edited by
3,397 views

4 Answers

Best answer
5 votes
5 votes

$F(A,B,C) = AB+BC+AC$

$= ABC' + ABC + ABC + A'BC + AB'C + ABC$

$= ABC' + ABC + A'BC + AB'C$

$=\Sigma m(6,7,3,5)$ (m - minterm and M-maxterm))

$=\pi M(0,1,2,4)$ (proof shown below)




$F(A,B,C) = \Sigma m(3,5,6,7)$

$\implies F'(A,B,C) = \Sigma m(0,1,2,4)$

$\implies F'(A,B,C)=m_0+m_1+m_2+m_4$

$\implies (F')'(A,B,C) = (m_0+m_1+m_2+m_4)'$

$\implies F(A,B,C) = (m_0' m_1' m_2' m_4)$ (DeMorgan's law)

$\implies F(A,B,C) = (M_0 M_1 M_2 M_4)$

$\implies F(A,B,C) = \pi M(0,1 2,4)$ (Complement of a minterm is its corresponding maxterm, for example,  $m_0 = A'B'C'$ and $m_0' = (A'B'C')' = A+B+C = M_0$)

Ref: http://www.cs.uiuc.edu/class/sp08/cs231/lectures/04-Kmap.pdf

selected by
2 votes
2 votes

First take the function, this is in SOP so find the min terms of the equations

so min terms are 3,5,6,7

For Finding the MAX TERMS subtract these  min terms from whole possible  terms with 3 variable which are \sum{0,1,2,3,4,5,6,7}- \sum{3,5,6,7} =\prod(0,1,2,4)

ans is option A

1 votes
1 votes
A will be the answer try to put this value on karnaugh map and remaining will be your answer.
0 votes
0 votes
option A is correct.

F(A,B,C)=AB+BC+AC=110+111+011+111+101+111=110+111+011+101=Σm(0,1,2,4)=πM(0,1,2,4)

Related questions

2 votes
2 votes
1 answer
1
1 votes
1 votes
2 answers
2
Jason asked Mar 21, 2018
2,466 views
Should we use Don't care terms while calculating POS expression?
0 votes
0 votes
1 answer
4
radha gogia asked Aug 16, 2015
13,613 views
R is not in 3NF R is in 3NF but not in BCNF R is in BCNF but not in 4NF R is in 4NF How to approach this ques ?