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 |
So ans is (C) part.
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