Tugas 4B

Pembuktian Hukum Aljabar Boolean

T1. Hukum Komutatif

(b) (A+B) (A+B’) =A

(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	AB	BA
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	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)

A	B	C	AB	BC	(AB)C	A(BC)
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

A	B	C	B +C	AB	AC	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)

A	B	C	BC	A+B	A+C	A+(BC)	(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

A	A	A+A
0 1	0 1	0 1

(b) A A = A

A	A	A A
0 1	0 1	0 1

T5.

(a) AB + AB’ = A

A	B	B’	AB	AB’	AB+AB’
0 0 1 1	0 1 0 1	1 0 1 0	0 0 0 1	0 0 1 0	0 0 1 1

(b) (A+B) (A+B')= A

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

T6. Hukum Redudansi

(a)A + A B = A

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

(b) A (A + B) = A

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

T7

(a) 0 + A = A

0	A	0+A
0 0	0 1	0 1

(b) 0 A = 0

0	A	0 A
0 0	0 1	0 0

T8

(a) 1 + A = 1

1	A	1+A
1 1	0 1	1 1

(b) 1 A = A

1	A	1 A
1 1	0 1	0 1

T9

(a)A’ + A = I

A	A’	I	A’+A
0 1	1 0	1 1	1 1

(b) A’ A = 0

A	A’	0	AA’
0 1	1 0	0 0	0 0

T10

(a ) A+ A’B = A+B

A	B	A’	A’B	A+B	A+A’B
0 0 1 1	0 1 0 1	1 1 0 0	0 1 0 0	0 1 1 1	0 1 1 1

(b) A (A’+B) = A B

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

T11. TheoremaDe Morgan's

(a) (A’+B’)= A’B’

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’

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

Tugas 4

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
Answer : A * 0 = 0

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
Answer : The complement 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
Answer : A + B + C

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
Answer : x'(x + y') = x'y'

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
Answer : Z + XYZ

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'
Answer : F = xy

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'
Answer : A'B'

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
Answer : F'= (A+B)CDE

9.An equivalent representation for the Boolean expression A' + 1 is
1. A
2. A'
3. 1
4. 0
Answer : 1

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))
Answer : AB

Tugas 3

Membuat tabel kebenaran gerbang XOR dengan 3,4,5 input

Tabel kebenaran untuk gerbang XOR dengan 3 input

A B C Q
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1

Tabel kebenaran untuk gerbang XOR dengan 4 input

A B C D Q
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 0
0 1 1 1 1
1 0 0 0 1
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0


Tabel kebenaran untuk gerbang XOR dengan 5 input

A B C D E Q
0 0 0 0 0 0
0 0 0 0 1 1
0 0 0 1 0 1
0 0 0 1 1 0
0 0 1 0 0 1
0 0 1 0 1 0
0 0 1 1 0 0
0 0 1 1 1 1
0 1 0 0 0 1
0 1 0 0 1 0
0 1 0 1 0 0
0 1 0 1 1 1
0 1 1 0 0 0
0 1 1 0 1 0
0 1 1 1 0 1
0 1 1 1 1 0
1 0 0 0 0 1
1 0 0 0 1 0
1 0 0 1 0 0
1 0 0 1 1 1
1 0 1 0 0 0
1 0 1 0 1 1
1 0 1 1 0 1
1 0 1 1 1 0
1 1 0 0 0 0
1 1 0 0 1 1
1 1 0 1 0 1
1 1 0 1 1 0
1 1 1 0 0 1
1 1 1 0 1 0
1 1 1 1 0 0
1 1 1 1 1 1

Kesimpulan yang didapat:
1. Bahwa dalam gerbang XOR outputnya akan 0 apabila masukkan nya (input) sama
2. Dan akan menghasilkan output 1 apabila masukkannya (input)berbeda

Lampu jalan akan menyala jika setiap kali switch ON, atau setiap kali TIMER ON dan Hari Gelap

Tugas 2

1. Tugas 2A

Desimal	Biner	Hexadesimal	BCD
125	1111101	7D	000100100101
59	111011	3B	01011001
111	01101111	6F	000100010001
89	1011001	59	10001001
169	10101001	A9	000101101001
215	11010111	D7	001000010101
972	1111001100	3CC	100101110010
856	1101011000	358	100001010110

2.Tugas 2B

Penemu desimal adalah Al Khawarizmi.

3.Tugas 2C
Kode ASCII merupakan suatu standar internasional dalam kode huruf dan simbol seperti Hex dan Unicode tetapi ASCII lebih bersifat universal. biasanya kode ASCII selalu digunakan oleh komputer dan alat komunikasi lain.

Nama	Biner	Hexa
H	1001000	48
E	1000101	45
R	1010010	52
L	1001100	4C
I	1001001	49
N	1001110	4E
Space	0100000	20
R	1010010	52
A	1000001	41
H	1001000	48
M	1001101	4D
I	1001001	49
L	1001100	4C