Bubble logic equivalence An OR gate symbol with inversion bubbles on its inputs is functionally equivalent to which standard gate (with normal, non-inverted inputs)?

Difficulty: Easy

Correct Answer: NAND

Explanation:


Introduction / Context:
Bubble (inversion) notation on logic symbols is a compact way to show complemented signals. Being able to translate bubble logic into standard Boolean expressions allows quick recognition of gate equivalences without redrawing entire schematics.


Given Data / Assumptions:

  • An OR gate has inversion bubbles at each input (but not at the output).
  • Inputs are the original variables A and B (or more), assumed to be active-high.
  • Positive logic conventions apply.


Concept / Approach:

Bubbles on inputs mean those inputs are inverted before entering the gate. For two inputs, the Boolean function is Y = (A’ + B’). Apply De Morgan's theorem to rewrite this as Y = (A · B)’ which is exactly the NAND function. Thus, an OR with complemented inputs is equivalent to a NAND with non-inverted inputs.


Step-by-Step Solution:

Write the bubble-form expression: Y = (A’ + B’).Apply De Morgan: (A’ + B’) = (A · B)’.Recognize: (A · B)’ is a NAND. Therefore, the symbol is functionally a NAND gate.


Verification / Alternative check:

Construct truth tables for Y1 = (A’ + B’) and Y2 = (A · B)’ and compare row by row; they match exactly. Logic simulators will also show identical waveforms for the same inputs, confirming the equivalence.


Why Other Options Are Wrong:

  • NOR corresponds to an OR with a bubble at the output, not at the inputs.
  • OR without bubbles is simply A + B, not (A’ + B’).
  • NOT is a single-input operator.
  • AND does not match the complemented form after applying De Morgan.


Common Pitfalls:

  • Confusing input bubbles with an output bubble. Input bubbles invert inputs; an output bubble inverts the final result.


Final Answer:

NAND

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