Question:

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

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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.
Updated On: May 22, 2025
  • \( 2^{n-1} \) min terms
  • \( 2^n \) min terms
  • \( 2^{n+1} \) min terms
  • \( 2^{n-1} \) max terms
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The Correct Option is B

Solution and Explanation

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. Each output line of the decoder corresponds to one of the \(2^n\) possible minterms of the n input variables. For example, 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 \(m_3\)) So, 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). Given the options are about minterms, it generates \(2^n\) minterms. \[ \boxed{2^n \text{ min terms}} \]
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