0
$\begingroup$

I have a truth table as follows

X Y Z A B C

0 0 0 0 0 1

0 0 1 0 1 0

0 1 0 1 0 1

0 1 1 1 0 0

1 0 0 0 1 1

1 0 1 1 0 0

1 1 0 1 0 1

1 1 1 1 1 0

where x, y and z are inputs and a, b and c are outputs. Normally, simplifying this into an expression would be easy with one output. I would just use k-maps and get the simplified Boolean expression from it. But with three outputs, I don't know how.

$\endgroup$
8
  • $\begingroup$ What are you trying to do with the outputs, are you trying to construct logic gates ? $\endgroup$
    – asddf
    Oct 11, 2017 at 21:34
  • $\begingroup$ I need to get the simplified boolean expression for the outputs so I can draw a circuit diagram $\endgroup$ Oct 11, 2017 at 21:41
  • $\begingroup$ with 3 outputs you need 3 expressions. $\endgroup$
    – asddf
    Oct 11, 2017 at 21:43
  • 1
    $\begingroup$ So would I have to make 3 k-maps, one for each output? $\endgroup$ Oct 11, 2017 at 21:50
  • $\begingroup$ yes or you can just take the dnf. $\endgroup$
    – asddf
    Oct 11, 2017 at 21:53

1 Answer 1

2
$\begingroup$

It's pretty clear by inspection that $c = \neg z$, so we can get that out of the way.

The simplest way I can see to specify $a$ is that $a = y \vee (x \wedge z)$. Again by inspection.

$b$ has no simple expression that I can see; you're not doing much better than just saying "x and not y and not z, or z and not y and not x, or x and y and z".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.