Table 3.5 shows the truth table for a “4-to-2 priority encoder.’’
(a) Write the complete truth table for S1 and S0 (all 16 combinations of D3, D2, D1, and D0).
(b) Draw the corresponding Karnaugh map.
(c) Obtain simplified Boolean expressions for them.
The Correct Answer and Explanation is :
(a) Truth Table for S1 and S0 (4-to-2 Priority Encoder):
The priority encoder converts a set of four binary inputs (D3, D2, D1, D0) to two binary outputs (S1, S0), representing the binary value of the highest-priority input. The priority order is D3 > D2 > D1 > D0. This means if multiple inputs are high, only the highest-priority input determines the outputs.
Below is the complete truth table for S1 and S0, considering all 16 combinations of D3, D2, D1, and D0.
| D3 | D2 | D1 | D0 | S1 | S0 |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 |
(b) Karnaugh Map:
We now map the outputs S1 and S0 to a Karnaugh map. To simplify the expressions, we’ll create separate Karnaugh maps for S1 and S0.
For S1:
| D3 \ D2 | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 0 | 0 | 0 | 0 |
| 01 | 0 | 0 | 0 | 1 |
| 11 | 1 | 1 | 1 | 1 |
| 10 | 1 | 1 | 1 | 1 |
For S0:
| D3 \ D2 | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 0 | 0 | 0 | 0 |
| 01 | 0 | 0 | 0 | 0 |
| 11 | 1 | 1 | 1 | 1 |
| 10 | 1 | 1 | 1 | 1 |
(c) Simplified Boolean Expressions:
We can now obtain the simplified Boolean expressions for S1 and S0 by observing the Karnaugh maps.
For S1:
From the Karnaugh map for S1, we see that it is 1 for all cases where D3 is 1, regardless of D2, D1, and D0. Therefore, the simplified Boolean expression for S1 is:
[
S1 = D3
]
For S0:
From the Karnaugh map for S0, we see that it is 1 when both D3 and D2 are 1, regardless of D1 and D0. Therefore, the simplified Boolean expression for S0 is:
[
S0 = D3 \cdot D2
]
Explanation:
- Truth Table: This table shows all possible input combinations for the 4-bit inputs (D3, D2, D1, D0), and the corresponding outputs (S1, S0) based on the priority encoding logic. The highest-priority input determines the output.
- Karnaugh Map: The Karnaugh map simplifies the Boolean expressions by visually grouping cells with the same output (1’s in this case). The grouping helps in finding the minimal expressions for the outputs.
- Simplified Boolean Expressions: By examining the Karnaugh map, we find that:
- S1 is simply the value of D3, as it is 1 when D3 is high, regardless of the other inputs.
- S0 is 1 when both D3 and D2 are high, so the Boolean expression is (S0 = D3 \cdot D2).
These simplified expressions allow us to construct efficient logic circuits for the priority encoder.