CS 3402:A.Berrached1 Chapter 2: Boolean Algebra and Logic Functions CS 3402-- Digital Logic Design.

CS 3402:A.Berrached 1

Chapter 2: Boolean Algebra and Logic Functions

CS 3402-- Digital Logic Design


Boolean AlgebraAlgebraic structure consisting of:

• a set of elements B• operations {AND, OR}

Notation: X AND Y X • Y XY X OR Y X+Y

• B contains at least two elements a & b such that a b Note: switching algebra is a subset of Boolean algebra where

B={0, 1}Axioms of Boolean Algebra1. Closure a,b in B, (i) a + b in B (ii) a • b in B2. Identities: 0, 1 in B (i) a + 0 = a (ii) a • 1 = a3. Commutative Laws: a,b in B, (i) a + b = b + a (ii) a • b = b • a

4. Associative Laws: (i) a + (b+c) = (a+b)+c = a+b+c (ii) a. (b.c) = (a.b).c = a.b.c5. Distributive Laws: (i) a + (b • c) = (a + b) • (a + c) (ii) a • (b + c) = (a • b) + (a • c)

6. Existence of the Complement: exists a’ unique in B (i) a + a’ = 1 (ii) a • a’ = 0a’ is complement of a


Principle of Duality

Definition of duality: a dual of a Boolean expression is derived by replacing AND

operations by ORs, OR operations by ANDs, constant 0s by 1s, and 1s by 0s (everything else is left unchanged).

Principle of duality: if a statement is true for an expression, then it is also true for the dual of the expression

Example: find the dual of the following equalities

1) XY+Z = 02) a(b+c) = ab + ac


Boolean Functions A Boolean function consists of an algebraic expression formed

with binary variables, the constants 0 and 1, the logic operation symbols, parenthesis, and an equal sign.

Example:

F(X,Y,Z) = X + Y’ Z or F = X + Y’ Z

X, Y and Z are Boolean variables A literal: The appearance of a variable or its complement in a

Boolean expression

A Boolean function can be represented with a truth table

A Boolean function can be represented with a logic circuit diagram composed of logic gates.


More than one way to map an expression to gatesE.g., Z = A' • B' • (C + D) = (A' • (B' • (C + D)))

From Boolean Expression to Gates

A

B

C

D T

2

T 1

Z

Z

A

B

C

D

For each Boolean function, there is only one unique truth table representation

=>Truth table is the unique signature of a Boolean function


Boolean FunctionsPossible Boolean Functions of Two variables

NAND

NOR

Description Z = 1 if X is 0 or Y is 0

Gates T ruth T able

X 0 0 1 1

Y 0 1 0 1

Z 1 1 1 0

X Y

Z

Description Z = 1 if both X and Y are 0

Gates T ruth T able X

Y Z X

0

0 1 1

Y 0

1 0 1

Z 1

0 0 0

F0 0 0 0 0

F1 0 0 0 1

F2 0 0 1 0

F3 0 0 1 1

F4 0 1 0 0

F5 0 1 0 1

F6 0 1 1 0

F7 0 1 1 1

F8 1 0 0 0

F9 1 0 0 1

F10 1 0 1 0

F1 1 1 0 1 1

F12 1 1 0 0

F13 1 1 0 1

F14 1 1 1 0

F15 1 1 1 1

X 0 0 1 1

Y 0 1 0 1

0 X • Y X Y X + Y X Y

1


Basic Logic Functions: NAND, NOR

NAND, NOR gates far outnumber AND, OR in typical designs

• easier to construct in the underlying transistor technologies

• they are functionally complete

Functionally Complete Operation Set:

• A set of logic operations from which any Boolean function can be realized (also called universal operation set)

• E.g. {AND, OR, NOT} is functionally complete

The NAND operation is also functionally complete

• => any Boolean function can be realized with one type of gate (the NAND gate).

The NOR operation is also functionally complete


Basic Logic Functions: XOR, XNOR XOR: X or Y but not both ("inequality", "difference")

XNOR: X and Y are the same ("equality", "coincidence")

(a) XOR (b) XNOR

Description Z = 1 if X has a different value than Y

Gates

T ruth T able

X

Y Z

X 0 0 1 1

Y 0 1 0 1

Z 0 1 1 0

Description Z = 1 if X has the same value as Y

Gates

T ruth T able

X

Y Z

X 0 0 1 1

Y 0 1 0 1

Z 1 0 0 1


Logic Functions: Rationale for Simplification Logic Minimization: reduce complexity of the gate level implementation

• reduce number of literals (gate inputs, circuit inputs)• reduce number of gates• reduce number of levels of gates

fewer inputs implies faster gates in some technologies

fan-ins (number of gate inputs) are limited in some technologies

Fewer circuit inputs implies fewer I/O pins

fewer levels of gates implies reduced signal propagation delays

number of gates (or gate packages) influences manufacturing costs

In general, need to make tradeoff between circuit delay and reduced gate count.


Simplification Using Boolean Algebra

Useful Theorems of Boolean Algebra:1. Idempotency Theorem

a. X + X = X b. X • X = X2. Null elements for + and • operators

a. X + 1 = 1 b. X . 0 = 03. Involution Theorem

(X’)’ = X4. Absorption Theorem

a. X + XY = X b. X.(X+Y) = X5. Simplification Theorem

a. XY + XY’ = X b. (X+Y).(X+Y’) = X6. Another Simplification Theorem

a. X + X’Y = X + Y b. X.(X’ + Y) = X.Y


DeMorgan's Theorems7. DeMorgan’s Theorem

a. (X+Y)’ = X’ . Y’ b. (X.Y)’ = X’ + Y’

The complement of the sum is the product of the complements The complement of the product is the sum of the complements

In general

a. (A+B+….+Z)’ = A’ . B’ . … .Z’

b. (A.B.C….Z)’ = A’ + B’ + ….+Z’


DeMorgan's Theorem

(X + Y)' = X' • Y'

(X • Y)' = X' + Y'

NOR is equivalent to ANDwith inputs complemented

NAND is equivalent to ORwith inputs complemented

X 0 0 1 1

Y 0 1 0 1

X 1 1 0 0

Y 1 0 1 0

X + Y 1 0 0 0

X•Y 1 0 0 0

X 0 0 1 1

Y 0 1 0 1

X 1 1 0 0

Y 1 0 1 0

X + Y 1 1 1 0

X•Y 1 1 1 0

DeMorgan’s Law can be used to get the complement of an expression

{F(X1,X2,...,Xn,0,1,+,•)}' = {F(X1',X2',...,Xn',1,0,•,+)}

Example:F = A B' C' + A' B' C + A B' C + A B C'

F' = (A' + B + C) • (A + B + C') • (A' + B + C') • (A' + B' + C)


Truth Table (Unique representation) Boolean Expressions Logic Diagrams

From TO

Boolean Expression ==> Logic Diagram

Logic Diagram ==> Boolean Expression

Boolean Expression ==> Truth Table

Truth Table ==> Boolean Expression

Function Representations


Function Representations

F(X,Y,Z) = X + Y’Z + X’Y’Z + X’Y’Z’


Deriving Boolean Expression from Truth Table


Definitions: Literal: A boolean variable or its complement

X X’ A B’

Product term: A literal or the logical product (AND) of multiple literals:X XY XYZ X’YZ’ A’BC

Note: X(YZ)'

Sum term: A literal or the logical sum (OR) of multiple literals:

X X’+Y X+Y+Z X’+Y+Z’ A’+B+C

Note: X+(Y+Z)'

Product and Sum Terms --Definitions


SOP & POS -- Definitions Sum of products (SOP) expression: The logic sum (OR) of multiple

product terms:

AB + A’C + B’ + ABC

AB’C + B’D’ + A’CD’

Product of sums (POS) expression: The logic product (AND) of multiple sum terms:

(A+B).( A’+C).B’.( A+B+C)

(A’ + B + C).( C’ + D)

Note:

SOP expressions ==> 2-level AND-OR circuit

POS expressions ==> 2-level OR-AND circuit


Minterms & Maxterms -- DefinitionsDefinitions A Minterm: for an n variable function, a minterm is a product term that

contains each of the n variables exactly one time in complemented or uncomplemented form.

Example: if X, Y and Z are the input variables, the minterms are:

X’Y’Z’ X’Y’Z X’YZ’ X’YZ XY’Z’ XY’Z XYZ’ XYZ

A Maxterm: for an n variable function, a maxterm is a sum term that contains each of the n variables exactly one time in complemented or uncomplemented form

Example: if X, Y and Z are the input variables, the maxterms are:

X’+Y’+Z’ X’+Y’+Z X’+Y+Z’ X’+Y+Z X+Y’+Z’ X+Y’+Z X+Y+Z’ X+Y+Z


Minterms For functions of three variables: X, Y, and Z

Minterm Designation Bit CombinationX’Y’Z’X’Y’ZX’YZ’X’YZXY’Z’XY’ZXYZ’XYZ

m0m1m2m3m4m5m6m7

000001010011100101110111

• The bit combination associated with each minterm is the only bit combination for which the minterm is equal to1.

•Example: X'Y'Z' = 1 iff X=0, Y=0, and Z=0

• Each bit represents one of the variables ( order is important) :

Un-complemented variable ==> 1

Complemented variable ==> 0


MaxtermsMaxterm Bit Combination DesignmationX + Y + ZX + Y + Z’X + Y’ + ZX + Y’ + Z’X’ + Y + ZX’ + Y + Z’X’ + Y’ + ZX’ + Y’ + Z’

000001010011100101110111

M0M1M2M3M4M5M6M7

• The bit combination associated with each Maxterm is the only bit combination for which the Maxterm is equal to 0.

Note: The ith Maxterm is the complement of the ith minterm; That is M

i = m

i


Standard (Canonical) forms of an expression A switching function can be represented by several

different, but equivalent, algebraic expressions. The standard form is a unique algebraic

representation of each function.• Standard SOP: sum of minterm form of a switching

function• Standard POS: the product of maxterm form ofa

switching function

Each switching function has a unique standard SOP and a unique standard POS.


Deriving Boolean Expression from Truth Table Input Output Minterm

A B C F term designation

0 0 0 1 A’B’C’ m0

0 0 1 0 A’B’C m1

0 1 0 0 A’BC’ m2

0 1 1 1 A’BC m3

1 0 0 0 AB’C’ m4

1 0 1 0 AB’C m5

1 1 0 0 ABC’ m6

1 1 1 0 ABC m7

F is 1 iff (A=0 AND B=0 AND C=0) or (A=0 AND B=1 AND C=1)

F is 1 iff (A’=1 AND B’=1 AND C’=1) or (A’=1 AND B=1 AND C=1)

F is 1 iff A’.B’.C’ = 1 OR A’.B.C= 1

F is 1 iff A’B’C’ + A’BC = 1

=> F = A’B’C’ + A’BC => F = m0 + m3

Short-hand notation: F = m ( 0, 3)


Sum of minterms form

A Boolean function is equal to the sum of minterms for which the output is one.

=> the sum of minterms (also called the standard SOP) form

Example: F = m ( 0, 3)


Deriving Boolean Expression from Truth Table Input Output Minterm Maxterm

A B C F term Designation term Designation

0 0 0 1 A’B’C’ m0 A + B + C M0

0 0 1 0 A’B’C m1 A + B + C’ M1

0 1 0 0 A’BC’ m2 A + B’ + C M2

0 1 1 1 A’BC m3 A + B’ + C’ M3

1 0 0 0 AB’C’ m4 A’ + B + C M4

1 0 1 0 AB’C m5 A ‘ + B + C’ M5

1 1 0 0 ABC’ m6 A’ + B’ + C M6

1 1 1 0 ABC m7 A’ + B’ + C’ M7

F is 0 iff (A+B+C’) = 0 AND (A+B’+C)=0 AND (A’+B+C)=0 AND (A’+B+C’) =0

AND (A’+B’+C) = 0 AND (A’+B’+C’) = 0

=> F = (A+B+C’) . (A+B’+C) . (A’+B+C) AND (A’+B+C’) . (A’+B’+C) . (A’+B’+C’)

=> F = M1.M2.M4.M5.M6.M7 ==> F = M(1,2,4,5,6,7)


Product of Maxterms

A Boolean function is equal to the product of Maxterms for which the output is 0.

=> the product of Maxterms (also called the standard Product of Sums) form

Example: F = M(1,2,4,5,6,7)


Examples:

Find the truth table for the following switching functions: F(A,B,C) = ABC’ + AB’C F(A,B,C) = AB + A’B’ + AC F(X, Z) = X + Z’ F(A,B,C,D) = A(B’ + CD’) + A’BC’

For each of the above functions, find their Standard SOP and POS.


Getting Standard Forms of a Switching Function F(A,B,C) = AB + A’B’ + AC

get standard SOP and POS forms of F

Method 1:

1. Derive Truth Table for F

2. Get SOP and POS from truth table

Method 2:

Use Shannon’s Expansion Theorem


Shannon’s Expansion Theorem

a) f(x1,x2,…,xn) = x1.f(1,x2,….,xn) + x1.f(0,x2,…,xn)

b) f(x1,x2,…,xn) = [ x1+ f(0,x2,….,xn)] . [ x1.f(1,x2,…,xn)]


Incompletely Specified Functions The output for certain input combination is not important

(I.e. we don't care about it). Certain input combinations never occur

Example: Design a circuit that takes as input a BCD digit and outputs a 1 iff the parity of the input is even.

Note: A BCD digit consists of 4 bits


Incompletely Specified Functions

W

FX

Y

Z

Block Diagram


Incompletely Specified Functions Truth Table

WXYZ F

0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1

1 0 0 1 0 1 1 0 0 1 d d d d d d

F = m ( 0, 3, 5, 6, 9)+d(10…15)

F = M(1,2,4,7, 8)+ d(10…15)

CS 3402:A.Berrached1 Chapter 2: Boolean Algebra and Logic Functions CS 3402-- Digital Logic Design.

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