tugas 4

Hukum Aljabar Boolean

1. Hukum Komutatif

(a) A + B = B + A

Tabel Kebenaran:

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

Tabel Kebenaran:

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

2. Hukum Asosiatif

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

Tabel Kebenaran:

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)

Tabel Kebenaran:

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

3. Hukum Distributif

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

Tabel Kebenaran:

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)

Tabel Kebenaran:

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

4. Hukum Identity

(a) A + A = A

Tabel Kebenaran:

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

(b) A A = A

Tabel Kebenaran:

A	A A
0	0
0	0
1	1
1	1

5.

(a) AB + A B’

Tabel Kebenaran:

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’)

Tabel Kebenaran:

A	B	B'	A+B	A+B'
0	0	1	0	1
0	1	0	1	0
1	0	1	1	1
1	1	0	1	1

6. Hukum Redudansi

(a) A + A B = A

Tabel Kebenaran:

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

Tabel Kebenaran:

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

7

(a) 0 + A = A

Tabel Kebenaran:

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

(b) 0 A = 0

Tabel Kebenaran:

A	0 A	0
0	0	0
0	0	0
1	0	0
1	0	0

8

(a) 1 + A = 1

Tabel Kebenaran:

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

(b) 1 A = A

Tabel Kebenaran:

A	1 A
0	0
0	0
1	1
1	1

9

(a) A’ + A = 1

Tabel Kebenaran:

A	A'	A'	1
0	1	1	1
0	1	1	1
1	0	1	1
1	0	1	1

(b) A’ A=0

Tabel Kebenaran:

A	A'	A'A	0
0	1	0	0
0	1	0	0
1	0	0	0
1	0	0	0

10

(a) A + A’ B =A + B

Tabel Kebenaran:

A	B	A'	A' B	A+B	A+A' 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) A (A’ + B) = AB

Tabel Kebenaran:

A	B	A'	A'+B	A B	A(A'+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

11. TheoremaDe Morgan's

(a) (A’+B’)= A’B’

Tabel Kebenaran:

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’

Tabel Kebenaran:

A	B	A'	B'	A B	(AB)'	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

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))

tugas 3

soal
Untuk 2 Input
Output akan berlogika 0 jika Kedua Input sama dan Output berlogika 1 jika Kedua Input Berbeda

Untuk Input >2 …………………………?
Buat tabel kebenaran untuk gerbang XOR 3 , 4 dan 5 input, jelaskan kesimpulan anda

tabel kebenaran 3 input


tabel kebenaran 4 input


tabel kebenaran 5 input


Kesimpulan
Gerbang XOR memiliki sifat, Output akan berlogika 1 jika Input 1 dalam jumlah Ganjil


soal
-Lampu jalan akan menyala jika
-Setiap kali switch On, atau setiapkali Timer On dan Hari gelap


tabel kebenaran

tugas 2c

tugas 2a:

penemu bilangan desimal:Nama lengkapnya adalah Abu Ja’far Muhammad bin Musa al-Khawarizmi. Lahir di Khawarizmi, Uzbeikistan, pada 194 H/780 M. Kepandaian dan kecerdasannya mengantarkannya masuk ke lingkungan Dar al-Hukama (Rumah Kebijaksanaan), sebuah lembaga penelitian dan pengembangan ilmu pengetahuan yang didirikan oleh Ma’mun Ar-Rasyid, seorang khalifah Abbasiyah yang terkenal.(http://citlafhdel.blogspot.com/2010/04/penemu-angka-no-dan-bilangan-desimal.html).

tugas 2b:

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