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Best answer

(a) The output of the circuit given as :-

Q_{n}=aQ_{n-1}+ab+bQ_{n-1}

Hence,

Q_{n}=Q_{n-1}(a+b)+ab

00 => Q_{n-1}(0+0) + 0.0 = Q_{n-1} (0) + 0 = 0+0 = 0

01 => Q_{n-1}(0+1) + 0.1 = Q_{n-1} (1) + 0 = Q_{n-1}+0 = Q_{n-1}

10 => Q_{n-1}(1+0) + 1.0 = Q_{n-1} (1) + 0 = Q_{n-1}+0 = Q_{n-1}

11 => Q_{n-1}(1+1) + 1.1 = Q_{n-1} (1) + 1 = Q_{n-1}+1 = 1

a | b | Qn |
---|---|---|

0 | 0 | 0 |

0 | 1 | Q_{n-1} |

1 | 0 | Q_{n-1} |

1 | 1 | 1 |

(c)

All the flip flops are operated by same clock , together all takes one propagation delay .

All the AND gates consumes one propagation delay individually.

Total propagation delay = T_{CLK} >= T_{ flip-flop} +T _{AND gates}

= 10ns+(10+10+10)ns = 40ns

Maximum clock frequency =1/T_{CLK} = 1/40ns= 10^{9}/40 = 25 MHz

So,Maximum clock frequency at which the counter can operate is 25 MHz

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