1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Digital Design - decoders and multiplexer help

  1. Jun 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Let f(x3; x2; x1; x0) = (y1; y0) such that y1 = 1 if the number of 1's in x3x2x1x0 is even, 0 otherwise, and, output y0 = 1 if the number of 1's in x3x2x1x0 is odd, 0 otherwise.

    1. Implement f with four 2-to-4 decoders.
    2. Implement f with a 4-to-1 multiplexer with selection inputs x0; x1 (in this order)


    2. Relevant equations



    3. The attempt at a solution

    How do I even start solving this? This was a bonus as it was not really taught to me so I have to learn myself.

    (IN THE CASE OF 0, 0, 0, 0, both y1 and y0 are 0 right? because there are no 1's at all)

    I begin by making a truth table of:
    Code (Text):

    x3 x2 x1 x0   y1 y0
     0  0  0  0    0  0
     0  0  0  1    0  1
     0  0  1  0    0  1
     0  0  1  1    1  0
     0  1  0  0    0  1
     0  1  0  1    1  0
     0  1  1  0    1  0
     0  1  1  1    0  1
     1  0  0  0    0  1
     1  0  0  1    1  0
     1  0  1  0    1  0
     1  0  1  1    0  1
     1  1  0  0    1  0
     1  1  0  1    0  1
     1  1  1  0    0  1
     1  1  1  1    1  0
    Where do I go from here?
     
    Last edited by a moderator: Jun 18, 2013
  2. jcsd
  3. Jun 18, 2013 #2

    lewando

    User Avatar
    Gold Member

    Zero is an even number.

    If x3..x0 | y1,y0 = 0,0,0,0 | 0,0 then y1 would indicate: "number of ones is not even" y0 would indicate "number of ones is not odd" --a contradiction.
     
  4. Jun 18, 2013 #3
    So that would mean the output of y1 and y0 in my case is 1, 0 for 0, 0 ,0 0 in my case because the number of 1's is even and not odd?
     
  5. Jun 18, 2013 #4

    lewando

    User Avatar
    Gold Member

    Yes. And it should make your job somewhat easier as well.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Digital Design - decoders and multiplexer help
  1. Digital Design (Replies: 2)

Loading...