Decoder behavior — number of active outputs: For standard binary-to-one-of-N decoders, only one output line should be active for any valid input code (one-hot). Assess the claim: “It is normal for more than one decoder output to be active at the same time.”

Introduction / Context:Decoders map an n-bit input into 2^n mutually exclusive outputs. Proper operation requires exactly one active output at a time for valid codes, enabling clean chip selects, address decoding, and resource enabling in digital systems. This question checks your understanding of the “one-hot” nature of decoders.

Given Data / Assumptions:

• Binary input is valid and stable within setup/hold windows.
• Outputs are active-low or active-high but mutually exclusive.
• No enable pin or glitching is considered during transitions.

Concept / Approach:For a device like the 74xx138 (3-to-8 decoder), each input combination selects one and only one output (with enable asserted). Multiple active outputs would break address decoding, causing bus contention and undefined system behavior. While hazards or transition glitches can momentarily assert multiple outputs if inputs change asynchronously, this is not “normal” design intent; careful timing or synchronized inputs prevent that.

Step-by-Step Solution:Examine truth table: for each input code, exactly one output is asserted.Enable disabled → all outputs inactive (still not multiple active).Glitch-free design ensures exclusivity under synchronous changes.Therefore, the statement is false for correct decoder use.

Verification / Alternative check:Datasheets specify “one-of-eight” or “one-of-sixteen” behavior; timing diagrams emphasize hazard-free changes when inputs are synchronized.

Why Other Options Are Wrong:“Correct” contradicts the definition. Gray code and priority encoders are unrelated: Gray code minimizes bit changes; priority encoders perform the inverse function and still produce one code, not multiple active decoder outputs.

Common Pitfalls:Ignoring enables; changing multiple inputs asynchronously leading to temporary glitches; misinterpreting active-low outputs as multiple actives.

Final Answer:Incorrect