NOR Gates, Boolean Algebra and much head-scratching

Discussion in 'Homework Help' started by kone, Feb 8, 2007.

  1. kone

    Thread Starter New Member

    Nov 19, 2006
    Hi all,

    I have to create a circuit using NOR gates which gives an ouput of 1 if 3 or 4 inputs agree

    I created the truth table and got the following

    f = (a'b'c'd')+(a'b'c'd)+(a'b'cd')+(a'bc'd')+(a'bcd)+(ab'c'd')+(ab'cd)+(abcd')+(abcd)

    I then used a Karnaugh map and got

    f = (a'b'c')+(a'b'd')+(b'c'd')+(a'c'd')+(bcd)+(abd)+(abc)+(acd)

    I am struggling and cannot seem to find a way to build the circuit.

    Can anyone point me in the right direction?

    Any help or pointers would be greatly appreciated.
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    Make a truth table. You will find something interesting when counting how many conditions satisfy your requirement and how many do not. You might also see a pattern if you isolate one group from the other. I did.;)
  3. rootboy


    Jan 4, 2007
    I had to take a little self-guided refresher course to answer this one. :)

    It's pretty simple, make yourself a truth table like thingmaker3 suggests. And while you will find that there is a pattern, I would do a "Product-of-Sums" equation instead.

    This is because there are only six cases of 0's to ten cases of 1's, and also because you want to use NOR gates. This makes using "Product-of-Sums" easier than using "Sum-of-Products".

    Then take just one of these six conditions and solve for that (the others will fall into place from there).

    To solve for this one condition, get yourself a four input NOR gate and using NOT gates condition your inputs and solve for the first "0" in your truth table.

    Two hints from here:

    1) Since you have six conditions in your truth table that result in a "0", it will take six four input NOR gates, one for each "0" in the table (actually, since you apparently have to use just NOR gates it will obviously take many more than this to create the entire circuit, but you should get the hint).

    2) Condition your four inputs (A,B,C,D) by complementing each input as inverted and non-inverted inputs.

    It helps to position these vertically above your six four input NOR gates so that you can pick either the inverted or the non-inverted inputs as needed for each of the NOR gates.

    Okay, three hints:

    I got my refresher out of my battered old copy of Clive Maxfield's "Bebop to the Boolean Boogie" which I can't recommend enough.

    Buy his book and turn to pages 90 & 91 (in the first edition, I'm not sure where it is in the later editions). You will thank me later. :)
  4. kone

    Thread Starter New Member

    Nov 19, 2006
    Thanks a lot for all the help guys, I am going to give this another go later today.

    I've just got a copy of Bebop to the Boolean Boogie (2nd edition) the Boolean Algebra Chapter (Chapter 9) looks very promising

    I'll let you both know how I get on with it.


  5. Dave

    Retired Moderator

    Nov 17, 2003