AND gate behavior — conditional pass-through: For a 2-input AND gate, if one input is fixed HIGH (logic 1), the output value equals the other input's logic level. Classify this statement.

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
Logic gate identities are foundational. This item checks whether learners recognize that a 2-input AND with one input tied HIGH acts as a buffer (passes the other input unchanged).


Given Data / Assumptions:

  • Boolean definition: Y = A * B (AND).
  • One input is fixed at logic 1 (HIGH).
  • Standard positive logic is assumed.


Concept / Approach:
If Y = A * B and A = 1, then Y = 1 * B = B. The gate becomes a conditional pass-through for B. This is a basic Boolean identity and underlies many enable/disable designs (an AND as an enable gate).


Step-by-Step Solution:

Start with Y = A * B.Fix A = 1 (HIGH).Compute Y = 1 * B = B.Therefore, the output reflects the other input.


Verification / Alternative check:
Truth table inspection with A = 1: If B = 0 → Y = 0; if B = 1 → Y = 1. Output equals B in all cases.


Why Other Options Are Wrong:

Incorrect / negative-logic qualifiers: The identity is true in standard positive logic; NAND or negative logic changes the function.Frequency qualifier: Logical identity is independent of clock rate (assuming proper timing and no hazards).


Common Pitfalls:
Confusing AND with OR (where tying one input HIGH forces HIGH). Mixing up AND and NAND effects.


Final Answer:
Correct

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