lect3 - CSE261 Spring 2008 Boolean (`switching') algebra (`...

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Unformatted text preview: CSE261 Spring 2008 Boolean (`switching') algebra (` switching' Combinational Logic Design Principles Boolean Boolean values: 0, 1 values: Positive-logic convention: analog voltages LOW, HIGH Positiveconvention: Signal values: denoted by variables (X, Y, FRED, etc.) (X FRED, values: Complement: X (opposite of X) Complement: AND: X Y AND: OR: OR: X+Y 0, 1 (Switching) Algebra Circuit Analysis Circuit Synthesis Combinational Combinational Literal: a variable or its complement: X, X, FRED Literal: Expression: literals combined by AND, OR, parentheses Expression: (A B C + Q5) RESET Equation: Variable = expression Equation: P = ((FRED Z) + (A B C + Q5) RESET 1 2 Basic Axioms A variable can take only one of two values (0,1) (0,1) (A1) Theorems - Single Variable Identity elements: Null elements: Idempotency: Involution: Complements: (T1) (T2) (T3) (T4) (T5) X+0=X X+1=1 X+X=X (X) = X (X X + X = 1 (T1') (T1' (T2') (T2' (T3') (T3' X1=X X0=0 XX=X X = 0 if X 1 (A1') (A1' X = 1 if X 0 NOT operation (The complement Operation): (A2) If X = 0 then X 1 (A2') (A2' If X = 1 then X 0 (T5') (T5' X X = 0 AND and OR operations (Multiplication and Addition): (A3) 0 0 = 0 (A3') 0 + 0 = 0 (A3' (A4) 1 1 = 1 (A4') 1 + 1 = 1 (A4' (A5) 0 1 = 1 0 = 0 (A5') 1 + 0 = 0 + 1 = 1 (A5' 3 Induction Proof: Show that the theorems are true for both X = 0 and X = 1 4 Combinational Logic Principles 1 CSE261 Spring 2008 Theorems - Multiple Variables Commutativity: Commutativity: (T6) Theorems - Multiple Variables Distributivity: Distributivity: X (Y + Z) = X Y + X Z (T8) (T8') X + Y Z = (X + Y) (X + Z) (T8' X+Y=Y+X (T6') (T6' XY=YX The inputs of AND and OR gates can be interchanged. Associativity: (T7) (X + Y) + Z = X +(Y + Z) (T7'') (X Y) Z = X (Y Z) (T7 Associativity: Multiplication distributes over addition; Addition distributes over over multiplication! Y X Z Y X The order of the input variables could be rearranged. Z X Y Z X Y Z Y X X Y Z Z Y X 5 X Y Z 6 Z Theorems - Multiple Variables Covering: Proof: (T9): (T9): (T9) Theorems - Multiple Variables Combining: Proof: (T10) X+XY=X (T9') (T9' X (X + Y) = X X Y + X Y = X (T10') (T10' (X + Y) (X + Y) = X X+XY=X1+XY = X (1 + Y) =X1 =X X (X + Y) = (X + 0) (X + Y) = X + (0 Y) =X+0 =X (theorem T1') T1' (theorem T8- Distributivity) T8- Distributivity) (theorem T2) (theorem T1') T1' (theorem T1) (theorem T8'- Distributivity) T8' Distributivity) (theorem T2') T2' (theorem T1) (T10) X=X1 = X (Y + Y) = X Y + X Y X=X+0 = X + (Y Y) =(X + Y) (X + Y) (theorem T1') T1' (theorem T5) (theorem T8 - Distributivity) Distributivity) (theorem T1) (theorem T5') T5' (theorem T8' - Distributivity) T8' Distributivity) (T10') (T10' (T9'): (T9' Covering and combining are used in minimizing logic functions. 7 8 Combinational Logic Principles 2 CSE261 Spring 2008 Theorems - Multiple Variables Consensus: (T11) (T11') (T11' (T13) Theorems - Multiple Variables DeMorgans Theorems (X1 X2 ... Xn) = X1 + X2 + ... + Xn Xn) (X1 + X2 + ... + Xn) = X1 X2 ... Xn Xn) X Y + X Z + Y Z = X Y + X Z (X + Y) (X + Z) (Y + Z) = (X + Y) (X + Z) (T13') (T13' Example: two-variable case Example: twoX1 MO ST IM PO RT AN T! ! ! Generalized Idempotency: Idempotency: (T12) (T12') (T12' X+X+...+X=X XX...X=X (X1 X2) X2 X1 X2 (X1 + X2) X1 X2 (X1 + X2) X1 X2 (X1 X2) 9 10 DeMorgan Symbol Equivalence DeMorgan Symbols 11 12 Combinational Logic Principles 3 CSE261 Spring 2008 Generalized DeMorgan's Theorem DeMorgan' (T14) Duality Swap 0 & 1, AND & OR Result: [ F (X1, X2, ... , Xn, + , ) ] = F (X1, X2, ... , Xn, , +) Xn, ] (X1 X2 Xn Example: F = (X1 X2) + (X2 + X3) X1 Theorems still true Why? Each axiom (A1-A5) has a dual (A1'-A5') (A1(A1' A5' F = [(X1 X2) + (X2 + X3)] Counterexample: X + X Y = X (T9) X X + Y = X (dual) Mathematical definition : F is a Boolean Function; FD the dual function is: FD (X1,X2, ... ,Xn, +, , ) = F (X1,X2, ... ,Xn, , +, ) Xn, Xn, X2 X3 X1 F = (X1 + X2) (X2 X3) (X1 X2 (X2 X3 X2 X3 13 14 Duality Example: FX1,X2,X3 = X1 + X2 X3 FDX1,X2,X3 = X1 (X2 + X3) FDX1,X2,X3 = (X1 (X2 + X3)) = X1 + X2 X3 FD X3)) X1 X2 X3 = FX1, X2, X3 X1 X2 X3 FDX1,X2,X3 = X1' (X2' + X3') X1' (X2' X3' X1 ,X2 ,X3 FDX1,X2,X3 = (X1 (X2 + X3) = X1 + X2 X3 FD X1 ,X2 ,X3 (X1 (X2 X3 X1 X2 X3 = X1 + X2 X3 X1 X2 Duality and DeMorgan's Theorem DeMorgan' or: or: FX1,X2, ... ,Xn = FDX1, X2, ... ,Xn FDX1 X2 Xn FX1,X2, ... ,Xn = FDX1, X2, ... ,Xn X1 X2 Xn X1 F X2 X3 X2 X3 FDX1',X2',X3' FD X1 X2 X3 16 Example: X1 F FX1,X2,X3 = FDX1,X2,X3 FDX1 ,X2 ,X3 15 X3 Combinational Logic Principles 4 CSE261 Spring 2008 Exercise: DeMorgan/Duality DeMorgan/Duality Find: F, FD, FDA,B,C, FDA,B,C F = [AB + ABC + BC] = (A + B)(A + B + C)(B + C) FA,B,C = AB + ABC + BC FD = (A + B)(A + B + C)(B + C) FDA,B,C = (A + B)(A + B + C)(B + C) FDA,B,C = [(A + B)(A + B + C)(B + C)] C)] = AB + ABC + BC 17 Representation of Logic Functions Truth table with 2n rows, n: the number of variables Literal: a variable or its complement Literal: X, Y n-variable minterm: product term with n literals minterm: X Y Z n-variable maxterm: sum term with n literals maxterm: X + Y + Z 18 Canonical Representation: Canonical Sum (Sum of Products - SOP): Sum Canonical Sum Implementation F = X Y Z + X Y Z + X Y Z XYZ of minterms corresponding to input combinations for which the function produces a 1 output. FX,Y,Z = (1,3,6) 1,3 F = X Y Z + X Y Z + X Y Z X Y Z XYZ F XYZ Canonical Product (Product of Sums - POS): Product of maxterms corresponding to input combinations for which the function produces a 0 output. FX,Y,Z = (0,2,4,5,7) 0,2 F = (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X (X (X 19 20 Combinational Logic Principles 5 CSE261 Spring 2008 Canonical Product Implementation F = (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X + Y + Z) (X (X (X (X + Y + Z) X Y Z (X + Y + Z) (X (X + Y + Z) (X F Logic Function Simplification F = X Y Z + X Y Z + X Y Z = X (Y Z + Y Z) + X Y Z = X ((Y + Y) Z) + X Y Z = X (1 Z) + X Y Z = X Z + X Y Z X Z (X + Y + Z) X Y (X + Y + Z) (X Z 21 X Y Z F 22 Exercise: F(A,B,C) = (0,2,4,7); Write: The Truth Table; The Canonical Sum (SOP); The Canonical Product (POS) Row 0 1 2 3 4 5 6 7 Combinational Analysis X 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 F Minterms Maxterms SOP: POS: F = A B C + A B C + A B C + A B C F = (A + B + C) (A + B + C) (A + B + C) (A + B + C) 23 Logic Circuit Logic Function Truth Table 24 Combinational Logic Principles 6 CSE261 Spring 2008 Example 1: Logic Circuit A B C (A + C) (A B) Example 1: Logic Function (A B) + C F Truth Table Logic Function Simplification F = (A + C) ((A B) + C) F = (A + C) ((A B) + C) = A (A B + C) + C (A B + C) =AAB+AC+CAB+CC =AB +AC+ABC+C = (A B + A B C) + (A C + C) = A B (1 + C) + ((A + 1) C) =AB+C A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 AB 0 0 0 0 0 0 1 1 AB+C 0 1 0 1 0 1 1 1 A+C 0 1 0 1 1 1 1 1 F 0 1 0 1 0 1 1 1 25 Same function, different circuit A B C AB F=AB+C 26 Example 2: A B Example 2: (A + B) B) F A B (A + B) B) F Using Demorgan's Demorgan' theorem: F = [(A + B) (B C)] B) (B C) C (B C) C A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 F 0 0 1 1 1 1 1 1 A B (A + B) B) F (B C) FA,B,C= (2,3,4,5,6,7) C OR FA,B,C = (0,1) A B (A + B) F = A + B + B C' (B C) C 27 28 Combinational Logic Principles 7 CSE261 Spring 2008 Example 2: Example 3: F = [(A + B)' (B C')']' B) = (A + B)'' + (B C')'' B) = A + B + B C' = A + B (1 + C') =A+B1 =A+B A B F using Demorgan's theorem Demorgan' (Multiply out) F = ((X + Y) Z) + (X Y Z) ((X (X F = (X Z) + (Y Z) + (X Y Z) (X 29 30 Example 3: Example 3: (Add out) New circuit, same function F = (X Z) + (Y Z) + (X Y Z) (Y (X F = ((X + Y) Z) + (X Y Z) ((X = ((X + Y + X) (X + Y + Y) (X + Y + Z) (Z + X) (Z + Y) (Z + Z) ((X = 1 1 (X + Y + Z) (X + Z) (Y + Z) (X 1 = (X + Y + Z) (X + Z) (Y + Z) (X 31 32 Combinational Logic Principles 8 CSE261 Spring 2008 Shortcut: Symbol substitution Different circuit, same function 33 34 Example 4: GW,X,Y,Z = W X Y + Y Z 35 Combinational Logic Principles 9 ...
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This note was uploaded on 05/08/2008 for the course CSE 261 taught by Professor Ercanli during the Spring '08 term at Syracuse.

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