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4-1 to 2-1 Multiplexer Help

  1. Oct 18, 2006 #1
    I cannot seem to understand how in the attached diagram, they went from the 4-1 multiplexer to the 2-1 multiplexer.
    The main part is the modified truth table. I cant understand what is going on for the life of me!

    If someone could please explain this, it would be much apprecieated.

    Thanks!
     

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    Last edited: Oct 18, 2006
  2. jcsd
  3. Oct 18, 2006 #2

    berkeman

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    Staff: Mentor

    They just simplified the generation of the output. The first implementation with the 4:1 is a bit overkill anyway, so the simplification to the 2:1 and using one of the inputs to generate the two input terms is pretty natural. Just look at the output function that is desired, and ask youself how you would generate it using only a 2:1 MUX. The solution shown is the same way you would come up with.
     
  4. Oct 18, 2006 #3
    I totally get the 4:1...but ahh I cant seem to get my head around the 2:1 for some reason:mad:

    I guess my main problem is I dont see how the modifed truth table for the 2:1 acutally results in the XOR truth table...
     
  5. Oct 18, 2006 #4

    berkeman

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    But you do see how the 2:1 MUX generates the desired output function, right? I didn't read any more than that into it. Is your text trying to generalize some method for simplifying circuits with this example? As I said, the 4:1 starting example is pretty contrived. You wouldn't normally use a MUX to switch between static signals.
     
  6. Oct 18, 2006 #5
    Well the text gets into Shannons Theorm in the next section. But acutally, I dont understand how the 2:1 generates the function.

    With the 4:1 its straightforward, whatever w1 and w2 correspond too, the value of f is listed (XOR truth table)

    with the 2:1, it says this in the text
    "w1 is 0, f has the same value as input w2"
    "w1 is 1, f has the same value as input w2' "

    I dont understand what the text is saying at all.

    ahhh sorry about this. Im really not sure why I dont understand this, but I dont want to move on until I do.
     
  7. Oct 18, 2006 #6

    berkeman

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    But if you look at the full truth table, you should see what the text is saying. Look at the first two lines, where w1=0. For the first line, f=0, which matches the value of w2=0. For the 2nd line, f=1, which happens to match the value of w2=1.

    Then look at the 3rd and 4th lines, where w1=1. On the 3rd line, f=1 but w2=0. And on the 4th line, f=0 but w2=1. So when w1=1, the value of f is the opposite of the value of w2.

    So all of that means that you can implement f with a 2:1 MUX, controlled by w1, and with w2 and w2' as the inputs. Does that help?
     
  8. Oct 18, 2006 #7
    Ohhhhhhhh! I see it now!

    Thankyou SOOOO much!
     
  9. Oct 18, 2006 #8
    Just one more question.

    Consider function G(A,B,C) = AB + BC. Let F(A,B,C) be the complement of G(A,B,C). Then F(A,B,C) can be implemented using A as the select input to a 2:1 multiplexer. The correct implmentation of F(A,B,C) is shown in...

    The question then lists five choices with diagrams.

    I just would like to know how to approach this.

    I found the complment to be (A' + B')(B' + C'), but what to do from here is where I am stuck.

    Thanks!
     
  10. Oct 18, 2006 #9

    berkeman

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    Hah! That's interesting. I don't know what the underlying theorem is (it's been a long time since I did low-level logic manipulations), but at least this problem turns out really simple. Cool.

    Start by drawing the Karnaugh map for G(A,B,C). Draw it with A on the left and BC on the top (so the map is 2 rows and 4 columns). The two rows are labeled at the left as 0,1, and the 4 columns are labeled along the top as 00, 01, 11, 10. Put 1's in the map where the function G=1.

    Now invert the map to show the F function, changing 0<-->1 in each square. Now gather terms, and write the F function minimally. You get something like B' + A'BC'. Now think about how you can make a very simple 2:1 MUX inplementation of this using A as the selector signal. What do you put into the A=1 input? What do you put into the A=0 input?

    Pretty cool, eh?
     
  11. Oct 18, 2006 #10
    Thanks again! Makes perfect sense. Just seems like practice is needed to look at the truth table and figure out how the select input relates to the data inputs.

    Thanks!
     
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