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

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Presentation on theme: "Logical Systems Synthesis. Logical Synthesis Basic Gate Synthesis f = a. b f = a + b f = a’"— Presentation transcript:

1 Logical Systems Synthesis

2 Logical Synthesis

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

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

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

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

7 An example of Step by Step Synthesis Single Output Circuits

8 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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10 F = A + MN

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

12 F = A + (B’ + C)(D’ + BE’)

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14 An example of Multi-Output Synthesis

15 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

16 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

17 Symbols

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

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

20 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 ’

21 () Operator Precedence ’. + Precedence Sequence

22 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)

23 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.

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

25 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

26 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

27 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

28 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

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

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

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

32 Function Manipulation Using Axioms and Theorems of Boolean Algebra

33 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

34 Find DeMorgan’s Equivalent of the Following

35 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

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

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

38 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

39 F = ((((B.E’)’. D)’. (B.C’)’)’. A’)’

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

41 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

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

43 Forms of Logic Expressions

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

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

46 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

47 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

48 SOP Expression

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50 POS Expression

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52 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)

53 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

54 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

55 More Terminology Minterm Maxterm Canonical Product Canonical Sum

56 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

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

58 C-SOP Expression

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60 C-POS Expression

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62 Writing Algebraic Expressions From Truth Tables

63 OR = Decomposing into Minterms Implementing 1’s

64 a.b’ + a’.b = f OR = = m2 + m1 = f

65 Minterms Notation

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67 Expressions involving Minterms

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73 AND = Decomposing into Maxterms Implementing 0’s

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

75 Maxterms Notation

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77 Expressions involving Maxterms

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

84 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

85 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’ = ?

86 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

87 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

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

89 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

90 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)

91 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

92 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

93 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

94 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

95 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)

96 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

97 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)

98 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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