@Arjun sir, what if about a and b equal to 1? S1 does not have any transitions defined for b=1 and S2 does not have any transitions defined for a=1.

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+11 votes

If the state machine described in figure should have a stable state, the restriction on the inputs is given by

- $a.b=1$
- $a+b=1$
- $\bar{a} + \bar{b} =0$
- $\overline{a.b}=1$
- $\overline{a+b} =1$

+6 votes

Best answer

If $a = 0$ state changes from $S_1$ to $S_2$ and if $b = 0$ state changes from $S_2$ to $S_1.$

So, $a = 0, b = 0$ is surely not a stable state as then the states will be oscillating. So, the condition for stability is that both $a$ and $b$ should not be $0$ together which is given by $a+b = 1$ or $\overline {ab} = 0.$

Options A and C are equivalent and both ensures stable states albeit by enforcing stricter than required conditions.

Correct Answer: Option B.

So, $a = 0, b = 0$ is surely not a stable state as then the states will be oscillating. So, the condition for stability is that both $a$ and $b$ should not be $0$ together which is given by $a+b = 1$ or $\overline {ab} = 0.$

Options A and C are equivalent and both ensures stable states albeit by enforcing stricter than required conditions.

Correct Answer: Option B.

+1 vote

+1

@Swapnil Naik stable state means after certain time(after all delays) it gets stable and doesn't change. For above diagram to make sure output does not change, at S1 we should apply a=1 and at S2 we should apply b=1. Any other input will make our output fluctuate between S1 and S2.

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