Boolean Simplification Help

Discussion in 'Homework Help' started by Pariah, Oct 5, 2008.

  1. Pariah

    Thread Starter New Member

    Oct 5, 2008
    1. The problem statement, all variables and given/known data
    We have been given a task to develop a circuit which displays the square of a binary number on a 3 x 7 seq displays.

    I have already gone through and done up the Karnaugh Maps for the task and have identified the minterms. However, I believe that these can still be simplified even more.

    2. Relevant equations
    Karnaugh Maps Output = A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'

    3. The attempt at a solution
    A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'
    Factor out A' and D from minterms 1 and 2
    A'D(b' + b) (c + c')
    = A'D + A'BC'D + ABCD + B'C'D' + AB'C'
    Factor out B and D from minterms 2 and 3
    bd(a' + a)(c' + c)
    = A'D + BD + B'C'D' + AB'C'

    This is where I am getting stuck. Is it possible to further simplify the equation or is this the final solution???
  2. mik3

    Senior Member

    Feb 4, 2008
    Well, i think you didnt get the Karnaugh map correct because you simplified it again using the boolean algebra. This is not correct because if you do solve correct a Karnaugh map you get the simplest solution. However, i can't see a simpler solution to your final answer.
  3. Dave

    Retired Moderator

    Nov 17, 2003
    Common factor simplification will reduce the gate count:

    A'D + BD = D(A' + B)

    This reduces from 2x 2I/P AND-gates and 1x 2I/P OR-gate to 1x 2I/P AND-gate and 1x 2I/P OR-gate

    B'C'D' + AB'C' = B'C'(A + D')

    This reduces from 2x 3I/P AND-gates and 1x 2I/P OR-gate to 1x 3I/P AND-gate and 1x 2I/P OR-gate.

    So the simplified expression would be: D(A' + B) + B'C'(A + D')

  4. Pariah

    Thread Starter New Member

    Oct 5, 2008
    Thanks alot for the help.

    Also, I understand that the Karnaugh Maps usually produce the simplest forms but I saw that the equation could be further simplified because of the common factors.

    So Am I right in thinking that the equation that Dave gave is correct or should I not try to further minimize the equation???

    I am confused :(
  5. hgmjr

    Retired Moderator

    Jan 28, 2005
    Take a look at the material on boolean algebra in the AAC ebook.

  6. Dave

    Retired Moderator

    Nov 17, 2003
    You may be able to DeMorganise the expression to less gates, however this may not translate to a more simplified expression in terms of transistors or terms used. You would need to play around with DeMorgans Theorems to see if you can actually simplify the expression further - it isn't obvious from the above expression if it would work.

    Your current expression utilises only one of each term and its compliment (actually there is no C); I would commend that this expression is fully simplified (note there are probably several simplifications of the original expression which would be equally simplified).

  7. Pariah

    Thread Starter New Member

    Oct 5, 2008
    Dave, so your saying that my original equation should not be further simplified???

    Also, thanks alot for the help guys.

    I think you are correct in saying that the original form is already simplified and I am only complicating everything by overthinking :)

    Once again thanks alot :)

    If I have some more questions I will be sure to ask :)
  8. Dave

    Retired Moderator

    Nov 17, 2003
    I cannot see that it can be logically simplified further. Like I said previously, you may be able to DeMorganise the expression to remove a term and its compliment, however this is not obvious nor would it yield a more simplified expression depending on how you make the judgement of simplification.

    I would commend that your expression is in its simplified form.

    No problems, that is why this site is here. Be sure to check the e-book and worksheets to help you with your studies.

  9. Ratch

    New Member

    Mar 20, 2007

    Sorry I did respond to this sooner, but I just got back from vacation.

    I did the simplification using the Quine-McCluskey tabulation method , and verified that you cannot simplify the above expression any more. You should be able to verify that by using a K-map.

    Your Boolean algebra is in error. There is no way that Minterms(3,5) are going to reduce to a two variable term (A'D).

    Neither, you already had the simplest reduction.

    See the last entry of this thread.