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Logic Gates with XOR gate need to verify solution

  1. Feb 16, 2012 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    Basically and most importantly... I wanted to see if I got my F function correct. That's at clause 2.

    At clause 3 I'm told

    a = b = 1
    c = d = 0

    and to calculate

    At clause 4 I'm asked if the logic value of P is 1, if the logic value of Q matters...

    3. The attempt at a solution

    http://img528.imageshack.us/img528/4162/annnnswer.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 16, 2012 #2

    I like Serena

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    What did you get for "p"?
     
  4. Feb 17, 2012 #3

    Femme_physics

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    Hey ILS

    To what question? Only in the last question they refer to "P", and I wrote my answer in this scanned paper.
     
  5. Feb 17, 2012 #4

    I like Serena

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    Hi :shy:

    In your diagram a point is marked with the letter "p".
    The corresponding part in your formula seems a little off.
     
  6. Feb 17, 2012 #5

    Femme_physics

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    It's just a point in the diagram, it has no relevance to the formula
     
  7. Feb 17, 2012 #6

    I like Serena

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    Your formula contains ##\overline{b}cd + \overline{d}##.
    This corresponds to the result of the OR gate "P" on the signals at p and at q, which is ##p+q##.

    This means that effectively you have ##p=\overline{b}cd## and ##q=\overline{d}## in your formula.
    The expression for q is correct, but the expression for p is not.
     
  8. Apr 1, 2012 #7
    N=Logic NOR
     
  9. Apr 2, 2012 #8

    Femme_physics

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    Sorry I must've missed this post.

    Is there any other posts I've missed for over a week?

    To reply: I don't think the actual signals in q and p matter at all. Only if they are one or zero. if we know one of them is one, then the other's irrelevant. Doesn't it make sense?
     
  10. Apr 2, 2012 #9

    NascentOxygen

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    In part (2) you have not recognized that Z is an inverting gate, it's NAND.
     
  11. Apr 2, 2012 #10

    I like Serena

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  12. Apr 2, 2012 #11
    but we dont know what they are, as they are dependent on the input, so we can't write either one off as irrelevant. whether or not you care about what the signal is, in order to state that the output of P = /bcd + /d, as you have done in your answer to clause 2, you need to be sure that those equations are correct.
    (which they are not, and that has caused you to answer part 3 incorrectly)

    (note P the gate ≠ p the signal)

    edit: just realised you may be talking about clause 4, in which case you are correct, however I Like Serena was pointing out a mistake in clase 2.
    the labels p and q are the signal inputs to the gate P.
     
  13. Apr 4, 2012 #12

    Femme_physics

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    Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh!!!!!!!!!!

    NOW I understand :)

    So let me get back to the first reply :P


    That would be
    http://img442.imageshack.us/img442/9803/peepee.jpg [Broken]


    Ahh...I see :) I forgot a big tag over everything. *smacks forehead*

    And now to correct the next question based on that:

    http://img855.imageshack.us/img855/1918/abc800.jpg [Broken]



    Thanks, I'll get on it :)
     
    Last edited by a moderator: May 5, 2017
  14. Apr 4, 2012 #13

    I like Serena

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    All good now.


    You still seem to have missed the last one!
     
  15. Apr 5, 2012 #14

    Femme_physics

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    You rock!
    Heh, on yea...that.. as soon as I fix my hair :) Thank you
     
  16. Apr 5, 2012 #15
    In X we get b' = 0
    In Y we get a.b = 1.1 =1
    In Z we get (b'.(c.d))'=b+(c.d)'=b+c'+d'=1+1+1=1 DeMorgan's Law (a.b)'=a'+b'
    In N we get (d.d)'=d'=1
    In P we get b+c'+d'+d'=b+c'+d'=1+1+1=1
    In R we get 1+1 , now since inputs are even in number , XORing gives result 0 , i-e f=0
    Hope it helps you ...
     
  17. Apr 5, 2012 #16
    Note here you mentioned N here to be NAND , its NOR but for this case the answere will come the same ...
    In N we get (d+d)'=d'=1
     
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