Question

A 4-bit priority encoder has inputs \( D_3, D_2, D_1, \) and \( D_0 \) in descending order of priority. The two-bit output \( AB \) is generated as 00, 01, 10, and 11 corresponding to inputs \( D_3, D_2, D_1, \) and \( D_0 \), respectively. The Boolean expression of the output bit \( B \) is \(\_\_\_\_\).

• A. \( \overline{D_3} D_2 \)
• B. \( \overline{D_3} D_2 + \overline{D_3} D_1 \)
• C. \( D_3 \overline{D_2} + \overline{D_3} D_1 \)
• D. \( \overline{D_3} \overline{D_1} \)

Hint:

To derive the output of a priority encoder, ensure higher-priority inputs are inactive when calculating the contribution of lower-priority inputs.

Answer 1

The Correct Option is B

Step 1: Understand the operation of the priority encoder.
A priority encoder generates a binary output based on the highest-priority active input. For this 4-bit encoder, the priority order is: \[ D_3>D_2>D_1>D_0. \] Step 2: Analyze when \(B\) is active.
The bit \(B\) represents part of the binary output and becomes high under these conditions: \(B = 1\) if \(D_2 = 1\), provided \(D_3 = 0\). \(B = 1\) if \(D_1 = 1\), provided \(D_3 = 0\). Step 3: Derive the Boolean expression for \(B\).
The above conditions can be expressed using Boolean logic: \[ B = \overline{D_3} D_2 + \overline{D_3} D_1. \] Here: \(\overline{D_3} D_2\): \(D_2\) is active, and \(D_3\) is inactive. \(\overline{D_3} D_1\): \(D_1\) is active, and \(D_3\) is inactive. Thus, the correct Boolean expression for \(B\) is: \[ \overline{D_3} D_2 + \overline{D_3} D_1. \] Final Answer: \[\boxed{\overline{D_3} D_2 + \overline{D_3} D_1}\]