chapter 4 boolean algebra rev2010 stud

8/12/2019 Chapter 4 Boolean Algebra Rev2010 Stud

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Basic Postulates

Duality

Fundamental Theorems

Simplification technique (Boolean Algebra)

Standard Forms of Boolean Expressions

Conversion to NAND/NAND and NOR/NOR network

1

CHAPTER 4

BOOLEAN ALGEBRA


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Background2

Boolean algebra is the mathematics of digital systems.

Terms used in Boolean algebra

A variableis a symbol used to represent a logical quantity.Any single variable can have the value of 1 or 0.

The complementis the inverse of a variable and is indicatedby a bar over the variable.

The complement of the variable . If , then .Sometimes a prime symbol is used to denote a complement ofa variable, .

A literalis a variable or the complement of the variable.


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Background3

Boolean Additionis the equivalent to the OR operation.

A sum termis the sum of literals. In logic circuits, a sumterm is produced by an OR operation with no AND

operations involved. Ex. A + B , A + B + C + D

A sum term is equal to 1 when one or more of the literalsin the term are 1.

A sum term is equal to 0 only if each of the literals is 0


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Background4

Boolean Multiplicationis the equivalent to the ANDoperation.

Aproduct termis the product of literals. In logic

circuits, a product term is produced by an AND operationwith no OR operations involved.

Ex. AB , ABCD

A product term is equal to 1 only if each of the literals

in the term is 1.

A product term is equal to 0 when one or more of theliterals are 0.


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Basic Postulates (Laws)5

Postulate 1: Definition A Boolean algebra is a closed algebraic system containing a set K of

two or more elements and the two operators . and +

Postulate 2: Existence of 1 and 0 elementsa) a + 0 = a

b) a . 1 = a

Postulate 3: Commutativitya) a + b = b + a

b) a . b = b . a


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Basic Postulates6

Postulate 4: Associativitya) a + ( b + c ) = ( a + b ) + c

b) a . ( b . c ) = ( a . b ) . c

Postulate 5: Distributivitya) a + ( b . c ) = ( a + b ) . ( a + c )

b) a . ( b + c ) = ( a . b ) + ( a . c )

Postulate 6: Existence of the complementa) a + a = 1

b) a . a = 0


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Duality7

The principle of duality states that , if an expression isvalid in Boolean algebra, the dual of the expression isalso valid.

The dual expression is found by replacing all + operatorswith . , all . operators with +, all 1s with 0s, and all 0swith 1s.


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Fundamental Theorems (Rules)8

There are 9 fundamental theorems of Boolean algebra,each of which rooted from the basic postulates.

Theorem 1: Idempotency

a) a + a = a

b) a . a = a

Theorem 2: Null elements for + and . operators

a) a + 1 = 1

b) a . 0 = 0


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9

Theorem 3

a) a'' = a

Theorem 4: Absorptiona) a + ab = a

b) a (a + b) = a

Theorem 5a) a + ab= a + b

b) a (a + b) = ab



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10

Theorem 6

a) ab + ab = a

b) (a+b)(a+b) = a

Theorem 7

a) ab + abc= ab + ac

b) (a+b)(a+b+c) = (a+b)(a+c)



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11

Theorem 8: DeMorgans theorem

a) (a+b) = ab

b) (ab) = a + b

Theorem 9: Consensus

a) ab + ac +bc = ab + ac

b) (a+b)(a+c)(b+c) = (a+b)(a+c)



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Simplification of Boolean Expression12


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13

Truth Table

A B C

0 0 0 1

0 0 1 10 1 0 1

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 0


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14

Simplification of Boolean Expression


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15

Simplification of Boolean Expression


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16

Truth Table

A B C

0 0 0 1

0 0 1 10 1 0 1

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 0

CBA


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Standard Forms of Boolean Expressions17

We can express the algebraic function in the form ofalgebraic expressions, truth tables, Venn diagrams,SOP/POS form and also the standard forms (canonicalforms).

All Boolean expression, regardless of their form, can beconverted into either one of the two standard forms:

sum-of-products(SOP) form, or

product-of-sum(POS)form.

Standardization makes the evaluation, simplification, andimplementation of Boolean expressions more systematic.


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Domain of a Boolean expression18

The domainof a general Boolean expression is the setof variables contained in the expression either incomplemented or uncomplemented form. Example, thedomain of the expression ABC + ACD +BD is the set

of variables A, B, C, D.

f ( A, B, C, D )= ABC + ACD +BD


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The Sum-of-Products (SOP) form19

SOP : A function where each product term is formed byadding a number of complemented or uncomplementedvariables (literal).

Ex: f(A, B, C, D) = ABC + BD + ACD

In an SOP expression, a single overbar cannot extend

more than one variable, although more than one variablein a term can have an overbar. For example, an SOPexpression can have the term but not .


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The Sum-of-Products (SOP) form20

Any logic expression can be changed into SOP form byapplying Boolean algebra techniques.

Ex.


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Canonical SOP21

Also known as the Standard Sum-of-Products form

A standard SOP expression is one in which allthevariables in the domain appear in each product term in

the expression. Any nonstandard SOP can be convertedto the standard SOP form using Boolean algebra.

Ex. Represent the following Boolean function in thecanonical SOP form:


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Canonical SOP22

=

=

Now look at each product terms and find for missing domain,

AB (missing C & D). Thus, add these domains into the term

AB = AB (C+C)(D+D)

= (A B C + A B C)(D + D)

= ABCD + ABCD + ABCD + ABCD


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Canonical SOP23

CD (missing A & B).

CD =CD (A+A)(B+B)

= (ACD + ACD)(B+B)

= ABCD + ABCD + ABCD + ABCD

Next, combine all these product terms,


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Canonical SOP24

AB + ABCD + CD

= ABCD + ABCD + ABCD + ABCD + ABCD+ ABCD+

ABCD + ABCD+ ABCD

= ABCD+ ABCD + ABCD + ABCD + ABCD + ABCD +

ABCD


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Canonical SOP25

For a switching function with n variables, if a productterm contains each of the n variables exactly one time incomplemented or uncomplemented form, the productterm is called a minterm.

Minterm function is called as canonical sum-of-productsform (canonical SOP) or the standard sum-of-products.Ex1: f(A,B,C) = ABC + ABC +ABC +ABC

An SOP expression is equal to 1 only if one or more ofthe product terms in the expression is equal to 1.


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The Product-of-Sums (POS) form26

A function by taking the product of ORing (sum terms)where each sum term is formed by ORing a number ofcomplemented or uncomplemented variables.

Ex: f(A, B, C, D) = (A+B+C)(B+C+ D)(A+ C+ D)

In a POS expression, a single overbar cannot extendmore than one variable, although more than one variablein a term can have an overbar. For example, a POSexpression can have the term but not.


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Canonical POS28

The terms are already in pos form. So inspect for missingdomain from each sum terms and add the domain.

(A + B) is missing domain C ;A + B = A + B + C C

= (A + B + C)(A + B + C)

( B + C) is missing domain A ;

B + C = AA + B + C

= (A + B + C)(A + B + C)


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Therefore, (A+B)(B+C) in standard POS form is

For a switching function with n variables, if a sum termcontains each of the n variables exactly one time incomplemented or uncomplemented form, the sum term iscalled a maxterm.

Thus, maxterm function is called as canonical product-of-sums form (canonical POS) or the standard product-of-sums.

Ex: f(A,B,C)= (A+B+C)(A+B+C)(A+B+C)(A+B+C)

Canonical POS29


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Canonical Forms30

A POS expression is equal to 0 only if one or more

of the sum terms in the expression is equal to 0.

An SOP expression is equal to 1 only if one or moreof the product terms in the expression is equal to 1.


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Decimal

A B Cf

(A,B,C) Minterms Maxterms

0 0 0 0 0 ABC m0 A+B+C M0

1 0 0 1 0 ABC m1 A+B+C M1

2 0 1 0 1 ABC m2 A+B+C M23 0 1 1 1 ABC m3 A+B+C M3

4 1 0 0 0 ABC m4 A + B+C M4

5 1 0 1 0 ABC m5 A+B+C M5

6 1 1 0 1 ABC m6 A+B+C M6

7 1 1 1 1 ABC m7 A+B+C M7

Ex3: Truth table for a function of 3 variables,


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From the truth table, determining standard SOP formfor the functionf:

Compact form: f(A,B,C) = m2+ m3 + m6 + m7

Minterm list form: f(A,B,C) =

Standard SOP form: f(A,B,C) =

32

)7,6,3,2(m

CBACBACBACBA

Determining standard POS form for the functionf:Compact form: f(A,B,C) = M0 M1 M4 M5

Maxterm list form: f(A,B,C) =

Standard POS form: f(A,B,C) =

)5,4,1,0(M

))()()(( CBACBACBACBA

chapter 4 boolean algebra rev2010 stud

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