Basic boolean logic

Discussion in 'Homework Help' started by fran1942, Mar 25, 2013.

  1. fran1942

    Thread Starter Member

    Jul 26, 2010
    58
    0
    Hello, I am trying to grasp this topic.
    I have the following expression (see attached image).
    I need to
    i) simplify using deMorgans theorem
    ii) then show how the expression could be implemented using only 2 input NOR gates.

    I "think" I have the deMorgans simplification correct, but I am not sure how to do part ii.
    Any help would be much appreciated.
     
    • bool.jpg
      bool.jpg
      File size:
      259.8 KB
      Views:
      83
  2. lunapt

    New Member

    Feb 2, 2013
    12
    0
    write the equation so it has two OR gates, then negate the entire equation
     
  3. fran1942

    Thread Starter Member

    Jul 26, 2010
    58
    0
    thanks, but that is the problem.
    I am not sure how to rewrite my deMorgan simplification as two NOR gates.

    Can anyone please help with this ?
     
  4. lunapt

    New Member

    Feb 2, 2013
    12
    0
    double negate the entire equation so you can deMorgans again to switch the gates
     
  5. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
    1,328
    You used DeMorgan's theorem incorrectly. When you use DeMorgan's theorem, you split up the "not" bar and change the operation. For example:

    (A+B)' = A'*B'

    The ' symbolizes a "Not" bar. As you can see, the entire first part (A+B) is inverted, so you have a bar above all of it. You split the bar so that each part has the bar over it, and then you change the operation from OR to AND (+ to *).

    EDIT: Sorry, you are correct. I misread the expression :p
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    You actually only need a single 2-input NOR gate.

    Your simplification is wrong.

    This is what you started with

    J = [(G'*H')' + (H*I')]'

    Applying DeMorgan's gets you

    J = (G'*H')*(H*I')'

    Applying DeMorgan's to the second term gets you

    J = (G'*H')*(H'+I)

    But at this point you just removed the parens around the second factor. You have to distribute one over the other.

    Now take a step back and look at your original expression

    J = [(G'*H')' + (H*I')]'

    This is ALREADY a NOR gate!

    J = NOR[(G'*H')', (H*I')]

    So focus on the guts of this expression and in three easy steps it will reduce to the OR of just two terms.
     
  7. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
    1,328
    That is not correct. The bar goes over the entire G*H expression. It is NOT G'*H', it is (G*H)'.

    The OP's simplification is done correctly.
     
  8. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Oops. That's what happens when a nearly blind man that sees double and triple images has to switch between windows and remember what lines he saw where.

    It's a shame, too, because my incorrect version simplifies so beautifully!

    But the OP's simplification is still wrong. The mistake I made does not affect the need to distribute one factor over the other.

    So hopefully this is the correct starting point.

    J = [(G*H)' + (H*I')]'

    This is still a NOR gate at the outermost level.

    Applying DeMorgan's

    J = (G*H)*(H*I')'

    Applying DeMorgan's to the second term

    J = (G*H)*(H'+I)

    The OP has this

    J = G*H*H'+I

    Which is wrong, since AND takes precedence over OR and this would simplify to J=I

    Instead, what you have is

    J = (G*H)*(H'+I)

    J = (G*H)*H'+(G*H)*I

    J = G*H*I
     
  9. DerStrom8

    Well-Known Member

    Feb 20, 2011
    2,428
    1,328
    Oops, you're right. Good catch!
     
  10. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    A quick count, assuming I am correct that J=GHI, is that it will take six NOR2 gates.
     
  11. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Bwtween us we'll get it right! :D
     
    DerStrom8 likes this.
Loading...