0
\$\begingroup\$

I came across a question regarding logic gates and I can't understand how to solve it.

Q) Construct the truth table and write the boolean output equation.

i) Realize using NOR gates only: $$f(A,B,C,D) = \prod (0, 1, 2, 7, 8, 9, 10) + \Phi(12, 13)$$

I tried to get Boolean equation using KMAP method and I got this Boolean equation $$(B + D).(B+C)$$ Is this correct? If not how to approach this problem? What does phi of 12, 13 mean here? Are they don't care values?

Any help is appreciated.

\$\endgroup\$
  • \$\begingroup\$ It may be two separate control inputs apart from the A,B,C,D. \$\endgroup\$ – Peter Smith Nov 14 '19 at 14:27
  • \$\begingroup\$ @PeterSmith How can I approach this problem? Firstly I'm just looking to derive a boolean equation from given values. I can't understand how to solve it with Kmap \$\endgroup\$ – user_12 Nov 14 '19 at 14:31
  • 4
    \$\begingroup\$ Possible duplicate of Boolean expression to NOR-gates .. and many more.. \$\endgroup\$ – Eugene Sh. Nov 14 '19 at 14:52
  • \$\begingroup\$ @EugeneSh. I've already seen that question and many other and it doesn't seem to be help me. I would really appreciate if you could help on how to derive Boolean equation from given values? Also what does phi refer here? I'm looking for a way to solve it using K-Map which I'm comfortable with? \$\endgroup\$ – user_12 Nov 14 '19 at 15:02
  • 1
    \$\begingroup\$ \$\Phi\$ is denoting the combinations which are don't cares. \$\endgroup\$ – Eugene Sh. Nov 14 '19 at 15:03
2
\$\begingroup\$

Wont be appropriate to post full solution, but will point out your errors.

  • Your Kmap is unfortunately flawed. $$f(A,B,C,D) = \prod (0, 1, 2, 7, 8, 9, 10) + \Phi(12, 13)$$

Here, maxterms \$(12, 13)\$ are dont cares and you have to fill all other maxterms with 0s in Kmap and round the biggest groups of zeroes. But you have filled them with 1s as like minterms.

  • After rounding process, you will get an SOP expression for \$f'(A,B,C,D)\$. You have to take its complement to get the POS expression for \$f(A,B,C,D)\$ .
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Even though I misplaced 1 in place of zeros I still got the boxes correct isn't it? \$\endgroup\$ – user_12 Nov 14 '19 at 17:38
  • \$\begingroup\$ Yea...but the expression is still wrong .. \$\endgroup\$ – Mitu Raj Nov 14 '19 at 17:42
  • 1
    \$\begingroup\$ Should I also consider that single one in the middle? I'm basically getting (B + D) * (B + C) * (A + B' + C' + D')? Is this correct? \$\endgroup\$ – user_12 Nov 14 '19 at 17:47
  • \$\begingroup\$ yep....you got it ;) \$\endgroup\$ – Mitu Raj Nov 14 '19 at 17:48
  • 1
    \$\begingroup\$ thank you for the help. :). \$\endgroup\$ – user_12 Nov 14 '19 at 18:01

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.