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.
I get it
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\$\begingroup\$ What is the name of the software that you are showing in the images? Thanks. \$\endgroup\$– codeaviatorCommented Nov 30, 2016 at 19:09
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3 Answers
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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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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\$ Commented May 23, 2015 at 17:29
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\$\begingroup\$ Ok, so it's F = ((S + X')' + (S' + Y')'). If you need only NOR, use another trick. A = (A' + 1')' \$\endgroup\$– user76844Commented May 23, 2015 at 17:36
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\$\begingroup\$ In fact, it works nice without the trick. \$\endgroup\$– user76844Commented May 23, 2015 at 17:40
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\$\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\$ Commented May 23, 2015 at 17:45
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\$\begingroup\$ You don't need it. Cant you see, it was only nor one step before? Nor is (. + .)' you had three nors there. \$\endgroup\$– user76844Commented May 23, 2015 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.
answered May 24, 2015 at 3:27
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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\$ Commented May 24, 2015 at 5:01
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\$\begingroup\$ Man, what's not clear with my answer? \$\endgroup\$– user76844Commented May 24, 2015 at 6:57
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\$\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\$ Commented May 24, 2015 at 7:38
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\$\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\$ Commented May 24, 2015 at 8:26
