plz help

Discussion in 'Homework Help' started by zulfiqariqbal, Jun 14, 2013.

  1. zulfiqariqbal

    Thread Starter New Member

    Jul 24, 2012
    5
    0
    I am new in digital electronic plz can you simplify this expression
    F=A'.B'(A.B+B')
     
  2. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    Why yes, yes I can, however, you are in homework HELP, so you will not get a turnkey solution. Post your attempt and we will have something to work with.
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,804
    There are two pretty obvious ways to tackle it, so don't get hung up because you more than one possible path. Pick on and follow it. All valid paths will lead to the same result.
     
  4. t06afre

    AAC Fanatic!

    May 11, 2009
    5,939
    1,222
    You can also find many "cheat sheets" on how to use Boolean algebra rules on the net. And also in your books. This one is quite simple so just follow the basic rules :)
     
  5. zulfiqariqbal

    Thread Starter New Member

    Jul 24, 2012
    5
    0
    Dear all I am new in circuits simplification plz can some one simplify this boolean
    expression

    F=A'.B'(A.B+B')
    I TRIED TO SOLVE IT AS

    F=A'.B'((A'+B')+B') BY USING THE DEMORGAN LAW
    F=A'.B'(A'+(B'+B')
    F=A'.B'(A'+1)
    F=A'.B'
    F=(A+B)'
    PLZ CHECK IT IS IT CORRECT?
     
    Last edited: Jun 15, 2013
  6. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    Thank you for finally posting your work, now we have something to start off of...


    Does B' + B' really equal 1?

    e.g:
    1 + 1 = ?
    0 + 0 = ?

    These can't be the same value!
     
  7. zulfiqariqbal

    Thread Starter New Member

    Jul 24, 2012
    5
    0
    I am not sure but i know A.B=(A'+B')'
    AND B'+B'=B'
     
  8. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    Is that what you put!?
     
  9. screen1988

    Member

    Mar 7, 2013
    310
    3
    Also you need to check the Demorgan law.
    A.B ≠ A' + B'
    I think it is more easier for you to use distributive property. Take a look here:http://www.allaboutcircuits.com/vol_4/chpt_7/4.html
     
  10. zulfiqariqbal

    Thread Starter New Member

    Jul 24, 2012
    5
    0
    but B'+B' is not equal to one it is equal to B'
    and A.B=(A'+B')'
     
  11. screen1988

    Member

    Mar 7, 2013
    310
    3
    Hi, do you know the rules below?
    1) distributive property
    A(B +C ) = AB + AC
    2) commutative property
    A + B = B + A
    AB = BA
    And how about: A.A' = ?
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,777
    4,804
    Agreed. But that's not what you indicated when you finally posted your work:

    So where you are claiming that, by DeMorgan's Law, that

    A.B = (A'+B')

    Since you have now come back and stated that

    A.B=(A'+B')'

    Don't you think it might be good to go back and rework your problem with that in mind.

    Similarly, your work used

    Where you are clearly claiming that (B'+B')=1, yet when you later say that (B'+B')=B', that again should be a big hint that you need to go back and work the problem over.

    Finally:

    You should almost never have to ask someone to check if your answer is correct. In most engineering problems, particularly in logic, you can readily prove whether or not the answer is correct, regardless of how it was arrived at, from the answer itself. You need to get in the habit of doing so.

    In this case, you are asking if

    F=(A+B)'

    is equivalent to

    F=A'.B'(A.B+B')

    Well, both are functions of two variables which means that there are only four possible cases that have to be checked. So check them.

    And remember, even if they are equivalent does not mean that the work supports the answer. If the work is invalid, then a correct answer means either that you cheated or that you got lucky and the errors either cancelled or were non-critical. In this case it is likely the latter unless your "work" was arrived at by trying to force the answer to a known good solution by whatever magical means necessary. In this case I suspect it is the 'lucky' route.
     
Loading...