4a. Quiz Aljabar Boolean 1. Give the relationship that represents the dual of the Boolean property A + 1 = 1? 1. A * 1 = 1 2. A * 0 = 0 3. A + 0 = 0 4. A * A = A 5. A * 1 = 1 2. Give the best definition of a literal? 2. The complement of a Boolean variable 3. 1 or 2 4. A Boolean variable interpreted literally 5. The actual understanding of a Boolean variable 3. Simplify the Boolean expression (A+B+C)(D+E)' + (A+B+C)(D+E) and choose the best answer. 1. A + B + C 2. D + E 3. A'B'C' 4. D'E' 2. x(x'y) = xy 5. Given the function F(X,Y,Z) = XZ + Z(X'+ XY), the equivalent most simplified Boolean representation for F is: 1. Z + YZ 2. Z + XYZ 3. XZ 4. X + YZ 6. Which of the following Boolean functions is algebraically complete? 1. F = xy 2. F = x + y 3. F = x' 4. F = xy + yz 5. F = x + y' 7. Simplification of the Boolean expression (A + B)'(C + D + E)' + (A + B)' yields which of the following results? 1. A + B 2. A'B' 3. C + D + E 4. C'D'E' 5. A'B'C'D'E' 8. Given that F = A'B'+ C'+ D'+ E', which of the following represent the only correct expression for F'? 2. F'= ABCDE 5. F'= (A+B)CDE 9. An equivalent representation for the Boolean expression A' + 1 is 1. A 2. A' 3. 1 4. 0 10. Simplification of the Boolean expression AB + ABC + ABCD + ABCDE + ABCDEF yields which of the following results? 1. ABCDEF 2. AB 3. AB + CD + EF |
T1. Hukum Komutatif
(a) A + B = B + A
A | B | A+B | B+A | A+B = B+A |
0 | 0 | 0 | 0 | benar |
0 | 1 | 1 | 1 | benar |
1 | 0 | 1 | 1 | benar |
1 | 1 | 1 | 1 | benar |
A | B | AB | BA | AB = BA |
0 | 0 | 0 | 0 | Benar |
0 | 1 | 0 | 0 | Benar |
1 | 0 | 0 | 0 | Benar |
1 | 1 | 1 | 1 | benar |
(a) (A + B) + C = A + (B + C)
A | B | C | A+B | B+C | (A+B)+C | A+(B+C) | (A+B)+C=A+(B+C) | ||||
0 | 0 | 0 | 0 | 0 | 0 | 0 | Benar | ||||
0 | 0 | 1 | 0 | 1 | 1 | 1 | Benar | ||||
0 | 1 | 0 | 1 | 1 | 1 | 1 | Benar | ||||
0 | 1 | 1 | 1 | 1 | 1 | 1 | Benar | ||||
1 | 0 | 0 | 1 | 0 | 1 | 1 | Benar | ||||
1 | 0 | 1 | 1 | 1 | 1 | 1 | Benar | ||||
1 | 1 | 0 | 1 | 1 | 1 | 1 | Benar | ||||
1 | 1 | 1 | 1 | 1 | 1 | 1 | Benar | ||||
(b) (A B) C = A (B C
A | B | C | A B | B C | (A B) C | A (B C) | (A B) C = A (B C) |
0 | 0 | 0 | 0 | 0 | 0 | 0 | Benar |
0 | 0 | 1 | 0 | 0 | 0 | 0 | Benar |
0 | 1 | 0 | 0 | 0 | 0 | 0 | Benar |
0 | 1 | 1 | 0 | 1 | 0 | 0 | Benar |
1 | 0 | 0 | 0 | 0 | 0 | 0 | Benar |
1 | 0 | 1 | 0 | 0 | 0 | 0 | Benar |
1 | 1 | 0 | 0 | 0 | 0 | 0 | Benar |
1 | 1 | 1 | 1 | 1 | 1 | 1 | Benar |
(a) A (B + C) = A B + A C
A | B | C | B+C | A B | A C | A (B+C) | A B+A C |
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 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
(b) A + (B C) = (A + B) (A + C)
A | B | C | B C | A+B | A+C | A+(B C) | (A+B)(A+C) |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
T4. Hukum Identity
(a) A + A = A
A | A + A | A + A = A |
0 | 0 | Benar |
0 | 0 | Benar |
1 | 1 | Benar |
1 | 1 | Benar |
A | A A | A A = A |
0 | 0 | Benar |
0 | 0 | Benar |
1 | 1 | Benar |
1 | 1 | Benar |
T5.
(a
A | B | B(invers) | A B | A B(invers) | |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 |
(b)
A | B | B(invers) | A+B | A+B(invers) | |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 1 |
T6. Hukum Redudansi
(a) A + A B = A
A | B | A B | A + A B |
0 | 0 | 0 | 0 |
0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 |
1 | 1 | 1 | 1 |
(b) A (A + B) = A
A | B | A + B | A (A + B) |
0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 |
T7
(a) 0 + A = A
A | 0 + A |
0 | 0 |
0 | 0 |
1 | 1 |
1 | 1 |
(b) 0 A = 0
A | 0 A | 0 |
0 | 0 | 0 |
0 | 0 | 0 |
1 | 0 | 0 |
1 | 0 | 0 |
T8
(a) 1 + A = 1
A | 1 + A | 1 |
0 | 1 | 1 |
0 | 1 | 1 |
1 | 1 | 1 |
1 | 1 | 1 |
(b) 1 A = A
A | 1 A |
0 | 0 |
0 | 0 |
1 | 1 |
1 | 1 |
T9
(a)
A | A(invers) | 1 | |
0 | 1 | 1 | 1 |
0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 |
1 | 0 | 1 | 1 |
(b)
A | A(invers) | 0 | |
0 | 1 | 0 | 0 |
0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 |
(a)
A | B | A(invers) | A(invers) B | A+B | A+A(invers) B |
0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 | 1 |
(b)
A | B | A(invers) | A(invers)+B | A B | A(A(invers)+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. TheoremaDe Morgan's
(a)
A | B | A(invers) | B(invers) | A+B | (A+B)invers | A(invers) B(invers) |
0 | 0 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 0 | 0 |
(b)
A | B | A(invers) | B(invers) | A B | (AB)invers | A(invers)+B(invers) |
0 | 0 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 1 |