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Consider the circuit given below with initial state $Q_0=1, Q_1=Q_2=0$. The state of the circuit is given by the value $4Q_2+2Q_1+Q_0$ Which one of the following is correct state sequence of the circuit?

1. $1, 3, 4, 6, 7, 5, 2$
2. $1, 2, 5, 3, 7, 6, 4$
3. $1, 2, 7, 3, 5, 6, 4$
4. $1, 6, 5, 7, 2, 3, 4$

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$$\begin{array}{|c|c|c|} \hline \mathbf{Q_0 = Q_{1prev}\oplus Q_{2prev}}& \mathbf{Q_1 = Q_{0prev}} &\mathbf {Q_2 = Q_{1prev}} \\\hline 1& 0& 0 \\ 0&1&0\\ 1& 0 & 1\\ 1&1 & 0\\ 1 & 1 & 1 \\0&1&1\\ 0&0&1 \\ 1&0&0\\\hline \end{array}$$State $= 4Q_2+ 2Q_1+Q_0$
So, state sequence $= 1, 2, 5, 3, 7, 6, 4$

Correct Answer: $B$
by

How to get this table ??
Table is from the figure. I have now added the formula for each column.
Thnx .
what is the significance of state eq. ?? in this question ??
what does state equation represent ?

I got the table, but didn't get this 4Q2 + 2Q1+Q0.

it is decimal equivalent of binary state.

sir i didnt understand how u apply 4Q2 + 2Q1+Q0 to column please explain .....till the table i understood .....

$Q_0Q_1Q_2 = 100,\;\; \text{state =}\;4\times 0+2\times 0+ 1 = 1$
It represents the initial state...I think
4Q2 + 2Q1+Q0 WHT IS THE USE OF THIS
it represents the lsb as q0 and msb as q2 , no other significance
Use of equation is to distinguish between LSB and MSB.
4Q2 + 2Q1+Q0 is just to tell us which is MSB and which is LSB.

I hope it helps.
@everyone from which topic this question belongs too.

Jeet,  it is counter in digital logic, in prerequisites, you must know Flip-flops, behavior, characteristic equations and excitation table.
Thanks a lot, Sir.
$Q_{2N} = D_{2} = Q_{1}$

$Q_{1N} = D_{1}=Q_{0}$

$Q_{0N}=D_{0}=Q_{1}\oplus Q_{2}$

and $4Q_{2}+2Q_{1}+Q_{0}$, just differentiate the $\textbf{MSB}$ and $\textbf{LSB}$ bit.
I read somewhere "In case of synchronous counters choosing msb and lsb does not matter , it can be from any side because the operation is done in parallel per clock pulse but in case of asynchronous counter it matters because it is done serially ." is this statement correct? it doesn't work for this problem if you change MSB and LSB.

Q0n = Q1n-1 exor Q2n-1

Q1n = Q0n-1

Q2n = Q1n-1

by option (B)

How did u get that present states of q0,q1,q2
3 bits maximum 8 possible values (2^3) and initially I started with the given condition in question.

The diagram of FF is: According to figure, we can write next state as:

$Q_{0N}=Q_1 \oplus Q_2, Q_{1N}=Q_0,Q_{2N}=Q_1$

Transition table can be given as:

Q2 Q1 Q0 Q2N Q1N Q0N
0 0 0 0 0 0
0 0 1 0 1 0
0 1 0 1 0 1
0 1 1 1 1 1
1 0 0 0 0 1
1 0 1 0 1 1
1 1 0 1 0 0
1 1 1 1 1 0

$4Q_2 + 2Q_1+Q_0$ is simply the decimal equivalent of state. Now, transition diagram can be drawn from above table as: Correct State Sequence: 1,2,5,3,7,6,4.

1. intial value of Q0,Q1,Q2 is 1 0 0  and clk is common for all flip flop b is the correct answer for this.