F(A, B, C, D) = B'D + A'D + BD = (B’ + A’ + B)D = (1 + A’)D = D [B’+B = 1 ; 1+A’ = 1]
=> F(A,B,C,D) = D = $\sum (1,3,5,7,9,11,13,15)$
Now let’s see how D = $\sum (1,3,5,7,9,11,13,15)$
A |
B |
C |
D |
minterms |
0 |
0 |
0 |
1 |
(0001)$_{2}$ = (1)$_{10}$ (A’B’C’D) |
0 |
0 |
1 |
1 |
(0011)$_{2}$ = (3)$_{10}$ (A’B’CD) |
0 |
1 |
0 |
1 |
(0101)$_{2}$ = (5)$_{10}$ (A’BC’D) |
0 |
1 |
1 |
1 |
(0111)$_{2}$ = (7)$_{10}$ (A’BCD) |
1 |
0 |
0 |
1 |
(1001)$_{2}$ = (9)$_{10}$ (AB’C’D) |
1 |
0 |
1 |
1 |
(1011)$_{2}$ = (11)$_{10}$ (AB’CD) |
1 |
1 |
0 |
1 |
(1101)$_{2}$ = (13)$_{10}$ (ABC’D) |
1 |
1 |
1 |
1 |
(1111)$_{2}$ = (15)$_{10}$ (ABCD) |
(We consider A’ as 0 and A as 1)
So, D = $\sum (1,3,5,7,9,11,13,15)$