Question

An n-to-m line decoder is used to generate (Note: "n-to-m line decoder" where \(m = 2^n\))

• A. \( 2^{n-1} \) min terms
• B. \( 2^n \) min terms
• C. \( 2^{n+1} \) min terms
• D. \( 2^{n-1} \) max terms

Hint:

A decoder with \(n\) inputs has \(2^n\) outputs.
Each output corresponds to one unique combination of the \(n\) inputs.
These outputs are typically the \(2^n\) minterms of the \(n\) input variables.

Answer 1

The Correct Option is B

Decoder Analysis 

This document describes the functionality of an n-to-m line decoder.

Topic	Description
Definition	An n-to-m line decoder, where \(m = 2^n\), takes an n-bit binary input and activates exactly one of its \(m = 2^n\) output lines for each unique input combination.
Output Lines	Each output line of the decoder corresponds to one of the \(2^n\) possible minterms of the n input variables.
Example: 2-to-4 Decoder	A 2-to-4 line decoder has 2 input lines (say A, B) and \(2^2 = 4\) output lines. The outputs could be: \(Y_0 = \overline{A}\overline{B}\) (minterm \(m_0\)) \(Y_1 = \overline{A}B\) (minterm \(m_1\)) \(Y_2 = A\overline{B}\) (minterm \(m_2\)) \(Y_3 = AB\) (minterm \(m_3\))
Minterm Generation	An n-to-\(2^n\) line decoder is used to generate \(2^n\) minterms (or maxterms if designed with inverted outputs or different logic, but typically minterms).
Conclusion	Given the context is about minterms, it generates \(2^n\) minterms.  Result: \(\boxed{2^n \text{ min terms}}\)