74148 (8-to-3 priority encoder): When data input I4 is the asserted (active) input on a 74148 priority encoder, the data outputs are stated as A0 = 1, A1 = 1, A2 = 0. Decide if this statement is correct (assume standard active-low inputs/outputs).

Introduction / Context:
The 74148 priority encoder outputs a 3-bit code (active-low) that represents the highest-priority active-low input. Correctly interpreting the active-low convention is key to matching expected outputs with datasheet truth tables.

Given Data / Assumptions:

• Input I4 is asserted (low), with higher-priority inputs I5–I7 inactive (high).
• Outputs A2, A1, A0 are active-low.
• Bit order: A2 = MSB, A0 = LSB.

Concept / Approach:
Decimal index 4 corresponds to binary 100. Because the outputs are active-low, the output code is the bitwise inversion of 100, namely 011. In terms of A2, A1, A0, that is A2 = 0, A1 = 1, A0 = 1. The statement given—A0 = 1, A1 = 1, A2 = 0—matches this pattern and is therefore correct.

Step-by-Step Solution:

Verification / Alternative check:
Refer to 74148 truth tables: for I4 asserted (low) with no higher-priority active inputs, the outputs read 0,1,1 (active-low) and the group signal pins behave accordingly.

Why Other Options Are Wrong:

• Incorrect: Conflicts with the active-low mapping.
• Only correct if active-high: Not required because outputs are natively active-low and already match.
• Ambiguous: Enable/GS pins do not alter the fundamental code here.

Common Pitfalls:
Misreading the MSB/LSB order; forgetting inversion due to active-low outputs.

Final Answer:
Correct