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For a state machine with the following state diagram the expression for the next state $S^+$ in terms of the current state $S$ and the input variables $x$ and $y$ is

  1. $S^+ = S' . y' + S . x$
  2. $S^+ = S. x . y' + S' . y . x'$
  3. $S^+ = x . y'$
  4. $S^+ = S' . y + S . x'$
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4 Answers

Best answer
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86 votes

$$\begin{array}{|c|cc|c|} \hline \textbf{Present State}  &&{\textbf {Inputs}}&     \textbf {Next State} \\\hline \textbf{S} & \textbf{x}  & \textbf{y} & \mathbf{S^+} \\\hline 0 & \text{X} & 0&1  \\\hline  0& \text{X} & 1 & 0  \\\hline  1& 0& \text{X}  & 0 \\\hline 1& 1 & \text{X} &1\\\hline \end{array}$$

From above table:

$S^+ = S'y'+Sx$
Correct Answer: $A$
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Answer is (A)

For S is 1 only when:      Either (S=1 and x=1) OR (s=0 and y=0)

Therefor  S(next) = S'y'+Sx

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For those who think A & D both should be the answer. First of all, this is a Digital Logic Design question because they have given options in term of SOP form and we write expressions only for the result obtained =1 in Truth table.
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Deriving SOP form is a good approach, but it didn’t clicked for me. I tried this approach. It’s basically elimination method.

Give inputs as S=0 and y=0 to all options, for these inputs S+ should be 1.

(A) S+ = 1

(B) S+ = 0 , eliminated.

(C) S+ = x.(1) = x, here the output depends on x but as (S = 0) state don’t have production for input ‘x’ , that’s why it’s eliminated too.

(D) S+ = 0 , eliminated.

So, option (A) is correct.
Answer:

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