De Morgans Law . . . .NAND Gates

Discussion in 'Homework Help' started by Gem, Sep 22, 2011.

  1. Gem

    Thread Starter New Member

    Aug 25, 2011
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    0
    Hi All,
    I'm stuck on the following question, any advice would be greatly received (thank you).

    Starting with the following statement:
    (A+B).(A+C)
    Show how this can be implemented using NAND gates using De Morgans law.
    So far I have . . . .
    The Boolean expression for a NAND gate is Q=(AB)' which is to say the inverse of AB and so the truth table would be as per below:



    (It won't let me add the table??)


    From this table we can see that the gate only gives a high output if both its inputs are low.
    If we look at our required output Q, we find the expression in brackets is an OR one, as we are adding together, and the whole expression is an AND, as we are multiplying.
    When we apply De Morgan's Law we can make the whole thing an expression of OR statements and construct an NAND circuit around that.
    As we know De Morgan's Law consists of two theorems that state, and when using a computer it is sometimes difficult to get the symbols correct therefore it is common to write:
    (A+B)'= (AB)' and (AB)' = (A+B)'
    Applying De Morgan's Law to (A+B)(A+C) gives us (A+B)' + (A+C)' and turns the whole thing into an OR. However if we negate only once we get the inverse of what we need and so we double negate. . . . . don't we??
     
    Last edited: Sep 22, 2011
  2. Gem

    Thread Starter New Member

    Aug 25, 2011
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    0
    A b q
    0 0 1
    0 1 0
    1 0 0
    1 1 0
     
  3. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    It is a good practice to review any written work you do, before submitting it.
    In you first post, the correct expressions are
    (A+B)'=A'B' and (AB)'=A'+B'
    not what you have written.

    Also in your second post, the table corresponds to a NOR gate, not a NAND one.

    In terms of your problem, let's name F=(A+B)(A+C).
    One can know that if you have an expression which is a Sum of Products, using double negation, a NAND equivalent can be easily taken.

    So try to turn F in a Sum of Products for starters and post your efforts. We'll help you further afterwards.
     
  4. Gem

    Thread Starter New Member

    Aug 25, 2011
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    Sorry about the mistakes, this is all really new to me, I'm an administrator for an electronics company, thought I'd be a good idea to get some basic electronic engineering understanding but really out of my depth now. But thank you for your help!
    The table for the NAND should be:
    A B Output
    0 0 1
    0 1 1
    1 0 1
    1 1 0
     
  5. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    It's not so hard if you take it in little by little. Let's take it slowly:

    F=(A+B)(A+C)
    =(A+B)A+(A+B)C
    =A*A+AB+AC+BC
    =A+AB+AC+BC
    =A(1+B+C)+BC
    =A+BC

    Can you turn it into a NAND circuit now?
     
    Gem likes this.
  6. Gem

    Thread Starter New Member

    Aug 25, 2011
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    0
    Thank you for your help, I'm still confused, I think I've been looking at the same question for so long I have hit a brick wall. I will have a fresh look on Monday and let you know how I get on - many thanks
     
  7. Georacer

    Moderator

    Nov 25, 2009
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    1,266
    Remember, (AB)' is a NAND gate and so is A', which isn't anything else than (AA)'. We can see in that way that a 1-input NAND gate (the same variable in both inputs actually) is a NOT gate.

    So take ((A+BC)')'' and try to make some NAND gates.
     
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