0
\$\begingroup\$

Function F as: $$ F=\overline{C} \space \space \overline{D} +A\overline{D} + A\overline{B}$$ is a SoP (sum of products) which is given by grouping K-map's 1's.

I tried to complement the function \$F\$ to get PoS (product of sums), like: $$ \overline{F}= (C+ D) \cdot (\overline{A}+D) \cdot (\overline{A}+B) $$ but I checked the K-map's 0's to see if it's a correct PoS, but it is not even a PoS because it contains max terms that the correct PoS shouldn't have.

Does any one know where I'm wrong?

\$\endgroup\$
5
  • \$\begingroup\$ Multiply it out, reduce and invert a second time. You've done DeMorgan's once. You have to do it twice to go from PoS to SOP. \$\endgroup\$ Commented Nov 16, 2018 at 17:31
  • \$\begingroup\$ @StainlessSteelRat does it work? I did and it didn't work \$\endgroup\$
    – parvin
    Commented Nov 16, 2018 at 17:32
  • \$\begingroup\$ It works. I did it. Answer was the same as K-map. \$\endgroup\$ Commented Nov 16, 2018 at 17:32
  • \$\begingroup\$ @StainlessSteelRat the answer is (C'+A')*(D'+A)*(B'+D') and when I draw the K-map it has wrong max terms. \$\endgroup\$
    – parvin
    Commented Nov 16, 2018 at 17:34
  • 1
    \$\begingroup\$ Correction: I get \$ f= (\overline B + \overline{D}) (A + \overline C) (A + \overline D ) \$ \$\endgroup\$ Commented Nov 16, 2018 at 18:17

1 Answer 1

2
\$\begingroup\$

$$ F=\overline{C} \space \overline{D} +A\overline{D} + A\overline{B}$$

DeMorgan's

$$ \overline F=\overline{\overline{C} \space \overline{D} +A\overline{D} + A\overline{B}}$$

$$ \overline F= (C + D) (\overline A + D) (\overline A + B)$$

First two terms: $$ \overline F= (C\overline A + D\overline A + CD + DD) (\overline A + B)$$

Simplify: $$ \overline F= (C\overline A + D) (\overline A + B)$$

Finish multiplying out last term, reduce and take DeMorgan's.

That should get you to your answer.


wide demo pic

\$\endgroup\$
2
  • \$\begingroup\$ +1 for 3k+ rep. Yay... you earn new privileges! \$\endgroup\$
    – user103380
    Commented Nov 16, 2018 at 18:32
  • \$\begingroup\$ Thanks. For the couple times a month I lurk I here. Rapidly down votes a couple of questions. \$\endgroup\$ Commented Nov 16, 2018 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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