Recovery and clarification: Assume a shift register with an active-LOW shift enable input (SE_L). When SE_L is LOW, the data path advances one bit on each active clock pulse. Under this common convention, does SE_L = 0 cause one-bit shifting per clock?

Introduction / Context:
The original stem omitted the control signal name. Applying the recovery-first policy, we interpret the intent as describing an active-LOW shift-enable input (often labeled SE_L or /SE). Many shift-register ICs use an enable that, when asserted, allows shifting to occur on each qualifying clock event.

Given Data / Assumptions:

• Assume a registered, clocked shift register.
• Assume a dedicated active-LOW shift enable (SE_L).
• Shifting occurs on the specified clock edge when enabled.

Concept / Approach:
Enable pins gate the action of the internal flip-flops. With SE_L asserted (LOW), each active clock edge transfers data from one stage to the next, achieving a one-bit-per-clock shift. When SE_L is deasserted (HIGH), the contents hold (no shift). Note that some parts invert this convention (enable HIGH); always check the function table.

Step-by-Step Solution:
1) Tie SE_L = 0 to assert shift enable.2) Apply a sequence of clock edges; on each edge, stages advance by one position.3) When SE_L = 1, hold state; no shifting occurs.4) Confirm behavior against the device timing diagram.

Verification / Alternative check:
Examine example datasheets where /SE or /CE gates shifting; simulation or bench tests show one-bit advancement per clock when the enable is active.

Why Other Options Are Wrong:
“Incorrect” would imply enable LOW prevents shifting, which contradicts the assumed active-LOW definition. Options that bind behavior to level clocks or asynchronous operation misstate typical edge-triggered, synchronous shifting.

Common Pitfalls:
Forgetting enable polarity; assuming shifting occurs on clock level rather than edges; ignoring hold behavior when enable is inactive.

Final Answer:
Correct