Homework Help: Electrical Engineering MUX help?

1. Oct 22, 2008

MagLok

Here's an image of the assignment: http://img363.imageshack.us/img363/9125/1331011mg1.jpg [Broken]

1 and 2 specifically..I think I can get the others knowing 1 and 2, ALSO: I don't want the answers, just the method of obtaining it. I'm completely lost with MUX.

-M

EDIT:

I understand the concept of inputs, the selector inputs, and the final values, but what happens inside of a multiplexer? Thanks in advance (again)

-M

Last edited by a moderator: May 3, 2017
2. Oct 22, 2008

jhicks

The logic of a MUX is pretty simple. Say there are 2 inputs X0 and X1 and one select bit given by S. If S is 0, X0 appears on the output. If S is 1, X1 appears on the output. The logic of this MUX is therefore OUT = SX1 + (~S)X0 because in order for OUT to be true, the select bit must be selecting an input that is true. I have used ~ to mean NOT.

Last edited: Oct 22, 2008
3. Oct 22, 2008

MagLok

Thanks jhicks, but unfortunately I still don't quite get it.

My notebook was lost earlier this week with examples, and now I just have this problem off of an old test to go by: http://img525.imageshack.us/img525/8118/456456le3.jpg [Broken]

I understand the k-map and the truth table, but I don't understand how he goes from those to a multiplexer (or how the multiplexer can go to the function.)

-M

Last edited by a moderator: May 3, 2017
4. Oct 22, 2008

jhicks

Well really the truth table goes to the k-map and the k-map gives you the MSOP (or MPOS) form of the output. You can't view the MUX as some sort of mechanical device that has a little switch that connects to a particular input based on the select bits, it's simply a piece of logic that lets the select bits control which output is expressed.

Ok I'm writing up an explanation for what's happening now. No doubt you come up with the MUX before you come up with the k-map and truth table:

F=(~a)(~c) + ab. a controls select bit 1 and c controls select bit 0. Let's look at the ab term. If a is true, then ~a is false and therefore the value of c doesn't matter because (~a)(~c) will never be true. If b is true, ab is true. If b is false, ab is false. Therefore, if a is true then we wish b to be tied to both inputs that can be selected when a is true because c doesn't matter, so b should be tied to selects 10 and 11

For 00 and 01, on the other hand, consider the (~a)(~c) term. If ~a is true, then (~a)(~c) evaluates to whatever (~c) is and ab will always be false. If ~c is 1 then (~a)(~c) evaluates to true, therefore 00 should be tied to 1. If ~c is 0 (i.e. c=1) then (~a)(~c) evaluates to false always, so tie the input chosen when a=0 and c=1 to 0.

Last edited: Oct 22, 2008