Decoder Help

Discussion in 'Homework Help' started by malas, May 16, 2013.

  1. malas

    Thread Starter New Member

    May 16, 2013
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    0
    Hi,
    I've been stuck on this problem for a while and just can't figure it out. I was given a function Ʃ(4,6,7,13,14) and told to apply it using a 3 to 8 decoder. The way I thought of doing it was that one of the variables- y in my case- is never adjusted for the function and therefore could be taken out and used essentially as some kind of enabler. I just can't seem to get past this and get the simulation to work. Any help would be greatly appreciated.

    Thank you
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    4,800
    What do you mean "apply it" to a 3-to-8 decoder?

    Do you mean implement this function USING a 3-to-8 decoder?

    What do you mean that one of your variables is never "adjusted" for the function?

    You say this variable is y. Well, unfortunately my mind reading skills are a bit on the fritz. It's good to know that you have a variable named y, but that tells me absolutely nothing.

    I think you are actually on the right track, but you need to learn to communicate your thoughts and ideas clearly and concretely.
     
  3. malas

    Thread Starter New Member

    May 16, 2013
    3
    0
    Thank you for replying- I am sorry about that- I am taking the course in a different language so I get the terms mixed up sometimes.

    What I mean is that I previously implemented the function using a 4 to 16 decoder and through the truth table I noticed that one of the variables never needed to be changed to give the correct output. Therefore I thought to take that variable out to reduce it to a 3 to 8 decoder. Then I would somehow use the variable that I just took and have it always be in HIGH status.

    I am having trouble figuring out how to do this- specifically with how to include 13 and 14 as part of the function.

    I hope this was a little clearer?

    Thank you again for the help.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,761
    4,800
    That was much clearer.

    You are going down the right track most of the way. But you have to remember that you don't have control over any of the inputs and therefore you can't just declare that one of them will always be HI. What when it isn't?

    But you had mentioned using that fourth variable as some kind of an enable. That's the right idea.

    As far as the 3-to-8 decoder is concerned, you have a function of three variables and the fourth is a don't care. But after you combine those together, you now have a function of two variables - the output of the decoder circuit and the remaining variable -- and that function is trivial to implement. Give a go and post some kind of diagram.
     
  5. malas

    Thread Starter New Member

    May 16, 2013
    3
    0
    Hi,
    this is what I have so far- basically i used the x as the msb to add to '6' and '5' so it would light up for 13 and 14. I'm still working on the one we talked about with the one variable being used as a 'enabler'. thank you again for your help.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,761
    4,800
    It would be nice if you stopped making us read your mind and guess what maps to what. You have x, x1, x2, and x3 but you spec your function in terms of multibit values. So which signal maps to which bit in the value. Don't make people reverse engineer your design!

    The function you are trying to implement has three 4-input terms. Your schematic has, in the 3-input portion of it, eight terms decoded but the top three are identical and the next two after that are identical. Why? What purpose do they serve?

    At this point, you are making this FAR more complicated than it is.
     
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