Clocked S–R flip-flop response with S = 0 and R = 0 If both synchronous inputs of an S–R flip-flop are LOW, what happens at the active clock edge?

Introduction / Context:
Clocked (synchronous) S–R flip-flops sample inputs only at a clock edge, unlike level-sensitive latches. Knowing the action associated with each input combination at the sampling instant is essential for designing state machines and avoiding forbidden operations.

Given Data / Assumptions:

• Synchronous S and R inputs are both LOW (S = 0, R = 0).
• A valid clock edge occurs (positive or negative depending on device).
• Asynchronous inputs (if any) are inactive.

Concept / Approach:

For a clocked S–R flip-flop, the behavior at the active edge mirrors the basic latch logic: S = 1, R = 0 → set; S = 0, R = 1 → reset; S = 0, R = 0 → no change; S = 1, R = 1 → invalid. With both inputs LOW, the flip-flop simply holds its prior state through the clock event.

Step-by-Step Solution:

Verification / Alternative check:

Consult a standard S–R FF characteristic table: the entry for S = R = 0 at the active clock indicates Q_next = Q_previous (no change).

Why Other Options Are Wrong:

Invalid: occurs only when S and R are both asserted (1,1) at the sampling moment.

Toggle: that is characteristic of a J–K FF with J = K = 1, not of an S–R FF.

Reset: would require R = 1 at the sampling instant.

Common Pitfalls:

Confusing the clocked S–R FF with a level latch; assuming that a clock edge forces a change even when inputs request hold; mixing S–R behavior with J–K toggling rules.

Final Answer:

No change will occur in the output.