Question

In a positive edge-triggered $JK$ flip-flop, the $J$ input is connected to $\bar{Q}$ (inverted output) and the $K$ input is connected to $Q$ (normal output). If the flip-flop is initially in the RESET state ($Q=0, \bar{Q}=1$) and is subjected to 4 consecutive positive clock pulses, what will be the final state of $Q$?

• A. $1$
• B. $0101$ (oscillates)
• C. $0$
• D. Indeterminate

Hint:

When inputs are wired as $J=\bar{Q}$ and $K=Q$, the configuration mimics a basic T-type or D-type structure configured to toggle every clock period. Even numbers of clock pulses return the state back to its initial value, while odd numbers yield the inverted initial value.

Answer 1

The Correct Option is C

Understanding the Concept: The operational behavior of a standard edge-triggered $JK$ flip-flop is governed strictly by its characteristic equation: \[ Q_{next} = J\bar{Q} + \bar{K}Q \] By mapping external feedback paths directly onto the control inputs, we substitute the given logic constraints into the characteristic equation to trace state transitions sequentially across each subsequent active clock edge.

Step 1: Formulate the specialized state transition equation.
The problem explicitly dictates the following permanent cross-connections from the outputs back to the input terminals: \[ J = \bar{Q} \] \[ K = Q \] Let us substitute these constraints into the standard $JK$ flip-flop characteristic equation: \[ Q_{next} = J\bar{Q} + \bar{K}Q \quad \Rightarrow \quad Q_{next} = (\bar{Q})\bar{Q} + (\bar{Q})Q \] Using Boolean algebraic simplification rules: - Since $\bar{Q} \cdot \bar{Q} = \bar{Q}$ - Since $\bar{Q} \cdot Q = 0$ We get the simplified state equation: \[ Q_{next} = \bar{Q} + 0 = \bar{Q} \] This reveals that at every single incoming positive clock edge, the next state will simply equal the inverse of the current output state.

Step 2: Trace the output state chronologically across 4 clock cycles.
We are given that the initial state of the system prior to any clock interaction is the RESET state: \[ \text{Initial State: } Q_0 = 0 \] Let us trace each positive edge transition step-by-step:
• First active clock pulse (Pulse 1): \[ Q_1 = \bar{Q}_0 = \bar{0} = 1 \]
• Second active clock pulse (Pulse 2): \[ Q_2 = \bar{Q}_1 = \bar{1} = 0 \]
• Third active clock pulse (Pulse 3): \[ Q_3 = \bar{Q}_2 = \bar{0} = 1 \]
• Fourth active clock pulse (Pulse 4): \[ Q_4 = \bar{Q}_3 = \bar{1} = 0 \] After exactly 4 discrete positive clock pulses, the system completes two full toggling cycles, resulting in a final state of $Q = 0$.