Not quite right.Would I be correct in saying the simplest I could write the equation for these logic gates would be ABC+A'B'C=X
Close. When you use DeMorgan's rule, you'll have to be careful to keep your parentheses. A'+B'C is very different from (A'+B')C.I had the A'B' wrote down on paper like (AB)'. I just wasn't sure how to indicate that on the computer. So if I use DeMorgan's rule my next step would be ABC+A'+B'C=X Would that be correct?
I suppose you could do it that way, but I have to ask--do you know how to use Karnaugh Maps? Or must this be done purely through boolean algebra?Ok, Once the equation looks like ABC+(A'+B')C=X the next step would be ABC+A'C+B'C=X.
Ah, good to know. Well, you can always put it into a K-map to check your work. I believe that other than that A'B' vs (AB)', and the "factorization", you had the correct answer.It can be done through k-maps. I'm just trying to have a better understanding of boolean simplification. I am not learning very much from the way my teacher is explaining it, and Exams are tomorrow. This was just a logic gate circuit I made, it isn't homework.
I've noted that, at a few points, you appear to struggle with proper expression notation. You might want to practice by drawing the truth tables for things like A'B' vs (AB)' and (A+B)C vs A+BC.Would I be correct in saying the simplest I could write the equation for these logic gates would be ABC+A'B'C
Oops! I think I had a symbol mixed up Yes, I believe you can reduce that down to X=C. My apologiesAre you sure I can't reduce that down to just C=X? I put it in a k-map and it gives me just C=X, that is if I am doing the k-map right.
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