edited by
1,282 views
0 votes
0 votes

The following sequential circuit has initial state $QAQB = 00$ with one input $X$ and one output $Z.$

What is the minimum input sequence which takes the machine to state $11?$

$(A) 00$

$(B) 10$

$(C) 11$

$(D)$ State $11$ is not reachable

edited by

1 Answer

Best answer
2 votes
2 votes

Here actually u have to go through the circuit diagram and through options.Basically it is asking the first 2 input sequence which would result in transition to state 11 which is 00 initially..

So if we take input = 0 initially..Hence x = 0 initially..

Now if u just go through the circuit we will get Ja  =  1  and Ka  =  Qb  = 0

Hence next state of A flip flop which is a JK flip flop  = 1 

Similarly  Jb    =   0   and   Kb  =  1 

Hence next state of B flip flop will be  0..

Hence after 1st cycle  we get   Qa Qb  =  1 0

Hence these inputs along with x = 0 once again (if we go with option A) , on applying similar to the procedure as above will lead us to have  Qa Qb  =  1 1

The key thing is minimum input is being asked..Hence always preferable to start with 00 i.e. in first cycle we applied x = 0 then in 2nd cycle we have applied x = 0 here as well..

Hence A) is the correct answer

selected by

Related questions

0 votes
0 votes
0 answers
3
Harsh Kumar asked Sep 16, 2018
741 views
If there are total 16 different states in a system and there is one external input, then how many flip flops are required? Is it 3 or 4 ?Thank you.
0 votes
0 votes
1 answer
4
Sambhrant Maurya asked Jan 6, 2019
451 views
The answer I got is 3 but it’s given as 5 and in the solution they are mentioning that the no. of flip flops here are defined by the sequence not the no. of states??