1 vote

Consider a $3$-bit counter, designed using $T$ flip-flops, as shown below:

Assuming the initial state of the counter given by $\text{PQR}$ as $000$, what are the next three states?

- $011,101,000$
- $001,010,111$
- $011,101,111$
- $001,010,000$

2 votes

P | Q | R |
---|---|---|

0 | 0 | 0 |

0 | 1 | 1 |

1 | 0 | 1 |

0 | 0 | 0 |

If P is 0, then Q will toggle in next clock cycle.

If Q is 0, then R will toggle in next clock cycle.

If R is 1, then P will toggle in next clock cycle.

Using these rules, the transition is shown in the table.

**Option A**

2 votes

From the given 3 state T flip flop the next input sequence are as follows:

- $T_P=R$
- $T_Q=\bar P$
- $T_R=\bar Q$

Initial State | Current input | Next State | ||||||

P | Q | R | $T_P$ | $T_Q$ | $T_R$ | $P^+$ | $Q^+$ | $R^+$ |

0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |

1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |

in T flip flop for low input (0)the next state is $Q_n$(current state), for high input(1) it toggle/complement the present state($\bar Q_n$)

$011,101,000$ option A is corret.