GATE CSE 1987 | Question: 13-a

Question

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

$$\begin{array}{|l|c|c|c|c|} \hline \text{State} & Q_{A} & Q_{B} & Q_{C} & Q_{D} \\\hline \text{Initial} & 1 & 1 & 1 & 1 \\ \text{After the first clock} & 0 & 1 & 1 & 1 \\ \text{After the second clock} & 0 & 0 & 1 & 1 \\ \text{After the third clock} & 0 & 0 & 0 & 1 \\\hline  \end{array}$$The state $Q_{A} Q_{B} Q_{C} Q_{D}$ after the fourth clock pulse is

• A. $0000$
• B. $1111$
• C. $1001$
• D. $1000$

Answer 1

Option (D) $\mathbf{1000}$
$Q_{An} =Q_{C} \oplus  Q_{D}, \ Q_{Bn} = Q_{A}, Q_{Cn}=Q_{B} \ \text{and} \ Q_{Dn}=Q_{C}$
$$\begin{array}{|c|c|c|c|} \hline \bf{Q_A} & \bf {Q_B}& \bf {Q_C} &  \bf{Q_D }\\\hline1&1&1&1\\ 0&1&1&1\\0&0&1&1\\ 0&0&0&1 \\\hline \mathbf{1}&\mathbf{0} &\mathbf{0}&\mathbf{0}\\\hline\end{array}$$

Comments

Bro here after 4 the clock pulse it should be 0000 na because in the question he gave that table after 1 clock cycle and so on so why did u take here 5 clock output plz clear

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Bro, u can see this...

Answer 2

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

Answer 3

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

Correct Answer (D)

Answer 4

A classic example of Linear Feedback Shift Register, Where we generate random numbers on diff clock cycles