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Best answer

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

$Q=aQ_{n-1}+ab+bQ_{n-1}$

Hence,

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

$00 \implies Q_{n-1}(0+0) + 0.0 = Q_{n-1}(0) + 0 = 0+0 = 0$

$01 \implies Q_{n-1}(0+1) + 0.1= Q_{n-1} (1)+ 0 = Q_{n-1}+0 = Q_{n-1}$

$10 \implies Q_{n-1}(1+0) + 1.0 = Q_{n-1} (1) + 0 = Q_{n-1}+0= Q_{n-1}$

$11 \implies Q_{n-1}(1+1)+ 1.1 =Q_{n-1}(1) + 1 =Q_{n-1}+1 = 1$

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 $= T_{\text{CLK}} \geq T_{\text{flip-flop}} +T_{\text{ AND gates}}$

$\qquad \qquad = 10ns+(10+10+10)ns = 40ns$

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

So, maximum clock frequency at which the counter can operate is $25 \text{ MHz}.$

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