Cascaded modulus counters — lowest output frequency A 12 MHz clock is applied to a cascaded counter chain containing a modulus-5 stage, a modulus-8 stage, and a modulus-10 stage. What is the lowest output frequency available from the chain?

Introduction / Context:Cascading counters with different moduli multiplies the overall division factor. Determining the final output frequency after multiple stages is a routine step in clock-generation and timing applications.

Given Data / Assumptions:

• Input clock fin = 12 MHz.
• Three cascaded stages with moduli 5, 8, and 10.
• Stages are cascaded so each divides the frequency presented by the previous one.

Concept / Approach:The total division factor equals the product of the individual moduli. Therefore, fout = fin / (5 * 8 * 10). Convert units carefully to present the result in kilohertz for comparison to the options.

Step-by-Step Solution:

Verification / Alternative check:Stepwise: 12 MHz → /5 = 2.4 MHz → /8 = 0.3 MHz → /10 = 0.03 MHz = 30 kHz. Matches the calculation.

Why Other Options Are Wrong:

• 10 kHz and 20 kHz: correspond to stronger division than the chain provides.
• 60 kHz: corresponds to halving the correct division factor (200 instead of 400).

Common Pitfalls:Adding moduli or mis-converting units; forgetting to multiply all stage moduli.

Final Answer:30 kHz