hi guys i am new to this forum and i am currently undertakin an electronic engineering degree and i thought it would be a good idea to join, anyway at the moment i am having a big problem with boolean algebra and all i can find on the internet is simple stuff. anyway if anyone can help me figure all this out it would be much appreciated. for example: __ _ _ (ABC + ABC + ABC + ABC)(A+B) and ___ _ _ _ __ _ _ ABC + ABC + ABC + ABC + ABC + ABC thanks heaps guys
Sorry guys disregard those last problems these are the correct ones, (' represents not) A'B'C' + A'BC' + A'BC + AB'C' + AB'C + ABC' and (AB'C' + AB'C + ABC + ABC')(A+B) thanks again
The first one A'B'C' + A'BC' + A'BC + AB'C' + AB'C + ABC': Take out common factors B'C' ; BC' ; C B'C'(A'+A) + BC'(A'+A) + C(A'B+AB') (A'+A) = 1 Therefore: B'C' + BC' + C(A'B+AB') Take out common factor C' C'(B'+B) + C(A'B+AB') (B'+B) = 1 Therefore your simplified expression is: C' + C(A'B+AB') The second one will use the same techniques as the first but look at taking out common factors AB and AB' from the first bracket, don't multiple out the two brackets. If you want further advice, or wish to run your answer past us, post it up and I'll be glad to check it. I am also moving this to the Homework Help forum. Dave
Do you care to embelish on how you simplied the first expression to C' + A'B + AB' - where did the C go? Dave
hi Dave... i used the karnaugh map to simplify both the problems if u need a help with karnaugh maps i'll be glad to help
Dave, Sure, use the identity X +X'Y = X+Y where X=C', X'=C, and Y=A'B+AB' Proof of The Identity: The dual of X +X'Y is X(X'+Y) = XY The dual of XY is X+Y, so the identity is proved. Other ways exist to prove the identity. Ratch
beck, Nope, not when the number of variables is greater than 5 or 6. It blows your mind to try to interpret that many variables variables on a K-map. Try: P(A,B,C,D,E,F) = SOP(6,9,13,18,19,25,27,29,41,45,57,61) Tabulation methods like the Quine-McCluskey method guarantee the best simplification and are applicable to computer operations. The answer to the first problem I got from the Quine-McCluskey method was A'B+AB'+C' . For the second problem, by inspection it is easily seen that (AB'C' + AB'C + ABC + ABC')(A+B)=(AB'C' + AB'C + ABC + ABC') . Then = (AB'C' + AB'C + ABC + ABC')= AB'(C'+C)+AB(C+C') = AB'+AB = A(B+B') = A Ratch