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The figure above implements the Boolean function:

  1. $\mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\Sigma(0,2,3,5,6,8,12,13)$
  2. $\mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\Sigma(1,3,4,6,7,8,12,13)$
  3. $\mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\Sigma(0,2,4,5,7,8,9,11)$
  4. $\mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\Sigma(1,3,4,6,7,8,9,11)$

2 Answers

2 2 votes

$A, B, C$ are Select Lines of the given 8 $\times$ 1 MUX , where $S2$ is MSB and $S0$ is LSB

Now,

$F = A’B’C’D + A’B’CD + A’BC’D’ + A’BC(1) + AB’C’(1) + AB’CD + ABC’(0) + ABC(0)$

     $= A’B’C’D + A’B’CD + A’BC’D’ + A’BC + AB’C’ + AB’CD$

     $= \ 0001 , \ 0011 ,  \ 0100 ,\  (0110 ,\  0111) , \  (1000 , 1001) ,\  1011$

     $=$ { $1, 3, 4, 6, 7, 8, 9, 11$ }

 

Correct Ans $:$ Option D

1 1 vote
$\text{F(A, B, C, D)}$ is $1$ for minterms $1,9 ,$ also $F$ is $0$ for minterms $0,12.$

So D is correct.
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