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Total distribution when measuring

  1. Nov 21, 2015 #1
    The meters used in measurement have some level of accuracy. There is a probability distribution of measuring the true value, and that distribution curve for the accuracy of the meter has some standard deviation. And the component being measured also has it's own manufacturing distribution that give a probability distribution curve about some ideal value.

    My question is what will the total distribution curve look like when taking into account both sources of variance in the measurement process? What is this effect called? And how do I calculate it? Thanks.
     
  2. jcsd
  3. Nov 22, 2015 #2
    Hi friend:

    I am not sure what the question is you are asking.

    What is the variable x for which the "total distribution curve" p(x) will give the probability of occurrence of a value within the range [x, x+dx]?

    The variable x presumably some physical association with the combination the variables of the other two distributions, say y and z. The variable y could be some kind of measurement error distribution, possibly the difference m-r between the "measured" value m and the "real" value r, or y could be the relative error (m-r)/r. The variable z could be some kind of construction error, either actual or relative, between the "real" value r and the "wanted" value w.

    You may want the distribution of x=m-w, but I have difficulty understanding why this would be useful.

    If that is what you want, and you assume that
    y = m-r, and
    z = r-w, and
    the distribution of y is pm(y), and
    the distribution of z is pc(z),​
    then,
    p(x) = ∫-∞-∞ pm(y) pc(z) (y-z) dz dy​

    I hope this is helpful.

    Regards,
    Buzz
     
  4. Nov 22, 2015 #3

    mfb

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    In general, you get a convolution. The convolution of two Gaussian distributions is a Gaussian distribution again, which simplifies many problems significantly.
     
  5. Nov 22, 2015 #4
    Yes, I sort of thought if was a convolution and that the final result would be another gaussian. So now I wonder if the standard deviation of the total distribution (the convolution) is a simple expression of the two different standard deviations that are in the two different gaussians in the convolution integral? I suppose I could work it out. But maybe someone knows this off the top of their head.

    I thought this might be a simple question since it would occur in quality control issues in measuring electronic circuit boards. Your measuring device has one distribution of accuracy giving us one source of error with some standard deviation. And the tolerances of the components on the printed circuit board would add up to give another source of error with some distribution with a different standard deviation. In quality control you want to pass product that stays within tolerance. And this brings up the question about which source of error are you seeing in the overall distribution curve that you are getting. How much of what you are seeing is due to the accuracy of the meter, and how much is due to the components you are testing?
     
  6. Nov 22, 2015 #5

    mfb

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    The square root of the summed squared standard deviations.
    The wikipedia article about Gaussian distributions (and various other articles) should explain that. It is a very basic rule for the convolution of distributions with a standard deviation.
     
  7. Nov 22, 2015 #6
    Does this mean that if you know the standard deviation of the overall gaussian distribution and the standard deviation of your measuring device (meter), you can easily calculate the standard deviation of just the component tolerances? (Assuming that everything has a gaussian distribution) Or is it more complicated than that?
     
  8. Nov 22, 2015 #7

    mfb

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    You can get an estimate, sure. You have to understand your measurement device of course - the larger its uncertainty, the better you have to understand this uncertainty. If your measurement uncertainty is 1/10 of the spread of the devices, it is fine. If it is 10 times the spread, forget it. Everything in between is interesting.
     
  9. Nov 23, 2015 #8
    Hi friend:

    I apologize for my senior moment. The integral I gave in my post #2 for was for the mean of the distribution for x, not the distribution itself.
    I gather from mfb's posts that it is reasonable to assume the distribution for y and z are both Gaussian. That does simplify the calculation.

    Regards,
    Buzz
     
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