13 votes

The below figure shows four D-type flip-flops connected as a shift register using a $XOR$ gate. The initial state and three subsequent states for three clock pulses are also given.

State | $Q_{A}$ | $Q_{B}$ | $Q_{C}$ | $Q_{D}$ |

Initial | $1$ | $1$ | $1$ | $1$ |

After the first clock | $0$ | $1$ | $1$ | $1$ |

After the second clock | $0$ | $0$ | $1$ | $1$ |

After the third clock | $0$ | $0$ | $0$ | $1$ |

The state $Q_{A} Q_{B} Q_{C} Q_{D}$ after the fourth clock pulse is

- $0000$
- $1111$
- $1001$
- $1000$

11 votes

Best answer

Option (D) $\mathbf{1000}$

$Q_{AN} =Q_{C} \bigoplus Q_{D}, \ Q_{ABN} = Q_{A}, Q_{CN}=Q_{B} \ \text{and} \ Q_{DN}=Q_{C}$

$$\begin{array}{|c|c|c|c|} \hline \bf{Q_A} & \bf {Q_B}& \bf {Q_C} & \bf{Q_D }\\\hline1&1&1&1\\ 0&1&1&1\\0&0&1&1\\ 0&0&0&1 \\ \mathbf{1}&\mathbf{0} &\mathbf{0}&\mathbf{0}\\\hline\end{array}$$

$Q_{AN} =Q_{C} \bigoplus Q_{D}, \ Q_{ABN} = Q_{A}, Q_{CN}=Q_{B} \ \text{and} \ Q_{DN}=Q_{C}$

$$\begin{array}{|c|c|c|c|} \hline \bf{Q_A} & \bf {Q_B}& \bf {Q_C} & \bf{Q_D }\\\hline1&1&1&1\\ 0&1&1&1\\0&0&1&1\\ 0&0&0&1 \\ \mathbf{1}&\mathbf{0} &\mathbf{0}&\mathbf{0}\\\hline\end{array}$$

1 vote

All have D flip-flops:

Q_{B}, Q_{C} and Q_{D} are depending on the previous states of Q_{A}, Q_{B} and Q_{C} respectively.

only Q_{A} is depending on previous XOR of Q_{C} and Q_{D}

So, the next state:

Q_{A} = XOR (prev Q_{C} and prev Q_{D})= 1

Q_{B} = prev of Q_{A} = 0

Q_{C} = prev of Q_{B} = 0

Q_{D} = prev of Q_{C} = 0

**Ans: 1000**

0 votes

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \bf{CP} & \bf {Q_A}& \bf {Q_B} & \bf{Q_C }& \bf{Q_D }& \bf{Q_A^+=Q_C\oplus Q_D }& \bf{Q_B^+=Q_A }& \bf{Q_C^+=Q_B }& \bf{Q_D^+=Q_C }& \bf{CP}\\\hline0&1&1&1&1&0&1&1&1&1\\ 1&0&1&1&1&0&0&1&1&2\\ 2&0&0&1&1&0&0&0&1&3 \\ 3&0&0&0&1&\mathbf{1}&\mathbf{0}&\mathbf{0}&\mathbf{0}&\mathbf{4}\\\hline\end{array}$$

**Correct Answer (D)**