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I can make the expression using nand gates but how can I rewrite as products of sum because of making with nor gates. $$F=S'X+SY$$

I try to take not of not but I can't.

How can I make the circuit just using nor gates.

enter image description here

I get it

enter image description here

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  • \$\begingroup\$ What is the name of the software that you are showing in the images? Thanks. \$\endgroup\$ – codeaviator Nov 30 '16 at 19:09
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s'x+sy=[s'x+sy]''=[(s'x)'(sy)']'=[(x'+s)(s'+y')]'=[x's'+x'y'+ss'+sy']'= [(x+s)'+(x+y)'+(s'+y)']'= [(x+s)'+(x+y)'+((0+s)'+y)']'= it's-ok :)

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  • \$\begingroup\$ it's the same. (0+s)'=s'. :) \$\endgroup\$ – Mark May 24 '15 at 7:17
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There is the DeMorgan's law. (A' + B') = (AB)'. Use it wisely.

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  • \$\begingroup\$ yeah i've mentioned it but i can't do. \$\endgroup\$ – AskQuestion May 23 '15 at 17:29
  • \$\begingroup\$ Ok, so it's F = ((S + X')' + (S' + Y')'). If you need only NOR, use another trick. A = (A' + 1')' \$\endgroup\$ – Gregory Kornblum May 23 '15 at 17:36
  • \$\begingroup\$ In fact, it works nice without the trick. \$\endgroup\$ – Gregory Kornblum May 23 '15 at 17:40
  • \$\begingroup\$ Yeah i've found same result. But, what about $$F = (S'+Y)(S+X)$$ ? isn't it right ? how can I find out the result? \$\endgroup\$ – AskQuestion May 23 '15 at 17:45
  • \$\begingroup\$ You don't need it. Cant you see, it was only nor one step before? Nor is (. + .)' you had three nors there. \$\endgroup\$ – Gregory Kornblum May 23 '15 at 17:50
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Demorgan Law can be applied in following steps :

1) Change all And's to Or's and vice versa. Be sure to give preference to AND while
   conversion when situation in not clear.

So F = S'X + SY becomes (S'+ X).(S + Y)

2) Next compliment each individual variable or group of variables which you assumed as a 
single entity in step 1.

For eg: (A+B)' would be a single entity, so compliment entire thing. But A + B' would be 2 entities and you would do A' + B to them.

So F = (S + X').(S' + Y')

3) Finally compliment the entire function.

So F = [(S + X').(S' + Y')] '

Now in order to implement your function as a POS or SOP, you could use NOR or NAND respectively. Also you can use single input gates (with both inputs shorted) in order to get them to function as a NOT gate.

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  • \$\begingroup\$ I try to make but i have used not gates. How can I get ride of them ? Just using nor gates. i.imgur.com/16BtSdN.png \$\endgroup\$ – AskQuestion May 24 '15 at 5:01
  • \$\begingroup\$ Man, what's not clear with my answer? \$\endgroup\$ – Gregory Kornblum May 24 '15 at 6:57
  • \$\begingroup\$ @AskQuestion. I think I mentioned it "Also you can use single input gates (with both inputs shorted) in order to get them to function as a NOT gate." So you can use NOR or NAND with both inputs same. They will function as NOT gate. \$\endgroup\$ – Plutonium smuggler May 24 '15 at 7:38
  • \$\begingroup\$ could you draw? \$\endgroup\$ – AskQuestion May 24 '15 at 8:22
  • \$\begingroup\$ @AskQuestion.The same as top left nor gate in your 2nd image. It is (s + s)' or S' ; clearly it functions as a NOT gate. \$\endgroup\$ – Plutonium smuggler May 24 '15 at 8:26

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