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2 Answers

Best answer
15 votes
15 votes
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.
3 votes
3 votes

Let’s take $S_1$ as $0$ and $S_2$ as 1

we can draw the truth table as follows:-

Characteristic Table
a b $Q_n$ $Q_{n+1}$
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1

 

The states in $Q_n$ column of the above table are $0$ which represent $S_1$ and $1$ which represent $S_2$.

From the above table we can find the following table :-

Derived table
a b $Q_{n+1}$
0 0 $\overline{Q}$
0 1 1
1 0 0
1 1 $Q$

For stable state following is the equation:-

                 (if in $S_1$ stay in S1) $\lor$ (if in $S_2$ stay in $S_2$ )

$\implies$ ((if a=1 don’t care about b just remain in $S_1$ ) $\lor$ (a=1 and b=1)) $\lor$ ((if b=1 don’t care about a just                    remain in $S_2$ ) $\lor$ (a=1 and b=1))

$\implies$ (a+a.b)+(b+a.b)

$\implies$ (a(1+b))+(b(a+1))

$\implies$ a+b

so, answer is $(b)\ a+b=1$

Answer:

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