations of 1 st term i.e. AB : ABC, ABC Write all the combinations of 2 nd term i.e. C : ABC, A BC, A B C, AB C Add all the terms AB + C = ABC + ABC + ABC + A BC+ A B C + AB C Remove the duplicates AB + C = ABC + ABC + A BC + A B C + AB C

31 Minterm Designation Shorthand Find the minterms designation of Z Sol. Write the terms Z Substitute 1 and Find the Decimal equivalent 1 x x x 2 0 = = 6 Express decimal subscript of m = m 6 Thus Z = m 6 Find the minterms designation of AB CD Sol. Write the terms A B C D Substitute 1 and Find the Decimal equivalent 1 x x x x 2 0 = = 10 Express decimal subscript of m = m 10 Thus AB CD = m 10

32 Maxterm This is the sum of all literals (with or without bar) within the logic system. It is the opposite of Minterm, here Bar represent to 1 and non-bar to 0 means =0 = 1 Find the maxterms designation of AB CD Sol. Write the terms A B C D Substitute 1 and Find the Decimal equivalent 0 x x x x 2 0 = = 5 Express decimal subscript of m = M 5 Thus AB CD = M 5

33 Canonical Expression Boolean expression composed of either Minterm or Maxterms called Canonical Expression. It can be represented in two forms : (i) Sum-of-Product (ii) Product-of-Sum Sum-of-Product SOP of Two variables and and output is Z : Output (Z) Product Now add all the product term having output Z=1 : + + = Z So this is purely sum of minterm called Canonical Sum-of-Product

34 Sum-of-Product SOP of Three variables, and Z output is F : Output [F] will be 1 if the inputs of 1 will be odd instead of all 1s Z F Product Z Z Z Z Now add all the product term having output Z=1 : Z+ Z + Z + Z = F So this is purely sum of minterm called Canonical Sum-of-Product

35 Convert (( ) + )) into canonical SOP. Sol. : (( ) + )) = ( ). ( ) [(A+B) = A. B ] = ( + ) ( + Z) [(AB) = A + B ] = + ( Z) = (+ ) (Z+Z ) + (+ ) Z = (+ ) (Z+Z ) + Z + Z = Z(+ ) + Z (+ ) + Z + Z = Z+Z + Z +Z + Z + Z = Z + Z + Z + Z + Z + Z = Z + Z + Z + Z + Z De morgn s Fill variable with missing variables with or without bar Convert F= (0,1,2,5) into Canonical SOP suing Short Hand. Sol : F = m 0 + m 1 + m 2 + m 5 m 0 = 000» Z m 1 = 001» Z m 2 = 010» Z m 5 = 101» Z So Canonical form of expression is Z + Z + Z + Z

36 Product-of-Sum POS of Three variables, and Z output is F : Output [F] will be 1 if the inputs of 1 will be odd. Z F Product Z Z Z Z Now add all the product term having output Z=1 : ( + +Z). ( ++Z ) + (+ +Z ).(++Z) = F So this is purely sum of minterm called Canonical Sum-of-Product

37 Convert F= (0,1,2,5) into Canonical POS(Maxterm) using Short Hand. Sol : F = M 0. M 1. M 2. M 5 M 0 = 000» ++Z M 1 = 001» ++Z M 2 = 010» + +Z M 5 = 101» ++Z So Canonical form of expression is (++Z). (++Z ). (+ +Z).( ++Z ) Note : Above is Canonical where 1s and 0s is fixed at odd places but it can be asked only SOP or POS where the function may be any no. of 1s. eg. POS : F= (1,2,3,4,6) or SOP : F= (1,2,3,4,6) etc.

38 Qus. Simplify AB CD + AB CD + ABCD + ABCD = AB C(D +D) + ABC(D +D) = AB C.1 + ABC.1 [D +D=1] = AB C + ABC [.1=] = AC(B +B) = AC.1 = AC De Morgan s Qus. Simplify () + + = [() = + ] = + + [eliminate common] = + ( + ) [ = ] = + + = ( + ) + = 1 + [1 + = 1] = 1

39 Karnaugh Map [ K-Map] K-Map is the graphical representation of the fundamental products in a truth table. Where each squire represents the Minterm or Maxterm. K-Map of 02 variables &

40 K-Map of 03 Variable, & Z Z Z Z Z Z Z Z Z 0 1 Z 2 Z 3 Z Z 4 5 Z 6 Z 7

41 K-Map of 04 Variable W,, & Z W Z

42 K-Map of 04 Variable W,, & Z W Z Z Z Z Z W W Z W Z 0 1 W Z 3 W Z 2 W W Z W Z 4 5 W Z 7 W Z 6 W W Z W Z W Z 15 W Z 14 W W Z W Z 8 9 W Z 11 W Z 10

43 Reducing functions Through K-Map Taking example of 4 variables SOP *SOP + POS *POS +

44 Example : Reduce F(W,,,Z) = (0,2,7,8,10,15) W Z Pair : m 7 + m 15 = Z Quad : m 0 + m 2 + m 8 + m 10 m = Z so Z +

45 Now solve same func. With algebraic method Reduce F(W,,,Z) = (0,2,7,8,10,15) m0 = 0000 = W Z m2 = 0010 = W Z m7 = 0111 = W Z m8 = 1000 = W Z m10 = 1010 = W Z m15 = 1111 = WZ F= W Z + W Z + W Z + W Z + W Z + WZ = (WZ+W Z) +(W Z +W Z )+(W Z +W Z ) = Z(W+W ) + Z (W +W) + Z (W + W) = Z + Z + Z = Z + Z ( + ) = Z + solved

46 Example : Reduce F(W,,,Z) = (0,2,7,8,10,15) W+ +Z Pair : m 7. m 15 = ++ Z Quad : m 0. m 2. m 8. m 10 = +Z so (+ + Z). ( + )