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I am trying to create a binary adder. I have this truth table:

enter image description here

This is the Karnaugh map that I made for this function (see that there is a mistake: instead of c, it had to be "c-in"):

enter image description here

Now, as the map looks like that, it gives me that s = c + c but that is certainly false. Can anyone help me with this?

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    \$\begingroup\$ Erm, I see a^b^c... \$\endgroup\$ Oct 20, 2013 at 18:39
  • \$\begingroup\$ yeah, you-re right. that's how I found it on the website, but I don't know how to get it. please, post an answer with the explanation. I am student, and I haven't been working with this for a long time \$\endgroup\$
    – Victor
    Oct 20, 2013 at 18:40
  • \$\begingroup\$ How did you draw the Karnaugh maps? \$\endgroup\$
    – Revious
    Oct 30, 2014 at 11:01

2 Answers 2

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If we look at the top left corner, where a=0 and c=0, we see that s follows b. This gives us a partial result of s=b. Moving down to the bottom left corner, we see that s is inverted when a=1. This is the XOR operation, and gives us s=a^b. Moving over to the right side, we see that the output is inverted when c=1. This gives us our final result of s=a^b^c.

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Karnaugh maps are usually drawn in Grey Code (with one bit changing at a time, so 000, 001, 011, 010 rather than with decimal sequence*) and if you do that, you get: enter image description here.

This corresponds to minterms

C'A'B + C'AB' + CA'B' + CAB = C' (A'B + AB') + C(A'B' + AB) The XOR function is A ^B=A'B + AB' and A XNOR B is A'B' + AB So f= C'(A XOR B) + C (A XNOR B) = C' (A XOR B) + C(A XOR B)' = C XOR (A XOR B) = A XOR B XOR C

  • Note: You can only group together cells and reduce using a K-Map when it is in Grey Code sequence, because grouping two elements together (and discarding the variable that changes) correspons to BC(A + A') = BC which only makes sense when one bit is changing.
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