Decoder selection with active-LOW outputs: A 4-line-to-16-line decoder (inputs weighted 1, 2, 4, 8) has active-LOW outputs. If the input is 1110, the claim is that output line 7 is driven LOW. Decide if this is correct.

Introduction / Context:
Decoders convert binary inputs to one-of-N outputs. With active-LOW outputs, exactly one output line is asserted LOW for any valid input. Correctly interpreting bit weights and active-level conventions is essential to determine which line is selected.

Given Data / Assumptions:

• 4-to-16 decoder with inputs weighted 1, 2, 4, 8 (LSB to MSB).
• Active-LOW outputs: selected line goes LOW; others remain HIGH.
• Input presented as 1110 (bit order 8 4 2 1 → 1 1 1 0).

Concept / Approach:
Translate the input code to its decimal equivalent using the given weights. The selected output index equals the binary input value. Active-LOW simply flips the asserted level, not the selected index.

Step-by-Step Solution:

Verification / Alternative check:
Cross-check with truth tables for 74xx154-style decoders (dual-enable aside). The index equals the binary input value regardless of output polarity.

Why Other Options Are Wrong:
“Correct” wrongly maps 1110 to 7. Output polarity does not change selected index, only logic level. Enable pins, while necessary, do not remap bit weights in normal operation.

Common Pitfalls:
Reversing bit order, misreading LSB/MSB, or confusing active-LOW assertion with index inversion.

Final Answer:
Incorrect