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.
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ACorrect
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BIncorrect
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COnly true for NAND gates
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DOnly true in negative logic
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EOnly at low frequencies
Answer
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