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hello guys, I have a question on digtal stuff,

Why is this happening to me, when I impliment the half adder in NOR Gates if I do it int as int the upper formula in the .jpg, the result is different if i do it as in the bottom formula. Both the 1'st formulas mean the same but the outbuts are different. can someone explain please???


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I am not too good at using boolean algebra but I can tell you that the half adder has four possible combinations at the input (0+0,0+1,1+0,1+1) but only three possible outputs. The reason being is that inputting 0+1 will give the same output as inputting 1+0. the output that will be given is 1 at the sum, and 0 at the carry in both cases.

I suspect that one of your equations corresponds to 0+1 at the input and the other to 1+0. heres a good website for understanding what i mean:


hope that helped, monoman

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maybe i understand you?....

in boolean algebra the 'Bar' above a letter means that it is the inverse value, therefore by having two bars above a letter it is the inverse of the inverse value, meaning the normal value. hence the bars can be cancelled out in the same kind of way that two minus's make a positive in conventional algebra, i.e. A - (-B) = A+B. using this fact, i think that you can simplify both of your output formulas, and will find that in fact they are both the same.

hehehe is that what you meant?? :)

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the way I have always built a half adder is to use an XOR gate and an AND gate.
usually I will build the XOR gate by combining an AND gate, a NAND gate, and an OR gate.

i dont think it is possible to build a half adder purely out of NOR gates and if it is I imagine that the circuit would be quite complicated. Surely there must be some kind of 'AND' function? if not surely it would make the circuit less complicated? my guess is that the problem lies in the way that you have built the circuit. I might be wrong about this but please post your circuit so we can have a look.

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