search
Log In
0 votes
608 views
Given f=f1.f2,  where f1=∑m(0,1,5)+d(2,3,7)

and f2=∑m(1,2,4,5)+d(0,7)

f=??
in Digital Logic 608 views

3 Answers

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
how 2 is in minterm of f

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 .
0
sourav u did it wrong . plz make a truth table and verify. plz edit or delete.
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)

0
Thanks for the solution.
0 votes

f1.f2 = $\sum$m (1, 5) + $\oslash$( 0, 2, 7)                     using :             1 . $\oslash$ = $\oslash$

where  $\oslash$  is dont care.                                                          0. $\oslash$ = 0

                                                                                                  $\oslash$ . $\oslash$ = $\oslash$

Related questions

1 vote
1 answer
2
668 views
According to me first we sort the array in O(nlogn) time and then in O(k) time , find the product , so total time complexity is O(nlogn) , so am I right or can it be done in lesser time ?
asked Apr 10, 2016 in Algorithm Challenges radha gogia 668 views
1 vote
1 answer
4
277 views
when we take the cartesian product of two DFA's then what happens to the dead state of two DFA's , do we combine it also in the cartesian product ?
asked Nov 14, 2015 in Theory of Computation radha gogia 277 views
...