Digital electronics – Ring counter state evolution An 8-bit ring counter has the initial state 10111110 (bit 7 to bit 0). After four clock pulses (circular shifting each clock), what will be the new 8-bit state?

Introduction / Context:Ring counters are sequential circuits formed by connecting flip-flops in a loop so that the bit pattern circulates with each clock pulse. This question assesses understanding of circular shifting and careful bit tracking over multiple clock edges.

Given Data / Assumptions:

• Counter width: 8 bits (bit7 … bit0).
• Initial state: 10111110.
• On each clock, the register performs a circular shift (ring) so the pattern rotates one position.
• No resets or additional injections of 1s or 0s occur.

Concept / Approach:In a ring counter, the least significant bit feeds back to the most significant bit (or vice versa, depending on the implementation). A circular shift preserves the number of 1s and 0s, merely moving their positions. We track the pattern after each clock pulse until the fourth pulse is reached.

Step-by-Step Solution:

Verification / Alternative check:Another way is to note that a 4-position circular shift of an 8-bit word equals a rotate by 4. Performing either four right-rotates or four left-rotates on 10111110 yields 11101011, confirming the result.

Why Other Options Are Wrong:

• 00010111: Not reachable in exactly four pure circular shifts from the given pattern.
• 11110000: Changes the count of 1s and 0s (violates preservation under rotation), so it cannot be a rotation of the original pattern.
• 00000000: Would require clearing bits; rotation does not change bit counts.

Common Pitfalls:

• Confusing a ring counter (rotate) with a simple shift register (shifts in zeros or ones).
• Forgetting that rotations preserve the number of 1s and 0s.
• Shifting the wrong direction; here, four rotates left or right both yield the same final state due to symmetry.

Final Answer:11101011