Chapter 2

Boolean Algebra and Logic Gates

2.1 Introduction

• To provide a basic vocabulary and a brief foundation in Boolean algebra– Boolean algebra

• To optimize simple circuits• To understand the purpose of algorithms

for optimizing complex circuits with millions of gates.

2.2 Basic Definitions

• Closure • Associative law• Commutative law• Identity element• Inverse• Distributed law

Closure

• Closure – A set S is closed with respect to a binary operator if,

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

– Example The set of natural number N = {1,2,3,4,…} is closed

with the binary operator plus(+) by rules of arithmetic addition

Associative law, Commutative law, etc.

• Associative law : (x*y)*z = x*(y*z) for all x,y,z∈S

• Commutative law : x*y = y*x for all x,y∈S

• Identity elements: A set is said to have identity elements with respect to a binary operation * on S if there exists an element e ∈S with the property for all x ∈S, e*x = x*e =xExample: set of integers I={…, -3, -2, -1, 0, 1, 2,

3, …}, x + 0 = 0 + x = x

Associative law, Commutative law, etc.

• Inverse : A set S having the identity elements e with a binary operator * is said to have an inverse, for all x∈S , y∈S, x * y = e example: In the set of integers, I, and the

operator + with 0, the inverse of an element a is (-a), since a + (-a) = 0

• Distributive law : x*(y.z)=(x*y).(x*z)

2.3 Axiomatic definitions of Boolean algebra

• George Boole– Developed Boolean algebra in 1854.

• C. E. Shannon– Introduced a two-valued Boolean algebra (s

witching algebra) in 1938.• E. V. Huntington

– Formulated the postulates in 1904. (refer to page 54)

Two-valued Boolean Algebra

• Two_valued Boolean algebra is defined on a set of two elements, B ={0,1}, with rules for the two binary operators + and .

• Closure• Two identity elements: 0 for + and 1 for • Distributed law

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

• x + x’=1 x x’ = 0

2.4 Basic theorems and properties of Boolean

Algebra• Duality

– If the dual of an expression is desired, we simply interchange OR and AND operators and replace 1’s by 0’s and 0’s by 1’s.

Basic theorems (Table 2.1)

The theorems, postulates in Table 2.1 are listed in pairs; each relation is the dual of the one paired with it.

A Proof of theorems

• x + xy = x x + xy = x.1 + xy postulate 2(b) = x(1+y) postulate 4(a) = x(y+1) postulate 3(a) = x.1 theorem 2(a) = x postulate 2(b)

2.4 Basic theorems and properties of Boolean Algebra (continued)

• Operator Precedence1. Parentheses2. NOT3. AND4. OR

Boolean algebra is an algebra that deals with binary variables and logic operations

An example: F1 = x + y'z

Its truth table is shown in Table 2.2 and its logic-circuit diagram is shown in Figure 2.1

2.5 Boolean Functions

Equivalent logics with the same truth table An example: F2 in Table

F2 = x’y’z + x’yz + xy’ = x’z ( y’+ y ) + xy’ = x’z + xy’

2.5 Boolean Functions (con.)

Implementation of F2

Algebraic manipulation

• A literal: a single variable within the term that requires a logic gate.

• Minimization: obtaining a simpler circuit

• Designers of digital circuits use computer minimization programs

1. x(x'+y) = xx' + xy = 0 + xy = xy. 2. x +x'y = (x+x')(x+y) = 1(x+y) = x + y. 3. (x+y)(x+y') = x + xy + xy' + yy' = x(1+y+y') = x. 4. xy + x'z + yz = xy + x'z + yz(x+x') = xy + x'z + xyz + x'yz = xy(1+z) + x'z(1+y) = xy + x'z 5. (x+y)(x'+z)(y+z) = (x+y)(x'+z) : by duality from f

unction 4. 4 and 5 are known as consensus theorem

Example 2.1

Complement of a function

The complement of a function F is F’ F’ is obtained from an exchange of 0’s for 1’s and 1’s

for 0’s of function F The complement of a function may derived algebraical

ly through DeMorgan’s theorem.

Complement of a function (cont.)

DeMorgan’s theorem (Table 2.1) can be extended to three variables.

(A + B + C)'= (A+x)' let B+C=x = A'x' by theorem 5(a)(DeMorgan) = A'(B+C)' substitute B+C=x = A'(B'C') by theorem 5(a)(DeMorgan) = A'B'C' by theorem 4(b)(associative)

DeMorgan’s theorem can be generalized. (A+B+C+D+…+F)' = A'B'C'D'…F' (ABCD…F)' = A' +B'+ C' + D' + … + F'

Example 2.2

Find the complement of the functions F1=x'yz'+x'y'z, F2=x(y'z'+yz).

<Answer> F1' = (x'yz'+x'y'z)' = (x'yz')'(x'y'z)'

= (x+y'+z)(x+y+z') F2' = [x(y'z'+yz)]' = x'+(y'z'+yz)'

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

Example 2.3

Find the complement of the functions F1 And F2 Example 2.2 by taking their duals and complementing each literal.

<Answer> 1. F1 = x'yz' + x'y'z. The dual of F1 is (x'+y+z')(x'+y'+z) Complement each literal : (x+y'+z)(x+y+z')=F1' 2. F2 = x(y'z'+yz). The dual of F2 is x+(y'+z')(y+z) Complement each literal : x'+(y+z)(y'+z')=F2'

2.6 Canonical and Standard Forms

• Minterms and Maxterms

Functions of Three Variables

• Example: f1, f2 , f1’?

Functions of Table 2.4

f1 = x’y’z+xy’z’+xyz = m1 + m4 + m7

f2 = x’yz+xy’z+xyz’+xyz = m3 + m5 + m6 + m7

f1’ = x’y’z’+ x’yz’ + x’yz + xy’z + xyz’ (f1’)’= (x’y’z’+ x’yz’+x’yz+xy’z + xyz’)’ = (x+y+z)(x+y’+z)(x+y+z)(x’+y+z’)(x’+y’+z) = M0.M2.M3.M5.M6 = f1

Similarly f2 = M0.M1.M2.M4

Sum of Minterms

• Any Boolean functions can be expressed as sum of minterms.

Example 2.4

Express F = A + B’C in sum of minterms.<Answer> A = A(B+B') = AB +AB' = AB(C+C') + AB'(C+C') = ABC + ABC' + A

B'C +AB'C‘B'C = B'C(A+A') = AB'C + A'B'C F = A + B'C = A' B'C + AB'C' + AB'C + ABC' + ABC

= m1 + m4 + m5 + m6 + m7

= ∑(1, 4, 5, 6, 7)

Product of Maxterms

• Any Boolean functions can be expressed as product of maxterms.

Example 2.5

Express the Boolean function F = xy + x'z in a product of maxterm form.

<Answer> F = xy + x'z = (xy+x')(xy+z)

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

= M0M2M4M5

= ∏(0, 2, 4, 5)

Conversion between Canonical Forms

• F(A,B,C) = (1,4,5,6,7) F’(A,B,C) = (0,2,3) = m0+m2+m3

If F, (F’)’ is taken by DeMorgan’s theorem

F=(m0+m2+m3)’=m0’.m2’.m3’ = M0M2M3 = (0,2,3)

• It is clear that mj’=Mj

Example: F(x,y,z)=xy+x’z

• F=xy+x’z = (1,3,6,7) = (0,2,4,5)

Standard Forms

• Sum of Products(SOP): a Boolean expression containing AND terms

• Product of Sums(POS): a Boolean expression containing OR terms

F1= y’ + xy + x’yz’ F2= x(y’+z)(x’+y+z)

Nonstandard forms: F3=AB+C(D+E)

• Figure 2-4(a): Nonstandard forms• Figure 2-4(b): Conversion to its standard

form

• 3-level 2- level

Incompletely Specified Functions

• Assume that the output of N1 does not generate all possible combinations of values for A,B, and C.

The function F is incompletely specified.

The minterm A’B’C and ABC’ are “don’t care terms”.

F = m(0,3,7)+ d(1,6)

2.7 Other Logic Operations

• 16 functions of two binary variables

Boolean expressions of the 16 functions

2.8 Digital Logic Gates

Extension to Multiple Inputs

• NAND and NOR operators are not associative (xy)z x(yz)

<Proof> (x↓y)↓z= [(x+y)'+z]' = (x+y)z'= xz' + yz' x↓(y↓z)= [x+(y+z)'] ' = x'(y+z)= x'y + x'z

3-input Gates• By definition xyz = (x+y+z)’ xyz = (xyz)’• F = [(ABC)'(DE)']' = ABC + DE

3-input XOR gate

Positive and Negative Logic

2-8 Integrated Circuits

• SSI• MSI• LSI• VLSI

Digital Logic Families• Typical digital IC families

– TTL– ECL– MOS– CMOS

• Evaluation parameters– Fan-out– Fan-in– Power of dissipation– Propagation delay– Noise margin