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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:

enter image description here

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 $$C = A \cdot B$$

Dave Tweed
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xxxtentacion
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  • Some of what you are looking for might already be found [here](https://electronics.stackexchange.com/questions/256263/how-do-you-implement-the-following-function-using-nothing-but-21-mux/256278#256278) on this site. – jonk Dec 05 '18 at 22:48

2 Answers2

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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\$.

user110971
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This should do it:

schematic

simulate this circuit – Schematic created using CircuitLab

Nightmerker
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