GATE CS 2014 Set 2 Q.17 (CS Q.7): Sequential Circuits, Counters and Registers

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

$k$ -bit binary up counter.

$k$ -bit binary down counter.

$k$ -bit ring counter.

$k$ -bit Johnson counter.

Official answer and solution

Official answer: C
From: Official GATE 2014 answer key

Solution

Trace the active output line of an n-to-2^n decoder driven by an n-bit binary counter.

An $n$ -bit binary counter counts cyclically through $2^n = k$ distinct binary states:
$0, 1, 2, \dots, k-1, 0, \dots$
An $n$ -to- $2^n$ decoder accepts an $n$ -bit binary input and activates exactly ONE of its $k = 2^n$ output lines at any time (the line matching the input index).
Connecting the $n$ -bit counter outputs to the decoder inputs causes the single active output line to rotate sequentially through positions $0, 1, 2, \dots, k-1$ and wrap back to $0$ :
$\text{Cycle: } 100\dots00 \to 010\dots00 \to 001\dots00 \to \dots \to 000\dots01 \to 100\dots00.$
A register/counter of length $k$ where a single circulating '1' bit shifts through $k$ unique states is by definition a $k$ -bit ring counter (option C).

Final answer

Common trap

Choosing Johnson counter, which produces 2k states using inverted feedback rather than a one-hot circulating token.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 2
Question: 17 of 65, printed as CS Q.7
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source