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Imagine we are measuring the period $T$ of a pendulum using a stopwatch

The stopwatch has an inerent instrumental uncertainty (e.g. 0.1s) let's keep this fact in mind

Due to random error (let's assume that there is no systematic error in this example) we end up with a series of values for the period of the pendulum:

$ T_1, T_2, ..., T_n $

After perfoming a "statistical analysis" on this sample of measurements, we found the mean of the sample, $\bar{T}$, and the standard error (standard deviation of the mean) $ \alpha = \dfrac{\sigma}{\sqrt{n}} $

Some authors (like Hughes & Hase in the book "Measurements and their Uncertainties) would report (with the appropriate decimal digits) the value of the measurement as:

$ T = (\bar{T} \pm \alpha) $

But shouldn't we also include the instrumental uncertainty of the stopwatch when reporting this value?

I thought that is because this uncertainty (0.1s) is somehow already in $\alpha $ but this is just a guess

And if we should include this uncertainty how should one do it?

Like this:

$\sqrt{\alpha^2 + (0.1s)^2}$?

Thanks in advance and sorry for any grammar mistake.

*Progress update: Currently reading 'An Introduction to Uncertainty in Measurement' by Les Kirkup and Bob Frenkel in an attempt to answer my own question due to the lack of answers

**Thanks for all the answers

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Lets consider the result of combining two Gaussian distributions of the measuring period $T$ (assumed the exact value $T_o$) with deviation $\sigma_1$ and $\sigma_2$: \begin{align} P_1(T) =& N_1 \exp\left(-\frac{(T-T_o)^2}{2\sigma_1^2}\right); \tag{1}\\ P_2(T) =& N_2 \exp\left(-\frac{(T-T_o)^2}{2\sigma_2^2}\right). \tag{2} \end{align} Where the $N_1$ and $N_2$ are the normalization constant, $N_1 = \frac{1}{\sqrt{2\pi}\sigma_1}$ and $N_2 = \frac{1}{\sqrt{2\pi}\sigma_2}$.

To investigate the combined effect of these two Gaussians, we consider a measurement rendering $T = t$ from Eq.(1), and then the Gaussian of Eq.(2) gives the final measurement $T$ with average at $t$. The total correlated probability:

\begin{align} P(T) =& \int_{-\infty}^\infty P_1(t-T_o) P_2(T-t) dt,\\ =& N_1 N_2 \int_{-\infty}^\infty dt \exp\left(-\frac{(t-T_o)^2}{2\sigma_1^2}\right) \exp\left(-\frac{(T-t)^2}{2\sigma_2^2}\right); \\ =& N_1 N_2 e^{-\left(\frac{T_o^2}{2\,\sigma_1^2}+\frac{T^2}{2\sigma_2^2}\right)} \int_{-\infty}^\infty dt \exp\left(-\frac{(\sigma_1^2+\sigma_2^2)t^2-2\sigma_2^2 t T_o -2\sigma_1^2 t T}{2\sigma_1^2\sigma_2^2}\right); \\ =& N_1 N_2 e^{-\left(\frac{T_o^2}{2\,\sigma_1^2}+\frac{T^2}{2\sigma_2^2}\right)} \int_{-\infty}^\infty dt \exp\left\{-\left[\frac{\sigma_1^2+\sigma_2^2}{2\sigma_1^2\sigma_2^2}\right] \left(t^2-2t \frac{\sigma_2^2 T_o + \sigma_1^2 T}{\sigma_1^2+\sigma_2^2}\right) \right\}; \\ =& N_1 N_2 e^{-\frac{T_o^2\, \sigma_2^2 + T^2\,\sigma_1^2}{2\,\sigma_1^2\, \sigma_2^2} + \frac{(\sigma_2^2 \, T_o + \sigma_1^2\, T)^2}{2\sigma_1^2\,\sigma_2^2 (\sigma_1^2+\sigma_2^2)} } \int_{-\infty}^\infty dt \exp\left\{-\left[\frac{\sigma_1^2+\sigma_2^2}{2\sigma_1^2\sigma_2^2}\right] \left(t- \frac{\sigma_2^2 T_o + \sigma_1^2 T}{\sigma_1^2+\sigma_2^2}\right)^2 \right\}; \\ =&\frac{1}{\sqrt{2\pi}\sigma_1}\frac{1}{\sqrt{2\pi}\sigma_2} \exp\left\{-\frac{T_o^2\, \sigma_2^2 + T^2\,\sigma_1^2}{2\,\sigma_1^2\, \sigma_2^2} + \frac{(\sigma_2^2 \, T_o + \sigma_1^2\, T)^2}{2\sigma_1^2\,\sigma_2^2 (\sigma_1^2+\sigma_2^2)} \right\} \sqrt{\pi\frac{2\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2}}\\ =&\sqrt{\frac{1}{2\pi\left(\sigma_1^2+\sigma_2^2\right)}} \exp\left\{-\frac{\sigma_1^2\sigma_2^2\left( T - T_o\right)^2}{2\,\sigma_1^2\, \sigma_2^2 \left(\sigma_1^2+\sigma_2^2\right)} \right\} \\ =&\sqrt{\frac{1}{2\pi\left(\sigma_1^2+\sigma_2^2\right)}} \exp\left\{-\frac{\left( T - T_o\right)^2}{2\, \left(\sigma_1^2+\sigma_2^2\right)} \right\} \\ \end{align}

The confluent integral renders a Gaussian distribution with a deviation

$$ \sigma^2 = \sigma_1^2 + \sigma_2^2. $$

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  • $\begingroup$ So that means that we should add these two fonts of error in quadrature? Also, does it mean that instrumental error is modeled by a gaussian? $\endgroup$ – armoredchihuahua May 26 at 17:16
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    $\begingroup$ (1) yes. (2) A gaussian is usually employed for model of random error. Instrument error is considered as an random error, if there are no personal effects. $\endgroup$ – ytlu May 29 at 3:59
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Engineer's view, proceed with caution!

When reporting uncertainty, you want to report every contribution together into a single value; but sometimes there is a need to distinguish between instrument limitations and uncertainty measured from repeated measurements. If you were an ideal measurer, you could simply say $1.3 \pm 0.05 \text{ s}$ where the stopwatch is measuring in 0.1s increments. This would be good enough for most applications, even if there's a lot more to say about it probabilistically. It's matter of resolution, so if you wanted better performance, simply use a better stopwatch; otherwise, combine the uncertainty in quadrature and report that figure.

Now, technically you should consider that every measurement has an uncertainty associated with it due to the instrument, and so there might be some propagation of error terms to consider. I'll have to check my own references when I get back to my bookshelf.

The most important thing is to ensure that anyone reading your work will understand how and why you calculated uncertainty the way you did. This is the purpose of things like GUM (Guide to the Expression of Uncertainty in Measurement). I highly recommend using GUM when e.g. publishing. For example, instrumental errors would fall under type B errors with GUM.

I'd be interested to hear other peoples' opinions on this too, by the way; I'm no experimental expert myself!

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I would do the following

  1. Once I collect all the measures for $T, \bar{T}$ and $\sigma$, I would check what the value of $3\sigma$ is, since this way you would have a chance of 99.73% that all the possible error values is likely to be contained in that range. If you use only one sigma, you would be accounting only 68.27%. Note that in CERN and other big research laboratories there is this famous criterion that 5 sigmas are needed in order to consider an event a discovery.
  2. If that value of your human error is bigger than the uncertainty of your stopwatch, I would definitely use $3\sigma$
  3. If that value is not bigger, which is unlikely, then I would use the uncertainty of your device as the error.

In general if you have error from different and unrelated sources, you are interested in taking the greatest of them. If you are sure that those sources are related and affecting each other, then for sure you should combined both deviations just as you mentioned $$ \Delta=\sqrt{(3\sigma)^2+\sum \Delta _{sources}^2} $$

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  • $\begingroup$ I thought one can only add the errors in this way when they are uncorrelated $\endgroup$ – armoredchihuahua May 28 at 2:14

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