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sevenshaded

  • 2 years ago

Find the minimum sum-of-products expression f(a, b, c, d) = (maxterm numbers)(1, 2, 3, 4, 9, 15) When I made a Karnaugh map for the maxterms, I came up with (B+C+D')(A+B +C')(A+B'+C+D)(A'+B'+C'+D') a) I think that's incorrect, perhaps due to my groupings of the 0's. b) If my answer is in fact correct, I can't figure out how to put it into a product of sums form. I just applied DeMorgan laws on it, but I'm not sure that's correct either. Please help? This is basically a maxterm expansion from a Karnaugh map (which I may or may not have grouped incorrectly) that I need in SOP

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  1. brahmaec
    • 2 years ago
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    Maxterm(1,2,3,4,9,15)=Minterm(5,6,7,8,10,11,12,13) so as u solve POS(Maxterm) for logic zero in K map , just solve SOP(Minterm) for logic 1 for the above stated equation to get the result in SOP or get the POS form in minimized form and get its dual relation that would convert the POS to the SOP...

  2. sevenshaded
    • 2 years ago
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    Okay, so are you saying to map the ones out, and take the minterm off of them? Doing that, I got B'C'D' + ABC' + BC'D + A'BC + ABC' + AB'C. Is that the correct answer to the problem?

  3. sevenshaded
    • 2 years ago
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    Also, I think you forgot to include 0 and 14 in your list of minterms...?

  4. sevenshaded
    • 2 years ago
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    Actually, I think I made a mistake. Is it B'C'D' + ABC' + BC'D + A'BC + ACD' + AB'C ?

  5. brahmaec
    • 2 years ago
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    ya i forgot to put in there there wud be 5 min term B'C'D'+ABC'+A'BD+BCD'+AB'C

  6. brahmaec
    • 2 years ago
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    also the dual is not going to give us the correct expression we can find out the SOP by the same relation Maxterm(1,2,3,4,9,15)=Minterm(0,5,6,7,8,10,11,12,13,14)

  7. sevenshaded
    • 2 years ago
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    I don't think I see where A'BD would be. But I think I can see BCD' as a pair of ones. Can there be more than one correct answer?

  8. brahmaec
    • 2 years ago
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    does ur answer match |dw:1349874605579:dw| check the grouping..

  9. sevenshaded
    • 2 years ago
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    Oh, I mapped it so that AB is on the top and CD is on the side. Lemme compare...

  10. sevenshaded
    • 2 years ago
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    I'm still not sure how you got those pairings. I don't even have ones listed in some of those areas. We're mapping the minterms, right?

  11. brahmaec
    • 2 years ago
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    yup...the min terms (0,5,6,7,8,10,11,12,13,14)

  12. sevenshaded
    • 2 years ago
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    |dw:1349875400799:dw|

  13. sevenshaded
    • 2 years ago
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    That's the map I came up with. In a case like the bottom right horizontal grouping in the lowest row across, would it matter whether I grouped that one with the one on its left or right? Since it's in the middle of two 1s.

  14. sevenshaded
    • 2 years ago
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    |dw:1349875843305:dw| I numbered my rows/columns.

  15. brahmaec
    • 2 years ago
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    thats why u r getting wrong answer this should be done like this .....|dw:1349875916540:dw|

  16. sevenshaded
    • 2 years ago
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    So you want to avoid overlapping groups if you can?

  17. brahmaec
    • 2 years ago
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    ya if there wud be overlapping the expression obtained is not in its minimized form ....and there wud be more number of minterms in the expressions..

  18. sevenshaded
    • 2 years ago
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    Okay, so the final answer is B'C'D' + ABC' + A'BD + BCD' + AB'C?

  19. sevenshaded
    • 2 years ago
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    Ohhh, I see. I had that backward. That overlapping was minimizing the expression.

  20. sevenshaded
    • 2 years ago
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    Yeah, that makes sense. Thank you very much for your help. I wasn't sure if I was mapping correctly or not. Makes a huge difference, lol. :)

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