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 |
Gerbang Logika XOR
Gerbang Logika XOR
Gerbang logika ini hanya akan mengeluarkan hasil output bernilai 1 jika hanya salah satu input saja yang bernilai 1. Maksudnya jika kedua input bernilai 1, maka hasil output-nya tetaplah 0.
KESIMPULAN : Jika Semua Input bernilai SAMA maka Output logika bernilai 0, Output akan benilai 1 jika input yang dimasukkan BERBEDA.output akn belogika 1 jika input berlogika ganjil,dan output akn berlogika 0 jika input berlogika genap.
Tugas 3.B
Lampu jalan akan menyala jika à Setiap kali switch On, atau setiapkali Timer On dan Hari gelap
Tabel Kebenaran :
A | B | C | C’ | Q1 = C’ . B | Q2 = A +( C’ . 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’ = Komplemen C (logika NOT)
Q1 = Output logika AND dari C’ dan B
Q 2 = Output logika Q1 + A
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