J–K pulse-triggered flip-flop with J = K = 1: When both inputs of a J–K pulse-triggered flip-flop are HIGH and a valid clock edge occurs, what is the resulting output action?

Introduction / Context:
The hallmark of the J–K flip-flop is its toggling behavior when J and K are both asserted. This property allows efficient construction of binary counters and frequency dividers. Knowing the exact outcome under J = K = 1 at the active clock edge is core knowledge in sequential logic.

Given Data / Assumptions:

• Edge-triggered J–K flip-flop (positive or negative edge as specified by device).
• J = 1 and K = 1 are stable around the triggering edge, satisfying setup/hold.
• No asynchronous inputs are asserted.

Concept / Approach:
From the characteristic table, J = K = 1 produces Q_next = NOT(Q_current). This is toggling, meaning the output flips state every valid edge. The behavior is deterministic so long as timing constraints are respected and the inputs are not changing during the aperture window.

Step-by-Step Solution:

Verification / Alternative check:
Simulate a divide-by-2 counter stage using a J–K with J = K tied HIGH; the output frequency is half the input clock, confirming toggling at each edge.

Why Other Options Are Wrong:

• Invalid: The condition is well-defined, not invalid.
• Not change / remain unchanged: That corresponds to J = K = 0 (hold), not 1,1.
• Tri-state: Standard flip-flop outputs are driven high/low, not tri-stated.

Common Pitfalls:
Violating setup/hold times causing race-like symptoms; mistakenly assuming level-sensitive behavior; overlooking asynchronous inputs that could override the synchronous path.

Final Answer:
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