0
\$\begingroup\$

So I am practicing for an exam in digital electronics. I am doing some exercises on writing expressions from truth tables, and I came across a description for the columns that should be true and false in the truth table I have never seen before.

It is the + d(0,4) part in the end that I don't know what it means. How does it apply when I write the truth table? I have googled for quite some time and I cannot find it.

This is what it looks like:

enter image description here

So my task was to write a truth table and then a sum of products expression from it. I'm confused and don't know if the d(0,4) in the end mean that columns 0 and 4 also should be true.

|improve this question
\$\endgroup\$
  • \$\begingroup\$ Guessing now, but I think these are the "don't care" terms. \$\endgroup\$ – Eugene Sh. Mar 28 at 17:49
  • \$\begingroup\$ Thank you, so I guess they are just to be ignored then and are just useless? \$\endgroup\$ – Jean Doe Mar 28 at 17:59
  • 1
    \$\begingroup\$ Not at all. "Don't care" are usually the combinations that are "illegal" for this specific circuit. They are very useful for the circuit optimization as you might chose the value which is making the circuit minimal/most optimal. \$\endgroup\$ – Eugene Sh. Mar 28 at 18:00
2
\$\begingroup\$

These are the "don't care" terms. The rows 0 and 4 can be either true or false. This specific notation seem not to be very common, but I found it's description in this paper.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Thank you. Does it matter if they are true or false for the outcome or can I just take the other numbers (1,5,8,11,13) and sum them while ignoring the d(0,4)? \$\endgroup\$ – Jean Doe Mar 28 at 18:00
  • \$\begingroup\$ A function with "don't cares" is basically a family of functions. All these functions should have the same behavior for the inputs within the \$\Sigma\$ (and complements), but can differ for the inputs for which it is "don't care". \$\endgroup\$ – Eugene Sh. Mar 28 at 18:02

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.