GATE 2024 EC Q.42: solution and answer

GATE 2024 · ECQ.42 · 2 marks · MCQ

The sequence of states $(Q_{1} Q_{0})$ of the given synchronous sequential circuit is ________.

GATE 2024 EC question 42: Two positive-edge-triggered T flip-flops share a common clock line CLK (the clock inputs are drawn with a triangle and no… — Figure from the GATE 2024 EC paper. It is part of the question.

(A)
00 → 10 → 11 → 00
(B)
11 → 00 → 10 → 01 → 00
(C)
01 → 10 → 11 → 00 → 01
(D)
00 → 01 → 10 → 00

Answer and solution

ANSWER

The state sequence is $11 \to 00 \to 10 \to 01 \to 00$ , so the answer is option (B).

Official answer: BSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Read the flip-flop inputs off the wiring. Complement is written with \overline{} (no prime is used). OR gate 1 takes $Q_{1}$ (fed back from the junction dot) and $\overline{Q_{0}}$ (the wire along the bottom of the drawing). OR gate 2 takes $Q_{1}$ (from the same dot) and $Q_{0}$ (the wire along the top). So $T_{1} = Q_{1} + \overline{Q_{0}}, T_{0} = Q_{1} + Q_{0} .$
Use the T flip-flop characteristic equation $Q^{+} = T \oplus Q$ . The flip-flop toggles when $T = 1$ and holds when $T = 0$ .
State $00$ : $T_{1} = 0 + 1 = 1$ , so $Q_{1}$ toggles to 1. $T_{0} = 0 + 0 = 0$ , so $Q_{0}$ holds at 0. Next state: $10$ .
State $10$ : $T_{1} = 1 + 1 = 1$ , so $Q_{1}$ toggles to 0. $T_{0} = 1 + 0 = 1$ , so $Q_{0}$ toggles to 1. Next state: $01$ .
State $01$ : $T_{1} = 0 + 0 = 0$ , so $Q_{1}$ holds at 0. $T_{0} = 0 + 1 = 1$ , so $Q_{0}$ toggles to 0. Next state: $00$ .
State $11$ : $T_{1} = 1 + 0 = 1$ and $T_{0} = 1 + 1 = 1$ , so both flip-flops toggle. Next state: $00$ .
Putting it together: $11 \to 00 \to 10 \to 01 \to 00$ . The circuit cycles through $00 \to 10 \to 01 \to 00$ , and state $11$ drops into that cycle at $00$ . This is option (B). Options (A) and (C) both send $10$ to $11$ (and (C) also sends $01$ to $10$ ), and (D) sends $00$ to $01$ . None of these transitions is what the circuit does.

KEY IDEA

Trace the wiring to get the excitation equations $T_{1} = Q_{1} + \overline{Q_{0}}$ and $T_{0} = Q_{1} + Q_{0}$ . Then use $Q^{+} = T \oplus Q$ to find the next state for all four present states. Because $Q_{1}$ feeds both OR gates, both flip-flops toggle whenever $Q_{1} = 1$ . That fact alone gives $10 \to 01$ and $11 \to 00$ .

WATCH OUT FOR

Treating the T input as if it were a D input ( $Q^{+} = T$ ) instead of a toggle ( $Q^{+} = T \oplus Q$ ). Doing that gives $10 \to 11$ , the wrong transition in options (A) and (C). A second common slip is misreading the long feedback wires: taking $Q_{0}$ into gate 1 instead of $\overline{Q_{0}}$ , or missing the $Q_{1}$ feedback into gate 1. Students also often skip state $11$ . It is outside the main cycle, but option (B) starts from it, so you have to check it.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 30 Sequential questions, with notes

GATE 2024 EC Q.42: solution and answer

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