Some simple(or not hehe) boolean algebra

Discussion in 'Homework Help' started by chaplin27, Sep 18, 2005.

  1. chaplin27

    Thread Starter New Member

    Sep 18, 2005
    2
    0
    Hey everyone,
    I've never posted here before, but I just stumbled upon it and thought I'd see if seomeone could help me out. I have to simplify the following expression:
    (BC' + A'D)(AB' +CD')

    Now, of course I don't want the answer (not that you'd give it anyways :p) hehe, I would just like a first step. I'm VERY new to this and I feel like I'm making some of the rules I'm using up haha, so that's not a good position to be in :). Any suggestions? Thank you very much for any replies. -Chap
     
  2. techduq

    New Member

    Aug 31, 2005
    7
    0
    A rule you'll need: AA' = 0, or more generally AA'BCDEFG...=0. Do you see why? Multiply out your expression and I think you're in for a surprise...
     
  3. chaplin27

    Thread Starter New Member

    Sep 18, 2005
    2
    0
    I'm an absolute idiot and looked right past the whole fact that I had to multiply the two terms!! Jesus, thanks for that delightful insight as well as the AA' = 0 rule...makes total sense...curiously, what does it mean if an expression evaluates to 0? In terms of the circuitry I should say...
    Thanks again!
     
  4. techduq

    New Member

    Aug 31, 2005
    7
    0

    It's easy (if rather pointless) to make a circuit that has a permanently low output. Consider an AND gate with four inputs (just as an example). Let the inputs be A, B, C, and A' (A run through an inverter). You can see that with A and A' both on the input, the gate will never have the condition necessary to output a high (i.e., you'll never have all the inputs high).
     
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