SIPO register contents after loading a nibble A serial-in/parallel-out (SIPO) 4-bit shift register initially contains 1111. The incoming data nibble is 0111. After four clock pulses (one per bit), what contents appear in the register?

Introduction / Context:
Predicting a shift register's state after a defined number of clock pulses is a staple timing exercise. Here, a 4-bit SIPO starts at 1111 and then receives exactly four serial bits: 0, 1, 1, 1 (a nibble 0111).

Given Data / Assumptions:

• Register width = 4 bits.
• Initial contents = 1111.
• Incoming data nibble = 0 then 1 then 1 then 1 (serial order).
• One bit is shifted in per clock; after 4 clocks the initial bits have been fully overwritten.

Concept / Approach:

In a SIPO, each clock pushes the incoming bit into the first stage and shifts prior bits toward the last stage. After N = 4 clocks, the register contains exactly the 4 bits that arrived, in the same order they were clocked (assuming conventional left/right mapping consistent with the device).

Step-by-Step Solution:

Verification / Alternative check:

Regardless of the initial 1111, after four fresh bits have been introduced, all original bits have shifted out. The final state must equal the incoming nibble 0111.

Why Other Options Are Wrong:

• 0000: would require four zeros.
• 1111: unchanged; impossible after injecting new bits.
• 1000: would occur if the nibble were 1000.

Common Pitfalls:

• Reversing bit order; ensure you use the order presented by the incoming serial stream.

Final Answer:

0111