Bubble-to-bubble connection intuition (De Morgan view): If a signal leaves a gate through an inverted (bubbled) output and enters the next gate on an inverting (bubbled) input, the two inversions cancel, producing an overall non-inverting transfer rather than a single remaining inversion. Evaluate this statement.
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AIncorrect
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BCorrect
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COnly true for CMOS, not TTL
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DDepends on fan-out
Answer
Correct Answer: Incorrect
Explanation
Introduction / Context: Logic symbols often show inversion as a small bubble. Understanding how “bubble-to-bubble” connections behave is crucial for reading schematics quickly and for using De Morgan transformations to simplify or restyle logic without changing function.
Given Data / Assumptions:
- One gate provides an inverted output (bubble on its output).
- The next gate has an inverting input (bubble on its input pin).
- We are reasoning functionally, assuming ideal logic levels.
Concept / Approach: A bubble denotes inversion. Two inversions in series cancel each other, yielding a non-inverting connection. This is a visual manifestation of De Morgan equivalences and the idea that NOT(NOT(X)) = X. Therefore, bubble-to-bubble effectively passes the signal without net inversion.
Step-by-Step Solution:
Let the original (internal) signal be S.First inversion at the producer: output = NOT(S).Second inversion at the consumer’s inverting input yields: effective input = NOT(NOT(S)) = S.Net result: no inversion overall (zero inversions remain).Verification / Alternative check: Substitute concrete logic levels: if S = 1, the producing gate output is 0; the inverting input interprets that as NOT(0) = 1 at its internal logic. If S = 0, the chain delivers 0. This confirms cancellation.
Why Other Options Are Wrong: “Correct” repeats the original claim that one inversion remains; that is the opposite of the cancellation result. “Only true for CMOS” and “Depends on fan-out” are red herrings; the cancellation is logical, not process-dependent.
Common Pitfalls: Mistaking symbol bubbles as delay markers; forgetting that each bubble is a NOT; assuming bubbles affect only thresholds instead of logic sense.
Final Answer: Incorrect