19,588 views
52 votes
52 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.

6 Answers

Best answer
42 votes
42 votes
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
12 votes
12 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 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 votes
1 votes
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.
Answer:

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