tugas 4

nama : DERI FITRAH SARI
no.bp : 0901082051

1. pembuktian mengenai hukum-hukum dengan menggunakan tabel kebenaran... yang mana hukum tersebut adalah..:
hukum aljabar bolean
1.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

2.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 A.B B.C (A.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 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
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)
A B C A+B A+C B.C A+(B.C) (A+B)(A+C)
0 0 0 0 0 0 0 0
0 0 1 0 1 0 0 0
0 1 0 1 0 0 0 0
0 1 1 1 1 1 1 1
1 0 0 1 1 0 1 1
1 0 1 1 1 0 1 1
1 1 0 1 1 0 1 1
1 1 1 1 1 1 1 1

4.Hukum Identity
(a) A + A = A
A A A+A <=> A
0 0 0 0
1 1 1 1

(b) A A = A
A A A.A <=> A
0 0 0 0
1 1 1 1

5.
(a) A B b.inv A.B A.Binv <=> A
0 0 1 0 0 0 0
0 1 0 0 0 0 0
1 0 1 0 1 1 1
1 1 0 1 0 1 1

(b)
A B B(inv) A+B A+B(inv) <=> A
0 0 1 0 1 0 0
0 1 0 1 0 0 0
1 0 1 1 1 1 1
1 1 0 1 1 1 1

6.Hukum Redudansi
(a) A + A B = A
A B A.B A+A.B <=> A
0 0 0 0 0
0 1 0 0 0
1 0 0 1 1
1 1 1 1 1

(b)A (A + B) = A
A B A+B A(A+B) <=> A
0 0 0 0 0
0 1 1 0 0
1 0 1 1 1
1 1 1 1 1

7.
(a) 0 + A = A
A 0 A+0 <=> A
1 0 1 1
0 0 0 0

(b) 0 A = 0
A 0 A.0 <=> 0
1 0 0 0
0 0 0 0

8.
(a) 1 + A = 1
A 1 A+1 <=> 1
0 1 1 1
1 1 1 1

(b)1 A = A
A 1 1.A <=> A
0 1 0 0
1 1 1 1

9.
(a)
A A(inv) 1 <=> 1
1 0 1 1
0 1 1 1

(b)
A A(inv) <=> 0
0 1 0 0
1 0 0 0

10.
(a)
A B A(inv)A(inv)B A+A(inv) B A+B
0 0 0 0 0
0 1 1 1 1
1 0 1 0 1
1 1 1 0 1

(b)
A B A(inv) A+B A.B A(A(inv)+B)
0 0 1 1 0 1 1 0 1 1 1 0
0 1 1 0 1 0 0 0 1 1 1 0
1 0 0 1 1 0 0 0 1 1 0 0
1 1 0 0 1 0 0 1 0 0 1 1

11.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

tugas 3

Tabel kebenaran gerbang XOR dengan 3 input

INPUT				OUTPUT
A		B	C		Q
0		0	0		0
0		0	1		1
0		1	0		1
0		1	1		1
1	0		0		1
1	0		1		1
1	1		1		0

Tabel kebenaran gerbang XOR dengan 4 input

INPUT				OUTPUT
A	B	C	D	Q
0	0	0	0	0
0	0	0	1	1
0	0	1	0	1
0	0	1	1	1
0	1	0	0	1
0	1	0	1	1
0	1	1	0	1
0	1	1	1	1
1	0	0	0	1
1	0	0	1	1
1	0	1	0	1
1	0	1	1	1
1	1	0	0	1
1	1	0	1	1
1	1	1	0	1
1	1	1	1	0

Tabel kebenaran XOR dengan 5 input

INPUT					OUTPUT
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	1
0	0	1	0	0	1
0	0	1	0	1	1
0	0	1	1	0	1
0	0	1	1	1	1
0	1	0	0	0	1
0	1	0	0	1	1
0	1	0	1	0	1
0	1	0	1	1	1
0	1	1	0	0	1
0	1	1	0	1	1
0	1	1	1	0	1
0	1	1	1	1	1
1	0	0	0	0	1
1	0	0	0	1	1
1	0	0	1	0	1
1	0	0	1	1	1
1	0	1	0	0	1
1	0	1	0	1	1
1	0	1	1	0	1
1	0	1	1	1	1
1	1	0	0	0	1
1	1	0	0	1	1
1	1	0	1	0	1
1	1	0	1	1	1
1	1	1	0	0	1
1	1	1	0	1	1
1	1	1	1	0	1
1	1	1	1	1	0

kesimpulan ;
grerbang Xor mengeluarkan input satu bila input satunya ganjil..
input bernilai nol bila input yang dim....asuk kan berbeda

3b.. lampu lalu lintas....