Gerbang XOR disebut juga gerbang EXCLUSIVE OR dikarenakan hanya
mengenali sinyal yang memiliki bit 1 (tinggi) dalam jumlah ganjil untuk
menghasilkan sinyal keluaran bernilai tinggi (1).
Tugas 3.A
1. Buat tabel kebenaran untuk gerbang XOR 3 , 4 dan 5 input, jelaskan kesimpulan anda
1. Tabel Kebenaran untuk gerbang XOR 3
Input A | Input B | Input C | Output 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 |
2. Tabel Kebenaran untuk gerbang XOR 4
Input A | Input B | Input C | Input D | Output 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 |
3. Tabel Kebenaran untuk gerbang XOR 5
Input A | Input B | Input C | Input D | Input E | Output 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 | 1 |
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 |
Tugas 3 B
Lampu jalan akan menyala jika
Switch dalm keadaan On,atau Timer dalam keadaan On dan Hari gelap,sepert gambar :
Tabel Kebenaran :
A | B | C | C(not) | D = C(not) . B | E = A +( C(not) . B) |
0 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 1 |
Ket :
A = Input Switch
B = Input Timer
C = Input light sensor
C(not) = Komplemen C (logika NOT)
D = Output logika AND dari C(not) dan B
E = Output logika D + A
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