5 variable Karnaugh maps

Discussion in 'Homework Help' started by katherine1117, Oct 31, 2009.

  1. katherine1117

    Thread Starter New Member

    Oct 28, 2009
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    i got a problem in solbing this:

    using a 5-variable karnaugh map, find the minimal pos for the boolean function,G(M,N,O,P,Q)=m(1,3,5,8,10,12,16,20,21,25)+d(0,4,7,28,30)

    i draw two maps when M=0and M=1 separatly and i got:M=0:(M+N’+Q’)(M+O’+P+Q’)(M+N+Q) when M=1:(M’+P)(M’+N+Q)(M’+N+Q’)(M’+N’+O’+Q)

    then how can i combine these together to come up with the anwser?

    thanks very much!
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Combine the two into a truth table, then go from there. A truth table will show relationships between the numbers. Certains gate functions will be obvious.
     
  3. katherine1117

    Thread Starter New Member

    Oct 28, 2009
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    thanks very much

    but we were supposed to solve the problem thought the karnaugh map, and i did draw two maps separately, should i multiply those together? or sum it?
     
  4. Wendy

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    Mar 24, 2008
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    The truth table is a Karnaugh map. Check out the Wikipedia entry, http://en.wikipedia.org/wiki/Karnaugh_maps.

    If I understand your first post the first equation defines the zeros, the second one defines the ones. You should be able to lay this out as a large 5 variable truth table.

    Why not show us what you've come up with so far?
     
  5. katherine1117

    Thread Starter New Member

    Oct 28, 2009
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    i understand. my teacher did not mention how to deal with such kind of problem though a big 5 variable truth table so i don't know wether i should use only one karnaugh map~
     
  6. Wendy

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    Mar 24, 2008
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    The answer is yes, you should use one map. The wikipedia article goes through how to make such maps, though they stop at 4. 5 variables is not unreasonably large, and is manageable.
     
  7. katherine1117

    Thread Starter New Member

    Oct 28, 2009
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    thank you so much i have been thinking about this question for quite a long time...
     
  8. Ratch

    New Member

    Mar 20, 2007
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    katherine1117,

    I am going on vacation for a couple of weeks tomorrow, so I won't have too much time to devote to your problem. You have both "do care" and "don't care" terms. A truth table is not going to help you much to simplify those minterms you listed. I assume that is what you want to do. You first have to get a 5 term K-map. Preferably one that shows the minterms on it. Then mark the care terms one way and the don't care terms another way or with a different color. Use the don't care terms to simplifying the do care terms, but don't do any simplification for only the don't care terms. I worked out the solution for you by both a K-map and the Quine-McCluskey tabulation method. You should still try use a K-map to gain practice.

    MNO'P'Q + N'P'Q' + M'NO'Q' + M'N'Q + N'OP' + M'P'Q' (or OP'Q' for last term)

    Notice that minterm #25 does not combine with any other term and the last term has two choices to cover minterm #12.

    If you have any questions, ask them before tomorrow.

    Ratch
     
  9. Wendy

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    Mar 24, 2008
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    Never the less, that was the assignment. You go with what was given.
     
  10. Ratch

    New Member

    Mar 20, 2007
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    katherine1117,

    I see that I did not read your request carefully. You wanted the POS, not the SOP. In that case, you want to make the function negative by taking the missing minterms. So F' = m(2,6,9,11,13,14,15,17,18,19,22,23,24,26,27,29,31) and d(0,4,7,28,30) . Simplying by a K-map we get F' = N'PQ' + M'NQ + OP + MNQ' + MP + MNO + MNO'Q . Then a simple application of DeMorgan's Theorem gives F = (N+P'+Q)(M+N'+Q')(O'+P')(M'+N'+Q)(M'+P')(N'+N'+O')(M'+N'+O+Q')

    Ratch
     
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