 # Chapter 2: Boolean Algebra and Logic Functions

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

Boolean Algebra Algebraic 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 Algebra 1. Closure a,b in B, (i) a + b in B (ii) a • b in B 2. Identities: 0, 1 in B (i) a + 0 = a (ii) a • 1 = a 3. 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.c 5. 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’ = 0 a’ is complement of a CS 3402:A.Berrached

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 = 0 2) a(b+c) = ab + ac CS 3402:A.Berrached

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. CS 3402:A.Berrached

From Boolean Expression to Gates
More than one way to map an expression to gates E.g., Z = A' • B' • (C + D) = (A' • (B' • (C + D))) A Z B T 1 C T 2 D A B Z C D For each Boolean function, there is only one unique truth table representation =>Truth table is the unique signature of a Boolean function CS 3402:A.Berrached

Boolean Functions Possible Boolean Functions of Two variables NAND NOR
Description Z = 1 if X is 0 or Y Gates T ruth T able NAND X X 1 Y 1 Z 1 Z Y Description Z = 1 if both X and Y are 0 Gates T ruth T able NOR X X 1 Y 1 Z 1 Z Y CS 3402:A.Berrached

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 CS 3402:A.Berrached

Basic Logic Functions: XOR, XNOR
XOR: X or Y but not both ("inequality", "difference") XNOR: X and Y are the same ("equality", "coincidence") CS 3402:A.Berrached

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. CS 3402:A.Berrached

Simplification Using Boolean Algebra
Useful Theorems of Boolean Algebra: 1. Idempotency Theorem a. X + X = X b. X • X = X 2. Null elements for + and • operators a. X + 1 = 1 b. X . 0 = 0 3. Involution Theorem (X’)’ = X 4. Absorption Theorem a. X + XY = X b. X.(X+Y) = X 5. Simplification Theorem a. XY + XY’ = X b. (X+Y).(X+Y’) = X 6. Another Simplification Theorem a. X + X’Y = X + Y b. X.(X’ + Y) = X.Y CS 3402:A.Berrached

DeMorgan's Theorems 7. 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’ CS 3402:A.Berrached

DeMorgan's Theorem (X + Y)' = X' • Y' NOR is equivalent to AND
with inputs complemented (X • Y)' = X' + Y' NAND is equivalent to OR with inputs complemented 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) CS 3402:A.Berrached

Function Representations
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 CS 3402:A.Berrached

Function Representations
F(X,Y,Z) = X + Y’Z + X’Y’Z + X’Y’Z’ CS 3402:A.Berrached

Deriving Boolean Expression from Truth Table
CS 3402:A.Berrached

Product and Sum Terms --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)' CS 3402:A.Berrached

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 CS 3402:A.Berrached

Minterms & Maxterms -- Definitions
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 CS 3402:A.Berrached

Minterms For functions of three variables: X, Y, and Z
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 CS 3402:A.Berrached

Maxterms 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 Mi = mi CS 3402:A.Berrached

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. CS 3402:A.Berrached

Deriving Boolean Expression from Truth Table
Input Output Minterm A B C F term designation A’B’C’ m0 A’B’C m1 A’BC’ m2 A’BC m3 AB’C’ m4 AB’C m5 ABC’ m6 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) CS 3402:A.Berrached

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) CS 3402:A.Berrached

Deriving Boolean Expression from Truth Table
Input Output Minterm Maxterm A B C F term Designation term Designation A’B’C’ m A + B + C M0 A’B’C m A + B + C’ M1 A’BC’ m A + B’ + C M2 A’BC m A + B’ + C’ M3 AB’C’ m A’ + B + C M4 AB’C m A ‘ + B + C’ M5 ABC’ m A’ + B’ + C M6 ABC m 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) CS 3402:A.Berrached

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) CS 3402:A.Berrached

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. CS 3402:A.Berrached

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 CS 3402:A.Berrached

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)] CS 3402:A.Berrached

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 CS 3402:A.Berrached

Incompletely Specified Functions
Block Diagram W X F Y Z CS 3402:A.Berrached

Incompletely Specified Functions
Truth Table F =  m ( 0, 3, 5, 6, 9)+d(10…15) F =  M(1,2,4,7, 8)+ d(10…15) CS 3402:A.Berrached

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