# 5 variable Karnaugh maps

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

1. ### katherine1117 Thread Starter New Member

Oct 28, 2009
9
0
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=1M+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
20,772
2,540
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
9
0
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 Moderator

Mar 24, 2008
20,772
2,540
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
9
0
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 Moderator

Mar 24, 2008
20,772
2,540
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.

Oct 28, 2009
9
0

8. ### Ratch New Member

Mar 20, 2007
1,068
3
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 Moderator

Mar 24, 2008
20,772
2,540
Never the less, that was the assignment. You go with what was given.

10. ### Ratch New Member

Mar 20, 2007
1,068
3
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