# Logic Gate Help

Discussion in 'Homework Help' started by BruceBly, Dec 3, 2012.

1. ### BruceBly Thread Starter New Member

Jul 26, 2012
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Would I be correct in saying the simplest I could write the equation for these logic gates would be ABC+A'B'C=X

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2. ### DerStrom8 Well-Known Member

Feb 20, 2011
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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

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3. ### BruceBly Thread Starter New Member

Jul 26, 2012
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Thank you. That worked.

4. ### MrChips Moderator

Oct 2, 2009
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Not quite right.

5. ### DerStrom8 Well-Known Member

Feb 20, 2011
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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

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6. ### BruceBly Thread Starter New Member

Jul 26, 2012
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Would the AB and the A'B' cancel out and just give a final result of C=X?

7. ### MrChips Moderator

Oct 2, 2009
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But before you consider that,

(AB)' is not the same as A'B'.

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8. ### DerStrom8 Well-Known Member

Feb 20, 2011
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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.

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9. ### BruceBly Thread Starter New Member

Jul 26, 2012
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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?

10. ### DerStrom8 Well-Known Member

Feb 20, 2011
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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.

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11. ### BruceBly Thread Starter New Member

Jul 26, 2012
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Ok, Once the equation looks like ABC+(A'+B')C=X the next step would be ABC+A'C+B'C=X.

12. ### DerStrom8 Well-Known Member

Feb 20, 2011
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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....

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13. ### BruceBly Thread Starter New Member

Jul 26, 2012
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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.

14. ### DerStrom8 Well-Known Member

Feb 20, 2011
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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.

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15. ### BruceBly Thread Starter New Member

Jul 26, 2012
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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.

16. ### WBahn Moderator

Mar 31, 2012
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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

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17. ### BruceBly Thread Starter New Member

Jul 26, 2012
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I don't suppose I could get any better of a breakdown than that. Thanks for your help.

18. ### WBahn Moderator

Mar 31, 2012
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You're welcome.

19. ### DerStrom8 Well-Known Member

Feb 20, 2011
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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

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20. ### MrChips Moderator

Oct 2, 2009
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Maybe you forgot to hit Post.