Algebraic Manipulation Problem

Discussion in 'Homework Help' started by Eternal Sky, Sep 11, 2008.

  1. Eternal Sky

    Thread Starter New Member

    Sep 11, 2008
    2
    0
    I'm trying to find the minimum sum-of-products expression for the function:

    f = x1x2'x3' + x1x2x4 + x1x2'x3x4'

    I've read over the boolean rules for simplification on this site, but it didn't really help me understand what I should be doing. I attempted to simplify the function by pulling out the common x1 term, but this just produced the function:

    f = x1(x2'x3' + x2x4 + x2'x3x4')

    Doing that didn't seem to make things any easier.

    If someone could just point me in the right direction, I would greatly appreciate it.
     
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    I'm confused between multiplication sign (x) and indexing for a variable (x1, x2, x3 etc)
     
  3. Eternal Sky

    Thread Starter New Member

    Sep 11, 2008
    2
    0
    Sorry. I'll try to make it easier to read:

    f = x1*x2'*x3' + x1*x2*x4 + x1*x2'*x3*x4'

    where * = multiplication
    and x1, x2, x3, x4 = input variables
    also, the ' character indicates NOT
     
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
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    Try using some the following identities; you may need to use some of them in reverse.
    I have used letters instead of x1, x2 etc as its easier on the eye.
    You can insert barackets anywhere and compare with my expressions.



    b+1=1
    a+ab=a
    a(a+b)=a
    a(a'+b)=ab
    a+a'b=a+b
    a+a'=1
    a+bc=(a+b)(a+c)
    (ab)'=a'+b'
    a'b'=(a+b)'

    the last two together are known as De Morgan's theorem.
     
    Last edited: Sep 11, 2008
  5. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Eternal Sky,

    Let me make it even easier to read.

    A ==> x1
    B ==> x2
    C ==> x3
    D ==> x4

    F = AB'C' + ABD +AB'CD'

    First we expand the terms.

    F = AB'C' + ABD +AB'CD'

    F = AB'C'(D+D') + AB(C+C')D + AB'CD'
    = AB'C'D + AB'C'D' +ABCD + ABC'D + AB'CD'

    There are no duplicate terms to remove.

    AB'C'D = 9
    AB'C'D' = 8
    ABCD = 15
    ABC'D = 13
    AB'CD' = 10

    So the Boolean expression can be described by the minterms as (8,9,10,13,15)

    Now simply using a Karnaugh map, or the Quine-McCluskey tabulation method we easily get

    F = AB'D' + ABD + AB'C'
    or
    F = AB'D' + ABD + AC'D

    You might want to read http://forum.allaboutcircuits.com/showthread.php?t=12279

    and especially http://forum.allaboutcircuits.com/showthread.php?t=13854 , the second paragraph of post #13 concerning using algebraic methods to simplify Boolean expressions.

    Ratch
     
    Last edited: Dec 28, 2009
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