Segmenting datasets: Difference between revisions
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* Statistical significance of the resulting spectra | * Statistical significance of the resulting spectra | ||
The shorter the segment, the higher the temporal resolution of the final <math>\varepsilon</math> time series. However, the spectrum's lowest resolved frequency and frequency resolution depends on the duration of the signal used to construct the spectrum. Therefore, the segment must remain sufficiently long such that the lowest wavenumber (frequencies) of the [[Velocity inertial subrange|inertial subrange]] are resolved by the spectra. This is particularly important when measurement noise drowns the highest wavenumber (frequencies) of the inertial subrange. Thus, using too short segments may inadvertently render the resulting spectra unusable for deriving <math>\varepsilon</math> from the [[Velocity inertial subrange|inertial subrange]]. | The shorter the segment, the higher the temporal resolution of the final <math>\varepsilon</math> time series and the more likely the segment will be [[Stationarity|stationary]]. However, the spectrum's lowest resolved frequency and frequency resolution depends on the duration of the signal used to construct the spectrum. Therefore, the segment must remain sufficiently long such that the lowest wavenumber (frequencies) of the [[Velocity inertial subrange|inertial subrange]] are resolved by the spectra. This is particularly important when measurement noise drowns the highest wavenumber (frequencies) of the inertial subrange. Thus, using too short segments may inadvertently render the resulting spectra unusable for deriving <math>\varepsilon</math> from the [[Velocity inertial subrange|inertial subrange]]. | ||
Revision as of 15:34, 30 November 2021
Once the raw observations have been quality-controlled, then you must split the time series into shorter segments by considering:
- Time and length scales of turbulence
- Stationarity of the segment and Taylor's frozen turbulence hypothesis
- Statistical significance of the resulting spectra
The shorter the segment, the higher the temporal resolution of the final [math]\displaystyle{ \varepsilon }[/math] time series and the more likely the segment will be stationary. However, the spectrum's lowest resolved frequency and frequency resolution depends on the duration of the signal used to construct the spectrum. Therefore, the segment must remain sufficiently long such that the lowest wavenumber (frequencies) of the inertial subrange are resolved by the spectra. This is particularly important when measurement noise drowns the highest wavenumber (frequencies) of the inertial subrange. Thus, using too short segments may inadvertently render the resulting spectra unusable for deriving [math]\displaystyle{ \varepsilon }[/math] from the inertial subrange.
Application to measured velocities
Measurements are typically collected in the following two ways:
- continuously, or in such long bursts that they can be considered continuous
- short bursts that are typically at most 2-3x the expected largest turbulence time scales (e.g., 10 min in ocean environments)
This segmenting step dictates the minimum burst duration when setting up your equipment. The act of chopping a time series into smaller subsets, i.e., segments, is effectively a form of low-pass (box-car) filtering. How to segment the time series is usually a more important consideration than detrending the time series since estimating [math]\displaystyle{ \varepsilon }[/math] relies on resolving the inertial subrange in the final spectra computed over each segment.
Notes
- ↑ Zhaohua Wu, Norden E. Huang, Steven R. Long and and Chung-Kang Peng. 2007. On the trend, detrending, and variability of nonlinear and nonstationary time series. PNAS. doi:10.1073/pnas.0701020104