0
$\begingroup$

So I had a truth table and using a Karnaugh map I simplified a function. I obtained.

$ f = \overline{A_3}A_2\overline{A_1} + \overline{A_2}\overline{A_0} + A_3\overline{A_0} $

Then using the distributive property of boolean algebra:

$ f = \overline{A_3}A_2\overline{A_1} + \overline{A_0} (\overline{A_2} + A_3) $

Ok, with this we have the minimum of logic gates to use.

Now I need to convert this to NAND. What seemed easier to me was to take the logigram (or electrical scheme) and directly change the gates to their equivalents with NAND. I obtained:

$ f = \overline { \overline{\overline{A_3}A_2\overline{A_1}} + \overline{\overline{A_0} \overline{A_2 \overline{A_3}}} }$

Now I have two questions about this:

  • How should I proceed algebraically to pass to one expression to another (I know I need to apply DeMorgan laws or use the complement of the function... I would thank if someone can give like a tip or something on how to start doing it.

  • I'm thinking if there is any way to simplify the NAND expression... I see that I have \$ A_2 \$ twice and also \$ \overline{A_3} \$ twice... I'm thinking if there is any way to go even further in the simplification... If someone could also give a tip on how to proceed to not ruin the rule of using NAND only. Thanks very much!

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

You can apply DeMorgan's Law and not distribute your NOTs.

for example:

$ab+c \longleftrightarrow \overline {\overline{a b} \overline c} $

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .