Ask Question
29 May, 13:32

A three-line-to-eight line decoder has inputs cba where the a input is the lsb (least significant bit). the output and gates are labeled q0 through q7. one would normally expect the output of and gate q4 to be active when the inputs are:

+3
Answers (1)
  1. 29 May, 16:14
    0
    a = false b = false c = true Since the outputs are labeled q0 through q7, it's logical to assume that when the value represented by cba is 0, for q0 to be enabled, when cba represents 1, then q1 to be enabled, all the way until cba represents 7 which causes q7 to be enabled. Since q4 is enabled, that would mean that cba has to represent the value 4. And since this is binary and "a" is the lsb, it would be logical to assume that c is the msb (Most significant bit) so a has the value of 1 b has the value of 2 c has the value of 4 And we need to determine some combination the a, b, c that adds up to 4. And that's obviously where c is true (value 4) and both b and a are false (values of 0). So a = false b = false c = true.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A three-line-to-eight line decoder has inputs cba where the a input is the lsb (least significant bit). the output and gates are labeled q0 ...” in 📘 Business if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers