Shift Registers — Load and Unload Cycles For a serial-in/serial-out 8-bit shift register, how many clock pulses are required to completely load the 8-bit word and then completely unload it from the register output?

Introduction / Context:
Shift registers move data one bit per clock edge. Knowing how many clock pulses are required to fill and then drain a serial register is fundamental for interface timing, buffering, and serial communication chores.

Given Data / Assumptions:

• Register length: 8 bits.
• Serial-in/serial-out operation (one bit shifted per clock).
• Definition: “Load” means shifting an entire 8-bit word into the register; “Unload” means shifting that 8-bit word out to the serial output.

Concept / Approach:
Each clock shifts one bit. To load N bits serially, N pulses are required. After loading, the word occupies the register stages; to shift the same N bits out, another N pulses are needed unless there is parallel readback.

Step-by-Step Solution:
Loading: 8 bits → 8 clock pulses.Unloading: the stored 8 bits must be shifted to the output → 8 additional pulses.Total pulses = 8 + 8 = 16.Therefore, 16 clock pulses are required.

Verification / Alternative check:
Simulate with an 8-stage shift register: after the first 8 pulses, the word is fully inside; after an additional 8, it has fully exited at the serial output.

Why Other Options Are Wrong:

• 2 or 4: Far too few for serial shifting of 8 bits.
• 8: Enough to load or unload, but not both.

Common Pitfalls:
Confusing serial and parallel load/unload modes; assuming data appears at the output immediately without counting the full shift length.

Final Answer:
16