suppose i have A'B'D'+A'BD+AC, how do i implement this on a multiplexer?

 

Implementing A'B'D'+A'BD+AC with a multiplexer.
You have got 4 inputs so 2^4 = 16 input multiplexer is needed.

One way to work out this is to write a truth table:-

IN ABCD
0 0000
1 1000
2 0100
3 1100
4 0010
5 1010
6 0110
7 1110
8 0001
9 1001
A 0101
B 1101
C 0011
D 1011
E 0111
F 1111

OK, looking at the terms in your equation, the output
needs to be true (1) if:-

not A and not B and not D don't care C = 00X0 = inputs 0,4

not A and B and D don't care C = 01X1 = inputs A,E

A and C don't care B, don't care D = 1X1X = inputs 5,7,D

Therefore tie these inputs of your 16-input multiplexer high and all the others low.

Actually as I'm not supposed to give you the answer
I HAVE DELIBERATELY MADE ONE MISTAKE IN THE ABOVE - For you to spot!

 

ok. lets see. my lecturer told me that the truth table was supposed to be like this,
IN ABCD
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
A 1010
B 1011
C 1100
D 1101
E 1110
F 1111
or either way would be fine? anyways, thanks

 

Hello Alexia

A multiplexer i a wonderful component to solve boolean problem. The principle is easy.

As JDT says you have four variables and this give 16 combinations, binary from 0000 to 1111. Sorry JDT, but the table is reversed.

The MUX is a digitaly controlled switch. With the inputs A to D you control the swich arm to move from in 0 to in F (Hex not).

The control input A to D is your inputs and Out is output for the expression. In 0 - F is programming inputs and must be set 0 or 1 for false and true.

At first you write a true table with all combinations from 0000 to 1111. If the input value meet the boolean expression, the output shall be "1" otherwise "0". Not that A´B´D´ meet two input values, C can bee either 0 or 1. A´B´D´ can bee A´B´C´D´ or A´B´C D´. AC meet four.

When you have found all input combinations that give a "1" on output you have to put corresponding programming inputs high ( to +5V). All other programming inputs puts to ground.

I hope this will help you. I test to attach a picture, but failed. If you want you can mail mee g.a.gustavsson@tele2.se .

g.a.gustavsson@tele2.se

Best regards
Gustav

 

By the way, a good way of checking if you answer is right is to notice that:-
If the term defines all four inputs, example ABCD then this only happens once (= F).
If the term has one "don't care", example ABC then this happens twice (= 7,F).
If the term has two "don't cares" example AB then this happens 4 times (=3,7,B,F).
And of course if only one input is defined, 3 don't cares, then 8 inputs are involved.

I drew my table my way because I normally think of D as the most significant bit =2^3 which is correct on the commercially available CD4514B (for example) but the answer is basically the same as long as you define how your multiplexer works.

See: http://focus.ti.com/lit/ds/symlink/cd4514b.pdf

 

Hello,

@Gustav180: never put your email adress directly at a forum.
They will be collected by spambots that will spam you.

@Alexia Noir:
Here is a page from the EDUCYPEDIA on Digital Logic:
http://www.educypedia.be/electronics/digitallogic.htm

There are more pages like that over there:

Digital electronics:
Arithmetic circuits D/A-A/D converters Number systems Codes and decoders Flip flops Technology Counters and registers General overview Technology-CMOS Digital logic Memories Technology-TTL Timers and oscillators

Greetings,
Bertus

 

hi,

thank you all for the info. my i ask, when designing the multiplexer, how do we separate from high voltage and low voltage?

 

Do you mean how do you put low voltage and high voltage into the inputs?

Low voltage normally = 0 (or false)
High voltage normally = 1 (or true)

This is called "positive logic". High voltage is usually the supply voltage of the logic circuit (or a voltage above a certain threshold). Low voltage is usually 0V (or ground).

 

oh i see. now i get it. thank you very much.