BOOLEAN ALGEBRA

BOOLEAN ALGEBRA◦ A Mathematical notation used to represent

the function of the Digital circuit.◦ A notation that allows variables & constants

to have only 2 possible values 0 & 1.◦ The Term Boolean Algebra honors a

fascinating English mathematician; George Boole

BOOLEAN FUNCTIONS-- An expression formed with Binary variables the two operators OR & AND & a UNARY operator not parenthesis & equal sign for the given variables the function can either be One or Zero.

EQUATION No. BOOLEAN EQUATION DESCRIPTION

1X + Y = Y + X

Commutative PropertyXY = YX

2X + (Y + Z) = (X + Y) + Z

Associative Property X(YZ) = (XY)Z

3 X X = X

Idempotent PropertyX + X = X

4 X 1 = X

Identity PropertyX + 1 = 1

5X 0 = 0

Null PropertyX + 0 = X

6 X (Y + Z) = XY + XZ

Distributive Property(XY) + (XZ) = X + YZ

7X X’ = 0

Negation PropertyX + X’ = 1

8 (X’)’ = X Double Negation Property

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X + XY = X

Absorption PropertyX (X + Y) = X

X + (X’Y) = X + YX’ + (XY) = X’ + Y

X’ + (XY’) = X’ + Y’

BOOLEAN ALGEBRA

The Rules of Boolean Algebra :AND Operations (·) 0·0 = 0 A·0 = 01·0 = 0 A·1 = A0·1 = 0 A·A = A1·1 = 1 A·A' = 0

OR Operations (+) 0+0 = 0 A+0 = A1+0 = 1 A+1 = 10+1 = 1 A+A = A1+1 = 1 A+A' = 1

NOT Operations (') 0' = 1 A'' = A

1' = 0

Associative Law (A·B)·C = A·(B·C) = A·B·C(A+B)+C = A+(B+C) = A+B+C

Distributive Law A·(B+C) = (A·B) + (A·C)A+(B·C) = (A+B) · (A+C)

Commutative Law A·B = B·A

A+B = B+A

Precedence AB = A·B

A·B+C = (A·B) + CA+B·C = A + (B·C)

DeMorgan's Theorem (A·B)' = A' + B' (NAND)(A+B)' = A' · B' (NOR)

DE MORGAN’S THEOREMA mathematician named De

Morgan developed a pair of important rules regarding group terms complementation in Boolean algebra.

He states that inversion or negation of a Boolean expression may be accomplished by negating each variable and changing each AND to OR each OR to AND.

De Morgan’s Theorem Equations

Is equivalent to

(xy)’ = x’ + y’

(x+y )’ = x’y’

This statement can be further summarized to:

The First Theorem states that a NOR gate is Equivalent to a bubbled AND gate.

The second Theorem States that a NAND Gate is equivalent to a Bubbled OR Gate.

In these rule, two procedures are used to form a complement.

The (+) symbol are replaced with (·) symbol or absence of operation, and (·)symbols with (+) symbol.

Each term in the expression is then complemented.(A·B)' = A' + B' (NAND)

(A+B)' = A' · B' (NOR)

NAND and NOR GATE IMPLEMENTATION

NAND and NOR GATE IMPLEMENTATIONAnother way of simplifying the

circuit by using only the universal gates, NAND and NOR.

NAND and NOR are known to be Universal Gates because they can function as other gates

BASIC GATES

NAND Gate ImplementationNOR Gate

Implementation

NOT Gate

OR Gate

AND Gate

Forms of Boolean Function

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FORMS OF FUNCTIONS:

CANONICAL Forms• Boolean function expressed as Sum of

Minterms or Product of Maxterms

1.MINTERM (denoted by lower case m)• A single variable or product of several

variables which may or may not be complemented

• A boolean expression containing ANDed Terms

• Example: A, A’B, A’B’C

F= A + A’B + A’B’C

MAXTERM (denoted by the uppercase M)

A single variable or sum of several variables

A Boolean expression containing OR Terms.

• Example: A, A’+B, A’+B’ + C (A) (A’+B) (A’+B’) (A’ + B’ + C) (A’ + B’ + C’)

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FORMS OF EQUATIONS:

FORMS OF EQUATIONS:

MINTERM:◦F = A’BC + AB’C’ + A’ B’ C◦F(A,B,C) =Σm3 + m4 + m1

MAXTERM:◦F = (A’+B+C) ( A+B’+C’ )(A’+ B’

+C)◦F(A,B,C) =ΠM4 + M3 + M6

MINTERM & MAXTERM Samples:

Sample Problem:

1. Amy watches TV, if it is Monday night & she has finished her homework

SOLUTION:A B X0 0 00 1 01 0 01 1 1

Minterm: F(A,B) =Σm3

Maxterm: F(A,B)=ПM0+M1+M2

MINTERM & MAXTERM Samples:

Standard Form of Equation:◦ The term that form the function may be

one or any number or literal.TWO TYPES:Sum of Product (SOP)

◦ A Boolean expression containing AND terms called PRODUCT of TERMS (one or more literals)

◦ Ex: F=Y’ + XY+X’YZ’PRODUCT OF SUM (POS)

◦ A Boolean expression containing OR terms called sum term.

◦ Each term may have any number of literals◦ Ex: F=(X+Y) + (Y’+Z)(X’+Y’Z)

SOP & POS Equations:

The Map MethodK-MAPPING (Karnaugh)

- Map a diagram made up of squares represents one minterm

KARNAUGH MAPA chart or grid containing boxes called cells; each which represents one minterm.

TYPES of MAP1.Two Variable Map

Consists of two variables2. Three Variable Map

A three variable map plotted in a map3. FOUR Variable Map

K-MAPPING (Karnaugh)

Steps:◦ Construct the K-Map & place 1’s in the squares

corresponding to 1’s in truth table;place zeros in the other squares.

◦ Examine the map for adjacent 1’s & loop those 1’s which are not adjacent to any other one’s.

◦ Looping continue as there are pairs octet or quad that contains 1. You can still loop the one that is already looped if there are still other 1’s left.

◦ Form the OR sum of all terms generated by each loop.

K-MAPPING (Karnaugh)

2 Variable Karnaugh Map2 INPUT OR GATE

The values of one variable appear across the top of the map, defining the column values, while the values of the other variable appear at the side, defining the values of the variable in each row.The Karnaugh map for the OR gate is completed by entering a '1' in each of the appropriate cells. Usually, you don't write in the '0's'. Within the map, adjacent cells containing 1's are grouped together in twos, fours, or eights. In this case, there is one horizontal and on vertical group of two. You indicate these groupings by drawing a circle round each one.The horizontal group corresponds to a B value of 1. In the left hand cell, A=0 and in the right hand cell, A=1. In other words, the value of A does not affect the outcome of the Boolean expression for these cells.

K-MAPPING (Karnaugh) 3-variable Karnaugh maps 3-variable examples

Here is the truth table for a 3-input system 1. Simplify the following expression using a Karnaugh map:

input C input B input A output0 0 0 00 0 1 0 You may be able to tell what is going to happen by completing the truth table for this expression.0 1 0 00 1 1 1 The Boolean statement is:1 0 0 01 0 1 11 1 0 11 1 1 1 The truth table is:

This is converted into a Karnaugh map, as follows: input C input B input A output0 0 0 00 0 1 10 1 0 10 1 1 01 0 0 01 0 1 11 1 0 11 1 1 0

The Karnaugh map is:

This is the exclusive OR function. The value of C is irrelevant.

K-MAPPING (Karnaugh)

2 variable K Map

B' B m0 00A' m0 m1 m1 01A m2 m3 m2 10

3 variable K Map 2n =23

00 01 11 100 m0 m1 m3 m21 m4 m5 m7 m6

4 variable K MapC'D' C'D CD CD'

A'B' m0 m1 m3 m2A'B m4 m5 m7 m6AB m12 m13 m16 m15AB' m8 m9 m11 m10

K-MAP Types:

K-MAPPING (Karnaugh)

TRUTH TABLE 3-variable:B'C' B'C BC BC'

A B C X 0 1 0 10 0 0 0 1 0 1 00 0 1 1

0 1 0 1 X = AB'C' + A'B'C+ABC+A'BC'0 1 1 01 0 0 11 1 0 01 1 1 1

TRUTH TABLE 4-variable:C'D' C'D CD CD'

DEST A B C D X A'B' 0 1 1 11 0 0 0 0 0 A'B 0 0 1 02 0 0 0 1 1 AB 0 1 0 03 0 1 1 0 1 AB' 0 1 1 04 0 1 0 0 15 0 1 0 0 06 0 0 1 1 07 0 1 1 1 0 X = B'D + C'D+A'CD+A'B"C8 1 0 0 1 19 1 0 0 0 1

10 1 0 0 1 111 1 0 1 1 012 1 1 0 1 013 1 1 0 1 114 1 1 0 1 115 1 1 1 1 0

Don’t Care conditions: It is represented by a letter (X) in the truth table instead of 0

or 1 The X’s are treated either as 0 or 1’s whichever leads to a

simpler circuit. Five variable Map: Consists of 5 variables .It uses 32 squares.

5-variable K-Map:

C'D'E' C'D'E C'DE' CD'E' CDE' C'DE CDE CD'E

A'B' 0 1 1 1 0 1 1 0A'B 0 1 0 0 0 0 1 0AB 0 0 0 0 1 1 0 0AB' 0 1 1 0 1 0 0 0

= AB'C'E + ACDE + BCD'E + A'B'C'D + A'C'DE + A'B'E

A B C D E

AB

K-MAPPING (Karnaugh)

Example Don’t Care!

DON’T CARE "X"

C'D' C'D CD CD'

A'B' 1 1 1 1A'B 0 X 1 XAB 0 0 X 0AB' 1 0 X 1

X=B'D' + A'C + A'B'

CD

AB

K-MAPPING (Karnaugh)