CH3 Boolean Algebra (cont d)access.ee.ntu.edu.tw/course/logic_design_94first/LectureNotes/CH03... · ACCESS IC LAB Graduate Institute of Electronics Engineering, NTU CH3 Boolean Algebra

ACCESS IC LAB

Graduate Institute of Electronics Engineering, NTU

CH3 Boolean Algebra (contCH3 Boolean Algebra (cont’’d)d)

Lecturer：吳安宇

Date：2005/10/7


pp. 2

IntroductionIntroductionv Today, you’ll know:

1. Guidelines for multiplying out/factoring expressions2. Exclusive-OR and Equivalence operations3. Positive logic and negative logic4. More about consensus theorem5. Algebraic simplification of switching expressions6. Approach to prove validity of an equation7. The difference between ordinary algebra and

Boolean algebra


pp. 3

Guidelines for Multiplying Out and Guidelines for Multiplying Out and FactoringFactoring

vUse X(Y+Z) = XY + XZ ...(1)(X+Y)(X+Z) = X + YZ ...(2)(X+Y)(X’+Z) = XZ + X’Y ...(3)

vFor multiplying out, (2) and (3) should be generally applied before (1) to avoid generating unnecessary termsvFor factoring, apply (1), (2), (3) from right terms to

left terms


pp. 4

Multiplying Out ExpressionMultiplying Out Expression

EX. F = (Q + AB)(C’D + Q’) = QC’D + Q’ABor F = QC’D + QQ’ + AB’C’D + AB’Q’

EX. (A+B+C’)(A+B+D)(A+B+E)(A+D’+E)(A’+C)= (A+B+C’D)(A+B+E)[AC+A’(D’+E)]

= (A+B+C’DE)(AC+A’D’+A’E)= AC+ABC+A’BD’+A’BE+A’C’DE (SOP form)

=> By brute force => 162 terms

Distributed Law


pp. 5

Factoring ExpressionFactoring Expression

v EX.AC + A’BD’ + A’BE + A’C’DE

= AC + A’(BD’ + BE + C’DE)XZ + X’Y = (X + Y)(X’ + Z)

= (A + BD’ + BE + C’DE)(A’ + C)= [ A + C’DE + B (D’ + E) ](A’ + C)

X + YZ = (X+Y)(X+Z)= (A + C’DE + B)(A + C’DE + D’ + E)(A’ + C)= (A + B + C’)(A + B + D)(A + B + E)(A + D’ + E)(A’ + C)


pp. 6

3.2 Exclusive3.2 Exclusive--OR OperationsOR OperationsvExclusive-OR (XOR)

X Y X Y0 00 11 01 1

0110

Truth TableSymbol

Boolean Expression : X Y = X’Y + XY’


pp. 7

ExclusiveExclusive--OR OperationsOR Operations

vUseful Theorems :

X 0 = X X Y = Y X (commutative)

X 1 = X’ (X Y) Z = X (Y Z) (associative)

X X = 0 X(Y Z) = XY XZ (distributive)

X X’= 1 (X Y)’ = X Y’ = X’ Y = XY + X’Y’


pp. 8

Proof of Distributive LawsProof of Distributive LawsvXY XZ = XY(XZ)’ + (XY)’XZ

= XY(X’ + Z’) + (X’ + Y’)XZ= XYZ’ +XY’Z= X(YZ’ + Y’Z)= X(Y Z)


pp. 9

Equivalence Operations Equivalence Operations (Exclusive NOR)(Exclusive NOR)

X Y X Y (X Y)’0 00 11 01 1

1001

1001

X Y = XY + X’Y’


pp. 10

Simplification of XOR and XNORSimplification of XOR and XNORvX Y = X’Y + XY’

X Y = X’Y’ + XY

n EX (see p.62). F = (A’B C) + (B AC’)

= [(A’B)C + (A’B)’C’] + [B’(AC’) + B(AC’)’]= A’BC + (A+B’)C’ + AB’C’ + B(A’ +C)= B(A’C + A’ + C) + C(A + B’ + AB’)= B(A’ + C) + C’(A + B’) ( can be further simplified)


pp. 11

3.3 Consensus Theorem3.3 Consensus TheoremXY + X’Z + YZ = XY + X’Z (YZ is redundant )

Proof : XY + X’Z + YZ = XY + X’Z + (X + X’)YZ

= (XY + XYZ) + (X’Z + X’YZ)= XY(1 + Z) + X’Z(1 + Y)= XY + X’Z


pp. 12

How to Find Consensus Term?How to Find Consensus Term?1. Find a pair of terms, one of which contains a

variable and the other contains its complement

A’C’D + A’BD + BCD + ABC + ACD’ (A ↔ A’)

2. Ignore the variable and its complement, the left variables composite the consensus term

(A’BD) + (ABC) → BD·BC = BCD (redundant term)


pp. 13

Consensus TheoremConsensus TheoremvApplication to eliminate redundant terms from

Boolean Expressions

a’b’ + ac + bc’ + b’c +ab = a’b’ + ac + bc’


pp. 14

Consensus TheoremConsensus Theorem

Example (others are on p.67) :

(a + b + c’)(a + b + d’)(b + c + d’)= (a + b + c’)(b + c + d’)

n Simplification of Boolean Expression can reduce the cost of realizing the network using gates

nDual form of consensus theorem(X + Y)(X’ + Z)(Y + Z) = (X + Y)(X’ + Z)

(a+ b + c’) + (b + c +d’) → a+b + b+d’ = a+b+d’


pp. 15

Algebraic Simplification of Switching Algebraic Simplification of Switching ExpressionExpression

vA. Combining TermsXY + XY’ =X(Y + Y’) = X

n EX.1 abc’d’ + abcd’ = abd’ (X = abd’, Y = c)n EX.2 ab’c + abc + a’bc

= ab’c + abc + abc + a’bc = ac + bc

n EX.3 (a + bc)(d + e’) + a’(b’ + c’)(d + e’)= d + e’


pp. 16


vRule B -- Eliminating Terms : X + XY = XXY + X’Z + YZ = XY + X’Z

n EX.1 a’b + a’bc = a’b (X = a’b)

a’bc’ + bcd + a’bd = a’bc’ + bcd (X = c, Y = bd, Z = a’b)


pp. 17


vRule C -- Eliminating Literals : X + X’Y = (X + X’)(X + Y) = X + Y

n EX. A’B + A’B’C’D’ + ABCD’= A’(B + B’C’D’) + ABCD (common term - A’)= A’(B + C’D’) + ABCD (Rule C)= B(A’ + ACD) + A’C’D’ (common term - B)= B(A’ + CD) + A’C’D’ (Rule C)= A’B + BCD’ + A’C’D’ (final terms)


pp. 18


vRule D -- Adding Redundant Terms

vAdd XX’ = 0

vMultiply by (X + X’) = 1

vAdd YZ to (XY + X’Z) (reverse of Consensus)

ØBecause XY + X’Z + YZ = XY + X’Z

vAdd XY to X


pp. 19


vEX.1 of “Adding Redundant Terms”

WX + XY + X’Z’ + WY’Z = WX + XY + X’Z’ + WY’Z’ + W’Z

(add W’Z by Consensus Theorem)= WX + XY + X’Z’ + WZ’

(eliminate WY’Z’ by WZ’)= WX + XY + X’Z’


pp. 20


n EX.2 A’B’C’D’ + A’BC’D’ + A’BD + A’BC’D + ABCD + ACD’+ B’CD’

= A’C’D’ + A’BD + B’CD’ + ABC(A, B, C, D methods are applied)n No easy way to determine when a Boolean

Expression has a min. no. of terms or literalsn Systematic way is presented in Ch.5 & CH.6


pp. 21

Proving Validity of an EquationProving Validity of an Equation

vApproach : vConstruct a Truth TablevManipulate one side of the equation till it’s

identical to the other sidevReduce both sides independently to the same

equationv(a) Perform same operation on both sides

(b) Cannot Subtract or Divide both sides(Subtraction, Division NOT defined)


pp. 22


vUsually :vReduce both sides to Sum of Products (SOP)vCompare both sidesvTry to Add or Delete terms by using Theorems


pp. 23

Proving Validity of an EquationProving Validity of an EquationvEX.1 Show that

A’BD’ + BCD + ABC’ + AB’D = BC’D’ + AD + A’BC

By Consensus Theorem : A’BD’ + BCD + ABC’ + AB’D + BC’D’ + A’BC + ABD

= AD + A’BD’ + BCD + ABC’ + BC’D’ + A’BC

= AD + BC’D’ + A’BC

1 2 31 + 2 1 + 3 2 + 3


pp. 24


vEX.2 Show A’BC’D + (A’ + BC)(A + C’D’) + BC’D + A’BC’

= ABCD + A’C’D’ + ABD + ABCD’ + BC’D

nReducing the left side

A’BC’D + (A’ + BC)(A + C’D’) + BC’D + A’BC’

= (A’ + BC)(A + C’D’) + BC’D + A’BC’

= ABC + A’C’D’ + BC’D + A’BC’

= ABC + A’C’D’ + BC’D


pp. 25


vEX.2(cont.)vReducing the left side

ABCD + A’C’D’ + ABD + ABCD’ + BC’D

= ABC + A’C’D’ + ABD + BC’D

= ABC + A’C’D’ + BC’DnBecause both sides were independently reduced

to the same expression, the original equation is valid


pp. 26

Boolean Algebra & Ordinary AlgebraBoolean Algebra & Ordinary AlgebravBoolean Algebra ≠ Ordinary AlgebraEX.1 X + Y = X + Z => Y = Z (?)

X = 1, Y = 0 => 1 + 0 = 1 + 1But 0 ≠ 1

EX.2 “If XY = XZ then Y = Z”True : when X ≠ 0False : when X = 0


pp. 27

Boolean Algebra & Ordinary AlgebraBoolean Algebra & Ordinary Algebra

vEX.3 if Y = Z then X + Y = X + Z (V)if Y = Z then XY = XZ (V)

n Add/Multiply the same term => Validn Subtract/Divide the same term => Not Valid

n Check programmed exercise 3.1, 3.2,…,3.5 for practice

CH3 Boolean Algebra (cont d)access.ee.ntu.edu.tw/course/logic_design_94first/LectureNotes/CH03... · ACCESS IC LAB Graduate Institute of Electronics Engineering, NTU CH3 Boolean Algebra

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