Boolean simplification

Discussion in 'Homework Help' started by mcewendavid, Aug 8, 2007.

  1. mcewendavid

    Thread Starter New Member

    Aug 8, 2007
    7
    0
    I have been looking at these problems for 2 days now......the text books I have just don't cover boolean simplification in enough detail. They only have examples must simpler than these. Any help would be much appreciated :D

    1) ABC' + AB' + AC + AB'C

    here's what I got (don't laugh!)
    = A(BC' + B' + C + B'C) - Factor out A
    = A(B' + C' + B'C)
    = A(B' + B'C')
    = A(B' + C')


    2) ABC'D + A'B'CD + A'BD + AC'D' + AB'C' + A'C'D

    = ABC'D + A'B'CD + A'BD + AB'C' + C'
    = A(BC'D + B'C') + A'(B'CD + BD) + C'
    wellll I had no idea where to go from there....but I'm pretty sure what I got was wrong? Any help please....this is stressing me out. :confused:

    Thankyou in Advance
    Dave:cool:
     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
  3. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Dealing with them one by one.

    This is correct even if it is unclear from your workings how you got to this answer. You will need to show all workings and state identities as and when you use them. Can you should the complete workings here?

    Dave
     
  4. mcewendavid

    Thread Starter New Member

    Aug 8, 2007
    7
    0
    ummmm.....that is my complete workings!??

    I....
    ABC' + AB' + AC + AB'C
    1)Factor out A
    2)C + C' = C'
    3)C' + C = C'
    4)B' + B' = B'

    1)= A(BC' + B' + C + B'C)
    2)= A(B' + C' + B'C)
    3)= A(B' + B'C')
    4)= A(B' + C')

    .....and that's it!??
    I dont know any other way to do it :)
     
  5. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    try looking at some laws ;
    then take a look at some common examples u wud find them on the net as well;
    some theorems,
    after a bit of practice u shud get a hang of it,

    but there is a easier way out,
    make a truth table for inputs a,b,c
    then learn a bit of kernaugh map;
    u wont have to go thru manipulations for simplification and k map really comes in handy.
     
  6. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Apologies for saying you got the correct answer, I should have looked at the question more thoroughly. For future reference:

    C + C' = 1

    Ok, the correct workings for the first one is:

    A(BC' + B' + C + B'C) - Factor out A

    Use the identity: C + CB' = C then

    A(BC' + B' + C)

    Use the identity: B' + BC' = B' + C' then

    A(B' + C' + C)

    And from before C + C' = 1

    So: AB' + A

    Using the identity: A + AB' = A

    Therefore the answer is: A

    If you look at the expression A(BC' + B' + C + B'C) you should be able to see that the answer is independant of the values of B, C or either of their compliments.

    The identities are available from: http://www.allaboutcircuits.com/vol_4/chpt_7/5.html

    Using the identities on this page, try the second question and we will give you some pointers.

    Dave
     
  7. mcewendavid

    Thread Starter New Member

    Aug 8, 2007
    7
    0
    Thanks for that first part guys, I'm starting to grasp it a little better but I still dont even know where to start on that question
    2) ABC'D + A'B'CD + A'BD + AC'D' + AB'C' + A'C'D

    ....simply because there are 3 and 4 AND's between every OR
    Should I start by factoring out? maybe A and A' ?
    I just have no idea...



    Another question if I may....
    3) Prove (A+C)(A+D)(B+D) = AB + CD + AD
    I got,
    (A+C)(A+D) = A + CD
    =(A+CD)(B+D)
    Multiply the brackets
    =AB + AD + CDB + CDD

    Now....does (CDB + CDD) somehow = CD?
    I can't find a rule to prove this but it somehow seems plausible...

    Thankyou
    Dave
     
  8. mcewendavid

    Thread Starter New Member

    Aug 8, 2007
    7
    0
    one more thing......does BD + D' + B' = 1?
    because B + B' = 1 and D + D' = 1

    If it does I think I'm set....
    Cheers



    ....Using this logic..
    2) ABC'D + A'B'CD + A'BD + AC'D' + AB'C' + A'C'D
    Factor out A and A'
    = A(BC'D + C'D' + B'C') + A'(B'CD + BD + C'D)
    Factor out C' and D
    = A(C'(BD + D' + B')) + A'(D(B'C + B + C'))
    logic here...BD + D' + B' = 1 and B'C + B + C' = 1
    = A(C'(1)) + A'(D(1))

    =AC' + A'D

    Well I'm fairly proud of myself anywayz :-D
     
  9. mcewendavid

    Thread Starter New Member

    Aug 8, 2007
    7
    0
    anybody?????
     
  10. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    well BD + D' + B' = 1
    considering all possibilities of b and d they do equal 1
    or
    BD + B' + D'= ((BD)')' + B' + D'=
    = (B' +D')'+ (B' + D')=

    and A + A'=1
    is this what u wanted?
     
  11. RiJoRI

    Well-Known Member

    Aug 15, 2007
    536
    26
    Try "DeMorgan-ing":
    (xy)' = (x' + y')
    (x + y)' = x'y'

    and
    (x'y)' = x''+y' = x+y'

    Again, check out Karnaugh mapping -- the answers pop right out!

    --Rich
     
  12. mcewendavid

    Thread Starter New Member

    Aug 8, 2007
    7
    0
    Thanks for the help guys :)
    It's awesome.....

    Dave
     
Loading...