3. boolean algebra

8/2/2019 3. Boolean Algebra

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Chapter# 2

Combinational Logic Circuits


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Boolean Algebra


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Binary (Boolean) Logic

Deals with binary variables and binary logic

functions

Has two discrete values 0 -> Open, False

1 -> Closed, True

Three basic logic operators

AND (.)

OR (+)

NOT ()


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Boolean Algebra

A branch of Algebra used for describing anddesigning two valued state variables

Introduced by George Boole in 19th Century Shannon used it to design switching circuits (1938)

A convenient and systematic way of expressing andanalyzing logic circuit operations

Mathematics of digital systems

Used to define a digital circuit and evaluate itsoperation

Used to simplify the logic circuits


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Boolean Algebra Terms

Variable

a symbol used to represent a logical quantity

A variable can have a 0 or 1 value

Example:

Complement

Inverse of a variable. Indicated by a bar over the variable

Example:

Literal Instance of a variable or its complement

X

X

A B


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Boolean Addition

Boolean Addition performed by OR gate

Sum Term is a sum of literals:

Sum term = 1 if any literal = 1

Sum term = 0 if all literals = 0

In logic circuits, a sum term is produced by anOR operation with no AND operationsinvolved

BA BA CBA


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Boolean Multiplication

Boolean Multiplication performed by AND gate

Product Term is a product of literals:

Product term = 1 if all literals = 1

Product term = 0 if any one literal = 0

In logic circuits, a product term is produced by anAND operation with no OR operations involved

BA. BA. CBA ..


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Precedence of Operators


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Function Evaluation


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Boolean Algebra - Postulates

Postulates are facts that can be taken as true

They do not require proof

There are certain Laws, Rules & Theorems

that govern the Boolean Algebra


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Laws, Rules & Theorems

of Boolean Algebra Commutative Law

for Addition and Multiplication

Associative Law for Addition and Multiplication

Distributive Law

Rules of Boolean Algebra

DeMorgan's Theorems


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Commutative Law

Commutative Law for Addition

A + B = B + A

Commutative Law for Multiplication

A.B = B.A


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Associative Law

Associative Law for Addition

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


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Associative Law

Associative Law for Multiplication

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


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Distributive Law

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


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Basic Identities of Boolean Addition

CAABBCCAAB

BABAA

AABA

B.ABA

ACAB)CB(A

C)BA()CB(A

ABBA

AA

1AA

AAA

11A

A0A

)heoremConsensusT(

)tionSimplifica(

)Absorption(

)sTheorem'DeMorgan(

)veDistributi(

)eAssociativ(

)eCommutativ(

)Involution()Complement(

)Idempotent(

)Identity(


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Duality Principle

A Boolean equation remains valid if we take

the dual of the expressions on both sides of the

equal sign

Dual of Expressions

Interchange 1s and 0s

Interchange AND (.) and OR (+)


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Basic Identities of Boolean

Multiplication

ZXXYYZZXXY

YXYXX

XXYX

YXYX

XZXY)ZY(X

XYYX

1XXXXX

11X

X0X

)ZX).(YX()ZY)(ZX)(YX(

Y.X)YX.(X

X)YX.(X

YXY.X

)ZX).(YX(Z.YX

X.YY.X

0X.X

XX.X

00.X

X1.X


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Laws & Rules of Boolean Algebra

(Floyds Summarized Version)

1. A + 0 = A

2. A + 1 = 1

3. A.0 = 0

4. A.1 = A

5. A + A = A

6. A + = 1

7. A.A = A

8.

9. = A

10.A + A.B = A

11. A + = A + B

12.(A+B).(A+C)

= A+B.C

A

A

BA.

0A.A


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Derivation of Absorption

.}{

}{1.

}){1(

rIdentityFoA

rIdentityFoA

veLawDistributiBA

ABA


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Derivation of Simplification

};{

}{)(

}{

}{)1.(

IdentityComplementYX

onDistributiYXXX

onDistributiYXXYX

IdentityYXYX

YXX


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Derivation of Consensus Theorem

},2{

};){1()1(

}{}){(

IdentityRuleZXXY

DistAssocYZXZXY

onDistributiYZXXYZZXXYIdentityXXYZZXXY

YZZXXY


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First Theorem

Second Theorem

Veracity of DeMorgan's Theorems

BABA .

BABA .

B.A BA

BA B.A


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Application of DeMorgan's Theorems

Apply to any number of variables

Apply to combination of variables

ZYXZ.Y.X

Z.Y.XZYX

BCACBA ).().(

).().().).(.( BCACBABCACBA

BCACBA .).().(.

CBBACABA ....

CBCABA ...


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Applying DeMorgans Theorem to

Logic Design Facilitate design using NAND/NOR gates

SHORTCUT METHOD: NOT all variables

Change all . to + and all + to . NOT the final result

Example:

)Z.Y).(Z.X).(Y.X(F

)ZY).(ZX).(YX(F

Z.YZ.XY.XF


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DeMorgans Theorem in circuit domain

Original circuit NAND equivalent circuit


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Boolean Analysis of Logic Circuits

A logic circuit is formed by a combination of the logic gates

Boolean algebra provides a concise way to express the

operation of a logic circuit

Boolean expression for a logic circuit Starting from the left, write the expression for each gate output, in terms

of its input variables and move towards the final output

CCAB

DCAB )(


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Boolean Analysis of Logic Circuits

Evaluating the expression

Evaluate the Boolean expression for all possiblecombinations of values for the input variables that make theexpression equal to 1

Putting the results in truth table

List all combinations of input variables in a binarysequence

Place 1 in the output column for each combination of input

variables that was determined in the evaluation Place a 0 in the output column for all other combinations of

input variables


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Constructing a Truth Table

From the expression, the output is a 1 if variableD = 1and

if AB=1or C=0

AB=1 if A=1 and B=1

CCAB

DCAB )(

1)( CAB

1)( CAB


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Constructing a Truth Table

Inputs Output Inputs Output

A B C D F A B C D F

0 0 0 0 0 1 0 0 0 0

0 0 0 1 1 1 0 0 1 1

0 0 1 0 0 1 0 1 0 0

0 0 1 1 0 1 0 1 1 0

0 1 0 0 0 1 1 0 0 0

0 1 0 1 1 1 1 0 1 1

0 1 1 0 0 1 1 1 0 0

0 1 1 1 0 1 1 1 1 1


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Simplification Using Boolean Algebra

Example 1:


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Can the following expression be

Simplified???

AB + A(B+C) + B(B+C)


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Simplification Using Boolean

Algebra

Why? Determines fewest gates to implement a logic

circuit


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AB + A(B+C) + B(B+C)

= AB + AB + AC + BB +BC, (AB+AB=AB, B.B=B)

= AB + AC + B + BC (C+1=1)

= AB + AC + B (A+1=1)

= B + AC


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Implementation with simplified expression

Equivalent circuit

Fewer gates

Reduced circuitry and interconnects


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Example 2:

BBA.

DC.

DCBA ...

A

.C.DB.ABA.


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Simplify using rules, laws and theorems

.C.DB.ABA. ).C.DB.A).(BA.(

.C.D)B.A).(BA(

.C.D)B.AB).(A(

.C.DB.B.A.C.D)B.A.A(

.C.DB.A


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Simplified equivalent circuit

.C.DB.A


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Example 3:

A

C

CBA ..

DC

D)C).(C.B.A(


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Simplify

)DC()C.B.A( D)C).(C.B.A(

)D.C()CBA(

)D(C.C)B(A

)DC(1BA

CBA


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Simplified equivalent circuit

CBA


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Types of Boolean Expressions

Define Domain of an expression

set of all variables (complemented or otherwise)

Boolean expressions may be expressed as: Sum-of-Products (SOP) Form

Product-of-Sums (POS) Form

Each form may contain single variable terms

May contain complemented and un-complementedterms

A SOP and POS expression cant have a term of morethan one variable having an over bar extending over theentire term


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Sum of Product (SOP) Form


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Sum-of-Products (SOP) Form

Two or more product terms summed by

Boolean addition

Any expression -> SOP using Boolean algebra Examples:

* A + BC

* A (B + CD) = AB + ACDDCBAADDCADBADCB*

CBACBAABC*


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Sum-of-Products (SOP) Form

BEFBCDAB)EFCD(BAB*

BDBCBADACAB)DCB)(BA(*

BADAC

CBCAC)BA(C)BA(C)BA(*

Conversion to SOP Form:


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Implementation of SOP Expression

B+AC+AD


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Standard SOP Form

SOP expressions containing all Variables in the

Domain in each term are in Standard Form.

Standard product terms (containing all Variables

in the Domain) are also called Minterms.

Any non-standard SOP expression may be

converted to Standard form by applying Boolean

Algebra Rule 6 to it.

Example:

)1AA(

CBACABCBA

)BB(CACBA

CACBA


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Standard SOP Form

Example: Determine Standard SOP expression

CBACBACABABCCBA

)CBCBCBBC(ACBA)CC)(BB(ACBA

ACBA

SHORTCUT: Introduce all possible combinations of themissingvariablesANDed with theoriginal term


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Standard SOP Form

Example: Determine Standard SOP expression

CBACBACABABCCBA

)CBCBCBBC(ACBA)CC)(BB(ACBA

ACBA

SHORTCUT: Introduce all possible combinations of themissingvariablesANDed with theoriginal term


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Binary Representation of a SOP Term A standard product term is equal to 1 for only one

combination of variable values.

A standard SOP expression is equal to 1 if one or more of

the product terms in the expression is equal to 1. A standard SOP expression corresponds to the input

combination of the truth table for which the functionproduces a 1 output.

Example: determine binary values which make the

standard SOP expression equal to 1

111100111010

ABCDDCABDCBA


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Binary Representation of a SOP Term A standard product term is equal to 1 for only one

combination of variable values.

A standard SOP expression is equal to 1 if one or more of

the product terms in the expression is equal to 1. A standard SOP expression corresponds to the input

combination of the truth table for which the functionproduces a 1 output.

Example: determine binary values which make the

standard SOP expression equal to 1

111100111010

ABCDDCABDCBA


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Product-of-Sums (POS) Form Two or more sum terms multiplied by Boolean

multiplication

Any expression -> POS using Boolean algebra Examples:

(A+B)(B+C)(A+B+C)

)ON)(ONL)(ML(

)ZYX)(ZY)(X(

)DCBD)(B(A)C(A


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)DCA(B)DC(BAB*

)DB)(CB(A)CDB(AACDAB*

C)BA(C)BA(C)BA(*

Conversion to POS Form:


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Implementation of POS expression

(A+B)(B+C+D)(A+C)


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Standard POS Form POS expressions containing all Variables in the

Domain in each term are in Standard Form.

Standard sum terms (containing all Variables in the

Domain) are also called Maxterms.

A Maxterm is a NOT Minterm. {A Maxterm is not aMinterm }

Any non-standard POS expression may be converted toStandard form by applying Boolean Algebra Rule 8

and Rule 12 {A+BC=(A+B)(A+C)} to it.}0.{ AA


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Standard POS Form

Example:

{Rule 8}

{Rule 12}

)CBA)(CBA)(CBA(

)BCA)(BCA)(CBA(

)BBCA)(CBA()CA)(CBA(

SHORTCUT: Introduce all possible combinations of themissing variables ORed with the original term


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Standard POS Form

Example:

{Rule 8}

{Rule 12}

)CBA)(CBA)(CBA(

)BCA)(BCA)(CBA(




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Standard POS Form

Example:

{Rule 8}

{Rule 12}

)CBA)(CBA)(CBA(

)BCA)(BCA)(CBA(




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Standard POS Form

Example:

{Rule 8}

{Rule 12}

)CBA)(CBA)(CBA(

)BCA)(BCA)(CBA(




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Standard POS Form

Example:

{Rule 8}

{Rule 12}

)CBA)(CBA)(CBA(

)BCA)(BCA)(CBA(


SHORTCUT: Introduce all possible combinations of themissingvariablesORed with theoriginal term


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Binary Representation of a POS Term

A standard sum term is equal to 0 for only onecombination of variable values.

A standard POS expression is equal to 0 if one or more ofthe product terms in the expression is equal to 0.

A standard POS expression corresponds to the inputcombination of the truth table for which the functionproduces a 0 output.

Example: determine binary values which make thestandard POS expression equal to 0

)1100()0101()0000(

)DCBA)(DCBA)(DCBA(


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Binary Representation of a POS Term

A standard sum term is equal to 0 for only onecombination of variable values.

A standard POS expression is equal to 0 if one or more ofthe product terms in the expression is equal to 0.

A standard POS expression corresponds to the inputcombination of the truth table for which the functionproduces a 0 output.

Example: determine binary values which make thestandard POS expression equal to 0

)1100()0101()0000(

)DCBA)(DCBA)(DCBA(


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Why Standard SOP and POS Forms?

Minimal Circuit implementation by switching

between Standard SOP or POS

Direct mapping of Standard Form expressions

and Truth Table entries.

Alternate Mapping methods for simplification

of expressions

PLD based function implementation


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Converting Standard SOP to Standard POS

Determine binary representation for all

Standard Product terms

Form new set of binary numbers from full

domain which is not part of the above

Form Standard Sum terms for the new binary

set


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Example of converting from standard SOP to

standard POS

ABCCBABCACBACBA

C)BAC)(BA)(CB(A

111101011010000 75320

110100001


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standard POS

ABCCBABCACBACBA

C)BAC)(BA)(CB(A

111101011010000 75320

110100001


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standard POS

ABCCBABCACBACBA

C)BAC)(BA)(CB(A

111101011010000 75320

110100001


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standard POS

ABCCBABCACBACBA

C)BAC)(BA)(CB(A

111101011010000 75320

110100001


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standard POS

ABCCBABCACBACBA

C)BAC)(BA)(CB(A

111101011010000 75320

110100001


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Converting Standard POS to Standard SOP

Reverse previous procedure:

Determine binary representation for all

Standard Sum terms in POS

Form new set of binary numbers from full

domain which is not part of the above

Form Standard Product terms for the new

binary set in SOP form


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Minterms, Maxterms, Binary Reps & TT

Minterms: Product terms in Standard SOP

form

Allow binary representation of Standard SOP product

terms in a truth table

Maxterms: Sum terms in Standard POS form

Allow binary representation of Standard POS sum terms

in a truth table


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CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

A B CMin-

terms

Max-

terms

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1


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CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

CBA .. CBA

A B CMin-

terms

Max-

terms

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1


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Converting SOP to TT Format

List all possible combinations of binary values forthe domain of the expression in TT format

Convert SOP expression to standard form

Place a 1 in the output column for each binaryvalue that makes the standard SOP a 1 and place a

0 for all remaining Minterms

Example next


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Converting SOP to TT FormatBCAABCCBACBABCBA

Input Output

A B C F0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 0

1 1 1 1


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Converting POS to TT Format

List all possible combinations of binary values forthe domain of the expression in TT format

Convert POS expression to standard form

Place a 0 in the output column for each binaryvalue that makes the standard POS a 0 and place a

1 for all remaining Maxterms

Example next


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Converting POS to TT Format

)CB)(BA(

)CBA)(CBA)(CBA)(CBA(

Input Output

A B C F0 0 0 1

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 1


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Determining Standard Expressions from TT

Determining Standard SOP Expression List each row in the Truth Table where the output is a 1 in Minterm

form

Sum all the standard SOP terms to get the Standard SOP expression

Example

Determining Standard POS Expression List each row in the Truth Table where the output is a 0 in Maxterm

form Sum all the standard POS terms to get the Standard POS expression

Example

DCBA1010

DCBA1001

3. boolean algebra

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