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Boolean Algebra and Logic Gates
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 =x Example: 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 (switching algebra) in 1938. E. V. Huntington Formulated the postulates in (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 theorem 2(a) = x postulate 2(b)

2.4 Basic theorems and properties of Boolean Algebra (continued)
Operator Precedence Parentheses NOT AND OR

2.5 Boolean Functions 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 (con.)
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’

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

Example 2.1 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 function 4. 4 and 5 are known as consensus theorem

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 algebraically 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 F1=x'yz'+x'y'z, F2=x(y'z'+yz). <Answer>
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' + AB'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 (xy)z  x(yz) <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 xyz = (x+y+z)’ xyz = (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

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