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

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

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

9. F = A + Z

10. F = A + Z
F = A + MN

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

12. F = A + (B’ + C)(D’ + BE’)

13. F = A + (B’ + C)(D’ + BE’)
F = A + (B’ + C)(D’ + BE’)

14. An example of Multi-Output Synthesis

15. Sharing Gates 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.
f = ab + a’c
g = ab + a’c’

16. Idea of sharing of Gates
f = a’c + ab
g = a’c’ + ab
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.

17. Symbols

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

19. Manipulation of Switching Functions
Boolean Algebra
Manipulation of Switching Functions
Boolean/Logic/Switching Expression/Function

20. Boolean Functions
Variables and constants take on only one of the two values : 0 or 1
There are three operators
OR written as a + b
AND written as a . b
NOT written as a ’

21. Operator Precedence () ’ . +
Sequence ’ . +

22. Examples s = ((x . y’ + x’ . y)’ . z) + ((x . y’ + x’ . y) . z’)
c = (x . y’ + x’ . y) . z + x . y
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)

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

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

25. Theorems (Single Variable)
Null Elements
Identities
Idempotency
Complements
Involution
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

26. Theorems (2- & 3- Variable)
Commutativity
Associativity
Distributivity
Covering
Combining
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

27. Theorems (2- & 3- Variable)
DeMorgan’s
Theorem
Consensus Theorem
(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)

28. Theorems (n-Variable)
Generalized
Idempotency
DeMorgan’s Theorem
x . x x = x (x1 . x xn)’ = x1’ + x2’ xn’
x + x x = x (x1 + x xn)’ = x1’ . x2’ xn’

29. Theorems (n-Variable)
Generalized DeMorgan’s Theorem
[F(x1, x2, -----, xn, . , + )]’ = F(x1’, x2’, -----, xn’, + , . )

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

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

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.b + a.c = a.(b + c)
1 + a = 1
a.1 = a

34. Find DeMorgan’s Equivalent of the Following

35. DeMorgan’s Equivalent
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’

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

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

38. Using DeMorgan’s Theorem NAND NOR Inter-Conversion
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
The Converse of this operation is also true

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

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

41. Using DeMorgan’s Theorem
Proof by Example
Using DeMorgan’s Theorem
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)’

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

43. Forms of Logic Expressions

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

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

46. Definitions Literal Sum Term Product Term
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 Sum of Products
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

50. POS Expression

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

53. Not an SOP or POS Expression
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)
nr = ((((b’ + e)’ + d’)’ + (b’ + c)’)’ + a)’
nd = ((((b.e’)’ . d)’ . (b.c’)’)’ . a’)’

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

55. More Terminology
Minterm
Maxterm
Canonical Sum
Canonical Product

56. Definitions Minterm aka Standard Product Term
A Product Term composed of all the variables of the given problem
Maxterm aka Standard Sum Term
A Sum Term composed of all the variables of the given problem

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

58. C-SOP Expression

60. C-POS Expression

62. Writing Algebraic Expressions From Truth Tables

63. Decomposing into Minterms
Implementing 1’s OR =

64. OR = OR =
a.b’ + a’.b = f
m2 + m1 = f

65. Minterms Notation

67. Expressions involving Minterms

73. Decomposing into Maxterms
Implementing 0’s
AND =

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

75. Maxterms Notation

77. Expressions involving Maxterms

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
Alternative Notation
f = ∑ (2, 5, 6) = ∏ (0, 1, 3, 4, 7)
f’ = ∏ (2, 5, 6) = ∑ (0, 1, 3, 4, 7)

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

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 = m1 + m5 + m6 + m8 + m9 +
m10 + m14 + m15
f’ = m0 + m2 + m3 + m4 + m7 +
m11 + m12 + m13
f = M0 . M2 . M3 . M4 . M7 .
M11 . M12 . M13
f’ = M1 . M5 . M6 . M8 . M9 .
M10 . M14 . M15

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’ = 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. Note : Observe the Duality inherent in this Generalization
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
F(x1, x2, …. , xn) = ∑mR = ∏MS
[F(x1, x2, …. , xn)]’ = ∑mS = ∏MR
Where,
T = {Set of all indices} = {0 to 2n - 1}
R = {Set of indices of required to implement C-SOP of F(x1, x2, …., xn)}
S = T – 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 = m1 + m5 + m6 + m8 + m9 +
m10 + m14 + m15
f’ = m0 + m2 + m3 + m4 + m7 +
m11 + m12 + m13
f = M0 . M2 . M3 . M4 . M7 .
M11 . M12 . M13
f’ = M1 . M5 . M6 . M8 . M9 .
M10 . M14 . M15

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