0
\$\begingroup\$

I have solved full adder and it gives me carry:

\$A\$\$.\$\$Cin\$+\$B\$\$.\$\$Cin\$+\$A\$\$.\$\$B\$

But in some books I found that carry is written as:

\$A\$\$.\$\$B\$\$+\$\$(\$\$A\$\$\bigoplus\$\$B\$\$)\$\$.\$\$Cin\$

How to derive second equation from first?

\$\endgroup\$
2
  • \$\begingroup\$ Have you tried Karnaugh mapping? \$\endgroup\$
    – user16324
    Sep 16, 2016 at 12:21
  • \$\begingroup\$ Yes, I have tried Karnaugh mapping \$\endgroup\$
    – Abhishek
    Sep 16, 2016 at 13:08

2 Answers 2

0
\$\begingroup\$

First off, lets make an observation. The following is true (if you don't believe me, I'll prove it later):

\$ A + B = AB + \bar AB + A \bar B \$

Now, by the definition of XOR we have:

\$ A \oplus B = \bar AB + A \bar B \$

Combining the two expressions we get:

\$ A + B = AB + A \oplus B \$

With that, lets begin:

AB+AC+BC = AB + C(A+B)        // Factor out C
         = AB + C(AB + A⨁B)  // Substitute the above expression
         = AB + CAB + C(A⨁B) // Factor out AB
         = AB + C(A⨁B)       // Invoke the absorption rule to get rid of CAB

Voila, that's what we were looking for. I don't claim that is the best way to do it, but it seems to work well enough.

Now, I'm going to show \$ A + B = AB + \bar AB + A \bar B \$ using sweet, pure Boolean logic.

\$ A+B = A(\bar B + B) + B(\bar A + A) = AB + \bar AB + A \bar B \$

\$\endgroup\$
0
\$\begingroup\$

Let's just call \$C_{in} = C\$, because it's easier to type and read. You want to show that:

$$A\cdot B + A\cdot C + B\cdot C = A\cdot B + \left(A \oplus B\right)\cdot C$$

Let's take the right side and show how it goes:

$$\begin{align*} A\cdot B &+ \left(A \oplus B\right)\cdot C \\ A\cdot B &+ \left(A\cdot \bar{B} + \bar{A}\cdot B\right)\cdot C \\ A\cdot B &+ A\cdot \bar{B}\cdot C + \bar{A}\cdot B\cdot C \\ A\cdot B\cdot C + A\cdot B\cdot \bar{C} &+ A\cdot \bar{B}\cdot C + \bar{A}\cdot B\cdot C \end{align*}$$

I think you can see how that expansion went, so far. It's now the left-most term that is interesting. We are allowed to replicate it as many times as we want, as it won't change the result. So the following expressions are exactly the same expression as the last one shown above:

$$A\cdot B\cdot C + A\cdot B\cdot \bar{C} + A\cdot \bar{B}\cdot C + \bar{A}\cdot B\cdot C \\ A\cdot B\cdot C + A\cdot B\cdot C + A\cdot B\cdot C + A\cdot B\cdot \bar{C} + A\cdot \bar{B}\cdot C + \bar{A}\cdot B\cdot C \\ \left(A\cdot B\cdot C + A\cdot B\cdot \bar{C}\right) + \left(A\cdot B\cdot C + A\cdot \bar{B}\cdot C\right) + \left(A\cdot B\cdot C + \bar{A}\cdot B\cdot C\right) \\ \left(A\cdot B\right) + \left(A\cdot C\right) + \left(B\cdot C\right) \\ A\cdot B + A\cdot C + B\cdot C$$

That should be convincing, I hope.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.