boolean algebra and logic gates -...

Digital Circuit Design

Boolean Algebra and Logic Gates

Lan-Da Van (范倫達), Ph. D.

Department of Computer Science

National Chiao Tung University

Taiwan, R.O.C.

Spring, 2018

[email protected]

http://www.cs.nctu.edu.tw/~ldvan/

Lecture 2


Lan-Da Van DCD-02-2

Outlines

Basic Definitions of Algebra

Axiomatic Definitions of Boolean Algebra

Basic Theorems and Properties of Boolean Algebra

Boolean Function

Canonical and Standard Forms

Other logic Operations

Digital Logic Gates

Lecture 2


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History

In 1854, George Boole developed an

algebraic system

Now, called Boolean algebra

In 1904, E. V. Huntington formulated

several postulates for Boolean

algebra.

Widely used in research nowadays.

In 1938, C. E. Shannon introduced a

two-valued Boolean algebra

Called switching algebra

https://en.wikipedia.org/wiki/George_Boole

http://www.bjinnovate.com/archives/515105.html

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Basic definitions: 1. Closure: A set S is closed with respective to a binary

operator if, for every pair of elements of S, the binary operator specifies a rule for obtaining a unique element of S.

2. Associative law: A binary operator * on a set S is said to be associative whenever (x*y)*z= x*(y*z) for all x, y, z S.

3. Commutative law: A binary operator * on a set S is said to be commutative whenever x*y= y*x for all x, y S.

4. Identity element: A set S is said to have an identity element with respect to a binary operation * on S if there exists an element eS with the property that x*e = e*x =x for every x S.

5. Inverse: A set S having the identity element e with respect to a binary operation * is said to have an inverse whenever x S, there exists an element yS such that x*y = e.

6. Distributive law: If * and ‧are two binary operators on a set S, * is said to be distributive over ‧ whenever x*(y‧z)=(x*y)‧(x*z)

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Axiomatic Definition of Boolean Algebra

Boolean Algebra defined by a set of elements B and two binary

operators + and ‧ and has the following postulates:

Postulate 1:

(a) The structure is closed with respect to the operator +.

(b) The structure is closed with respect to the operator ‧.

Postulate 2:

(a) The element 0 is an identity element with respect to +.

0+x = x+0 = x

(b) The element 1 is an identity element with respect to ‧.

1‧x = x‧1= x

Postulate 3:

(a) The structure is commutative with respect to +.

x+y = y+x

(b) The structure is commutative with respect to ‧.

x‧y = y‧x

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Axiomatic Definition of Boolean Algebra

Postulate 4:

(a) The operator ‧ is distributive over +:

x‧(y+z)=(x‧y)+(x‧z)

(b) The operator + is distributive over‧

x+(y‧z)=(x+y)‧(x+z)

Postulate 5:

(a) For every element x B, there exists an element x' B (complement of x) such that (a) x+x'=1 and (b) x‧x'=0.

Postulate 6: There exist at least two elements x, y B such that x ≠ y.

Difference with ordinary algebra

The operator + is distributive over‧ is valid for Boolean algebra, but not for ordinary algebra.

Boolean algebra does not have additive and multiplicative inverses; therefore, there are not subtraction or division operations.

Complement is valid for Boolean algebra, but not for ordinary algebra.

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Two-Valued Boolean Algebra

Definition: A two-valued Boolean algebra is

defined on a set of two elements, B = {0,1},

with rules for the two binary operators + and ‧as shown in the following operator tables.

The rules of operations

x y xy x y x+y x x

0 0 0 0 0 0 0 1

0 1 0 0 1 1 1 0

1 0 0 1 0 1

1 1 1 1 1 1

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Postulate 1: The structure is closed with respect to the two

operators is obvious from the tables since the result of each

operation is either 1 or 0 and {1, 0} B.

Postulate 2: From the table, we see that

0+0=0 0+1=1+0=1;

1‧1=1 1‧0=0‧1=0

=> Two identity elements

Postulate 3: The commutative laws are obvious from the

symmetry of the binary operator tables.

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Postulate 4: The distributive law can be shown to

hold from the truth table of all possible values of x, y,

and z.

Postulate 5: Complement

x+x'=1: since 0+0'=0+1=1 and 1+1'=1+0=1

x‧x'=0: since 0‧0'=0‧1=0 and 1‧1'=1‧0=0

Postulate 6: Has two distinct elements 1 and 0, with 0

≠ 1

Note

A set of two elements

+ : OR operation; ‧ : AND operation

A complement operator: NOT operation

Binary logic is a two-valued Boolean algebra

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Basic Theorems and Properties

Duality

The binary operators are interchanged; AND OR

The identity elements are interchanged; 1 0

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• Theorem 1(a): x+x = x

– x+x = (x+x) 1 by postulate 2(b)

= (x+x) (x+x') 5(a)

= x+xx' 4(b)

= x+0 5(b)

= x 2(a)

– Theorem 1(b): x‧x = x

– x‧x = x‧ x + 0 by postulate 2(a)

= xx + xx' 5(b)

= x (x + x') 4(a)

= x‧1 5(a)

= x 2(b)


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Theorem 2(a): x + 1 = 1

x + 1 = 1 (x + 1)

= (x + x')(x + 1)

= x + x' 1

= x + x‘ = 1

Theorem 2(b): x ‧0 =0

By duality

Theorem 3: (x')' = x

Postulate 5 defines the complement of x, x + x' = 1 and x‧x' = 0

The complement of x' is x is also (x')'


Lecture 2



Theorem 6 (a): x + xy = x

x + xy = x 1 + xy

= x (1 +y)

= x‧1= x

Theorem 6 (b): x(x+y) = x

By duality

By means of truth table

x y xy x + xy

0 0 0 0

0 1 0 0

1 0 0 1

1 1 1 1


Lecture 2



DeMorgan's Theorems

(x+y)' = x' y'

(x y)' = x' + y‘

Operator Precedence

– Parentheses

– NOT

– AND

– OR

x y x+y (x+y) x y xy

0 0 0 1 1 1 1

0 1 1 0 1 0 0

1 0 1 0 0 1 0

1 1 1 0 0 0 0


Lecture 2



Boolean Functions

A Boolean function

binary variables

binary operators OR and AND

unary operator NOT

parentheses

Examples

F1= x y z'

F2 = x + y'z

F3 = x' y' z + x' y z + x y'

F4 = x y' + x' z

F3=F4 Why and How to Verify???

Lecture 2



Boolean Functions The truth table of 2n entries

Two Boolean expressions may specify the same function

F3 = F4

x y z F1 F2 F3 F4

0 0 0 0 0 0 0

0 0 1 0 1 1 1

0 1 0 0 0 0 0

0 1 1 0 0 1 1

1 0 0 0 1 1 1

1 0 1 0 1 1 1

1 1 0 1 1 0 0

1 1 1 0 1 0 0

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• Implementation with logic gates

• F4 is more economical

F3

F4

F2

Boolean Functions

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Algebraic Manipulation

To minimize Boolean expressions literal: a primed or unprimed variable (an input to a gate)

term: an implementation with a gate

The minimization of the number of literals and the number of terms => a circuit with less equipment

It is a hard problem (no specific rules to follow)

Ex1: x(x'+y) = xx' + xy = 0+ xy = xy

Ex2: x+x'y = (x+x')(x+y) = 1 (x+y) = x+y

Ex3: (x+y)(x+y') = x+xy+xy'+yy' = x(1+y+y') = x

Ex4: x'y'z + x'yz + xy' = x'z(y'+y) + xy'= x'z + xy'

Ex5: xy + x'z + yz = xy + x'z + yz(x+x')= xy + x'z + yzx + yzx'= xy(1+z) + x'z(1+y) = xy +x'z

Ex6: (x+y)(x'+z)(y+z) = (x+y)(x'+z) by duality from the previous result

Lecture 2



Complement of a Function

Complement Procedures:

An interchange of 0's for 1's and 1's for 0's in the value of F

By DeMorgan's theorem

(A+B+C)' = (A+X)' let B+C = X

= A'X‘ by DeMorgan's

= A'(B+C)' by DeMorgan's

= A'(B'C') by associative

= A'B'C'

Generalizations

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

(ABC ... F)' = A'+ B'+C'+ ... +F'

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Complement Examples:

(x'yz' + x'y'z)' = (x'yz')' (x‘y'z)'

= (x+y'+z) (x+y+z')

[x(y'z'+yz)]' = x' + ( y'z'+yz)'

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

= x' + (y+z) (y'+z')

A simpler procedure

Take the dual of the function and complement each literal

Complement of a Function

Lecture 2




Minterms

A minterm (also called a standard product): an AND term

consists of all literals in their normal form or in their

complement form

For example, two binary variables x and y,

xy, xy', x'y, x'y'

n variables has 2n minterms

Maxterms

A maxterm (also called a standard sum): an OR term

For example, two binary variables x and y,

x+y, x+y', x‘+y, x‘+y'

n variables has 2n maxterms

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Each maxterm is the complement of its corresponding

minterm, and vice versa

x y z term Term

0 0 0 x‘y’z‘ m0 x+y+z M0

0 0 1 x‘y’z m1 x+y+z‘ M1

0 1 0 x‘yz‘ m2 x+y‘+z M2

0 1 1 x‘yz m3 x+y‘+z‘ M3

1 0 0 xy’z‘ m4 x‘+y+z M4

1 0 1 xy’z m5 x‘+y+z‘ M5

1 1 0 xyz‘ m6 x‘+y‘+z M6

1 1 1 xyz m7 x‘+y‘+z‘ M7


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A Boolean function can be expressed by

a truth table

sum of minterms

f1 = x'y'z + xy'z' + xyz = m1 + m4 +m7 =S(1, 4, 7)

f2 = x'yz+ xy'z + xyz'+xyz = m3 + m5 +m6 + m7 =S(3, 5, 6, 7)

x y z f1 f2

0 0 0 0 0

0 0 1 1 0

0 1 0 0 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 0

Canonical Forms

Lecture 2



Canonical Forms

The complement of a Boolean function

Find the minterms that produce a 0

f1' = m0 + m2 +m3 + m5 + m6

= x'y'z'+x'yz'+x'yz+xy'z+xyz'

f1 = (f1')'

= (x+y+z)(x+y'+z) (x+y'+z') (x'+y+z')(x'+y'+z)

= M0 M2 M3 M5 M6

Any Boolean function can be expressed as

A sum of minterms

A product of maxterms

Both are called “canonical form”

Lecture 2



Canonical Forms

Sum of minterms

F = A+B'C

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

= AB +AB' + B'C

= AB(C+C') + AB'(C+C') + (A+A')B'C

=ABC+ABC'+AB'C+AB'C'+A'B'C

F(A,B,C)

= A'B'C +AB'C' +AB'C+ABC'+ ABC

= m1 + m4 +m5 + m6 + m7

= S(1, 4, 5, 6, 7)

Or, built the truth table first

Lecture 2



Canonical Forms

Product of maxterms F1(x,y,z) =F1 = (x+y+z)(x+y'+z)(x'+y+z)(x'+y+z')

= M0M2M4M5 = P(0,2,4,5)

F2= x + yz = (x + y)(x + z) = (x+y+zz')(x+z+yy')=(x+y+z)(x+y+z’)(x+y'+z)

F3 = x'+y = x' + y + zz'= (x'+y+z)(x'+y+z')

F4 = xy + x'z= (xy + x') (xy +z)= (x+x')(y+x')(x+z)(y+z)= (x'+y)(x+z)(y+z)

=(x+y+z)(x+y'+z)(x'+y+z)(x'+y+z')=M0M2M4M5

Lecture 2



Conversion between Two Canonical Forms

Conversion between Canonical Forms

F(A,B,C) = S(1,4,5,6,7)

F'(A,B,C) = S(0,2,3)

By DeMorgan's theorem

F(A,B,C) = P(0,2,3)

mj' = Mj

Interchange the symbols S and P and list those numbers

missing from the original form

S of 1's

P of 0's

Lecture 2



Conversion between Two Canonical Forms

Example

F = xy + xz

F(x, y, z) = S(1, 3, 6, 7)

F(x, y, z) = P (0, 2, 4, 5)

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Standard Forms

Standard Forms

Canonical forms are seldom used

sum of products

Example 1: F1 = y' + xy+ x'yz'

product of sums

Example 2: F2 = x(y'+z)(x'+y+z)

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Standard Forms

Multi-level implementation

Example 3: F3 = AB+C(D+E)=AB+CD+CE

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Other Logic Operations

2n rows in the truth table of n binary variables

22nfunctions for n binary variables

16 functions of two binary variables

All the new symbols except for the exclusive-OR symbol are not in

common use by digital designers

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Other Logic Operations

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Digital Logic Gates

Boolean expression: AND, OR and NOT operations

Constructing gates of other logic operations the feasibility and economy

the possibility of extending gate's inputs

the basic properties of the binary operations

the ability of the gate to implement Boolean functions

Consider the 16 functions two: null and identity are equal to a constant

four: inhibition, implication, transfer, complement are repeated twice

inhibition and implication are not commutative or associative.

eight: transfer, complement, AND, OR, NAND, NOR, XOR, and equivalence are used as standard gates

complement: inverter

transfer: buffer (increasing drive strength)

equivalence: XNOR

Lecture 2



Digital Logic Gates

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Extension to multiple inputs

A gate can be extended to multiple inputs

if its binary operation is commutative and associative

AND and OR are commutative and associative

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

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

Digital Logic Gates

Lecture 2



NAND and NOR are commutative but not associative

=> they are not extendable

Digital Logic Gates

z

Lecture 2



Multiple NOR = a complement of OR gate

Multiple NAND = a complement of AND

The cascaded NAND operations = sum of products

The cascaded NOR operations = product of sums

Digital Logic Gates

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The XOR and XNOR gates are commutative and associative

Multiple-input XOR gates are uncommon?

XOR is an odd function: it is equal to 1 if the inputs variables have

an odd number of 1's

Digital Logic Gates

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Positive and Negative Logic

Positive and Negative Logic

two signal values <=> two logic

values

positive logic: H=1; L=0

negative logic: H=0; L=1

Consider a TTL gate

a positive logic NAND gate

a negative logic OR gate

the positive logic is used in this book

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Lecture 2



Conclusions

From this lecture, you have learned the follows:


Axiomatic Definitions of Boolean Algebra

Basic Theorems and Properties of Boolean Algebra

Boolean Function


Other logic Operations

Digital Logic Gates

boolean algebra and logic gates -...

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