Ring counters — correct initialization to start operation: To operate correctly, how must a ring counter be initialized at start-up? Select the statement that describes the required preset/clear pattern.

Introduction / Context:
A ring counter is a shift register with its serial output fed back to the input, circulating a fixed pattern through the stages. For correct operation, it must begin with a valid one-hot state (exactly one flip-flop set and the rest cleared). This question checks whether you know the necessary start-up condition to avoid lock-up or all-zero states that would never generate a circulating 1.

Given Data / Assumptions:

• n-stage ring counter implemented from edge-triggered flip-flops.
• Feedback path feeds Q_last back to the serial input of the first stage.
• No self-correcting bias unless additional logic is provided.

Concept / Approach:
In a simple ring counter, the only valid circulating pattern is a single 1 (or a single 0, depending on design) moving around the register. If all flip-flops are cleared, the pattern remains all zeros and will not evolve into a one-hot state. Therefore, the register must be initialized by presetting exactly one flip-flop and clearing all others, creating a one-hot word that will circulate with each clock pulse.

Step-by-Step Solution:

Verification / Alternative check:
Timing diagrams show the single 1 marching through successive stages; simulation confirms that the all-zero state is absorbing without extra injection logic, requiring explicit initialization in hardware or during reset.

Why Other Options Are Wrong:

• Clearing one and presetting the rest: Produces an (0111…1) state, not one-hot.
• Clearing all: Leaves an all-zero lock state that will not start the ring.
• Presetting all: Creates all ones; also not a one-hot state.
• Leaving floating: Unsafe and non-deterministic; may violate design rules.

Common Pitfalls:
Assuming rings self-start; forgetting to incorporate a reset network; confusing ring counters with Johnson counters, which can self-correct under some feedback schemes.

Final Answer:
Presetting one flip-flop and clearing all the others