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Ive got a boolean-expression

(~a and ~b or ~b and c)

And now i will convert that expression to a NOR-Gate

I tried it many times. I know the solution but i dont know how do i get this.

my last solution was: (~(~(~b (~a + c))) => (~(b+~(~a + c)))

solution: (b nor(c nor(a nor b)))

Thanks for help

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3 Answers 3

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There is a trick for converting sum-of-products to nand gates and a trick for converting product-of-sums to nor gates. (In diagram below).

But you want nor gates and you are starting in sum-of-products form. So how do you convert sum-of-products to product-of-sums? (Also this one.)

Well, in your case the product-of-sums form is trivially derived just by factoring out the \$\overline{B}\$ to get \$\overline{B}(\overline{A}+C)\$. More generally though, the plug-n-chug way you convert sum-of-products to product-of-sums is

  1. double negate your formula. \$\overline{\overline{\overline{A} \, \overline{B} + \overline{B} \, C}} \$
  2. Use Demorgan to push the first not through the sum. \$\overline{(\overline{\overline{A}\,\overline{B}})\,(\overline{\overline{B}\,C})} \$
  3. And one more Demorgan to turn the products into sums. \$ \overline{(A+B)(B+\overline{C})} \$
  4. Now multiply the product-of-sums to get a negated sum-of-products. \$ \overline{AB+A\overline{C}+B+B\overline{C}} = \overline{A\overline{C} + B}\$
  5. Almost there! A negated sum of products is a product of sums by two more Demorgans: \$ (\overline{A\,\overline{C}})(\overline{B}) = (\overline{A}+C)(\overline{B}).\$

Now you've got a product of sums.

Finally, here's how you convert product-of-sums to nor gates. It's sometimes called "pushing bubbles".

schematic

simulate this circuit – Schematic created using CircuitLab

So I get what you get (and what @jippie got). (\$\overline{A}\$ NOR C) NOR B. \$\overline{\overline{\overline{A+B}+C}+B}\$ is logicially equivalent, of course, but requires 3 2-input nor gates instead of 2 2-input nor gates and a "1-input nor gate" (not gate).

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  • \$\begingroup\$ +1 for what appears to be a method that works most of the time (if not all the time). \$\endgroup\$
    – jippie
    Commented Jul 10, 2013 at 6:24
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your given expression: (~B.C)+(~A.~B)

this can be written as: ~B.{C+(~A.~B)} please note that the presence of ~B inside the parenthesis is taken only for our convenience, in fact we will get our initial equation when we take the ~B outside the curly brackets into it because ~B.~B=~B

on applying demorgan's law to "(~A.~B)" we will get ~B.{C + ~(A+B)}

on applying demorgan's law to the entire expression above we will get ~[B+{~(C+ ~(A+B))}] this is nothing but

(B nor(C nor(A nor B)))

. stands for logical AND + stands for logical OR ~ stands for logical NOT

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(~a and ~b or ~b and c)

Allow me to rewrite your formula as follows: \$ F = \overline{A} \cdot \overline{B} + \overline{B} \cdot C = (\overline{A}+C)\overline{B}\$

NOR equivalent

A NOR gate is equivalent with an AND gate with inverted inputs: \$ G = \overline{P+Q} = \overline{P} \cdot \overline{Q} \$

Replace P with \$(\overline{A}+C)\$ and Q with \$\overline{B}\$:

\$ \Rightarrow F = \overline{\overline{A}+C} \cdot \overline{\overline{B}} = \overline{\overline{\overline{A}+C}+B} \$

NOT equivalent

An NOT is a NOR-gate of which both inputs are identical. If you really only want to use NOR gates you would: \$ H = \overline{R+R} = \overline{R} \$

\$ \Rightarrow F = \overline{\overline{\overline{A+A}+C}+B} \$

In your original notation: F = {[(A nor A) nor C] nor B}

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