Help with Boolean Algebra Simplification

Discussion in 'Homework Help' started by Emile_Uk, Nov 17, 2003.

  1. Emile_Uk

    Thread Starter New Member

    Nov 17, 2003
    1
    0
    Hi,

    If anyone can help me simplify the following two boolean expressions whilst listing the laws used at each stage then that would be hugely appreciated!

    The two expressions are:

    ([P]*Q*[X]*Y)+([P]*Q*X*[Y])+(P*[Q]*X*[Y])+(P*[Q]*X*Y)+(P*Q*[X]*Y)+(P*Q*X*Y)

    and

    (P*[Q]*[X]*Y)+(P*Q*[X]*Y)+(P*Q*X*[Y])

    Where * is AND and [] is an inverted input.

    Any help would be greatly appreciated!

    Cheers,

    Emile
     
  2. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    143
    Firstly the second one (because its easier):

    (P*[Q]*[X]*Y)+(P*Q*[X]*Y)+(P*Q*X*[Y])

    Deal with the first two brackets and take out P*[X]*Y as common factors

    Therefore: P*[X]*Y([Q]+Q) where [Q]+Q = 1 (Complementary Law)

    So that gives (P*[X]*Y) + (P*Q*X*[Y]) (with the third bracket reintroduced)

    Now take out P as a common factor:

    So you get P*{([X]*Y) + (Q*X*[Y])}

    Thats it in its simplest form.

    The first one is considerably more difficult:

    ([P]*Q*[X]*Y)+([P]*Q*X*[Y])+(P*[Q]*X*[Y])+(P*[Q]*X*Y)+(P*Q*[X]*Y)+(P*Q*X*Y)

    Look at brackets 3 and 4 and take out P*[Q]*X as common factors

    So P*[Q]*X([Y]+Y) where [Y]+Y = 1

    Do the same with brackets 5 and 6 taking P*Q*Y out as common factors

    So P*Q*Y([X]+X] where [X]+X = 1

    Brackets 1 and 2 are just take out [P]*Q as common factors and get:

    [P]*Q*{([X]*Y)+(X*[Y])}

    So your answer should now look something like:

    [P]*Q*{([X]*Y) + (X.[Y])} + (P*[Q]*X) + (P*Q*X)

    Take out common factors of P*X in the last two expresions to get P*X([Q]+Q) where [Q] + Q = 1 and just gives P*X

    Your final answer should look like:

    [P]*Q*{([X]*Y) + (X.[Y])} + P*X

    I haven't had chance to look at this with another method but will post back if the answer comes out any different. I'd also appreciate comments on any errors I have made or if you'd like any further help :)
     
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