Tugas 4

Tugas 4.B

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

A * 1 = 1

A * 0 = 0

A + 0 = 0

A * A = A

A * 1 = 1

2. Give the best definition of a literal?

A Boolean variable

The complement of a Boolean variable ( Jawabannya )

1 or 2

A Boolean variable interpreted literally

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.

A + B + C

D + E

A'B'C'

D'E'

None of the above

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

x'(x + y') = x'y'

x(x'y) = xy

x*x' + y = xy

x'(xy') = x'y'

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:

Z + YZ

Z + XYZ

XZ

X + YZ

None of the above

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

F = xy

F = x + y

F = x'

F = xy + yz

F = x + y'

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

A + B

A'B'

C + D + E

C'D'E'

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'?

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

F'= ABCDE

F'= AB(C+D+E)

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

F'= (A+B)CDE

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

A

A'

1

0

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

ABCDEF

AB

AB + CD + EF

A + B + C + D + E + F

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

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+A
0	0	0
1	0	1

(b) 0 A = 0

A		0 A
0	0	0
1	0	0

T8.

(a) 1 + A = 1

A		1+A
0	1	1
1	1	1

(b) 1 A = A

A		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