Selasa, 20 April 2010
TUGAS 4 A HUKUM KOMUTATIF 1. A + B = B + A A B A+B B+A 0 0 0 0 0 1 1 1 1 0 1 1 1 1 1 1 2. A . B = B . A A B A B B A 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 1. (A + B) + C = A + (B + C) A B C A+B (A+B)+C B+C A+(B+C) 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 2. (A B) C = A (B C) A B C A B (A . B) C B C A (B . C) 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 1. A B C B+C A (B+C) A . B A . B + A 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 1 1 1 0 0 0 1 0 1 1 1 0 1 1 1 1 1 1 1 1 2. A B C B . C A+ (B . C) A+B A+C (A+B)(A+C) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 T5. 1. A . B + A . B' = A A B B' A B A B' AB+AB' 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 1 1 1 1 0 1 0 1 2. (A+B) (A+B') = A A B B' A+B A+B' (A+B)(A+B') 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 1 1 1 1 1 0 1 1 1 1. A B A . B A+AB 0 0 0 0 0 1 0 0 1 0 0 1 1 1 1 1 2. A B A+B A(A+B) 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 T7. 1. 0 + A = A A 0+A 0 0 0 1 0 1 2. 0 . A = 0 A 0 0 0 1 0 0 T8. 1. 1 + A = 1 A 1+A 0 1 1 1 1 1 2. 1 . A = A A 0 1 0 1 1 1 T9. 1. A' + A = 1 A A' A'+A 0 1 1 1 0 1 2. A' A = 0 A A' A' A 0 1 0 1 0 0 T10. 1. A + A' B = A + B A B A' A' B A+A'B A+B 0 0 1 0 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 1 0 0 1 1 2. A ( A' + B) = A B A B 'A' A'+B A(A'+B) A . B 0 0 1 1 0 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 0 1 1 1 T11.THEOREMA De MORGAN'S 1. A B A' B' (A+B)' A' . B' 0 0 1 1 1 1 0 1 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 2. A B A' B' (A . B)' A'+B' 0 0 1 1 1 1 0 1 1 0 1 1 1 0 0 1 1 1 1 1 0 0 0 0 TUGAS 4 B The complement of a Boolean variable Which of the following relationships represents the dual of the Boolean property x + x'y = x + y?
A (B + C) = A . B + A
A + (B . C) = (A + B) (A + C)
A + A . B = A
A (A + B) = A
( A + B)' = A' B'
( A B )' = A' + B'
(Note: * = AND, + = OR and ' = NOT)
0 komentar:
Posting Komentar