eee122a boolean algebra and logic gates

8/10/2019 EEE122A Boolean Algebra and Logic Gates

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DIGITAL CIRCUITDIGITAL CIRCUIT

DESIGNDESIGN

EEE122 AEEE122 ARef. Morris MANO & Michael D. CILETTIRef. Morris MANO & Michael D. CILETTI

DIGITAL DESIGN 4DIGITAL DESIGN 4thth editionedition

Fatih University- Faculty of Engineering- Electric and Electronic Dept.


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Chapter 2Chapter 2

Boolean Algebra andBoolean Algebra and

Logic GatesLogic Gates


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Postulates & AxiomsPostulates & AxiomsPostulates form the basic assumptions from which itis possible to deduce the rules, theorems, and

properties of the system

DefinitionDefinitionBoolean algebra is an algebraic structure defined by a set of

elements, , together with two binary operators + and . Providedthat the following postulates are satisfied

1. x + y B x y B Closure2. x + 0 = x x 1 = x Identity

3. x + y = y + x x y = y x Commutativity

4. x(y + z) = xy + xz x + yz = (x + y)(x + z) Distributivity

5. x + x = 1 x x = 0 Complement6. At least 2 elements: x,y B such that x y Cardinality


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Postulate 1: ClosurePostulate 1: Closure

A Boolean algebra is a closedA Boolean algebra is a closed

algebraic system containing a setalgebraic system containing a set of two or more elements and twoof two or more elements and two

operatorsoperators and

AND Operator OR Operator

(Dot/Multiplication) (Plus/Addition)

++++


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Postulate 2: Identity PropertyPostulate 2: Identity Property

There exists unique elements 1 (one)There exists unique elements 1 (one)

and 0 (zero) such that:and 0 (zero) such that:

where 0 is the identity element for the OR operator

and 1 is the identity element for the AND operator.

(a) x+0=x

(b) x.1=x


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Postulate 3: CommutativePostulate 3: Commutative

For every x and y in set

(a) x+y=y+x(b) x.y=y.x

Postulate 4: AssociativePostulate 4: Associative

For every x , y and Z

(a) x+(y+z)=(x+y)+z

(b) x.(y.z) =(x.y).z


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Postulate 5: Distributive PropertyPostulate 5: Distributive Property

(a) x.(y+z)=x.y+x.z

(b) x+y.z =(x+y).(x+z)

For every x, y and Z in set

Postulate 6: Complement PropertyPostulate 6: Complement Property

For every x in there exists a unique element x such that:x+x=1 x.x =0

Properties of 0 and 1 elementsProperties of 0 and 1 elements

1=01=0x.1=xx.1=xx+1=1x+1=1

0=10=1x.0=0x.0=0x+0=xx+0=x

ComplementComplementANDANDOROR

1=01=0x.1=xx.1=xx+1=1x+1=1

0=10=1x.0=0x.0=0x+0=xx+0=x

ComplementComplementANDANDOROR


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Duality PropertyDuality PropertyIf a Boolean expression is true, the dual ofthe expression is also true. To find the dual,

Replace all ORs with ANDs, all ANDs withORs, all ones with zeros and all zeros withones.

In equation form this can expressed as:

If f1=F(x1,x2,x3,,0,1,+,)the duality of f1

is F(x1,x2,x3,.,1,0, ,+)

Find the dual of: x+yz =(x+y).(x+z)

Sol. The dual is x.(y+z)=x.y+x.z


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Basic BooleanBasic BooleanTheoremsTheorems


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Theorem 1:Theorem 1:

( )( )( )

1 P2b

P6a

P5a

0 P6b

P2a

a a a a

a a a a

a aa

a

a

+ = += + +

= +

= +

=

a a a+ =

( )( )( )

( )

1 P2b

P6a

T1b

P5b

0 P6b

P2a

aa aa

aa a a

a a a

a aa

a

a

== +

= +

= +

= +

=

a.a=a


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Theorem 2: Null elements forTheorem 2: Null elements for

OR and AND operatorsOR and AND operators

1 1a + =

Theorem 3:Theorem 3:

a.0=0

a= a


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Theorem 4: AbsorptionTheorem 4: Absorption

( )( )

( )

( )

1 2

1 5

6

1

6

a ab a ab P b

a b P b

a b b b P a

a b b T a

a P a

+ = +

= +

= + +

= +

=

a ab a+ = ( )a a b a+ =

Theorem 5: AbsorptionTheorem 5: Absorption


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Theorem 6Theorem 6

ab abc ab ac+ = +

ab ab a+ = ( )

( )a b a b a+ + =

Theorem 7Theorem 7

( ) ( ) ( ) ( )a b a b c a b a c+ + + = + +


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Theorem 8:Theorem 8: DeMorganDeMorganss theoremtheorem

ab ac bc ab ac+ + = +

a b a b

a b a b

+ =

= +

Theorem 9: ConsensusTheorem 9: Consensus

( ) ( ) ( ) ( ) ( )a b a c b c a b a c+ + + + + = + + +


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Logic SimplificationLogic Simplification

SimplifySimplify

( )( )( )

F ab ab ab

a a b ab

b ab b a b b

b a

= + +

= + +

= + = + +

= +

1

1


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Boolean FunctionsBoolean Functions


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Boolean FunctionsBoolean FunctionsDefineDefine ( )1 2 3, , , , nf a a a aK

Where:f = Boolean function with value of 0 or 1

a1, a2, a3, . = Boolean variables (0 or 1)

Logical functions can be represented in twoLogical functions can be represented in two

ways:ways: A finite, but nonA finite, but non--uniqueunique Boolean expressionBoolean expression..

A uniqueA unique andand finitefinite truth tablestruth tables


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Boolean FunctionsBoolean Functions

Example of Boolean function or expression:Example of Boolean function or expression:

f(x,y,z) = (x + yf(x,y,z) = (x + y)z + x)z + x

Notations:Notations:

ff is the name of the function.is the name of the function.

(x,y,z)(x,y,z) are theare the input variablesinput variables, each representing 1 or 0., each representing 1 or 0.Listing the inputs is optional, but sometimes helpful.Listing the inputs is optional, but sometimes helpful.

AA literalliteral is any occurrence of an input variable or itsis any occurrence of an input variable or its

complement. The function above has four literals:complement. The function above has four literals: x, yx, y, z, z,, xx..

Precedence is important,Precedence is important,

NOT has the highest precedence, followed by AND, andNOT has the highest precedence, followed by AND, and

then OR.then OR.


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Boolean FunctionsBoolean FunctionsNotes:With n variables there are 2n input combinations.

1. Recall each function can have two outputs 0 or 1, so thereare different functional combinations22

n

00

11

F1F1

11

00

F2F2

00

00

F0F0

11

00

a0a0

11

11

F3F3

0 1 0 2 30, , , 1oF F a F a F = = = =

ForFor n=3, 2n=3, 233=8=8 input combinput combinationinationssandand 2288=256 output=256 output combcombinationinationss

For n= 2 there are 4 possiblefunctons

Possible functions with n= 1.there are 4 possible functons


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Truth tablesTruth tables AA truth tabletruth table shows all possible inputs and outputs of a function.shows all possible inputs and outputs of a function.

Remember that each input variable represents either 1 or 0.Remember that each input variable represents either 1 or 0.

Because there are only a finite number of values (1 and 0),Because there are only a finite number of values (1 and 0),

truth tables themselves are finite.truth tables themselves are finite.

A function with n variables has 2A function with n variables has 2nnpossible combinations ofpossible combinations of

inputs.inputs.

Inputs are listed in binary orderInputs are listed in binary orderin this example, from 000 toin this example, from 000 to

111.111.

F A B AC A C = + +

11

00

11

00

11

00

11

00

CC

11

11

00

00

00

00

00

00

ABAB

00

00

00

00

11

00

11

00

11001111

11111111

00000011

11110011

11

11

00

00

BB

110000

000000

110000

000000

FFAA

11

00

11

00

11

00

11

00

CC

11

11

00

00

00

00

00

00

ABAB

00

00

00

00

11

00

11

00

11001111

11111111

00000011

11110011

11

11

00

00

BB

110000

000000

110000

000000

FFAA A CAC


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Implementation of Boolean function with GatesImplementation of Boolean function with Gates


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Complement of a functionComplement of a function

The complement of a function always outputs 0The complement of a function always outputs 0

whewhenn the original function outputthe original function outputss 1, and 11, and 1 whewhenn

the original producethe original producess..

In a truth table, we can just exchange 0s and 1sIn a truth table, we can just exchange 0s and 1s

in the output column(s)in the output column(s)

f(x,y,z) = x(yz + yz)

x y z f(x,y,z)

0 0 0 00 0 1 0

0 1 0 00 1 1 01 0 0 11 0 1 01 1 0 01 1 1 1

x y z f(x,y,z)

0 0 0 10 0 1 1

0 1 0 10 1 1 11 0 0 01 0 1 11 1 0 1

1 1 1 0


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Complementing a function algebraicallyComplementing a function algebraically DeMorganDeMorganss law canlaw can bebe useusedd toto findfind the complementsthe complements of functionsof functions

You can also take the dual of the function, and then complementYou can also take the dual of the function, and then complement

each literaleach literal If f(x,y,z) =If f(x,y,z) = x(yx(yzz ++ yzyz))

the dual of f isthe dual of f is x + (yx + (y + z+ z)(y + z))(y + z)

then complementing each literal givesthen complementing each literal gives xx + (y + z)(y+ (y + z)(y + z+ z))

so fso f(x,y,z) = x(x,y,z) = x + (y + z)(y+ (y + z)(y + z+ z))

f(x,y,z) = x(yz + yz)

f(x,y,z) = ( x(yz + yz) ) [ complement both sides ]

= x + (yz + yz) [ because (xy) = x + y ]= x + (yz) (yz) [ because (x + y) = x y ]= x + (y + z)(y + z) [ because (xy) = x + y, twice]


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Canonical and StandardCanonical and StandardFormsForms

Sum of Products (SOP)Sum of Products (SOP)

Product of Sums (POS)Product of Sums (POS)


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SOPSOP & POS& POS

SOPSOP: are Boolean: are Boolean functionsfunctions thatthat formed by SUMMINGformed by SUMMING ANDedANDedterms.terms.

Example:Example: ( ), , ,F A B C D ABC BD AC D= + +

literals

(product terms)

POSPOS: are Boolean: are Boolean functionsfunctions thatthat formed by taking theformed by taking the PRODUCT ofPRODUCT of

ORedORed terms.terms.Example:Example:


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Canonical SOP FormCanonical SOP Form MintermMinterm::

A product term which contains exactly one complemented orA product term which contains exactly one complemented oruncomplementeduncomplemented form of each variable.form of each variable.

Canonical SOP formCanonical SOP form

A function which is a sum of onlyA function which is a sum of only mintermsminterms

Decimal

No

a b c minterms Notation

0 0 0 0abc

m0

1 0 0 1abc

m1

2 0 1 0abc

m2

3 0 1 1abc

m3

4 1 0 0abc

m4

5 1 0 1abc

m5

6 1 1 0abc

m6

7 1 1 1 abc m7

Minterms are equalto 1 when true.

is the decimal equivalent of binary code (a,b,c)


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Canonical SOP FormCanonical SOP Form:Example:Example

00111111

00001111

11110011

1100001100111100

00001100

11110000

00000000

FFCCBBAA

( ) ( ), , 1, 4,5F A B C m=

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

1m A BC =

4m A B C =

5m A B C =

Truth Table Find F

All 0s in Truth Table

This can be expressed as: m1+m4+m5


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ThreeThreeVariableVariable

MaxtermMaxterm

a b c Maxterms Notation

0 0 0 a b c+ + M00 0 1 a b c+ + M1

0 1 0 a b c+ + M2

0 1 1 a b c+ + M3

1 0 0 a b c+ + M4

1 0 1 a b c+ + M5

1 1 0 a b c+ + M6

1 1 1 a b c+ + M7

Maxterms areequal to 0when true

MaxtermsMaxterms

A sum term which contains exactly one complemented orA sum term which contains exactly one complemented or

uncomplementeduncomplemented form of eachform of each of inputof input variable.variable.

Canonical POS formCanonical POS form A function which is a product of onlyA function which is a product of only maxtermsmaxterms

Canonical POS FormCanonical POS Form

C i lC i l POS E lPOS E l


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CanonicalCanonical POS ExamplePOS Example

11111111

00001111

11110011

1100001111111100

00001100

00110000

11000000

FFCCBBAA

1M A B C= + +

Truth Table

( ) ( ), , 1, 2,6F A B C M =

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

2M A B C= + +

M6=A+B+C

F=M1.M6.M2

Or shortly as:

Find F

1s in the truth table

GeneralGeneral:: Maxterms are

complements of minterms

C i SOP f i f


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Example:Example: Expand the following function to canonical SOP formExpand the following function to canonical SOP form

( ) ( ) ( ), ,F A B C A B C C ABC BC A A= + + + +

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

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

11111111

11001111

00110011

0000011

11111100

00001100

11110000

11000000

FFCCBBAA

Converting SOP function from nonConverting SOP function from non--

canonical to canonical formcanonical to canonical form


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Converting POS function from nonConverting POS function from non--

canonical to canonical formcanonical to canonical form RecallRecall

So, to convert functionSo, to convert function

1.1. SumSum thethe POS literal with missing term(s) in thePOS literal with missing term(s) in theform ofform of

2.2. ExpandExpand thethe functionfunction

( )( ) and 0a bc a b a c aa+ = + + =

a a

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

( ) ( )( )( ), ,F A B C A B CC A B C AA B C = + + + + + +

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

EX:EX: Expand the followingExpand the followingfunction to canonical POSfunction to canonical POS

0011111100001111

11110011

0000011

11111100

11001100

00110000

00000000

FFCCBBAA


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Other Logic GatesOther Logic Gates


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Other LogicOther Logic OperationsOperations


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Other Logic GatesOther Logic Gates


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Extension to MultipleExtension to Multiplenputnput

Nonassociativity of the NOR operator


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Positive and Negative LogicPositive and Negative Logic

eee122a boolean algebra and logic gates

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