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+2 votes
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Kindly have a try .... IN canonical POS form following equation is written as

F(A,B,C)=AB+BC+AC  

(A)πM(0,1,2,4)  

(B)πM(3,5,6,7)

(C)πM(0,1,2,3)  

(D)πM(4,5,6,7)

asked in Digital Logic by (23 points)
edited by | 526 views

4 Answers

+5 votes
Best answer

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

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

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

$=\Sigma m(6,7,3,5)$ (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

answered by Veteran (413k points)
selected by
+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

answered by (115 points)
+1 vote
A will be the answer try to put this value on karnaugh map and remaining will be your answer.
answered by (213 points)
0
The terms have 2 variables.How to convert it into 3 variables terms so as to map it into a K-map.
0
Ask yourself how to get AB when dealing with karanugh map. You have to put two 1 and try to get the desired variable. After you put all the 1 fill remaining with 0 and try to map them. You get your result.
0
AB = ABC' + ABC

Like this you can do.
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)
answered by Active (1.3k points)

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