1. Not finding help here? Sign up for a free 30min 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!

Decoder output logic

  1. Sep 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Given that the variables A[2:0] form a three bit grey code to represent numerical values:

    i) Write down all binary combinations for A[2:0], in ascending order of numerical value

    ii) A three bit grey code to decimal decoder is to be designed, where the outputs X[0:7] are active low. Write down the boolean functions of each output terminal.

    2. Relevant equations



    3. The attempt at a solution

    I think the first part is easy enough, it's just grey code (only change one bit at a time)

    A2:A0

    0 000
    1 001
    2 011
    3 010
    4 110
    5 111
    6 101
    7 100

    The second part is confusing me though, and it's to do with the "active low part". Initially I got what's below but how is it any different to if the outputs weren't active low ?!

    X0 = A2'A1'A0'
    X1 = A2'A1'A0
    X2 = A2'A1A0
    X3 = A2'A1A0'
    X4 = A2A1A0'
    X5 = A2A1A0
    X6 = A2A1'A0
    X7 = A2A1'A0'
     
  2. jcsd
  3. Sep 5, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If, for example, 2 would be A2'A1A0 if it was active high, you want the complement for active low. Use DeMorgan's law. The complement of a product is not the product of the complements.
     
  4. Sep 5, 2012 #3
    Hmm ok so for X0 I had A2' . A1' . A0' when I should have all that [A2' . A1' . A0']'

    So that becomes A2 + A1 + A0 ?
     
  5. Sep 5, 2012 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You tell me. Is that low only when they are all 0?
     
  6. Sep 5, 2012 #5
    Yes, that function is low only when they are all 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook