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Homework Help: 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)


    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


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    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


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    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.
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