JK Flip Flop J=K=1 Operation Explained

This question concerns the behavior of a JK flip-flop when both the J and K inputs are held at logic 1, under a specific clock input frequency.

Understanding JK Flip Flop Modes

A JK flip-flop is a fundamental sequential logic circuit. Its output (Q) changes based on the J and K inputs and the clock signal. The behavior is summarized below:

• If J=0 and K=0, the output Q retains its previous state.
• If J=0 and K=1, the output Q is reset to 0 on the clock edge.
• If J=1 and K=0, the output Q is set to 1 on the clock edge.
• If J=1 and K=1, the output Q toggles (changes to the opposite state) on the clock edge. This is often called the "toggle mode".

Analyzing the Toggle Mode

In this specific problem, we are given that J = 1 and K = 1. This means the JK flip-flop is configured to operate in its toggle mode.

When a flip-flop is in toggle mode, its Q output inverts its state every time a clock pulse occurs. If the Q output is currently 0, it becomes 1 after the clock edge. If it's 1, it becomes 0.

Calculating the Output Frequency

The clock input frequency ($f_{clk}$) is given as 100 kHz.

Since the flip-flop toggles its state on each clock pulse, the output Q will change its state twice during each clock cycle. This effectively halves the frequency of the output signal compared to the clock input.

The output frequency ($f_{out}$) can be calculated using the formula:

$$ f_{out} = \frac{f_{clk}}{2} $$

Substituting the given clock frequency:

$$ f_{out} = \frac{100 \text{ kHz}}{2} $$

$$ f_{out} = 50 \text{ kHz} $$

The output signal alternates between 0 and 1, creating a square wave. Therefore, the Q output will be a 50 kHz square wave.

Conclusion

Based on the analysis, when a JK flip-flop has J=1, K=1, and receives a 100 kHz clock input, the Q output becomes a 50 kHz square wave because the flip-flop repeatedly toggles its state.