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.
Carry input is present in which of the following?
The time delay in a look-ahead carry adder is independent of
The characteristics of the combinational circuits are :
A. Output at any time is function of inputs at that time
B. Contains memory elements
C. Do not have feedback paths
D. Clock is used to trigger the circuits to obtain outputs
Choose the correct answer from the options given below :
Match the terms in List - I with the options given in List - II :
List - I | List – II | ||
(a) | Decoder | (i) | 1 line to 2 nlines |
(b) | Multiplexer | (ii) | n lines to 2 nlines |
(c) | De multiplexer | (iii) | 2 nlines to 1 line |
(iv) | 2 nlines to 2 n−1 lines |
Which of the following represents the borrow in a half subtraction?