J–K flip-flop behavior: When both J and K inputs are HIGH (1) during the active clock, what happens to the output state Q of a J–K flip-flop?

Introduction / Context:
A J–K flip-flop is a versatile synchronous bistable device widely used in counters, frequency dividers, and memory elements. Understanding its response to different input combinations is essential for reliable digital design and timing analysis.

Given Data / Assumptions:

• Synchronous operation with a clock edge controlling state changes.
• Inputs J = 1 and K = 1 are applied and are stable during the active clock transition.
• Ideal flip-flop behavior (no setup/hold violations, no metastability).

Concept / Approach:
The characteristic table of a J–K flip-flop states: J = 0, K = 0 → Q holds; J = 0, K = 1 → Q resets; J = 1, K = 0 → Q sets; J = 1, K = 1 → Q toggles (complements). Thus, when both inputs are asserted, the device complements its present state on each active clock edge.

Step-by-Step Solution:

Verification / Alternative check:
Derive from excitation equations: For toggle action, J = K = 1 satisfies transitions Q → Q̄. Simulation or timing diagram confirms that repeated clock edges produce alternating Q levels 0,1,0,1,…

Why Other Options Are Wrong:

• be invalid: The J–K specifically resolves the “invalid” J = K = 1 problem found in S–R latches; it is defined to toggle.
• not change / change: “Change” is ambiguous; the defined action is a complement (toggle) on each active edge, not an arbitrary or one-time change. “Not change” is the J = K = 0 case.

Common Pitfalls:
Confusing J–K with an S–R latch (where S = R = 1 is invalid), or forgetting that toggling only occurs on the clock event, not continuously.

Final Answer:
toggle