2
\$\begingroup\$

Here is the expression I am trying to represent in the Karnaugh Map:

enter image description here

This is what I have done: Can somebody confirm if this is right? I have done the truth table right, however I am having doubts on whether the map has been done correctly.

enter image description here

Any help would be appreciated

\$\endgroup\$
  • 1
    \$\begingroup\$ Seems correct. \$\overline{(xy)}\$ simplifies to \$\overline{x}+\overline{y}\$. Then withinin the left part you get a \$\overline{x}+x\$ which is always 1, so only the right part is of interest. The right part is exactly the same as your K-map. \$\endgroup\$ – jippie Feb 28 '15 at 23:07
  • \$\begingroup\$ @jippie, you should post your comment as an answer so that this question does not remain unanswered. \$\endgroup\$ – Joel Reyes Noche Mar 1 '15 at 0:25
  • \$\begingroup\$ @jippie - I understood what you mean, but do i have to change anything on the picture above or is that right? \$\endgroup\$ – Mathematica Mar 1 '15 at 9:40
  • \$\begingroup\$ Also asked at math.stackexchange.com/q/1169384/18398 \$\endgroup\$ – Joel Reyes Noche Mar 1 '15 at 9:56
  • \$\begingroup\$ @Mathematica not in my opinion. \$\endgroup\$ – jippie Mar 1 '15 at 11:40
1
\$\begingroup\$

Your K-Map is correct.

R = (X + X' + Y' + Z)(XY + X'Y'Z+YZ')

R = (1 + Y' + Z)(XY + X'Y'Z+YZ')

R = (1)(XY + X'Y'Z+YZ')

R = XY + X'Y'Z+YZ'

R = XYZ' + XYZ + X'Y'Z + XYZ' + X'YZ'

R = XYZ + X'Y'Z + XYZ' + X'YZ'

respective terms of last equation will be logic one in K-Map

\$\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.