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I have the following problem:

Express the following boolean expressions as sums of products and simplify as much as possible using a Karnaugh map

enter image description here

I drew the Karnaugh map and then placed my values in the table as true (First one, B non D meaning 10 and non B and D meaning 01) We then have the following values: 0100,0110,1100,1110 (as A and C can be either 0 or 1). So we get:

enter image description here

We notice we only have one group (which is circled in blue) and then we have:

0100
0110
1100
1110

We see that the only variables that don't modify their values are B and D and therefore we get the following simplified version:

B non D

But this is the answer only for the expression in brackets, without the minus. Any ideas how I can solve it in case I have a minus in front of the expression? How does it change my expression?

My second question is how I should solve it when I have a double negation like this one enter image description here

When mapping does the first one mean 1111 and the rest 0101, 1101, 0101 and then I solve it the same way? Any ideas? Thank you!

Diana
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  • Have you covered [de Morgan](https://en.wikipedia.org/wiki/De_Morgan%27s_laws) in your course yet ? – Paul R May 18 '15 at 15:20
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    You are using a Karnaugh map for four variables. However, your expression only depends on two variables. `A` and `C` do not occur. – Axel Kemper May 19 '15 at 17:19

2 Answers2

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enter image description hereFor the first question, the expression without the negation can be called (B XOR D) so XOR with a negation is basically XNOR. it can be represented in sums of products as (BD + B'D')

Ps Akshay
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(A.C.!(!B.!D)) + (!A.C.!(!B.!D)) + (A!C!(!B.!D))+(!A!C!(!B.!D)) Steps

  1. = C.!(!B.!D).(A+!A) + !C.!(!B.!D).(A+!A)
  2. = C.!(!B.!D).1 + !C.!(!B.!D).1
  3. = C.!(!B.!D) + !C.!(!B.!D)
  4. = !(!B.!D).(C+!C)
  5. = !(!B.!D)
  6. = !B+!D
Andy3B
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