Cascaded J–K flip-flops as a binary divider — output frequency Four J–K flip-flops are cascaded (J and K of each tied HIGH so they toggle). If the input frequency to the first stage is fin = 32 kHz, what is the output frequency after the fourth stage (fout)?

Introduction / Context:Cascading toggle flip-flops creates a binary frequency divider or ripple counter. Each stage divides the incoming frequency by 2, so chaining N stages yields division by 2^N. Recognizing this behavior is crucial for clock generation, baud-rate derivation, and timing state machines in digital design.

Given Data / Assumptions:

• Four J–K flip-flops in series, each with J = K = 1 (toggle mode).
• Input frequency fin = 32 kHz at the first stage.
• We observe the output of the fourth stage.

Concept / Approach:For toggle flip-flops, each positive edge (or designated clock event) causes the Q output to toggle, effectively halving the frequency. Thus, fout = fin / 2^N, where N is the number of cascaded stages.

Step-by-Step Solution:

Verification / Alternative check:Track stage-by-stage: 32 kHz → 16 kHz → 8 kHz → 4 kHz → 2 kHz. The fourth output is 2 kHz, confirming the calculation.

Why Other Options Are Wrong:

• 1 kHz or 4 kHz: correspond to division by 32 or 8 (wrong stage count).
• 16 kHz: only a single division (first stage) rather than four.

Common Pitfalls:Forgetting that each stage divides by 2 or miscounting how many stages the signal has passed through.

Final Answer:2 kHz