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
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
  3. 11SnowFlakes

    Thread Starter New Member

    Dec 19, 2011
    6
    0
    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
    5,142
    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
    5,142
    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
    0
    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
    5,142
    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
    5,142
    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.
     
    11SnowFlakes likes this.
  11. prescott2006

    Active Member

    Nov 8, 2008
    72
    1
    Hint: x = f1, so you left two functions to be implemented by 2 MUXs.
     
    11SnowFlakes likes this.
  12. 11SnowFlakes

    Thread Starter New Member

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