8-bit shift register initially clear – serial entry (right-most bit first) The sequence 10011100 is shifted in (LSB first) to an 8-bit parallel-out shift register initially 00000000. After four clock pulses, what are the Q outputs?

Introduction / Context:
Serial loading order and shift direction must be interpreted correctly to predict the intermediate register contents. With an initially cleared register, early states reveal exactly how data propagates inward from the serial input.

Given Data / Assumptions:

• Initial contents: 00000000 (MSB→LSB).
• Input pattern: 10011100, right-most bit first → serial input sequence = 0, 0, 1, 1, 1, 0, 0, 1.
• Register shifts right each clock; serial input enters the MSB.

Concept / Approach:
Track the first four input bits and states. Because the register starts at all zeros, the first two inputs are zero and do not change the visible pattern. Once ones arrive, they collect at the upper bits.

Step-by-Step Solution:
Clock 1 (in=0): 00000000.Clock 2 (in=0): 00000000.Clock 3 (in=1): 10000000.Clock 4 (in=1): 11000000.Therefore, after four pulses, Q = 11000000.

Verification / Alternative check:
Simulate with a quick script or hand table; both agree on 11000000 after four clocks with this loading order.

Why Other Options Are Wrong:
10011100: would require the full eight clocks to load all bits.00001100 / 00000011: imply LSB-entry or left-shift direction; not the stated operation.11110000: corresponds to a different initial state or more clocks.

Common Pitfalls:
Assuming the serial input enters the LSB, or forgetting that the first two zeros leave the register unchanged from its cleared state.

Final Answer:
11000000