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Homework Help: OR NAND to NAND

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Hi guys I need to implement [(B+C)*D]' in NAND only. can someone please help.

    2. Relevant equations

    3. The attempt at a solution
    this is what i have.
    [(B+C)*D]'=(B+C)'+D'=(B'C')+D' and i get stuck...can anyone please explain and help.

    any good resources to teach/practice this?

    I just built a table and the result [(B'C')'*D]' worked. can anyone explain the algebra behind this?
    I don't get what rule lets you go from B'C'+D' OR to AND since B'C' is not inverted all together (BC)' so it cant be treated as X' right?
    Last edited: Oct 9, 2011
  2. jcsd
  3. Oct 9, 2011 #2


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    Staff: Mentor

    You can always prove the basic conversions of AND and OR to NAND and NOR versions of those gates using enumerated logic tables. Then use the those results as a basis for making algebraic substitutions. It's equivalent to memorizing a list of identities and knowing when to apply them.

    You may find that the "algebra" is easier to do visually using the circuit symbols for the gates. You can pick out a few rules of thumb for transforming gates from one type to another by pondering the attached crib sheet.


    Attached Files:

  4. Oct 9, 2011 #3
    That Helps a lot. Thank you
  5. Oct 24, 2011 #4
    A couple of things to remember that helped me with these kinds of problems:

    1 - Any AND or OR logic gate (including NAND and NOR) can be replaced with the opposite symbol if you invert ALL of its inputs and outputs (put an inverting circle at each point that doesn't already have one and remove any existing inverting circles).

    2 - Inverting circles can be moved anywhere along the line it is connected to. For instance, if the output of a NAND gate feeds the input of another gate, the inverting circle can be "slid" away from the NAND (making it into an AND gate) and placed at the input of the gate it is connected to.

    3 - A NOT gate (inverter) can be replaced by any NAND or NOR gate with all of its inputs tied together.

    Hope that helps.
  6. Oct 24, 2011 #5


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    Staff: Mentor

    De Morgan's Theorem: A or B = not (notA and notB)
    You may find this easier to remember if you NOT both sides, leaving
    not (A or B) = notA and notB
    ..... this is the NOR operation

    You really only need to memorize one of De Morgans theorems. The other one is similar, but with operators swapped.

    A and B = not (notA or notB)
    or, taking NOT of both sides, you have:
    NOT (A and B) = notA or notB
    ..... that's the NAND operation
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