Jumat, 16 April 2010

II. Soal :



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 (Jawabanya)

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 (Jawabannya)

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 (Jawabannya)

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’ (Jawabannya)

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 (Jawabannya)

3. XZ

4. X + YZ

5. None of the above



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

1. F = xy (Jawabannya)

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’ (Jawabannya)

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 (Jawabannya)



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

1. A

2. A’

3. 1 (Jawabannya)

4. 0



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

1. ABCDEF

2.AB (Jawabannya)

3.AB +CD +EF

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

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

tugas 4A sistem logika

Hukum Aljabar Boolean

Hukum Aljabar Boolean

T1. Hukum Komutatif

(a) A + B = B + A

Pembuktian:

A

B

A+B

B+A

A+B = B+A

0

0

0

0

Ö

0

1

1

1

Ö

1

0

1

1

Ö

1

1

1

1

Ö

(b) A B = B A

Pembuktian:

A

B

AB

BA

AB = BA

0

0

0

0

Ö

0

1

0

0

Ö

1

0

0

0

Ö

1

1

1

1

Ö

T2. Hukum Asosiatif

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

Pembuktian:

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

Ö

0

0

1

0

1

1

1

Ö

0

1

0

1

1

1

1

Ö

0

1

1

1

1

1

1

Ö

1

0

0

1

0

1

1

Ö

1

0

1

1

1

1

1

Ö

1

1

0

1

1

1

1

Ö

1

1

1

1

1

1

1

Ö

(b) (A B) C = A (B C)

Pembuktian:

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

Ö

0

0

1

0

0

0

0

Ö

0

1

0

0

0

0

0

Ö

0

1

1

0

1

0

0

Ö

1

0

0

0

0

0

0

Ö

1

0

1

0

0

0

0

Ö

1

1

0

0

0

0

0

Ö

1

1

1

1

1

1

1

Ö

T3. Hukum Distributif

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

Pembuktian:

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)

Pembuktian:

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

Pembuktian:

A

A + A

A + A = A

0

0

Ö

0

0

Ö

1

1

Ö

1

1

Ö


(b) A A = A

Pembuktian:

A

A A

A A = A

0

0

Ö

0

0

Ö

1

1

Ö

1

1

Ö

T5.

(a)

Pembuktian:

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)

Pembuktian:

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

Pembuktian:

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

Pembuktian:

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

Pembuktian:

A

0 + A

0

0

0

0

1

1

1

1

(b) 0 A = 0

Pembuktian:

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

Pembuktian:

A

1 A

0

0

0

0

1

1

1

1

T9

(a)

Pembuktian:

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

T10

(a)

Pembuktian:

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)

Pembuktian:

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

0

0