Digital Electronics

Discussion in 'Homework Help' started by coo0004, Dec 6, 2006.

  1. coo0004

    Thread Starter New Member

    Dec 6, 2006
    4
    0
    hi guys
    i am new to this forum and i am currently undertakin an electronic engineering degree and i thought it would be a good idea to join, anyway at the moment i am having a big problem with boolean algebra and all i can find on the internet is simple stuff. anyway if anyone can help me figure all this out it would be much appreciated.

    for example:
    __ _ _
    (ABC + ABC + ABC + ABC)(A+B)

    and

    ___ _ _ _ __ _ _
    ABC + ABC + ABC + ABC + ABC + ABC

    thanks heaps guys
     
  2. coo0004

    Thread Starter New Member

    Dec 6, 2006
    4
    0
    Sorry guys disregard those last problems

    these are the correct ones, (' represents not)

    A'B'C' + A'BC' + A'BC + AB'C' + AB'C + ABC'

    and

    (AB'C' + AB'C + ABC + ABC')(A+B)

    thanks again
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    What are you trying to do, simplify the expressions?

    Dave
     
  4. coo0004

    Thread Starter New Member

    Dec 6, 2006
    4
    0
    yeh i was trying to minimise the equation. any help is appreciated
     
  5. m4yh3m

    Senior Member

    Apr 28, 2004
    186
    39
    I always threw those things into Electronics Workbench and let it simplify it :p
     
  6. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    The first one A'B'C' + A'BC' + A'BC + AB'C' + AB'C + ABC':

    Take out common factors B'C' ; BC' ; C

    B'C'(A'+A) + BC'(A'+A) + C(A'B+AB')

    (A'+A) = 1

    Therefore:

    B'C' + BC' + C(A'B+AB')

    Take out common factor C'

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

    (B'+B) = 1

    Therefore your simplified expression is:

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

    The second one will use the same techniques as the first but look at taking out common factors AB and AB' from the first bracket, don't multiple out the two brackets.

    If you want further advice, or wish to run your answer past us, post it up and I'll be glad to check it. I am also moving this to the Homework Help forum.

    Dave
     
  7. dina

    Member

    Nov 12, 2006
    19
    0
    Hi
    here's a simplification for the first problem:
    C' + A'B + AB'
    for the second one:
    it's simply A
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Do you care to embelish on how you simplied the first expression to C' + A'B + AB' - where did the C go?

    Dave
     
  9. dina

    Member

    Nov 12, 2006
    19
    0
    hi Dave...
    i used the karnaugh map to simplify both the problems
    if u need a help with karnaugh maps i'll be glad to help
     
  10. beck

    Member

    Nov 27, 2006
    26
    0
    yup,
    K-Maps are the best.
     
  11. dragan733

    Senior Member

    Dec 12, 2004
    152
    0
    A'+AB=A'+B
    A'(B+B')+AB
    A'B+A'B'+AB+A'B
    A'B=A'B+A'B
    A'(B+B')+B(A+A')=A'+B
     
  12. Sonofjustice

    New Member

    Jul 3, 2008
    3
    0
    Yeah There Is Also This Formula Saying X'+XY= X'+Y
     
  13. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Dave,

    Sure, use the identity X +X'Y = X+Y where X=C', X'=C, and Y=A'B+AB'

    Proof of The Identity:

    The dual of X +X'Y is X(X'+Y) = XY

    The dual of XY is X+Y, so the identity is proved. Other ways exist to prove the identity. Ratch
     
  14. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    beck,

    Nope, not when the number of variables is greater than 5 or 6. It blows your mind to try to interpret that many variables variables on a K-map. Try:

    P(A,B,C,D,E,F) = SOP(6,9,13,18,19,25,27,29,41,45,57,61)

    Tabulation methods like the Quine-McCluskey method guarantee the best simplification and are applicable to computer operations.

    The answer to the first problem I got from the Quine-McCluskey method was A'B+AB'+C' .

    For the second problem, by inspection it is easily seen that
    (AB'C' + AB'C + ABC + ABC')(A+B)=(AB'C' + AB'C + ABC + ABC') .

    Then = (AB'C' + AB'C + ABC + ABC')= AB'(C'+C)+AB(C+C') = AB'+AB = A(B+B') = A

    Ratch
     
    Last edited: Jul 20, 2008
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