Tugas 5

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Full Adder 

Rangkaian Full-Adder, pada prinsipnya bekerja seperti Half-Adder, tetapi mampu menampung bilangan Carry dari hasil penjumlahan sebelumnya. Jadi jumlah inputnya ada 3: A, B dan Ci, sementara bagian output ada 2: S dan Co. Ci ini dipakai untuk menampung bit Carry dari penjumlahan sebelumnya.

penjumlahan bilangan-bilangan biner hanya dengan penjumlahan bilangan desimal dimana hasil penjumlahan tersebut terbagi menjadi 2 bagian,yang SUMMARY(SUM) dan CARRY out(CARRY),apabila hasil penjumlahan pada suatu tingkat atau kolom melebihi nilai maksimumnya maka output CARRY akan berada pada keadaan logika 1.

perhatikan contoh berikut.

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A. Hukum Aljabar Boolean

1.Hukum komutatif

          A+B = B+A

          AB   = BA

A	B	A+B	B+A	AB	BA
0	0	0	0	0	0
0	1	1	1	0	0
1	0	1	1	0	0
1	1	1	1	1	1

2. Hukum Asosiatif

          (A+B)+C = A+(B+C)

          (AB)C      = A(BC)

A	B	C	A+B	B+C	AB	BC	(A+B)+C	A+(B+C)	(AB)C	A(BC)
0	0	0	0	0	0	0	0	0	0	0
0	0	1	0	1	0	0	1	1	0	0
0	1	0	1	1	0	0	1	1	0	0
0	1	1	1	1	0	1	1	1	0	0
1	0	0	1	0	0	0	1	1	0	0
1	0	1	1	1	0	0	1	1	0	0
1	1	0	1	1	1	0	1	1	0	0
1	1	1	1	1	1	1	1	1	1	1

3. Hukum Distributif

          A(B+C) = AB+AC

          A+(BC) = (A+B)(A+C)

A	B	C	B+C	A(B+C)	AB	AC	AB+AC	BC	A+(BC)	A+B	A+C	(A+B)(A+C)
0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	1	1	0	0	0	0	0	0	0	1	0
0	1	0	1	0	0	0	0	0	0	1	0	0
0	1	1	1	0	0	0	0	1	1	1	1	1
1	0	0	0	0	0	0	0	0	1	1	1	1
1	0	1	1	1	0	1	1	0	1	1	1	1
1	1	0	1	1	1	0	1	0	1	1	1	1
1	1	1	1	1	1	1	1	1	1	1	1	1

4. Hukum Identity

          A+A = A

          AA = A

A	A+A	AA
0	0	0
1	1	1

5.

          AB+AB’ = A

          (A+B)(A+B’) = A

A	B	B’	AB	AB’	AB+AB’	A+B	A+B’	(A+B)(A+B’)
0	0	1	0	0	0	0	1	0
0	1	0	0	0	0	1	0	0
1	0	1	0	1	1	1	1	1
1	1	0	1	0	1	1	1	1

6. Hukum Redudansi

          A+AB = A

          A (A + B) = A

A	B	AB	A+AB	A+B	A(A+B)
0	0	0	0	0	0
0	1	0	0	1	0
1	0	0	1	1	1
1	1	1	1	1	1

7.

          0+A = A

          0A = 0

A	0+A	0A
0	0	0
1	1	0

8.

          1+A = 1

          1A = A

A	1+A	1A
0	1	0
1	1	1

9.

          A’+A = 1

          A’A = 0

A	A’	A’+A	A’A
0	1	1	0
1	0	1	0

10.

          A+A’B = A+B

          A(A’+B) = AB

A	B	A’B	A+A’B	A+B	A’	A’+B	A(A’+B)	AB
0	0	0	0	0	1	1	0	0
0	1	1	1	1	1	1	0	0
1	0	0	1	1	0	0	0	0
1	1	0	1	1	0	1	1	1

11. TheoremaDe Morgan's

          (A+B)’ = A’B’

          (AB)’ = A’+B.

A	B	A+B	(A+B)’	A’B’	AB	(AB)’	A’+B’
0	0	0	1	1	0	1	1
0	1	1	0	0	0	1	1
1	0	1	0	0	0	1	1
1	1	1	0	0	1	0	0

B. 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))