De Morgan's first theorem: a NOR gate is equivalent to a bubbled _____ gate. Choose the correct gate to complete the statement.

Difficulty: Easy

Correct Answer: AND

Explanation:


Introduction / Context:
De Morgan's theorems provide equivalences between logic gate implementations, enabling circuit simplification and technology mapping. One classic identity states that a NOR operation can be realized using an AND gate with input inversion (bubbles) and output inversion representation.


Given Data / Assumptions:

  • NOR is the complement of OR: A NOR B = NOT(A OR B).
  • Bubble notation indicates inversion on inputs or outputs of a logic symbol.
  • Gate 'equivalence' is functional equivalence.


Concept / Approach:

  • De Morgan 1: NOT(A OR B) = (NOT A) AND (NOT B).
  • This means a NOR gate behaves as an AND gate whose inputs are inverted (bubbles on inputs), and typically the symbol shows inversion at inputs rather than an explicit inverter block.


Step-by-Step Solution:

Start from NOR: Y = NOT(A OR B).Apply De Morgan: Y = (NOT A) AND (NOT B).Interpretation: Use an AND gate; place bubbles on both inputs to indicate NOT A and NOT B.Hence: A NOR B is equivalent to a bubbled AND gate.


Verification / Alternative check:

Truth table check for A, B ∈ {0,1} shows equality of outputs for NOR and bubbled-input AND.


Why Other Options Are Wrong:

  • XAND: Not a standard logic gate.
  • XOR: NOT(A XOR B) is XNOR, unrelated to NOR equivalence above.
  • NOR: Statement asks for gate equivalence with bubbling; the gate is AND with bubbles, not NOR itself.
  • None of the above: Incorrect because AND is correct.


Common Pitfalls:

  • Placing the bubble on the output instead of inputs for this identity.
  • Confusing NOR equivalence with NAND equivalence (which relates to OR with input bubbles).


Final Answer:

AND

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