Boolean expressions

Discussion in 'Math' started by Johnie, Feb 14, 2006.

  1. Johnie

    Thread Starter New Member

    Feb 14, 2006
    1
    0
    How do you simplify these four boolean expressions step by step?

    1. Y = A'B'C' + A'BC + ABC + AB'C' + AB'C

    2. Y = (B + C')(B') + (A' + B + C')'

    3. Y = (C + D)' + A'CD' + AB'C' + A'B'CD + ACD'

    4. Y = AB(C'D)' + A'BD + B'C'D'
     
  2. CoulombMagician

    Active Member

    Jan 10, 2006
    37
    0
    What are the associative, distributive, and commutative properties of complement with respect to + and * in boolean logic? Your textbook has a table somewhere.

    You know that A+A' = 1 and A*A'=0 for both states of A, start applying these identities to those equations.
     
  3. ecjohnny

    Senior Member

    Jul 16, 2005
    142
    0
    Heres is my answer.. i might be wrong in some steps but still hope it will help you out. i got Q1,2&4. Q3 got confused by the "bar". here:

    1) Y = A'B'C' + A'BC + ABC + AB'C' + AB'C
    = A'(B'C'+BC)+ABC+AB'(C'+C)
    = A'(1)+ABC+AB'(1)
    = A'+ABC+AB'
    =A'+BC+B'
    =A'+B'+C

    2) Y = (B + C')(B') + (A' + B + C')'
    =B'B+B'C'+AB'C
    =0+B'C'+AB'C
    =B(C'+AC)
    =B(C'+A)
    =BC'+AB

    3)DUNNOE
    4) Y = AB(C'D)' + A'BD + B'C'D'
    =AB(C+D')+A'BD+B'C'D'
    =ABC+ABD'+A'BD+B'C'D'
    =ABC+B(AD'+A'D)+B'C'D'
    =ABC+B(1)+B'C'D'
    =ABC+B+B'C'D'
    =ABC+B+C'D'
     
  4. Papabravo

    Expert

    Feb 24, 2006
    9,898
    1,722
    The first simplification on this one is that
    A'CD' + ACD' = CD'
    and the second simplification is that
    ( C + D)' + CD' = D'
    So we are left with
    D' + AB'C' + A'B'CD
     
  5. Ebag

    New Member

    Dec 14, 2008
    1
    0
    AB'+C how can you draw this with the least number of two-input nor gates
    If you know how email me the answer at babolar4444@yahoo.com
     
  6. vvkannan

    Active Member

    Aug 9, 2008
    138
    11

    let me try.
    just implement as usual using NOT,AND and OR gate.
    adding NOT before AND will give you NOR and NOT after OR will give you NOR.
    Giving the same input to both terminals of NOR would invert the function (i.e) would act as NOT as (x+x)' = x'.
    I think 4 NOR gates would be required in this case
     
Loading...