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 $(b)\ a+b=1$