Minggu, 18 April 2010

TUGAS 4 SISTEM DIGITAL

HUKUM ALJABAR BOOLEAN

T1. Hukum Komutatif

a) A + B = B + A

Tabel Pembuktiannya :

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 Pembuktiannya :

A

B

A B

B A

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 Pembuktiaanya :

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 Pembuktiannya :

A

B

C

A B

B C

( A 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

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 Pembuktiannya :

A

B

C

B + C

A B

A C

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 Pembuktiannya :

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

Tabel Pembuktiannya :

A

A + A

0

0

0

0

1

1

1

1


b) A A = A

Tabel Pembuktiannya :

A

A A

0

0

0

0

1

1

1

1


T5.

a) A B + A B'

Tabel Pembuktiannya :

A

B

B’

AB

AB’

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 Pembuktiannya :

A

B

B’

A + B

AB’

( 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


T6. Hukum Redudansi

a) A + A B = A

Tabel Pembuktiannya :

A

B

A + B

A (A + B)

0

0

0

0

0

1

1

0

1

0

1

1

1

1

1

1


b) A (A + B) = A

Tabel Pembuktiannya :

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 Pembuktiannya :

A

0 + A

0

0

0

0

1

1

1

1


b) 0 A = 0

Tabel Pembuktiannya :

A

0 A

0

0

0

0

0

0

0

1

0

0

1

0

0


T8.

a) 1 + A = 1

Tabel Pembuktiannya :

A

1 + A

1

0

1

1

0

1

1

1

1

1

1

1

1


b) 1 A = A

Tabel Pembuktiannya :

A

1 A

0

0

0

0

1

1

1

1

T9

a) A’ + A = 1

Tabel Pembuktiannya :

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 Pembuktiannya :

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 Pembuktiannya :

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 Pembuktiannya :

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 Pembuktiannya :

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 Pembuktiannya :

A

B

A’

B’

A B

(A B)’

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




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))