0
\$\begingroup\$

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 $$C = A \cdot B$$

\$\endgroup\$
  • \$\begingroup\$ Some of what you are looking for might already be found here on this site. \$\endgroup\$ – jonk Dec 5 '18 at 22:48
1
\$\begingroup\$

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\$.

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

This should do it:

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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