Clocked S–R device behavior — both inputs LOW For a clocked S–R flip-flop, if both S and R inputs are LOW when the active clock edge arrives, what happens to the output state?

Difficulty: Easy

Correct Answer: No change will occur in the output.

Explanation:


Introduction / Context:
Understanding the truth table of clocked S–R storage elements is crucial for designing synchronous systems. The S–R function generalizes the idea of setting and resetting a latch, while the clock edge determines when sampling occurs in synchronous variants.


Given Data / Assumptions:

  • Clocked (edge-controlled) S–R flip-flop.
  • Active edge occurs while S = 0 and R = 0.
  • No asynchronous inputs (CLR/PR) are active.


Concept / Approach:
In the canonical S–R truth table, S = 0 and R = 0 at the sampling moment cause the device to hold its previous state (no change). Only S = 1 forces a set, and R = 1 forces a reset. The forbidden or indeterminate input for a level-sensitive SR latch is S = R = 1; in clocked variants, this is either disallowed or mapped to a defined behavior depending on implementation, but it is not the present case.


Step-by-Step Reasoning:

At the active clock edge, inputs are S = 0, R = 0.S/R both LOW ⇒ neither set nor reset is requested.Therefore, Qnext = Qprev (hold condition).


Verification / Alternative check:
Vendor timing diagrams show flat Q behavior across the sampling edge when S and R are both 0. Simulation of a behavioral SR model will also confirm that Q remains unchanged for this input pair at the clock event.


Why Other Options Are Wrong:

  • Invalid state: associated with S = R = 1 in basic SR latches, not with both inputs LOW.
  • Toggle: SR devices toggle only in special JK designs, not with S = R = 0.
  • Reset / Set: would require R = 1 or S = 1 respectively.


Common Pitfalls:

  • Confusing SR operation with JK behavior, where J = K = 1 causes toggling.


Final Answer:
No change will occur in the output.

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