Let ݇\(k = 2^n\). A circuit is built by giving the output of an ݊\(n\)-bit…
2014
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
- A.
\(k\)݇-bit binary up counter. - B.
\(k\)݇-bit binary down counter. - C.
\(k\)݇-bit ring counter. - D.
\(k\)݇-bit Johnson counter.
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Correct answer: C
Key insight: the n-bit counter cycles through 2^n binary values and the decoder converts each binary value into a single active output (one-hot).
The n-bit binary counter generates the sequence of binary numbers 0 to 2^n - 1.
The n-to-2^n decoder maps each binary number to exactly one of k = 2^n outputs being high (one-hot encoding).
As the counter increments, the single high output moves through the k outputs in a repeating sequence, which is the defining behaviour of a k-bit ring counter.
Short elimination of other choices:
A binary up or down counter outputs n-bit binary codes, not a one-hot 2^n-line sequence.
A Johnson counter produces 2n states with characteristic multi-bit patterns, not the 2^n one-hot rotation produced here.
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