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truthtable

I'm trying to solve two questions that I am unsure how to approach. I know boolean means true/false but that's about it.

  1. The question reads: "A boolean equation in sum-of-products can be minimized as Y = AB + BD + ACD. I'm not sure how to approach this. I see binary numbers. and the solution must be either true or false. In C, 0 usually means true (==0).

  2. The second parts asks to "draw the full-adder."

Edit:

As I am learning more about drawing logic circuits I realized I should have stated the equation is Y = |A|B + |B|D + AC|D. You can see this on the page I drew and attached. Can you confirm that the circuit drawingenter image description here is correct? Thank you very much.

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    \$\begingroup\$ "In C, 0 usually means true (==0)" What? Are you talking about C, the programming language? In C, 0 is the only value that is false, and any other value is true. \$\endgroup\$ – The Photon Dec 7 '14 at 5:14
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I see binary numbers. and the solution must be either true or false.

Yes. The answer will be true or false. But exactly for what combination of inputs ? The equation gives you this answer. It will be true ( = 1 )for a given combination of inputs and false ( = 0 ) for the rest. To test this, plug in random values of input, and solve the equation. The answer you get must match with that provided in the truth table for the same set of inputs. So in essence, you have "Reduced" that big table to a single equation and can use it to find the result rather than have to look the table again and again.

As far as drawing a circuit from sum of product form is concerned, you proceed as follows :

1) Get the simplified equation (You already have it I guess).

2) For all terms appearing as product ( like AB , which has the inputs A and B ANDed), use an AND gate for each such input.

3) Connect all the outputs of AND gates to inputs of an OR gate. In your case, you have AB, BD, ACD OR'd ( + means OR operation). So connect the outputs of AND gate from step 2 to an OR gate.

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  • \$\begingroup\$ I added the circuit drawing, is this correct? \$\endgroup\$ – Carlo Dec 7 '14 at 16:04
  • \$\begingroup\$ Yes, except the top labelling ; (A|) . (B|) not equals (AB)| . Circuit looks correct though. \$\endgroup\$ – Plutonium smuggler Dec 7 '14 at 16:10

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