1. **Problem Statement:** We have a sequential circuit with four flip-flops A, B, C, D described by the state equations: $$A(t+1) = (CD' + C'D)x + (CD + C'D')x'$$ $$B(t+1) = A$$ $$C(t+1) = B$$ $$D(t+1) = C$$ We need to find the sequence of states for two cases: - When $x=1$, starting from state $ABCD=0001$ - When $x=0$, starting from state $ABCD=0000$ 2. **Understanding the equations:** - $x'$ means NOT $x$. - $C'D$ means $C$ is 0 and $D$ is 1. - The next state of $A$ depends on $x$ and the current states of $C$ and $D$. - The next states of $B$, $C$, and $D$ are just the current states of $A$, $B$, and $C$ respectively. 3. **Simplify $A(t+1)$ for each $x$ value:** - For $x=1$: $$A(t+1) = CD' + C'D$$ - For $x=0$: $$A(t+1) = CD + C'D'$$ 4. **Calculate sequence for $x=1$, starting from $ABCD=0001$:** - Initial state: $A=0, B=0, C=0, D=1$ - Calculate $A(t+1)$: $$CD' + C'D = (0)(0) + (1)(1) = 0 + 1 = 1$$ - Next states: $$A(t+1) = 1$$ $$B(t+1) = A = 0$$ $$C(t+1) = B = 0$$ $$D(t+1) = C = 0$$ - Next state: $1000$ Repeat for next states: - State $1000$: $$A(t+1) = CD' + C'D = (0)(1) + (0)(0) = 0$$ Next states: $A=0, B=1, C=0, D=0$ → $0100$ - State $0100$: $$A(t+1) = (0)(1) + (1)(0) = 0$$ Next states: $A=0, B=0, C=1, D=0$ → $0010$ - State $0010$: $$A(t+1) = (1)(1) + (0)(0) = 1$$ Next states: $A=1, B=0, C=0, D=1$ → $1001$ - State $1001$: $$A(t+1) = (0)(0) + (1)(1) = 1$$ Next states: $A=1, B=1, C=0, D=0$ → $1100$ - State $1100$: $$A(t+1) = (0)(1) + (0)(0) = 0$$ Next states: $A=0, B=1, C=1, D=0$ → $0110$ - State $0110$: $$A(t+1) = (1)(0) + (0)(1) = 0$$ Next states: $A=0, B=0, C=1, D=1$ → $0011$ - State $0011$: $$A(t+1) = (1)(0) + (0)(1) = 0$$ Next states: $A=0, B=0, C=1, D=1$ → $0011$ (repeats) Sequence for $x=1$ starting from $0001$ is: $$0001 \to 1000 \to 0100 \to 0010 \to 1001 \to 1100 \to 0110 \to 0011 \to 0011 ...$$ 5. **Calculate sequence for $x=0$, starting from $ABCD=0000$:** - For $x=0$, $$A(t+1) = CD + C'D'$$ - Initial state: $0000$ $$A(t+1) = (0)(0) + (1)(1) = 1$$ Next states: $A=1, B=0, C=0, D=0$ → $1000$ - State $1000$: $$A(t+1) = (0)(0) + (1)(1) = 1$$ Next states: $A=1, B=1, C=0, D=0$ → $1100$ - State $1100$: $$A(t+1) = (0)(0) + (1)(0) = 0$$ Next states: $A=0, B=1, C=1, D=0$ → $0110$ - State $0110$: $$A(t+1) = (1)(0) + (0)(1) = 0$$ Next states: $A=0, B=0, C=1, D=1$ → $0011$ - State $0011$: $$A(t+1) = (1)(1) + (0)(0) = 1$$ Next states: $A=1, B=0, C=0, D=1$ → $1001$ - State $1001$: $$A(t+1) = (0)(1) + (1)(0) = 0$$ Next states: $A=0, B=1, C=0, D=0$ → $0100$ - State $0100$: $$A(t+1) = (0)(1) + (1)(0) = 0$$ Next states: $A=0, B=0, C=1, D=0$ → $0010$ - State $0010$: $$A(t+1) = (1)(0) + (0)(1) = 0$$ Next states: $A=0, B=0, C=0, D=1$ → $0001$ - State $0001$: $$A(t+1) = (0)(0) + (1)(1) = 1$$ Next states: $A=1, B=0, C=0, D=0$ → $1000$ (cycle repeats) Sequence for $x=0$ starting from $0000$ is: $$0000 \to 1000 \to 1100 \to 0110 \to 0011 \to 1001 \to 0100 \to 0010 \to 0001 \to 1000 ...$$ **Final answers:** - For $x=1$, starting from $0001$: $$0001, 1000, 0100, 0010, 1001, 1100, 0110, 0011, 0011, ...$$ - For $x=0$, starting from $0000$: $$0000, 1000, 1100, 0110, 0011, 1001, 0100, 0010, 0001, 1000, ...$$