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33 votes
33 votes

Consider the circuit in the diagram. The $\oplus$ operator represents Ex-OR. The D flip-flops are initialized to zeroes (cleared).

The following data: $100110000$ is supplied to the “data” terminal in nine clock cycles. After that the values of $q_{2}q_{1}q_{0}$ are:

  1. $000$
  2. $001$
  3. $010$
  4. $101$
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7 Answers

Best answer
46 votes
46 votes
$${\begin{array}{|c|c|c|c||}\hline
\textbf{Data}&    \bf{Q_0}&  \bf{Q_1}&\bf{ Q_{2}} \\\hline
1&1&\text{$Q0_{start}$ XOR $Q2_{start}$ = 0 XOR 0 = 0}&\text{$Q1_{start}  = 0$} \\\hline 0&0&\text{1 XOR 0 = 1}&0  \\ \hline    0&0&\text{0 XOR 0 = 0}&1 \\ \hline   1&1&\text{0XOR 1 = 1}&0 \\ \hline   1&1&\text{1 XOR 0 = 1}&1 \\ \hline   0&0&\text{1 XOR 1 = 0}& 1  \\ \hline   0&0& \text{0XOR 1 = 1}& 0 \\ \hline   0&0&\text{0 XOR 0 = 0} &1\\ \hline   0 & 0& \text{0 XOR 1 = 1}&0\\\hline
 \end{array}}$$

So, option C.
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4 votes
4 votes
The question is simple, but just because of delays of XOR gate q1 & q2 are working old datas of q0..  At a particular instance, value present at q0 is from data but q1 is getting from XOR gate, so the value calculated from previous bit. Thats it.
Just follow the data given to q0,
q0 value is the data bit.
q1,q2 will be evaluated from previous state.
C is the ans.
1 votes
1 votes
Clock pulse ↓   state (q2q1q0)↓

Starting                     000

1st clock                   001

2nd clock                 010

3rd clock                  100

4th clock                   011

5th clock                   111

6th clock                   100

7th clock                   010

8th clock                   100

9th clock                   010
Answer:

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