Serial loading time for a 5-bit shift register How many clock pulses are required to load all 5 bits serially into a 5-bit shift register?

Introduction / Context:Serial-in shift registers accept one bit per clock edge. To completely fill an n-bit register starting empty (or regardless of previous contents, if we define loading as bringing in n new bits), you must shift in n bits. This question checks that basic but crucial timing relationship.

Given Data / Assumptions:

• Register length n = 5 bits.
• One new input bit is captured per qualifying clock edge.
• No parallel load operation is being used.

Concept / Approach:Each clock captures the current serial input into the first stage and shifts existing contents one position. After k pulses, k of the most recent input bits occupy the first k stages. Therefore, to fully load 5 fresh bits, we need 5 clock pulses.

Step-by-Step Solution:Pulse 1 → 1st bit enters stage 0.Pulse 2 → 2nd bit enters stage 0; 1st bit shifts to stage 1.…Pulse 5 → 5th bit enters; previous 4 bits have shifted, filling all 5 stages.

Verification / Alternative check:Timing diagrams for SISO/SIPO registers demonstrate that n pulses are required to move n new bits fully into an n-bit structure.

Why Other Options Are Wrong:

• 2, 3, 4 pulses: Insufficient; fewer than n pulses yield fewer than n newly-loaded bits.

Common Pitfalls:

• Mistaking the time to output the first bit at the far end for the time to fully load the register.
• Assuming double-data-rate or multi-bit-per-clock behavior; standard shift registers are one bit per qualifying edge.

Final Answer:5