03-BooleanAlgebra (3)

8/11/2019 03-BooleanAlgebra (3)

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Boolean Algebra and Logic Gates 1

Binary Logic and Gates

Binary variables take on one of two values. Logical operators operate on binary values andbinary variables.

Basic logical operators are the logic functionsAND, OR and NOT.Logic gates implement logic functions. Boolean Algebra: a useful mathematical systemfor specifying and transforming logic functions. We study Boolean algebra as a foundation fordesigning and analyzing digital systems!


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Binary Variables

Recall that the two binary values havedifferent names: True/False On/Off Yes/No 1/0

We use 1 and 0 to denote the two values. Variable identifier examples: A, B, y, z, or X 1 for now RESET, START_IT, or ADD1 later


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Logical Operations

The three basic logical operations are: AND OR

NOTAND is denoted by a dot ().OR is denoted by a plus (+).

NOT is denoted by an overbar ( ), asingle quote mark (') after, or (~) beforethe variable.


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Examples: is read Y is equal to A AND B. is read z is equal to x OR y.

is read X is equal to NOT A.

Notation Examples

Note: The statement:1 + 1 = 2 (read one plus one equals two)

is not the same as1 + 1 = 1 (read 1 or 1 equals 1).

= B A Y y x z =

A X =


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Operator Definitions

Operations are defined on the values"0" and "1" for each operator:

AND 0 0 = 0 0 1 = 0

1 0 = 0 1 1 = 1

OR

0 + 0 = 00 + 1 = 1

1 + 0 = 11 + 1 = 1

NOT

1 0 0 1


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01

10X

NOTX Z

Truth Tables

Tabular listing of the values of a function for allpossible combinations of values on its argumentsExample: Truth tables for the basic logic operations:

111001010

000Z = XYYX

AND ORX Y Z = X+Y0 0 0

0 1 11 0 11 1 1


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Truth Tables Cont d

Used to evaluate any logic functionConsider F ( X , Y , Z ) = X Y + Y Z

X Y Z X Y Y Y Z F = X Y + Y Z0 0 0 0 1 0 00 0 1 0 1 1 10 1 0 0 0 0 00 1 1 0 0 0 01 0 0 0 1 0 01 0 1 0 1 1 11 1 0 1 0 0 11 1 1 1 0 0 1


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Example: Logic Using Switches

Light is on (L = 1) forL(A, B, C, D) =

and off (L = 0), otherwise.Useful model for relay and CMOS gate circuits,the foundation of current digital logic circuits

Logic Function Implementation cont d

B

A

D

C

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



Logic Gates

In the earliest computers, switches were openedand closed by magnetic fields produced byenergizing coils in relays . The switches in turn

opened and closed the current paths.Later, vacuum tubes that open and closecurrent paths electronically replaced relays.

Today, transistors are used as electronicswitches that open and close current paths.



Logic Gate Symbols and Behavior

Logic gates have special symbols:

And waveform behavior in time as follows :X 0 0 1 1

Y 0 1 0 1

X Y (AND) 0 0 0 1

X + Y (OR) 0 1 1 1

(NOT) X 1 1 0 0

OR gate

X Y

Z = X + Y X Y

Z = X Y

AND gate

X Z = X

NOT gate or inverter



Logic Diagrams and Expressions

Boolean equations, truth tables and logic diagrams describethe same function!Truth tables are unique, but expressions and logic diagramsare not. This gives flexibility in implementing functions.

X

Y F

Z

Logic Diagram

Logic Equation

Z Y XF

Truth Table

11 1 1

11 1 011 0 111 0 000 1 1

00 1 010 0 100 0 0

X Y Z Z Y XF



Gate Delay

In actual physical gates, if an input changes thatcauses the output to change, the output changedoes not occur instantaneously. The delay between an input change and theoutput change is the gate delay denoted by t G:

tG tGInput

Output

Time (ns)

0

0

1

1

0 0.5 1 1.5

tG = 0.3 ns



1.

3.

5.

7.

9.

11.

13.

15.

17.

CommutativeAssociative

Distributive

DeMorgan s

2.

4.

6.

8.

X . 1 X =X . 0 0=X . X X =

0=X . X

Boolean Algebra

10.

12.

14.

16.

X + Y Y + X =(X + Y ) Z + X + (Y Z )+=X (Y + Z ) XY XZ +=X + Y X . Y=

XY YX =(XY ) Z X (Y Z )=X + YZ (X + Y ) (X + Z )=X . Y X + Y=

X + 0 X =

+X 1 1=X + X X =

1=X + X X = X

Invented by George Boole in 1854An algebraic structure defined by a set B = {0, 1}, together with twobinary operators (+ and ) and a unary operator ( )

Idempotence

Complement

Involution

Identity element



Some Properties of Boolean Algebra

Boolean Algebra is defined in general by a set B that canhave more than two valuesA two-valued Boolean algebra is also know as SwitchingAlgebra. The Boolean set B is restricted to 0 and 1.

Switching circuits can be represented by this algebra.The dual of an algebraic expression is obtained byinterchanging + and and interchanging 0 s and 1 s. The identities appear in dual pairs. When there is only

one identity on a line the identity is self-dual, i. e., thedual expression = the original expression.Sometimes, the dot symbol (AND operator) is notwritten when the meaning is clear



Example: F = (A + C) B + 0dual F = (A C + B) 1 = A C + B

Example: G = X Y + (W + Z)

dual G =Example: H = A B + A C + B Cdual H =

Unless it happens to be self-dual, the dual of anexpression does not equal the expression itselfAre any of these functions self-dual?(A+B)(A+C)(B+C)=(A+BC)(B+C)=AB+AC+BC

Dual of a Boolean Expression

(X+Y) (W Z) = (X+Y) (W+Z)

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

H is self-dual


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Boolean Operator Precedence

The order of evaluation is: 1. Parentheses2. NOT3. AND4. OR

Consequence: Parentheses appeararound OR expressions Example: F = A(B + C)(C + D)


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Boolean Algebraic Proof Example 1

A + A B = A (Absorption Theorem)Proof Steps JustificationA + A B

= A 1 + A B Identity element: A 1 = A

= A ( 1 + B) Distributive= A 1 1 + B = 1= A Identity element

Our primary reason for doing proofs is to learn: Careful and efficient use of the identities and theorems ofBoolean algebra, and

How to choose the appropriate identity or theorem to applyto make forward progress, irrespective of the application.


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AB + AC + BC = AB + AC (Consensus Theorem)Proof Steps Justification= AB + AC + BC= AB + AC + 1 BC Identity element= AB + AC + (A + A) BC Complement= AB + AC + ABC + ABC Distributive= AB + ABC + AC + ACB Commutative

= AB 1 + ABC + AC 1 + ACB Identity element= AB (1+C) + AC (1 + B) Distributive= AB . 1 + AC . 1 1+X = 1= AB + AC Identity element

Boolean Algebraic Proof Example 2


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Truth Table to Verify DeMorgan s

X Y XY X+Y X Y X+Y X Y XY X+Y

0 0 0 0 1 1 1 1 1 10 1 0 1 1 0 0 0 1 11 0 0 1 0 1 0 0 1 11 1 1 1 0 0 0 0 0 0

X + Y = X Y X Y = X + Y

Generalized DeMorgan s Theorem: X1 + X 2 + + X n = X 1 X 2 X n

X1 X 2 X n = X 1 + X 2 + + X n


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Complementing Functions

Use DeMorgan's Theorem:1. Interchange AND and OR operators2. Complement each constant and literal

Example: Complement F =

F = (x + y + z)(x + y + z)Example: Complement G = (a + bc)d + e

G = (a (b + c) + d) e

x z y z y x


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Next Canonical Forms

Minterms and MaxtermsSum-of-Minterm (SOM) Canonical Form

Product-of-Maxterm (POM) Canonical Form

Representation of Complements of Functions

Conversions between Representations


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Minterms

Minterms are AND terms with every variablepresent in either true or complemented form.Given that each binary variable may appearnormal (e.g., x) or complemented (e.g., ), there

are 2n

minterms for n variables. Example: Two variables (X and Y) produce2 x 2 = 4 combinations:

(both normal)

(X normal, Y complemented) (X complemented, Y normal) (both complemented)

Thus there are four minterms of two variables.

Y X XY

Y X Y X

x


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Maxterms

Max terms are OR terms with every variable intrue or complemented form.Given that each binary variable may appearnormal (e.g., x) or complemented (e.g., x), there

are 2 n max terms for n variables. Example: Two variables (X and Y) produce2 x 2 = 4 combinations:

(both normal) (x normal, y complemented) (x complemented, y normal) (both complemented)

Y X Y X Y X Y X


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Two variable minterms and maxterms.

The minterm m i should evaluate to 1 for eachcombination of x and y.

The maxterm is the complement of the minterm

Minterms & Maxterms for 2 variables

x y Index Minterm Maxterm0 0 0 m0 = x y M0 = x + y

0 1 1 m1 = x y M1 = x + y 1 0 2 m2 = x y M2 = x + y 1 1 3 m3 = x y M3 = x + y


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Minterms & Maxterms for 3 variables

M3 = x + y + zm3 = x y z3110M4 = x + y + zm4 = x y z4001M5 = x + y + zm5 = x y z5101

M6 = x + y + zm6 = x y z6011 1

100y

1

000x

1

010z

M7 = x + y + z m7 = x y z7

M2 = x + y + z m2 = x y z2M1 = x + y + z m1 = x y z1M0 = x + y + zm0 = x y z0

MaxtermMintermIndex

Maxterm M i is the complement of minterm mi

M i = mi and mi = M i


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Purpose of the Index

Minterms and Maxterms are designated with an indexThe index number corresponds to a binary patternThe index for the minterm or maxterm, expressed as abinary number, is used to determine whether the variable

is shown in the true or complemented formFor Minterms: 1 means the variable is Not Complemented and

0 means the variable is Complemented.

For Maxterms: 0 means the variable is Not Complemented and

1 means the variable is Complemented.


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Standard Order

All variables should be present in a minterm ormaxterm and should be listed in the same order(usually alphabetically)

Example: For variables a, b, c: Maxterms (a + b + c), (a + b + c) are in standard order

However, (b + a + c) is NOT in standard order

(a + c) does NOT contain all variables

Minterms (a b c) and (a b c) are in standard order

However, (b a c) is not in standard order

(a c) does not contain all variables


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Sum-Of-Minterm (SOM)

Sum-Of-Minterm (SOM) canonical form:Sum of minterms of entries that evaluate to 1

x y z F Minterm0 0 0 0

0 0 1 1 m1 = x y z0 1 0 00 1 1 01 0 0 0

1 0 1 01 1 0 1 m6 = x y z1 1 1 1 m7 = x y z

F = m 1 + m 6 + m 7 = (1, 6, 7) = x y z + x y z + x y z

Focus on the1 entries


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+ a b c d

Sum-Of-Minterm Examples

F (a , b , c , d ) = (2, 3, 6, 10, 11)

F (a , b , c , d ) = m2 + m3 + m6 + m10 + m11

G (a , b , c , d ) = (0, 1, 12, 15)

G (a , b , c , d ) = m0 + m1 + m12 + m15

+ a b c da b c d + a b c d+ a b c d + a b c d

+ a b c da b c d + a b c d


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Product-Of-Maxterm (POM)Product-Of-Maxterm (POM) canonical form:Product of maxterms of entries that evaluate to 0

x y z F Maxterm0 0 0 1

0 0 1 10 1 0 0 M2 = ( x + y + z )0 1 1 11 0 0 0 M4 = ( x + y + z )

1 0 1 11 1 0 0 M6 = ( x + y + z )1 1 1 1

Focus on the0 entries

F = M 2M 4M 6 = (2, 4, 6) = ( x+y+z ) ( x+y+z ) ( x+y+z )


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F (a , b , c , d ) = (1, 3, 6, 11)

F (a , b , c , d ) = M1 M 3 M 6 M 11

G (a , b , c , d ) = (0, 4, 12, 15)

G (a , b , c , d ) = M0 M 4 M 12 M 15

Product-Of-Maxterm Examples

(a+b+c+d ) (a+b+c+d ) (a+b+c+d ) (a+b+c+d )

(a+b+c+d ) (a+b+c+d ) (a+b+c+d ) (a+b+c+d )


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Converting to Sum-of-Min terms Form

A function that is not in the Sum-of-Minterms formcan be converted to that form by means of a truth tableConsider F = y + x z

x y z F

Minterm0 0 0 1 m0 = x y z 0 0 1 1 m1 = x y z0 1 0 1 m2 = x y z0 1 1 01 0 0 1 m4 = x y z 1 0 1 1 m5 = x y z 1 1 0 01 1 1 0

F = (0, 1, 2, 4, 5) = m0 + m 1 + m 2 + m 4 + m 5 =

x y z + x y z + x y z +

x y z + x y z


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Converting to Product-of-Maxterms Form

A function that is not in the Product-of-Minterms formcan be converted to that form by means of a truth tableConsider again: F = y + x z

x y z F

Minterm0 0 0 10 0 1 10 1 0 10 1 1 0 M3 = ( x+ y+ z )1 0 0 11 0 1 11 1 0 0 M6 = ( x+ y+ z )1 1 1 0 M7 = ( x+ y+ z )

F = (3, 6, 7) = M3 M 6 M 7 =

( x+y+z ) ( x+y+z ) ( x+y+z )


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Algebraic Conversion to Sum-of-Minterms

Expand all terms first to explicitly list all mintermsAND any term missing a variable v with ( v + v)Example 1: f = x + x y (2 variables)

f = x ( y + y) + x y f = x y + x y + x yf = m 3 + m 2 + m 0 = (0, 2, 3) Example 2: g = a + b c (3 variables)

g = a (b + b)(c + c) + ( a + a) b c g = a b c + a b c + a b c + a b c + a b c + a b c g = a b c + a b c + a b c + a b c + a b c g = m 1 + m 4 + m 5 + m 6 + m 7 = (1, 4, 5, 6, 7)


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Algebraic Conversion to Product-of-Maxterms

Expand all terms first to explicitly list all maxtermsOR any term missing a variable v with v v Example 1: f = x + x y (2 variables)Apply 2 nd distributive law:

f = ( x + x) ( x + y) = 1 ( x + y) = ( x + y) = M 1Example 2: g = a c + b c + a b (3 variables)

g = ( a c + b c + a) (a c + b c + b) (distributive) g = ( c + b c + a) (a c + c + b) ( x + x y = x + y) g = ( c + b + a) (a + c + b) ( x + x y = x + y) g = ( a + b + c) (a + b + c) = M 5 . M 2 = (2, 5)


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Function Complements

The complement of a function expressed as asum of minterms is constructed by selecting theminterms missing in the sum-of-mintermscanonical form

Alternatively, the complement of a functionexpressed by a Sum of Minterms form is simplythe Product of Maxterms with the same indices

Example: Given F(x, y, z) = (1, 3, 5, 7) F(x, y, z) = (0, 2, 4, 6) F(x, y, z) = (1, 3, 5, 7)


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Standard Sum-of-Products (SOP) form:equations are written as an OR of AND termsStandard Product-of-Sums (POS) form:equations are written as an AND of OR termsExamples: SOP: POS:

These mixed forms are neither SOP nor POS

Standard Forms

BCBACBAC)CB(AB)(A

C)(AC)B(AB)(ACACBA

f


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Standard Sum-of-Products (SOP)

A sum of minterms form for n variables canbe written down directly from a truth table. Implementation of this form is a two-level

network of gates such that: The first level consists of n -input AND gates

The second level is a single OR gate

This form often can be simplified so that thecorresponding circuit is simpler.


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A Simplification Example:

Writing the minterm expression:F = A B C + A B C + A B C + ABC + ABC

Simplifying:F = A B C + A (B C + B C + B C + B C)F = A B C + A (B (C + C) + B (C + C))

F = A B C + A (B + B)F = A B C + AF = B C + ASimplified F contains 3 literals compared to 15

Standard Sum-of-Products (SOP)

) 7 , 6 , 5 , 4 , 1 ( ) C , B , A ( F S


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AND/OR Two-Level Implementation

The two implementations for F are shown below

F

ABC

A

BC

ABC

A

BC

BC

F

B

C

A

It is quite

apparent whichis simpler!


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SOP and POS Observations

The previous examples show that: Canonical Forms (Sum-of-minterms, Product-of-Maxterms),

or other standard forms (SOP, POS) differ in complexity

Boolean algebra can be used to manipulate equations intosimpler forms

Simpler equations lead to simpler implementations

Questions:

How can we attain a simplest expression?

Is there only one minimum cost circuit?

The next part will deal with these issues

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03-BooleanAlgebra (3)