I have difficulties with one equation. It's an XOR gate with 4 inputs.
F(A,B,C,D)=(A⊕B)C+(B⊕C)(A⊕C)
'B' from the second bracket has a line above it , so does the 'A' from the third.
I did the truth table,but once transferred to Karnaugh map, the result is very hard to work with:
0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0
Using it, it's very hard to find the minimal disjunctive normal form,since I can't combine any of the TRUE's . I'm not sure if doing each one seperately is the correct way to do it ,since we haven't studied that yet in uni.How to proceed?
Truth table:
a b c d | XOR | OP |
|---------+-----+----|
| 0 0 0 0 | 0 | 0 |
| 0 0 0 1 | 1 | 1 |
| 0 0 1 0 | 1 | 1 |
| 0 0 1 1 | 0 | 0 |
| 0 1 0 0 | 1 | 1 |
| 0 1 0 1 | 0 | 0 |
| 0 1 1 0 | 0 | 0 |
| 0 1 1 1 | 1 | 0 |
| 1 0 0 0 | 1 | 1 |
| 1 0 0 1 | 0 | 0 |
| 1 0 1 0 | 0 | 0 |
| 1 0 1 1 | 1 | 0 |
| 1 1 0 0 | 0 | 0 |
| 1 1 0 1 | 1 | 0 |
| 1 1 1 0 | 1 | 0 |
| 1 1 1 1 | 0 | 0 |
Best Answer
First, I can immediately tell that your K-map is wrong, since there is no $D$ in the formula, so we should be able to group everything in groups of two or more.
Second, so your formula is $(A \oplus B)C + (\bar{B} \oplus C)(\bar{A} \oplus C)$
OK, so for the first term we get a 1 when C is true and exactly one of $A$ and $B$ is true.
For the second term we get a 1 when two things are true: $\bar{B} \oplus C$ (and that is true when B and C have the same value) and when $\bar{A} \oplus C$ is true ( and that is true when A and C have the same value). Together, this means that the second term is a 1 when A, B, and C all have the same value.
here is the K-map I get (blue is for first term, and red for second):
\begin{array}{|c|c|c|c|c|} \hline & \bar{A} \bar{B} & A \bar{B} & AB & \bar{A}B\\ \hline \bar{C} \bar{D} & \color{red}1 & 0&0&0\\ \hline C \bar{D} & 0&\color{blue}1&\color{red}1&\color{blue}1\\ \hline CD &0&\color{blue}1&\color{red}1&\color{blue}1\\ \hline \bar{C}D &\color{red}1&0&0&0\\ \hline \end{array}