Exclusive-OR behavior (2-input XOR): Assess the statement: “An exclusive-OR provides a LOW output if one input or the other input is HIGH.”

Difficulty: Easy

Correct Answer: Incorrect

Explanation:


Introduction / Context:
The exclusive-OR (XOR) gate is fundamental in digital electronics for parity, addition (sum bit), and data comparison. Its defining property is that the output is HIGH when the inputs are different and LOW when they are the same. The given statement claims the output is LOW when exactly one input is HIGH, which contradicts the XOR truth table.



Given Data / Assumptions:

  • XOR has two inputs A, B and output X.
  • Definition: X = 1 when A ≠ B; X = 0 when A = B.
  • Standard positive-logic symbols are assumed.


Concept / Approach:
Check the XOR truth table for the two cases where exactly one input is HIGH: (A, B) = (1, 0) and (0, 1). In both cases, inputs differ, so XOR must output HIGH. Therefore, the statement that it outputs LOW is incorrect.



Step-by-Step Solution:

Case 1: A = 1, B = 0 → A ≠ B → XOR = 1 (HIGH).Case 2: A = 0, B = 1 → A ≠ B → XOR = 1 (HIGH).Same-input cases (0,0) or (1,1) give LOW. The statement is the opposite of correct behavior.


Verification / Alternative check:
Use the Boolean form: X = ĀB + AB̄ (read as (NOT A)*B + A*(NOT B)). When exactly one input is 1, one of the two product terms equals 1, so X = 1.



Why Other Options Are Wrong:

Correct: Contradicts the XOR definition.Only true for active-LOW XOR symbols: Symbol polarity does not invert XOR function, it changes annotation for active levels.Not enough information: Standard XOR behavior is fully defined.


Common Pitfalls:
Confusing XOR with XNOR or with OR; forgetting XOR outputs HIGH when inputs differ.


Final Answer:
Incorrect

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