Question

Express the following SOP expression a a POS expression using a Karnaugh map and DeMorgan's theorems A middot B middot C middot D + A middot B middot C middot D + A middot B middot C middot D + A middot B middot C middot D + A middot B middot C middot D + A middot B middot C middot D + A middot B middot C middot D + A middot B middot C middot D.

Answer 1

Solution:

First of all let's do using DeMorgan's law,

So we have, A'B'C'D'+AB'C'D+ABCD+A'B'C'D+A'BC'D+AB'C'D'+ABCD'+A'BCD

Let's do souble complement

(A'B'C'D'+AB'C'D+ABCD+A'B'C'D+A'BC'D+AB'C'D'+ABCD'+A'BCD)''

Now DeMorgan's law says when the complement is there break the bar(complement) and change the sign

= ((A'B'C'D')'.(AB'C'D)'.(ABCD)'.(A'B'C'D)'.(A'BC'D)'.(AB'C'D')'.(ABCD')'.(A'BCD)')'

= ((A''+B''+C''+D'').(A'+B''+C''+D').(A'+B'+C'+D').(A''+B''+C''+D').(A''+B'+C''+D').(A'+'B''+C''+D'').(A'+B'+C'+D'').(A''+B'+C'+D'))'

= ((A+B+C+D).(A'+B+C+D').(A'+B'+C'+D').(A+B+C+D').(A+B'+C+D').(A'+B+C+D).(A'+B'+C'+D).(A+B'+C'+D'))'

After solving this final solution is

(A'+C).(A'+B+D').(A+B+C').

and using K-map is given below:I hope this helps. Don't forget to give a thumbs up if you like this

soluti3 卜1(2, 3,4) G, lo, 13, 12, 13) Lo Оф 0