I have this truth table:
CD | 00 | 01 | 11 | 10
AB | | | |
------+----+----+----+----
00 | 1 | 1 | 0 | 0
------+----+----+----+----
01 | 0 | 1 | 1 | 0
------+----+----+----+----
11 | 0 | 1 | 1 | 0
------+----+----+----+----
10 | 0 | 1 | 1 | 0
------+----+----+----+----
Now I need the f function... and I already have the result:
f=¬A¬B¬C+BD+AD
...but I really can't get there. I checked my calculations for hours and I can't get that result. Can you help me in understanding the logic? Thans a lot in advance!
I assume you are currently studying Karnaugh Maps (K-Maps). If you have not been introduced to them, I strongly suggest you do a search for them and look at some examples and explanations. Anywa, the idea here is to identify groups or blocks of 1's that represent a simple expression. Thus, for example, in your diagram the four 1's in the very middle corresponds to the term BD. And the two 1's in row 3 together with the two 1's in row four combine to the term AD. Finally, the two 1's in row 1 is your A'B'C' term. Add them together, and you have your function.
Addendum
OK, so you got to:
$$A'B'C' + BD + AB'D$$
So far so good! But you can continue:
$$= A'B'C' + (B + AB')D = A'B'D' + (B +A)D = A'B'C'+BD+AD$$
That step where I go from $B +AB'$ to $B+A$ is called Reduction ... Few texts mention it (because you can derive it from more basic principles) but it's a very common pattern when doing boolean algebra, so add it to your toolbox!