if f1 give the 2(0) since it is dont care and f2 gives(1) since its minterm then f will not contain the 2 as minterm in its function.

there is difference between dont care and minterm

minterm means value 1

dont care can be 1 and 0 .

The Gateway to Computer Science Excellence

0 votes

since f= f1.f2 so f will be equal to 1 iff f1=1 and f2 =1

f1 is 1 for the minterms 0,1,2,3,5,7 and f2 is 1 for the minterm 0,1,2,4,5,7

so f will be 1 for the minterm 0,1,2,5,7 where 7 is the dont care minterm .

so f = ∑m(1,5)+d(0,2,7)

0 votes

According to me F is AnD operation of f1 and f2 so in f minterm will be common min-term of both function.

since it is AND operation true is only for f1=1 f2=1.

and for dont care first take common and

then check the f1 dont care is present in minterm of f2 or not if yes then take it in(e.g. let 2 in f1 be 1 and 2 is in f2 so it will be 1) similarly check for f2 vice versa. this is done because . 1 and Dn't care is a don't care because the value will now depend on what i take the don't care to be . so we just take intersections of don't care with minterms of f2.

f1=∑m(0,1,5)+d(2,3,7)

f2=∑m(1,2,4,5)+d(0,7)

f=∑m(1,5)+d(0,2,7)

- All categories
- General Aptitude 1.9k
- Engineering Mathematics 7.5k
- Digital Logic 2.9k
- Programming and DS 4.9k
- Algorithms 4.3k
- Theory of Computation 6.2k
- Compiler Design 2.1k
- Databases 4.1k
- CO and Architecture 3.4k
- Computer Networks 4.1k
- Non GATE 1.5k
- Others 1.5k
- Admissions 595
- Exam Queries 576
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 17

50,648 questions

56,422 answers

195,196 comments

99,860 users