In a J – K flip flop triggered by the rising edge of the clock signal, what is output ‘Q’ when it is triggered and J = 1, K = 0?
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J-K Flip Flop Operation Explained
A J-K flip-flop is a fundamental sequential logic circuit used for storing a single bit of data. It's an edge-triggered device, meaning its state changes are synchronized with the clock signal's active edge. This question concerns a J-K flip-flop activated by the rising edge of the clock.
J-K Flip Flop Input States
The output state ($Q$) of a J-K flip-flop transitions based on the values of the $J$ and $K$ inputs at the moment the clock pulse's rising edge occurs. The behavior is summarized in the following table:
| Input $J$ | Input $K$ | Next State $Q_{next}$ |
| 0 | 0 | No Change (holds previous state $Q$) |
| 0 | 1 | 0 (Reset state) |
| 1 | 0 | 1 (Set state) |
| 1 | 1 | Toggle (opposite of previous state $Q$, i.e., $\overline{Q}$) |
Specific J=1, K=0 Case Analysis
We are given the specific condition where the inputs to the J-K flip-flop are $J=1$ and $K=0$. The flip-flop is triggered by the rising edge of the clock signal.
According to the established behavior of a J-K flip-flop:
- When $J=1$ and $K=0$, the flip-flop is in the 'Set' state.
- In the 'Set' state, the output terminal $Q$ is forced to become $1$.
- This change occurs precisely on the rising edge of the clock pulse.
Therefore, when the inputs are $J=1$ and $K=0$ and the clock signal triggers a rising edge, the output $Q$ will transition to $1$.
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