Sunday, July 4, 2010
Saturday, June 19, 2010
Tugas 7
Up Down Counter
dasar dari counter ialah toggle flip-flop, yakni flip-flop yang outputnya akan berubah jika mendeteksi perubahan sinyal pada input nya (ada dua mode, high-to-low atau low-to-high)
toggle flip-flop ini ialah JK Flip-flop dengan R=S=J=K=1
dari timing diagram tersebut, perhatikan jika J=K=1 maka Q (anggap sebagai output) akan berubah nilainya ketika mendeteksi perubahan sinyal low-to-high pada CLK, atau akan membagi frekuensi CLK dngan 2.
Wednesday, June 2, 2010
Subtractor
Komplemen 2= komplemen 1+1
Komplemen 1 : invers dari bilangan asal
Contoh:
A= 8 = 1000
B= 5 = 0101
Langkah 1:
komplemen 1 =0101
komplemen 1 =0101
Wednesday, April 28, 2010
Tugas 5
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.
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.
Sunday, April 18, 2010
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)
(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
3. 1 or 2
1. A + B + C
2. D + E
3. A'B'C'
4. D'E'
4. Which of the following relationships represents the dual of the Boolean property x + x'y = x + y?
2. 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
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'?
2. F'= ABCDE
5. F'= (A+B)CDE
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
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