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Computer Logic and Digital Design Chapter 2 Henry Hexmoor AND logic

Computer Logic and Digital Design Chapter 2 Henry Hexmoor AND logic. input. output. AND Gate. B. AB. S1 . S2. S1 S2 path? off off no on off no off on no

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Computer Logic and Digital Design Chapter 2 Henry Hexmoor AND logic

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  1. Computer Logic and Digital Design Chapter 2Henry HexmoorAND logic input output AND Gate B AB S1 . S2 S1 S2 path? off off no on off no off on no on on yes

  2. input OR logic S1 + S2 S1 S2 path? off off no on off yes off on yes on on yes S1 S2 output

  3. NOT logic S S 1 0 0 1 S S'

  4. X (X . Y)’ X Y X’ + Y’ Y X (X + Y)’ Y X X’ . Y’ Y Equivalent Symbols of NAND, NOR Gates NAND Symbols Normal Symbol Alternate NAND Symbol According to DeMorgan’s theorem T13: (X . Y)’ = X’ + Y’ NOR Symbols Alternate NOR Symbol Normal NOR Symbol According to DeMorgan’s theorem T13’: (X + Y)’ = X’ . Y’

  5. Literals (n) and terms • number of truth table rows = 2n Boolean Algebra X Y Z Y’Z X+Y’Z 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0 0 1 1 0 0 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 0 1 • See table 2-3 • Using demorgan’s laws OR and AND are interchanged… the duality principle

  6. Switching Algebra Axioms & Theorems (A1) X = 0 if X ≠ 1 (A1’) X = 1 if X ≠ 0 (A2) If X = 0, then X’ = 1 (A2’) if X = 1, then, X’ = 0 (A3) 0 . 0 = 0 (A3’) 1 + 1 = 1 (A4) 1 . 1 = 1 (A4’) 0 + 0 = 0 (A5) 0 . 1 = 1 . 0 = 0 (A5’) 1 + 0 = 0 + 1 = 1 (T1) X + 0 = X (T1’) X . 1 = X (Identities) (T2) X + 1 = 1 (T2’) X . 0 = 0 (Null elements) (T3) X + X = X (T3’) X . X = X (Idempotency) (T4) (X’)’ = X (Involution) (T5) X + X’ = 1 (T5’) X . X’ = 0 (Complements) (T6) X + Y = Y + X (T6’) X . Y = Y . X (Commutativity) (T7) (X + Y) + Z = X + (Y + Z) (T7’) (X . Y) . Z = X . (Y . Z) (Associativity) (T8) X . Y + X . Z = X . (Y + Z) (T8’) (X + Y) . (X + Z) = X + Y . Z (Distributivity) (T9) X + X . Y = X (T9’) X . (X + Y) = X (Covering) (T10) X . Y + X . Y’ = X (T10’) (X + Y) . (X + Y’) = X (Combining) (T11) X . Y + X’. Z + Y . Z = X . Y + X’ . Z (T11’) (X + Y) . ( X’ + Z) . (Y + Z) = (X + Y) . (X’ + Z) (Consensus) (T12) X + X + . . . + X = X (T12’) X . X . . . . . X = X (Generalized idempotency) (T13) (X1 . X2 . . . . . Xn)’ = X1’ + X2’ + . . . + Xn’ (T13’) (X1 + X2 + . . . + Xn)’ = X1’ . X2’ . . . . . Xn’ (DeMorgan’s theorems) (T14) [F(X1, X2, . . ., Xn, +, .)]’ = F(X1’, X2’, . . ., Xn’, . , +) (Generalized DeMorgran’s theorem)

  7. Switching Algebra Axioms • First two axioms state that a variable X can only take on only one of two values: (A1) X = 0 if X ¹ 1 (A1’) X = 1 if X ¹ 0 • Not Axioms, formally define X’ (X prime or NOT X): (A2) If X = 0, then X’ = 1 (A2’) if X = 1, then, X’ = 0 Note: Above axioms are stated in pairs with only difference being the interchange of the symbols 0 and 1.

  8. Three More Switching Algebra Axioms • The following three Boolean Algebra axioms state and formally define the AND, OR operations: (A3) 0 . 0 = 0 (A3’) 1 + 1 = 1 (A4) 1 . 1 = 1 (A4’) 0 + 0 = 0 (A5) 0 . 1 = 1 .0 = 0 (A5’) 1 + 0 = 0 + 1 = 1 Axioms A1-A5, A1’-A5’ completely define switching algebra.

  9. Switching Algebra: Single-Variables Theorems • Switching-algebra theorems are statements known to be always true (proven using axioms) that allow us to manipulate algebraic logic expressions to allow for simpler analysis. (e.g . X + 0 = X allow us to replace every X +0 with X) The Theorems: (T1-T5, T1’-T5’) (T1) X + 0 = X (T1’) X . 1 = X (Identities) (T2) X + 1 = 1 (T2’) X . 0 = 0 (Null elements) (T3) X + X = X (T3’) X . X = X (Idempotency) (T4) (X’)’ = X (Involution) (T5) X + X’ = 1 (T5’) X . X’ = 0 (Complements)

  10. Perfect Induction • Most theorems in switching algebra are simple to prove using perfect induction: Since a switching variable can only take the values 0 and 1 we can prove a theorem involving a single variable X by proving it true for X = 0 and X =1 Example: To prove (T1) X + 0 = X [X = 0] 0 + 0 = 0 true according to axiom A4’ [X = 1] 1 + 0 = 1 true according to axiom A5’

  11. Switching Algebra: Two- and Three-Variable Theorems (Commutativity) (T6) X + Y = Y + X (T6’) X . Y = Y . X (Associativity) (T7) (X + Y) + Z = X + (Y + Z) (T7’) (X . Y) . Z = X . (Y . Z) T6-T7, T6’ -T7’ are similar to commutative and associative laws for addition and multiplication of integers and reals.

  12. Two- and Three-Variable Theorems (Continued) (Distributivity) (T8) X . Y + X . Z = X . (Y + Z) (T8’) (X + Y) . (X + Z) = X + Y . Z • T8 allows to multiply-out an expression to get sum-of-products form (distribute logical multiplication over logical addition): For example: V . (W + X) . (Y + Z) = V .W . Y + V. W. Z + V. X . Y + V. X . Z sum-of-products form • T8’ allows to add-out an expression to get a product-of-sums form (distribute logical addition over logical multiplication): For example: (V . W . X) + (Y . Z ) = (V + Y) . (V + Z) . (W + Y) . (W + Z) . (X + Y) . (X + Z) product-of-sums form

  13. x y z y + z x.(y + z) x.y x.z x.y + x.z 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 Theorem Proof using Truth Table • Can use truth table to prove T8 by perfect induction. • i.e, Prove that: X . Y + X . Z = X . (Y + Z) (i) Construct truth table for both sides of above equality. (ii) Check that from truth table check that that X . Y + X . Z = X . (Y + Z) This is satisfied because output column values for X . Y + X . Z and output column values for X . (Y + Z) are equal for all cases.

  14. Two- and Three-Variable Theorems (Continued) (Covering) (T9) X + X . Y = X (T9’) X . (X + Y) = X (Combining) (T10) X . Y + X . Y’ = X (T10’) (X + Y) . (X + Y’) = X • T9-T10 used in the minimization of logic functions.

  15. Two- and Three-Variable Theorems (Continued) (Consensus) (T11) X . Y + X’. Z + Y . Z = X . Y + X’ . Z (T11’) (X + Y) . ( X’ + Z) . (Y + Z) = (X + Y) . (X’ + Z) • In T11 the term Y. Z is called the consensus of the term X . Y and the term X’ . Z: • If Y . Z = 1, then either X . Y or X’ . Z must also be 1. • Thus the term Y . Z is redundant and may be dropped.

  16. n-Variable Theorems (Generalized idempotency) (T12) X + X + . . . + X = X (T12’) X . X . . . . . X = X (DeMorgan’s theorems) (T13) (X1 . X2 . . . . . Xn)’ = X1’ + X2’ + . . . + Xn’ (T13’) (X1 + X2 + . . . + Xn)’ = X1’ . X2’ . . . . . Xn’ (T13), (T13’) are probably the most commonly used theorems of switching algebra.

  17. Examples Using DeMorgan’s theorems Example: Equivalence of NAND Gate: A two-input NAND Gate has the output expression Z = (X . Y)’ using (T13) Z = (X . Y)’ = (X’ + Y’) The function of a NAND gate can be achieved with an OR gate with an inverter at each input. Example: Equivalence of NOR Gate A two-input NOR Gate has the output expression Z=(X+Y)’ using (T13’) Z = (X + Y)’ = X’ . Y’ The function of a NOR gate can be achieved with an AND gate with an inverter at each input.

  18. n-Variable Theorems (Continued) (Generalized DeMorgran’s theorem) (T14) [F(X1, X2, . . ., Xn, +, .)]’ = F(X1’, X2’, . . ., Xn’, . , +) • States that given any n-variable logic expression its complement can be found by swapping + and . and complementing all variables. Example: F(W,X,Y,Z) = (W’.X) + ( X.Y) + (W.(X’ + Z’)) = ((W)’ . X) + (X. Y) + (W.((X)’ + (Z)’)) [F(W,X,Y,Z)]’ = ((W’)’ + X’) .(X’ + Y’).(W’ + ((X’)’.(Z’)’)) Using T4, (X’)’ = X simplifies it to: [F(W,X,Y,Z)]’ = (W + X’) . (X’ + Y’) . (W’ + (X . Z))

  19. Logic Function Representation Definitions • A literal: is a variable or a complement of a variable Examples: X, Y, X’, Y’ • A product term: is a single literal, or a product of two or more literals. Examples: Z’ W.Y.Y X.Y’.Z W’.Y’.Z • A sum-of-products expression: is a logical sum of product terms. Example: Z’ + W.X.Y + X.Y’.Z + W’.Y’.Z • A sum term: is a single literal or logical sum of two or more literals Examples: Z’ W + X + Y X + Y’ + Z W’ + Y’ + Z • A product-of-sums expression: is a logical product of sum terms. Example: Z’. (W + X + Y) . (X + Y’ + Z) . (W’ + Y’ + Z) • A normal term: is a product or sum term in which no variable appears more than once Examples of non-normal terms: W.X.X.Y’ W+W+X’+Y X.X’.Y Examples of normal terms: W . X . Y’ W + X’ + Y

  20. Logic Expression Algebraic Manipulation Example • Prove that the following identity is true using Algebraic expression Manipulation : (one can also prove it using a truth table) X .Y + X . Z = ((X’ + Y’) . (X’ + Z’))’ • Starting from the left hand side of the identity: Let F = X .Y + X . Z A = X . Y B = X . Z Then F = A + B • Using DeMorgan’s theorem T 13 on F: F = A + B = (A’ . B’)’ (1) • Using DeMorgan’s theorem T 13’ on A, B: A = X . Y = (X’ + Y’)’ (2) B = X . Z = (X’ + Z’)’ (3) • Substituting A, B from (2), (3), back in F in (1) gives: F = (A’ . B’)’ = ((X’ + Y’) . (X’ + Z’))’ Which is equal to the right hand side of the identity.

  21. Terminology: Minterms • A minterm is a special product of literals, in which each input variable appears exactly once. • A function with n variables has 2n minterms (since each variable can appear complemented or not) • A three-variable function, such as f(x,y,z), has 23 = 8 minterms: • Each minterm is true for exactly one combination of inputs: x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz Minterm Is true when… Shorthand x’y’z’ x=0, y=0, z=0 m0 x’y’z x=0, y=0, z=1 m1 x’yz’ x=0, y=1, z=0 m2 x’yz x=0, y=1, z=1 m3 xy’z’ x=1, y=0, z=0 m4 xy’z x=1, y=0, z=1 m5 xyz’ x=1, y=1, z=0 m6 xyz x=1, y=1, z=1 m7

  22. Terminology: Maxterms • A maxterm is a special sum of literals, in which each input variable appears exactly once. • A function with n variables has 2n maxterms (since each variable can appear complemented or not) • A three-variable function, such as f(x,y,z), has 23 = 8 maxterms: Maxterm Shorthand x’+ y’+ z’ M0 x’+ y’ + z M1 x’+ y + z’ M2 x’ + y + z M3 x + y’ + z’ M4 x + y’ + z M5 x + y + z’ M6 x + y + z M7

  23. Canonical Forms • Minterms and Maxterms • Index Representation of Minterms and Maxterms • Sum-of-Minterm (SOM) Representations • Product-of-Maxterm (POM) Representations • Representation of Complements of Functions • Conversions between Representations

  24. Logic Function Representation Definitions • Minterm An n-variable minterm is a normal product term with n literals. There are 2n such products terms. Example of 4-variable minterms: W.X’.Y’.Z’ W.X.Y’.Z W’.X’.Y.Z’ • Maxterm An n-variable maxterm is a normal sum term with n literals. There are 2n such sum terms. Examples of 4-variable maxterms: W’ + X’ + Y + Z’ W + X’ + Y’ + Z W’ + X’ + Y + Z • A minterm can be defined as as product term that is 1 in exactly one row of the truth table. • A maxterm can similarly be defined as a sum term that is 0 in exactly one row in the truth table.

  25. Minterms/Maxterms for A 3-variable function F(X,Y,Z) Row X Y Z F Minterm Maxterm 0 0 0 0 F(0,0,0) X’.Y’.Z’ X + Y + Z 1 0 0 1 F(0,0,1) X’.Y’.Z X + Y + Z’ 2 0 1 0 F(0,1,0) X’.Y.Z’ X + Y’ + Z 3 0 1 1 F(0,1,1) X’.Y.Z X + Y’ + Z’ 4 1 0 0 F(1,0,0) X.Y’.Z’ X’ + Y + Z 5 1 0 1 F(1,0,1) X.Y’.Z X’ + Y + Z’ 6 1 1 0 F(1,1,0) X.Y.Z’ X’ + Y’ + Z 7 1 1 1 F(1,1,1) X.Y.Z X’ + Y’ + Z’

  26. Canonical Sum Example Row X Y Z F 0 0 0 0 1 1 0 0 1 0 2 0 1 0 0 3 0 1 1 1 4 1 0 0 1 5 1 0 1 0 6 1 1 0 1 7 1 1 1 1 • The function represented by the truth table: has the canonical sum representation: F = SX,Y,Z m(0, 3, 4, 6, 7) = X’.Y’.Z’ + X’.Y.Z + X.Y’.Z’ + X.Y’.Z’ + X.Y.Z Minterm list using S notation Algebraic canonical sum of minterms

  27. Canonical Product Example Row X Y Z F 0 0 0 0 1 1 0 0 1 0 2 0 1 0 0 3 0 1 1 1 4 1 0 0 1 5 1 0 1 0 6 1 1 0 1 7 1 1 1 1 • The function represented by the truth table: has the canonical product representation: F = PX,Y,Z M(1,2,5) = (X + Y + Z’) . (X + Y’ + Z) . (X’ + Y + Z’)

  28. = m M = M m i i i i Minterm and Maxterm Relationship • Review: DeMorgan's Theorem and • Two-variable example: and M2 is the complement of m2 and vice-versa. • Since DeMorgan's Theorem holds for n variables, the above holds for terms of n variables • giving: and Thus Mi is the complement of mi. + = × = + x y x y x · y x y = = + m x· y M x y 2 2

  29. Conversion Between Minterm/Maxterm Lists • To convert between a minterm list and a maxterm list take the set complement. Example (page 44, 3rd ed.): F’(X,Y, Z) = S m(1,3, 4, 6) = m1 + m3 + m4 + m6 F(X,Y, Z) = P M(1,3,4,6) = M1M3M4M6

  30. Verbal Synthesis Example: An Alarm Circuit • A verbal logic description: • The ALARM output is 1 if the panic input is 1, or if the ENABLE input is 1, the EXISTING input is 0, and the house is not secure. • The house is secure if the WINDOW, DOOR, GARAGE inputs are all 1 • This can be put in logic expressions as follows: ALARM = PANIC + ENABLE . EXISTING’ . SECURE’ SECURE = WINDOW. DOOR. GARAGE ALARM = PANIC + ENABLE . EXISTING’. (WINDOW . DOOR . GARAGE)’ In sum of products form as (by using DeMorgan T13 and multiplying out) : ALARM = PANIC + ENABLE. EXISTING’ . WINDOW’ + ENABLE . EXISTING’. DOOR’+ ENABLE. EXISTING’. GARAGE’

  31. Literal cost • Gate input cost: literals + terms cost criteria2-4 OR Gate

  32. Combinational Circuit Minimization • Canonical sum and product logic expressions do not provide a circuit realization with the minimum number of gates. • Minimization methods reduce the cost of two level AND-OR, NAND-NAND, OR-AND, NOR-NOR circuits in three ways: • By minimizing the number of first level gates • By minimizing the number of inputs of each first-level gate. • Minimizing the inputs of the second level gate • Most minimization methods are based on the combining theorems T10, T10’:

  33. Karnaugh Maps • A Karnaugh Map or (K-map for short) is a graphical representation of the truth table of a logic function. • The K-map for an n-input logic function is an array with 2n cells or squares, one for each input combination or minterm. • The rows and columns are labeled so that the input combination for any cell is determined from the row and column headings. • The row and columns of the map are ordered in such a way that each cell differs from an adjacent cell in only one input variable: • Thus for an n-variable K-map, each cell has n adjacent cells. • The K-map for a function is filled by putting: • a ‘1’ in the square corresponding to a minterm • a ‘0’ otherwise (maybe omitted)

  34. Y Y Y Y 0 1 0 1 X X 0 1 0 1 X X 0 0 2 2 3 3 1 1 2-Variable K-map For a 2-variable logic function F(X,Y): Truth Table: Row X Y F Minterm 0 0 0 F(0,0) X’.Y’ 1 0 1 F(0,1) X’.Y 2 1 0 F(1,0) X.Y’ 3 1 1 F(1,1) X .Y K-map Example: For the function F(X,Y) = SX,Y m(1,2,3) Truth Table: K-map Row X Y F 0 0 0 0 1 0 1 1 2 1 0 1 3 1 1 1 1 1 1

  35. YZ 3 variable Karnaugh mapTextbook convention 00 01 11 10 0 1 X’ Y Z’ X’ Y Z X’ Y Z X’ Y Z X X Y Z’ XY’ Z’ X Y’Z X Y Z

  36. Y Y YZ YZ 00 01 11 10 00 01 11 10 X X 6 6 0 0 2 2 4 4 0 1 0 1 3 3 5 5 7 7 1 1 X X Z Y 3-Variable K-map For a 3-variable logic function F(X,Y,Z): Truth Table: K-map Row X Y Z F Minterm 0 0 0 0 F(0,0,0) X’.Y’.Z’ 1 0 0 1 F(0,0,1) X’.Y’.Z 2 0 1 0 F(0,1,0) X’.Y.Z’ 3 0 1 1 F(0,1,1) X’.Y.Z 4 1 0 0 F(1,0,0) X.Y’.Z’ 5 1 0 1 F(1,0,1) X.Y’.Z 6 1 1 0 F(1,1,0) X.Y.Z’ 7 1 1 1 F(1,1,1) X.Y.Z Example: For the function F(X,Y,Z) = SX,Y,Z (1,4,6,7) K-map Row X Y Z F 0 0 0 0 0 1 0 0 1 1 2 0 1 0 0 3 0 1 1 0 4 1 0 0 1 5 1 0 1 0 6 1 1 0 1 7 1 1 1 1 1 Truth Table: 1 1 1

  37. Y YZ 00 01 11 10 WX 12 8 0 4 00 01 11 10 13 5 9 1 X 15 7 11 3 W 14 2 10 6 Z 4-Variable K-map For a 4-variable logic function F(W,X,Y,Z): Row W X Y Z F Minterm 0 0 0 0 0 F(0,0,0,0) W’.X’.Y’.Z’ 1 0 0 0 1 F(0,0,0,1) W’. X’.Y’.Z 2 0 0 1 0 F(0,0,1,0) W’. X’.Y.Z’ 3 0 0 1 1 F(0,0,1,1) W’. X’.Y.Z 4 0 1 0 0 F(0,1,0,0) W’. X.Y’.Z’ 5 0 1 0 1 F(0,1,0,1) W’.X.Y’.Z 6 0 1 1 0 F(0,1,1,0) W’.X.Y.Z’ 7 0 1 1 1 F(0,1,1,1) W’.X.Y.Z 8 1 0 0 0 F(1,0,0,0) W.X’.Y’.Z’ 9 1 0 0 1 F(1,0,0,1) W.X’.Y’.Z 10 1 0 1 0 F(1,0,1,0) W.X’.Y.Z’ 11 1 0 1 1 F(1,0,1,1) W.X’.Y.Z 12 1 1 0 0 F(1,1,0,0) W.X.Y’.Z’ 13 1 1 0 1 F(1,1,0,1) W.X.Y’.Z 14 1 1 1 0 F(1,1,1,0) W.X.Y.Z’ 15 1 1 1 1 F(1,1,1,1) W.X.Y.Z Truth Table: K-map

  38. Minimization Using K-maps • Group or combine as many adjacent 1-cells as possible: • The larger the group is, the fewer the number of literals in the resulting product term. • Grouping 2 adjacent 1-cells eliminates 1 variable, grouping 4 1-cells eliminates 2 variables, grouping 8 1-cells eliminates 3 variables, and so on. In general, grouping 2n squares eliminates n variables. • Select as few groups as possible to cover all the 1-cells (minterms) of the function: • The fewer the groups, the fewer the number of product terms in the minimized function.

  39. 000 100 010 110 Z’ • 1cell Square overlap is OK. YZ Karnaugh mapsExample 2-4 (Figure 2-14b, page 54, 3rd ed.) 00 01 11 10 1 0 1 1 X 1 1 1 XY’ 100 101 F2(X,Y,Z) = m(0,2,4,5,6) = Z’ + XY’

  40. YZ 4 variable Karnaugh mapexample 2-5 (Figure 2-19, page 57, 3rd ed.) 00 01 11 10 0000 0001 0100 0101 0010 0110 0000 0001 1100 1101 1000 1001 00 01 11 10 1 0100 1100 0110 1110 1 1 1 1 WX 1 1 1 1 1 1 F(W,X,Y,Z) = S m(0,1,2,4,5,5,8,9,12,13,14) = Y’ + W’Z’ + XZ’

  41. W’Z’ and XZ’ are prime implicants (each rectangle needs all its cells) Y’ is an essential prime implicant… (the rectangle contains exclusive (i.e., nonshared) cells YZ prime/essential implicants 00 01 11 10 00 01 11 10 1 1 1 1 1 WX 1 1 1 1 1 1 F(X,Y,Z) = m(0,1,2,4,5,5,8,9,12,13,14) = Y’ + W’Z’ + XZ’ Selection rule: minimize overlap among prime implicants

  42. YZ product of sums: Maxterms 00 01 11 10 00 01 11 10 1 1 1 0 1 1 0 WX 1 1 0 1 1 1 0 1 0 F(X,Y,Z) = m(0,1,2,4,5,5,8,9,12,13,14) = YZ + WX’Y  (Y’ + Z’) (W’ + X + Y) Selection rule: minimize overlap among prime implicants

  43. More Examples… • x’z + y’z +xyz’. x’z’ + xy’z

  44. Manipulate equations algebraically G = AC’E + AC’F + AD’E + AD’F + BCDE’F’ = A (C’E + C’F + D’E + D’F) + BCDE’F’ = A (C’ + D’)(E + F) + BCDE’F’ multilevel optimization2-6 OR Gate

  45. See figures 2-22 and 2-23 in 4th edition other gate types OR Gate

  46. Demonstrate by means of truth tables the validity of the following identity: • (XYZ)’ = X’ + Y’ + Z’ • (Q 2-1) • 2. Optimize the following Boolean function by means a three-variable map: • F(X,Y,Z) = Sm(1,3,6,7) • (Q 2-14) HW 2 OR Gate

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