Difficulty: Easy
Correct Answer: active-HIGH inputs and an active-LOW output.
Explanation:
Introduction / Context:
A NAND (Not-AND) gate is one of the fundamental building blocks in digital electronics. It outputs the logical complement of the AND function. Understanding what “active-HIGH inputs” and “active-LOW output” mean clarifies how NAND behaves in truth tables, logic design, and real circuits.
Given Data / Assumptions:
Concept / Approach:
The AND gate produces HIGH only when all inputs are HIGH. The NAND gate inverts that result, producing LOW only when all inputs are HIGH. Thus, the inputs are interpreted in the normal active-HIGH sense, and the output is LOW (active-LOW) in the unique case that all inputs are asserted.
Step-by-Step Solution:
1) Consider an AND gate: for inputs A, B, the output is A * B.2) Invert that to obtain NAND: Y = (A * B)' = NOT(A * B).3) With active-HIGH inputs, the “all-ones” condition (A = 1 and B = 1) makes AND = 1.4) NAND then produces Y = 0 only in that condition; otherwise Y = 1.5) Therefore, inputs are active-HIGH, and the asserted condition at the output is a LOW (active-LOW) when all inputs are HIGH.
Verification / Alternative check:
Truth table: 00→1, 01→1, 10→1, 11→0 for 2-input NAND. The output is LOW exactly when both inputs are HIGH, matching the “active-HIGH inputs, active-LOW output” description.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing AND with NAND; overlooking that “active-LOW output” simply means the output’s asserted condition is LOW (the bubble at the output in logic symbols).
Final Answer:
active-HIGH inputs and an active-LOW output.
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