Shift registers — serially loading a nibble (4 bits) To shift a nibble (four bits) of data serially into a shift register, how many clock pulses are required?

Introduction / Context:Serial shift registers accept data one bit per clock edge. Knowing how many clock pulses are required to load a specified number of bits is fundamental to timing analysis in digital systems, serial interfaces, and logic simulation.

Given Data / Assumptions:

• The word length to be shifted in is one nibble, i.e., 4 bits.
• Data is applied to the serial input and clocked in bit by bit.
• Each active clock edge shifts the register by exactly one bit position.

Concept / Approach:

A serial-in register advances its contents by one bit for each clock. Therefore, loading N bits requires N clock pulses, independent of the bit values (0 or 1). The number of 1s in the data does not affect the number of clocks; 0s shift just like 1s.

Step-by-Step Solution:

Verification / Alternative check:

Simulate a 4-stage register (Q3..Q0) with any 4-bit pattern. After the first clock, one bit resides in Q0; after the second, two bits are present; after the fourth, all four bits occupy the register. This confirms the one-clock-per-bit rule.

Why Other Options Are Wrong:

• One clock pulse: loads only one bit, not four.
• Eight clock pulses: would be needed for a byte, not a nibble.
• One pulse for each 1 in the data: incorrect; clocks are independent of bit values.

Common Pitfalls:

• Confusing serial loading with parallel loading (which would take a single strobe, not multiple clocks).
• Assuming data value affects clock count.

Final Answer:

four clock pulses