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A function $F(A, B, C)$ defined by three Boolean variables $A$, $B$ and $C$ when expressed as sum
of products is given by

$F = \bar{A}⋅ \bar{B} ⋅ \bar{C}+ \bar{A}⋅ B ⋅\bar{ C}+ A ⋅ \bar{B} ⋅ \bar{C}$

where, $\bar{A}, \bar{B},$ and $\bar{C}$ are the complements of the respective variables. The product of sums
(POS) form of the function $F$ is

  1. $F = (A + B + C) ⋅ (A + \bar{B} + C) ⋅ (\bar{A}+ B + C)$
  2. $F = (\bar{A}+ \bar{B} + \bar{C}) ⋅ (\bar{A}+ B + \bar{C}) ⋅ (A + \bar{B} + \bar{C})$
  3. $F = (A + B + \bar{C}) ⋅ (A + \bar{B} + \bar{C}) ⋅ (\bar{A}+ B + \bar{C}) ⋅ (\bar{A}+ \bar{B} + C) ⋅ (\bar{A}+ \bar{B} + \bar{C})$
  4. $F = (\bar{A}+ \bar{B} + C) ⋅ (\bar{A}+ B + C) ⋅ (A + \bar{B} + C) ⋅ (A + B + \bar{C}) ⋅ (A + B + C)$
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2 Answers

3 votes
3 votes

Answer should be (C)

Here , F(A, B, C) = A̅⋅ B̅ ⋅ C̅+ A̅⋅ B ⋅ C̅+ A ⋅ B̅ ⋅ C̅  .Since it is written in Sum of Products(SoP) and  It contains minterms as m0 , m2 , m. So , We can write   F(A, B, C)  as Σ(0,2,4) .it means in truth table, F(A,B,C) is '1' when decimal equivalent of (A,B,C) is 0,2,4 . So, in other positions ie. 1,3,5,6,7  F(A, B, C) should be '0' .

So, Simply , F(A, B, C) = Σ(0,2,4) = π (1,3,5,6,7)

Here , π (1,3,5,7,8) means We have maxterms as M1 ,M3 ,M5 ,M7 ,M8 . So , F(A,B,C) can be written in Product of Sum(PoS) as  F = (A + B + C̅) ⋅ (A + B̅ + C̅) ⋅ (A̅+ B + C̅) ⋅ (A̅+ B̅ + C) ⋅ (A̅+ B̅ + C̅)

1 votes
1 votes

F(A,B,C) = A'B'C' + A'BC' + AB'C'

                = Σm(0 , 2, 4 ) 

                = πM(1 , 3 , 5 , 6 , 7  )

                = (A + B + C') ⋅ (A + B' + C') ⋅ (A'+ B + C') ⋅ (A'+ B' + C) ⋅ (A'+ B' + C')

option C is matching

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Please provide, how to make the truth table of such type of questions