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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.
+2

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

@Ashwin+Kulkarni    @joshi_nitish

+4

@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

0
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

In binary counter of n bits can count upto $2^n$ numbers..when this op from counter  is fed to decoder one of n out of $2^n$ will be activated. So, this arrangement of counter and decoder is behaving as $2^n(k)$ ring counter..
edited
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@Praveen_Saini , @Pooja_Palod

The arrangement mentioned in question will produce op as :

001

010

100

Ring counter op is

100
010
001

How these are similar  ?
+48

let us take a 2 bit binary counter, it gives counting as 00,01,10,11, ...so this output is given as input to 2*4 decoder, where we will be having 4 output lines, which only one will be active at any time, based on the input we get,

input to decoder            output of decoder

00                                   1000

01                                  0100

10                                     0010

11                                   0001

this is same as ring counter counting 4 states,

0
^^ exactly :)
0

@Shiva

Can u explain a little bit please..am not able  to understand that input to decoder 00 how we will get 1000..

Am not able to understand below part only whatever u write aboe in paragraph I understand..

input to decoder            output of decoder

00                                   1000

01                                  0100

10                                     0010

11                                   0001

this is same as ring counter counting 4 states,

+2

@aman

00 is associated with line 1 (P)of output of decoder

similarly 01 : line 2(q)

10 : line 3(r)

11 : line 4(s)

when input to decoder is 00 line 1 will be activated and all other will be deactivated so we can write it as p~q~r~s means p=1 and q=r=s=0 which is 1000

again

for input 01 only line number 2 is activated : so equation will be ~p q ~r ~s = 0100 ..

and same for next two states.

I hope you got it now.

+1
why not johnson counter ?
+2
if it is a jhonson counter then it has to count 2k states ..but actually it is counting k states..
+1
For Johnson Counter, nth bit is complemented before arriving at the 1st bit(counting from MSB to LSB). Here this is not happening.
+1 vote

k--bit ring counter.

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

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.