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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 –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 C. E. Shannon –Introduced a two-valued Boolean algebra (switching algebra) in 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.1 theorem 2(a) = x postulate 2(b)

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2.4 Basic theorems and properties of Boolean Algebra (continued) Operator Precedence 1.Parentheses 2.NOT 3.AND 4.OR

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Boolean algebra is an algebra that deals with binary variables and logic operations An example : F 1 = x + y'z Its truth table is shown in Table 2.2 and its logic-circuit diagram is shown in Figure Boolean Functions

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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.)

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

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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 Find the complement of the functions F 1 =x'yz'+x'y'z, F 2 =x(y'z'+yz). F 1 ' = (x'yz'+x'y'z)' = (x'yz')'(x'y'z)' = (x+y'+z)(x+y+z') F 2 ' = [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 F 1 And F 2 Example 2.2 by taking their duals and complementing each literal. 1. F 1 = x'yz' + x'y'z. The dual of F 1 is (x'+y+z')(x'+y'+z) Complement each literal : (x+y'+z)(x+y+z')=F 1 ' 2. F 2 = x(y'z'+yz). The dual of F 2 is x+(y'+z')(y+z) Complement each literal : x'+(y+z)(y'+z')=F 2 '

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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 = m 1 + m 4 + m 7 f2 = x yz+xy z+xyz +xyz = m 3 + m 5 + m 6 + m 7 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) = M 0.M 2.M 3.M 5.M 6 = f1 Similarly f2 = M 0.M 1.M 2.M 4

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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. 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 = m 1 + m 4 + m 5 + m 6 + m 7 = (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. 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') = M 0 M 2 M 4 M 5 = (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) = m 0 +m 2 +m 3 If F, (F ) is taken by DeMorgan s theorem F=(m 0 +m 2 +m 3 ) =m 0.m 2.m 3 = M 0 M 2 M 3 = (0,2,3) It is clear that m j =M j

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

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