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The Connection: Truth Tables to Functions

a b c F 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0. OR. The Connection: Truth Tables to Functions. Condition that a is 0, b is 0, c is 1. Function F is true if any of these and-terms are true!. Sum-of-Products form. = m 0. a b c F 0 0 0 0 0 0 1 1 0 1 0 1

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The Connection: Truth Tables to Functions

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  1. a b c F 0 0 0 00 0 1 10 1 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 OR The Connection: Truth Tables to Functions Condition that a is 0, b is 0, c is 1. Function F is true if any ofthese and-terms are true! Sum-of-Products form

  2. = m0 a b c F 0 0 0 00 0 1 10 1 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 = m1 = m2 = m3 = m4 = m5 = m6 = m7 Minterm Shorthand A minterm has one literal for each input variable, either in its normal or complemented form. Note: Binary ordering A canonical sum-of-products form of an expression consists only of minterms OR’d together

  3. Two variables: Four variables: a b minterm 0 0 a’b’ = m0 0 1 a’b = m1 1 0 a b’ = m2 1 1 a b = m3 a b c d minterm 0 0 0 0 a’b’c’d’ = m0 0 0 0 1 a’b’c’d = m1 0 0 1 0 a’b’c d’ = m2 0 0 1 1 a’b’c d = m3 0 1 0 0 a’b c’d’ = m4 0 1 0 1 a’b c’d = m5 0 1 1 0 a’b c d’ = m6 0 1 1 1 a’b c d = m7 1 0 0 0 a b’c’d’ = m8 1 0 0 1 a b’c’d = m9 1 0 1 0 a b’c d’ = m10 1 0 1 1 a b’c d = m11 1 1 0 0 a b c’d’ = m12 1 1 0 1 a b c’d = m13 1 1 1 0 a b c d’ = m14 1 1 1 1 a b c d = m15 Three variables: a b c minterm 0 0 0 a’b’c’ = m0 0 0 1 a’b’c = m1 0 1 0 a’b c’ = m2 0 1 1 a’b c = m3 1 0 0 a b’c’ = m4 1 0 1 a b’c = m5 1 1 0 a b c’ = m6 1 1 1 a b c = m7 Minterms of Different Sizes

  4. Sum-of-Products Minimization F in canonical sum-of-products form (minterm form): Use algebraic manipulation to make a simpler sum-of-products form Use commutativity to reorder to group similar terms Duplicate term - OK Use distributivity to factor out common terms Use x’+x = 1 identity We will find a better method (K-maps) later…

  5. A B C F F 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 Product-of-Sums from a Truth Table Find an expressionfor F’ (the complement) Complement both sides… Use DeMorgan’s Law to re-express as product-of sums

  6. A B 0 0 0 0 0 1 0 1 1 0 1 0 1 1 1 1 Maxterms C F F 1 0 0 1 1 0 1 0 0 0 1 1 0 0 1 0 1 1 Maxterms 0 0 1 0 1 1 • To find a Product-of-Sums form for a truth table • Make one maxterm for each row in which the function is zero • For each maxterm, each variable appears once • In its complemented form if it is one in the row • In its regular form if it is zero in the row

  7. A + B + C = M0 A + B + C = M1 A + B + C = M2 A + B + C = M3 A + B + C = M4 A + B + C = M5 A + B + C = M6 A + B + C = M7 Maxterm Shorthand Product of Sums Maxterms A B C 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 F in canonical maxterm form:

  8. Boolean operations and gates • Theorem: Any operation than can be represented by a truth table can be represented in Boolean algebra • All truth tables can be made out of only and, or, and notfunctions

  9. X X NAND/NOR expressions Any expression can be made of and ANDs, ORs and NOTs We can make ANDs and ORs from NANDs and NORs and NOTs Thus, we can make any expression out of NANDs, NORs, and NOTs We can make NOTs out of a single NAND gate So, we can make any expression out of just NANDs and NORs note: NANDs and NORs are easy to build with switches

  10. Using DeMorgan’s Law NORs can be made with NANDs! NANDs can be made out of NORs! We can make any Boolean expression out of only NOR Gates NAND-only circuits We can make any Boolean expression out of only NAND Gates

  11. Sum-of-Products Circuits with NANDs Introduce Double Inverters DeMorgan’sLaw Sum-of-Productsworks well with NANDs

  12. Product-of-Sums Circuits with NORs Introduce Double Inverters DeMorgan’sLaw Product-of-Sums works well with NORs

  13. Introduce Double Inverters to make NANDs: A B D C B A Represent inverters with NANDs C D B D Converting General Circuits to NANDs     Add inverters as needed to maintain correct polarity    

  14. a a c d f g a b c d e f b f g e c b c a c d e f g d a b d e g a b c a b c d g a b c d e f g b c f g a b c d f g Seven-Segment Example A seven-segment display is used to display numbers

  15. number A B C D a b c d e f g 0 0 0 0 0 1 1 1 1 1 1 0 1 0 0 0 1 0 1 1 0 0 0 0 2 0 0 1 0 1 1 0 1 1 0 1 3 0 0 1 1 1 1 1 1 0 0 1 4 0 1 0 0 0 1 1 0 0 1 1 5 0 1 0 1 1 0 1 1 0 1 1 6 0 1 1 0 1 0 1 1 1 1 1 7 0 1 1 1 1 1 1 0 0 0 0 8 1 0 0 0 1 1 1 1 1 1 1 9 1 0 0 1 1 1 1 1 0 1 1 10 1 0 1 0 x x x x x x x 11 1 0 1 1 x x x x x x x 12 1 1 0 0 x x x x x x x 13 1 1 0 1 x x x x x x x 14 1 1 1 0 x x x x x x x 15 1 1 1 1 x x x x x x x Invalid inputs, assume zero Seven Segment Truth Table Inputs: Four binary inputs, interpreted as a four-bit binary number Outputs: Seven outputs, for each of the seven segments segment a = A’B’C’D’ + A’B’CD’ + A’B’CD + A’BC’D + A’BCD’ + A’BCD + AB’C’D’ + AB’C’D (canonical SOP) segment a = A’C + A’BD + AB’C’ + B’C’D’ (minimal SOP) segment a = (A+B+C+D’) (A+B’+C+D) (canonical and minimal POS)

  16. a A B D C a A B A B C D C D Circuits for Segment a segment a = A’C + A’BD + AB’C’ + B’C’D’(minimal SOP) segment a = (A+B+C+D’) (A+B’+C+D) (canonical and minimal POS)

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