Rise time vs. propagation delay in digital waveforms: The interval for a square wave to move from 10% to 90% of its final level is called rise time, not propagation delay. Evaluate the correctness of calling it “propagation delay.”

Difficulty: Easy

Correct Answer: Incorrect

Explanation:


Introduction / Context:
Two frequently cited timing parameters are rise time (10% to 90% transition at the same node) and propagation delay (time from an input transition to the corresponding 50% output crossing). Mixing these definitions leads to errors when budgeting timing and edges in digital systems.



Given Data / Assumptions:

  • Rise time (tr): time for a waveform to go from 10% to 90% of its final value at a given node.
  • Fall time (tf): time for a waveform to go from 90% to 10%.
  • Propagation delay (tpd): delay from input event to output response, typically measured between 50% points (or specified points) at input and output.


Concept / Approach:
The question’s statement labels a 10%–90% interval as “propagation delay,” which is a definition error. That interval is rise time (or fall time). Propagation delay depends on both device internal delays and load effects and is measured between different nodes (input to output), not within a single edge at one node.



Step-by-Step Solution:

Identify the metric in the statement: 10%–90% transition.Map to correct term: rise time (or fall time for a falling edge).Contrast with propagation delay: input-to-output timing, commonly 50% crossings.Conclude the statement is incorrect.


Verification / Alternative check:
Timing diagrams and datasheets consistently separate tr/tf (edge rates) from tpd (input-to-output delay). Simulation tools and oscilloscopes use the same conventions.



Why Other Options Are Wrong:

Correct: Conflicts with standard timing definitions.Correct only for TTL outputs or high fan-out: Device family or loading does not change definitions.


Common Pitfalls:
Using propagation delay as a synonym for “how fast the edge looks”; ignoring that slow edges can still have short input-to-output delay, and vice versa.


Final Answer:
Incorrect

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