# multiplexers

Discussion in 'Homework Help' started by 11SnowFlakes, Dec 19, 2011.

1. ### 11SnowFlakes Thread Starter New Member

Dec 19, 2011
6
0
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

Nov 25, 2009
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3. ### 11SnowFlakes Thread Starter New Member

Dec 19, 2011
6
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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

4. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
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.

5. ### 11SnowFlakes Thread Starter New Member

Dec 19, 2011
6
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?

6. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I didn't say you could produce more than one function with only one multiplexer. I was only referring to this:
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?

7. ### 11SnowFlakes Thread Starter New Member

Dec 19, 2011
6
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f2=Ʃ(2,3,4,5)
f3=Ʃ(1,2,4,7)
I can only use two 8-to-1 MUX

8. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
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.

9. ### 11SnowFlakes Thread Starter New Member

Dec 19, 2011
6
0
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...

10. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Write down side-by-side the binary sequence and the grey sequence and you will notice that you only need two functions, not three.

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11. ### prescott2006 Active Member

Nov 8, 2008
72
1
Hint: x = f1, so you left two functions to be implemented by 2 MUXs.

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12. ### 11SnowFlakes Thread Starter New Member

Dec 19, 2011
6
0
)) I can see it now, thank you))