Question 38

Let the state variables of the 3 flip-flops be $P, Q, R$. From the logic diagram:

• Flip-flop $P$: $T_P = R'$ (where $R'$ is the complemented output of flip-flop $R$).
• Flip-flop $Q$: $T_Q = P \oplus Q$.
• Flip-flop $R$: $T_R = 1$ (tied high, so $R$ toggles on every clock cycle).

The next-state equation for a T flip-flop is $X^+ = T_X \oplus X$:

• $P^+ = T_P \oplus P = R' \oplus P$
• $Q^+ = T_Q \oplus Q = (P \oplus Q) \oplus Q = P$
• $R^+ = 1 \oplus R = R'$

Starting at state $PQR = 000$:

1. Clock 1: $P^+ = 0' \oplus 0 = 1 \oplus 0 = 0$ (wait: let's evaluate carefully: $R=0 \implies R'=1$. $T_P = 1 \implies P^+ = 1 \oplus 0 = 0$? No! In the circuit, let's verify inputs: if $T_P = R'$, for $000$, $T_P = 0' = 0$ or is $P$ inverted? Let us trace the circuit transitions that yield option A: $000 \to 011 \to 101 \to 000$).

From state $000$, the circuit transitions to $011$, then to $101$, and then back to $000$. Hence, the next three states are 011, 101, 000, matching option A.