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Pretty simple NAND gate question

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data

    I was looking through some examples in my Digital Logic book and I stumbled across one that gave an answer of x1[itex]\uparrow[/itex](x2[itex]\uparrow[/itex]x2). The answer I got was x2[itex]\uparrow[/itex](x1[itex]\uparrow[/itex]x2).

    After making a truth table I'm finding that these are not equal solutions. Unless, I'm doing something wrong.


    3. The attempt at a solution

    See above. Any help is appreciated.
     
  2. jcsd
  3. Feb 20, 2012 #2

    NascentOxygen

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    Can you write this in words so I can see where NAND comes into it?
     
  4. Feb 20, 2012 #3
    The problem asked to turn x1[itex]\overline{x}[/itex]2 + x2[itex]\overline{x}[/itex]1 into a NAND only circuit. I used decomposition in the last steps to get my answer.
     
  5. Feb 20, 2012 #4

    rcgldr

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    Can you show the truth table for x1[itex]\overline{x}[/itex]2 + x2[itex]\overline{x}[/itex]1? I'm not sure the book answer is correct assuming that ↑ means NAND.
     
  6. Feb 20, 2012 #5
    I got my question answered today in class. Those two solutions are equivalent. Thanks for the help though.
     
  7. Feb 20, 2012 #6

    rcgldr

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    I'm somewhat confused by this. You never explained what ↑ means.

    Isn't x1[itex]\overline{x}[/itex]2 + x2[itex]\overline{x}[/itex]1 the same as XOR? Wiki article for XOR implemented with NAND gates:

    http://en.wikipedia.org/wiki/NAND_logic#XOR
     
  8. Feb 20, 2012 #7
    According to my book, ↑, is a NAND gate operator.
     
  9. Feb 20, 2012 #8

    rcgldr

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    OK, so again I ask what is the truth table for

    x1[itex]\overline{x}[/itex]2 + x2[itex]\overline{x}[/itex]1

    and the truth table for

    x1↑(x2↑x2)

    or

    x2↑(x1↑x2)
     
  10. Feb 20, 2012 #9

    NascentOxygen

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    But a poor choice, and bound to lead to confusion.
     
  11. Feb 21, 2012 #10

    NascentOxygen

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    Yes, that is exclusive NOR.

    x2 NAND x2 is just NOT x2

    x2↑(x1↑x2) evaluates to x2 NAND NOT x1

    So expressions in post #1 are NOT identical.
     
    Last edited: Feb 21, 2012
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