# How multiplexer truth table works?

Discussion in 'Homework Help' started by Doraemon, May 21, 2015.

1. ### Doraemon Thread Starter New Member

May 21, 2015
13
0
Hello, I just read this artile: http://www.allaboutcircuits.com/vol_4/chpt_9/7.html

The truth table is:

The result is D I guess. And according to the diagram, the logic is ((I1 AND A) OR (I0 AND A)) I guess.
I'm new to electronics and this truth table is confusing, anyone can help?
How the calculation is carried out?

2. ### kubeek AAC Fanatic!

Sep 20, 2005
4,670
804
I think the correct equation is ((I0 AND A) OR (I1 AND (NOT A)))

Doraemon likes this.
3. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
I would argue that the truth table is wrong - or, at best, misleading.

Row 4 seems to suggest that when the select (A) is 0, the output (D) is input 1 (I1).

A 2-to1 multiplexer works according to the following: When the select is 0, the output is the value of input 0, the output is the same as input 1, otherwise.

This means the equation is $D=I_{1}*select+I_{0}*\overline{select}$

Doraemon likes this.
4. ### Doraemon Thread Starter New Member

May 21, 2015
13
0
Thank you I was trying to figure this out since last 2 hours XD.

5. ### WBahn Moderator

Mar 31, 2012
17,786
4,807
While the truth table is that of a 2:1 MUX, the naming of the signals is very misleading. Also, in the e-book, the next diagram that shows the implementation is all but worthless since it doesn't include any distinction between the two outputs of the 1:2 line decoder.

A much better presentation of the 2:1 MUX truth table, in my opinion, is to put the select lines to the left of the table.

Code (Text):
1.
2. S | A1 | A0 || Y
3. --+----+----++---
4. 0 |  0 |  0 || 0
5. 0 |  0 |  1 || 1
6. 0 |  1 |  0 || 0
7. 0 |  1 |  1 || 1
8. 1 |  0 |  0 || 0
9. 1 |  0 |  1 || 0
10. 1 |  1 |  0 || 1
11. 1 |  1 |  1 || 1
12.

Doraemon likes this.