Algebraic Boolean Simplification

Discussion in 'Homework Help' started by mikehoward82, Sep 11, 2010.

  1. mikehoward82

    Thread Starter New Member

    Sep 11, 2010
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    I was given several boolean algebraic problems. The instructions are to use the boolean algebraic identities to transform the left side of the equation into the right side.

    This one was simple
    x+wx' = x + w
    so I used the distributive identity
    x+wx'
    (x +w)(x + x')
    (x +w)(1)
    = x + w

    The next two have blowed my mind I have yet to have any formal classes on this topic and the instructor who I believe on a whim inherited the course for the semester did not appear to know how to do these either
    and stated he would get back to us through e-mail after looking into it. I have yet to see a mass e-mail sent as of today, hopefully he will actually send more information.

    Anyways here are the other two problems

    xy' + y'z + zx' = xy' + zx'

    I have worked on this basically all day without any actual pprogress everything I write ends up going nowhere any advice or information on how to start to reduce this boolean algebraic expression from the left to the right would be greatly appreciated.

    The other problem I have not even given a stab at(so I shouldn't say it has blowed my mind) and after lots of searching I have not found any information on how to go about it

    (a XOR b)' XOR c = a'b'c' + abc' a'bc + ab'c

    The XOR is actually symbolized in the problem.

    I have been going through my old Discrete Mathematics book in hopes to find further information, My discrete mathematics course covered truth tables throughly but only lightly covered boolean algebra problems similar to these, if I remeber correctly. So if anyone has some time and wants to help me out, please do, I'm stressing over it right now, I'm just hoping that it'll come to me and the light will flash but right now I am in the dark. :confused:
     
    Last edited: Sep 11, 2010
  2. Georacer

    Moderator

    Nov 25, 2009
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    For the third expression, you don't have to do any brainwork, just use the definition of XOR: a XOR b = a'b + ab'. Simple as that! Oh, and you forgot a "+" sign between the two middle terms of the result.

    As for the second expression, are you sure it is copied correctly? There's no reason for the y'z term to be eliminated. Textbooks are mistaken less often that we are, but nontheless, they can be.

    EDIT: It seems I was wrong: the second expression does get simplified into xy' + zx'. I did the carnot map to double check and it turns out there are all the right circumstances for the middle term to get eliminated. The carnot map is proof by itself, but for a boolean solution, you will have to wait a bit. F1 Monza race is waging right now!
     
    Last edited: Sep 12, 2010
  3. mikehoward82

    Thread Starter New Member

    Sep 11, 2010
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    you're right, there's a typo

    the third equation should include another + sign as

    (a XOR b)' XOR c = a'b'c' + abc' + a'bc + ab'c

    I'll look into what you said about the definition of XOR, thanks for the reply.
     
  4. mikehoward82

    Thread Starter New Member

    Sep 11, 2010
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    I believe that I now have the process of simplification for the second equation.

    so
    XY' + Y'Z + X'Z
    XY' + Y'Z(1) + X'Z
    XY' + Y'Z(X + X') + X'Z
    XY' + XY'Z + X'Y'Z + X'Z
    XY'(1 + Z) + X'Z(1 + Y')
    XY'(1) + X'Z(1)
    XY' + X'Z
     
  5. Georacer

    Moderator

    Nov 25, 2009
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    Couldn't be more correct! Well done! If only I wasn't too lazy to do it by hand and not test it through software (because I 've spent 2 years doing such operations and by now I 'm sick of them).
     
  6. mikehoward82

    Thread Starter New Member

    Sep 11, 2010
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    I'm confused on the third one.

    So A xor B = AB' + A'B that's how xor works

    but now (A xor B)' so this makes it A'B' + AB correct
    that confuses me

    but appears correct according to the truth table
    A-- B-- A xor B --(A xor B)'
    0-- 0 -----0-------- 1
    0 --1 -----1-------- 0
    1-- 0 -----1-------- 0
    1 --1 -----0 --------1
     
  7. Georacer

    Moderator

    Nov 25, 2009
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    We 'll examine this from two perspectives, the mathematical, and the intuitional:

    From the mathematical side, if we complement A'B+AB', we get
    (A'B+AB')'=
    (A'B)'(AB')'=
    (A+B')(A'+B)=
    AA'+AB+B'A'+BB'=
    AB+A'B' %Because either X or X' are 0

    From the intuitional side, if XOR is true when only one of A or B are true, then NOT XOR would be true when not only one of A or B are true. That is when they are both true, or when none of them are true. This is interpreted as AB+A'B'.
    You shouldn't think of your conventional logic where NOT XOR would dictate that the opposite of only one of A or B is true, is actually both of them is true. Everyday logic sometimes collides with boolean logic, wich works with set theory. Just stick to the rules and you 're in the clear.

    Am I clear?

    EDIT: Oh, and in case you want some additional reading material, I wouldn't suggest Discrete Math, but Boolean or Digital Logic. My favourit book is Digital Design by Morris Mano
     
    Last edited: Sep 12, 2010
  8. mikehoward82

    Thread Starter New Member

    Sep 11, 2010
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    Thanks, it is clear.

    Discrete Math is the closest to this type of Mathematics that I have had to date. I'm not sure why the instructor threw us these problems without really teaching the concepts and giving examples, and it appears he himself was unsure how to answer the last two ( as stated he said he inherited the material for the course )

    I looked at reviews for the book you mentioned and many state it's not a good book for self teaching, would you agree or disagree?

    I will look further into it though, a good book would be nice to have. ( but I'm pretty much outta book shelf space, time install more soon) The book for this course merely covers the topic in an appendix that is not even in the book rather on the disc included, lame.
     
  9. Georacer

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    Nov 25, 2009
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    Well, I wouldn't say so. The latest edition (4th I think) has plenty of examples and exercises to practice on. I studied it thoroughly and I can say I was satisfied with the result. Actually, I could answer all your questions thanks to this book. I would easily recomend it. It starts from the very basics of digital logic and covers even advanced topics like data buses configurations and Hardware Description Languages. It is accompanied by a CD-ROM with signal simulation software.
     
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