NAND circuit to NOR

Discussion in 'Homework Help' started by nanduko, Oct 31, 2009.

  1. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    Hello everyone !

    So i have to turn this circuit that is made of nand gates into the equivalent of this but this time only with NOR gates. This is the circuit :

    [​IMG]


    First i tryed to turn the circuit into some boolen algebra statement but nothing happned, actually it became harder =/ i am strungling trying to solve this loads hours ....help me please...
     
  2. hgmjr

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    Jan 28, 2005
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    Take a look at DeMorgan's Theorem.

    hgmjr
     
  3. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    Well ok De morgan's theorem, but i need an expression 1st for the circuit, there is where i am strungling , i am getting confused all the time....

    This is stupid but anyway... my conclusion is: (((((pp)'p)')'(rr)')')'

    this is not right, right?
     
  4. hgmjr

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    Jan 28, 2005
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    First observation is that there is no q in your boolean expression.

    Have you noticed that four of the nand gates are configured as simple inverters?

    hgmjr
     
  5. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    indeed theres no q, it was meant to be like this (((((pp)'q)')'(rr)')')'

    but still not sure if this is .
     
  6. hgmjr

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    HINT: If you feed the same signal into both inputs of a NAND gate, all that is doing is inverting the input.

    hgmjr
     
  7. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    yeah i know that, thats why so many '''''' xD

    ill try again and post
     
  8. hgmjr

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    Jan 28, 2005
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    Using your boolean notation, I get simply:

    (p'q)r'

    because (pp)' = p'

    hgmjr
     
  9. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    i tryed again in a bit different way,

    isnt it ( (p' NAND q)' NAND (r') )' ?
     
  10. hgmjr

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    Jan 28, 2005
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    That looks correct although it would be clearer if some simplification were performed.

    hgmjr
     
  11. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    hmmmm i tried to simplify it a bit, not sure again if its right

    ( (p' NAND q)' NAND (r') )' =

    ((p' q) nand (r') )' =

    (((p' q) (r') )' )' =

    ((p+q') + r)' =

    (p'q) r'

    i believe theres a mistake somewhere
     
  12. hgmjr

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    Jan 28, 2005
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    If you take a look at post #8 in this thread, you will see that I got the same simplified answer you have in post #11.

    I guess your next step is to implement this boolean expression using NOR gates. Right?

    hgmjr
     
  13. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    yeah as long as this result is ok for the given circuit, ah btw forgot to mention that i hd already proved that p'q = p nor (p nor q) , the problem is the r' now, should i call for example p'q=d and say r'd = r nor ( r nor d) ?

    Thanks in adance
     
  14. hgmjr

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    My thought was that if you needed an inversion, you would feed the same signal into both inputs of a NOR gate just as was done in the NAND gate version of the circuit.

    hgmjr
     
  15. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    indeed i could feed a nor gate with 2x r so the result is r' , but still theres a multiply , how can do that as a nor gate . i mean --> (p'q) multiply r'
     
  16. hgmjr

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    You do realize that there is not an actual multiply operation involved. That is simply the indication of an AND function.

    hgmjr
     
  17. hgmjr

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    See if the information at this website helps you out.

    hgmjr
     
  18. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    ehm lol yeah, i think i finally figured out, ill post the scematic as soon as possible
     
  19. nanduko

    Thread Starter New Member

    Oct 31, 2009
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    nor version of this (p'q) r' is :


    [​IMG]


    hopefully

    EDIT: i think that the second NOR (where r entrance is ) is not needed tho.
     
    Last edited: Oct 31, 2009
  20. hgmjr

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    Jan 28, 2005
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    Your diagram has a couple of errors. Take another stab.

    hgmjr
     
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