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37 votes
37 votes

Given $f_1$, $f_3$ and $f$ in canonical sum of products form (in decimal) for the circuit

$f_1 = \Sigma m(4, 5, 6, 7, 8)$

$f_3 = \Sigma m(1, 6, 15)$

$f = \Sigma m(1, 6, 8, 15)$

then $f_2$ is

  1. $\Sigma m(4, 6)$

  2. $\Sigma m(4, 8)$

  3. $\Sigma m(6, 8)$

  4. $\Sigma m(4, 6, 8)$

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5 Answers

43 votes
43 votes

$\begin{matrix} \text{Options} & f_1 \cdot f_2 & f_3 + (f_1 \cdot f_2) & \\ A & \{4,6\} & \{1,4, 6,15\} & {\color{Red}{fails}} \\ B & \{4,8\} & \{1,4,6, 8,15\} & {\color{Red} {fails}}\\ C & \{6,8\} & \{1,6,8,15\} & ok\\ D & \{4,6, 8\} & \{1,4,6, 8, 15\} & {\color{Red} {fails} } \end{matrix}$

answer = option C

6 votes
6 votes
option c is correct.

first of all the and term is one only when when both are one. and or is one when any one is one. so the terms in answer wll be there which comes either in f1.f2 or in f3 . as there is an "OR" between them . so the terms 1,6,15 are contributed by f3 while 8 should be contributed by f1.f2. so we want that f1.f2 should atleast result in 8. and we know for that both should contain 8. f1 is having 8 so f2 atleast should contain 8. now what if f2 contain 4 also. then 4 should apper in f. which is not. so 6,8 is the valid option
1 votes
1 votes
Expression Boils down to:

$f1.f2 + f3 = f$

$(f1 ⋂ f2) U f3 = f$

we use intersection for AND because if a f1 and f2 both are true for a minterm then only output will be true.

we use Union for OR because if one among  f1 and f2 is true for a minterm then output will be true.

${4,5,6,7,8}⋂ f2 U {1,6,15} = {1,6,8,15}$

${4,5,6,7,8}⋂ f2 = {8}$

$f2 = {8,any other elements preset in f }$

check for options c contains {6,8} 6 is in f

option (C)
Answer:

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