boolean algebra help.

Discussion in 'Homework Help' started by norules44, Oct 28, 2014.

  1. norules44

    Thread Starter New Member

    Oct 28, 2014
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    0
    i did in the past boolean algebra but i forgot it, can any1 try to help me out on this?
    problem 1:
    F(wxyz)= wx+xyz+x'yz+x'yz'+[wx+w' (x+z)]
    i did this but i got stacked.
    F(wxyz)= wx+xyz+x'yz+x'yz'+[wx+w' (x+z)]
    = wx+xyz+x'y(z+z')+[wx+w' (x+z)]
    =wx+xyz+x'y+(w'+x')wx.wz
    =wx+xyz+x'y

    can any1 tell me if i made a mistake? if its going to simplify more?

    also i got this one:
    problem 2: f(a,b,c)= a'b'c+a'bc'+ab'c+abc'+abc i did this:
    =a'b'c+a'bc'+ab'c+ab(c'+c)
    =a'b'c+a'bc'+ab'c+ab
    =a'b'c+a'bc'+ac+a
    =a'b'c+a'bc'+a

    can any1 tell me if i made a mistake? if its going to simplify more?

    tnx a lot for reading!
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    That operation does the period represent?

    Have you checked your work by seeing if the truth tables for the initial equation match your result? Have you tried to verify your design with a Karnaugh Map (assuming you have been introduced to those yet, if not don't worry).

    And thank you for showing your work!

    When checking for equality of two Boolean expressions, it is often a good first step to look for boundary cases. For instance, in your first one the first two terms are identical, so that means we can focus on whether the last term in your answer, x'y, covers all of the cases covered by the last four terms of the original problem that aren't covered by the first two, and no more. Look at the last term w'(x+z). If z=1, and w=0, then this term is 1 and the result is 1, regardless of the values of x or y. Does that agree with your proposed solution?

    In your second one, your answer is saying that if a=1, then the output is 1 regardless of the values of b or c. So look at the original equation and see if you can violate this, namely see if you can force the output to be 0 even when a=1. If we set a=1, then we are left with the last three terms reduced to

    f(1,b,c) = b'c + bc' + bc

    Well, what if b=1 and c=0 ?

    Look carefully at where you claimed that ab'c+ab = ac+a. Do you really believe this?
     
    norules44 likes this.
  3. norules44

    Thread Starter New Member

    Oct 28, 2014
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    well tnx for your reply and ur time, but im not allowed to use karnaught maps yet, i know how to do them but im not allowed, also i know how to make a truth table and find when is 1 and when is 0, but i have to do this exercise and reduce it to the smallest possible equation, showing the way im doing it, yes u are right about the second problem but how do i proceed further?
    f(a,b,c)= a'b'c+a'bc'+ab'c+abc'+abc i did this:
    =a'b'c+a'bc'+ab'c+ab(c'+c)
    =a'b'c+a'bc'+ab'c+ab
    after a research which i did i found out that
    if i got x+xy this equal to x+y
    and x'+xy= x+y
    and x+x'y= x'+y
    but i dont know what to do now, i mean i got 3 terms
     
    Last edited: Oct 28, 2014
  4. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Does that mean that you can't work the problem algebraically and then take out a sheet of paper and use a K-map to check your work?

    I'm not talking about solving the problem using anything other than pure Boolean algebra -- I am talking about developing the ability to CHECK your work. Think about it, I doubt that coming to an on-line forum and asking if your work is correct is MORE acceptable than doing a Karnaugh map or a truth table to check it yourself!

    As for proceeding, look at the part that I called into question and figure out what you did wrong and how to do it right? Feel free to post your musings on that part of it and we can delve into it more deeply if needed.

    And I can't evaluate your first problem until you explain what operation you are using the period for (i.e., in wx.wz).
     
    norules44 likes this.
  5. norules44

    Thread Starter New Member

    Oct 28, 2014
    4
    0
    in the problem 1 i used de morgan's law in this part
    F(wxyz)= wx+xyz+x'yz+x'yz'+[wx+w' (x+z)]
    = wx+xyz+x'y(z+z')+[wx+w' (x+z)]
    =wx+xyz+x'y+(w'+x')wx.wz
    =wx+xyz+x'y


    f(a,b,c)= a'b'c+a'bc'+ab'c+abc'+abc i did this:
    =a'b'c+a'bc'+ab'c+ab(c'+c)
    =a'b'c+a'bc'+ab'c+ab
    after a research which i did i found out that
    if i got x+xy this equal to x+y
    and x'+xy= x+y
    and x+x'y= x'+y
    but i dont know what to do now, i mean i got 3 terms
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    I'll say it again: I CANNOT evaluate your work UNTIL you EXPLAIN what operation you are using the PERIOD for!

    You have (w'+x')wx.wz

    What does wx.wz mean?

    This is a simple question. Why won't you answer it?
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Think about this. If y=1, then x+y is 1 regardless of the value of x. But is x+xy also 1? What if x=0.

    Always, always, ALWAYS ask if the answer makes sense!
     
    norules44 likes this.
  8. norules44

    Thread Starter New Member

    Oct 28, 2014
    4
    0
    i replied it i said i used the morgans law.




    tnx a lot but anyway i dont know what to do to proceed more
     
  9. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    Using DeMorgan's Law does not answer what the dot represents. Try describing the term in English, such as "w and x foo w and z", replacing "foo" with an English word.
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    So what does the dot represent?

    If I tell you f(a,b) = a.b, what is the truth table?

    DeMorgan's Theorems are either: ab = a' + b' or a + b = a'b'

    The next step is to look carefully at your claim that x+xy equals x+y.
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
    2,438
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    Hi,

    You still need to define the dot. Saying it is DM theorem is not helping. You need to write it out completely because using a dot like that is too unconventional. Failure to write it out in FULL will result in you not getting an answer for that problem again.

    Also, are you allowed to use exclusive or, ie "XOR", to simplify? This would help if you could because it can reduce the number of terms needed.
     
  12. amilton542

    Active Member

    Nov 13, 2010
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    Boolean algebra of that complexity presumably implies you should be familiar with k-maps where that of which has already been bought to your attention in post #2. Just do them anyway, even if you're not allowed. If you're required to use Boolean algebra methods alone, do the k-map anyway and use it as an aid (provided you got it right and you're striving for the largest groupings of 2^n) in order to ensure you know what the final simplified logic should look like.
     
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