Boolean Algebra (4)

8/11/2019 Boolean Algebra (4)

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RAM PRAVESH [email protected]

NIT DELHI

Faculty Of Electronics andCommunication Technology

Boolean Algebra


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Boolean Functions

Boolean algebraprovides the operations and

the rules for working with the set {0, 1}.

Electronic and optical switches can be

studied using this set and the rules ofBoolean algebra.

We are going to focus on three operations:

Boolean complementation,

Boolean sum, and

Boolean product


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Boolean Functions and Logic Gates

The complementisdenoted by a bar.It is defined by

0=1 1=0

The Boolean sum,denoted by +or by OR

(inclusive OR), has thefollowing values:1+1=1 1+0=10+1=1 0+0=0

Not GateInverter

OR Gate


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The Boolean product,

denoted by

or byAND, has thefollowing values:1

1=1 1

0=0

0

1=0 0

0=0

XOR (exclusive OR)1

1=0 1

0=10

1=1 0

0=0xy =xy+xy


AND Gate

XOR Gate


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NANDand NORare two very important gates.

Their symbols and truth tables are shown below


NOR

NAND

x

y = x y

x y= x + y

NAND

NOR


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Boolean Functions


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Boolean Functions

The complement, Boolean sum, and Booleanproductcorrespond the Logical operators

,,andrespectively, where

0 corresponds to F , and 1 corresponds to T.

1

0+ (0+1)=0

Its logical equivalent is

(T

F)

(F

T) F


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Boolean Expressions andBoolean Functions

Definition:Let B = {0, 1}. The variable xis calleda Boolean variableif it assumes values only fromB.

A functionfromBn, the set {(x1, x2, , xn) |xiB,

1 i n}, toB is called a Boolean functionofdegree n.

Boolean functionscan be represented using

expressions made up from the variables andBoolean operations.


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Boolean Expressions and Boolean Functions

Example: the function F(x,y)=xy from the setof ordered pairs of Boolean variable to theset {0, 1} is a Boolean function of degree 2with F(1,1)=0, F(1,0)=1, F(0,1)=0, and

F(0,0)=0.

this functionFrom B2To B

x y F(x,y)

0 0 0

0 1 0

1 0 1

1 1 0


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The Boolean expressionsin the variablesx1, x2, , xnare defined recursively as:

0,1,x1,x2,,xnare Boolean expressions.If E1and E2are Boolean expressions, thenE1, E1E2, and E1+E2are Boolean expressions.

Each Boolean expressionrepresents a Boolean

function. The values of this function areobtained by substituting 0 and 1 for thevariables in the expression.


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Example: Find the value of the Boolean functionF(x,y,z)=xy+z (this function From B3To B)

Definition: The Boolean functions Fand Gof nvariablesare equalif and only if

F(b1, b2, , bn) = G(b1, b2, , bn)

whenever b1, b2, , bnbelong to B.

Two different Boolean expressions that represent thesame function are called equivalent.

For example, the Boolean expressions xy, xy+0,and xy1 are equivalent


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Example: Find the value of the Boolean functionF(x,y,z)=xy+z (this function From B3To B)

A Boolean functionof degree ncan be represented by an

n-cube(hypercube) with the corresponding function valueat each vertex.


001

111110

011

100

010

000

101


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The complementof the Boolean function F is the

function F

where F(b1, b2, , bn) = F(b1, b2, , bn)

Let Fand Gbe Boolean functionsof degree n.The Boolean sumF+Gis defined by

(F+G)(b1,b2,,bn) = F(b1,b2, ,bn)+G(b1,b2,,bn)

and Boolean productFGis defined by

(FG)(b1,b2, ,bn) = F(b1,b2, ,bn) G(b1,b2, ,bn)


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Question: How many different Boolean functionsof degreenare there?

Solution:

There are 2ndifferent n-tuples of 0s and 1s.

A Boolean function is an assignment of 0 or 1 toeach of these 2ndifferent n-tuples.

Therefore, there are 22ndifferent Boolean

functions.


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Question: How many different Boolean functions

of degree4are there? 16


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Question: How many different Boolean functionsof degree 1are there?

Solution: There are four of them, F1,F2,F3, andF4

x F1 F2 F3 F4

0 0 0 1 1

1 0 1 0 1


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Identities of Boolean algebra


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Identities of Boolean algebra(Duality)

There are useful identities of Boolean expressions that

can help us to transforman expressionAinto anequivalentexpressionB(see previous table)

We can derive additional identities with the help of thedualof a Boolean expression.

Duality principle

The Boolean equation remains valid if we take the dualof the expression on both sides of the equals sign.

The dualof a Boolean expressionis obtained byinterchangingBoolean sumsand Boolean productsandinterchanging0sand 1s.


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Examples:

The dual of x(y + z)is x + yz

The dual of x1 + (y + z) is (x + 0) ( y . z).

The dual of a Boolean function Frepresented by aBoolean expression is the function represented bythe dual of this expression.

This dual function, denoted by Fd, does not dependon the particular Boolean expression used torepresent F.


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Therefore, an identity between functions

represented by Boolean expressions remains validwhen the duals of both sides of the identity aretaken.

We can use this fact, called the duality principle,to derive new identities.

For example, consider the absorption lawx(x + y) = x.

By taking the duals of both sides of this identity,we obtain the equation x + xy = x, which is also anidentity (and also called an absorption law).


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Definition of a Boolean Algebra

All the properties of Boolean functions and

expressions that we have discovered also apply toother mathematical structuressuch aspropositionsand setsand the operations defined onthem.

If we can show that a particular structure is aBoolean algebra, then we know that all resultsestablished about Boolean algebras apply to this

structure.For this purpose, we need anabstract definitionofa Boolean algebra.


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Definition of a Boolean Algebra

Definition:A Boolean algebra is a set B with two

binary operationsand , elements0and 1, and aunary operation such that the followingproperties hold for all x, y, and z in B:

x 0 = x and x 1 = x (identity laws)

x (x) = 1 and x (x) = 0 (Complement laws)

(x y) z = x (y z) and(x y) z = x (y z) and (associative laws)

x y = y x and x y = y x (commutative laws)

x (y z) = (x y) (x z) andx (y z) = (x y) (x z) (distributive laws)


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Representing Boolean Functions

Two important problems of Boolean algebra will

be studied in the next slides:

Given the value of a Boolean function, how can

a Boolean expression that represents thisfunction be found?

Is there a smaller set of operators that canbe used to represent all Boolean functions?


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Any Boolean functioncan be represented as a :

Sum of products (SOP)of variables and theircomplements. Disjunctive normal form (DNF)

Sum-of-products Expansions

Or

Product of sums (POS)of variables and their

complements. Conjunctive normal form (CNF)Product-of-sums Expansions


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Minterms andmaxterms

The table below illustrates the minterms andmaxterms forthree inputBoolean variables


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FindBoolean expressions that represent thefunction F(A,B,C) which is given in the followingtable.


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Sum-of-products Expansions (DNF)

Product -of- sums Expansions (CNF)


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For example


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Example Find a minterm that equal 1 ifx1= x3=0

andx2= x4=x5=1 and equals 0 otherwise.The minterm is x1x2x3 x4 x5

Example Find the sum-of-products expansion forthe function F(x,y,z)=(x+y)z (truth table)F(x,y,z)=(x+y)z

=xz+yz Distributive law

=x1z+1yz Identity law=x(y+y)z+(x+x)yz Unity law=xy z+xyz+xyz+xyz Distributive law=xy z+xyz+xyz Idempotent law


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Is there a smaller set of operators that can be used

to represent all Boolean functions?Every Boolean function can be represented using theBoolean operators +, . , and . So, we say that theset{+, ., } is functionally complete.

We can eliminate all the +using this entityThen set{ ., } is functionally complete.

We can eliminate all the .using this entityThen set{ +, } is functionally complete.


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Is there a smaller set of operators that can be used

to represent all Boolean functions?Both of the sets { } and {} arefunctionally complete.

x+y using NAND how?xy using NOR

x

y = x y

x y= x + y

NAND

NOR

x = x

x

xy = (x

y)

(x

y)

x = x x

x+y = (x y) (x y)


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Combinations of Gates

Logic gates can have one or more inputs

Constructthe circuitthat producesthe following output:

f


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Constructthe circuitthat producesthe following output:

b f G


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A committee of three individuals decides issues for

an organization. Each individualvotes either yesor nofor each proposal that arises. A proposal is passedifitreceives at leasttwo yesvotes. Designa circuit thatdetermines whether a proposal passes.

Sometimes light fixtures are controlled by more thanswitch. Circuits needed to be designed so that flippingany one of the switches for the fixture turns the light

on when it is off and turns the light off when it is on.Design circuits that accomplish this when there are twoswitches and when there are three switches.

C b f G ( )


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Combinations of Gates (half adder)

Combinational logic circuits give us many useful

devices.One of the simplest is the half adder, which findsthe sum of two bits.

We can gain some insight as to the construction of a

half adder by looking at its truth table

C bi i f G ( )


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Combinations of Gates (half adder)

xy

SumCarry

C bi i f G ( )


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Combinations of Gates (full adder)

We can change our half adder into to a full adder by

including gates for processing the carry bit.The truth table for a full adderis shown below.

C bi ti f G t (f ll dd )


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Combinations of Gates (full adder)

E i


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pp. (765-766)

1-6

10-11

13-19

Exercises

Mi i i ti f Ci it


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Minimization of Circuits

Karnaugh Maps

It is a graphical approach used to simplify a Booleanfunction (sum of products expansion).

Let F(A,B)be a sum-of-products Boolean expression.To minimize F, we use a rectangular array of two rows

and two columnsin which rows and columns are labeled asfollows:

Two variables

22possible

Minterms.

Each squareis called a cellcorresponds to a minterm.Cellsare called adjacentif the mintermsthat theyrepresent differin exactly one literal.

Mi i i ti f Ci it


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A two-variablesKarnaugh Map

If a mintermis presentin F(A,B), then we place a 1inthe cell corresponding to the minterm, otherwise thecell is left emptyor place a O.

The resulting array is called K-map corresponding tothe expression


Mi i i ti f Ci it


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A two-variablesKarnaugh Map

FindBoolean expressions that represent the functionF(A,B) which is given in the following table.


Dont care condition ? x

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A Three-variablesKarnaugh Map


Use K-Map to find a minimal expansion of thefunction, and draw the circuit diagram.

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A four-variablesKarnaugh Map


Use K-Map to find a minimalexpansion of the function, anddraw the circuit diagram.

Exercises


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pp. (779-780)

1-2

6-7

12

Exercises

Boolean Algebra (4)

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