110 views Assume the initial values of $K0$, $Q_0$ and $Q_1$ to be $1$.
Which of the following state transition tables correctly correspond to the circuit given above? (Note: $Q_{kN}$ and $Q_{kN+1}$ represent current and next state respctively)

1. $\begin{array}{|c|c|c|c|} \hline Q_{1N} & Q_{0N} & Q_{1N+1} & Q_{0N+1} \\ \hline 1 & 1 & 1 & 0 \\ \hline 1 & 0 & 0 & 0 \\ \hline 0 & 0 & 1 & 1 \\ \hline \end{array}$
2. $\begin{array}{|c|c|c|c|} \hline Q_{1N} & Q_{0N} & Q_{1N+1} & Q_{0N+1} \\ \hline 1 & 1 & 0 & 1 \\ \hline 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 1 \\ \hline \end{array}$
3. $\begin{array}{|c|c|c|c|} \hline Q_{1N} & Q_{0N} & Q_{1N+1} & Q_{0N+1} \\ \hline 1 & 1 & 1 & 0 \\ \hline 1 & 0 & 0 & 1 \\ \hline 0 & 1 & 1 & 1 \\ \hline \end{array}$
4. $\begin{array}{|c|c|c|c|} \hline Q_{1N} & Q_{0N} & Q_{1N+1} & Q_{0N+1} \\ \hline 1 & 1 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 \\ \hline 0 & 1 & 1 & 1 \\ \hline \end{array}$

in Others
edited | 110 views
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Here K0= 1,it is not mentioned.
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i am getting a is this correct
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Q0 = Q1'Q0'+K0'Q0'

Q1 = Q0'Q1'+Q0Q1
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@sandygate

yes, A is correct by Junior (977 points)