Understanding JK Flip-Flop Output Stability

The question asks for the possible combination of inputs (\(J_n\), \(K_n\)) for a JK flip-flop when its current output \(Q_n\) is 1, and this output does not change after a clock pulse is applied. This means the next output \(Q_{n+1}\) must also be 1, i.e., \(Q_{n+1}\) = \(Q_n\) = 1.

JK Flip-Flop Truth Table Analysis

To determine the correct input combination, let's refer to the characteristic truth table of a JK flip-flop:

Identifying Stable Output Conditions for \(Q_n\) = 1

We are given that the current output \(Q_n\) = 1 and the next output \(Q_{n+1}\) also remains 1. We need to find the rows in the truth table where both \(Q_n\) = 1 and \(Q_{n+1}\) = 1. Let's examine the rows:

• When J=0, K=0, and \(Q_n\)=1, the next state \(Q_{n+1}\) is 1 (Hold state). This combination (0, 0) satisfies the condition.
• When J=1, K=0, and \(Q_n\)=1, the next state \(Q_{n+1}\) is 1 (Set state). This combination (1, 0) also satisfies the condition.

Deriving the Input Combination (\(J_n\), \(K_n\))

From the analysis above, we have two possible (J, K) input combinations that result in \(Q_{n+1}\) = 1 when \(Q_n\) = 1:

• (0, 0)
• (1, 0)

In both of these valid combinations, the input K is consistently 0. The input J, however, can be either 0 or 1. In digital logic, when an input can be either 0 or 1 without affecting the desired outcome, it is referred to as a 'don't care' condition, denoted by 'X'.

Therefore, the possible combination of inputs (\(J_n\), \(K_n\)) that ensures \(Q_n\) = 1 remains unchanged (\(Q_{n+1}\) = 1) is (X, 0).

Option Analysis

Let's evaluate each given option based on our derivation:

• Option 1: (X, 1)
  • If K=1, and \(Q_n\)=1:
  • If J=0, (0, 1) with \(Q_n\)=1 results in \(Q_{n+1}\)=0 (Reset).
  • If J=1, (1, 1) with \(Q_n\)=1 results in \(Q_{n+1}\)=0 (Toggle).
  • In both cases, \(Q_{n+1}\) would be 0, which means the output changes from 1 to 0. So, this option is incorrect.
• Option 2: (X, 0)
  • If K=0, and \(Q_n\)=1:
  • If J=0, (0, 0) with \(Q_n\)=1 results in \(Q_{n+1}\)=1 (Hold).
  • If J=1, (1, 0) with \(Q_n\)=1 results in \(Q_{n+1}\)=1 (Set).
  • In both cases, \(Q_{n+1}\) remains 1, meaning the output does not change. So, this option is correct.
• Option 3: (0, 1)
  • With (J=0, K=1) and \(Q_n\)=1, the next state \(Q_{n+1}\) becomes 0 (Reset). This indicates a change in output, so this option is incorrect.
• Option 4: (1, 1)
  • With (J=1, K=1) and \(Q_n\)=1, the next state \(Q_{n+1}\) becomes 0 (Toggle). This indicates a change in output, so this option is incorrect.

Based on the JK flip-flop truth table and the given condition, the input combination that keeps the output \(Q_n\) = 1 unchanged is (X, 0).