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
- double negate your formula. \$\overline{\overline{\overline{A} \, \overline{B} + \overline{B} \, C}} \$
- Use Demorgan to push the first not through the sum. \$\overline{(\overline{\overline{A}\,\overline{B}})\,(\overline{\overline{B}\,C})} \$
- And one more Demorgan to turn the products into sums. \$ \overline{(A+B)(B+\overline{C})} \$
- 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}\$
- 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".
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).
First of all, these are the steps you should follow in order to solve for SOP:
- Write AND terms for each input combination which produce HIGH output.
- Write the input variable if it is 1, and write the complement if the variable value is 0.
- OR the AND terms to obtain the output function. In other words, add the AND terms together to produce your wanted output function.
- SOP will have this form from the truth table given in the question: $$F = \overline{A}BC + A\overline{B}C + AB\overline{C} + ABC$$
The first term:
$$\overline{A}BC$$
A is equal to 0 in the truth table when output F is equal to 1.
The second term:
$$A\overline{B}C$$
B is equal to 0 in the truth table when output F is equal to 1.
The third term: $$AB\overline{C}$$
C is equal to 0 in the truth table when output F is equal to 1.
The fourth term: $$ABC$$
A, B, C are all equal to 1 in the truth table when output F is equal to 1.
Secondly, these are the steps you should follow in order to solve for POS:
- Write OR terms when the output F is equal to 0.
- Write the input variable (A, B, C) if the value is zero, and write the complement if the input is 1.
- AND the OR terms to obtain the output function. In other words, multiple the OR terms together to get the final output logic equation.
- POS will have this form from the truth table given in the question:
$$F=(A+B+C)(A+B+\overline{C})(A+\overline{B}+C)(\overline{A} + B + C)$$
The first term:
$$(A+B+C)$$
A, B, and C are equal to zero and the output F is equal to zero as well.
The second term:
$$(A+B+\overline{C})$$
Output F is equal to zero but C is equal to 1. Hence why we take complement.
The third term:
$$(A+\overline{B}+C)$$
Output F is equal to zero but B is equal to 1. Hence why we take complement.
The fourth term:
$$(\overline{A}+B+C)$$
Output F is equal to zero but A is equal to 1. Hence why we take complement.
Essentially, you have to follow the three first steps in order to successfully for SOP and POS. You could have a lot more terms depending on your output results or the number of inputs you have.
Note that the results shown above are not in the reduced format. You could potentially reduce those Boolean-logic equations much more.
Best Answer
I think the easiest way is to convert to a k-map, and then get the POS. In your example, you've got:
In this case, excluding the left column gives (x+y), and excluding the two bottom middle boxes gives (z' + y'), giving an answer of (x + y)(z' + y')