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Digital logic design

  1. Nov 16, 2007 #1
    [SOLVED] digital logic design

    i want to simplify this function using four-variable map: F(w,x,y,z) = [tex]\sum[/tex](0,1,2,4,5,7,11,15) , the map contains one "1" at the upper right corner. my question is how can i take this 1 in a group? or what can i do with it???
  2. jcsd
  3. Nov 16, 2007 #2
    another question is if i have a 3-variable map ( i mean the k-map) full of 1s and Xs(representing the don't care conditions), can i take all as one group??? if yes then the function will be zero!??
  4. Nov 17, 2007 #3

    for your first question can you give us your truth table -

    what I understand is that you gave us a sum of minterms and stated that there's only a 1 in the logic table "at the upper right corner" - however this is for your k-map am I right?

    If that's the only case that gives you a logic of 1 - then what does that say about your other logic combinations? Which one should you be concerned with then?
  5. Nov 17, 2007 #4
    If your k-map is only filled with 1's an x's
    then the function will be true for all cases so to speak

    why would the function be 0?
    Last edited: Nov 17, 2007
  6. Nov 17, 2007 #5
    you don't need the truth table, and you can get it from the sum of minterms if you want....i ment that there is one 1 at the corner alone, and i have to simplify the function by taking 1's in a group (2 ones or 4 or 8...) but now i solved this, i took it alone and it worked....
  7. Nov 17, 2007 #6
    the function is with variables x, y, z; when taking all in a group and even in more groups, but all th literals are for sure taken in groups, then x and x' will cancel, and y will cancel with y', and z with z'
  8. Nov 17, 2007 #7
    hmm i see what you are trying to do

    but then for what cases will your system give you a logic of 1?

    if all cases are true then it should be just a logic 1 for your function.

    You may try PM'ing berkman for help - he helped many people with Digital design questions.
  9. Nov 17, 2007 #8
    sorry but what is PM'ing berkman?
  10. Nov 17, 2007 #9
    you send a personal message to a member called berkman - he's a member on these forums that helped me back when i was taking digital design a couple years back.

    I'm still certain that for a function if all cases are true or don't cares - then you get a f() = 1
    - double check with your professor if you have to.
  11. Nov 19, 2007 #10


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    Staff: Mentor

    I've read the thread, and sorry, I don't understand the question. Could you please re-state it? And if you could post a copy of your K-map along with your restatement of the question, that would help.
  12. Nov 29, 2007 #11
    Answer is << complete solution deleted by berkeman >>
    Last edited by a moderator: Nov 30, 2007
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