in ripple counter, to work properly, T_{clock} is in prior adjusted as below given eqn

T_{clock} >= N*T_{flip}_{_flop}

therefore entire input frequency will be frequency of input clock

therefore input frequency = $\frac{1}{C}$Ghz

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For a N-stage ripple carry counter which uses flipflops of propagation delay = P ns and clock period of C ns. Give the expression of the frequency of input signal?

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in ripple counter, to work properly, T_{clock} is in prior adjusted as below given eqn

T_{clock} >= N*T_{flip}_{_flop}

therefore entire input frequency will be frequency of input clock

therefore input frequency = $\frac{1}{C}$Ghz

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