GATE CSE 2008 | Question: 8

Question

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

• A. $\Sigma m(4, 6)$

• B. $\Sigma m(4, 8)$

• C. $\Sigma m(6, 8)$

• D. $\Sigma m(4, 6, 8)$

Comments

4,5,7 Can not present in f2. Because if we add 4,5,7 in f2 then f will also contain 4,5,7.

f2 must contain 8 because 8 is present in f but not present in f3 so 8 must be present in both f1 and f2 then only we will get 8 in final output.

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Detailed Video Explanation with Complete Analysis: https://youtu.be/r7V6uY_t7X4?t=6304

Must Watch the above Complete Lecture to learn Combining Boolean Fuctions wrt Minterms & Maxterms for all standard boolean operations.

Note: $2^{12}$ different functions are possible for $f_2$ (Watch above video solution), & Option C is just one such function possible for $f_2.$

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just put the option

Option C is correct

Answer 1

Answer is C.

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

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

Answer 2

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

answer = option C

Comments

this should be the best answer...

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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

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This is not the best answer because spoonfeeding won’t help in gate

Answer 3

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

Answer 4

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 5

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}

Answer 6

Detailed Video Explanation with Complete Analysis: https://youtu.be/r7V6uY_t7X4?t=6304

Must Watch the above Complete Lecture to learn Combining Boolean Fuctions wrt Minterms & Maxterms for all standard boolean operations.

Note: $2^{12}$ different functions are possible for $f_2$ (Watch above video solution), & Option C is just one such function possible for $f_2.$