GATE CS 2015 Set 1 Q37 — Sequential Circuits

Let $Q_D(n)$ and $Q_{JK}(n)$ represent the states of the $D$ and $JK$ flip-flops at clock cycle $n$, respectively. Based on the connections described:

• The input to the $D$ flip-flop is $D = Q_{JK}(n)$, so $Q_D(n+1) = Q_{JK}(n)$.
• The inputs to the $JK$ flip-flop are $J = K = Q_D(n)$. This means:
  • If $Q_D(n) = 1$, the $JK$ flip-flop toggles: $Q_{JK}(n+1) = \overline{Q_{JK}(n)}$.
  • If $Q_D(n) = 0$, the $JK$ flip-flop holds its state: $Q_{JK}(n+1) = Q_{JK}(n)$.

Given the initial state ($n=0$) is $Q_D(0) = 1$ and $Q_{JK}(0) = 0$, we can trace the states cycle-by-cycle:

The sequence generated at the $Q$ output of the $JK$ flip-flop (including the initial state) is $0, 1, 1, 0, 1, 1, 0, \ldots$, which corresponds to option A.