GATE CS 2015 Set 1 Q.47 (CS Q.37): Sequential Circuits, Counters and Registers

A positive edge-triggered D flip-flop is connected to a positive edge-triggered JK flip-flop as follows. The Q output of the D flip-flop is connected to both the J and K inputs of the JK flip-flop, while the Q output of the JK flip-flop is connected to the input of the D flip-flop. Initially, the output of the D flip-flop is set to logic one and the output of the JK flip-flop is cleared. Which one of the following is the bit sequence (including the initial state) generated at the Q output of the JK flip-flop when the flip-flops are connected to a free-running common clock? Assume that $J = K = 1$ is the toggle mode and $J = K = 0$ is the state-holding mode of the JK flip-flop. Both the flip-flops have non-zero propagation delays.

$0110110...$

$0100100...$

$011101110...$

$011001100...$

Official answer and solution

Official answer: A
From: Official GATE 2015 answer key

Solution

Trace the cycle-by-cycle state transitions of the coupled D and JK flip-flops from the initial state.

Flip-flop connections:

$D_{\text{in}} = Q_{JK}$
$J = K = Q_D$ (toggle mode when $Q_D = 1$ ; hold mode when $Q_D = 0$ )

Initial state at $t = 0$ : $Q_D = 1, Q_{JK} = 0$ .

Tracing successive clock cycles:

Cycle 1:
Inputs before clock: $D = 0$ , $J = K = 1$ .
Next states: $Q_D^+ = 0$ , $Q_{JK}^+ = \overline{Q_{JK}} = \overline{0} = 1$ .
Output: $Q_{JK} = 1$ .
Cycle 2:
Inputs before clock: $D = 1$ , $J = K = 0$ .
Next states: $Q_D^+ = 1$ , $Q_{JK}^+ = Q_{JK} = 1$ (holds).
Output: $Q_{JK} = 1$ .
Cycle 3:
Inputs before clock: $D = 1$ , $J = K = 1$ .
Next states: $Q_D^+ = 1$ , $Q_{JK}^+ = \overline{Q_{JK}} = \overline{1} = 0$ .
Output: $Q_{JK} = 0$ .
Cycle 4:
Inputs before clock: $D = 0$ , $J = K = 1$ .
State is $(Q_D = 1, Q_{JK} = 0)$ , exactly repeating Cycle 1.

The sequence of bits generated at $Q_{JK}$ starting from the initial state is:

0, 1, 1, 0, 1, 1, 0, \dots

This matches option A.

Final answer

Common trap

Omitting the initial state bit 0 from the printed output sequence.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 1
Question: 47 of 65, printed as CS Q.37
Section: Computer Science and Information Technology

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