ACE Basic test

Question

Test Digital logic ques 15

Answer 1

We know for a n bit Ring counter , mod value is n itself..So input frequency divides with n itself..

So after going through 1st counter  , we have , output frequency  =  Input / mod value

                                                                                              =  Input / 25

                                                                                              =  (2 * 10⁶) / 25

                                                                                              =  80000 Hz

Now mod value of n bit ripple counter      =     2ⁿ

So using 4 bit counter , we get mod value of 16..

Hence output frequency going through this counter , we have     =  80000 / 16

                                                                                             =  5000 Hz

Now we have mod 25 counter so input frequency to it divides by 25 ,

Hence output frequency       =      5000 / 25    =    200 Hz

Now we have Johnson counter finally , so mod value of n bit Johnson counter is 2n..

Hence output frequency       =      200 /  (2 * 4)

                                          =      25 Hz

Hence final output frequency   =   25 Hz

Answer 2

2000000/(25*16*25*8)=25 Hz