Register

GATE CSE 2014 Set 2 | Question: 7

19,715 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.
User Avatar

6 Answers

0 votes
0 votes

Let's consider each options

D)

K or 2^n bit JC will have 2^(n+1) states, but the decoder outputs 1 out of 2^n states of binary counter. So this option is clearly wrong

A,B)

Each of the counters here for k or 2^n bit, will have 2^(2^n) output, but decoder in the circuit has outputs 1 out of 2^n.

So, A,B are also wrong here

Now let's C) Ring counter

K or 2^n bit RC will have K or 2^n states, which reflects the given situation. It's the right option

User Avatar
Page:
Voting & Accepting an answer helps improve the community
Answer:
← PreviousNext →
← Previous in categoryNext in category →

Related questions

33 votes
33 votes
3 answers
6
go_editor asked Sep 28, 2014
9,485 views
GATE CSE 2014 Set 2 | Question: 8
Consider the equation $(123)_5=(x8)_y$ with $x$ and $y$ as unknown. The number of possible solutions is _____ .
Consider the equation $(123)_5=(x8)_y$ with $x$ and $y$ as unknown. The number of possible solutions is _____ .
User Avatar
38 votes
38 votes
4 answers
7
go_editor asked Sep 28, 2014
12,381 views
GATE CSE 2014 Set 2 | Question: 6
The dual of a Boolean function $F(x_1,x_2,\dots,x_n,+, .,')$, written as $F^D$ is the same expression as that of $F$ with $+$ and $⋅$ swapped. $F$ is said to be self-dual if $F = F^D$. The number of self-dual functions with $n$ Boolean variables is $2^n$ $2^{n-1}$ $2^{2^{n}}$ $2^{2^{n-1}}$
The dual of a Boolean function $F(x_1,x_2,\dots,x_n,+, .,')$, written as $F^D$ is the same expression as that of $F$ with $+$ and $⋅$ swapped. $F$ is said to be self-du...
User Avatar
Link Copied!