Chapter 3 (Boolean Algebra and Logic Design)

Principles OfPrinciples Of

Digital DesignDigital Design

Chapter 3Chapter 3

Boolean Algebra and Logic Design

 Boolean Algebra

 Logic Gates

 Digital Design

 Implementation Technology

ASICsASICs

Gate ArraysGate Arrays

Slides by Philip Pham, University of California, Irvine

A set is a collection of objects with a common property

If S is a set and x is a member of the set S, then x  SIf S is a set and x is a member of the set S, then x  S

A = {1, 2, 3, 4} denotes the set A, whose elements are 1, 2, 3, 4A = {1, 2, 3, 4} denotes the set A, whose elements are 1, 2, 3, 4

A binary operator on a set S is a rule that assigns to eachpair of elements in S another element that is in S

Axioms are assumption that are valid without proof

Basic Algebraic PropertiesBasic Algebraic Properties

Slides by Philip Pham, University of California, Irvine

Closure

A set S is closed with respect to a binary operator ● iff for all x, y  S,(x ● y)  SA set S is closed with respect to a binary operator ● iff for all x, y  S,(x ● y)  S

Z+ = {1, 2, 3, …} is closed to addition, because positive numbers are in Z+Z+ = {1, 2, 3, …} is closed to addition, because positive numbers are in Z+

Associativity

A binary operator ● defined on a set S is associative iff for allx, y, z  SA binary operator ● defined on a set S is associative iff for allx, y, z  S

(x  y)  z = x  (y  z)(x  y)  z = x  (y  z)

Identity Element

A set S has an identity element e for every x  SA set S has an identity element e for every x  S

e  x = x  e = xe  x = x  e = x

x + 0 = 0 + x = xx + 0 = 0 + x = x

Examples of AxiomsExamples of Axioms

Slides by Philip Pham, University of California, Irvine

Commutativity

A binary operator ● is commutative iff for all x, y  SA binary operator ● is commutative iff for all x, y  S

x  y = y  xx  y = y  x

Inverse Element

A set S has an inverse iff for every x  S, there exists an elementy  S such thatA set S has an inverse iff for every x  S, there exists an elementy  S such that

x  y = e x  y = e

Distributivity

If ● and □ are two binary operators on a set S, ● is said to bedistributive over □ if, for all x, y, z  SIf ● and □ are two binary operators on a set S, ● is said to bedistributive over □ if, for all x, y, z  S

x  (y □ z) = (x  y) □ (x  z)x  (y □ z) = (x  y) □ (x  z)

Examples of AxiomsExamples of Axioms

Slides by Philip Pham, University of California, Irvine

Axiom 1 (Closure Property): (a) B is closed with respect to the operator +; (b) B is alsoclosed with respect to the operator ∙

Axiom 2 (Identity Element): (a) B has an identity element with respect to +, designatedby 0; (b) B also has an identity element with respect ∙ , designated by 1

Axiom 3 (Commutativity Property): (a) B is commutative with respect to +; (b) B is alsocommutative with respect to ∙

Axiom 4 (Distributivity Property): (a) The operator ∙ is distributive over +; (b) similarly,the operator + is distributive over ∙

Axiom 5 (Complement Element): For every x  B, there exists an element x′  B suchthat (a) x + x′ = 1 and (b) x ∙ x′ = 0This second element x′, is called the complement of x

Axiom 6 (Lower Cardinality Bound): There are at least two elements x, y  B such thatx  y

Axiomatic Definition of Boolean AlgebraAxiomatic Definition of Boolean Algebra

Boolean algebra is a set of elements B with two binary operators, + and ∙, whichsatisfies the following six axioms:Boolean algebra is a set of elements B with two binary operators, + and ∙, whichsatisfies the following six axioms:

Slides by Philip Pham, University of California, Irvine

In ordinary algebra, + is not distributive ∙

Boolean algebra does not have inverses with respect to + and ∙ ;therefore, there are no subtraction or division operations in Booleanalgebra

Complements are available in Boolean algebra, but not in ordinaryalgebra

Boolean algebra applies to a finite set of elements, whereas ordinaryalgebra would apply to the infinite sets of real numbers

The definition above for Boolean algebra does not includeassociativity, since it can be derived from the other axioms

Axiomatic Definition of Boolean AlgebraAxiomatic Definition of Boolean Algebra

Differences between Boolean algebra and ordinary algebraDifferences between Boolean algebra and ordinary algebra

Slides by Philip Pham, University of California, Irvine

Two-valued Boolean AlgebraTwo-valued Boolean Algebra

Set B has two elements: 0 and 1

Algebra has two operators: AND and OR

x ∙ y

AND Operator

x + y

OR Operator

Slides by Philip Pham, University of California, Irvine

Two-valued Boolean AlgebraTwo-valued Boolean Algebra

Axiom 1 (Closure Property): Closure is evident in the AND/ORtables, since the result of each operation is an element of B.

Axiom 2 (Identity Element): The identity elements in this algebra are0 for the operator + and 1 for the operator ∙. From the AND/ORtables, we see that:

0 + 0 = 0, and 0 + 1 = 1 + 0 = 10 + 0 = 0, and 0 + 1 = 1 + 0 = 1

1 ∙ 1 = 1, and 1 ∙ 0 = 0 ∙ 1 = 01 ∙ 1 = 1, and 1 ∙ 0 = 0 ∙ 1 = 0

Axiom 3 (Commutativity Property): The commutativity laws followfrom the symmetry of the operator tables.

Two-valued Boolean algebra satisfies Huntington axiomsTwo-valued Boolean algebra satisfies Huntington axioms

Slides by Philip Pham, University of California, Irvine

Two-valued Boolean AlgebraTwo-valued Boolean Algebra

Axiom 4 (Distributivity): The distributivity of this algebra can bedemonstrated by checking both sides of the equation.

y + z

x ∙ (y + z)

(xy) + (xz)

Proof of distributivity of ∙

x + (yz)

x + y

x + z

(x + y)(x + z)

Proof of distributivity of +

x ∙ (y + z) = (x ∙ y) + (x ∙ z)

x + (y ∙ z) = (x + y)(x + z).

Slides by Philip Pham, University of California, Irvine

Two-valued Boolean AlgebraTwo-valued Boolean Algebra

Axiom 5 (Complement): 0 and 1 are complements of each other, since0 + 0′ = 0 + 1 = 1 and 1 + 1′ = 1 + 0 = 1; furthermore,0 ∙ 0′ = 0 ∙ 1 = 0 and 1 ∙ 1′ = 1 ∙ 0 = 0.

Axiom 6 (Cardinality): The cardinality axiom is satisfied, since thistwo-valued Boolean algebra has two distinct elements, 1 and 0, and1  0.

x′

NOT Operator

Slides by Philip Pham, University of California, Irvine

Boolean Operator ProcedureBoolean Operator Procedure

Boolean operators are applied in the following order.

Parentheses( )Parentheses( )

NOT′NOT′

AND∙AND∙

OR+OR+

Example: Evaluate expression (x + xy)′ for x = 1 and y = 0:Example: Evaluate expression (x + xy)′ for x = 1 and y = 0:

(1 + 1∙0)′ = (1 + 0)′ = (1)′ = 0(1 + 1∙0)′ = (1 + 0)′ = (1)′ = 0

Slides by Philip Pham, University of California, Irvine

Duality PrincipleDuality Principle

Any algebraic expression derived from axioms stays validwhen

OR and ANDOR and AND

0 and 10 and 1

are interchanged.

Example:Example:

IfIf

X + 1 = 1X + 1 = 1

thenthen

X ∙ 0 = 0X ∙ 0 = 0

by the duality principleby the duality principle

Slides by Philip Pham, University of California, Irvine

Theorem of Boolean AlgebraTheorem of Boolean Algebra

Theorem 1

(a)

x + x

(Idempotency)

(b)

Theorem 2

(a)

x + 1

(b)

x ∙ 0

Theorem 3

(a)

yx + x

(Absorption)

(b)

(y + x)x

Theorem 4

(x′)′

(Involution)

Theorem 5

(a)

(x + y) + z

x + (y + z)

(Associativity)

(b)

x(yz)

(xy)z

Theorem 6

(a)

(x + y)′

x′y′

(De Morgan’s Law)

(b)

(xy)′

x′ + y′

Basic Theorems of Boolean Algebra

Slides by Philip Pham, University of California, Irvine

Theorem Proofs in Boolean AlgebraTheorem Proofs in Boolean Algebra

Theorems can be proved by transformations based on axiomsand theorems

Duality

Example:Example:

Theorem 1(a) Idempotency: x + x = x.Theorem 1(a) Idempotency: x + x = x.

Proof:Proof:

x + x=(x + x) ∙ 1by identity (Ax. 2b)x + x=(x + x) ∙ 1by identity (Ax. 2b)

=(x + x) (x + x′)by complement (Ax. 5a)=(x + x) (x + x′)by complement (Ax. 5a)

=x + xx′by distributivity (Ax. 4b)=x + xx′by distributivity (Ax. 4b)

=x + 0by complement (Ax. 5b)=x + 0by complement (Ax. 5b)

=xby identity (Ax. 2a)=xby identity (Ax. 2a)

Example:Example:

Theorem 1(b) Idempotency: x ∙ x = x.Theorem 1(b) Idempotency: x ∙ x = x.

Proof:Proof:

x + x=xby Theorem 1(a)x + x=xby Theorem 1(a)

x ∙ x=xby Duality principle x ∙ x=xby Duality principle

Slides by Philip Pham, University of California, Irvine

Theorem Proofs in Boolean AlgebraTheorem Proofs in Boolean Algebra

Checking theorems for every combinations of variable value

Example:Example:

Theorem 6(a) DeMorgan’s Law: (x + y)′ = x′y′Theorem 6(a) DeMorgan’s Law: (x + y)′ = x′y′

x + y

(x + y)′

x′

y′

x′y′

Proof of Demorgan’s First Theorem

Slides by Philip Pham, University of California, Irvine

Note 1: When we evaluate Boolean expressions, we must follow aspecific order of operations, namely, (1) parentheses, (2) NOT, (3) AND,(4) OR.Note 1: When we evaluate Boolean expressions, we must follow aspecific order of operations, namely, (1) parentheses, (2) NOT, (3) AND,(4) OR.

Note 2: A primed or unprimed variable is usually called a literal.Note 2: A primed or unprimed variable is usually called a literal.

Boolean FunctionsBoolean Functions

Algebraic expression, which are formed from binaryvariables and Boolean operators AND, OR and NOT.

Example:Example:

F1 = xy + xy′z + x′yzF1 = xy + xy′z + x′yz

This function would be equal to 1 if x = 1 and y = 1, orThis function would be equal to 1 if x = 1 and y = 1, or

if x = 1 and y = 0 and z = 1, orif x = 1 and y = 0 and z = 1, or

if x = 0 and y = 1 and z = 1;if x = 0 and y = 1 and z = 1;

otherwise, F1 = 0.otherwise, F1 = 0.

Slides by Philip Pham, University of California, Irvine

Boolean FunctionsBoolean Functions

Truth tables which list the functional value for allcombinations of variable values.

Example:Example:

F1 = xy + xy′z + x′yzF1 = xy + xy′z + x′yz

RowNumbers

VariableValues

FunctionValues

Slides by Philip Pham, University of California, Irvine

Complement of a FunctionComplement of a Function

Complement of function F is function F′, where F′ can beobtained by:

Interchanging 0 and 1 in the truth table.Interchanging 0 and 1 in the truth table.

RowNumbers

VariableValues

FunctionValues

F1′

Example:Example:

F1 = xy + xy′z + x′yzF1 = xy + xy′z + x′yz

Slides by Philip Pham, University of California, Irvine

Complement of a FunctionComplement of a Function

Complement of function F is function F′, where F′ can beobtained by:

Repeatedly applying DeMorgan’s theorems.Repeatedly applying DeMorgan’s theorems.

Duality PrincipleDuality Principle

Example: F1′=(xy + xy′z + x′yz)′by definition of FExample: F1′=(xy + xy′z + x′yz)′by definition of F

=(xy)′ (xy′z)′ (x′yz)′by DeMorgan’s Th.=(xy)′ (xy′z)′ (x′yz)′by DeMorgan’s Th.

=(x′ + y′)(x′ + y + z′)(x + y′ + z′)by DeMorgan’s Th.=(x′ + y′)(x′ + y + z′)(x + y′ + z′)by DeMorgan’s Th.

Example: F1=(x ∙ y) + (x ∙ y′ ∙ z) + (x′ ∙ y ∙ z)Example: F1=(x ∙ y) + (x ∙ y′ ∙ z) + (x′ ∙ y ∙ z)

F1′=(x′ + y′) ∙ (x′ + y + z′) ∙ (x + y′ + z′)F1′=(x′ + y′) ∙ (x′ + y + z′) ∙ (x + y′ + z′)

Slides by Philip Pham, University of California, Irvine

Graphic Representation ofBoolean FunctionsGraphic Representation ofBoolean Functions

′

∙

AND-OR Expression

OR-AND Expression

F size=5 ANDsF size=5 ANDs

2 ORs2 ORs

2 NOTs2 NOTs

F size=2 ANDsF size=2 ANDs

5 ORs5 ORs

4 NOTs4 NOTs

F′F′

Two different expressions have different sizes

Slides by Philip Pham, University of California, Irvine

We can prove expression equivalence by algebraicmanipulation in which each transformation uses an axiomor a theorem of Boolean algebra.

Expression EquivalenceExpression Equivalence

xy + xy′z + x′yz

requires

ANDs

ORs

NOTs

xy + xz + yz

requires

ANDs

ORs

Difference:

ANDs

and

NOTs

Example: F1=xy + xy′z + x′yzExample: F1=xy + xy′z + x′yz

=xy + xz + yz=xy + xz + yz

Proof:Proof:

xy + xy′z + x′yz =xy + xyz + xy′z + x′yz by absorption=xy + x(y + y′)z + x′yz by distributivity=xy + x1z + x′yz by complement=xy + xz + x′yz by identity=xy + xyz + xz + x′yz by absorption=xy + xz + (x + x′)yz by distributivity=xy + xz + 1yz by complement=xy + xz + yz by identity xy + xy′z + x′yz =xy + xyz + xy′z + x′yz by absorption=xy + x(y + y′)z + x′yz by distributivity=xy + x1z + x′yz by complement=xy + xz + x′yz by identity=xy + xyz + xz + x′yz by absorption=xy + xz + (x + x′)yz by distributivity=xy + xz + 1yz by complement=xy + xz + yz by identity

Slides by Philip Pham, University of California, Irvine

MintermsMinterms

Minterms for Three Binary Variables

Minterm definition

If i = bn – 1…b0 is a binary number between 0 and 2n – 1, then aminterm of n variables xn – 1, xn – 2…,x0, could be represented as:If i = bn – 1…b0 is a binary number between 0 and 2n – 1, then aminterm of n variables xn – 1, xn – 2…,x0, could be represented as:

mi (xn – 1, xn – 2…,x0) = yn – 1…y0mi (xn – 1, xn – 2…,x0) = yn – 1…y0

where for all k such that 0 ≤ k ≤ n – 1,where for all k such that 0 ≤ k ≤ n – 1,

yk= { yk= {

xk if bk = 1xk if bk = 1

xk′ if bk = 0xk′ if bk = 0

Minterms

Designation

x′y′z′

x′y′z

x′yz′

x′yz

xy′z′

xy′z

xyz′

xyz

Slides by Philip Pham, University of California, Irvine

Sum-of-MintermsSum-of-Minterms

Any Boolean function can be expressed as a sum (OR) ofits 1-minterms:

F(list of variables) = Σ(list of 1-minterm indices)

RowNumbers

VariableValues

FunctionValues

F1′

Example:Example:

Equation Table

F1 = xy + xy′z + x′yz

F1′ = (x′ + y′)(x′ + y + z′)(x + y′ + z′)

F1(x, y, z)=Σ(3, 5, 6, 7)F1(x, y, z)=Σ(3, 5, 6, 7)

=m3 + m5 + m6 + m7=m3 + m5 + m6 + m7

=x′yz + xy′z + xyz′ + xyz=x′yz + xy′z + xyz′ + xyz

F1′(x, y, z)=Σ(0, 1, 2, 4)F1′(x, y, z)=Σ(0, 1, 2, 4)

=m0 + m1 + m2 + m4=m0 + m1 + m2 + m4

=x′y′z′ + x′y′z + x′yz′ + xy′z′=x′y′z′ + x′y′z + x′yz′ + xy′z′

Slides by Philip Pham, University of California, Irvine

Expansion to Sum-of-MintermsExpansion to Sum-of-Minterms

Any Boolean function can be expanded into a sum-of-minterms form be expanding each term with (x + x′) for eachmissing variable x.

Example:Example:

F = x + yzF = x + yz

=x(y + y′)(z + z′) + (x + x′)yz=x(y + y′)(z + z′) + (x + x′)yz

=xyz + xy′z + xyz′ + xy′z′ + xyz + x′yz=xyz + xy′z + xyz′ + xy′z′ + xyz + x′yz

After removing duplicates and rearranging the minterms inascending order:After removing duplicates and rearranging the minterms inascending order:

F = x′yz + xy′z′ + xy′z + xyz′ + xyzF = x′yz + xy′z′ + xy′z + xyz′ + xyz

=m3 + m4 + m5 + m6 + m7=m3 + m4 + m5 + m6 + m7

=Σ(3, 4, 5, 6, 7)=Σ(3, 4, 5, 6, 7)

Slides by Philip Pham, University of California, Irvine

Conversion to Sum-of-MintermsConversion to Sum-of-Minterms

F = m3 + m4 + m5 + m6 + m7F = m3 + m4 + m5 + m6 + m7

Each Boolean function can be converted into a sum-of-minterms form by generating the truth table and identifying1-minterms.

Example: F = x + yzExample: F = x + yz

Slides by Philip Pham, University of California, Irvine

MaxtermsMaxterms

Maxterms can be defined as the complement of minterms:

Mi = mi′ and Mi′ = miMi = mi′ and Mi′ = mi

Maxterms for Three Binary Variables

Maxterms

Designation

x + y + z

x + y + z′

x + y′ + z

x + y′ + z′

x′ + y + z

x′ + y + z′

x′ + y′ + z

x′ + y′ + z′

Slides by Philip Pham, University of California, Irvine

Product-of-MaxtermsProduct-of-Maxterms

Any Boolean function can be expressed as a product (AND)of its 0-maxterms:

F(list of variables) = Π(list of 0-maxterm indices)

RowNumbers

VariableValues

FunctionValues

F1′

Example:Example:

Equation Table

F1 = xy + xy′z + x′yz

F1′ = (x′ + y′)(x′ + y + z′)(x + y′ + z′)

F1(x, y, z)=Π(0, 1, 2, 4)F1(x, y, z)=Π(0, 1, 2, 4)

=M0 M1 M2 M4=M0 M1 M2 M4

=(x + y + z)(x + y + z′)(x + y′ + z)(x′ + y + z)=(x + y + z)(x + y + z′)(x + y′ + z)(x′ + y + z)

F1′(x, y, z)=Π(3, 5, 6, 7)F1′(x, y, z)=Π(3, 5, 6, 7)

=M3 M5 M6 M7=M3 M5 M6 M7

=(x + y′ + z′)(x′ + y + z′)(x′ + y′ + z)(x′ + y′ + z′)=(x + y′ + z′)(x′ + y + z′)(x′ + y′ + z)(x′ + y′ + z′)

Product-of-maxterms can also be obtained bycomplementing the sum-of-mintermsProduct-of-maxterms can also be obtained bycomplementing the sum-of-minterms

(F1)′=(x′yz + xy′z + xyz′ + xyz)′ (F1)′=(x′yz + xy′z + xyz′ + xyz)′

=(x + y′ + z′)(x′ + y + z′)(x′ + y′ + z)(x′ + y′ + z′)=(x + y′ + z′)(x′ + y + z′)(x′ + y′ + z)(x′ + y′ + z′)

=M3 M5 M6 M7=M3 M5 M6 M7

F1=(F1′)′F1=(F1′)′

=(x′y′z′ + x′y′z + x′yz′ + xyz′)′=(x′y′z′ + x′y′z + x′yz′ + xyz′)′

=(x + y + z)(x + y + z′)(x + y′ + z)(x′ + y′ + z)=(x + y + z)(x + y + z′)(x + y′ + z)(x′ + y′ + z)

=M0 M1 M2 M4=M0 M1 M2 M4

Slides by Philip Pham, University of California, Irvine

Expansion to Product-of-MaxtermsExpansion to Product-of-Maxterms

Any Boolean function can be expanded into a product-of-maxtermsform be expanding each term with xx′ for each missing variable x.

Example:Example:

ConvertConvert

F =x′y′ + xzF =x′y′ + xz

=(x′y′ + x)(x′y′ + z)=(x′y′ + x)(x′y′ + z)

=(x′ + x)(y′ + x)(x′ + z)(y′ + z)=(x′ + x)(y′ + x)(x′ + z)(y′ + z)

=(x + y′)(x′ + z)(y′ + z)=(x + y′)(x′ + z)(y′ + z)

ExpandExpand

x + y′=x + y′ + zz′=(x + y′ + z)(x + y′ + z′)x + y′=x + y′ + zz′=(x + y′ + z)(x + y′ + z′)

x′ + z=x′ + z + yy′=(x′ + y + z)(x′ + y′ + z)x′ + z=x′ + z + yy′=(x′ + y + z)(x′ + y′ + z)

y′ + z=y′ + z + xx′=(x + y′ + z)(x′ + y′ + z)y′ + z=y′ + z + xx′=(x + y′ + z)(x′ + y′ + z)

CombineCombine

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

=M2 M3 M4 M6=(2, 3, 4, 6)=M2 M3 M4 M6=(2, 3, 4, 6)

missingmissing

xixi

missingmissing

yiyi

missingmissing

cici

Slides by Philip Pham, University of California, Irvine

Conversion to Product-of-MaxtermsConversion to Product-of-Maxterms

Any Boolean expression can be converted into a sum-of-maxterms by generating the truth table and listing all the 0-maxterms.

Example: F = x′y′ + xzExample: F = x′y′ + xz

F(x, y, z)=Σ(0, 1, 5, 7)F(x, y, z)=Σ(0, 1, 5, 7)

F(x, y, z)=Π(2, 3, 4, 6)F(x, y, z)=Π(2, 3, 4, 6)

Slides by Philip Pham, University of California, Irvine

Canonical FormsCanonical Forms

Two canonical forms:

Sum-of-mintermsSum-of-minterms

Product-of-maxtermsProduct-of-maxterms

Canonical forms are unique.

Conversion between canonical forms is achieved by:

Exchanging Σ and ΠExchanging Σ and Π

Listing all the missing indicesListing all the missing indices

Slides by Philip Pham, University of California, Irvine

Standard FormsStandard Forms

Two standard forms

Sum-of-productsSum-of-products

Product-of-sumsProduct-of-sums

Standard forms are not unique.

Sum-of-products is an OR expression with product termsthat may have less literals than minterms

Example:Example:

F = xy + x′yz + xy′zF = xy + x′yz + xy′z

Slides by Philip Pham, University of California, Irvine

Standard FormsStandard Forms

Product-of-sums is an AND expression with sum terms thatmay have less literals than maxterms

Standard forms have fewer operators (literals) thancanonical forms

Standard forms can be derived from canonical forms bycombining terms that differ in one variable (this is, terms atdistance 1)

Example:Example:

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

Example:Example:

F1=xyz + xyz′ + xy′z + x′yzF1=xyz + xyz′ + xy′z + x′yz

=xyz + xyz′ + xyz + xy′z + xyz + x′yz=xyz + xyz′ + xyz + xy′z + xyz + x′yz

=xy(z + z′) + x(y + y′)z + (x + x′)yz=xy(z + z′) + x(y + y′)z + (x + x′)yz

=xy + xz + yz=xy + xz + yz

Slides by Philip Pham, University of California, Irvine

Non-standard FormsNon-standard Forms

Non-standard forms have fewer operators (literals) thanstandard forms.

They are obtained by factoring variables.

Example:Example:

xy + xy′z + xy′w=x(y + y′z + y′w)xy + xy′z + xy′w=x(y + y′z + y′w)

=x(y + y′(z + w))=x(y + y′(z + w))

Slides by Philip Pham, University of California, Irvine

Strategy for Operator (Literal)Reduction in Boolean ExpressionsStrategy for Operator (Literal)Reduction in Boolean Expressions

Algebraic Expression

Generate canonical form

Expand into truth table

Find functional subcubes

Factor sub-expressions

Non-standard form

Truth table

Canonical form

Standard form

Slides by Philip Pham, University of California, Irvine

Binary Logic OperationsBinary Logic Operations

There are 22n Boolean functions for n binary variables

Therefore, 16 Boolean functions for n = 2. They are

Name

OperatorSymbol

FunctionalValues for x,y =

AlgebraicExpression

Comment

Zero

F0 = 0

Binary constant 0

AND

x ∙ y

F1 = xy

x and y

Inhibition

x / y

F2 = xy′

x but not y

Transfer

F3 = x

Inhibition

y / x

F4 = x′y

y but not x

Transfer

F5 = y

XOR

x  y

F6 = xy′ + x′y

x or y but not both

x + y

F7 = x + y

x or y

NOR

x ↓ y

F8 = (x + y)′

Not-OR

Equivalence

x  y

F9 = xy + x′y′

x equals y

Complement

y′

F10 = y′

Not y

Implication

x  y

F11 = x + y′

If y, then x

Complement

x′

F12 = x′

Not x

Implication

x  y

F13 = x′ + y

If x, then y

NAND

x ↑ y

F14 = (xy)′

Not-AND

One

F15 = 1

Binary constant 1

There are two functions thatgenerate constants: Zero andOne. For every combinationof variable values, the Zerofunction will return to 0,whereas the One functionwill return to 1.

There are four functions ofone variable, which indicateComplement and Transferoperations. Specifically, theComplement function willproduce the complement ofone of the binary variables.The Transfer functions bycontrast will reproduce oneof the binary variables at theoutput.

There are ten functions thatdefine eight specific binaryoperations: AND, Inhibition,XOR, OR, NOR, Equivalence,Implication, and NAND.

Slides by Philip Pham, University of California, Irvine

Digital Logic GatesDigital Logic Gates

Name

GraphicSymbol

FunctionalExpression

Number oftransistors

Delayin ns

Inverter

F = x′

Driver

F = x

AND

F = xy

2.4

F = x + y

2.4

NAND

F = (xy)′

1.4

NOR

F = (x + y)′

1.4

XOR

F = x  y

4.2

XNOR

F = x  y

3.2

Basic Logic Library

(CMOS Technology Implementations)

Slides by Philip Pham, University of California, Irvine

Multiple-Input GatesMultiple-Input Gates

Name

GraphicSymbol

FunctionalExpression

Number oftransistors

Delayin ns

3–input AND

F = xyz

2.8

4–input AND

F = xyzw

3.2

3–input OR

F = x + y + z

2.8

4–input OR

F = x + y + z + w

3.2

3–input NAND

F = (xyz)′

1.8

4–input NAND

F = (xyzw)′

2.2

3–input NOR

F = (x + y + z)′

1.8

4–input NOR

F = (x + y + z + w)′

2.2

Multiple-Input Standard Logic Gates

Slides by Philip Pham, University of California, Irvine

Multiple-Operator (Complex) GatesMultiple-Operator (Complex) Gates

Name

GraphicSymbol

FunctionalExpression

Number oftransistors

Delayin ns

2–wide,2–inputAOI

F = (wx + yz)′

2.0

3–wide,2–inputAOI

F = (uv + wx + yz)′

2.4

2–wide,3–inputAOI

F = (uvw + xyz)′

2.2

2–wide,2–inputOAI

F = ((w + x)(y + z))′

2.0

3–wide,2–inputOAI

F = ((u + v)(w + x)(y + z))′

2.2

2–wide,3–inputOAI

F = ((u + v + w)(x + y + z))′

2.4

Multiple-Operator Standard Logic Gates

Slides by Philip Pham, University of California, Irvine

Full-adder Design Using XOR GatesFull-adder Design Using XOR Gates

ci + 1

Truth table

si=xi′yi′ci + xi′yici′ + xiyi′ci′ + xiyicisi=xi′yi′ci + xi′yici′ + xiyi′ci′ + xiyici

=(xi′yi + xiyi′)ci′ + (xi′yi′ + xiyi)ci=(xi′yi + xiyi′)ci′ + (xi′yi′ + xiyi)ci

=(xi  yi)ci′ + (xi  yi)ci=(xi  yi)ci′ + (xi  yi)ci

= (xi  yi)ci′ + (xi  yi)′ci= (xi  yi)ci′ + (xi  yi)′ci

=(xi  yi)  ci=(xi  yi)  ci

ci + 1= xiyici′ + xiyici + xi′yici + xiyi′ci ci + 1= xiyici′ + xiyici + xi′yici + xiyi′ci

=xiyi(ci′ + ci) + ci(xi′yi + xiyi′)=xiyi(ci′ + ci) + ci(xi′yi + xiyi′)

=xi yi + ci(xi  yi)=xi yi + ci(xi  yi)

Full–adder equation

xixi

yiyi

cici

ci + 1ci + 1

2.4

4.2

2.4

Logic Schematic (46 Transistors)

Input/Output

Path

Delay

(ns)

ci to ci + 1

4.8 ns

ci to si

4.2 ns

xi, yi to ci + 1

9.0 ns

xi, yi to si

8.4 ns

Full–adder delays

Slides by Philip Pham, University of California, Irvine

Full-adder Design Using Fast GatesFull-adder Design Using Fast Gates

xi  yi=xi yi + xi′yi′ xi  yi=xi yi + xi′yi′

=((xi yi)′ (xi′yi′)′ )′=((xi yi)′ (xi′yi′)′ )′

=((xi yi)′ (xi + yi))′=((xi yi)′ (xi + yi))′

si=(xi yi)ci′ + (xi  yi)cisi=(xi yi)ci′ + (xi  yi)ci

=(xi  yi)′ci′ + (xi  yi)ci=(xi  yi)′ci′ + (xi  yi)ci

=(xi  yi)  ci=(xi  yi)  ci

ci + 1= xi yi + ci(xi + yi) ci + 1= xi yi + ci(xi + yi)

=((xi yi)′ (ci(xi + yi))′)′=((xi yi)′ (ci(xi + yi))′)′

Full–adder equation

ci + 1

Truth table

Logic Schematic (36 Transistors)

Input/Output

Path

Delay

(ns)

ci to ci + 1

2.8 ns

ci to si

3.8 ns

xi, yi to ci + 1

5.2 ns

xi, yi to si

7.6 ns

Full–adder delays

xixi

yiyi

cici

sisi

ci + 1ci + 1

2.4

1.4

2.4

1.4

Slides by Philip Pham, University of California, Irvine

Full-adder Design withMultiple-input GatesFull-adder Design withMultiple-input Gates

ci + 1

Truth table

Full–adder equation

Logic Schematic (56 Transistors)

xixi

yiyi

cici

sisi

ci + 1ci + 1

2.2

1.8

1.4

1.8

Input/Output

Path

Delay

(ns)

ci to ci + 1

3.2 ns

ci to si

5.0 ns

xi, yi to ci + 1

4.2 ns

xi, yi to si

5.0 ns

Full–adder delays

si=xi′yi′ci + xi′yici′ + xiyi′ci′ + xiyicisi=xi′yi′ci + xi′yici′ + xiyi′ci′ + xiyici

=((xi′yi′ci)′ (xi′yici′)′ (xiyi′ci′)′ (xiyici)’)′=((xi′yi′ci)′ (xi′yici′)′ (xiyi′ci′)′ (xiyici)’)′

ci + 1= xiyi + cixi + ciyi ci + 1= xiyi + cixi + ciyi

=((xiyi)′ (cixi)′ (ciyi)′)′=((xiyi)′ (cixi)′ (ciyi)′)′

Slides by Philip Pham, University of California, Irvine

Full-adder Design with Complex GatesFull-adder Design with Complex Gates

Full–adder equation

ci + 1

Truth table

Input/Output

Path

Delay

(ns)

ci to ci + 1

3.4 ns

ci to si

4.4 ns

xi, yi to ci + 1

3.4 ns

xi, yi to si

4.4 ns

Full–adder delays

Logic Schematic (46 Transistors)

sisi

ci + 1ci + 1

xixi

yiyi

cici

1.4

2.4

2.0

si=xi′yi′ci + xi′yici′ + xiyi′ci′ + xiyicisi=xi′yi′ci + xi′yici′ + xiyi′ci′ + xiyici

=((xi′yi′ci+ xi′yici′)′ (xiyi′ci′ + xiyici)′)′=((xi′yi′ci+ xi′yici′)′ (xiyi′ci′ + xiyici)′)′

ci + 1= xiyi + cixi + ciyi ci + 1= xiyi + cixi + ciyi

=((xiyi)′ (cixi)′ (ciyi)′)′=((xiyi)′ (cixi)′ (ciyi)′)′

=((xi′ + yi′)(ci′ + xi′)(ci′ + yi′))′=((xi′ + yi′)(ci′ + xi′)(ci′ + yi′))′

=(xi′ yi′ + ci′ xi′ + ci′ yi′ )′=(xi′ yi′ + ci′ xi′ + ci′ yi′ )′

Slides by Philip Pham, University of California, Irvine

VLSI TechnologyVLSI Technology

Small-scale integration (SSI)

10 gates/package10 gates/package

Medium-scale integration (MSI)

10 – 100 gates/package (2 – 4 bit slices)10 – 100 gates/package (2 – 4 bit slices)

Large-scale integration (LSI)

100 – 1000 gates/package (controllers, datapaths, bit slices)100 – 1000 gates/package (controllers, datapaths, bit slices)

Very-large-scale integration (VLSI)

1000+ gates/package (systems on a chip)1000+ gates/package (systems on a chip)

Custom designs (Standard cells)

Gate arrays (GAs)

Field-programmable (FPGAs)

Slides by Philip Pham, University of California, Irvine

Custom DesignCustom Design

Each designed by hand

Standard cells

Same height, different widthsSame height, different widths

Routing in channels and over the cellsRouting in channels and over the cells

Two or more metal layersTwo or more metal layers

Standard Cell Approach

Standard Cells

Routing Channel

Slides by Philip Pham, University of California, Irvine

Example of Custom DesignExample of Custom Design

Full-Adder Implementation with Standard Cells

sisi

ci + 1ci + 1

xixi

yiyi

cici

yi ciyi ci

ci + 1ci + 1

xi cixi ci

xi yixi yi

xi′ yi′ ci′xi′ yi′ ci′

yi′yi′

ci′ci′

xi′xi′

sisi

xi′ yi ci′xi′ yi ci′

xi yi′ ci′xi yi′ ci′

xi yi cixi yi ci

Slides by Philip Pham, University of California, Irvine

Semi-Custom Approach withGate ArraysSemi-Custom Approach withGate Arrays

Gate arrays are prefabricated arrays of interconnected gates

All gates are the same type (3–input NAND, for example)

Two or more metal layers used to connect gates

Full-Adder Implementation in a Gate Array

xixi

yiyi

cici

sisi

ci + 1ci + 1

(xi yi)′(xi yi)′

yi′yi′

ci′ci′

xi′xi′

(yi ci)′(yi ci)′

(xi ci)′(xi ci)′

(xi yi ci)′(xi yi ci)′

(xi yi′ci′)′(xi yi′ci′)′

(xi′yi ci′)′(xi′yi ci′)′

(xi′yi′ci)′(xi′yi′ci)′

Slides by Philip Pham, University of California, Irvine

Field–Programmable Approach withFPGAField–Programmable Approach withFPGA

FPGAs are programmed by loading data into internal memory

Excellent for rapid prototyping

Low density and low speed

4-variableBooleanfunction

GQGQ

HQHQ

g1g1

g2g2

g3g3

g4g4

h1h1

h2h2

h3h3

h4h4

4-variableBooleanfunction

PLB

Array Structure

Programmable Logic Blocks (PLB)

Interconnect Point (IP)

Slides by Philip Pham, University of California, Irvine

Full-adder Implemented with FPGAFull-adder Implemented with FPGA

1 programmable logic block for each full-adder

3 out of 4 inputs are used for each Boolean function

ci + 1ci + 1

GQGQ

HQHQ

sisi

g1g1

g2g2

g3g3

g4g4

h1h1

h2h2

h3h3

h4h4

UnusedUnused

cici

xixi

yiyi

Slides by Philip Pham, University of California, Irvine

Chapter SummaryChapter Summary

Boolean Algebra

AxiomsAxioms

Basic theoremsBasic theorems

Boolean Functions

Specification of Boolean Functions

Truth tablesTruth tables

Algebraic expressionsAlgebraic expressions

Canonical forms

Standard forms

Non-standard forms

Algebraic Manipulation of Boolean Expressions

Logic Gates

Simple gatesSimple gates

Multiple-input gatesMultiple-input gates

Complex gatesComplex gates

Implementation Technology

SSI (Small-scale integration)SSI (Small-scale integration)

MSI (Medium-scale integration)MSI (Medium-scale integration)

LSI (Large-scale integration)LSI (Large-scale integration)

VLSI (Very-large-scale integration)VLSI (Very-large-scale integration)

Custom designs

Semi-custom designs

Field-programmable