chapter 2 boolean algebra and logic gates. 2.1 introduction to provide a basic vocabulary and a...
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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
• 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 =xExample: 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 (s
witching algebra) in 1938.• E. V. Huntington
– Formulated the postulates in 1904. (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 Precedence1. Parentheses2. NOT3. AND4. OR
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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
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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 f
unction 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 algebraical
ly 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 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' + 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)
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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