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Consider the synchronous sequential circuit in the below figure

Given that the initial state of the circuit is S$_4$, identify the set of states, which are not reachable.

in Digital Logic by Veteran (105k points) | 515 views

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  • $Q_{3N} = D_3 \Rightarrow Q_{3N} = Q_2$
  • $Q_{2N} = D_2 \Rightarrow Q_{2N} = Q_1$
  • $Q_{1N} = D_1 \Rightarrow Q_{1N} = Q_3 \oplus Q_2$

$$ \begin{array}{|ccc|ccc|} \hline \textbf{$Q_1$} & \textbf {$Q_2$} &\textbf {$Q_3$} & \textbf {$Q_{1N}$} & \textbf{$Q_{2N}$} & \textbf{$Q_{3N}$}\\\hline0&0&0&0&0&0 \\\hline 0&0&1&1 &0 &0\\\hline 0&1&0&1&0&1\\\hline0&1&1&0&0&1 \\\hline1&0&0&0&1&0 \\\hline 1&0&1&1 &1 &0\\\hline 1&1&0&1&1&1\\\hline1&1&1&0&1&1 \\\hline \end{array}$$Given that the initial state   $=S_4 = 100.$

Unreachable state is $S_0$

So, set of states which are not reachable $= \{S_{0}\}$

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How did you get the sequence of states as in the diagram?
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See the table. From there you can easily get the sequence of states.

For each (Q1, Q2, Q3) values, there is a state. The transition happens on seeing the(Q1, Q2, Q3) and to which that value is matched to (Q1N, Q2N, Q3N).

Like from (0,0,0) we go to (0,0,0).

From (1,0,0) we go to (0,1,0)
0
Understood. Thanks

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