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**Boolean Algebra and Logic Gates**

Chapter 2 Boolean Algebra and Logic Gates

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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.

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**2.2 Basic Definitions Closure Associative law Commutative law**

Identity element Inverse Distributed law

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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

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**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

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**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)

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**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)

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**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

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**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.

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**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.

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**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)

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**2.4 Basic theorems and properties of Boolean Algebra (continued)**

Operator Precedence Parentheses NOT AND OR

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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

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**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’

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Implementation of F2

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**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

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**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

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**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.

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**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'

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**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')

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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'

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**2.6 Canonical and Standard Forms**

Minterms and Maxterms

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**Functions of Three Variables**

Example: f1, f2 , f1’?

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**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

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Sum of Minterms Any Boolean functions can be expressed as sum of minterms.

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**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)

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Product of Maxterms Any Boolean functions can be expressed as product of maxterms.

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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)

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**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

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**Example: F(x,y,z)=xy+x’z**

F=xy+x’z = (1,3,6,7) = (0,2,4,5)

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Standard Forms Sum of Products(SOP): a Boolean expression containing AND terms Product of Sums(POS): a Boolean expression containing OR terms

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

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**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

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**Incompletely Specified Functions**

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

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**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)

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**2.7 Other Logic Operations**

16 functions of two binary variables

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**Boolean expressions of the 16 functions**

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2.8 Digital Logic Gates

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**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

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**3-input Gates By definition xyz = (x+y+z)’ xyz = (xyz)’**

F = [(ABC)'(DE)']' = ABC + DE

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3-input XOR gate

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**Positive and Negative Logic**

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2-8 Integrated Circuits SSI MSI LSI VLSI

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**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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A. Abhari CPS2131 Chapter 2: Boolean Algebra and Logic Gates Topics in this Chapter: Boolean Algebra Boolean Functions Boolean Function Simplification.

A. Abhari CPS2131 Chapter 2: Boolean Algebra and Logic Gates Topics in this Chapter: Boolean Algebra Boolean Functions Boolean Function Simplification.

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