Hello,
I need some help. The task is to design circuit that converts 3-bit Gray code into binary using two 8-in-1 multiplexers. So I have
input output The problem is that I don't know how to implement this with
000 000 2 8-in-1 multiplexers.
001 001
010 011
011 010
100 111
101 110
110 100
111 101

 

Hello and welcome the the AAC forums!

Take a look here: http://forum.allaboutcircuits.com/showthread.php?t=53799
and come back with your questions.

 

Let the inputs be x y z, and the outputs f1 f2 f3.
In 8-in-1 multiplexer there are 8 inputs and 3 selection inputs. The selection pins will be x y z. f1=Ʃ(4,5,6,7), so I0-I3 are 0 and I4-I7 are 1. If xyz is 000,001,010,011 the output will be zero, for the other combinations it will be 1. That is the first multiplexer. Is this ok? Now I have one more multiplexer and information of 2 bits. What about that? Can I send those 2bit combination from the inputs of the other multiplexer? Am i getting somewhere or missing the point of this? Thank you

 

Yes, your first MUX sounds fine. We can be extra sure if you post a schematic too.

For the second one, if you don't want to waste hardware, you can use a 4-to-1 MUX, with 2 select inputs. If you must use an 8-to-1 MUX, you can utilize the two LSB select inputs and manipulate only the 4 first MUX inputs. The last 4 will never be selected, provided that you tie the MSB select input to 0.

 

I'm sorry, but I don't understand how the outputs f2 and f3 will be generated from the second multiplexer. I understand the concept that you explained of working with two of the three selection inputs an therefore with 4 of the 8 inputs, but the output will be either 0 or 1, isn't that right?

 

I didn't say you could produce more than one function with only one multiplexer. I was only referring to this:

Now I have one more multiplexer and information of 2 bits. What about that? Can I send those 2bit combination from the inputs of the other multiplexer? Am i getting somewhere or missing the point of this? Thank you

You still need one MUX per function, if you want to have access to their output at the same time.

What are your other two functions anyway?

 

f2=Ʃ(2,3,4,5)
f3=Ʃ(1,2,4,7)
I can only use two 8-to-1 MUX

 

Do you need to see all three outputs simultaneously? Do you need to implement all three f1, f2 and f3 with 2 8-to-1 MUXs? I don't see any two variable functions here, as you mentioned before.

 

I have to design the whole circuit(that converts 3-bit Gray to binary code)with 2 8-to-1 MUXs. About the 2-bit information I made a mistake. Sorry if I am bothering...

 

Write down side-by-side the binary sequence and the grey sequence and you will notice that you only need two functions, not three.

 

Hint: x = f1, so you left two functions to be implemented by 2 MUXs.

 

)) I can see it now, thank you))