problem with some gate

Discussion in 'Homework Help' started by roeyhaim, Nov 22, 2010.

  1. roeyhaim

    Thread Starter New Member

    Nov 22, 2010
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    hello guys.
    I'm stuck with a problem and need help...

    This is the Q:
    I tried for all night and didn't get it....
    any suggestion?
     
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Where are you stuck?

    A\smile B= A\overline{B}

    Apply that to the function, have you had previous examples of applying logic to a function?
     
  3. roeyhaim

    Thread Starter New Member

    Nov 22, 2010
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    well i tried to simplify the function.
    I got F(x,y,z) = xy + xz'

    but i dont know how to continue...
     
  4. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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  5. Georacer

    Moderator

    Nov 25, 2009
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    You did well simplifying the function. Your task from now on is to convert any single operation to a smile operation. Start with a factorization.

    Here are some hints:
    a+b'=(a'b)'
    x'=(x'1)

    Can you figure it out?
     
  6. renotenz

    New Member

    Oct 25, 2010
    11
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    You can still simplify that function actually... there's one common letter there

    After that you shall see a pattern which Georacer already wrote.

    Also remember that : A'B = BA'

    Cheers :) <= Smile function for you :D
     
  7. roeyhaim

    Thread Starter New Member

    Nov 22, 2010
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    I'm sorry, but i still don't get it...

    I give up...

    My brain is gonna explode.

    Thanks for trying :)

    Now I scared from the yellow smile :D
     
  8. Georacer

    Moderator

    Nov 25, 2009
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    Don't give up yet. Try a bit more. You have problems doing the factorization? Finding the part to be substituted?

    Post any scribbles you have done, even if they are completely wrong. Don't let a good thread end without a closure.
     
  9. roeyhaim

    Thread Starter New Member

    Nov 22, 2010
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    OK...
    but please dont laugh at me...

    i thought maybe to put the x out and then to play a little with the y

    maybe to do some demorgan.

    but the Q is what to do with the +

    any hint / advise?
     
  10. renotenz

    New Member

    Oct 25, 2010
    11
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    There it is.

    Put X out, then you'll get X(Y + Z')

    See any resembelance with what Georacer has typed?

    Don't forget that your :) is A'B, same as BA', same as X'Y, YX', etc...
     
  11. roeyhaim

    Thread Starter New Member

    Nov 22, 2010
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    so, tell me if im right

    now i got X(y'z)'

    and put this in circuit :

    X:)1 = X'
    Y:)Z = Y'Z
    Y'Z:)1 = (Y'Z)'
    X':)Y'Z = X''(Y'Z)'

    X''(Y'Z)' = X(Y+Z')

    Am I right?
    This is the solution?
     
  12. Georacer

    Moderator

    Nov 25, 2009
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    Not quite.
    Y'Z:)1 = (Y'Z)' is correct only with parentheses: (Y'Z):)1 = (Y'Z)' and
    X':)Y'Z = X''(Y'Z)' is wrong.
    You have also written down some unneded equations. So let's sum it up:

    xy+xz'=
    x(y+z')=
    x(y'z)'=
    (y'z)'x=
    ...

    Can you take it from here and substitute for the :) operation? Make one step at a time and start from inside out.
     
  13. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    From the original question, the solution should have the :) operator only, so go through the steps to substitute it, and leave it (and ONLY the :)) in for the final solution.
     
  14. roeyhaim

    Thread Starter New Member

    Nov 22, 2010
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    sorry but im lust again...
    I feel so :eek:

    What i need is to figure out how to get
    X(Y'Z)' by input to a :) gate only two variables?

    with no other gates?
    just ONE AND ONLY :) gate?
     
  15. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    As I read it, you can use all the :) gates you want, but you aren't allowed to have an inverter.
     
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