GATE CS 2010 Q32 — Combinational Circuits

From the given sequential circuit diagram, we can determine the input equations for the two D flip-flops as follows:

• The input to the first flip-flop is $D_1 = Q_1 \odot Q_0$ (XNOR gate).
• The input to the second flip-flop is $D_0 = \overline{Q_0}$ (NOT gate).

Since these are D flip-flops, the next-state equations are $Q_1^+ = D_1$ and $Q_0^+ = D_0$. Starting with the initial state $Q_1Q_0 = 00$, we trace the transitions step-by-step:

1. First Clock Cycle:
   Current state $Q_1Q_0 = 00$
   $Q_1^+ = 0 \odot 0 = 1$ and $Q_0^+ = \overline{0} = 1 \implies Q_1Q_0 = 11$
2. Second Clock Cycle:
   Current state $Q_1Q_0 = 11$
   $Q_1^+ = 1 \odot 1 = 1$ and $Q_0^+ = \overline{1} = 0 \implies Q_1Q_0 = 10$
3. Third Clock Cycle:
   Current state $Q_1Q_0 = 10$
   $Q_1^+ = 1 \odot 0 = 0$ and $Q_0^+ = \overline{0} = 1 \implies Q_1Q_0 = 01$
4. Fourth Clock Cycle:
   Current state $Q_1Q_0 = 01$
   $Q_1^+ = 0 \odot 1 = 0$ and $Q_0^+ = \overline{1} = 0 \implies Q_1Q_0 = 00$

Thus, the next four values of $Q_1Q_0$ are $11$, $10$, $01$, $00$.