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Logical Systems Synthesis

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Logical Synthesis

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Basic Gate Synthesis f = a. b f = a + b f = a’

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Basic Gate Synthesis f = (a. b)’ f = (a + b)’ f = a

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F = A + B. C Correct Incorrect Operator Precedence in Synthesis

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F = (A + B). C Incorrect Correct Operator Precedence in Synthesis

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An example of Step by Step Synthesis Single Output Circuits

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F = A + (B’ + C)(D’ + BE’) Reducing it to a more abstract form P = BE’ M = B’ + C F = A + M(D’ + P) N = D’ + P F = A + MN Z = MN F = A + Z

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F = A + MN

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F = A + Z F = A + MN F = A + M(D’ + P)

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F = A + (B’ + C)(D’ + BE’)

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An example of Multi-Output Synthesis

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f = ab + a’c g = ab + a’c’ These are the common terms in both logic expressions, therefore, they can be used to as a common gate when synthesizing the complete digital system. Sharing Gates

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Idea of sharing of Gates Circuits for f and g have a common product term (Implicant) namely ab. We are going to make it a common factor i.e. we are going to use the corresponding gate to drive both outputs f & g. f = a’c + ab g = a’c’ + ab

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Symbols

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From Digital Design, 5th Edition by M. Morris Mano and Michael Ciletti

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Boolean Algebra Manipulation of Switching Functions Boolean/Logic/SwitchingExpression/Function

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Boolean Functions Variables and constants take on only one of the two values : 0 or 1 There are three operators – ORwritten as a + b – ANDwritten as a. b – NOTwritten as a ’

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() Operator Precedence ’. + Precedence Sequence

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Examples s = (((x. y’) + (x’. y))’. z) + (((x. y’) + (x’. y)). z’) c = (x. y’ + x’. y). z + x. y f = ( a.b + a.c + b.c’)’. (a + b + c) + a.b.c f = (((a.b) + (a.c) + (b.c’))’. (a + b + c)) + (a.b.c) s = ((x. y’ + x’. y)’. z) + ((x. y’ + x’. y). z’) c = (((x. y’) + (x’. y)). z) + (x. y)

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Metatheorem Duality Any Theorem or Identity in Switching Algebra remains true if 0 & 1 are swapped and. & + are swapped throughout the expression A Theorem about Theorems Note: Duality does not imply logical equivalence. It only states that one fact leads to another fact.

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Axioms / Postulates 0. 0 = = = 1. 0 = 0 x = 0 implies x’ = = = = = 1 x = 1 implies x’ = 0

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Theorems (Single Variable) x. 0 = 0 x. 1 = x x. x = x x. x’ = 0 (x’)’ = x x + 1 = 1 x + 0 = x x + x = x x + x’ = 1 (x’)’ = x Null Elements Identities Idempotency Complements Involution

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Theorems (2- & 3- Variable) x. y = y. x x. (y. z) = (x. y). z x. (y + z) = x. y + x. z x + (x. y) = x (x. y) + (x. y’) = x x + y = y + x x + (y + z) = (x + y) + z x + (y. z) = (x + y). (x + z) x. (x + y) = x (x + y). (x + y’) = x Commutativity Associativity Distributivity Covering Combining

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Theorems (2- & 3- Variable) (x. y)’ = x’ + y’ (x. y) + (x’. z) + (y. z) = (x. y) + (x’. z) (x + y)’ = x’. y’ (x + y). (x’ + z). (y + z) = (x + y). (x’ + z) DeMorgan’s Theorem Consensus Theorem

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Theorems (n-Variable) x. x x = x (x 1. x x n )’ = x 1 ’ + x 2 ’ x n ’ x + x x = x (x 1 + x x n )’ = x 1 ’. x 2 ’ x n ’ Generalized Idempotency DeMorgan’s Theorem

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Theorems (n-Variable) [F(x 1, x 2, -----, x n,., + )]’ = F(x 1 ’, x 2 ’, -----, x n ’, +,. ) Generalized DeMorgan’s Theorem

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f = ( a. b )’ = a’ + b’ f = ( a + b )’ = a’. b’ Implications of DeMorgan’s Theorem

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f = { [a. ( b’ + c ) ]’ }’ = { a’ + ( b’ + c )’ }’ = { a’ + b. c’ }’ f’ = a’ + ( b. c’ ) Implications of DeMorgan’s Theorem

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Function Manipulation Using Axioms and Theorems of Boolean Algebra

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Proof of Consensus Theorem x.y + x’.z + y.z = x.y + x’.z x.y + x’.z + y.z = x.y + x’.z + y.z.1 = x.y + x’.z + y.z.(x + x’) = x.y + x’.z + x.y.z + x’.y.z = x.y + x.y.z + x’.z + x’.y.z = x.y.1 + x.y.z + x’.z.1 + x’.y.z = x.y.(1 + z) + x’.z.(1 + y) = x.y.1 + x’.z.1 = x.y + x’.z Theorem Used a = a.1 1 = a + a’ a.(b + c) = a.b + a.c a = a.1 a.b + a.c = a.(b + c) 1 + a = 1 a.1 = a

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Find DeMorgan’s Equivalent of the Following

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F = ( ( B. E’ ) + D’ ). ( B’ + C ) + A = { [ ( ( ( B. E’ ) + D’ ). ( B’ + C ) ) + A ]’ }’ = { ( ( ( B. E’ ) + D’ ). ( B’ + C) )’. A’ }’ = { ( ( B. E’ ) + D’ )’ + ( B’ + C )’. A’ }’ = { ( B. E’ )’. D ) + ( B. C’ ). A’ }’ = { ( B’ + E ). D ) + ( B. C’ ). A’ }’ = { ( ( B’ + E ). D ) ) + ( ( B. C’ ). A’ ) }’ F = { ( B’ + E ). D ) + ( B. C’ ). A’ }’ F’ = ( B’ + E ). D ) + ( B. C’ ). A’ DeMorgan’s Equivalent

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F = ( ( B. E’ ) + D’ ). ( B’ + C ) + A DeMorgan’s Equivalent

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F’ = ( B’ + E ). D ) + ( B. C’ ). A’ DeMorgan’s Equivalent

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A circuit developed using NAND gates only can be converted into a circuit employing only NOR gates by directly replacing the NAND gates with the NOR gates and inverting all the inputs and outputs of the circuit NAND NOR Inter-Conversion The Converse of this operation is also true Using DeMorgan’s Theorem

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F = ((((B.E’)’. D)’. (B.C’)’)’. A’)’

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F’ = ((((B’ + E)’ + D’)’ + (B’ + C)’)’ + A)’

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Proof by Example F = ((((B.E’)’. D)’. (B.C’)’)’. A’)’ = ((((B’ + E). D)’. (B’ + C))’. A’)’ = ((((B’ + E)’ + D’). (B’ + C))’. A’)’ = ((((B’ + E)’ + D’)’ + (B’ + C)’). A’)’ = ((((B’ + E)’ + D’)’ + (B’ + C)’)’ + A) F’ = ((((B’ + E)’ + D’)’ + (B’ + C)’)’ + A)’ Using DeMorgan’s Theorem

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F = A + (B’ + C). (D’ + B.E’) F = A + (B’ + C)(D’ + BE’) Can also be written in the following notation

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Forms of Logic Expressions

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Generalized Forms of Logic Expressions SOP Sum of Products POS Product of Sums

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Terminology Literal Product Term Sum Term Product of Sums Sum of Products

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Definitions Literal – A variable in either it’s complemented or uncomplemented form Sum Term – One or more Literals connected by an OR operator Product Term – One or more Literals connected by an AND operator

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Definitions Product of Sums – One or more Sum Terms connected by an AND operator Sum of Products – One or more Product Terms connected by an OR operator

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SOP Expression

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POS Expression

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More SOP Expressions ABC + A’BC’ AB + A’BC’ + C’D’ + D A’B + CD’ + EF + GK + HL’ More POS Expressions (A + B’)(A’ + C’ + D)F’ (A + C)(B + D’)(B’+C)(A + D’ + E) (A + B’ + C)(A + C)

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nd = ((((b.e’)’. d)’. (b.c’)’)’. a’)’ nr = ((((b’ + e)’ + d’)’ + (b’ + c)’)’ + a)’ s = (((x. y’) + (x’. y))’. z) + (((x. y’) + (x’. y)). z’) f = (((a.b) + (a.c) + (b.c’))’. (a + b + c)) + (a.b.c) c = (((x. y’) + (x’. y)). z) + (x. y) Not an SOP or POS Expression

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Canonical Forms of Logic Expressions C-SOP Sum of Standard Product Terms C-POS Product of Standard Sum Terms aka Canonical Sum aka Canonical Product

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More Terminology Minterm Maxterm Canonical Product Canonical Sum

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Definitions Mintermaka Standard Product Term – A Product Term composed of all the variables of the given problem Maxtermaka Standard Sum Term – A Sum Term composed of all the variables of the given problem

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Definitions Canonical Sum – One or more Minterms connected by an OR operator Canonical Product – One or more Maxterms connected by an AND operator

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C-SOP Expression

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C-POS Expression

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Writing Algebraic Expressions From Truth Tables

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OR = Decomposing into Minterms Implementing 1’s

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a.b’ + a’.b = f OR = = m2 + m1 = f

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Minterms Notation

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Expressions involving Minterms

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AND = Decomposing into Maxterms Implementing 0’s

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AND = ( a + b’ ) ( a’ + b ) = f AND = M1. M2 = f

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Maxterms Notation

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Expressions involving Maxterms

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Write Expressions using Minterm/Maxterm Terminology for the following Truth Table Using Minterms f = ? Using Maxterms f = ?

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Write Expressions using Minterm/Maxterm Terminology for the following Truth Table Using Minterms f = m2 + m5 + m6 Using Maxterms f = M0. M1. M3. M4. M7

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Write Expressions using Minterm/Maxterm Terminology for the following Truth Table Using Minterms f = m2 + m5 + m6 f’ = ? Using Maxterms f = M0. M1. M3. M4. M7 f’ = ?

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Write Expressions using Minterm/Maxterm Terminology for the following Truth Table Using Minterms f = m2 + m5 + m6 f’ = m0 + m1 + m3 + m4 + m7 Using Maxterms f = M0. M1. M3. M4. M7 f’ = M2. M5. M6

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Write Expressions using Minterm/Maxterm Terminology for the following Truth Table Using Minterms f = m2 + m5 + m6 f’ = m0 + m1 + m3 + m4 + m7 Using Maxterms f = M0. M1. M3. M4. M7 f’ = M2. M5. M6 f = ∑ (2, 5, 6) = ∏ (0, 1, 3, 4, 7) f’ = ∏ (2, 5, 6) = ∑ (0, 1, 3, 4, 7) Alternative Notation

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Write Expressions using Minterm/Maxterm Terminology for the following Truth Table Using Minterms f = ? f’ = ? Using Maxterms f = ? f’ = ?

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Write Expressions using Minterm/Maxterm Terminology for the following Truth Table Using Minterms f = m1 + m5 + m6 + m8 + m9 + m10 + m14 + m15 f’ = m0 + m2 + m3 + m4 + m7 + m11 + m12 + m13 Using Maxterms f = M0. M2. M3. M4. M7. M11. M12. M13 f’ = M1. M5. M6. M8. M9. M10. M14. M15

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Write Expressions using Minterm/Maxterm Terminology for the following Truth Table f = ∑ (1, 5, 6, 8, 9, 10, 14, 15) f = ∏ (0, 2, 3, 4, 7, 11, 12, 13) f’ = ∑ (0, 2, 3, 4, 7, 11, 12, 13) f’ = ∏ (1, 5, 6, 8, 9, 10, 14, 15)

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Can you intuitively establish a relationship within the following set of results? f = m2 + m5 + m6 f’ = m0 + m1 + m3 + m4 + m7 f = M0. M1. M3. M4. M7 f’ = M2. M5. M6 f = M0. M2. M3. M4. M7. M11. M12. M13 f’ = M1. M5. M6. M8. M9. M10. M14. M15 f = m1 + m5 + m6 + m8 + m9 + m10 + m14 + m15 f’ = m0 + m2 + m3 + m4 + m7 + m11 + m12 + m13

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Relationship between Minterms and Maxterms f = m0 + m3 f = x’ y’ + x y = ( x’ x + y’ x ) ( x’ y + y’ y ) = ( x y’ ) ( x’ y ) f’ = [ (x y’) (x’ y) ]’ = x’ y + x y’ = m1 + m2 f’ = m1 + m2 f = m0 + m3 f = x’ y’ + x y f’ = ( x’ y’ + x y )’ = ( x’. y’ )’ ( x. y )’ = ( x + y ) ( x’ + y’ ) f’ = ( x + y ) ( x’ + y’ ) = M0. M3 f’ = M0. M3 f = m0 + m3 f’ = m1 + m2 = x’ y + x y’ f = ( x’ y + x y’ )’ = ( x’ y )’ ( x y’ )’ = ( x + y’ ) ( x’ + y) = M1. M2 f = M1. M2

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To implement a function we can use the logic 1’s in the output to obtain a Minterms’ SOP or use the logic 0’s to obtain a Maxterms’ POS Where, T = {Set of all indices} = {0 to 2 n - 1} R = {Set of indices of required to implement C-SOP of F(x1, x2, …., xn)} S = T – R F(x1, x2, …., xn) = ∑m R = ∏M S [F(x1, x2, …., xn)]’ = ∑m S = ∏M R Note : Observe the Duality inherent in this Generalization

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Use Boolean Algebra to establish the relationship between the following set of Expressions f = m2 + m5 + m6 f’ = m0 + m1 + m3 + m4 + m7 f = M0. M1. M3. M4. M7 f’ = M2. M5. M6 f = M0. M2. M3. M4. M7. M11. M12. M13 f’ = M1. M5. M6. M8. M9. M10. M14. M15 f = m1 + m5 + m6 + m8 + m9 + m10 + m14 + m15 f’ = m0 + m2 + m3 + m4 + m7 + m11 + m12 + m13

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Converting SOP to POS f = x y + x’ z(SOP Form) Using Distributive Law f = x y + x’ z f = ( x y + x’ ) (x y + z) f = ( x + x’ ) ( y + x’ ) ( x + z ) ( y + z ) f = ( y + x’ ) ( x + z ) ( y + z ) f = ( y + x’ ) ( x + z ) ( y + z )(POS Form)

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Converting POS to SOP f = ( x + y ) ( x’ + z )(POS Form) Using Distributive Law f = ( x + y ) ( x’ + z ) f = x ( x’ + z ) + y ( x’ + z ) f = x x’ + x z + y x’ + y z f = x z + y x’ + y z f = x z + y x’ + y z(SOP Form) Note : To obtain the expression f = x y + x’ z as in the previous slide apply Consensus Theorem

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Converting SOP to C-SOP F = A + B’C(SOP Form) Expanding both terms separately using {x = x. 1} & {1 = (x + x’)} A = A (B + B’) = AB + AB’ = AB (C + C’) + AB’ (C + C’) A = ABC + ABC’ + AB’C + AB’C’ B’C = B’C (A + A’) = AB’C + A’B’C Combining the expressions F = ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C = ABC + ABC’ + AB’C + AB’C + AB’C’ + A’B’C Using {x + x = x} F = ABC + ABC’ + AB’C + AB’C’ + A’B’C F = ABC + ABC’ + AB’C + AB’C’ + A’B’C (C-SOP Form)

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Converting POS to C-POS f = ( x’ + y ) ( x + z ) ( y + z ) (POS Form) Expanding the three terms separately using {x = x + 0} & {0 = x x’} ( x’ + y ) = x’ + y + z z’ = ( x’ + y + z ) ( x’ + y + z’ ) ( x + z ) = x + z + y y’ = ( x + z + y ) ( x + z + y’ ) = ( x + y + z ) ( x + y’ + z ) ( y + z ) = y + z + x x’ = ( y + z + x ) ( y + z + x’ ) = ( x + y + z ) ( x’ + y + z ) Combining the expressions f = ( x’ + y + z ) ( x’ + y + z’ ) ( x + y + z ) ( x + y’ + z ) ( x + y + z ) ( x’ + y + z ) f = ( x’ + y + z ) ( x’ + y + z ) ( x + y + z ) ( x + y + z ) ( x’ + y + z’ ) ( x + y’ + z ) Using {x. x = x} f = ( x’ + y + z’ ) ( x + y + z ) ( x’ + y + z’ ) ( x + y’ + z ) f = ( x’ + y + z’ ) ( x + y + z ) ( x’ + y + z’ ) ( x + y’ + z )(C-POS Form)

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