4-bit even parity generator - am I right?

Discussion in 'Homework Help' started by mpaska85, Mar 10, 2011.

  1. mpaska85

    Thread Starter New Member

    Mar 4, 2011
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    Hi, just like some feedback if I've tackled a problem from my homework from the right angle in my introductory computer engineering unit.

    We've been given the problem to design a 4-bit even-parity generator using only AND, OR and NOT gates.

    The logic diagram I have come up with is:
    [​IMG]

    And here is my truth table:
    [​IMG]

    Does this look correct? Any help will be appreciated! This is my first assignment, and we have to implement it on a breadboard in class and I'm a little hesitant at how many gates my circuit requires so I have a feeling I'm on the wrong path.
     
  2. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    I am just now taking digital logic, but, for what its worth, given the gate restrictions, I don't see any simplifications. Well, you do realize you only need four NOT gates, right?
     
  3. mpaska85

    Thread Starter New Member

    Mar 4, 2011
    4
    0
    Ah, yes. Thank you! I've now got it down to 4 NOT gates and a bit more confident of actually implementing this now.

    [​IMG]
     
  4. Arbitrator

    New Member

    Mar 10, 2011
    16
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  5. Arbitrator

    New Member

    Mar 10, 2011
    16
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    I couldn't get it to upload in Logisim but the attached is pdf and jpeg of the 4 bit even parity generator. It should be correct.
     
  6. mpaska85

    Thread Starter New Member

    Mar 4, 2011
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    Thanks Arbitrator, that's definitely a more simplified version then my attempts. However before I move on I really want to get to that answer by myself.

    My boolean expression that I ink out from my k-map is ABCD' + ABC'D + AB'CD + AB'C'D' + A'B'CD' + A'B'C'D

    Is there a further simplification rule that I am missing?
     
  7. Arbitrator

    New Member

    Mar 10, 2011
    16
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    this has labels on inputs and output.
     
  8. Arbitrator

    New Member

    Mar 10, 2011
    16
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    i do it now , i think there should be 8 Boolean expressions wait 5 minutes
     
  9. Arbitrator

    New Member

    Mar 10, 2011
    16
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    yes i got a 8 expressions of groups of a,b,c,d . i don't know if you can group anything from k-map since they all diagonal.
    i got A'B'CD' + A'BC'D' + ABCD' + ABC'D + AB'CD + AB'C'D' + A'BCD + A'B'C'D

    how did you simplify out AB'C'D' and A'BCD?
     
  10. mpaska85

    Thread Starter New Member

    Mar 4, 2011
    4
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    Yeap, that's the expression I've been getting too. In the interest of actually learning this, what method do I use to get the expression used in your circuit diagram?
     
  11. Arbitrator

    New Member

    Mar 10, 2011
    16
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    i got it from the simplest XOR expression then just changed gates :S that doesn't help you much
     
  12. justtrying

    Active Member

    Mar 9, 2011
    329
    335
    we did this in a lab a few weeks back, there is no way to use k-maps to simplify Boolean. You can only simplify this further if you use Boolean logic to get XOR expressions by combining (in pairs) A'B'CD'+A'B'C'D to get A'B'(C XOR D) - eventually it is a simple implementation with XOR gates only, if XORs are unavailable, they can be implemented with regular gates, as is done in the diagram posted above.
     
  13. Arbitrator

    New Member

    Mar 10, 2011
    16
    0
    i understand it now you explained it like that. :)
     
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