Senin, 19 April 2010

tugas 4

Tugas 4.A

Hukum Aljabar Boolean & Tabel Kebenarannya

T1. Hukum Komutatif

A) 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

B) 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

T2. Hukum Asosiatif

A) (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

B) (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

T3. Hukum Distributif

A) A (B + C) = A B + A



A

B

C

B+C

A(B+C)

A B

AB+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

B) A + (B C) = (A + B) (A + C)



A

B

C

B C

A+(BC)

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

T4. Hukum Identity

A) A + A = A



A

A+A

0

0

1

1


B) A A = A



A

AA

0

0

1

1

T5.

A) AB + AB’ = 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


B) (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

T6. Hukum Redudansi

A) A + A B = A

A

B

A B

A+AB

0

0

0

0

0

1

0

0

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

0+A

0

0

0

1

0

1

B) 0 A = 0

A

0

0A

0

0

0

1

0

0

T8.

A) 1 + A = 1

A

1

1+A

0

1

1

1

1

1

B) 1 A = A

A

1

1 A

0

1

0

1

1

1

T9.

A) A’ + A = 1

A

A’

A’+A

0

1

1

1

0

1

B) A’ A = 0

A

A’

A’ A

0

1

0

1

0

0

T10.

A) 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

B) A ( A’ + B) = A B

A

B

A’

A’+B

A(A’+B)

AB

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

A) ( A + B)’ = A’ B’

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

B) ( A B )’ = A’ + B’

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

1. Give the relationship that represents the dual of the Boolean property A + 1 = 1?
(Note: * = AND, + = OR and ' = NOT)

2. A * 0 = 0 ( Jawabannya )

2. Give the best definition of a literal?

2. The complement of a Boolean variable ( Jawabannya )

3. Simplify the Boolean expression (A+B+C)(D+E)' + (A+B+C)(D+E) and choose the best answer.

1. A + B + C ( Jawabannya )

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

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

5. Given the function F(X,Y,Z) = XZ + Z(X'+ XY), the equivalent most simplified Boolean representation for F is:

2. Z + XYZ ( Jawabannya )

6. Which of the following Boolean functions is algebraically complete?

1. F = xy ( Jawabannya )

7. Simplification of the Boolean expression (A + B)'(C + D + E)' + (A + B)' yields which of the following results?

2. A'B' ( Jawabannya )

8. Given that F = A'B'+ C'+ D'+ E', which of the following represent the only correct expression for F'?

5. F'= (A+B)CDE ( Jawabannya )

9. An equivalent representation for the Boolean expression A' + 1 is

3. 1 ( Jawabannya )

10. Simplification of the Boolean expression AB + ABC + ABCD + ABCDE + ABCDEF yields which of the following results?

2. AB ( Jawabannya )

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