Tugas digital 3.
I. Konversikanlah : A=B7 H, B=8A H kedalam
1. Y=A.B =82 H.
1011 0111
1000 1010
1000 0010
2. Y=A+B =BF H
1011 0111
1000 1010
1011 1111
3. Y=A.B =7D H
1011 0111
1000 1010
0111 1101
4. Y=A+B =40 H
1011 0111
1000 1010
0100 0000
5. 
Y=A+ B =3D H
1011 0111
1000 1010
0011 1101
II. Buat tabel kebenaran untuk gerbang XOR 3,4,5 jelaskan kesimpulan
3 Inputan
| A | B | C | 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 | 0 |
4 Inputan
| A | B | C | D | 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 | 0 |
| 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 |
5 Inputan
| 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 | 1 | 1 | 1 | 0 | 0 | |
| 0 | 1 | 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 | 0 | |
| 1 | 0 | 1 | 0 | 0 | 1 | |
| 1 | 0 | 1 | 0 | 1 | 0 | |
| 1 | 0 | 1 | 1 | 0 | 0 | |
| 1 | 0 | 1 | 1 | 1 | 0 | |
| 1 | 0 | 0 | 0 | 0 | 0 | |
| 1 | 0 | 0 | 0 | 1 | 1 | |
| 1 | 1 | 0 | 1 | 0 | 1 | |
| 1 | 1 | 0 | 1 | 1 | 0 | |
| 1 | 1 | 1 | 0 | 0 | 0 | |
| 1 | 1 | 1 | 0 | 1 | 0 | |
| 1 | 1 | 1 | 1 | 0 | 0 | |
| 1 | 1 | 1 | 1 | 1 | 0 | |
III. Lampu jalan akan menyala , jika setiap kali switch On, atau setiap kali timer On dan hari gelap.
| Switch | Timer | Light sensor | Lampu |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
Jadi kesimpulanya adalah bila semua bernilai nol maka lampu tidak akan menyala tetapi sebaliknya bila salah satu bernilai satu maka lampu akan menyala.
