To find the essential prime implicants of F(A,B,C,D)=∑m(0,1,5,7,10,14,15), we can use a Karnaugh map:
CD\AB 00 01 11 10
-------------
00| 1 1 1 0
01| 0 1 0 0
11| 1 1 0 1
10| 0 0 0 0
From the K-map, we can see that there are four prime implicants:
- A'BC'D'
- AB'C'D'
- AB'CD
- ABCD
To determine the essential prime implicants, we need to check if any minterms can only be covered by a single prime implicant. We can use the Petrick's method for this:
F(A,B,C,D) = A'BC'D' + AB'C'D' + AB'CD + ABCD
= (A'BC'D' + AB'C'D') + (AB'CD + ABCD)
= (A'BC' + AB'C')D' + (AB'+AB)CD
The first term has the product of sums form (A'BC' + AB'C')
, which can be written as (A'+B')C'
. The second term has the sum of products form (AB'+AB)CD
, which can be written as AB'C+ABC'
.
Therefore, the simplified expression is:
F(A,B,C,D) = (A'+B')C'D' + AB'C+ABC'
From this expression, we can see that the essential prime implicants are A'BC'D'
and AB'C'D'
. The other two prime implicants are not essential because they can be covered by a combination of the essential prime implicants. Therefore, the number of essential prime implicants is 2.