# Multiplexer diagram of $C = A \cdot B$

I'm having some trouble figuring out how to illustrate, as a diagram, a multiplexer. I realise that a multiplexer takes in a number of input signals and depending on the control sign, outputs the selected signal as a single output. In my course a normal Multiplexer looks like the following:

Before I did anything, I set $$se1 \equiv S$$ and $$out \equiv X$$. Therefore the Boolean expression, for the following will be $$(\neg SI_0)+(SI_1)=X$$. For me that's not the hard part but what if I have the following Boolean expressions and I want to convert it into a Multiplexer diagram:

$$C = A + B$$ $$C = A \cdot B$$

How would I go about doing this? My initial thought was that since the Boolean expression is in the form $$C=(\neg S \cdot A)+(S \cdot B)$$ and the only explanation I could come up with, is that $$S \equiv A$$ therefore the control sign would be A and not S, therefore the it has the same diagram as the one above but with $$I_0 \equiv A$$ $$I_1 \equiv B$$ $$S \equiv A$$ . But then how could I do that with $$C = A \cdot B$$

• Some of what you are looking for might already be found here on this site. – jonk Dec 5 '18 at 22:48

You can do it by putting 0 in the first input, $$\A\$$ in the second input, and $$\B\$$ in the select bit. Hence $$C = \bar{B} 0 + BA = AB.$$
If you set the first input to $$\A\$$, the second input to $$\B\$$, and the select bit to $$\\bar{A}\$$ you get $$A + \bar{A}B = A(B + 1) + \bar{A} B = A + AB +\bar{A} B = A + (A + \bar{A}) B = A + B.$$
It should be noted that if you set the select bit to $$\A\$$, as you have done, you get $$\bar{A}A + AB = AB,$$ not $$\A + B\$$.