0
$\begingroup$

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  |

1 Answers 1

1

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}

  • 0
    Yes, that's my formula. How would you do my K-map? – 2017-02-20
  • 0
    @Tisho Never tried a K-map in Mathjax before .... Let me try ... – 2017-02-20
  • 0
    Thanks a lot, I need this work done for tomorrow and I've ran out of ideas and knowledge myself. – 2017-02-20
  • 0
    @Tisho Ok, it's up! You know how to group, right? – 2017-02-20
  • 0
    Would you mind a short explation as to why that is the correct K-map? After all ,i'd like to understand, not just copy you. :) And aren't usually A and B inputs vertical, while C and D - horizontal in the K-map? I believe i do ,yea. Maybe my truth table is wrong as well, do you include D in it? – 2017-02-20
  • 0
    Added truth table to main post. – 2017-02-20
  • 0
    @Tisho what is the XOR and OP in your truth table? I did add explanation for my K-map. And yes, I have A and B for the verticals, and C and D for the horizontals (not that it matters, really!) – 2017-02-20
  • 0
    XOR is what i have as logical value of the function aka what I type in the k-map at it's correct position. THat's how we're teached to be done. With XOR gate, uneven number of 1's equals 1 as the result of the function. – 2017-02-20