ch3 boolean algebra (cont...
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Graduate Institute of Electronics Engineering, NTU
CH3 Boolean Algebra (contCH3 Boolean Algebra (cont’’d)d)
Lecturer:吳安宇
Date:2005/10/7
Graduate Institute of Electronics Engineering, NTU
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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
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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
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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
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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)
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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’
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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’
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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)
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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’
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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)
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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
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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)
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Consensus TheoremConsensus TheoremvApplication to eliminate redundant terms from
Boolean Expressions
a’b’ + ac + bc’ + b’c +ab = a’b’ + ac + bc’
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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’
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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’
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Algebraic Simplification of Switching Algebraic Simplification of Switching ExpressionExpression
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)
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Algebraic Simplification of Switching Algebraic Simplification of Switching ExpressionExpression
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)
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Algebraic Simplification of Switching Algebraic Simplification of Switching ExpressionExpression
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
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Algebraic Simplification of Switching Algebraic Simplification of Switching ExpressionExpression
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’
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Algebraic Simplification of Switching Algebraic Simplification of Switching ExpressionExpression
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
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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)
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Proving Validity of an EquationProving Validity of an Equation
vUsually :vReduce both sides to Sum of Products (SOP)vCompare both sidesvTry to Add or Delete terms by using Theorems
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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
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Proving Validity of an EquationProving Validity of an Equation
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
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Proving Validity of an EquationProving Validity of an Equation
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
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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
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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