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Boolean expression needs simplifying

  1. Oct 12, 2012 #1
    I have attached the problem and my working. It would be much appreciated if someone could check through my working and let me know if I have slipped up anywhere.

    Thanks in advance

    Steve
     

    Attached Files:

    Last edited: Oct 12, 2012
  2. jcsd
  3. Oct 12, 2012 #2

    lewando

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    Gold Member

    Attachment not attached.
     
  4. Oct 12, 2012 #3
    I realised that I had made a couple of errors so I deleted the attachment. I have now reattached it above.

    Also, I’ve just realised that A’+A=1… (the 0 I have put is incorrect)

    So
    A’+(AC) = (A’+A)+(A’+C) = 1 + (A’+C) = 1
    B(1) = B

    C(AB’+AC’+B) = AB’C+ACC’+BC = AB’C+BC = AB’C+BC (which somehow is the same answer as before)


    Cheers

    Steve
     
    Last edited: Oct 12, 2012
  5. Oct 12, 2012 #4

    lewando

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    4th line:
    ~A+(AC) = (~A+A)(~A+C) = 1(~A+C) = (~A+C) --got lucky.

    last line:
    BC(1+~A) != BC
     
  6. Oct 12, 2012 #5
    ok I see,

    1 + (A’+C) is actually 1(~A+C)

    Thanks for your help lewando.
     
  7. Oct 12, 2012 #6

    NascentOxygen

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    Staff: Mentor

    What is your final simplification of the expression, Steve Collins?
     
  8. Oct 12, 2012 #7
    ac+bc

    I've been staring at this one for a while:

    ~a~bc+~ab~c+a~b~c+abc

    Can this be simplified at all?
     
    Last edited: Oct 12, 2012
  9. Oct 12, 2012 #8

    NascentOxygen

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    Is this your answer to my question? If so, write it as c(a+b), though that is still wrong. The easiest way to check is to draw up two truth-tables, one for the original expression, and another for your simplification. They should agree.

    Please explain what rule you used to go from the LHS to the RHS here:
    It can be simplified using exclusive-OR operations.
     
  10. Oct 13, 2012 #9
    I've been keeping my results in sum of products form as I find it easier to read when simulating in Proteus.

    A'+(AC)=(A'+A)+(A'+C) Should have read A'+(AC)=(A'+A)(A'+C)

    I have constructed a truth table for the original problem and my answer and they seem to concur.
     

    Attached Files:

  11. Oct 13, 2012 #10

    NascentOxygen

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    My mistake, I overlooked a closing parenthesis in your handwriting.
     
  12. Oct 13, 2012 #11

    NascentOxygen

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    If you need encouragement: it simplifies to entirely exclusive-ORs. :smile:

    ¬xy + x¬y = x⊕y
     
  13. Oct 15, 2012 #12
    Thanks for the help and sorry for the late reply… Not enough hours in the day!

    With NascentOxygen’s direction I’ve discovered that this Boolean expression is for a full counter so my ultimate goal is to achieve:
    a xor b xor c

    I have got to my working below, but I am struggling to proceed. I think that I may be going down the wrong path or there is a rule that I cannot find after much searching.

    ~a~bc+~ab~c+a~b~c+abc
    ~a(~bc+b~c)+a(~b~c+bc)
    ~a(b xor c)+a~(b xor c)
     
  14. Oct 15, 2012 #13

    NascentOxygen

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    If you are able to get this far, then you are almost done.

    Your last expression matches ~x.y + x.~y
     
  15. Oct 16, 2012 #14
    Thank you NascentOxygen... Your help has been greatly appreciated.
     
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