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Tolerance Question

Discussion in 'Electronic Design' started by Joshua Guthrie, Apr 14, 2005.

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  1. I have a tolerance statistics question:
    It is probably best if I describe it with an example, even though I
    don't have one specificlly in mind. It just bothers me that I either
    forgot how to do this or was never taught.

    Suppose I have a sensor measuring something. For my example I have a
    thermocouple in water reading 211F. The tolerance on the sensor is

    My understanding is that for the T/C that I have a normal distribution
    with three times the standard deviation being 2F centered around 211F.
    That is to say that I have a 99.73% chance that the true reading is
    within +/-2F of my indicated value. Am I on the right page here?

    I add another thermocouple, it reads 213F with the same tolerance. I
    know the water is uniform in temp.

    Now with my second sensor included in the mix, the most probable true
    value would become 212F (the avg of the two sensors as they have the
    same PDF, just centered different).

    My question is how do i compute my new standard deviation of the
    combination of the two sensors? I want to know my new tolerance. Or
    am I way off shore in my thinking.

  2. John  Larkin

    John Larkin Guest

    Most electronic device tolerances are guaranteed limits (guaranteed
    worst-case error), not standard deviation.

  3. Bob Eldred

    Bob Eldred Guest

    My understanding of tolerance is that all, 100% of the sensor must be within
    the tolerance range, not some statistical grouping like 99.73%. Furthermore
    any that are not within the stated range may be rejected, not sold or not
    used in service usually with a money back guarantee. The tolerance is
    usually stated as a percentage of some value, often the full scale value. It
    is sometimes stated as the accuracy of any reading; i.e., within a
    percentage or number of units of true. In your example I would interpret it
    as meaning the reading is within +/- 2deg F of the actual temperature. If
    the reference is boiling water, the readings would "always" be within +/- 2
    degF of 212 degF. BTW, that's a lousy tolerance for a thermocouple.
  4. I read in that Joshua Guthrie
    You are getting off-beam answers because you are asking a statistics
    problem in an electronics newsgroup.

    The combined distribution due to two sensors is no longer
    Gaussian/normal but bimodal. I think you need to ask on a math or stats
  5. John  Larkin

    John Larkin Guest

    A single batch of sensors may have a bimodal distribution if the
    better ones have been culled from the batch. Sort of a gaussian with
    the edges clipped and a big notch in the middle.

  6. I read in that Joshua Guthrie
    OK, I'm going to have another go at this, because a small dim light
    appeared in what I laughingly call my brain.

    You may be OK to assume for this example that the tolerance band is
    between the 3-sigma points of the distribution, but it may not be in a
    practical case.
    In this example, yes, but in general, there is a valley in the
    distribution between the two nominal values.
    Standard deviations are like r.m.s. values - they add as the square root
    of the sum of the squares. To determine the new sigma, we can subtract
    212 (the 'd.c. component') to give two 'r.m.s.' values, -(1)^2 +
    0.667^2 and +1^2 + 0.667^2. The new SD or r.m.s. value is them
    sqrt(2*0.667^2 + 2*1^2) = 2.211.

    That 2 degrees difference make a LOT of difference to the 3-sigma
    tolerance, going from +/- 2 degrees to +/- 6.634 degrees. But I suppose
    if you think about the two distribution curves added together, the
    result has a much broader peak than either individual curve.

    Please understand that I may not have worked this out correctly. Stats
    are a bit out of my field. It's only the thing about sigma = r.m.s.
    value that tempts me to try for a solution.
  7. I read in that John Larkin
    Yes, the OP's question is about two sensors with a 2 degree difference
    when reading the same temperature.
  8. This is very similar to the path I was wandering down. But the thing
    that stumped me was this.

    The more sensors I put in the water and average the results, the less
    confident I would get in the results. This seems counter intuitive.

    I would think, as I added more sensors measuring the same phenomena,
    the more my tolerances in the system as a whole should tightened up.
    Besides, I have additional information, and know the PDF of the data
    points. Kinda like if I added an infinite number of sensors, the more
    infinitesimal my error would get.

    But you all are probably right, I'm probably off topic in this group.
  9. John Larkin

    John Larkin Guest

    If the entire production lot of thermistors were skewed, adding more
    sensors would merely converge on the wrong point. If the mfr is
    selling 2% devices, why would he invest in 0.1% standards? You can buy
    a reel of laser-trimmed 5% resistors, for example, that are all very
    close to one another and are all about 2% high.

    The only safe thing to do is buy a good, certified platinum RTD and
    use it with a good, 4-wire DVM to measure the actual temperature and
    calibrate your thermistor setup, or get some equivalent accurate
    Not at all.

  10. Fred Bloggs

    Fred Bloggs Guest

    It is counterintuitive because it is wrong.
    And that is exactly right.
    The PDF is not on the temperature measurement, it is on the error
    function of the population of available sensors from which you have
    selected a sample of two. The average of two fixed sensors from that
    population is a repeatable and deterministic function, it will not
    change as long as those two sensors are fixed. As you add more sensors
    and average, the error function approaches that of the sensor population
    as a whole and the STD, or root mean square error, of the sample
    function deviation from population mean approaches zero. If the sensor
    population mean error function with temperature is zero, then the
    tolerance will tighten up and also approach zero.
    Yes- you are.
  11. Joshua,
    You aren't quite thinking about this right.

    The typical tolerance on a part like this isn't gaussian, or bi-modal
    (at least not anymore.) Instead, they are typically uniform. When you
    add two sensors, what you get is a possible distribution that is more
    likely to be correct, but there is a slightly greater chance it is be
    beyond the original spec. As you add sensors, the distribution gains a
    peak at your specified value, and a gaussian distribution around that
    point occurs. As you add sensors, your average is more likely to be the
    correct value, and the utility of doing more rigourous statistical
    analysis on your measurements increases. You can then do things like,
    reject values that differ from the average by an amount greater than
    'x', etc.

  12. I read in that Joshua Guthrie
    Not if they are all giving you significantly different answers. In your
    example, the two sensors were giving results 3 sigmas apart. If they had
    been 0.1 sigma apart, the effect on the combined sigma would have been
    much less, but it would still have been larger.
  13. Gareth

    Gareth Guest

    I think the central limit theorem may apply here.

    The central limit theorem states that given a distribution with a mean m
    and standard deviation sd, the sampling distribution of the mean
    approaches a normal distribution with a mean m and a standard deviation
    sd/sqrt(N) as N, the sample size, increases.

    So if you take the average of N readings the standard deviation of the
    average will be reduced by a factor of sqrt(N).

    Note that the average will approach a normal distribution even if the
    original distribution is not normal.

    As others have already pointed out, this average may not converge on the
    right answer, for example if all your sensors came from the same batch
    they could well all read a bit high or a bit low.


  14. Gareth

    Gareth Guest

    There is an applet here which shows the central limit theorem:

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