digital-logic
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
So Complement Law says \$\overline{\overline{X}} = X\$
We start with AND - OR. $$BD + \overline{A}B\overline{C} + \overline{A}CD$$ Double Complement. $$\overline{\overline{BD + \overline{A}B\overline{C} + \overline{A}CD}}$$ Use DeMorgan's Theorem to remove lower complement. $$\overline{\overline{BD} • \overline{\overline{A}B\overline{C}} • \overline{\overline{A}CD}}$$ AND - OR has become NAND - NAND. Use DeMorgan's on terms. $$\overline{(\overline{B} + \overline{D}) • (A + \overline {B} + C) • (A + \overline{C} + \overline{D})}$$ NAND - NAND has become OR - NAND. Use DeMorgan's Theorem to remove complement. $$\overline{(\overline{B} + \overline{D})} + \overline{(A + \overline {B} + C)} + \overline{(A + \overline{C} + \overline{D})}$$ OR - NAND has become NOR - OR. Double Complement again. $$\overline{\overline{\overline{(\overline{B} + \overline{D})} + \overline{(A + \overline {B} + C)} + \overline{(A + \overline{C} + \overline{D})}}}$$ NOR - OR become NOR - NOR. With an extra NOR connected as a NOT gate.
Following is the truth table for a sixteen-to-one demultiplexer.
A0 through A3 are the output address select inputs, and are used to steer the digital input signal on E1 bar to the output corresponding to the port pointed to by the address select inputs, as long as E2 bar is low.
If you need to construct the Boolean equation for each state, then all you have to do is to collect the terms required to make that state unique and relate them properly.
Best Answer
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
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).