Serial in / Serial out (SISO) 8-bit shift register timing An 8-bit SISO shift register is clocked at 150 kHz. What is the time delay from the serial input to the Q3 tap (after four stages)?

Introduction / Context:Shift registers propagate data one stage per clock. The delay to a particular tap equals the number of stages to that tap multiplied by the clock period. This is a core timing idea in digital design and serial interfaces.

Given Data / Assumptions:

• Clock frequency f = 150 kHz.
• Tap Q3 is four stages from the serial input.
• Each clock advances data by one stage.

Concept / Approach:Clock period T = 1 / f. Time to reach Q3 = 4 * T, because the bit must traverse four flip-flops.

Step-by-Step Solution:

Verification / Alternative check:Dimensional check: kHz implies microsecond periods. Four microsecond-scale periods give a few tens of microseconds, which matches 26.67 µs and not milliseconds.

Why Other Options Are Wrong:

• 1.67 µs: Too small; equals roughly T/4, not 4T.
• 26.7 ms / 267 ms: Millisecond values are three orders of magnitude too large for a 150 kHz clock.

Common Pitfalls:

• Mistaking the number of stages (three vs. four) for Q3.
• Mixing kHz with ms instead of µs during unit conversion.

Final Answer:26.67 µs