digital-logickarnaugh map
Given this function F(A,B,C,D) = Product-M(4,8,9,11,12) . D(2,3,6,7,10,14)
I know how to get a simplified SOP expression using K-map. However, I'm not sure about simplifying this function until it is K-map ready.
My attempt:
(F(A,B,C,D))' = Sum-m(0,1,2,3,5,6,7,10,13,14,15) + d(0,1,4,5,8,9,11,12,13,15)
There are overlapping minterms. So I don't know how to proceed.
Just simply make a truth table that shows values of F for each of the combinations of A, B and C. Then transfer this information to the K-Map format table to do the minimization.
The point of minimizing a boolean function is that you want to output the same 0s and 1s with the smallest number of terms. A don't care can be either a 0 or a 1 - whichever makes the function simpler.
In your case, there's no reason to treat the entire CD' block of don't cares as 1s: if it doesn't make your SOP function simpler, don't add it.
I think the simplest SOP expression is AC' + A'C. Here's a picture of those two terms.
Best Answer
I'm assuming that your function is in POS form with D() being the don't-cares: $$ F(A,B,C,D) = M_4 \cdot M_8 \cdot M_9 \cdot M_{11} \cdot M_{12} \cdot D(2,3,6,7,10,14) $$ If so, since you already know how to compute the simplified version of SOP functions using K-Maps remember that the POS form is a product of maxterms. A maxterm tells you when a boolean function is zero (it is the dual of a minterm, which tells you when the boolean function is one). Therefore plugin a zero in your K-Map for each maxterm: $$ M_i $$ and mark your don't cares as you would normally. The empty cells will then be when your function is one, ore your minterms, and since you already know how to simplify the K-Map using those it should be easy.