MadeEasy Workbook: Digital Logic - Digital Counter

Question

How to solve this? 

Comments

As it is saying counters are cleared then present state Q0=0 Q1=0 Q2=0. Now it is pass through a full adder and sum is connected with J. So J=$Q0\oplus Q1\oplus Q2$ and k=Q0*Q1+Q1*Q2+Q2*Q0.

So J=0 and K=0 so J-K flip flop will Hold the current state.

A should be correct.

Answer 1

I didn't completely understand the meaning of the question , but i'll still try to answer it.

I assume it to be a 3 bit binary up counter , because nothing's been specified.

Now it has been said that just before the counter is cleared. That is just before outputs show 000 , the last state would be , as i assume , 111.

Now 111 is passed through a full adder.

sum = 1 xor 1 xor 1 = 1.

and , carry  = 1.

So J = 1 ,and K=1;

Qnext in JK flip flop is,

Qn = JQ'+K'Q = 1.Q' + 0.Q = Q'

Thus , it can be seen that the JK flip flop is in toggle state.

I might be wrong , kindly correct me if I went wrong somewhere.