in Digital Logic
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46 votes
46 votes

Let $k=2^n$. A circuit is built by giving the output of an $n$-bit binary counter as input to an $n\text{-to-}2^n$ bit decoder. This circuit is equivalent to a 

  1. $k$-bit binary up counter. 
  2. $k$-bit binary down counter.
  3. $k$--bit ring counter.
  4. $k$-bit Johnson counter.
in Digital Logic
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4 Comments

I think it's D)option .We make johnson counter with such combination. This is from Morris Mano

@Ashwin+Kulkarni    @joshi_nitish

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@Shivam,

in ring counter(modn), only one bit is high(1) at every count out of n bits and here also due to decoder, only one bit is high at every count out of k bits.

and we can sum up like this -> "the above circuit is counting k distinct states and at each state only one bit is high out of k bits".......this is property of ring counter

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Thanks @joshi_nitish

I got it.
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Can anyone plz explain me what is meant by k bit up counter and k bit down counter and in which case we could have got those  as answers ?

 Kindly help
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5 Answers

39 votes
39 votes
Best answer
Binary counter of $n$ bits can count up to $2^n$ numbers. When this output from counter is fed as input (n bit) to decoder one out of $2^n$ output lines will be activated. So, this arrangement of counter and decoder is behaving as $2^n$ or $k\text{-bit}$ ring counter.

Correct Answer: $C$
edited by

4 Comments

What is meant by

  1. kk-bit binary up counter. 
  2. kk-bit binary down counter   . Can someone plz explain with an example.
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Why not binary  Down counter?
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edited by

@Shiva Chaitanya Gaju If there are 2 bits then ring counter will be able to count only 2  states right how can it count 2power n states to count 2 power n states it should be a ripple counter right

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11 votes
11 votes

The $n$-bit binary counter takes in $n$ bits, and can output one of $2^n$ possible states.

Out of these $2^n$ states, $n$ states are taken in by the decoder, which outputs one of the $2^n$ possible states.

 

We're mapping $2^n$ states to $2^n$ states.

=> $k$ states to $k$ states.

This is the property of ring counters.

 

Option C

2 Comments

how k states to k states, initial input in n-bit(for binary counter) , final output is 2^n-bit(decoder),

so we have mapped n states to 2^n states.
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Initial input n is to decoder but the option says k bit counter and k = 2^n . Decoder will give 2^n unique outputs and a 2^n ring counter too will give 2^n counts, hence the answer.
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2 votes
2 votes

A circuit is built by giving the output of an $n-bit$ binary counter as input to an $n-to-2^n$ bit decoder.

Output of $n-bit$ binary counter gives $2^n$ output. Number of variable in this $2^n$ outputs is $n$. Decoder generates minterms of the input, for $n$ input there are $2^n$ minterms $(0, 1, ..., 2^{n-1})$ which are nothing but states. Now, since output of binary counter generates $2^n$ outputs in sequence, the outputs of decoder are also in sequence. So, these $2^{n}$ states are minterms & are in sequence which is the output of $2^n$ bit ring counter. 

1 vote
1 vote
For output of a decoder , only single output will be ‘1’ and remaining will be ‘0’ at the same time. So high output  will give the count of the ring counter. Hence Ans is ( C) part.

1 comment

if suppose assuming that somehow the arrangement was in a way that low output gives answer then it would be johnsons counter?
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Answer:

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