In a 4-to-2 priority encoder, if both D1 and D3 are HIGH, which input determines the output and why?
D3 determines the output because it has higher priority
A priority encoder differs from an ordinary encoder in one crucial way: it is designed to handle the situation where more than one input is active at the same time. Instead of producing an invalid output, it applies a fixed priority ranking and encodes only the highest-priority active input, ignoring all lower ones.
In a 4-to-2 priority encoder with inputs D0, D1, D2, D3, the convention is that the higher-numbered input has higher priority:
With D1 = 1 and D3 = 1 simultaneously, the encoder compares their priorities. D3 outranks D1, so D3 determines the output, and the 2-bit output is the binary code for 3, i.e. 11.
This is exactly why the output is not undefined for multiple active inputs — resolving that ambiguity is the entire purpose of a priority encoder. The idea that both D1 and D3 set the output together is impossible, because the two-bit output can represent only one input number at a time. And D1 determining the output because it is the lower number reverses the rule: it is the higher-numbered, higher-priority input D3, not the lower one, that wins.
Which of the following represents the borrow in a half subtraction?
Carry input is present in which of the following?
The number of control lines for 8-to-1 multiplexer is:
X/Y (where X represents the input code and Y represents the output code) as a qualifying symbol does NOT relate to:
Consider 4-bit gray code representation of numbers. Let h3 h2 h1 h0 be the gray code representation of a number n and g1g2g3g0 be the gray code representation of the number (n + 1) modulo 16. Which one of the following functions is correct?