Two-input AND gate behavior – if one input is HIGH (logic 1), does the output mirror the other input?

Introduction / Context:
Understanding how an AND gate behaves under partial assertion is essential for gating and enabling signals. This question asks whether forcing one input HIGH causes the output to equal the remaining input, a useful identity in practical designs.

Given Data / Assumptions:

• Standard 2-input AND gate with ideal Boolean behavior.
• Active-HIGH logic convention: 1 = asserted, 0 = deasserted.
• No hazards or metastability considered for the logical statement.

Concept / Approach:
The AND operation is multiplicative in Boolean algebra: Y = A * B. If A = 1, then Y = 1 * B = B, so the output mirrors B. This is the enabling property of AND: a HIGH input acts like an enable that passes the other signal through unchanged (subject to normal gate delays and thresholds).

Step-by-Step Solution:

Verification / Alternative check:
Truth table rows confirm the identity: (1,0) → 0 and (1,1) → 1, so when A = 1 the output follows B exactly. This is why AND is used for clock enables, conditional data paths, and masking logic where a control line gates another signal.

Why Other Options Are Wrong:
Technology (CMOS vs TTL) and propagation delay do not change the logical identity; tying inputs together is irrelevant to the stated condition.

Common Pitfalls:
Confusing AND enabling with OR enabling; for OR, tying one input HIGH forces the output HIGH regardless of the other input. With AND, HIGH enables and LOW disables.

Final Answer:
Correct