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+17 votes
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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. $\Sigma m(4, 6, 8)$

asked in Digital Logic by Veteran (59.4k points) | 1.7k views
0
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.

3 Answers

+26 votes
Best answer
answer - C

with AND gates we will choose intersection of min-terms

with OR gates we will take union of min-terms
answered by Loyal (9.2k points)
selected by
+15 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

answered by Boss (30.6k points)
+1
this should be the best answer...
0

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

+5 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
answered by Boss (15.7k points)


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