Would I be correct in saying the simplest I could write the equation for these logic gates would be ABC+A'B'C=X
Try uploading the image directly to AAC by clicking the "Go Advanced" button and select "attach" (the paperclip symbol). See if that helps. Regards, Matt
Great! So the first thing I see with your work is that you have A'B' when it should be (AB)'. You'll have to use DeMorgan's rule to break the bar. That is, invert each input and change the operation from AND to OR. Try that and see where you can go from there
No it would not. I'm afraid there's much more to it than that. And take a look at my previous post. A'B' is incorrect.
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?
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 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? EDIT: Looking back over that, it appears that what you have written IS the simplest form....
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.
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.
Are 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.
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. As has already been established, the basic equation for this circuit is Y = ABC + (AB)'C Factor out the C to get Y = [AB + (AB)']C (This is actually the form that is very obvious from the circuit itself) But we know that (X+X') = 1 (here, X = AB), so this is just Y = C If you missed this, then you can still get there the long way: Y = ABC + (AB)'C expands directly to Y = ABC + (A' + B')C Y = ABC + A'C + B'C At this point, you can do the following Y = (AB + A' + B')C Y = (AB + A'(B+B') + (A+A')B')C Y = (AB + A'B + A'B' + AB' + A'B')C Y = (AB + A'B + AB' + A'B')C You could note, at this point, that the factor in parens has all four possible combinations of A and B, or you could keep plugging away Y = ((A + A')B + (A + A')B')C Y = (B + B')C Y = C
You're welcome. Normally I would not just post a detailed answer, but rather ask leading questions to help you get their on your own. But in this case a number of factors made me decide that just providing the answer was reasonable, not the least of which is the fact that you had already determined the correct answer on your own and that that was a problem you made up for yourself (taking your word for that, but seems likely).
Oops! I think I had a symbol mixed up Yes, I believe you can reduce that down to X=C. My apologies EDIT: This is strange. I posted this last night, but it never showed up! I hit "quote" this morning to fix it and the text I typed last night was still here