Here the knowing of basic meaning of minterm and dont care is needed..Following 3 points are useful here :
a) d + 0 = d [Useful while doing the OR ing of functions]
b) d . 1 = d [Useful while doing the AND ing of functions ]
c) Dont care terms are not affected by NOT gate(negation of a function) [ Simply bcoz negation of dont care term will be dont care only ; it cannot have definite truth value ]
Given , F1(ABCD) = Σm ( 1,4,5,8,10,13,15) + d(3,12)
F2(ABCD) = Σm ( 0,2,3,5,9,10,14,15) + d(6,13)
So for :
a) F3 = F1 + F2 :
If a minterm is either in F1 or F2 , then it will be in F3 ..The reason being for the combination of variables as indicated by minterm , the function value is 1..So if there is such combination of variables in either F1 or F2 , that will suffice for its inclusion in F3..
Hence minterm part of F3 = Σm ( 0,1,2,3,4,5,8,9,10,13,14,15 )
Now as mentioned earlier dont care should remain dont care only after combining functions..
Now say in F1 we have '3' as a dont care term..But it is a part of F3 , meaning that 'd + 1 = 1' which violates dont care property..
Hence '3' is not a dont care term in F3..
Checking for other dont care terms of the original functions similarly , we conclude as :
F3 = Σm ( 0,1,2,3,4,5,8,9,10,13,14,15 ) + d (6 , 12)
b) For
F4 = F1 . F2 :
For this as far as minterm part is concerned , the minterm say 'x' is a part of F4 iff 'x' is a minterm of both F1 and F2..
F1(ABCD) = Σm ( 1,4,5,8,10,13,15) + d(3,12)
F2(ABCD) = Σm ( 0,2,3,5,9,10,14,15) + d(6,13)
So minterm part of F4 = Σm ( 5 , 10 , 15 )
Now for dont care terms we need to preserve 'd' by ensuring : 'd . 1 = d'
So for dont care term of F1 , we check with minterms of F2 [ as they cannot be minterms of F1 ]..
And we are looking for minterms here bcoz we want ; d . 1 = d..If that is a maxterm of other function [Say F2 when the dont care term belongs to F1] then d . 0 = 0 which does not preserve its property..
So say , '3' is a minterm of F2 and dont care term of F1 ..Hence '3' is dont care term of F1.F2..But '12' will not come as it is a maxterm of F2..
Hence proceeding for other dont care terms , we get :
F4 = Σm ( 5 ,10,15 ) + d ( 3 , 13 )
c) For
F5 = F1 . F2'
So F2' = 1 - (Σm ( 0,2,3,5,9,10,14,15) + d(6,13))
= Σm ( 1,4,7,8,11,12 ) + d(6,13) [ The maxterms in the original function will become minterm now ; dont care terms will remain as it is ]
F1 = Σm ( 1,4,5,8,10,13,15 ) + d(3,12)
So proceeding in the similar manner as the b) part , we have :
F5 = Σm ( 1,4,8 ) + d(12,13)
