JK flip-flop as a divide-by-two A JK flip-flop is clocked at 20 kHz with inputs J = 1 and K = 1. What is the resulting Q output waveform?

Introduction / Context:One classic application of the JK flip-flop is clock division. When both inputs are asserted (J = K = 1), the device toggles state at each active clock edge, creating a square wave at half the input frequency. This forms the basis of ripple counters and frequency dividers.

Given Data / Assumptions:

• JK flip-flop with J = 1 and K = 1 continuously.
• Clock frequency: 20 kHz.
• Positive edge triggering assumed (behavior is analogous for negative edge with appropriate phase shift).

Concept / Approach:With J = K = 1, the JK truth table enters the toggle mode. Each active clock edge complements Q. Thus, after one edge Q flips; after the next, it flips back—completing one full cycle every two clock edges, which halves the frequency and produces a 50% duty square wave (ideal case).

Step-by-Step Solution:

Verification / Alternative check:Scope measurement will show Q transitions on every other clock edge, confirming frequency division by two. Chaining n such stages yields division by 2^n.

Why Other Options Are Wrong:

• Constantly LOW/HIGH: Would imply asynchronous forcing or disabled clock, not the toggle mode.
• 20 kHz square wave: That would indicate pass-through or buffering, not divide-by-two toggling.

Common Pitfalls:Forgetting that propagation delays introduce phase lag; also, mixing up duty cycle if asynchronous clears/presets occur.

Final Answer:a 10 kHz square wave