6,034 views

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

Observation: As f1 has 4 if f2 will have 4 then f should also have 4,

Except for option C, all have 4 in f2.
Option C is right choice for F2

### Subscribe to GO Classes for GATE CSE 2022

Wth AND gates we will choose intersection of min-terms.

With OR gates we will take union of min-terms.

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

this should be the best answer...

From option we can decide in optimal time!

Now, there is only 3 choices 4, 6, and 8 and in final f we can see that 6 and 8 is there but 4 is not! and in last stage we are having 'OR' gate which means it includes everything(union)!

so, option without 4 is only our anser. Hence, C

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
by
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)
by
After solving the Logic Gate, we got equation –

f = f1.f2 + f3

which means what is present in f3 is always present in f because we are taking union with f3.

as f3 is (1,6,15) so f must have these , and instead of these we have one extra i.e. 8( which means 8 is comes from intersection of f1.f2)

and 6 is already be there in f3, so may be it is not present in f2, But 4 is also not present in f2 because , if it is present then f1.f2 = { 4 } which also comes in f & it is not present in f.

That’s why options which have 4 is not correct answer.

and we find answer as {6 , 8}