Consider a sequential digital circuit consisting of T flip-flops and D flip-flops as shown in the figure. CLKIN is the clock input to the circuit. At the beginning, Q1, Q2 and Q3 have values 0, 1 and 1, respectively. Which one of the given values of (Q1, Q2, Q3) can NEVER be obtained with this digital circuit?

From the given synchronous sequential circuit diagram, the inputs to the flip-flops are:

• For the first T flip-flop ($Q_1$): $T_1 = \overline{Q_3}$

• For the D flip-flop ($Q_2$): $D_2 = Q_1$

• For the second T flip-flop ($Q_3$): $T_3 = Q_2$

The characteristic equations for the next states are:

• $Q_1^+ = Q_1 \oplus T_1 = Q_1 \oplus \overline{Q_3}$

• $Q_2^+ = D_2 = Q_1$

• $Q_3^+ = Q_3 \oplus T_3 = Q_3 \oplus Q_2$

Starting from the initial state $(Q_1, Q_2, Q_3) = (0, 1, 1)$:

1. Clock 1:

   • $Q_1^+ = 0 \oplus \overline{1} = 0 \oplus 0 = 0$

   • $Q_2^+ = 0$

   • $Q_3^+ = 1 \oplus 1 = 0$

   • Next state: $(0, 0, 0)$

2. Clock 2:

   • $Q_1^+ = 0 \oplus \overline{0} = 0 \oplus 1 = 1$

   • $Q_2^+ = 0$

   • $Q_3^+ = 0 \oplus 0 = 0$

   • Next state: $(1, 0, 0)$ *(Option B)*

3. Clock 3:

   • $Q_1^+ = 1 \oplus \overline{0} = 1 \oplus 1 = 0$

   • $Q_2^+ = 1$

   • $Q_3^+ = 0 \oplus 0 = 0$

   • Next state: $(0, 1, 0)$

4. Clock 4:

   • $Q_1^+ = 0 \oplus \overline{0} = 0 \oplus 1 = 1$

   • $Q_2^+ = 0$

   • $Q_3^+ = 0 \oplus 1 = 1$

   • Next state: $(1, 0, 1)$ *(Option C)*

5. Clock 5:

   • $Q_1^+ = 1 \oplus \overline{1} = 1 \oplus 0 = 1$

   • $Q_2^+ = 1$

   • $Q_3^+ = 1 \oplus 0 = 1$

   • Next state: $(1, 1, 1)$ *(Option D)*

6. Clock 6:

   • $Q_1^+ = 1 \oplus \overline{1} = 1 \oplus 0 = 1$

   • $Q_2^+ = 1$

   • $Q_3^+ = 1 \oplus 1 = 0$

   • Next state: $(1, 1, 0)$

7. Clock 7:

   • $Q_1^+ = 1 \oplus \overline{0} = 1 \oplus 1 = 0$

   • $Q_2^+ = 1$

   • $Q_3^+ = 0 \oplus 1 = 1$

   • Next state: $(0, 1, 1)$ *(loops back to initial state)*

The cycle of states is:

The state $(0, 0, 1)$ is never reached in this state trajectory.

Therefore, the correct option is (A).