Clearing without reset: An 8-bit serial-in/parallel-out (SIPO) shift register has no asynchronous clear. You apply a logic 0 at the serial input to flush the register. After how many clock cycles will that injected 0 reach the QH (8th) output tap?

Introduction / Context:
When a shift register lacks a dedicated clear pin, designers sometimes “clear” it by clocking in zeros until all stages are overwritten. Knowing exactly how many cycles are needed ensures predictable recovery from unknown states and clean start-up sequences.

Given Data / Assumptions:

• Register: 8-bit SIPO with taps QA..QH (QA is the first stage, QH the eighth).
• No asynchronous/synchronous clear available.
• A logic 0 is applied to the serial data input.
• We want the cycle count until that specific 0 reaches QH.

Concept / Approach:
Each clock advances the serial input bit one stage deeper into the register. After k clocks, the injected bit resides at stage k. Therefore, to reach stage 8 (QH), the bit must traverse 8 stages and thus needs 8 clock cycles from the moment of injection.

Step-by-Step Solution:

Verification / Alternative check:
General rule: to push a known bit from the input to stage n, apply exactly n clocks. Simulation waveforms of standard 8-bit SIPO devices validate this propagation behavior.

Why Other Options Are Wrong:

• 1 or 7 clocks: Insufficient to reach the 8th stage.
• 9 clocks: One extra clock would push the original 0 out while a new input bit enters.

Common Pitfalls:
Counting from zero instead of one; confusing stage index with tap label; overlooking pipeline latency when back-to-back zeros are clocked in.

Final Answer:
8