350 Solved Electrical Engineering Problems: For the FE/PE Exams in Electrical Engineering

The following truth table defines the input-output relationship of the logic block diagram shown in Figure 7.1. Derive the corresponding minterm canonical form.
| A | B | C | D | f |
| 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 0 |
Solution:
Use a "1" to represent an uncomplemented literal and a "0" to represent a complemented literal.
f(A, B, C, D) = ? (m 0, m 2, m 5, m 7, m 9, m 11, m 12, m 14)
f(A, B, C, D) = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD
The following truth table defines the input-output relationship of the logic block diagram in Figure 7.2. Derive the corresponding maxterm canonical form.
| A | B | C | D | g |
| 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 |
| 0 | 1 |