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The lowest frequency in kHz, if a 6 MHz clock frequency is applied to a cascaded counter of modulus 2 counter and modulus 3 counter are ________. is it correct ?

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+1
yes given answer is correct .

Yes in case of cascading counter which is nothing but an standard asynchronous counter u can multiply individual mod values to get resultant mod value..

So  let us see how it works as well..Here after the input signal goes through 1st counter which is a mod 2 counter , so f1 = finp / 2

Now this output works as an input signal for 2nd flip flop of counter which is mod 3

So

fout   =  f1 / 3

==>    fout   =  finput / 6

So net result is that output frequency gets divided by 6 w.r.t input frequency.

Thus this is nothing but the mod value of counter..

Hence output frequency here =  6 MHz / 6

=  1 MHz

As a shortcut , if u r given 2 counters each of mod value m and n then frequency gets divided by mn as a whole..

by Veteran (102k points)
edited
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just look at the 2nd part of  that question is it similiar to that question ?

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Will there be multiple output signals corresponding to same frequency when they come out of the first counter?
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Can you provide any link from where I can understand all these? I mean not any coaching material with formulas, but some ppts of professors or any assignment given in iits or better universities.
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rishabh gupta refer to neso academy videos they r the best on youtube digital logic
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what happens when  synchronus counter is used