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Propagation of uncertainty for no-load voltage

  1. Jul 5, 2013 #1
    Need help propagating the uncertainty of Es = Ex ×(x+xo/xs+xo) . I understand all of the rules and can do it for a formula such as vo+ at. But I am having trouble with the (A+B)(/A+C)
     
  2. jcsd
  3. Jul 5, 2013 #2

    gneill

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    Staff: Mentor

    Can you clarify that? Is the sub-expression

    (x+(xo/xs)+xo)

    or

    (x+xo)/(xs+xo)

    and what is /A in (A+B)(/A+C)? How does it relate to your first expression? Or did you mean (A+B)/(A+C)?

    How you attack the problem will depend upon your current level of study.

    There are two main ways to propagate uncertainty through an expression. The first is to take each operation one at a time and calculate a result and uncertainty for it, thus replacing the value pair with a single new value with its own uncertainty, then move on to the next operation and do similarly, then the next, and so on until the whole expression has been resolved to a single result with uncertainty. The other method involves using a bit of calculus (partial derivatives) to calculate the uncertainty all at once.
     
  4. Jul 5, 2013 #3
    Thank you for your reply.

    Es = Ex (x+xo/xs+xo).

    The first method you described is more towards what I am going for. The example I gave of (A+B/C+B) was a mistake.

    I have done complex propagation such as this one, but I am having trouble with the addition divided by addition part. I am looking for a propagation that eventually simplifies to a simple uncertainty.

    Such as q1q2/r2 becoming a more complex uncertainty.
     
  5. Jul 5, 2013 #4

    gneill

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    Okay, so I must assume that xo/xs is a term by itself then, as in the first of my suggested choices. Parentheses are important for removing ambiguity in order of operations when dealing with ASCII-rendered equations.
    I don't see an addition divided by an addition in your expression, if it really is

    Es = Ex [x + (xo/xs) + xo]

    Or do you really mean:
    $$E_s = E_x \frac{(x + x_o)}{(x_s + x_o)}$$
     
  6. Jul 5, 2013 #5
    Yes I mean the second expression.
     
  7. Jul 5, 2013 #6
    How were you able to produce that using these functions in the keyboard? or did you import that from online? I was unable to produce that on here, although I see more parentheses as being more useful.
     
  8. Jul 5, 2013 #7

    gneill

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    Okay :smile:
    I wrote it as a LaTeX expression. Physics Forums has built-in LaTeX interpretation. If you use the "Quote" button to quote the post as if to reply to it, you'll see the LateX commands as I wrote them. Have a look here for an introduction to using it.

    For the first approach, first calculate the addition pairs separately, arriving at values and uncertainties for each. Then perform the division and arrive at a single value with an associated uncertainty. Finally, tackle the remaining multiplication.

    Is this a one-time calculation with one set of values for the variables, or do you have multiple sets of values to tackle?
     
  9. Jul 5, 2013 #8
    Well, this is to be calculated for 3 sets in total. However, it is for a lab report. And in this report I have to show the propagation of this uncertainty and the steps involved.
     
  10. Jul 5, 2013 #9

    gneill

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    I see. You can probably detail the calculation symbolically, step by step, in a stand-alone paragraph, and just tabulate the results in a table.
     
  11. Jul 5, 2013 #10
    See I get lost once I get to
     
  12. Jul 5, 2013 #11
    Well.. screwed up that quote thing.
     
  13. Jul 5, 2013 #12
  14. Jul 5, 2013 #13

    gneill

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    I see an OpenOffice spreadsheet with a figure embedded, but that figure has no content. Probably better to upload an image directly anyways; I get leery opening documents with "smarts" from the net...
    Myself, I'll generally paste a picture into Windows Paint, save it as a .gif file and upload that.

    Anyways, you might take note of the fact that you can go back and edit your own post if you feel that you've hit "Post" before it was tidied up for show and tell...
     
  15. Jul 5, 2013 #14
    It seems there was a little tiny note in the question... It includes: Assume E(sub S) is constant. How does that change the propagation?
     
  16. Jul 5, 2013 #15
    Well.. I came up with an answer, not sure how I could post it on here, but thanks for your help gneill. I'll try in a bit to post it.
     
  17. Jul 5, 2013 #16

    gneill

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    In this context, presumably it means that ##E_s## is a fixed value with no error (i.e. a perfectly known, exact value). But that would be strange, given that it's on the LHS of the equation...
     
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