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Boolean Algebra Hard

  1. Aug 7, 2005 #1
    Boolean Algebra Hard!!!

    hey all, got stuck on some boolean algebra just wondering if you all can check my working out thanks :)
    basically the ` represent bar's and in the example no. 1 the p is barred, r is barred,and the overall function is barred hope you get the gist of things thanks

    1. [(p` + qr)(pq+r`)]`

    my working out is, using de morgan's rule
    = [(p` + qr)]` + [(pq + r`)]
    = p``*(qr)` + r``*(pq)`
    =p*(qr)` + r*(pq)` \\ De morgan's rule again
    = P( q` + r` ) + r(P` + q`) \\ expanded out
    = pq` + r`p + rp` + q`r
    \\ i use the rule that r`p + rp` = `
    thus = q`(R + p)

    is that right hopefully i did it correctly :)

    2. (z + (x*y`)) + yx + (x*(y` + z))
    thats the function my working out is i expanded it out.
    thus
    = (z +xy`) + yx + xy` + zx
    \\ then i left it as it is as use many of the boolean rules and got
    xy` +Z + YX
    X(Y` + y)+ z
    =X + z
    \\i used the karnaugh maps and got x + z to be the simpliest function as well but was wondering ifanyone can check this out for me thanks.


    3. [x` + (y`*z`)][yz` + x`][y`+z`]
    = [x` + z`y`][x` + yz`][y` + z`]
    \\then i used the rule that P(P+Q)=P so that means taking x` as a common factor takinga look at the first 2 functions out ofthe 3
    = x`(yz` + x`)
    =x` \\ now we have (y` + z`) left as a function thus
    = x`(y`+z`)
    \\ using de morgan's rule
    =(x+yz)`

    is that the simpliestform and how would i draw this as a simplified switchinig circuit because i belive the whole function barred you cant draw it???

    thats all hope this isnt a load of gibberish :)
     
  2. jcsd
  3. Aug 7, 2005 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure of this rule. Suppose r = 0, p = 1, then r'p = 1*1 = 1, so 1 + rp' = 1 which is not equal to 0 or '.

    Number 2 is correct. You will just use the fact that x + xA where A can be anything will just be x, and you'll find you have an x swallowing up everything, there's no need for Karnaugh maps. 3 is right. You can always bar a whole circuit. How you bar it depends on what gates you have available to you, but the following two ways work:

    Use an AND and an OR gate to make x+yz, then use an inverter gate to bar it. Otherwise, use a NAND gate and NAND x+yz with itself. If you can bar a single thing like x, y, or z, then you can bar any larger expression. And I believe this way will be better then doing x'(y' + z') because you'd need to bar 3 things.
     
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