Question

Consider a sequential digital circuit consisting of T flip-flops and D flip-flops as shown. At the beginning, \(Q_1=0\), \(Q_2=1\), \(Q_3=1\). Which one of the given values of \((Q_1,Q_2,Q_3)\) can NEVER be obtained with this circuit?

• A. \((0,0,1)\)
• B. \((1,0,0)\)
• C. \((1,0,1)\)
• D. \((1,1,1)\)

Hint:

Translate the diagram into next–state equations first. Then either use algebraic constraints or a short state walk from the initial condition to spot unreachable tuples quickly.

Answer 1

The Correct Option is A

Step 1: Write next–state equations
For a T–FF: \(Q^+ = Q \oplus T\). For a D–FF: \(Q^+ = D\).

From the figure (using the labels in the question): \[ \begin{aligned} T_1 &= \overline{Q_3}, & Q_1^+ &= Q_1 \oplus \overline{Q_3}, \\ D_2 &= Q_1, & Q_2^+ &= Q_1, \\ T_3 &= Q_2, & Q_3^+ &= Q_3 \oplus Q_2. \end{aligned} \]
Step 2: Iterate from the given initial state \((Q_1,Q_2,Q_3)=(0,1,1)\)
On each rising edge (all FFs clocked together) apply the above equations: \[ \begin{array}{c|c} \text{Clock} & (Q_1,Q_2,Q_3) \\ \hline 0 & (0,1,1)\ \text{(given)} \\ 1 & (0,0,0) \\ 2 & (1,0,0) \\ 3 & (0,1,0) \\ 4 & (1,0,1) \\ 5 & (1,1,1) \\ 6 & (1,1,0) \\ 7 & (0,1,1)\ \text{(cycle repeats)} \end{array} \]
Step 3: Read off the reachable set
The reachable states are \(\{(0,1,1), (0,0,0), (1,0,0), (0,1,0), (1,0,1), (1,1,1), (1,1,0)\}\). The tuple \((0,0,1)\) never appears and the sequence repeats thereafter, so it is unreachable. \[ \boxed{\text{State }(0,0,1)\ \text{can never be obtained.}} \]