The main contents of the chapter consist of the following: Expressing boolean functions; relationships between algebraic equations, symbols, and truth tables; simplification of boolean expressions; minterms and maxterms; AND-OR representations.
Lecture More Boolean Algebra A B Overvie w ° Expressing Boolean functions ° Relationships between algebraic equations, symbols, and truth tables ° Simplification of Boolean expressions ° Minterms and Maxterms ° AND-OR representations • Product of sums • Sum of products Axioms and Graphical representation of DeMorgan's Law 10A) X Y Y X 10B) X 11A) X YZ 11B) X 12A) XY 12B) X Y W 13A) X XY X Y 13B) X XY X Y 13C) X XY X Y 13D) X XY X Y 14A) XY 14B) X Y Y X XY Z Y Z Z X Y Commutative Law X Y XY Z Y X Y Z Associative Law XZ XW XZ YW Consensus Theorem YZ Distributiv e Law Simplification Using the Laws Boolean Functions ° Boolean algebra deals with binary variables and logic operations ° Function results in binary or x 0 0 1 1 y 0 1 0 1 z 1 1 F 0 0 1 x y z z’ y+z’ F = x(y+z’) F = x(y+z’) Boolean Functions ° Boolean algebra deals with binary variables and logic operations ° Function results in binary or x 0 0 1 1 y 0 1 0 1 z 1 1 xy 0 0 0 1 yz 0 0 G 0 0 1 x xy y G = xy +yz z yz We will learn how to transition between equation, symbols, and truth table Representation Conversion ° Need to transition between boolean expression, truth table, and circuit (symbols) ° Converting between truth table and expression is easy ° Converting between expression and circuit is easy ° More difficult to convert to truth table Boolean Expression Circuit Truth Table Truth Table to Expression ° Converting a truth table to an expression • Each row with output of becomes a product term • Sum product terms together x 0 0 1 1 y 0 1 0 1 z 1 1 G 0 0 1 Any Boolean Expression can be represented in sum of products form! xyz + xyz’ + x’yz ° Complementing Functions Step 1: assign temporary names • b + c -> z G = a + b+ c G’ = (a + b + c)’ • (a + z)’ = G’ ° Step 2: Use DeMorgans’ Law • (a + z)’ = a’ z’ ° Step 3: Resubstitute (b+c) for z • a’ z’ = a’ (b + c)’ ° Step 4: Use DeMorgans’ Law • a’ (b + c)’ = a’ (b’ c’) ° Step 5: Associative rule • a’ (b’ c’) = a’ b’ c’ G = a + b+ c G’ = a’ b’ c’ = a’b’c’ Complementation Example ° Find complement of F = x’z + yz • F’ = (x’z + yz)’ ° DeMorgan’s • F’ = (x’z)’ (yz)’ ° DeMorgan’s • F’ = (x’’+z’)(y’+z’) ° Reduction -> eliminate double negation on x • F’ = (x+z’)(y’+z’) This format is called product of sums Conversion Between Canonical Forms ° Easy to convert between minterm and maxterm representations ° For maxterm representation, select rows with 0’s x 0 0 1 1 y 0 1 0 1 z 1 1 G 0 0 1 G = xyz + xyz’ + x’yz G = m7 + m6 + m3 = Σ(3, 6, 7) G = M0M1M2M4M5 = Π(0,1,2,4,5) G = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)(x’+y+z’) Representation of Circuits ° All logic expressions can be represented in 2level format ° Circuits can be reduced to minimal 2-level representation ° Sum of products representation most common in industry Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Example 1: AB + A(B+C) +B(B+C) = Solution: AB + A(B+C) +B(B+C) = AB+AB+AC+BB+BC = AB+AB+AC+B+BC = AB+AC+B+BC = AB+AC+B = B+AC Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Gate Network for Example 1: AB+A(B+C)+B(B+C) A AB B A B A(B+C) B+C AB+A(B+C)+B(B+C) C B B(B+C) Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Gate Network for Example 1: B+AC B A C AC B+AC Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Gate Network for Example 1: AB+A(B+C)+B(B+C) ≡ B+AC A AB B B A A(B+C) B B+C C B ≡ AB+A(B+C)+B(B+C) B+AC A C AC B(B+C) (a) (b) Equivalent Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Example 2: Using Boolean algebra techniques, simplify the following expression: [AB(C + BD) + AB]C Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Example 2: Solution: [ A B(C BD ) A B]C ( A BC A BBD A B)C ( A BC A.0.D A B )C A B )C ( A BC ( A BC A B )C Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Example 2: Solution: ( A BC A B )C A BCC A BC A BC A BC BC ( A A) BC.1 BC Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Example 3: Using Boolean algebra techniques, simplify the following expression: ABC + ABC + ABC + ABC + ABC ABC + ABC + ABC + ABC + ABC + ABC ABC + ABC + ABC + ABC + ABC + ABC BC(A + A) + BC(A + A) + AC(B + B) BC + BC + AC C(B + A) + BC Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Example 4: Using Boolean algebra techniques, simplify the following expression: AB AC ABC AB AC + A BC ( A + B).( A + C ) + A BC A A + AC + A B + BC + A BC A + AC + A B + BC + A BC A + A BC + A B + BC A(1 + BC ) + A B + BC A + A B + BC A(1 + B) + BC A + BC A + BC Boolean Algebra and Logic Simplification Example 4: Simplify the following Boolean functions T1 and T2 to a minimum number of literals: T1 = ∑ (0,1, 2) = ∏ (3, 4, 5, 6, 7) T2 = ∑ (3, 4, 5, 6, 7) = ∏ (0,1, 2) Summary ° Truth table, circuit, and boolean expression formats are equivalent ° Easy to translate truth table to SOP and POS representation ° Boolean algebra rules can be used to reduce circuit size while maintaining function ° All logic functions can be made from AND, OR, and NOT ° Easiest way to understand: Do examples! ° Next time: More logic gates! ... Equivalent Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Example 2: Using Boolean algebra techniques, simplify the following expression: [AB(C + BD) + AB]C Boolean Algebra. .. Using Boolean Algebra Example 2: Solution: ( A BC A B )C A BCC A BC A BC A BC BC ( A A) BC.1 BC Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Example 3: Using Boolean. .. Boolean Algebra and Logic Simplification Simplification Using Boolean Algebra Gate Network for Example 1: AB+A(B+C)+B(B+C) A AB B A B A(B+C) B+C AB+A(B+C)+B(B+C) C B B(B+C) Boolean Algebra and Logic