Forskjell mellom versjoner av «Combining measurements of time periods with unequal number of periods»

Revisjonen fra 4. feb. 2013 kl. 18:17

Suppose you want to estimate the mean period [math]\bar{T}[/math] and its uncertainty with N measurements of the time [math]t_i[/math] for a number of periods per measurement n_i.

We assume that each time measurement has the same uncertainty [math]\sigma_t[/math].

We estimate the mean period by minimizing the [math]\chi^2[/math] of the measurements with respect to the mean period[math]\bar{T}[/math],

[math]\chi^2 = \sum^N_{i=1} {(t_i-n_i \bar{T})^2\over \sigma_t^2},[/math]

by setting the first derivative to zero:

[math]{\partial\chi^2\over {\partial \bar{T}}} = 0 = \sum^N_{i=1}{-2n_i(t_i-n_i\bar{t})\over \sigma_t^2}.[/math]

We solve for the mean period: [math]\bar{T}={\sum^N_{i=1}n_it_i\over \sum^N_{i=1}n_i^2}.[/math]

To find the uncertainty on [math]\bar{T}[/math] we use the second derivative of the [math]\chi^2[/math]:

[math]{1\over \sigma^2} = {1\over 2}{\partial^2\chi^2\over\partial\bar{T}^2} = \sum^N_{i=1}{n_i^2\over\sigma_t^2},[/math]

which gives us

[math]\sigma^2={\sigma_t^2\over \sum^N_{i=1}{n_i^2}}.[/math]

We can estimate [math]\sigma^2[/math] from the standard deviation of the measured times with respect to the estimated mean period times the number of periods measured in measurement [math]i[/math]: [math]\sigma^2 \simeq S^2 ={1\over N-1} \sum^N_{i=1}(t_i-n_i\bar{T})^2.[/math]

These formulae give simple, well-known results for two examples:

1) All [math]n_i=1[/math] (measure 1 period at a time):
[math]\bar{T}={\sum^N_{i=1}t_i\over N}[/math]
[math]\sigma ={\sigma_t\over \sqrt{n}}[/math]

2) All [math]n_i=m[/math] (measure a constant number m periods at a time):
[math]\bar{T}={1\over m}{\sum^N_{i=1}t_i\over N}[/math]
[math]\sigma ={1\over m}{\sigma_t\over\sqrt{n}}[/math]