GATE 2018 IN Q.44: solution and answer

GATE 2018 · INQ.44 · 2 marks · MCQ

A 2-bit synchronous counter using two $J$ - $K$ flip flops is shown. The expressions for the inputs to the $J$ - $K$ flip flops are also shown in the figure. The output sequence of the counter starting from $Q_{1} Q_{2} = 00$ is

GATE 2018 IN question 44: Two J-K flip-flops (each box labelled J, K, Q, Q-bar, SET, CLR) share one Clock line, so the counter is synchronous. — Figure from the GATE 2018 IN paper. It is part of the question.

(A)
$00 \to 11 \to 10 \to 01 \to 00 \dots$
(B)
$00 \to 01 \to 10 \to 11 \to 00 \dots$
(C)
$00 \to 01 \to 11 \to 10 \to 00 \dots$
(D)
$00 \to 10 \to 11 \to 01 \to 00 \dots$

Answer and solution

ANSWER

Option (C): starting from 00, the counter goes $00 \to 01 \to 11 \to 10 \to 00 \dots$

Official answer: CSource: Official answer key, GATE 2018 IN (IIT Guwahati)

Read the inputs off the figure (overbar = complement): $J_{1} = Q_{1} + Q_{2}$ , $K_{1} = \overline{Q_{1}} + \overline{Q_{2}}$ , $J_{2} = \overline{Q_{1}} + Q_{2}$ , $K_{2} = Q_{1} + \overline{Q_{2}}$ . Both flip-flops share one clock, so all four inputs are worked out from the present state, and both outputs change on the same edge.
Use the J-K characteristic: $J K = 00$ holds, $01$ resets to 0, $10$ sets to 1, $11$ toggles.
Present state $Q_{1} Q_{2} = 00$ : $J_{1} K_{1} = 01 \Rightarrow Q_{1} = 0$ (reset). $J_{2} K_{2} = 11 \Rightarrow Q_{2}$ toggles $0 \to 1$ . Next state $01$ .
Present state $01$ : $J_{1} K_{1} = 11 \Rightarrow Q_{1}$ toggles $0 \to 1$ . $J_{2} K_{2} = 10 \Rightarrow Q_{2} = 1$ (set). Next state $11$ .
Present state $11$ : $J_{1} K_{1} = 10 \Rightarrow Q_{1} = 1$ (set). $J_{2} K_{2} = 11 \Rightarrow Q_{2}$ toggles $1 \to 0$ . Next state $10$ .
Present state $10$ : $J_{1} K_{1} = 11 \Rightarrow Q_{1}$ toggles $1 \to 0$ . $J_{2} K_{2} = 01 \Rightarrow Q_{2} = 0$ (reset). Next state $00$ , which is where the counter started.
Cross-check with the characteristic equation $Q^{+} = J \overline{Q} + \overline{K} Q$ . Since $\overline{K_{1}} = Q_{1} Q_{2}$ and $\overline{K_{2}} = \overline{Q_{1}} Q_{2}$ : $Q_{1}^{+} Q_{2}^{+} = (Q_{1} + Q_{2}) \overline{Q_{1}} + Q_{1} Q_{2} = \overline{Q_{1}} Q_{2} + Q_{1} Q_{2} = Q_{2} = (\overline{Q_{1}} + Q_{2}) \overline{Q_{2}} + \overline{Q_{1}} Q_{2} = \overline{Q_{1}} \overline{Q_{2}} + \overline{Q_{1}} Q_{2} = \overline{Q_{1}}$ So the next state is $Q_{1}^{+} Q_{2}^{+} = Q_{2} \overline{Q_{1}}$ : $00 \to 01 \to 11 \to 10 \to 00$ , the same as the state-by-state table.
Sequence: $00 \to 01 \to 11 \to 10 \to 00 \dots$ . This is a 2-bit Gray-code (twisted-ring) counter, which is option (C).

KEY IDEA

In a synchronous counter, work out every J and K from the present state, then apply the J-K rules (00 hold, 01 reset, 10 set, 11 toggle) to both flip-flops at once. You can also check it in one line with $Q^{+} = J \overline{Q} + \overline{K} Q$ . Here the inputs simplify to $Q_{1}^{+} = Q_{2}$ and $Q_{2}^{+} = \overline{Q_{1}}$ , a 2-bit twisted-ring counter, and that gives the Gray sequence 00, 01, 11, 10.

WATCH OUT FOR

Updating $Q_{1}$ first and then using the new $Q_{1}$ to work out $J_{2}$ and $K_{2}$ within the same clock edge. That is ripple-style thinking, and it is wrong for a synchronous counter, where every input comes from the old state. Other slips: mixing up $J K = 01$ (reset) with $10$ (set), and missing an overbar on the input labels. For example, $J_{1} = Q_{1} + Q_{2}$ has no bars, while $K_{1} = \overline{Q_{1}} + \overline{Q_{2}}$ has two.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 30 Sequential questions, with notes

GATE 2018 IN Q.44: solution and answer

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