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4-Bit Ring Counter — A ring counter is loaded with a single 1 (one-hot). What is the output frequency at any given stage relative to the clock?

Difficulty: Easy

Correct Answer: One-fourth the clock frequency

Explanation:


Introduction:
In a ring counter with a single circulating 1 (one-hot encoding), only one stage is HIGH at a time, and that HIGH token advances by one position each clock. Understanding the per-stage output frequency helps with timing, decoding, and LED scanning applications.

Given Data / Assumptions:

  • 4 stages; synchronous shifting on each clock edge.
  • Exactly one stage is HIGH in any state.
  • Outputs are observed at the Q pins of the stages.


Concept / Approach:
In four stages, the HIGH appears at a given stage once every 4 clocks. Therefore that stage’s output is HIGH for one clock out of four and LOW for the other three, yielding a frequency that is 1/4 of the clock frequency (and a 25% duty cycle, assuming one-cycle-wide HIGH).

Step-by-Step Solution:

Label stages S0..S3; initialize 1000.Sequence per clock: 1000 → 0100 → 0010 → 0001 → 1000 ...S0 is HIGH once every 4 clocks ⇒ f_out(S0) = f_clk / 4.


Verification / Alternative check:

Count rising edges between consecutive HIGH occurrences at one stage; the interval is 4 clocks.


Why Other Options Are Wrong:

Same as the clock: Would require the bit to toggle every clock, which it does not.Twice the clock / Half the clock: Not consistent with one-hot circulation among four positions.


Common Pitfalls:

Confusing per-stage frequency with the shift frequency of the token.Assuming multiple 1s (two-hot) as in Johnson counters; this is one-hot.


Final Answer:

One-fourth the clock frequency
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