GATE 2023 EC Q.63: solution and answer

GATE 2023 · ECQ.63 · 2 marks · NAT

In a given sequential circuit, initial states are $Q_{1} = 1$ and $Q_{2} = 0$ . For a clock frequency of $1 MHz$ , the frequency of signal $Q_{2}$ in kHz, is ________ (rounded off to the nearest integer).

GATE 2023 EC question 63: Two positive-edge-triggered D flip-flops (clock input shown as a plain triangle with no bubble) share one clock, labelled… — Figure from the GATE 2023 EC paper. It is part of the question.

Answer and solution

ANSWER

The circuit cycles through the four states $10 \to 00 \to 01 \to 11$ , so $Q_{2}$ repeats every 4 clock periods and $f_{Q_{2}} = 1 MHz /4 = 250 kHz$ .

Official answer: 250, or 500Source: Official answer key, GATE 2023 EC (IIT Kanpur)

Read the connections from the figure: $D_{1} = Q_{2}$ (the feedback wire over the top) and $D_{2} = \overline{Q_{1}}$ (the diagonal wire from the bubbled output; the bar marks the complement). The $Q_{1}$ and $\overline{Q_{2}}$ outputs are unused stubs. Both flip-flops take the same clock, so they change state together on each clock edge.
D flip-flop characteristic equation: $Q^{+} = D$ . So $Q_{1}^{+} = Q_{2}$ and $Q_{2}^{+} = \overline{Q_{1}}$ .
Step the state $(Q_{1} Q_{2})$ forward from the given start $10$ . Clock 1: $Q_{1}^{+} = Q_{2} = 0$ , $Q_{2}^{+} = \overline{1} = 0$ , giving $00$ .
Clock 2: $Q_{1}^{+} = 0$ , $Q_{2}^{+} = \overline{0} = 1$ , giving $01$ . Clock 3: $Q_{1}^{+} = 1$ , $Q_{2}^{+} = \overline{0} = 1$ , giving $11$ . Clock 4: $Q_{1}^{+} = 1$ , $Q_{2}^{+} = \overline{1} = 0$ , giving $10$ , which is the starting state again.
The sequence $10 \to 00 \to 01 \to 11 \to 10$ repeats every 4 clock pulses. The loop through the two flip-flops has exactly one inversion, so this is a 2-bit twisted-ring (Johnson) counter, which has $2 n = 4$ states for $n = 2$ .
Over one cycle $Q_{2}$ reads $0, 0, 1, 1$ . That is a square wave (50% duty cycle) whose period is 4 clock periods.
$f_{Q_{2}} = \frac{f _{c l k}}{4} = \frac{1 MHz}{4} = 250 kHz$

KEY IDEA

Write $Q^{+} = D$ for each flip-flop and step the state table from the given initial state until it repeats. The number of states in the cycle is the division ratio: an output that repeats every $N$ clocks has frequency $f_{c l k} / N$ . A shift loop with exactly one inversion (a twisted-ring, or Johnson, counter) of $n$ flip-flops has $2 n$ states, so here $N = 4$ .

WATCH OUT FOR

Reading the diagonal wire as coming from $Q_{1}$ instead of $\overline{Q_{1}}$ . With no inversion the loop is a plain ring counter, $10 \to 01 \to 10$ , which gives a 2-state cycle and 500 kHz. (The official key accepts both 250 and 500, but the figure as printed gives 250.) A second slip is to count state changes of $Q_{2}$ rather than full periods: $Q_{2}$ changes every 2 clocks, but one full period is 4 clocks.

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 30 Sequential questions, with notes

GATE 2023 EC Q.63: solution and answer

Answer and solution

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