Boolean Algebra...Boolean Algebra Page 1 of 18 Prepared By : Jaya Sood Boolean Algebra The Algebra...

Boolean Algebra

Page 1 of 18 Prepared By : Jaya Sood

Boolean Algebra The Algebra which deals with only two values that is either TRUE or FALSE is

called Boolean Algebra. It was developed by George Boole in 1854.

Boolean Variable : The variable which can have only two values i.e. TRUE or FALSE are called

Boolean or logical variables.

Boolean Constants : There are two Boolean constants ; TRUE i.e. 1 or FALSE i.e. 0.

Boolean Function : Boolean function consists of Boolean variables, Boolean constants, Boolean

operators, parenthesis and equal to operator.

Truth Table : A Truth Table is a tabular representation of all possible values of logical variables

with all the possible results. A Truth Table of n input variables will have 2n rows.

Boolean Operators : The operators which operate on Boolean variables and constants are

called Boolean Operators.

a. NOT : It is represented by ’(compliment), ~(negation) or -(bar). It inverts or reverses the

function value. (Truth Table for NOT operator)

X X’ ( ~X) ( X ) 0 1

1 0

b. AND : It is represented by a .(dot). It performs logical multiplication. The output is true if

all the inputs are true otherwise the output is false. (Truth Table of AND operator)

X Y X.Y

0 0 0

0 1 0

1 0 0

1 1 1

c. OR : It is represented by a +(plus). It is used to perform logical addition. The output is

True if either of the inputs is true. (Truth Table of OR operator)

X Y X+Y

0 0 0

0 1 1

1 0 1

1 1 1

d. NAND : It is compliment of AND and derived from NOT-AND. The output is false when all

the inputs are true. (Truth Table of NAND operator)

X Y (X.Y)’ 0 0 1

0 1 1

1 0 1

1 1 0

e. NOR : It is compliment of OR and derived from NOT-OR. The output is false if either of

the inputs is true. (Truth Table of NOR operator)

X Y (X+Y)’ 0 0 1

0 1 0

1 0 0

1 1 0

Boolean Algebra


Order Of Evaluation of logical operators in a Truth table from high to low is as follows:

1. ( )

2. NOT 3. AND

4. OR

Ques: Verify using Truth Table that X + X.Y = X for each X, Y in {0,1}.

This is a 2 variable expression, so we require 22 = 4 rows (combinations) of input.

X Y X.Y X+X.Y

0 0 0 0

0 1 0 0

1 0 0 1

1 1 1 1 Comparing the column X and X+X.Y, we find the contents of both the columns are identical. Hence proved.

Ques: Prepare the Truth Table for the following function :

F = A.(B’ + C) + C’

This is a 3 variable expression, so number of input rows = 23 = 8.

A B C B’ (B’+C) A.(B’+C) C’ F

0 0 0 1 1 0 1 1

0 0 1 1 1 0 0 0

0 1 0 0 0 0 1 1

0 1 1 0 1 0 0 0

1 0 0 1 1 1 1 1

1 0 1 1 1 1 0 1

1 1 0 0 0 0 1 1

1 1 1 0 1 1 0 1

Ques: Prepare a Truth Table for the following function : F = x’y’z w + x’ y z w

This is a 3 variable expression, so number of input rows = 24 = 16

X Y Z W X’ Y’ X’Y’ZW X’YZW F

0 0 0 0 1 1 0 0 0

0 0 0 1 1 1 0 0 0

0 0 1 0 1 1 0 0 0

0 0 1 1 1 1 1 0 1

0 1 0 0 1 0 0 0 0

0 1 0 1 1 0 0 0 0

0 1 1 0 1 0 0 0 0

0 1 1 1 1 0 0 1 1

1 0 0 0 0 1 0 0 0

1 0 0 1 0 1 0 0 0

1 0 1 0 0 1 0 0 0

1 0 1 1 0 1 0 0 0

1 1 0 0 0 0 0 0 0

1 1 0 1 0 0 0 0 0

1 1 1 0 0 0 0 0 0

1 1 1 1 0 0 0 0 0

Boolean Algebra


Duality Principle : It states that starting from a Boolean relation, you can device another

Boolean relation by :

1. changing each OR to AND and vice versa 2. changing each 0 to 1 and vice versa

Ques: Write the Dual of the following :

a. a + (a’.b) = a + b

dual : a.(a’+b) = a.b b. a + 0 = a

dual : a.1 = a

c. (a + 0).(b + 1).(c + 1 + 0)

dual : a.1 + b.0 + c.0.1

d. (u + w).(uv’ + w) dual : u.w + (u + v’).w

AXIOMS & THEOREMS OF BOOLEAN ALGEBRA

1. Closure Property : It states that Boolean operations (+ and .) on Boolean variables

and constants will always result into a Boolean value.

2. Identity law : It states that sum of anything and 0 or product of anything and 1 is

same as original.

A + 0 = A

A.1 = A

3. Commutative Law : It states that we can reverse the order of variables that are ORed

or ANDed together without changing truth of operation.

A + B = B + A

A.B = B.A

4. Distributive Law : ORing variables and then ANDing the result with a single variable

is equivalent to ANDing each variable with single variable and then Oring the products.

A. (B + C) = A.B + A.C

A + (B.C) = (A + B).(A + C)

5. Complement Law : It states that sum of Boolean variable with its complement is

always true and product of Boolean variable with its compliment is always false.

A + A’ = 1 (Tautology)

A.A’ = 0 (Fallacy)

6. Null Element : No matter what is the value of Boolean variable, its sum with one is

always 1 and its product with 0 is always 0.

A + 1 = 1 (Tautology)

A.0 = 0 (Fallacy)

7. Idempotency : It states that sum or product of Boolean quantity to itself results in

original.

A + A = A

A.A = A

8. Involution Law : It states that complementing a variable twice or even any even

number of times results in original value.

(A’)’ = A

Boolean Algebra


9. Associative Law : It states that we can associate, group of sum or product variables

together with parenthesis without altering the truth of expression.

A + (B + C) = (A + B) + C A.(B.C) = (A.B).C

10. Absorption Law : It is also known as covering and has three forms.

1st Law A + A.B = A

A.(A + B) = A 2nd Law A + A’B = A + B

A.(A’ + B) = A.B

3rd Law (A + B)(A’ + C)(B + C) = (A + B)(A’ + C)

A.B + A’.C + B.C = A.B + A’.C

Ques. Verify the Absorption Laws algebraically.

1st Law :

a. A + A.B = A

LHS = A.1 + A.B by Identity A.1=A = A.(1 + B) by Distributive A.(B+C)= A.B+A.C

= A.1 by Null element A+1=1

= A = RHS by Identity A.1=A

b. A.(A + B) = A LHS = A.A + A.B by Distributive A.(B+C)= A.B+A.C

= A + A.B by Idempotency A.A=A

= A.1 + A.B by Identity A.1=A

= A.(1 + B) by Distributive A.(B+C)= A.B+A.C

= A.1 by Null element A+1=1

= A = RHS by Identity A.1=A 2nd Law :

a. A + A’.B = A + B

LHS = (A+ A') . (A+B) by Distributive A.(B+C)= A.B+A.C

= 1. (A+B) by Complement A+A'=1

= A+B by Identity A.1=A

b. A.(A’ + B) = A.B

LHS = A.A’ + A.B by Distributive A.(B+C)= A.B+A.C

= 0 + A.B by Complement A.A’= 0

= A.B by Identity A+0=A 3rd Law :

a. (A + B)(A’ + C)(B + C) = (A + B)(A’ + C)

LHS= (A + B)(A’ + C)(B + C)(B+C) by Idempotency A.A=A

= (A + B)(B + C)(A’ + C)(B+C) by Commutative A.B=B.A

= ((B + A)(B + C)(C + A’)(C + B)) by Commutative A+B=B+A

= (B + A.C)(C + A’.B) by Distributive A+(B.C)=(A+B).(A+C) = (B + A.C).C + (B + A.C).A’.B by Distributive A.(B+C)=A.B+A.C

= B.C + A.C.C + A’B.B + A’.A.B.C by Distributive A.(B+C)=A.B+A.C

= B.C + A.C + A’B + 0 by Idempotency A.A=A, Complement A’A=0

= (A + B).C + A’.(A + B) by Distributive A.(B+C)=A.B+A.C

= (A + B)(A’ + C) by Distributive A+(B.C )=(A+B).(A+C) = RHS

Boolean Algebra


b. A.B + A’.C + B.C = A.B + A’.C

LHS = A.B + A’.C + B.C

= A.B + A’.C + B.C.1 by Identity A.1=A = A.B + A’.C + B.C.(A +A’) by Complement A+A'=1

= A.B + A’.C + A.B.C + A’.B.C by Distributive A.(B+C)=A.B+A.C

= A.B + A.B.C + A’.C + A’.B.C by Commutative A+B=B+A

= A.B.(1 + C) + A’.C.(1 + B) by Distributive A.(B+C)=A.B+A.C

= A.B.1 + A’.C.1 by Null element A+1=1 = A.B + A’.C by Identity A.1=A

(A + B)(A’ + C)(B + C) = (A + B)(A’ + C)

LHS= (A + B)(A’ + C)(B + C+ 0) by Identity A+0=A

= (A + B)(A’ + C)(B + C+ A.A’) by Complement A.A’= 0 = (A + B)(A’ + C)(B + C+ A) (B + C+ A’) by Distributive A+(B.C)=(A+B).(A+C)

= (A + B) (B + C+ A) (A’ + C) (B + C+ A’) by Commutative A.B=B.A

= (A + B) (A + B + C) (A’ + C) (A’ + C + B) by Commutative A+B=B+A

= (A + B + 0) (A + B + C) (A’ + C +0) (A’ + C + B) by Identity A+0=A

= ((A + B) + (0. C)). ( (A’ + C) + (0.B)) by Distributive A+(B.C )=(A+B).(A+C)

= ((A + B) + 0). ( (A’ + C) + 0) by Null Element A.0 = 0 = (A + B). (A’ + C) by Identity A+0=A

= RHS

Ques. Prove Algebraically: X’.Y +Z = (X’ + Y’ + Z)(X’ + Y + Z)(X + Y + Z)

RHS = (X’ + Y’ + Z)(X’ + Y + Z)(X + Y + Z) = (X’ + Y’ + Z)(X’.X + (Y + Z)) by Distributive A+(B.C )=(A+B).(A+C)

= (X’ + Y’ + Z)(0 + (Y + Z))

= (X’ + Y’ + Z).(Y + Z)

= (Y + Z).(X’ + (Y’ + Z))

= (Y + Z).X’ + (Y + Z).(Y’ + Z) by Distributive A.(B+C)=A.B+A.C = (Y + Z).X’ + (Y + Z).Y’ + (Y+Z).Z by Distributive A.(B+C)=A.B+A.C

= X’.Y + X’Z + Y.Y’ + Y’Z + Y.Z + Z.Z by Distributive A.(B+C)=A.B+A.C

= X’.Y + X’Z + 0 + Y’Z + Y.Z + Z

= X’.Y + X’Z + Z.(Y’ + Y) + Z

= X’.Y + X’Z + Z.1 + Z = X’.Y + Z(1 + X’)

= X’.Y + Z

LHS = X’.Y + Z

= (X’ + Z)(Y + Z)

= (X’ + Z + 0)(Y + Z + 0) = (X’ + Y.Y’ + Z)(X.X’ + Y + Z)

= ((X’ + Z) + Y.Y’).(X.X’ + (Y + Z))

= (Y + X’ + Z)(Y’ + X’ + Z)(X + Y + Z)(X’ + Y + Z)

= (X’ + Y + Z)(Y’ + X’ + Z)(X + Y + Z)

Q. Prove Algebraically : X’.Y + Y’.Z = X’.Y.Z + X’.Y.Z’ + X.Y’.Z + X’.Y’.Z

RHS = X’.Y.Z + X’.Y.Z’ + X.Y’.Z + X’.Y’.Z

= X’.Y.(Z + Z’) + Y’.Z.(X + X’)

= X’Y.1 + Y.Z.1

= X’Y + Y’Z

LHS = X’.Y + Y’.Z

= X’.Y.1 + Y’.Z.1

= X’.Y.(Z + Z’) + Y’.Z.(X + X’)

= X’.Y.Z + X’.Y.Z’ + X.Y’.Z + X’.Y’.Z

Boolean Algebra


Q. Prove Algebraically. (A + B).(A’ + C) = (A + B + C)(A + B + C’)(A’ + B + C)(A’ + B’ + C)

LHS = (A + B).(A’ + C)

= (A + B + 0).(A’ + 0 + C) = (A + B + C.C’).(A’ + B.B’ + C)

= ((A + B) + C.C’).((A’ + C) + B.B’)

= (A + B + C)(A + B + C’)(A’ + B + C)(A’ + B’ + C)

RHS = (A + B + C)(A + B + C’)(A’ + B + C)(A’ + B’ + C) = (A + B + C.C’).(A’ + B.B’ + C)

= (A + B + 0).(A’ + 0 + C)

= (A + B).(A’ + C)

11. De Morgan Theorem : states that the complement of a sum/product equals the product /sum of the complements.

(A + B)’ = A’. B’

(A . B)’ = A’ + B’

Break the line - change the sign (A + B) = A . B

Complement law states that - X + X’ = 1 also X.X’ = 0

Assuming that De Morgan theorem is true, then by proving the complement laws

through it. we shall also establish De Morgan laws : Now, for 1st De Morgan Law. Let X = (A+B), So by Complement law:

(i) (A + B) + (A + B)’ = 1 (ii) (A + B).(A + B)’ = 0

(i) (A + B) + (A + B)’ = 1

LHS = (A + B) + (A + B)’

= (A + B) + A’.B’ by De Morgan = (A + B + A’)(A + B + B’) by Distributive

= (1 + B)(1 + A) by Null element

= 1 . 1 = 1

(ii) (A + B).(A + B)’ = 0 LHS = (A + B).(A + B)’

= (A + B). A’.B’ by De Morgan

= A.A’.B + B.A’.B’ by Distributive

= 0.B’ + A’.0 by Null element

= 0 + 0 = 0

Now, for 2nd De Morgan Law. Let X = (A.B), So by Complement law:

(iii) (A.B) + (A.B)’ = 1

(iv) (A.B).(A.B)’ = 0

(iii) (A.B) + (A.B)’ = 1

LHS = (A.B) + (A.B)’

= A.B + (A’ + B’) by De Morgan

= (A + A’ + B’).(B + A’ + B’) by Distributive

= (1 + B’)(1 + A’)

= 1 . 1 = 1

(iv) (A.B).(A.B)’ = 0

LHS = (A.B).(A.B)’

= (A.B).(A’ + B’) by De Morgan

= A.B.A’ + A.B.B’ by Distributive

= 0.B + A.0 = 0 + 0 = 0

Boolean Algebra


Q. Find the complement of the Boolean function: F(A,B) = (A.B)’ + A’ + A.B

F’ = ( (A.B)’ + A’ + A.B )’

= ( (A’ + B’) + A’ + A.B )’ = ( (A’ + B’) + (A’ + A.B) )’

= ( A’ + B’ )’ . ( A’ + A.B )’

= (A’’ . B’’) . (A’’ . (A.B) ’)

= (A . B) . (A . (A’ + B’))

= A . B . ( A.A’ + A.B’) = A . B . ( 0 + A.B’)

= A . B . ( A.B’ )

= A . A . B . B’

= A .0 = 0

Q. Prove Algebraically. (a + b)’.(a’ + b’) = a’.b’

LHS = (a + b)’.(a’ + b’)

= (a’.b’). (a’ + b’)

= a’.b’.a’ + a’.b’.b’

= a’.b’ + a’.b’

= a’.b’

Literal : A literal is a variable or its complement eg. x, x’, y, y’ etc

MINTERM : Minterm is a Product of all literals with or without the bar within the logic

system. For any combination of inputs in the Truth Table if the value of a variable is 0 the corresponding literals are complemented in the product (Minterm).

Minterms for a 3 variable Function F(X, Y, Z)

X Y Z Minterm Shorthand

Notation

0 0 0 X’.Y’.Z’ m0

0 0 1 X’.Y’.Z m1

0 1 0 X’.Y.Z’ m2

0 1 1 X’.Y.Z m3

1 0 0 X.Y’.Z’ m4

1 0 1 X.Y’.Z m5

1 1 0 X.Y.Z’ m6

1 1 1 X.Y.Z m7

MAXTERM : Maxterm is the Sum of all the literals with or without the bar within the logic

system. For any combination of inputs in the Truth Table if the variable is 1, the corresponding literals are complemented in the sum (Maxterm)

Maxterms for a 3 variable Function F(X, Y, Z)

X Y Z Minterm Shorthand

Notation

0 0 0 ( X+Y+Z ) M0

0 0 1 ( X+Y+Z’ ) M1

0 1 0 ( X+Y’+Z ) M2

0 1 1 ( X+Y’+Z’ ) M3

1 0 0 ( X’+Y+Z ) M4

1 0 1 ( X’+Y+Z’ ) M5

1 1 0 ( X’+Y’+Z ) M6

1 1 1 ( X’+Y’+Z’ ) M7

Boolean Algebra


Finding MINTERM/MAXTERM shorthand notation

Q. Find the Minterm designation of X’.Y.Z’

Minterm = X’.Y.Z’

Binary word = 010

Decimal equivalent = 0*22 + 1*21 + 0*20

= 0 + 2 + 0

= 2 shorthand notation = m2

Q. Find Maxterm designation of (X’ + Y + Z’) Maxterm = (x’ + y + z’ )

Binary word = 101

Decimal Equivalent = 1*22 + 0*21 + 1*20

= 4 + 0 + 1

= 5 Shorthand notation = M5

CANONICAL EXPRESSION

A Boolean expression composed entirely either of Minterms or Maxterms is called canonical

expression. It is of two forms : 1. Canonical Sum-Of-Products Form (SOP)

2. Canonical Product-Of-Sum Form (POS)

Canonical SOP : Canonical expression composed of sum of Minterms (products of all literals) is

in canonical SOP form

Steps To Derive Canonical SOP step 1 : Prepare a Truth Table for Boolean function

step 2 : Add a new column and write Minterms ( Product of Literals where 0 is complemented)

for all those combinations where output is 1

step 3 : Summation (+) of above Minterms gives you the desired Canonical SOP

Q. Express in SOP the Boolean function F(X, Y, Z). The Truth Table for which is given below.

A B C F Minterm Shorthand

0 0 0 0

0 0 1 1 x’.y’.z m1

0 1 0 1 x’.y.z’ m2

0 1 1 0

1 0 0 1 x.y’.z’ m4

1 0 1 0

1 1 0 0

1 1 1 1 x.y.z m7

F = m1 + m2 + m4 + m7 F = ∑(1, 2, 4, 7)

F = x’.y’.z + x’.y.z’ + x.y’.z’ + x.y.z

Canonical POS : Canonical expression composed of product of Maxterms (sum of all literals) is

in canonical POS form.

Steps To Derive Canonical POS step1 : Prepare a Truth Table for Boolean function

step2 : Add a new column and write Maxterms ( Sum of Literals where 1 is complemented) for

all those combinations where output is 0

step3 : Product (.) of above Maxterms gives you the desired Canonical POS

Boolean Algebra


Q. Express in POS the Boolean function F(A, B, C). The Truth Table for which is given below.

A B C F Maxterms Shorthand

0 0 0 0 (x + y + z) M0

0 0 1 1

0 1 0 0 (x + y’ + z) M2

0 1 1 1

1 0 0 0 (x’ + y + z) M4

1 0 1 1

1 1 0 0 (x’ + y’ + z) M6

1 1 1 1

F = M0 + M2 + M4 + M6

F = π(0, 2, 4, 6)

F = (x + y + z)(x + y’ + z)(x’ + y + z)(x’ + y’ + z)

Algebraic Method To Derive Canonical SOP

step1 : Convert the given Boolean function into sum of products form

step2 : For every term where a variable is missing multiply with

(missingVar + (missingVar)’) and expand

step3 : Remove duplicate terms if any

Q. Find canonical SOP of the following function

F = (A + C)(B + C)

Above function is product of sum so we need to convert it to sum of products form.

F = C + A.B By distributive

F = 1.1.C + A.B.1 Now we need to introduce the missing variables in each term

F = (A + A’)(B + B’).C + A.B.(C + C’)

= ((A + A’).B + (A + A’).B’).C + A.B.C + A.B.C’

= (A.B + A’.B + A.B’ + 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’ = A.B.C + A’.B.C + A.B’.C + A’.B’.C + A.B.C’

= 111 + 011 + 101 + 001 + 110

= m7 + m3 + m5 + m1 + m6

F = ∑(1, 3, 5, 6, 7)

Algebraic Method To Derive Canonical POS

step1 : Convert the given Boolean function into product of sum form

step2 : For every term where a variable is missing multiply with

(missingVar.missingVar)’) and expand

step3 : Remove duplicate terms if any

Q. Find the canonical POS of the following function

F = A.C + B.C

Above function is in sum of products so we need to convert it to product of sums form.

F = C . ( A + B )

= ( 0 + 0 + C ) . ( A + B +0 ) Now we need to introduce missing variable

= (A.A’ + B.B’ + C).(A + B + C.C’)

= (A + B.B’ + C)(A’ + B.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’)

= (A + B + C)(A + B’ + C)(A’ + B + C)(A’ + B’ + C)(A + B + C’) = 000 . 010 . 100 . 110 . 001

= M0 . M2 . M4 . M6 . M1

= π(0, 1, 2, 4, 6)

Boolean Algebra


KARNOUGH MAP is a graphical representation of information in truth table. Karnough map is

extensively used for simplifying Boolean expression upto six variables. The map consist of 2n squares for n variable function.

Drawing K-MAP for SOP (Minterms)

Note : The numbering scheme for squares is in Grey Code in which only one bit differs in

successive binary numbers. 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000

0 1 3 2 6 7 5 4 12 13 15 14 10 11 9 8

Steps/Rules for SOP reduction using KMAP

1. Draw an empty KMAP for Minterms ( Product of Literals where 0 is complemented).

2. Map the given function by entering 1’s as the output in the squares whose Minterms are

present in Boolean function

3. First encircle all the possible octets, then the quads, next the pairs and finally singlets. Do not forget to roll the map both vertically and horizontally and overlap groups if the need be.

4. Remove redundant groups if any.

5. Write the reduced expressions for all groups and (+) them.

Overlapping : We are allowed to use a Minterm (1) in more than one group.

Octets : are groups of eight adjacent Minterms (1’s). In a quad three variables that change their state are removed.

Quads : are groups of four adjacent Minterms (1’s). In a quad two variables that change

their state are removed.

Pairs : are groups of two adjacent Minterms (1’s) . In a pair one variable that changes its

state is removed. Singlets : is only one Minterm (1). No variable is removed from singlet.

Map Rolling : while encircling the groups of octets, quads and pairs, the map is

considered as if its left and right edges and top and bottom edges are adjacent.

Redundant group : It is a group whose all Minterms (1’s) are overlapped by other

groups. Redundant groups are ignored.

Q. Obtain a simplified form a Boolean expression -

F(U, V, W, Z) = ∑(0, 1, 3, 5, 7, 9, 10, 11, 12, 13, 14, 15)

Octet 1 : Z

Quad 2 : UV

Quad 3 : UW

Pair 4 : U’V’W’ Singlet : Nil

Map Rolling: Done

Redundant Check: Done F(U, V, W, Z) = U’V’W’ + UW + UV + Z

(00) W’Z’

(01)

W’Z (11)

WZ (10)

WZ’

(00) U’V’ 1 0 1 1 1 3 2 (01)

U’V 4 1 5 1 7 6

(11)

UV 1 12 1 3 1 15 1 14 (10)

UV’ 8 1 9 1 11 1 10

3

4

1 2

Boolean Algebra


Q. Obtain a simplified form for a Boolean expression.

F(U, V, W, Z) = ∑(0, 1, 2, 4, 5, 7, 9, 10, 13, 15)

Octet : Nil

Quad 1: W’Z

Quad 2: U’W’ Quad 3 : VZ

Pair 4 : V’WZ’

Singlet : Nil , Map Rolling: Done, Redundant Check: Done F(U, V, W, Z) = V’WZ’ + VZ + W’Z + U’W’

Q. Reduce the following using KMAP -

a. F(U, V, W, Z) = ∑(0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 14)

Octet 1: V’ Octet 2: Z’

Quad : Nil , Pair: Nil , Singlet : Nil,

Map Rolling: Done, Redundant Check: Done F(U, V, W, Z) = V’ + Z’

b. F(U, V, W, Z) = ∑(0, 1, 2, 6, 8, 9, 10, 14)

Redundant Quad

obtained by rolling the Map

Vertically and Horizontally

simultaneously.

(00) W’Z’

(01)

W’Z (11)

WZ (10)

WZ’

(00) U’V’ 1 0 1 1

3 1 2 (01)

U’V 1 4 1 5 1 7 6 (11)

UV 12 1 13 1 15 14

(10)

UV’ 8 1 9 11 1 10

(00) W’Z’

(01)

W’Z (11)

WZ (10)

WZ’

(00) U’V’ 1 0 1 1 1 3 1 2 (01)

U’V 1 4 5 7 1 6 (11)

UV 1 12 13 15 1 14 (10)

UV’ 1 8 1 9 1 11 1 10

(00) W’Z’

(01)

W’Z (11)

WZ (10)

WZ’

(00) U’V’ 1 0 1 1 3 1 2 (01)

U’V 4 5 7 1 6

(11)

UV 12 13 15 1 14

(10)

UV’ 1 8 1 9 11 1 10

1

2

1

2

3

4

2

3

1

Boolean Algebra


Quad 1: V’W’

Quad 2: WZ’

Quad : Nil , Pair: Nil , Singlet : Nil, Map Rolling: Done, Redundant Check: Done

F(U, V, W, Z) = V’W’ +WZ’

c. F(A, B, C, D) = ∑(3, 4, 5, 13, 15)

Octet : Nil

Quad : Nil

Pair 1 : UVZ Pair 2 : U’VW’

Singlet 3 : U’V’WZ

Map Rolling: Done, Redundant Check: Done F(U, V, W, Z) = UVZ + U’VW’+ U’V’WZ

Drawing K-MAP for POS (Maxterms)

Steps/Rules for POS reduction using KMAP 1. Draw an empty KMAP for Maxterms ( Sum of Literals where 1 is complemented).

2. Map the given function by entering 0’s as the output in the squares whose Maxterms are

present in Boolean function

3. First encircle all the possible octets, then the quads, next the pairs and finally singlets. Do

not forget to roll the map both vertically and horizontally and overlap groups if the need be. 4. Remove redundant groups if any.

5. Write the reduced expressions for all groups and (.) them.

Q. Find the simplified POS expression of given function

F(A, B, C, D) = π (0, 1, 3, 5, 6, 7, 10, 14, 15)

Octet : Nil Quad 1: (A +D’)

Quad 2: (B’+C’)

Pair 3 : (A+B+C)

Pair 4: (A’+C’+D)

Singlet : Nil, Map Rolling: Done, Redundant Check: Done

F(A,B,C,D)= (A +D’). (B’+C’). (A+B+C). (A’+C’+D)

(00) W’Z’

(01)

W’Z (11)

WZ (10)

WZ’

(00) U’V’

0 1 1 3 2 (01)

U’V 1 4 1 5 7 6 (11)

UV 12 1 13 1 15 14

(10)

UV’ 8 9 11 10

(00) (C+D)

(01)

(C+D’) (11)

(C’+D’) (10)

(C’+D)

(00) (A+B) 0 0 0 1 0 3 2

(01)

(A+B’) 4 0 5 0 7 0 6

(11)

(A’+B’) 12 13 0 15 0 14

(10)

(A’+B) 8 9 11 0 10

1

2

3

4

1

2

3

Boolean Algebra


Changing POS into reduced SOP using K-Map

Q. Obtain a simplified form for the Boolean expression using KMAP

F(U, V, W, Z) = π(1, 3, 5, 7, 12, 14)

Changing into SOP form

F(U, V, W, Z) = ∑(0, 2, 4, 6, 8, 9, 10, 11, 13, 15)

Octet : Nil

Quad 1: A’.D’

Quad 2: A.D Quad 3: A.B’

Singlet : Nil, Map Rolling: Done, Redundant Check: Done

F(A, B, C, D) = A’D’ + AD + A.B’

Q. Reduce the following 3-variable functions using K-MAP 1. F(A,B,C) = ∑(0, 1, 2, 4, 6, 7)

Quad 1 : C’

Pair 2 : A’B’

Pair 3 : A.B

F(U,V.) = C’ + A’.B’ + A.B

2. F(U,V,W) = π(1, 3, 6, 7)

Pair 1: (A+C’) Pair 2 : (A’+B’)

F(A,B,C) = (A+C’). (A’+B’)

(00) W’Z’

(01)

W’Z (11)

WZ (10)

WZ’

(00) U’V’ 1 0

1 3 1 2 (01)

U’V 1 4 5 7 1 6 (11)

UV 12 1 13 1 15 14

(10)

UV’ 1 8 1 9 1 11 1 10

(00) B’C’

(01)

B’C (11)

BC (10)

BC’

(0) A’ 1 0 1 1 3 1 2 (1)

A 1 4 5

1 7 1 6

(00) (B+C)

(01)

(B+C’) (11)

(B’+C’) (10)

(B’+C)

(0) (A)

0 0 1 0 3

2

(1)

(A’) 4 5

0 7 0 6

1

3

2

1

2

1

2 3

Boolean Algebra


LOGIC GATES : A gate is an electronic circuit which operates on one or more signals to produce

an output signal. Gates are digital circuits because the inputs and outputs can take only two

values either low voltage (denoted by 0) or high voltage (denoted by 1). They are also called logic circuits because they can be analysed with Boolean algebra.

There are following types of gates :

NOT f( x) = x’

AND f(x, y) = x.y

OR f(x, y) = x+y

NAND f(x, y) = (x.y)’

NOR f(x, y) = (x+y)’

XOR f(x, y) = x’y +x.y’

Exclusive OR (XOR) gate produces a

high output if odd number of input

signals are high.

Truth Table for XOR Gate

x y x’ y’ x’y xy’ f= x xor y

0 0 1 1 0 0 0

0 1 1 0 1 0 1

1 0 0 1 0 1 1

1 1 0 0 0 0 0

f

Boolean Algebra


Q. Design a Logic circuit to realise the following :

F(A, B, C, D) = A.B + A.C’ + A’.B’.C

= (A AND B) OR (A AND (NOT C)) OR ((NOT A) AND (NOT B) AND C)

Q. Draw the logic circuit diagram for the given function.

F(X, Y, Z) = (X + Y) . (X’ + Z’) . (Y + Z)

= (X OR Y) AND ((NOT X) OR (NOT Z)) AND (Y OR Z)

Q. Write the equivalent Boolean function for the logic circuit : 1

Ans : F = (X + Y’) . (X’ + Y) . (X’ + Y’)

2.

Ans : F = U.V’ + U’.V + U’.V’

Boolean Algebra


Universal Gates: NAND and NOR gates are called universal gates because all the other switching functions like NOT, AND, OR can be easily implemented using either NAND or NOR.

NAND - to - NAND logic

Tip: Convert the Given expression into Sum of Products (SOP) and complement twice to get

NAND-to-NAND logic

1. NOT operation using NAND’s

F = X’

F = (X.X)’

2. AND operation using NAND’s F = X.Y

F = X.Y + X.Y

F’’= ( ( X.Y + X.Y )’ )’

= ( (X.Y)’ . (X.Y)’ )’

= (X NAND Y) NAND (X NAND Y)

3. OR operation using NAND’s

F = X + Y

F = X.X + Y.Y

F’’= ( (X.X + Y.Y )’ )’

F = ( (X.X)’ . (Y.Y)’ )’ F = (X NAND X) NAND (Y NAND Y)

NOR - to - NOR logic

Tip: Convert the Given expression into Product of Sums(POS) and complement twice to get NOR - to - NOR logic

1. NOT operation using NOR’s

F = X’ F = (X + X)’

2. AND operation using NOR’s

F = X.Y

F = (X+X) . (Y+Y)

F = ( ( (X+X) . (Y+Y) )’ )’

F = ( (X + X)’ + (Y + Y)’ )’ F = (X NOR X) NOR (Y NOR Y)

3. OR operation using NOR’s F = X + Y

F = (X + Y) . (X + Y)

F = ( ( (X + Y) . (X + Y) )’ )’

F = ( (X + Y)’ + (X + Y)’ )’

F = (X NOR Y) NOR (X NOR Y)

Boolean Algebra


Q. Draw the LCD using NAND gates only.

F(X, Y, Z) = Y.Z + X.Z

F’’ = ( (Y.Z + X.Z)’ )’

F = ( (Y.Z)’ . (X.Z)’ )’

= (Y NAND Z) NAND (X NAND Z)

Q. Draw the LCD using NAND gates only.

F(A, B, C) = A.B’.C + B.C’

F’’ = ( ( A.B’.C + B.C’ )’ )’

F = ( (A.B’.C)’ . (B.C’)’ )’

= ( A.(B.B)’.C)’ . (B.(C.C)’)’ )’

= ( A NAND (B NAND B) NAND C) NAND ( B NAND (C NAND C))

A B C

Q. Represent (X + Y’).Z with the help of NAND’s only.

F = (X + Y’).Z

Converting the given function into SOP form

by distributive law : F = X.Z + Y’.Z

F’’ = (( X.Z + Y’Z )’)’

F = ( (X.Z)’ . ((Y.Y)’.Z)’ )’

= ( X NAND Z) NAND (( Y NANDY) NAND

Z)

Q. Draw LCD for given function using NOR gates

F = A.(B’ + C)

F’’ = (( A . (B’+C) )’)’

F = ( (A)’ + (B’+C)’ )’

F = ( (A+A)’ + ( (B+B)’ + C)’ )’

F = (A NOR A) NOR ((B NOR B) NOR C)

Boolean Algebra


Q. Represent the given function in NOR - to - NOR LCD

F = (A’ + B’ + C’).(A + B’+ C’).(A + B + C’)

F’’=(( (A’+B’+C’) . (A+B’+C’) . (A+B+C’) )’)’ F’’=( (A’+B’+C’)’ + (A+B’+C’)’ + (A+B+C’)’ )’

F=(((A+A)’+(B+B)’+(C+C)’)’+(A+(B+B)’+(C+C)’)’+(A+B+(C+C)’)’)’

= ((A NOR A) NOR (B NOR B) NOR( C NOR C)) NOR ( A NOR ( B NOR B) NOR ( C NOR C)) NOR ( A NOR B NOR ( C NOR C))

Q. Draw LCD using NOR gates.

F = X.Y’ + Z

Convert the given function into POS form using distributive law :

F = (X+Z) . (Y’+Z)

F’’= (( (X+Z) . (Y’+Z) )’)’

F = ( (X+Z)’ + (Y’+Z)’ )’ F = ((X+Z)’ + ( (Y+Y)’ + Z)’)’

= ( X NOR Z) NOR ((Y NOR Y) NOR Z)

Boolean Algebra...Boolean Algebra Page 1 of 18 Prepared By : Jaya Sood Boolean Algebra The Algebra...

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