Minggu, 06 Juni 2010

Tugas 4

Hukum Aljabar Boolean

T1. Hukum Komutatif
(a) A + B = B + A
Tabel Kebenaran:
A
B
A + B
B + A
0
0
0
0
0
1
1
1
1
0
1
1
1
1
1
1
(b) A B = B A
Tabel Kebenaran:
A
B
AB
BA
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1
T2. Hukum Asosiatif
(a) (A + B) + C = A + (B + C)
Tabel Kebenaran:
A
B
C
A + B
B + C
(A+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
0
1
0
1
1
1
0
1
1
1
1
1
1
1
0
1
1
1
1
1
1
1
1
1
1
1
(b) (A B) C = A (B C)
Tabel Kebenaran:
A
B
C
AB
BC
(AB)C
A(BC)
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
1
0
0
1
0
0
0
0
0
0
1
0
1
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
1
1
1
T3. Hukum Distributif
(a) A (B + C) = A B + A C
Tabel Kebenaran:
A
B
C
B +C
AB
AC
A(B+C)
(AB)+(AC)
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)
Tabel Kebenaran:
A
B
C
BC
A+B
A+C
A+(BC)
(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
Tabel Kebenaran:
A
A + A
0
0
0
0
1
1
1
1
(b) A A = A
Tabel Kebenaran:
A
A A
0
0
0
0
1
1
1
1
T5.
(a) AB + A B’
Tabel Kebenaran:
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



(b) (A+B)(A+B’)
Tabel Kebenaran:
A
B
B'
A+B
A+B'
0
0
1
0
1
0
1
0
1
0
1
0
1
1
1
1
1
0
1
1
T6. Hukum Redudansi
(a) A + A B = A
Tabel Kebenaran:
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
Tabel Kebenaran:
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
Tabel Kebenaran:
A
0 + A
0
0
0
0
1
1
1
1
(b) 0 A = 0
Tabel Kebenaran:
A
0 A
0
0
0
0
0
0
0
1
0
0
1
0
0
T8
(a) 1 + A = 1
Tabel Kebenaran:
A
1 + A
1
0
1
1
0
1
1
1
1
1
1
1
1



(b) 1 A = A
Tabel Kebenaran:
A
1 A
0
0
0
0
1
1
1
1
T9
(a) A’ + A = 1
Tabel Kebenaran:
A
A'
A'
1
0
1
1
1
0
1
1
1
1
0
1
1
1
0
1
1



(b) A’ A=0
Tabel Kebenaran:
A
A'
A'A
0
0
1
0
0
0
1
0
0
1
0
0
0
1
0
0
0
T10
(a) A + A’ B =A + B
Tabel Kebenaran:
A
B
A'
A' B
A+B
A+A' 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 (A’ + B) = AB
Tabel Kebenaran:
A
B
A'
A'+B
A B
A(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. TheoremaDe Morgan's
(a) (A’+B’)= A’B’
Tabel Kebenaran:
A
B
A'
B'
A+B
(A+B)'
A' B'
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’ + B’
Tabel Kebenaran:
A
B
A'
B'
A B
(AB)'
A'+B'
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
0
0





Quiz Aljabar Boolean




1. Give the relationship that represents the dual of the Boolean property A + 1 = 1?

(Note: * = AND, + = OR and ' = NOT)

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?

1. A Boolean variable

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'

5. None of the above



4. Which of the following relationships represents the dual of the Boolean property x + x'y = x + y?

1. x'(x + y') = x'y'

2. x(x'y) = xy

3. x*x' + y = xy

4. x'(xy') = x'y'


5. 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

5. None of the above



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'?

1. F'= A+B+C+D+E

2. F'= ABCDE

3. F'= AB(C+D+E)

4. F'= AB+C'+D'+E'

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

4. A + B + C + D + E + F

5. A + B(C+D(E+F))


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