Last Updated :
19 Nov, 2018
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The above sequential circuit is built using JK flip-flops is initialized with Q2Q1Q0 = 000. The state sequence for this circuit for the next 3 clock cycle is
(A) 001, 010, 011
(B) 111, 110, 101
(C) 100, 110, 111
(D) 100, 011, 001
Answer: (C)
Explanation: JK ff truth table—
| j |
k |
Q |
| 0 |
0 |
Q0 |
| 1 |
0 |
1 |
| 0 |
1 |
0 |
| 1 |
1 |
Q0’ |
Initially Q2Q1Q0=000
Present state FF input Next state
| Q2 |
Q1 |
Q0 |
J2 |
K2 |
J1 |
K1 |
J0 |
K0 |
Q2 |
Q1 |
Q0 |
| 0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
| 1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
| 1 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
| |
|
|
|
|
|
|
|
|
|
|
|
So ans is ( C) part.
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