GATE | GATE-CS-2014-(Set-3) | Question 65
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 |
|
|
|
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|
|
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|
So ans is (C) part.
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Last Updated :
29 Sep, 2021
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